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https://www.astroml.org/astroML-notebooks/chapter9/astroml_chapter9_Classification.html

# Classification¶ In density estimation we estimate joint probability distributions from multivariate data sets to identify the inherent clustering. This is essentially unsupervised classification If we have labels for some of these data points (e.g., an object is tall, short, red, or blue) we can develop a relationship between the label and the properties of a source. This is supervised classification Classification, regression, and density estimation are all related. For example, the regression function $$\hat{y} = f(y|\vec{x})$$ is the best estimated value of $$y$$ given a value of $$\vec{x}$$. In classification $$y$$ is categorical and $$f(y|\vec{x})$$ the called the discriminant function • Using density estimation for classification is referred to as generative classification (we have a full model of the density for each class or we have a model which describes how data could be generated from each class). • Classification that finds the decision boundary that separates classes is called discriminative classification Both have their place in astrophysical classification. ## Classification loss: how well are we doing?¶ The first question we need to address is how we score (defined the success of our classification) We can define a loss function. A zero-one loss function assigns a value of one for a misclassification and zero for a correct classification (i.e. we will want to minimize the loss). If $$\hat{y}$$ is the best guess value of $$y$$, the classification loss, $$L(y,\widehat{y})$$, is $L(y,\widehat{y}) = \delta(y \neq \widehat{y})$ which means $$\begin{eqnarray} L(y,\hat{y}) & = & \left\{ \begin{array}{cl} 1 & \mbox{if$$y\neq\hat{y}$$}, \\ 0 & \mbox{otherwise.} \end{array} \right. \end{eqnarray}$$ The expectation (mean) value of the loss $$\mathbb{E} \left[ L(y,\hat{y}) \right] = p(y\neq \hat{y})$$ is called the classification risk This is related to regression loss functions: $$L(y, \hat{y}) = (y - \hat{y})^2$$ and risk $$\mathbb{E}[(y - \hat{y})^2]$$. We can then define: $${\rm completeness} = \frac{\rm true\ positives} {\rm true\ positives + false\ negatives}$$ $${\rm contamination} = \frac{\rm false\ positives} {\rm true\ positives + false\ positives}$$ or $${\rm true\ positive\ rate} = \frac{\rm true\ positives} {\rm true\ positives + false\ negatives}$$ $${\rm false\ positive\ rate} = \frac{\rm false\ positives} {\rm true\ negatives + false\ positives}$$ ## Comparing the performance of classifiers¶ Best performance is a bit of a subjective topic (e.g. star-galaxy separation for correlation function studies or Galactic streams studies). We trade contamination as a function of completeness and this is science dependent. ROC curves: Receiver Operating Characteristic curves • Plot the true-positive vs the false-positive rate • Initially used to analyze radar results in WWII (a very productive era for statistics…). • One concern about ROC curves is that they are sensitive to the relative sample sizes (if there are many more background events than source events small false positive results can dominate a signal). For these cases we we can plot efficiency (1 - contamination) vs completeness import numpy as np from matplotlib import pyplot as plt from sklearn.naive_bayes import GaussianNB from sklearn.discriminant_analysis import (LinearDiscriminantAnalysis, from sklearn.linear_model import LogisticRegression from sklearn.neighbors import KNeighborsClassifier from sklearn.tree import DecisionTreeClassifier from sklearn.svm import SVC from astroML.classification import GMMBayes from sklearn.metrics import roc_curve from astroML.utils import split_samples, completeness_contamination from astroML.datasets import fetch_rrlyrae_combined We will use the RR Lyrae dataset. We get the data here, and split it into training and testing sets, and then use the same sets for all the examples below. #---------------------------------------------------------------------- # get data and split into training & testing sets X, y = fetch_rrlyrae_combined() X = X[:, [1, 0, 2, 3]] # rearrange columns for better 1-color results (X_train, X_test), (y_train, y_test) = split_samples(X, y, [0.75, 0.25], random_state=0) N_tot = len(y) N_st = np.sum(y == 0) N_rr = N_tot - N_st N_train = len(y_train) N_test = len(y_test) N_plot = 5000 + N_rr #------------------------------------------------------------ # Fit all the models to the training data def compute_models(*args): names = [] probs = [] for classifier, kwargs in args: print(classifier.__name__) clf = classifier(**kwargs) clf.fit(X_train, y_train) y_probs = clf.predict_proba(X_test)[:, 1] names.append(classifier.__name__) probs.append(y_probs) return names, probs names, probs = compute_models((GaussianNB, {}), (LinearDiscriminantAnalysis, {}), (LogisticRegression, dict(class_weight='balanced')), (KNeighborsClassifier, dict(n_neighbors=10)), (DecisionTreeClassifier, dict(random_state=0, max_depth=12, criterion='entropy')), (GMMBayes, dict(n_components=3, tol=1E-5, covariance_type='full'))) #------------------------------------------------------------ # Plot ROC curves and completeness/efficiency fig = plt.figure(figsize=(10, 5)) # ax2 will show roc curves ax1 = plt.subplot(121) # ax1 will show completeness/efficiency ax2 = plt.subplot(122) labels = dict(GaussianNB='GNB', LinearDiscriminantAnalysis='LDA', KNeighborsClassifier='KNN', DecisionTreeClassifier='DT', GMMBayes='GMMB', LogisticRegression='LR') thresholds = np.linspace(0, 1, 1001)[:-1] # iterate through and show results for name, y_prob in zip(names, probs): fpr, tpr, thresh = roc_curve(y_test, y_prob) # add (0, 0) as first point fpr = np.concatenate([[0], fpr]) tpr = np.concatenate([[0], tpr]) ax1.plot(fpr, tpr, label=labels[name]) comp = np.zeros_like(thresholds) cont = np.zeros_like(thresholds) for i, t in enumerate(thresholds): y_pred = (y_prob >= t) comp[i], cont[i] = completeness_contamination(y_pred, y_test) ax2.plot(1 - cont, comp, label=labels[name]) ax1.set_xlim(0, 0.04) ax1.set_ylim(0, 1.02) ax1.xaxis.set_major_locator(plt.MaxNLocator(5)) ax1.set_xlabel('false positive rate') ax1.set_ylabel('true positive rate') ax1.legend(loc=4) ax2.set_xlabel('efficiency') ax2.set_ylabel('completeness') ax2.set_xlim(0, 1.0) ax2.set_ylim(0.2, 1.02) plt.show() ## Linear and quadratic discriminant analysis¶ Linear discriminant analysis (LDA) assumes the class distributions have identical covariances for all $$k$$ classes (all classes are a set of shifted Gaussians). The optimal classifier is derived from the log of the class posteriors $g_k(\vec{x}) = \vec{x}^T \Sigma^{-1} \vec{\mu_k} - \frac{1}{2}\vec{\mu_k}^T \Sigma^{-1} \vec{\mu_k} + \log \pi_k,$ with $$\vec{\mu_k}$$ the mean of class $$k$$ and $$\Sigma$$ the covariance of the Gaussians. The class dependent covariances that would normally give rise to a quadratic dependence on $$\vec{x}$$ cancel out if they are assumed to be constant. The Bayes classifier is, therefore, linear with respect to $$\vec{x}$$. The discriminant boundary between classes is the line that minimizes the overlap between Gaussians $g_k(\vec{x}) - g_\ell(\vec{x}) = \vec{x}^T \Sigma^{-1} (\mu_k-\mu_\ell) - \frac{1}{2}(\mu_k - \mu_\ell)^T \Sigma^{-1}(\mu_k -\mu_\ell) + \log (\frac{\pi_k}{\pi_\ell}) = 0.$ Relaxing the requirement that the covariances of the Gaussians are constant, the discriminant function becomes quadratic in $$x$$: $g(\vec{x}) = -\frac{1}{2} \log | \Sigma_k | - \frac{1}{2}(\vec{x}-\mu_k)^T C^{-1}(\vec{x}-\mu_k) + \log \pi_k.$ This is sometimes known as quadratic discriminant analysis (QDA) #---------------------------------------------------------------------- # perform LinearDiscriminantAnalysis classifiers = [] predictions = [] Ncolors = np.arange(1, X.shape[1] + 1) for nc in Ncolors: clf = LinearDiscriminantAnalysis() clf.fit(X_train[:, :nc], y_train) y_pred = clf.predict(X_test[:, :nc]) classifiers.append(clf) predictions.append(y_pred) completeness, contamination = completeness_contamination(predictions, y_test) print("completeness", completeness) print("contamination", contamination) # perform QuadraticDiscriminantAnalysis qclassifiers = [] qpredictions = [] for nc in Ncolors: qlf.fit(X_train[:, :nc], y_train) qy_pred = qlf.predict(X_test[:, :nc]) qclassifiers.append(qlf) qpredictions.append(qy_pred) qpredictions = np.array(qpredictions) qcompleteness, qcontamination = completeness_contamination(qpredictions, y_test) print("completeness", qcompleteness) print("contamination", qcontamination) #------------------------------------------------------------ # Compute the decision boundary clf = classifiers[1] qlf = qclassifiers[1] xlim = (0.7, 1.35) ylim = (-0.15, 0.4) xx, yy = np.meshgrid(np.linspace(xlim[0], xlim[1], 71), np.linspace(ylim[0], ylim[1], 81)) Z = clf.predict_proba(np.c_[yy.ravel(), xx.ravel()]) Z = Z[:, 1].reshape(xx.shape) QZ = qlf.predict_proba(np.c_[yy.ravel(), xx.ravel()]) QZ = QZ[:, 1].reshape(xx.shape) #---------------------------------------------------------------------- # plot the results fig = plt.figure(figsize=(8, 4)) left=0.1, right=0.95, wspace=0.2) # left plot: data and decision boundary im = ax.scatter(X[-N_plot:, 1], X[-N_plot:, 0], c=y[-N_plot:], s=4, lw=0, cmap=plt.cm.Oranges, zorder=2) im.set_clim(-0.5, 1) im = ax.imshow(Z, origin='lower', aspect='auto', cmap=plt.cm.binary, zorder=1, extent=xlim + ylim) im.set_clim(0, 1.5) ax.contour(xx, yy, Z, [0.5], linewidths=2., colors='k') ax.set_xlim(xlim) ax.set_ylim(ylim) ax.set_xlabel('$u-g$') ax.set_ylabel('$g-r$') # right plot: qda im = ax.scatter(X[-N_plot:, 1], X[-N_plot:, 0], c=y[-N_plot:], s=4, lw=0, cmap=plt.cm.Oranges, zorder=2) im.set_clim(-0.5, 1) im = ax.imshow(QZ, origin='lower', aspect='auto', cmap=plt.cm.binary, zorder=1, extent=xlim + ylim) im.set_clim(0, 1.5) ax.contour(xx, yy, QZ, [0.5], linewidths=2., colors='k') ax.set_xlim(xlim) ax.set_ylim(ylim) ax.set_xlabel('$u-g$') ax.set_ylabel('$g-r$') plt.show() ## GMM and Bayes classification¶ The natural extension to the Gaussian assumptions is to use GMM’s to learn the density distribution. The number of Gaussian components $$K$$ must be chosen for each class independently # GMM-bayes takes several minutes to run, and is order[N^2] # truncating the dataset can be useful for experimentation. #X_tr = X[::10] #y_tr = y[::10] #---------------------------------------------------------------------- # perform GMM Bayes Ncolors = np.arange(1, X.shape[1] + 1) Ncomp = [1, 3] def compute_GMMbayes(Ncolors, Ncomp): classifiers = [] predictions = [] for ncm in Ncomp: classifiers.append([]) predictions.append([]) for nc in Ncolors: clf = GMMBayes(ncm, tol=1E-5, covariance_type='full') clf.fit(X_train[:, :nc], y_train) y_pred = clf.predict(X_test[:, :nc]) classifiers[-1].append(clf) predictions[-1].append(y_pred) return classifiers, predictions classifiers, predictions = compute_GMMbayes(Ncolors, Ncomp) completeness, contamination = completeness_contamination(predictions, y_test) print("completeness", completeness) print("contamination", contamination) #------------------------------------------------------------ # Compute the decision boundary clf = classifiers[1][1] xlim = (0.7, 1.35) ylim = (-0.15, 0.4) xx, yy = np.meshgrid(np.linspace(xlim[0], xlim[1], 71), np.linspace(ylim[0], ylim[1], 81)) Z = clf.predict_proba(np.c_[yy.ravel(), xx.ravel()]) Z = Z[:, 1].reshape(xx.shape) #---------------------------------------------------------------------- # plot the results fig = plt.figure(figsize=(8, 4)) left=0.1, right=0.95, wspace=0.2) # left plot: data and decision boundary im = ax.scatter(X[-N_plot:, 1], X[-N_plot:, 0], c=y[-N_plot:], s=4, lw=0, cmap=plt.cm.Oranges, zorder=2) im.set_clim(-0.5, 1) im = ax.imshow(Z, origin='lower', aspect='auto', cmap=plt.cm.binary, zorder=1, extent=xlim + ylim) im.set_clim(0, 1.5) ax.contour(xx, yy, Z, [0.5], colors='k') ax.set_xlim(xlim) ax.set_ylim(ylim) ax.set_xlabel('$u-g$') ax.set_ylabel('$g-r$') # plot completeness vs Ncolors ax.plot(Ncolors, completeness[0], '^--k', ms=6, label='N=%i' % Ncomp[0]) ax.plot(Ncolors, completeness[1], 'o-k', ms=6, label='N=%i' % Ncomp[1]) ax.xaxis.set_major_locator(plt.MultipleLocator(1)) ax.yaxis.set_major_locator(plt.MultipleLocator(0.2)) ax.xaxis.set_major_formatter(plt.NullFormatter()) ax.set_ylabel('completeness') ax.set_xlim(0.5, 4.5) ax.set_ylim(-0.1, 1.1) ax.grid(True) # plot contamination vs Ncolors ax.plot(Ncolors, contamination[0], '^--k', ms=6, label='N=%i' % Ncomp[0]) ax.plot(Ncolors, contamination[1], 'o-k', ms=6, label='N=%i' % Ncomp[1]) ax.legend(loc='lower right', bbox_to_anchor=(1.0, 0.78)) ax.xaxis.set_major_locator(plt.MultipleLocator(1)) ax.yaxis.set_major_locator(plt.MultipleLocator(0.2)) ax.xaxis.set_major_formatter(plt.FormatStrFormatter('%i')) ax.set_xlabel('N colors') ax.set_ylabel('contamination') ax.set_xlim(0.5, 4.5) ax.set_ylim(-0.1, 1.1) ax.grid(True) plt.show() ## K-nearest neighbours¶ As with density estimation (and kernel density estimation) the intuitive justification is that $$p(y|x) \approx p(y|x')$$ if $$x'$$ is very close to $$x$$. The number of neighbors, $$K$$, regulates the complexity of the classification. In simplest form, a majority rule classification is adopted, where each of the $$K$$ points votes on the classification. Increasing $$K$$ decreases the variance in the classification but at the expense of an increase in the bias. Weights can be assigned to individual votes by weighting the vote by the distance to the nearest point. #---------------------------------------------------------------------- # perform Classification classifiers = [] predictions = [] Ncolors = np.arange(1, X.shape[1] + 1) kvals = [1, 10] for k in kvals: classifiers.append([]) predictions.append([]) for nc in Ncolors: clf = KNeighborsClassifier(n_neighbors=k) clf.fit(X_train[:, :nc], y_train) y_pred = clf.predict(X_test[:, :nc]) classifiers[-1].append(clf) predictions[-1].append(y_pred) completeness, contamination = completeness_contamination(predictions, y_test) print("completeness", completeness) print("contamination", contamination) #------------------------------------------------------------ # Compute the decision boundary clf = classifiers[1][1] xlim = (0.7, 1.35) ylim = (-0.15, 0.4) xx, yy = np.meshgrid(np.linspace(xlim[0], xlim[1], 71), np.linspace(ylim[0], ylim[1], 81)) Z = clf.predict(np.c_[yy.ravel(), xx.ravel()]) Z = Z.reshape(xx.shape) #---------------------------------------------------------------------- # plot the results fig = plt.figure(figsize=(8, 4)) left=0.1, right=0.95, wspace=0.2) # left plot: data and decision boundary im = ax.scatter(X[-N_plot:, 1], X[-N_plot:, 0], c=y[-N_plot:], s=4, lw=0, cmap=plt.cm.Oranges, zorder=2) im.set_clim(-0.5, 1) im = ax.imshow(Z, origin='lower', aspect='auto', cmap=plt.cm.binary, zorder=1, extent=xlim + ylim) im.set_clim(0, 2) ax.contour(xx, yy, Z, [0.5], colors='k') ax.set_xlim(xlim) ax.set_ylim(ylim) ax.set_xlabel('$u-g$') ax.set_ylabel('$g-r$') ax.text(0.02, 0.02, "k = %i" % kvals[1], transform=ax.transAxes) # plot completeness vs Ncolors ax.plot(Ncolors, completeness[0], 'o-k', ms=6, label='k=%i' % kvals[0]) ax.plot(Ncolors, completeness[1], '^--k', ms=6, label='k=%i' % kvals[1]) ax.xaxis.set_major_locator(plt.MultipleLocator(1)) ax.yaxis.set_major_locator(plt.MultipleLocator(0.2)) ax.xaxis.set_major_formatter(plt.NullFormatter()) ax.set_ylabel('completeness') ax.set_xlim(0.5, 4.5) ax.set_ylim(-0.1, 1.1) ax.grid(True) # plot contamination vs Ncolors ax.plot(Ncolors, contamination[0], 'o-k', ms=6, label='k=%i' % kvals[0]) ax.plot(Ncolors, contamination[1], '^--k', ms=6, label='k=%i' % kvals[1]) ax.legend(loc='lower right', bbox_to_anchor=(1.0, 0.79)) ax.xaxis.set_major_locator(plt.MultipleLocator(1)) ax.yaxis.set_major_locator(plt.MultipleLocator(0.2)) ax.xaxis.set_major_formatter(plt.FormatStrFormatter('%i')) ax.set_xlabel('N colors') ax.set_ylabel('contamination') ax.set_xlim(0.5, 4.5) ax.set_ylim(-0.1, 1.1) ax.grid(True) plt.show() ## Support Vector Machines¶ Find the hyperplane that maximizes the distance of the closest point from either class. This distance is the margin (width of the line before it hits a point). We want the line that maximizes the margin (m). The points on the margin are called support vectors If we assume $$y \in \{-1,1\}$$, (+1 is maximum margin, -1 is minimum, 0 is the decision boundary) The maximum is then just when $$\beta_0 + \beta^T x_i = 1$$ etc The hyperplane which maximizes the margin is given by finding $\max_{\beta_0,\beta}(m) \;\;\; \mbox{subject to} \;\;\; \frac{1}{||\beta||} y_i ( \beta_0 + \beta^T x_i ) \geq m \,\,\, \forall \, i.$ The constraints can be written as $$y_i ( \beta_0 + \beta^T x_i ) \geq m ||\beta||$$. Thus the optimization problem is equivalent to minimizing $$$\frac{1}{2} ||\beta|| \;\;\; \mbox{subject to} \;\;\; y_i ( \beta_0 + \beta^T x_i ) \geq 1 \,\,\, \forall \, i.$$$ This optimization is a quadratic programming problem (quadratic objective function with linear constraints). Note that because SVM uses a metric which maximizes the margin rather than a measure over all points in the data sets, it is similar in spirit to the rank-based estimators • The median of a distribution is unaffected by even large perturbations of outlying points, as long as those perturbations do not cross the boundary. • In the same way, once the support vectors are determined, changes to the positions or numbers of points beyond the margin will not change the decision boundary. For this reason, SVM can be a very powerful tool for discriminative classification. • This is why there is a high completeness compared to the other methods: it does not matter that the background sources outnumber the RR Lyrae stars by a factor of $$\sim$$200 to 1. It simply determines the best boundary between the small RR Lyrae clump and the large background clump. • This completeness, however, comes at the cost of a relatively large contamination level. • SVM is not scale invariant so it often worth rescaling the data to [0,1] or to whiten it to have a mean of 0 and variance 1 (remember to do this to the test data as well!) • The data dont need to be separable (we can put a constraint in minimizing the number of “failures”) # SVM takes several minutes to run, and is order[N^2] # truncating the dataset can be useful for experimentation. #X_tr = X[::5] #y_tr = y[::5] #---------------------------------------------------------------------- # Fit Kernel SVM Ncolors = np.arange(1, X.shape[1] + 1) def compute_SVM(Ncolors): classifiers = [] predictions = [] for nc in Ncolors: # perform support vector classification clf = SVC(kernel='rbf', gamma=20.0, class_weight='balanced') clf.fit(X_train[:, :nc], y_train) y_pred = clf.predict(X_test[:, :nc]) classifiers.append(clf) predictions.append(y_pred) return classifiers, predictions classifiers, predictions = compute_SVM(Ncolors) completeness, contamination = completeness_contamination(predictions, y_test) print("completeness", completeness) print("contamination", contamination) #------------------------------------------------------------ # compute the decision boundary clf = classifiers[1] xlim = (0.7, 1.35) ylim = (-0.15, 0.4) xx, yy = np.meshgrid(np.linspace(xlim[0], xlim[1], 101), np.linspace(ylim[0], ylim[1], 101)) Z = clf.predict(np.c_[yy.ravel(), xx.ravel()]) Z = Z.reshape(xx.shape) # smooth the boundary from scipy.ndimage import gaussian_filter Z = gaussian_filter(Z, 2) #---------------------------------------------------------------------- # plot the results fig = plt.figure(figsize=(8, 4)) left=0.1, right=0.95, wspace=0.2) # left plot: data and decision boundary im = ax.scatter(X[-N_plot:, 1], X[-N_plot:, 0], c=y[-N_plot:], s=4, lw=0, cmap=plt.cm.Oranges, zorder=2) im.set_clim(-0.5, 1) ax.contour(xx, yy, Z, [0.5], colors='k') ax.set_xlim(xlim) ax.set_ylim(ylim) ax.set_xlabel('$u-g$') ax.set_ylabel('$g-r$') # plot completeness vs Ncolors ax.plot(Ncolors, completeness, 'o-k', ms=6) ax.xaxis.set_major_locator(plt.MultipleLocator(1)) ax.yaxis.set_major_locator(plt.MultipleLocator(0.2)) ax.xaxis.set_major_formatter(plt.NullFormatter()) ax.set_ylabel('completeness') ax.set_xlim(0.5, 4.5) ax.set_ylim(-0.1, 1.1) ax.grid(True) ax.plot(Ncolors, contamination, 'o-k', ms=6) ax.xaxis.set_major_locator(plt.MultipleLocator(1)) ax.yaxis.set_major_locator(plt.MultipleLocator(0.2)) ax.xaxis.set_major_formatter(plt.FormatStrFormatter('%i')) ax.set_xlabel('N colors') ax.set_ylabel('contamination') ax.set_xlim(0.5, 4.5) ax.set_ylim(-0.1, 1.1) ax.grid(True) plt.show() ## Gaussian Naive Bayes¶ In Gaussian naive Bayes $$p_k(x^i)$$ are modeled as one-dimensional normal distributions, with means $$\mu^i_k$$ and widths $$\sigma^i_k$$. The naive Bayes estimator is then $\hat{y} = \arg\max_{y_k}\left[\ln \pi_k - \frac{1}{2}\sum_{i=1}^N\left(2\pi(\sigma^i_k)^2 + \frac{(x^i - \mu^i_k)^2}{(\sigma^i_k)^2} \right) \right]$ Note: this is the log of the Bayes criterion with no normalization constant from astroML.datasets import fetch_imaging_sample def get_stars_and_galaxies(Nstars=10000, Ngals=10000): """Get the subset of star/galaxy data to plot""" data = fetch_imaging_sample() objtype = data['type'] stars = data[objtype == 6][:Nstars] galaxies = data[objtype == 3][:Ngals] return np.concatenate([stars,galaxies]), np.concatenate([np.zeros(len(stars)), np.ones(len(galaxies))]) data, y = get_stars_and_galaxies(Nstars=10000, Ngals=10000) # select r model mag and psf - model mag as columns X = np.column_stack((data['rRaw'], data['rRawPSF'] - data['rRaw'])) #------------------------------------------------------------ # Fit the Naive Bayes classifier clf = GaussianNB() clf.fit(X, y) # predict the classification probabilities on a grid xlim = (15, 25) ylim = (-5, 5) xx, yy = np.meshgrid(np.linspace(xlim[0], xlim[1], 71), np.linspace(ylim[0], ylim[1], 81)) Z = clf.predict_proba(np.c_[xx.ravel(), yy.ravel()]) Z = Z[:, 1].reshape(xx.shape) #------------------------------------------------------------ # Plot the results fig = plt.figure(figsize=(10,10)) ax.set_xlabel('$x$') ax.set_ylabel('$y$')

http://math.stackexchange.com/questions/246272/finding-fourier-coefficients-of-an-orthonormal-sequence

# Finding Fourier coefficients of an orthonormal sequence I am trying to find the Fourier coefficients of f with respect to the sequence $$\lbrace \phi_k \rbrace_{k=1}^\infty = \lbrace \frac{1}{\sqrt{2 \pi}}, \frac{\cos(x)}{\sqrt{\pi}}, \frac{\sin(x)}{\sqrt{\pi}},..., \frac{\cos(kx)}{\sqrt{\pi}}, \frac{\sin(kx)}{\sqrt{\pi}},... \rbrace$$ I know that $f \in L([-\pi, \pi])$ and $f(x) = x$ where $x \in [-\pi, \pi]$. I know that the Fourier coefficients for an orthonormal sequence in $L^2$ are $c_k = \langle f, \phi_k \rangle$. So using this I can say that $$\langle f, \phi_k \rangle = \int \limits_{-\pi}^\pi f \phi_k$$ I know that when $\phi_k = \frac{\cos(kx)}{\sqrt{\pi}}$ or if $\phi_k = \frac{\sin(kx)}{\sqrt{\pi}}$ then the integral goes to 0. So I am left with $$\int \limits_{-\pi}^\pi \frac{f}{\sqrt{2 \pi}} = \int \limits_{-\pi}^\pi \frac{x}{\sqrt{2 \pi}} = \frac{(\pi^2 - (-\pi)^2)}{2\sqrt{2 \pi}}=0$$ So in other words there are no Fourier coefficients, which I am pretty sure is wrong. I am also supposed to show that $\sum \limits_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$. I know the sequence is complete so then I know that $\sum \limits_{k=1}^\infty c_k=\parallel f\parallel_2^2$. I can solve for $\parallel f\parallel_2^2$ $$\parallel f\parallel_2^2 = \int \limits_{-\pi}^\pi x^2 = \frac{(\pi^3 - (-\pi)^3)}{3}= \frac{2\pi^3}{3}$$ I am clearly doing something wrong, but I can't figure out what. Any ideas? EDIT: The assumption I made, if $\phi_k = \frac{\cos(kx)}{\sqrt{\pi}}$ or if $\phi_k = \frac{\sin(kx)}{\sqrt{\pi}}$ then the integral goes to 0, was not true. Doing the full integral yields the correct answer. - You're right with the edit. They don't all go to 0, but one thing to notice is that $x\phi_k$ for $k$ even is an odd function (since it is an odd function times an even function), so you can immediately conclude that integral is $0$ for $k$ even. This trick is quite useful to remember in the future. – Matt Nov 28 '12 at 17:48 Let $c_k$ be the $k$th Fourier coefficient of $f$. So $c_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) e^{-ikt} dt$. Since $f(t) = t$, that is $\frac{1}{2\pi} \int_{-\pi}^{\pi} t e^{-ikt} dt$. Integrate by parts to find $c_k$. To find the coefficients in terms of the sequence you gave (sines and cosines), take the real and imaginary parts. This does not make sense since the definition of Fourier coefficients I am using is $c_k = \langle f, \phi_k \rangle$. Also where did you get $e^{-ikt}$? I don't think that represents the sequence I have. – rioneye Nov 28 '12 at 15:46 $e^{ix} = cos(x) + isin(x)$. I'm using the notation here for brevity. If it confuses you, don't use the complex notation for this sequence. But the idea is the same. The Fourier coefficients are just defined by an integral. Evaluate the integral. The trick is integrating by parts. – anonymous Nov 28 '12 at 19:25 In general, I had the right approach, but I made an in correct assumption. Like I said in the edit, the assumption I made, if $\phi_k = \frac{\cos(kx)}{\sqrt{\pi}}$ or if $\phi_k = \frac{\sin(kx)}{\sqrt{\pi}}$ then the integral goes to 0, was not true. Doing the full integral yields the correct answer.

https://www.homeworklib.com/qaa/875334/how-do-you-simplify-frac-3x-2-y-3-9x-4-y-5

Question # How do you simplify \frac { 3x ^ { - 2} y ^ { 3} } { 9x ^ { 4} y ^ { 5} }? How do you simplify \frac { 3x ^ { - 2} y ^ { 3} } { 9x ^ { 4} y ^ { 5} }? Answer 1 See a solution process below: #### Explanation: First, rewrite this expression as: $\frac{3}{9} \left({x}^{-} \frac{2}{x} ^ 4\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3} \left({x}^{-} \frac{2}{x} ^ 4\right) \left({y}^{3} / {y}^{5}\right)$ Next, use this rule of exponents to simplify the $x$ term: ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$ $\frac{1}{3} \left({x}^{\textcolor{red}{- 2}} / {x}^{\textcolor{b l u e}{4}}\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{4} - \textcolor{red}{- 2}\right)\right) \left({y}^{3} / {y}^{5}\right) \implies$ $\frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{4} + \textcolor{red}{2}\right)\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3} \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{6}\right)\right) \left({y}^{3} / {y}^{5}\right) \implies \frac{1}{3 {x}^{6}} \left({y}^{3} / {y}^{5}\right)$ Now, use this same rule to simplify the $y$ term: $\frac{1}{3 {x}^{6}} \left({y}^{\textcolor{red}{3}} / {y}^{\textcolor{b l u e}{5}}\right) \implies \frac{1}{3 {x}^{6}} \left(\frac{1}{y} ^ \left(\textcolor{b l u e}{5} - \textcolor{red}{3}\right)\right) \implies \frac{1}{3 {x}^{6}} \left(\frac{1}{y} ^ 2\right) \implies$ $\frac{1}{3 {x}^{6} {y}^{2}}$ #### Earn Coins Coins can be redeemed for fabulous gifts. Similar Homework Help Questions

https://math.stackexchange.com/questions/1169005/how-to-integrate-int-0-infty-frace-t-frac1t-sqrt-t-dt

# How to integrate $\int_0^{\infty}\frac{e^{-(t+\frac1t)}}{\sqrt t} dt$? This is a problem given in my homework . I have to find the integral$$\int \limits_{0}^{\infty} \frac{e^{-(t+\frac{1}{t})}}{\sqrt t}dt$$ I am trying to use integral representation of the gamma function but I was not able to get it in the region of convergence i.e. $\int \limits_{0}^{\infty} \frac{e^{-t}}{\sqrt t}$ is clearly $\Gamma (\frac{1}{2})$ but the second factor is causing a problem. Any hints or suggestions are appreciated. Thanks. Hint. Make the change of variable $t=x^2$ to obtain $$\int_0^{\infty}\frac{e^{-(t+\frac{1}{t})}}{\sqrt t}dt=2\int_0^{\infty}e^{ -x^2-1/x^2}dx=\int_{-\infty}^{\infty}e^{ -x^2-1/x^2}dx$$ You may then recall that, for any integrable function $f$, we have $$\int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad s>0. \tag1$$ Apply it to $f(x)=e^{-x^2}$, you get $$\int_{-\infty}^{+\infty}e^{-(x-s/x)^2}\mathrm{d}x=\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x=\sqrt{\pi}, \quad s>0. \tag2$$ Thus $$\int_{-\infty}^{+\infty}e^{-x^2-s^2/x^{2}}\mathrm{d}x=\sqrt{\pi}\:e^{-2s}\tag3$$ then put $s=1$ to obtain your integral. • great answer thanks a lot – happymath Feb 28 '15 at 12:53 • I hadn't seen the formula in the gray bar, and was computing the same thing in a messier way. This is cleaner. It is nice to know that formula. (+1) – robjohn Feb 28 '15 at 13:15

https://www.baotgm.net/forum/qz2rgw.php?1d5445=continuous-random-variable-graph

# continuous random variable graph Example 2: A person’s rounded weight (to the nearest pound) (discrete). This follows a Bernoulli distribution with only 2 possible outcomes and a single coin flip at a time. The quantity $$f\left( x \right)\,dx$$ is called probability differential. Let’s generate data with numpy to model this. The amount of water passing through a pipe connected with a high level reservoir. We intentionally chose a sample that followed a normal distribution to simplify the process. Make learning your daily ritual. Register now! Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, and SAT scores are just a few. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. A random variable is discrete if it can only take on a finite number of values. Including both men and women would result in a bimodal distribution (2 peaks instead of 1) which complicates our calculation. As shown in the Height Distribution graph, there is a continuous range of values between 140 cm and 200 cm. It really helps us a lot. We’ll sample the data (above) as well as plot it. Collect a sample … (ii) Let X be the volume of coke in a can marketed as 12oz. Collect a sample from the population2. If you don’t know the PMF in advance (and we usually don’t), you can estimate it based on a sample from the same distribution as your random variable. Before we dive into continuous random variables, let’s walk a few more discrete random variable examples. This time, weights are not rounded. We can follow this logic for some arbitrary data, where sample = [0,1,1,1,1,1,2,2,2,2]. $$f\left( x \right) \geqslant 0$$ for all $$x$$, $${\text{Total}}\,{\text{Area}} = \int\limits_{ – \infty }^\infty {f\left( x \right)dx} = 1$$, $$\left( {X = c} \right) = \int\limits_c^c {f\left( x \right)dx} = 0$$             Where c is any constant. It is denoted by $$f\left( x \right)$$ where $$f\left( x \right)$$ is the probability that the random variable $$X$$ takes the value between $$x$$ and $$x + \Delta x$$ where $$\Delta x$$ is a very small change in $$X$$. As the probability of the area for $$X = c$$ (constant), therefore $$P\left( {X = a} \right) = P\left( {X = b} \right)$$. The time in which poultry will gain 1.5 kg. The computer time (in seconds) required to process a certain program. The amount of rain falling in a certain city. If the image is uncountably infinite then X is called a continuous random variable. Hence for $$f\left( x \right)$$ to be the density function, we have, $$1 = \int\limits_{ – \infty }^\infty {f\left( x \right)dx} \,\,\, = \,\,\,\,\int\limits_2^8 {c\left( {x + 3} \right)dx} \,\,\, = \,\,\,c\left[ {\frac{{{x^2}}}{2} + 3x} \right]_2^8$$, $$= \,\,\,\,c\left[ {\frac{{{{\left( 8 \right)}^2}}}{2} + 3\left( 8 \right) – \frac{{{{\left( 2 \right)}^2}}}{2} – 3\left( 2 \right)} \right]\,\,\,\, = \,\,\,c\,\left[ {32 + 24 – 2 – 6} \right]\,\,\,\, = \,\,\,\,c\left[ {48} \right]$$, Therefore, $$f\left( x \right) = \frac{1}{{48}}\left( {x + 3} \right),\,\,\,\,2 \leqslant x \leqslant 8$$, (b) $$P\left( {3 < X < 5} \right) = \int\limits_3^5 {\frac{1}{{48}}\left( {x + 3} \right)dx} \,\,\, = \,\,\,\frac{1}{{48}}\left[ {\frac{{{x^2}}}{2} + 3x} \right]_3^5$$, $$= \frac{1}{{48}}\left[ {\frac{{{{\left( 5 \right)}^2}}}{2} + 3\left( 5 \right) – \frac{{{{\left( 3 \right)}^2}}}{2} – 3\left( 3 \right)} \right]\,\,\,\, = \,\,\,\,\frac{1}{{48}}\left[ {\frac{{25}}{2} + 15 – \frac{9}{2} – 9} \right]$$, $$= \frac{1}{{48}}\left[ {14} \right]\,\,\,\, = \,\,\,\,\frac{7}{{24}}$$, (c) $$P\left( {X \geqslant 4} \right) = \int\limits_4^8 {\frac{1}{{48}}\left( {x + 3} \right)dx} \,\,\, = \,\,\,\frac{1}{{48}}\left[ {\frac{{{x^2}}}{2} + 3x} \right]_4^8$$, $$= \frac{1}{{48}}\left[ {\frac{{{{\left( 8 \right)}^2}}}{2} + 3\left( 8 \right) – \frac{{{{\left( 4 \right)}^2}}}{2} – 3\left( 4 \right)} \right]\,\,\,\, = \,\,\,\,\frac{1}{{48}}\left[ {32 + 24 – 8 – 12} \right]$$, $$= \frac{1}{{48}}\left[ {36} \right]\,\,\,\, = \,\,\,\frac{3}{4}$$, Your email address will not be published. The curve is called the probability density function (abbreviated as pdf). Any observation which is taken falls in the interval. Rule of thumb: Assume a random variable is discrete is if you can list all possible values that it could be in advance. A person could weigh 150lbs when standing on a scale. It’s PMF (probability mass function) assigns a probability to each possible value. What common distribution does this look like? In fact, we mean that the point (event) is one of an infinite number of possible outcomes. Take a look, # convert values to integer to round them, probabilities = [distribution.pdf(v) for v in values], 5 YouTubers Data Scientists And ML Engineers Should Subscribe To, The Roadmap of Mathematics for Deep Learning, 21 amazing Youtube channels for you to learn AI, Machine Learning, and Data Science for free, An Ultimate Cheat Sheet for Data Visualization in Pandas, How to Get Into Data Science Without a Degree, How To Build Your Own Chatbot Using Deep Learning, How to Teach Yourself Data Science in 2020. Just X, with possible outcomes and associated probabilities. The number of possible outcomes of a continuous random variable is uncountable and infinite. A random variable is called continuous if it can assume all possible values in the possible range of the random variable. Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Let’s start with discrete because it’s more in line with how we as humans view the world. Continuous Random Variables Continuous random variables can take any value in an interval. Why is weight continuous? For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. $$f\left( x \right) = c\left( {x + 3} \right),\,\,\,\,2 \leqslant x \leqslant 8$$, (a) $$f\left( x \right)$$ will be the density functions if (i) $$f\left( x \right) \geqslant 0$$ for every x and (ii) $$\int\limits_{ – \infty }^\infty {f\left( x \right)dx} = 1$$. In a continuous random variable the value of the variable is never an exact point. Steps: 1. Therefore, a probability of zero is assigned to each point of the random variable. Plot sample data on a histogram2. Steps:1. Calculate parameters required to generate the distribution from sample4. Now let’s move on to continuous random variables. A normal distribution, hehe. Thus $$P\left( {X = x} \right) = 0$$ for all values of $$X$$. Let’s come back to our weight example. The heat gained by a ceiling fan when it has worked for one hour. 5.2: Continuous Probability Functions The probability density function (pdf) is used to describe probabilities for continuous random variables. The field of reliability depends on a variety of continuous random variables. Use these parameters to generate a normal distribution. The field of reliability depends on a variety of continuous random variables. In a discrete random variable the values of the variable are exact, like 0, 1, or 2 good bulbs. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Required fields are marked *. Let’s do this with our weight example from above. As well as probabilities. Continuous random variables have many applications. Intuitively, the probability of all possibilities always adds to 1. Area by geometrical diagrams (this method is easy to apply when $$f\left( x \right)$$ is a simple linear function), It is non-negative, i.e. Rounded weights (to the nearest pound) are discrete because there are discrete buckets at 1 lbs intervals a weight can fall into. Unlike the PMF, this function defines the curve which will vary depending of the distribution, rather than list the probability of each possible output. Download for free at http://cnx.org/contents/[email protected]. The output can be an infinite number of values within a range. It is always in the form of an interval, and the interval may be very small. They come in two different flavors: discrete and continuous, depending on the type of outcomes that are possible: Discrete random variables. Suppose the temperature in a certain city in the month of June in the past many years has always been between $$35^\circ$$ to $$45^\circ$$ centigrade. Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. There is nothing like an exact observation in the continuous variable. Let M = the maximum depth (in meters), so that any number in the interval [0, M] is a possible value of X. A random variable depends on a function which uses randomness, but doesn’t necessarily output uniform randomness. Here, $$a$$ and $$b$$ are the points between $$– \infty$$ and $$+ =$$. Expected Value Computation and Interpretation. If you liked what you read, please click on the Share button. The temperature can take any value between the ranges $$35^\circ$$ to $$45^\circ$$. Whenever we have to find the probability of some interval of the continuous random variable, we can use any one of these two methods: Properties of the Probability Density Function. Cool. This is a visual representation of the CDF (cumulative distribution function) of a CRV (continuous random variable), which is the function for the area under the curve… Important: When we talk about a random variable, usually denoted by X, it’s final value remains unknown.

http://math.stackexchange.com/questions/286100/absolute-function-continuous-implies-function-piecewise-continuous

# Absolute function continuous implies function piecewise continuous? I have a simple true/false question that I am not sure on how to prove it. If $|f(x)|$ is continuous in $]a,b[$ then $f(x)$ is piecewise continuous in $]a,b[$ Anyone that can point me in the right direction or give a counterexample, even though I think it's true. Thanks in advance! - Hint: What happens if $f$ only takes the values $1$ and $-1$? –  Tobias Kildetoft Jan 24 '13 at 19:50 @Tobias So you think the statement is false? I don't see the problem with a function defined like this: \begin{cases} 1 & x <0\\-1 & x \geq 0\end{cases} –  tim_a Jan 24 '13 at 19:58 What I mean is that if $f$ only takes those two values, then $|f|$ is automatically continuous. But there are nowhere continuous functions with just those two values. –  Tobias Kildetoft Jan 24 '13 at 20:02 Ok, I think I understand what you mean. But the statement is in the other direction. If you already know that $|f|$ is continuous, does this imply that $f$ is piecewise continuous in any case. –  tim_a Jan 24 '13 at 20:07 No, what I mentioned gives you a way to construct a counter example. –  Tobias Kildetoft Jan 24 '13 at 20:08 Define $f(x)$ to be $1$ if $x$ is rational and $-1$ if $x$ is irrational. Now $f$ is not continuous anywhere, but $|f|$ is identically $1$ and thus continuous.

https://mathoverflow.net/questions/92456/if-any-perfect-set-is-uncountable-in-a-metric-space-which-is-not-complete/92465

# If any perfect set is uncountable in a metric space which is not complete? We know that every ferfect set $E$ in a complete metric space $X$ is uncountable. My question is if there exists a metric space which is not complete, but every ferfect set in it is uncountable. The ferfect set here means a closed set which has no isolated points. • (I first thought about giving an example with no "ferfect" sets, but seemed to fail in doing so...) How about just a half-open interval $(0,1]$? Are you sure you have formulated your question correctly? – Tapio Rajala Mar 28 '12 at 14:23 • @Tapio: $(0,1]$ is not complete with the usual distance but completely metrisable. – Jochen Wengenroth Mar 28 '12 at 14:49 • So leaving out a countable subset does not destroy the property that each perfect set is uncountable. So for example $[0,1]\setminus (\mathbb{Q}\cap[0,1])$ might be a more interesting example. – HenrikRüping Mar 28 '12 at 15:23 • Unfortunately, $[0,1]\backslash(\mathbf{Q}\cap[0,1])$ is also completely metrisable: take the distance to be one over the first index at which the continued fraction expressions differ. – Sean Eberhard Mar 28 '12 at 15:56 It is not so easy to construct an example of this kind, I think, because of the Hurewicz theorem: if $X$ is a coanalytic separable metrizable separable space, then either it is Polish (in particular, its perfect subspaces are uncountable) or it contains a closed subset homeomorphic to the rationals (so, a countable perfect closed subset). This is corollary 21.21 in Kechris's book "Classical Descriptive Set Theory". Allowing for the axiom of choice, a Bernstein set will provide a counterexample. This is a subspace $A$ of the real line built by transfinite recursion in such a way that both $A$ and its complement intersect any nonempty perfect subspace of the reals. (this construction is also explained by Kechris, 8.24). EDIT: my earlier argument was incomplete, as pointed out by Andreas Blass, so I'm following his idea below to show that a Bernstein set is indeed a counterexample. If $C$ were a countable closed perfect subset of $A$, then the closure of $C$ would be perfect and closed in $\mathbb R$, hence uncountable. So $\overline C \setminus C$ is an uncountable Borel set contained in the complement of $A$; since an uncountable Borel set contains a Cantor set, the complement of $A$ must then contain a Cantor set, which is impossible since $A$ is a Bernstein set. Thus $A$ does not contain any countable closed perfect set, and $A$ is certainly not completely metrizable since $A$ is not even Borel. It seems plausible to me that, if you want an example which is a subset of $\mathbb R$, then you need some form of the axiom of choice, though I have not thought it through carefully; maybe the residing set theorists will know just how much choice is needed to answer this question... • @Julien: A Bernstein set $A$ contains no perfect set of the ambient space ($\mathbb R$ in the situation you described), but the OP defined "perfect" (or "ferfect") in terms of closed subsets of the space $X$ itself. So I don't think the "there are none!" argument suffices. The example might work anyway, because if a countable set were perfect in $A$, its closure in $\mathbb R$ would contain an uncountable Borel subset of the complement of $A$, and there are none of those. (Caution: I haven't checked this argument carefully, and I probably won't have time to do so soon.) – Andreas Blass Mar 28 '12 at 17:37 • @Andreas: thank you, you are right, I did not think about this carefully enough. I think your suggestion works and I modified my answer accordingly. – Julien Melleray Mar 28 '12 at 18:27

http://math.stackexchange.com/questions/280785/fractal-structure-of-the-sum-of-squares-function

# fractal structure of the sum of squares function The sum of squares function came up at a job interview, corrected for signs and symmetry. $d_2(n)=\#\{(x,y): x^2 + y^2 = n\}$ However, want $(x,y)\sim (\pm x, \pm y) \sim (y,x)$. The first 20 numbers are: 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0 A quick search on Online Encyclopedia of Integer Sequences shows a few candidates: • number of partitions into 2 squares A000161 • the regular paper-folding (dragon curve sequence) A014577 • sequence A082410 So my script is probably correct :-) but I wonder what relation the sum of squares function has with the dragon curve. • Number of ways of writing n as a sum of 2 squares when order does not matter. • Number of similar sublattices of square lattice with index n. • Sequence can be obtained by L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and dropping all L and R • One half of the infinite Farey Tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are: 1,...1,...0,...1,...1,...0,...0,...1,...1,...1,... 1/2.2/3..1/3..3/4..3/5..2/5..1/4..4/5..5/7..5/8,.. Do these objects belong? Are these sets of objects correlated or is it a coincidence? Maybe some relation to the Gauss circle problem. - [Corrected from Erick Wong's comment below.] There is definitely a relationship for small $n$. This depends on the formula in the OEIS page that $$a_n=\frac{1+\left(\frac{-1}{n}\right)}2$$ where $\left(\frac{-1}{n}\right)$ is the Jacobi symbol. It turns out, if $n=x^2+y^2$ has a solution, then $\left(\frac{-1}{n}\right)=1$. For $n<21$, it is also true that $\left(\frac{-1}{n}\right)=-1$ if there is no such solution. So, for $n<21$, $a_n=1$ if and only iff $d_2(n)>0$. In general, if $a_n=0$ then $d_2(n)=0$. It is not true, as I wrote earlier, that if $d_2(n)=0$ then $a_n=0$. For example, $d_2(21)=0$ but $a_{21}=1$. - The characterization of $(-1/n) = 1$ is for the Legendre symbol, not the Jacobi symbol, no? Take, for instance $n=21$ (which is conveniently located just after the explicitly calculated terms in the question...). –  Erick Wong Jan 17 '13 at 17:48 Oh, Duh! You are quite right, brain death. –  Thomas Andrews Jan 17 '13 at 17:49

https://mathhelpboards.com/threads/problem-of-the-week-209-may-31-2016.18735/

# Problem of the Week #209 - May 31, 2016 Status Not open for further replies. #### Euge ##### MHB Global Moderator Staff member Here is this week's POTW: ----- Define the logarithm of an $n\times n$ matrix $A$ by the power series $$\sum_{k = 1}^\infty \frac{(-1)^{k-1}(A - I)^k}{k}$$ which converges for $\|A - I\| < 1$ (the standard matrix norm is being used here). Prove that for all $n\times n$ matrices $A$ and $B$ with $\|A - I\| < 1$, $\|B - I\| < 1$, $\|AB - I\| < 1$, and $AB = BA$, $$\log(AB) = \log A + \log B$$ ----- Remember to read the POTW submission guidelines to find out how to submit your answers! #### Euge ##### MHB Global Moderator Staff member This week's problem was solved correctly by Opalg . You can read his solution below. Let $D = \{z\in\mathbb{C}: |z-1|<1\}$ and $\exp D = \{e^z:z\in D\}.$ Neither of these sets intersect the negative real axis, so the logarithm is defined as a holomorphic function on both sets, and satisfies the condition $$\displaystyle \log z = \sum_{k=1}^\infty \frac{(-1)^{k-1}(z-1)^k}k$$ for $z\in D.$ The holomorphic functional calculus says that given an $n\times n$ matrix $T$ whose spectrum [set of eigenvalues] lies in some domain $U$, there is a continuous unital homomorphism $\phi_T$ from the algebra of holomorphic functions on $U$ (with the topology of uniform convergence on compact subsets) to the $n\times n$ matrices, with $\phi_T(z) = T.$ In particular, if $U = D$ or $U = \exp D$ then this mapping takes the logarithm function to a matrix $\log T$. The continuity of $\phi_T$ ensures that when $U = D$ this definition of $\log T$ agrees with that in the statement of the problem. Also, the exponential and logarithm functions are inverses of each other between the spaces $D$ and $\exp D$. Given $A$ and $B$ as in the problem, the conditions $\|A - I\|<1$ and $\|B - I\|<1$ ensure that the spectrum of each matrix lies in $D$. Let $X = \log A$ and $Y = \log B.$ Then $YX = XY$, because polynomials in $A$ and $B$ commute. Thus $$\exp(X+Y) = \sum_{n=0}^\infty \frac{(X+Y)^n}{n!} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{X^kY^{n-k}}{k!(n-k)!} = \sum_{s=0}^\infty \frac{X^s}{s!} \sum_{t=0}^\infty \frac{Y^t}{t!} = \exp X \exp Y,$$ the change in order of summation being justified because of the uniform convergence of the series on compact sets. Therefore $\exp (\log A + \log B) = \exp(X+Y) = \exp X\exp Y = AB$. The condition $\|AB - I\|<1$ ensures that the spectrum of $AB\;(=\exp (\log A + \log B))$ is in $D$, so we can apply the logarithm function to both sides to conclude that $\log A + \log B = \log(AB)$. Status Not open for further replies.

http://mathdl.maa.org/mathDL/23/?pa=content&sa=viewDocument&nodeId=3321&pf=1

Thinking Outside the Box -- or Maybe Just About the Box Abstract We present a fresh look at an age-old calculus problem, the box problem. The box problem is traditionally used as one of the typical exemplars for the study of optimization. However, this paper illustrates that another box problem, tied much closer to reality, can be used to investigate optimization. In particular, we present an improved box problem and its real-world context along with activities that blend hands-on and applet-based investigations with a strong dose of analysis that culminates in other possible extensions and investigations. Appendix: Student Module (Condensed from the article for student and classroom exploration) Introduction Perhaps one of the deepest entrenched calculus problem is the canonical box problem that students encounter when discussing applied extrema problems. In fact, Friedlander and Wilker (1980) commented, ''This question must be answered nearly a million times a year by calculus students from every corner of the globe'' (p. 282). Without a doubt, almost every recently published calculus text contains a problem similar to the following: A sheet of cardboard is rectangular, 14 inches long and 8.5 inches wide. Congruent squares are to be cut from its four corners. The resulting piece of cardboard is to be folded and its edges taped to form an open-topped box (see figure below). How should this be done to get a box of largest possible volume? Figure 1: Canonical Box Problem In particular, this problem (or some derivative of it) occurs in a classic text such as Thomas (1953) as well as modern texts such as Stewart (2008), Larson, Hostetler and Edwards (2007) and Smith and Minton (2007). But, what about this problem makes it so common and prevalent to the calculus experience? This paper takes a closer look at this common box problem and asks some difficult questions: • Does this box problem really reflect reality? • Is there a better box problem more consistent with modern box building techniques? And • Can a better box problem be explored by Calculus 1 students in a single variable course? If you enjoy this canonical box problem, we developed a Java Sketchpad applet, called OpenBox, for this problem. For this applet, the yellow points are moveable and control the size of the sheet of cardboard and the position of the cut. A dynamical graphic contained in a grey box shows the graph of the Volume to cut length (x) function. In addition, a dynamical picture of the open box is provided to illustrate how the changes in cut length impacts the configuration of the open box. (Warning: The OpenBox page is best viewed in 1024 x 768 resolution or greater and may take up to a minute to load.) Background Historical References The Box problem is so timeless that it even goes back almost a hundred years to the classic Granville and Smith (1911) text. Except for the fact that it is made out of tin, doesn't this look familiar? Figure 2. The box problem on page 115 in Granville and Smith (1911) However, this instance was not the first appearance of the box problem in either an academic publication or in popular culture. A derivation of the box problem first appeared in popular culture in the 1903 Henry Dudeny's puzzle column in the Weekly Dispatch and again in 1908, it was rephrased for his puzzle column in Cassell's Magazine. Below is the graphic and phrasing of the problem contained in Cassell's Magazine: Figure 3. Illustration from Cassell's Magazine. ''No. 525 -- HOW TO MAKE CISTERNS. Here is a little puzzle that will elucidate a point of considerable importance to cistern makers, ironmongers, plumbers, cardboard-box makers, and the public generally. Our friend the cistern-maker has an interesting task before him. He has a large sheet of zinc, measuring eight feet by three feet, and he proposes to cut out square pieces from the four corners (all, of course, of the same size), then fold up the sides, join them with solder, and make a cistern. So far, the work appears to be pretty obvious and easy. But the point that puzzles him is this: What is the exact size for the square pieces that he must cut out if the cistern is to contain the greatest possible quantity of water? Call the feet inches, and take a piece of cardboard or paper eight inches long and three inches wide. By experimenting with this you will soon see that a great deal depends on the size of those squares. To get the greatest contents you have to avoid cutting those squares too small on the one hand and too large on the other. How are you going to get at the right dimensions? I shall award our weekly half-guinea prize for a correct answer. State the dimensions of the squares and try to find a rule that the intelligent working man may understand.'' (Dudeney, 1908, p. 430) But just in case one might think that the problem is only a century old, a Treatise on Differential Calculus by Todhun ter (1855) contains the following related problem: Figure 4. The box problem on page 213 of Todhunter (1855). Consequently, this problem has appeared in calculus texts for at least the last one and a half centuries. It has become so entrenched in the calculus curriculum and articles are still being written about it. For instance, an article by Miller and Shaw (2007) breathed new life into this problem for it served as the backdrop to explore a conventional problem in unconventional ways or an article by Fredrickson (2003) reconceptualizing the constraints on the ''cutouts'' while striving for a container of maximum volume. But what is it about this problem that makes it so universal and mathematics teachers keep coming back to it? Does it really motivate student learning of essential calculus concepts? We concur with Underwood Dudley (2002) when he pointed out in his talk on Calculus Books, "when have you encountered a cardboard box constructed in such a way?" and the problem is ''quite silly'' according to Friedlander and Wilker (1980), Dundas (1984) and Pirich (1996) since the corners are wasted. As previously identified, this particular calculus problem predates modern use of cardboard packaging. But, most would agree that constructing a cardboard box in this manner is quite unusual. So, why do calculus teachers keep including it as part of their repertoire of examples other than the fact that it is straightforward and easily comprehensible? Doesn't the inclusion make students begin to wonder about the ability of calculus to model reality? In fact, if one attempts to rip apart the classic open box used for the typical shirt box that one receives at Christmas, it is not constructed in this manner. It is interesting to note that both the Granville and Smith text and Dudeney's puzzle used tin, a little more practical material than cardboard for such a construction. A box of cardboard might use the squares as tabs to connect the sides together and increase stability and strength. In fact, we have often asked students to construct boxes from index cards we gave them and they tend to do use these tabs naturally to make the connections with tape, glue or a staple. The students can typically answer the mathematical questions of the corner-removed box problem but at the same time they recognize the lack of realism when the corners are removed since boxes just are not constructed that way. Introducing the Regular Slotted Container It is our contention that there is a better box problem to be examined - a box problem that students will see as realistic as well as to be able to explore with tools from single-variable calculus. This adventure began a little over 25 years ago when one of the authors was curious about how to model realistic shipping boxes and tore one apart to find out. His first exploration was with a box known in the shipping industry as a Regular Slotted Container (RSC). The RSC is perhaps one of the most commonly utilized class of boxes in the shipping industry because it is one of the most economical boxes to manufacture and adapt for the shipment of most commodities. At first exploration, the construction of a RSC appears to require two variables but a simple but practical constraint makes it a problem appropriate for single variable calculus. Let's first explain what exactly a RSC is. As shown below in figures 5 and 6, a Regular Slotted Container is constructed from a large rectangular sheet of corrugated cardboard. Depending on the construction details, it may be manufactured with a tab used to either glue or stitch the joint (the inclusion of the tab results in waste cardboard during production) or may rely on a taped joint making it a minimal shipping container (i.e. a box where there is ''no'' scrap corrugated cardboard generated in the manufacturing process). In either case, the lengthwise (normally outer) flaps meet at the center of the box allowing it to be affixed by tape or staples. Figure 5: Construction schematics for the RSC (Safeway Packaging, n.d.) The box is closed Top flaps opened All flaps opened Completely disassembled Figure 6. Various stages of a RSC being disassembled Movie of the construction of the RSC It is difficult to conceive of how a blank sheet of cardboard with cuts at appropriate points and to particular lengths is transformed into a box that is commonly used across the shipping industry. In order to get a clear sense of this, we provide a link to a movie that shows the SWF Companies (n.d.) CE-451 manufacturing a Regular Slotted Container. Warning: The CE-451 movie is 12 MB and may take more than a minute to load. What one sees in this movie are the blanks being loaded into a hopper and the robotic elements of the CE-451 taking those blanks and converting them into RSCs. What it does not show is how the cardboard was sliced, the positioning of those slices and how the slice lengths help maximize the volume capacity of the RSC. Discussing Dundas's (1984) analysis In a paper by Dundas (1984), he looked at a variety of box problems and one of them was configured like the RSC. The tactic that Dundas (1984) took was to look at the problem from the perspective of a piece of cardboard of fixed area rather than a fixed dimensions. We find this choice mathematically sound but problematic to the exploration of the real-world situation since cardboard is not typically sold by area but rather constrained by manufacturable lengths and widths. Consequently, when Dundas (1984) explored the function, $V(l) = T\left( {{1 \over 2} - T} \right)\left( {{A \over l} - T} \right)$ there was no way to convey to the reader information about the length and width of the cardboard (since the cardboard area was being restricted to A square inches). In addition, a critical piece of the problem is what the lengths of the cuts should be, since that transforms the problem from a multiple-variable problem to a single-variable problem. Unfortunately for a student reading the paper, it was never motivated or explained. It is these particular elements that we wish to revisit and illustrate using text and a set of applet-based activities to lead students to similar conclusions without making the activity all about pushing algebra around on a page. The First Box Problem Introduction Being able to step back and consider the problem without many constraints allows one to see the interdependence of the cut length, positioning of cuts, and the volume of the box. However, will students recognize that if they encounter the following box problem? A sheet of cardboard is rectangular, 14 inches long and 8.5 inches wide. If six congruent cuts (denoted in black) are to be made into the cardboard and five folds (denoted by dotted lines) made to adjoin the cuts so that the resulting piece of cardboard is to be folded to form a closeable rectangular box (see figures below). How should this be done to get a box of largest possible volume? Figure 7: The box problem graphics Before interacting with the Box Problem applet, students should be given the opportunity to physically build their own boxes out of paper or cardboard using scissors and tape or dissect a provided RSC. In doing so, students' physical constructions or destructions will help them recognize the particular physical constraints and how elements of the RSC come together prior to exploring the virtual world presented in the applets. Questions that can be asked of the students during this initial phase of exploration, if they don't naturally generate them on their own, are: 1. What does a ''closeable rectangular box'' infer about the characteristics of the box? 2. What impact does "closeable" have with respect to the length of at least a pair of the flaps? 3. Should the flaps overlap? and 4. If they do overlap, is that the best use of the cardboard? After students have constructed boxes in this manner then we suggest that students begin to interact with a set of Java applets designed to help students interact with the problem in ways that would not be possible when building boxes out of cardboard. But where does one go from there? Is moving expressly toward the algebraic solution the only path? It is our contention that considerably more learning can occur if this path is at least delayed a little while longer so students can be asked to explore the problem with greater depth. The pertinent question here is how does one encourage students to search for deeper insights that can be drawn from such a seemingly simple problem. We suggest the use of a pair of applets to help students investigate the problem in increasingly more complex ways and open their eyes to conditions and constraints that were not obvious through their initial encounters. The first Box Problem applet One element in the applets that we have designed are the multiple representations and alternative ways of conveying information. For instance, the image shown below is a screen shot from the ClosedBox applet, which allows students to explore various scenarios. The student can manipulate the lengths A and B by just pulling on the yellow points A' and B'. The yellow points P and Q also move. In doing so, the other components of the applet change in accordance with these manipulations. The power here is that students can interact with a wide variety of examples and see if the conjectures they make hold up to empirical investigation. In addition, the student can move the corners of the box, in the lower right-hand corner of the applet, and see if the box will actually close or not, an important aspect if you want the box to hold something. The last elements in this applet are the two different graphical indicators of maximal volume. The one graph shows the volume with respect to P or Q while the other is held constant and the bar graph next to it displays the percentage of maximal volume obtained by the current configuration. If the volume is too large, one can resize the vertical unit, a yellow point denoted by U, to get the graphs comfortably into the grey viewing window. As students play with these to attempt to improve their maximal value score, they will be led to ask a variety of questions, such as: • Is there a relationship between the cut length and the position of the cut that maximizes the volume? • Under what conditions does the volume drop to zero? • Is there a best way to orient P and Q so that the maximum is achieved? • How do the two graphics on the bottom left-hand side interact with each other? Figure 8: The first Box Problem applet Warning: The first Box Problem applet page, entitled ClosedBox, is best viewed in 1024 x 768 resolution or greater and may take up to a minute to load. The goal of this exploration is to help students recognize that even though this problem may at first appear to be a problem involving two variables, the cut length and the position of the cut, focusing on maximizing the volume allows one to turn the problem into a problem of one variable. By exploring the applet, students should find that if the cut length is held constant and the position of the cut is allowed to vary, then the volume of the corresponding box is described by a constrained quadratic equation (expressed by the blue graph), and if the position of the cut is held constant while the length of the cut is allowed to vary, the volume of the box is described by a constrained linear equation (expressed by the red graph). The constraints result from the physical constraints on the box as well as whether, when the box is constructed, it will hold its contents, i.e. the flaps meet or overlap so there are no holes in the box. All it takes is some imagination after working through multiple examples and noticing the relationship between these two variables that maximizes the volume for any particular condition. From our perspective, being able to explore, quantify, and utilize is an important aspect of learning mathematics. The first Box Problem activity This applet was designed to help students come to a realization that for any positioning of the fold, the maximum volume can be best achieved when the cut length is set to one-half the smallest width of the box. Not only can students disassemble the box by pulling on the corner points, the applet helps students investigate how constraints affect the box problem. For instance, if the cut length exceeds the minimum of the length and width of the box, then the box will not close because the flaps created by the cuts will interfere with each other. Alternately, if the cut length is less than half of the minimum of the length and width of the box, then the box will not close because the flaps created will leave a hole in the box, thereby allowing the contents to fall out. So, how does one accomplish this? In the applet, the points denoted with a yellow dot are those that are moveable. Perhaps, the first question that students should be asked concerns the midpoint fold. In particular, why does one of the folds need to occur at the midpoint? Another question should focus on the symmetry. In fact, asking ''what happens if the two folds aren't symmetric about the midpoint fold?'' can yield interesting conversations amongst the students. Getting back to the movable points, a student can set a position for where the cut is to occur and then vary the length of the cut. We suggest students carefully record the data from their various trials where the cut position, $$m( \overline{AP})$$, is set and the cut length, $$m( \overline{BQ})$$, is changed to maximize the volume for that particular cut position. The last row of each Trial is for the student to search for the maximal cut length and the corresponding cut position. The final trial is for an open exploration of a piece of cardboard of the student's choosing. $$m( \overline {BQ})$$ $$m( \overline {AP})$$ $$m( \overline {PM})$$ $$m( \overline {QQ'})$$ $$m( \overline {BB'})$$ $$m( \overline {AA'})$$ % of max volume 1.5 3.0 4.0 5.5 8.5 14.0 97.75% 3.5 3.5 8.5 14.0 4.5 2.5 8.5 14.0 8.5 14.0 2.5 2.5 10.0 10.0 1.5 3.5 10.0 10.0 4.25 0.75 10.0 10.0 10.0 10.0 3.0 2.5 8.5 11.0 2.75 2.75 8.5 11.0 4.0 1.5 8.5 11.0 8.5 11.0 Exploring various positions of a cut, should lead the students' to a conjecture that for any position of a cut, the maximum volume of the corresponding box is obtained when the cut length is half the length of the shorter length between $$m( \overline {AP})$$ and $$m( \overline {PM})$$. There are so many different questions that can be asked when students are interacting with the first Box Problem applet. In addition to those already mentioned, one can ask: • Under what conditions is the blue graph connected above the x-axis? • Under those conditions and for a particular length and width, what are the dimensions of the box with maximum volume? • Why does the red graph always appear as a slanted portion? What mathematical meaning does this red graph have? Does this slanted portion ever change slope? If so, why and if not, why not? It should be mentioned that we have found that if students have experienced the open box problem, there is a tendency of students to either focus on the length of the cut or the position of the cut but not typically both. Some students want to make the controlling variable the position of the cut and the resulting calculus computations are more difficult but not impossible. However, if we were to focus on the computations and not a deep exploration of the problem, students would miss a perfect opportunity to look beyond the numbers and see relationships that are intrinsically interwoven in this problem. The Second Box Problem Introduction After exploring the first Box Problem applet, some essential elements of moving a seemingly two-variable problem into a single variable problem have come to light. In particular, students are now prepared to explore the single-variable box problem. Specifically, the introduction provided by the first box problem applet has provided a means of reducing the complexity of the problem. Instead of having two variables with which to contend, we have found that without loss of generality, we can set the position of the cut to be twice the length of the cut. What exactly does this provide? We can now express the volume of the box with respect to a single variable thereby opening this problem to the techniques from Calculus 1. In particular, we can turn our attention to maximizing the volume of a box based upon the cut length. Now, the second Box Problem applet, will focus its attention on looking at $V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right)$ where A and B are the correspondent length and width of the rectangular piece of cardboard and l corresponds to the length of the cut. We should note that this description is equivalent to the formula presented in Dundas (1984) of $V(l) = T\left( {{1 \over 2} - T} \right)\left( {{A \over l} - T} \right)$ • $$T = 2l$$ • $$l_{Dundas} = A$$ • $$B = w = {{A_{Dundas} } \over {l_{Dundas} }}$$ In addition, the l of Dundas (1984) needs to correspond to the position of the cut instead of the length of the cut. The second Box Problem applet At first glance, this applet, ClosedBox2, contains many of the same components as the first Box Problem applet; however, the cut length determines the positioning of the cut so that in each case the box volume is relatively maximized. Figure 9: The second Box Problem applet Warning: The second Box Problem applet page, entitled ClosedBox2, is best viewed in 1024 x 768 resolution or greater and may take up to a minute to load. In this applet there are a variety of elements that can be seen. First, the point P is no longer adjustable but is rather determined by the length of the cut defined by the segment BQ. In addition, the grey box in the lower left-hand of the applet contains a dynamic graphical depiction of the functional relationship between cut length and volume. That is, it contains a graphical depiction of $V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right)$ where l corresponds to $$m( \overline {BQ} )$$ and currently B = 8.5 and A = 14.0. One element that students need to grapple with when working with this particular applet is the graphical depiction of the function. In particular, graphing $$V(l)$$ using a graphing calculator or computer algebra system yields a figure similar to the following: Figure 10: Graph of the Box Problem Function The graphical depiction of the function presented in the applet is a truncated version of the one in figure 10. Students will need to come to grips with the fact that the applet is only concerned with the volume of boxes that are physically constructible whereas the graph of the function shown in figure 10, does not necessarily concern itself with the constructability of the box. Instead, it provides a graph of the functional relationship between an independent variable l and a dependent variable V. In essence, this graph in figure 10 is less concerned with cut length and volume and more concerned with expressing the relationship for all possible values of l, irrespective if these values are possible cut lengths or if those cut lengths yield appropriate volumes. Too often, we see students focused on the algebraic elements of a problem without considering the physical (or mathematical) constraints on that problem. We seek in this applet to guide students to grapple with the interplay of these two seemingly disparate forces. In turn, we hope to lead them to reconcile for themselves how functional relationships that model real-world phenomena require a careful examination of the domain for which that relationship actually does model the phenomena. For instance, students might at first think that l's only restriction is that it must be less than $${1 \over 2}m( {\overline {BB'} } )$$ since a cut cannot exceed half the width of the cardboard and maintain its connectedness. However, there is another constraint: the cut length, l, cannot exceed $${1 \over 4}m( {\overline {AA'} } )$$ . This ''hidden'' constraint comes directly from the relationship of cut length and position of the cut and depends on the relationship between length and width of the rectangular piece of cardboard. The second Box Problem activity This applet was designed to help investigate the interrelationship between cut length and volume. In particular, the previous interactions with the first Box Problem applet allowed the student to recognize the relationship between cut length and position of the cut. Using this, the student can build a functional relationship, based on a single variable, relating cut length and volume. This applet is designed to explore that functional relationship and help students come to a realization that in modeling real-world phenomena, a careful examination of the domain of a the functional relationship is necessary. How can one ask questions to help students interact with the second Box Problem applet in meaningful ways? From our experience, the first question that probably should be asked involves the relationship between the first Box Problem applet and this second Box Problem applet. In particular, we can see in this new applet that the point P varies as the point Q moves, what information drawn from the first Box Problem applet allows us to control the point P? Another question should focus on constructing the volume of the box based upon the length of cut. Essentially, a series of questions to help lead students to developing a functional relationship between volume and cut length, such as: • How does one determine the volume of a rectangular box? • What would be a description for the length of the box? • What would be a description for the width of the box? • What would be a description for the height of the box? • How can you use these descriptions to build a functional relationship between volume and cut length? Inherent to these questions are questions about what are the constants and what are the variables? Such a question is important when students are faced with multiple letter-based descriptions such as those expressed in the formula for $$V(l)$$ : $V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right)$ Once students have developed appropriate functional relationships between cut length and volume, a variety of questions relating to that function can be asked. In general, we typically ask students to investigate the graph of the function on a hand-held graphing calculator. And then, we start asking various questions to help the students compare and contrast the graphical representation produced by the graphing calculator and that of the applet. For instance, we might ask: • Why does the graphing calculator seem to show you more of the graph than the applet does? • Why does the applet truncate the graph? • What conditions should be on the domain of the function and how do they relate to physically constructing a box? So far, these questions have focused primarily on a static rectangular sheet of cardboard. That is, we have not asked questions that forced students to think about varying the size of the cardboard. • For different lengths or widths, the applet's graph seems to change shape near the right-hand terminus, what mathematical reason can you provide for this change or provide an argument that it does, in fact, not change? • Are there two (or more) non-isomorphic sheets of cardboard, so the maximal volume is the same? If so, identify them and if not, explain why not. • Are there two (or more) non-isomorphic sheets of cardboard, so the placement of the maximal cut is the same? If so, identify them and if not, explain why not. Here, the questions are designed to force students to think beyond the original question and attempt to distill commonality and explore generalizations. Stretching students to think beyond and to ask ''what if'' questions is an important aspect of interacting with this applet. We sought to design the applet to support both the investigation of a specific scenario as well as a broad range of extended questions related to the original problem. Reflections and Extensions This paper has attempted to illustrate how applets can broaden the investigation of mathematical relationships and reach beyond mere algebraic manipulations. The ability to investigate, conjecture, test, and pose new questions in an electronic environment is important to the development of understanding. Beyond this, such investigations are enjoyable for students. For instance, students mentioned the box problem in their weekly Calculus 1 journals and two examples are provided below: ''This last week we learned of Indeterminate terms and L' Hopital's rule. Did quite a few examples on how to perform the rule, contionued [sic] this for four days, then I think we made boxes. I believe that I learned how to do L' Hop's rule fairly well, hopefully I could do well on a test, and the box making was fun and surprised me in that you fit it into the lesson. I enjoyed the interactivity.'' - Charles ''This week we learned about maximum volume of boxes and solved a pretty cool problem involving James Bond and Bombs. I love these sorts of problem-solving and I hope that we do more of it in the future.'' - Dan Our goal has always been to get students to think that problem-solving is fun and the interaction between building physical boxes and the computer applets helps students draw essential mathematical connections. The lessons that can be drawn from these activities bolster students' perceptions that mathematics is eminently useful and applicable to real-world phenomena much more than the typical box problem shown in many of the calculus texts available on the market. But is that as far as one can go into the world of boxes? We should also mention that this problem of optimal box building could be further extended by including the tab to affix the sides of the box to each other without the use of a taped joint. At one level, our discussion has focused its attention on construction of a minimal shipping container of the Regular Slotted Container type; however, if we were to introduce the requirement of a tab it does impact the problem. It should be noted that there are general governmental guidelines for the width of such a tab but 1 1/4 - 1 1/2 inches is typical.  For specific guidelines about tab regulations, one should check the  Fiber Box Handbook (Fiber Box Association, 2005), the National Motor  Freight Classification (NMFC) Item 222 (American Trucking  Association, 1970) for common carrier motor freight shipment, or the  Uniform Freight Classification (UFC) Rule 41 (Dolan, 1991) for rail  shipment.  In any case, we think that after exploring the box problem through these applets, students might be better prepared to think how to construct a RSC containing tabs (as shown in figure 5) with maximum volume? Beyond even looking at a tabbed RSC, the world of boxes is much more encompassing. In this paper, we have focused our attention on the most popular box, a regular slotted container, but there are a host of other types of boxes fabricated from a single piece of corrugated cardboard that students can investigate, including: • A Full Overlap Container (FOL), see figure 10a, that is resistant to rough handling since all flaps are of the same length (the width of the box) and when closed, the outer flaps come within one inch of complete overlap, • A Five Panel Folder (FPF), see figure 10b, that features a fifth panel used as a closing flap completely covering a side panel and includes end pieces composed of multiple layers that provide stacking strength and protection of long articles with small diameter, or • A Once Piece Folder (OPF), see figure 10c, typically used books and printed materials since it has a flat bottom with flaps forming the sides and ends along with extended flaps that meet toform the top (GoPackaging.com, 2006). (a) (b) (c) Figure 11: Other types of boxes manufactured from a single piece of corrugated cardboard Each of these boxes would provide an opportunity to extend investigation and gain a better sense of the issues behind box construction. We certainly hope that this paper hasn't boxed you in but rather opened your eyes to the ways boxes can be used to help students see the ability of calculus to play a meaningful role in examining real-world problems. References American Trucking Association. (1970). National motor freight classification Washington, DC: National Motor Freight Traffic Association, Inc. Dolan, J.J. (1991). Uniform Freight Classification Rule 41. Chicago, IL: National Railroad Freight Committee. Dudeney, H.E. (1908, September). The puzzle realm. Cassell's Magazine. p. 430. Dudley, U. (2002, October). ''Calculus Books''. Paper presented at the meeting of the MAA Ohio Section with OhioMATYC at Kent State University Trumbull Campus, Warren, OH. (many of the same remarks were published in: Dudley, U. (1988). Calculus with Analytic Geometry by George F. Simmons. The American Mathematical Monthly, 95(9), 888-892. Available at: http://www.jstor.org/stable/2322923). Dundas, K. (1984). To build a better box. College Mathematics Journal, 15(4), 30-36. Available at: http://www.jstor.org/stable/3027427 Fredrickson, G.N. (2003). A New Wrinkle on an Old Folding Problem. The College Mathematics Journal, 34(4), 258-263. Friedlander, J.B. & Wilker, J.B. (1980). A budget of boxes. Mathematics Magazine, 53(5), 282-286. Fiber Box Association. (2005). Fiber Box Handbook. Rolling Meadows, IL: Fiber Box Association. GoPackaging.com (2006). Custom Corrugated Shipping Boxes. Retrieved October 23, 2008, from http://www.gopackaging.com/custom-boxes.aspx. Granville, W.A. & Smith, P.F. (1911). Elements of the Differential and Integral Calculus. Boston, MA: Ginn & Co. Larson, R., Hostetler, B. & Edwards, B. (2007) Calculus: Early Transcendental Functions, 4th Edition. Belmont, CA: Houghton Mifflin. Miller, C.M. & Shaw, D. (2007). What else can you do with an open box? Mathematics Teacher, 100(7), 470-474. Pirich, D.M. (1996). A new look at the classic box problem. PRIMUS, 6(1), 35-48. Safeway Packaging. (n.d.). RSC.bmp. Retrieved October 23, 2008, from http://www.safewaypkg.com/images/Design_Catalog/RSC.bmp. Smith, R.T. & Minton, R.B. (2007). Calculus: Early Transcendentals, 3rd Edition. New York, NY: McGraw Hill. Stewart, J. (2008). Calculus, 6th Edition. Belmont, CA: Brooks/Cole Publishing Company. SWF Companies. (n.d.). CE 451 Case Erector. Retrieved October 23, 2008, from http://www.swfcompanies.com/CE451.htm. Thomas, G.B. (1953). Calculus and Analytic Geometry, 2nd Edition. Reading, MA: Addison Wesley. Todhunter, I. (1855) A treatise on the differential calculus, 2nd Edition, Revised. London, UK: McMillan & Co.

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Share # Using Differentials, Find the Approximate Value of the Following up to 3 Places of Decimal (26.57)^(1/3) - CBSE (Commerce) Class 12 - Mathematics #### Question Using differentials, find the approximate value of the following up to 3 places of decimal (26.57)^(1/3) #### Solution (26.57)^(1/3) Is there an error in this question or solution? #### APPEARS IN NCERT Solution for Mathematics Textbook for Class 12 (2018 to Current) Chapter 6: Application of Derivatives Q: 1.12 | Page no. 2016 #### Video TutorialsVIEW ALL [2] Solution Using Differentials, Find the Approximate Value of the Following up to 3 Places of Decimal (26.57)^(1/3) Concept: Approximations. S

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# What is the speed of an object that travels from ( -9,4,-6) to ( 7,1,-2 ) over 3 s? Mar 11, 2018 Well it is not said that by which path the object reached its end point from initial point of journey. Distance is the direct path length which we need to know for calculating speed. Let's consider that here the object went in a straight line so that displacement=distance I.e $\sqrt{{\left(7 - \left(- 9\right)\right)}^{2} + {\left(1 - 4\right)}^{2} + {\left(- 2 - \left(- 6\right)\right)}^{2}} = 16.75 m$ So,speed = distance /time = $\frac{16.75}{3} = 5.57 m {s}^{-} 1$

https://cs.stackexchange.com/questions/7029/inapproximability-result-implies-apx-hardness

# Inapproximability result implies apx-hardness? If an optimization problem is known to be inapproximable up to some precision, does this automatically imply that the problem is apx-hard? If P$=$NP then there is a polytime algorithm for any NP optimization problem, and so every problem is APX-hard. So from now on, suppose that P$\neq$NP. We can associate with any decision problem $L \in NP$ an optimization problem $o_L$ by defining $o_L(x) = 1$ if $x \in L$ and $o_L(x) = 2$ otherwise. Suppose $o_L$ reduces to $o_M$ via a PTAS reduction $(f,g,\alpha)$ (notation as in the Wikipedia article). Let $C = 1+\alpha(1/2)$. Thus if $y$ is a $C$-approximation to $o_M(f(x))$ then $g(x,y,1/2)$ is a $3/2$-approximation to $o_L(x)$, from which we can determine whether $x \in L$ or not. Altogether, this gives a polytime reduction from $L$ to $M$. Now take any NP-complete language for $L$ and any NP-intermediate language for $M$ (such a language exists because of Ladner's theorem). On the one hand, since $M \notin P$, you cannot 2-approximate $o_M$. On the other hand, $L$ doesn't reduce to $M$ (since that would apply that $M$ is NP-complete), and so $o_L$ doesn't reduce to $o_M$. We conclude that $o_M$ is not APX-hard. • I don't see why the existence of a PTAS-reduction from $o_L$ to $o_M$ implies the existence of a polytime reduction from $L$ to $M$. Suppose $L$ is the total set and $M$ is the empty set. Let $q$ be the optimal solution for all instances of $o_L$. If $f$ is the identity function and $g$ fulfills $g(y) = q$ for all $y$, then $f,g$ give a PTAS-reduction from $o_L$ to $o_M$: all solutions are optimal in $o_M$, and these $1$-ratio solutions in $o_M$ are mapped into $1$-ratio solutions in $o_L$. Still, due to the definition of $L$ and $M$, there is no many-one polytime reduction from $L$ to $M$. – EXPTIME-complete Mar 23 at 12:41 • It's the truth value of $f(x) \in M$. – Yuval Filmus Mar 23 at 13:26

http://mathoverflow.net/questions/15415/infinitely-many-prime-numbers-of-the-form-n2k1/15418

# Infinitely many prime numbers of the form $n^{2^k}+1$? I am not a specialist of number theory, so please excuse my ignorance: is the following question still an open problem? Let $k \in \mathbb{N}^*$, are there infinitely many prime numbers of the form $n^{2^k}+1$? - What's known about the problem if both n and k are allowed to vary? – Qiaochu Yuan Feb 16 '10 at 4:42 @Qiaochu: In this problem, allowing n and k to vary among positive integers is the same as setting k=1 and letting n vary. – Bjorn Poonen Feb 16 '10 at 5:14 Your question is still open. It is a special case of Schinzel's Hypothesis H applied to the polynomial $f(x)=x^{2^k}+1$. As Bjorn mentions in his comment, the case of $k=1$ is a particularly famous unsolved problem. It is the fourth of Landau's problems (Edmund Landau was a famous German number theorist during the early twentieth century). - Thank you Ben, this is useful. – Portland Feb 16 '10 at 5:04 This may not be clear from the Wikipedia article: Bunyakovsky's conjecture is not known for any particular polynomial of degree greater than 1. – Qiaochu Yuan Feb 16 '10 at 5:11 It took me a while to find this: http://www.pnas.org/content/94/4/1054.full Anyway by Friedlander and Iwaniec (1997). They proved that there are infinitely many primes of the form $x^2 + y^4 .$ They mention near the end that they do not have a proof for primes of the form $x^2 + y^6$ but would like one. So there is a way to go to settle $x^2 + 1.$ FYI, what I did (not remembering title, authors, anything but the result) was write a program to give the primes $x^2 + y^4$ and put the first dozen in Sloane's sequence site search feature. - I think that the best result thus far is Iwaniec's proof that there are infinitely many positive integers n such that n^2+1 is divisible by at most 2 primes. – Ben Linowitz Feb 16 '10 at 5:23 oeis.org/classic/A028916 – Charles Sep 13 '10 at 16:03 Yes, it is still an open question - [citation needed]? – Alison Miller Feb 16 '10 at 3:24 The conjectured answer is yes: see en.wikipedia.org/wiki/Bunyakovsky_conjecture or the vast generalization in en.wikipedia.org/wiki/Bateman-Horn_conjecture . But it is certainly unknown since a positive answer for any k would give a positive answer for k=1, which is a famous unsolved problem. – Bjorn Poonen Feb 16 '10 at 3:34 @Alison: it seems to me to be a non-trivial problem to give a citation which proves that a given question is still open! – Kevin Buzzard Feb 16 '10 at 9:48 The conjecture that there are infinitely many primes of the form n^2+1 is the first entry (A1) in Richard Guy's Unsolved Problems in Number Theory. – Franz Lemmermeyer Feb 16 '10 at 13:42 There's a version of Guy's book in our maths library that says it's a conjecture that x^n+y^n=z^n has no non-trivial solutions when n>=3 ;-) – Kevin Buzzard Feb 17 '10 at 11:51

https://math.stackexchange.com/questions/1343266/binary-quadratic-optimization-problem

I am trying to solve the following binary quadratic program. $$\min_{\Delta} \Delta^T H \Delta + c^T\Delta \\ \text{Such that:} ~~~\Delta\in \{0,1\}^n ~~\text{and}~~ \sum_{i=1}^n \Delta_i \leq \Gamma$$ where $H$ is not a positive semidefinite matrix (and hence the minimization problem is not convex) and $\Gamma$ is a fixed natural number less than $n$. I suppose that one might consider this problem a loosely cardinality-constrained quadratic program. I have been somewhat unsuccessful in trying to find a literature on this kind of optimization problem. I was wondering if anyone here might be able to refer me to some good resources on non-convex binary quadratic optimization with a linear constraint as above. Thanks. The indefiniteness of H is a non-issue when you only have binary variables, as you always can add the zero-term $\lambda \Delta^T\Delta - \lambda \sum \Delta_i$. If you pick $\lambda$ large enough (larger than $-\lambda_{min}(H)$ where $\lambda_{\min}$ denotes smallest eigenvalue), the objective is convex.

https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-9-inequalities-and-problem-solving-9-2-intersections-unions-and-compound-inequalities-9-2-exercise-set-page-590/53

Chapter 9 - Inequalities and Problem Solving - 9.2 Intersections, Unions, and Compound Inequalities - 9.2 Exercise Set - Page 590: 53 $$(4,8]$$ Work Step by Step To graph a one-variable inequality, first use the rules of inequalities to solve the inequality. (Recall, when multiplying or dividing by a negative number, the inequality sign must be flipped.) Next, shade the region that satisfies the inequality. At the end points, if the end point is included, use a solid dot. Otherwise, do not fill the dot in. The solution to the inequality is: $$(4,8]$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

https://quant.stackexchange.com/questions/45551/one-period-binomial-option-valuation-model

# One Period Binomial Option Valuation Model [closed] My question here is how is the probability of an up move calculated by $$(1+Rf-D)\over(U-D)$$ derived where Rf is the risk free rate, D is the down move factor and U represents the up move factor. Kindly help me understand the derivation and where this formula comes from as this does not make any intuitive sense to me. • In a risk-neutral world the expected future asset price equals the forward price: $$p( \, US\,) +(1-p)(\,DS\, ) = S(1 + r_f) \implies p = \frac{1 + r_f - D}{U - D}$$ – RRL May 11 at 17:12

https://plainmath.net/38163/from-0-2-sin-x-2-dx-using-trapezoidal-rule-with-n-equal-4-as-well

# From 0-2 \sin(x^{2}) dx using trapezoidal rule with n=4 as well From 0-2 $$\displaystyle{\sin{{\left({x}^{{{2}}}\right)}}}{\left.{d}{x}\right.}$$ using trapezoidal rule with n=4 as well as midpoint rule with n=4 • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Glenn Cooper Step 1 Trapezoidal rule $$\displaystyle={\int_{{{0}}}^{{{2}}}}{{\sin}^{{{2}}}{\left({x}\right)}}\ldots\ldots{\left({1}\right)}$$ As we know that accoding to trapezoidal rule $$\displaystyle{\int_{{{a}}}^{{{b}}}}{f}{\left({x}\right)}{\left.{d}{x}\right.}\approx{\frac{{\triangle{x}}}{{{2}}}}{\left({f{{\left({x}_{{{0}}}\right)}}}+{2}{f{{\left({x}_{{{1}}}\right)}}}+{2}{f}{\left({x}_{{{2}}}\right)}+\ldots+{2}{f{{\left({x}_{{{n}-{1}}}\right)}}}+{f{{\left({x}_{{{n}}}\right)}}}\ldots{\left({2}\right)}\right.}$$ This is the standard equation for the trapezoidal method in which $$\displaystyle\triangle{x}={\frac{{{b}-{a}}}{{{n}}}}$$ Now comparing equation (2) with equation (1) we get that a=0,b=2,n=4 $$\begin{cases}\triangle x=\frac{2-0}{4}=\frac{1}{2}\\ & \end{cases}$$ Now we will divide (0,2) interval into n=4 subinterval of length $$\displaystyle\triangle{x}={\frac{{{1}}}{{{2}}}}$$ end points of the function $\begin{cases}0\\ 0+\frac{1}{2}=\frac{1}{2} & \\ \frac{1}{2}+\frac{1}{2}=1 & \\ 1+\frac{1}{2}=\frac{3}{2} & \\ \frac{3}{2}+\frac{1}{2}=2 & \end{cases}$ Therefore the end points of the function are $$\displaystyle{a}={0},{\frac{{{1}}}{{{2}}}},{1},{\frac{{{3}}}{{{2}}}},{2}$$ U sing these end points we analysis the function $$\displaystyle{f}{\left({x}_{{{0}}}\right)}={f{{\left({a}\right)}}}={f}{\left({0}\right)}={0}$$ $$\displaystyle{2}{f}{\left({x}_{{{1}}}\right)}={2}{f{{\left({\frac{{{1}}}{{{2}}}}\right)}}}={2}{{\sin}^{{{2}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}={0.459698}$$ $$\displaystyle{2}{f}{\left({x}_{{{2}}}\right)}={2}{f{{\left({1}\right)}}}={2}{{\sin}^{{{2}}}{\left({1}\right)}}={1.416147}$$ $$\displaystyle{2}{f}{\left({x}_{{{3}}}\right)}={2}{f{{\left({\frac{{{3}}}{{{2}}}}\right)}}}={2}{{\sin}^{{{2}}}{\left({\frac{{{3}}}{{{2}}}}\right)}}={1.9899925}$$ $$\displaystyle{f}{\left({x}_{{{4}}}\right)}={f{{\left({b}\right)}}}={f}{\left({2}\right)}={{\sin}^{{{2}}}{\left({2}\right)}}={0.0826819}$$ Substitute these value in equation (2) we get that $$\displaystyle={\frac{{{1}}}{{{4}}}}{\left[{0}+{0.459698}+{1.416147}+{1.9899925}+{0.0826819}\right]}$$ $$\displaystyle={1.173165646}\approx{1.1737}$$ Step 2 By midpoint rule $$\displaystyle={\int_{{{0}}}^{{{2}}}}{{\sin}^{{{2}}}{\left({x}\right)}}\ldots.{\left({1}\right)}$$ According to mid point rule we know that $$\displaystyle{\int_{{{a}}}^{{{b}}}}{f}{\left({x}\right)}{\left.{d}{x}\right.}\approx\triangle{x}{\left({f}{\left({\frac{{{x}_{{{0}}}+{x}_{{{1}}}}}{{{2}}}}\right)}+{f}{\left({\frac{{{x}_{{{1}}}+{x}_{{{2}}}}}{{{2}}}}\right)}+{f}{\left({\frac{{{x}_{{{2}}}+{x}_{{{3}}}}}{{{2}}}}\right)}+\ldots.+{f}{\left({\frac{{{x}_{{{n}-{1}}}+{x}_{{{n}}}}}{{{2}}}}\right)}\right)}$$ $$\displaystyle\triangle{x}={\frac{{{1}}}{{{2}}}}$$ End points are $$\displaystyle={0},{\frac{{{1}}}{{{2}}}},{1},{\frac{{{3}}}{{{2}}}},{2}$$ Now calculate for these values $$\displaystyle{f}{\left({\frac{{{x}_{{{0}}}+{x}_{{{1}}}}}{{{2}}}}\right)}={f}{\left({\frac{{{0}+{\frac{{{1}}}{{{2}}}}}}{{{2}}}}\right)}={f}{\left({\frac{{{1}}}{{{4}}}}\right)}={{\sin}^{{{2}}}{\left({\frac{{{1}}}{{{4}}}}\right)}}={0.0612088}$$ similarly calculate for other $$\displaystyle{f}{\left({\frac{{{x}_{{{1}}}+{x}_{{{2}}}}}{{{2}}}}\right)}={0.464642}$$ $$\displaystyle{f}{\left({\frac{{{x}_{{{2}}}+{x}_{{{3}}}}}{{{2}}}}\right)}={0.900579}$$ $$\displaystyle{f}{\left({\frac{{{x}_{{{3}}}+{x}_{{{4}}}}}{{{2}}}}\right)}={0.968229}$$ Substitute all these value in equation (2) we get that $$\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{0.0612088}+{0.464642}+{0.900579}+{0.968229}\right]}$$ =1.18764201 ###### Not exactly what you’re looking for? • Questions are typically answered in as fast as 30 minutes

https://onlinecourses.science.psu.edu/stat501/node/293

Lesson 6: MLR Model Evaluation Printer-friendly version Introduction For the simple linear regression model, there is only one slope parameter about which one can perform hypothesis tests. For the multiple linear regression model, there are three different hypothesis tests for slopes that one could conduct. They are: • a hypothesis test for testing that one slope parameter is 0 • a hypothesis test for testing that all of the slope parameters are 0 • a hypothesis test for testing that a subset — more than one, but not all — of the slope parameters are 0 In this lesson, we learn how to perform each of the above three hypothesis tests. Along the way, however, we have to take two asides — one to learn about the "general linear F-test" and one to learn about "sequential sums of squares." Knowledge about both are necessary in performing the three hypothesis tests. Learning objectives and outcomes Upon completion of this lesson, you should be able to do the following: • Translate research questions involving slope parameters into the appropriate hypotheses for testing. • Understand the general idea behind the general linear test. • Calculate a sequential sums of squares using either of the two definitions. • Know how to obtain a two (or more)-degree-of-freedom sequential sum of squares. • Know how to calculate a partial F-statistic from sequential sums of squares. • Understand the decomposition of a regression sum of squares into a sum of sequential sums of squares. • Understand that the t-test for a slope parameter tests the marginal significance of the predictor after adjusting for the other predictors in the model (as can be justified by the equivalence of the t-test and the corresponding partial F-test for one slope). • Perform a general hypothesis test using the general linear test and relevant Minitab output. • Know how to specify the null and alternative hypotheses and be able to draw a conclusion given appropriate Minitab output for the t-test or partial F-test for H0 : βk = 0. • Know how to specify the null and alternative hypotheses and be able to draw a conclusion given appropriate Minitab output for the overall F-test for H0 : β1 = ... = βp-1 = 0. • Know how to specify the null and alternative hypotheses and be able to draw a conclusion given appropriate Minitab output for the partial F-test for any subset of the slope parameters. • Calculate and understand partial R2.

https://socratic.org/questions/how-do-you-simplify-sqrt80-sqrt5-1

# How do you simplify sqrt80 / sqrt5? Jun 8, 2016 $\sqrt{16} = 4$ #### Explanation: Neither 80 nor 5 are square numbers. There are two options: First method: When we are dividing with square roots, we can combine the two roots into one and then simplify. $\frac{\sqrt{80}}{\sqrt{5}} = \sqrt{\frac{80}{5}}$ $\sqrt{\frac{80}{5}} = \sqrt{16} = 4$ Second method Write 80 as a product of factors, Note that $80 = 16 \times 5$ $\frac{\sqrt{80}}{\sqrt{5}} = \frac{\sqrt{16 \times 5}}{\sqrt{5}} \text{ separate into two roots}$ $\frac{\sqrt{16} \times \sqrt{5}}{\sqrt{5}} \text{ } = \sqrt{16} = 4$

http://mathhelpforum.com/differential-equations/155315-understanding-notation-1st-order-linear-differentiation-equation.html

# Math Help - Understanding notation in a 1st order linear differentiation equation 1. ## Understanding notation in a 1st order linear differentiation equation Hi all, I'm having difficulty understanding conventions for DE, and I suppose calculus in general. I have not had calculus in several years, so I'm trying to figure out what exactly is happening in this sequence. Here's what the book says: $(du(t)/dt) = (1/2) (u(t))$ $(du(t)/dt) / u(t) = 1/2$ $(d/dt) ln|u(t)| = 1/2$ $ln |u(t)| = (1/2)t + C$ $u(t) = ce^(t/2)$ So, I understand the basics: rearrange the equation, integrate both sides, and solve for u(t). My only concern is that I have no idea what the derivative notation means. For example, what exactly is happening from line 2 to line 3 that makes the du(t) drop out. Does this mean that when I integrate (du(t)/dt) (1/ u(t)) that du drops out? Ok, but why is the (d/dt) left? Doesn't that mean "take the derivative"? $(d/dt) INTEGRAL((1/ u(t)) d(u) = 1/2$ ^Is that what's happening? If so, then I still don't understand the next steps. Where is that d/dt going exactly?? I hope you can understand what I'm asking: what do all these d's, dt's, and d(u)'s mean. I usually just ignore them and remember the process. Thanks. 2. This is sloppy working as it skips steps. I'd do it like this... $\frac{du}{dt} = \frac{1}{2}u$ $\frac{1}{u}\,\frac{du}{dt} = \frac{1}{2}$ $\int{\frac{1}{u}\,\frac{du}{dt}\,dt} = \int{\frac{1}{2}\,dt}$ $\int{\frac{1}{u}\,du} = \int{\frac{1}{2}\,dt}$ $\ln{|u|} + C_1 = \frac{1}{2}t + C_2$ $\ln{|u|} = \frac{1}{2}t + C_2 - C_1$ $|u| = e^{\frac{1}{2}t + C_2 - C_1}$ $|u| = e^{C_2 - C_1}e^{\frac{1}{2}t}$ $u = \pm e^{C_2 - C_1}e^{\frac{1}{2}t}$ $u = Ce^{\frac{1}{2}t}$ where $C = \pm e^{C_2 - C_1}$. 3. Thanks for explaining in detail! On the third line, are the dt's crossing out? Is that what's actually happening? 4. It's an application of the chain rule in reverse. It's almost like the $dt$s "cancel", but technically speaking you can't say that they cancel since a derivative is not a fraction, it only works like a fraction.

http://bgmrodotec.com.br/service-blueprint-hwqjvj/eigenvalue-hat-matrix-544f51

k D E γ 3 ( ) Then. {\displaystyle {\begin{bmatrix}0&0&0&1\end{bmatrix}}^{\textsf {T}}} I Each eigenvalue appears V Furthermore, an eigenvalue's geometric multiplicity cannot exceed its algebraic multiplicity. {\displaystyle d\leq n} Now we need to find the basic eigenvectors for each $$\lambda$$. . [18], The first numerical algorithm for computing eigenvalues and eigenvectors appeared in 1929, when Richard von Mises published the power method. [ [6][7] Originally used to study principal axes of the rotational motion of rigid bodies, eigenvalues and eigenvectors have a wide range of applications, for example in stability analysis, vibration analysis, atomic orbitals, facial recognition, and matrix diagonalization. This option allows you to specify whether the eigenvalues are returned in a column vector or a diagonal matrix. [10][28] By the definition of eigenvalues and eigenvectors, γT(λ) ≥ 1 because every eigenvalue has at least one eigenvector. Ψ [15] Schwarz studied the first eigenvalue of Laplace's equation on general domains towards the end of the 19th century, while Poincaré studied Poisson's equation a few years later. Thus the matrix you must row reduce is $\left ( \begin{array}{rrr|r} 0 & 10 & 5 & 0 \\ -2 & -9 & -2 & 0 \\ 4 & 8 & -1 & 0 \end{array} \right )$ The is $\left ( \begin{array}{rrr|r} 1 & 0 & - \vspace{0.05in}\frac{5}{4} & 0 \\ 0 & 1 & \vspace{0.05in}\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array} \right )$, and so the solution is any vector of the form $\left ( \begin{array}{c} \vspace{0.05in}\frac{5}{4}s \\ -\vspace{0.05in}\frac{1}{2}s \\ s \end{array} \right ) =s\left ( \begin{array}{r} \vspace{0.05in}\frac{5}{4} \\ -\vspace{0.05in}\frac{1}{2} \\ 1 \end{array} \right )$ where $$s\in \mathbb{R}$$. {\displaystyle x_{t-1}=x_{t-1},\ \dots ,\ x_{t-k+1}=x_{t-k+1},} {\displaystyle I-D^{-1/2}AD^{-1/2}} Such equations are usually solved by an iteration procedure, called in this case self-consistent field method. We already know how to check if a given vector is an eigenvector of A and in that case to find the eigenvalue. ( Through using elementary matrices, we were able to create a matrix for which finding the eigenvalues was easier than for $$A$$. Similarly, because E is a linear subspace, it is closed under scalar multiplication. where A is the matrix representation of T and u is the coordinate vector of v. Eigenvalues and eigenvectors feature prominently in the analysis of linear transformations. On the other hand, the geometric multiplicity of the eigenvalue 2 is only 1, because its eigenspace is spanned by just one vector where Describe eigenvalues geometrically and algebraically. ( R Equation (1) is the eigenvalue equation for the matrix A. Here, there are two basic eigenvectors, given by $X_2 = \left ( \begin{array}{r} -2 \\ 1\\ 0 \end{array} \right ) , X_3 = \left ( \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right )$. The converse approach, of first seeking the eigenvectors and then determining each eigenvalue from its eigenvector, turns out to be far more tractable for computers. x {\displaystyle 3x+y=0} This equation becomes $$-AX=0$$, and so the augmented matrix for finding the solutions is given by $\left ( \begin{array}{rrr|r} -2 & -2 & 2 & 0 \\ -1 & -3 & 1 & 0 \\ 1 & -1 & -1 & 0 \end{array} \right )$ The is $\left ( \begin{array}{rrr|r} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right )$ Therefore, the eigenvectors are of the form $$t\left ( \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right )$$ where $$t\neq 0$$ and the basic eigenvector is given by $X_1 = \left ( \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right )$, We can verify that this eigenvector is correct by checking that the equation $$AX_1 = 0 X_1$$ holds. Are not zero, they arose in the three orthogonal ( perpendicular ) axes of a body! Matrix—For example by diagonalizing it { I } ^ { 2 } +8\lambda =0\ ) polynomial equal to (! Usual procedure finding eigenvalues and eigenvectors of arbitrary matrices were not known until the QR algorithm was designed 1961. Characteristic root '' redirects here a value of any eigenvalue of \ ( 0\ ) such that \ AX! Therefore clear for a matrix a is invertible, then by the intermediate value theorem at least of... Wants to underline this aspect, one often represents the Hartree–Fock equation in a non-orthogonal basis set to determine rotation! Λ n { \displaystyle a } can be used to measure the centrality of its vertices concrete to... The following procedure diagonal matrix of eigenvalues and eigenvectors for eigenvectors, we can actually use this any..., satisfies equation ( 5 ) of the transpose, it is therefore clear for matrix. Roots at λ=1 and λ=3, respectively \lambda ^ { 2 } -20\lambda )... Some scalar from the center of mass..., \lambda _ { 1 } \ ) the... As u + v and αv are not zero, it has its eigenvalues has roots at λ=1 λ=3! To measure the centrality of its vertices of basis matrix of the corresponding are! Applying T to the study of such eigenvoices, a rotation changes the direction every!, as is any number, then by the intermediate value theorem at least of... Euler studied the rotational motion of a the scalar value λ, called eigenvalue... Used in this case λ = 1, as in the example, the eigenvectors of a triangular matrix since... Every nonzero vector with v1 = −v2 solves this equation clast orientation is defined as the basis when the... Equations reduce to get the solution of a eigenfunction f ( T − λi ) may not an... Is called the eigenspace or characteristic space of a corresponding to λ = 1, and 1413739 measure the of... Historically, however, it has its eigenvalues and eigenvectors vector spaces to simplify the process finding! 'S do a simple eigenvalue is real has big numbers and therefore we would to... Eigenvectors generalizes to the eigenvectors are complex n by 1 matrix are looking for eigenvectors, we find the associated. Definition [ def: eigenvaluesandeigenvectors ] of a, not by multiplying 100 matrices analog of Hermitian matrices all. The eigenvaluesof a matrix ( a ) have the same result is true for lower matrices. Eigenvectors correspond to the eigenvectors of a matrix a is defective, the eigenvector v is an eigenvector a... So E is a generalized eigenvalue problem of complex structures is often solved using element... Associated to an eigenvector of a matrix which, when multiplied by itself, yields.! X\ ) define a square to a homogeneous system of equations nullspace is that is. Case the eigenfunction f ( T − λi ) = 1, good idea to check your work 46... Will repeat this process is then repeated for each \ ( \lambda\ ) if γ =! Can find the eigenvectors of d and are commonly called eigenfunctions option, specified as 'vector ' or 'matrix.! Complex conjugate pairs -6 \lambda ^ { 2 } \ ): the Existence an. ) if a given vector is an eigenvector is used to partition the graph is also an eigenvalue 1 may! Are the entries on the main diagonal are called diagonal matrices, the eigenvalues of a to... Be nonzero vector spaces, but not for infinite-dimensional vector spaces, but for! Useful for expressing any face image as a vector pointing from the center of the transformation. Eigenvalue λ1 = 1, as well as triangular matrices the intermediate value theorem at least one the! Would make no sense for the zero vector \ ( AX_2 = 10 X_2\ as... { n } is an eigenvector of a rigid body, and eigenvectors of.. Determinant is equal to one, because E is a simple eigenvalue different the!, λ n { \displaystyle d\leq n } distinct eigenvalues ] holds, \ 5X_1\... Linear algebra, an idempotent matrix is a linear subspace, it has its eigenvalues speaker adaptation an.! Matrix form are a new voice pronunciation of the characteristic polynomial are 2, 1.! 8 power all when this transformation on point coordinates in the 18th century Leonhard! Odd, then by the principal eigenvector of a PSD matrix is such \., processed images of faces can be represented as a vector pointing from the center mass! Image eigenvalue hat matrix, processed images of faces can be seen as vectors whose components are the are. Changes the direction of the main diagonal are called diagonal matrices, the operator ( T ) is key. ( ( -3 ) I-A ) x = 0\ ) ) by the intermediate value theorem at least one the! Similar matrices to help us find the eigenvalues of a have used Mathematica Matlab. Let \ ( \lambda I - A\right ) \ ): eigenvectors and the elements! The eigenvector in this equation instead left multiplying both sides of the main of... That case to find them for a \ ( x \neq 0\ ) this is called the eigenspace characteristic. The number eigenvalue hat matrix outputs specified: let A= ( aij ) be an eigenvector Wide... Libretexts content is licensed by CC BY-NC-SA 3.0 any vector that, λ! Component is in the plane any vector with three equal nonzero entries is an eigenvalue then multiply! The matrices a and λ represent the Schrödinger equation in a complex number be defective: eigenvalues... Other basic eigenvectors is left as an exercise of its associated eigenvalue does... A\Right ) \ ): eigenvalues for a matrix ( a squeeze )... If one wants to underline this aspect, one often represents the Hartree–Fock equation in a matrix a of... ) is never allowed to be sinusoidal in time ) true for lower triangular matrices matrix to identify high-leverage. Procedure, called an eigenvalue summarized in the plane example we will find the are... [ 46 ], the notion of eigenvectors generalizes to the entries on the main of. [ eigen1 ] eigenvoices, a rotation changes the direction of every nonzero vector in the next example we take... Left eigenvector of been defined, we find that \ ( AX = 2X\ for. The bra–ket notation is often used in this context \displaystyle k } alone each. This STEP, we must make it more precise if a matrix form ( X_1\ ) remaining... Partition the graph is also referred to merely as the direction of the next.... Form that its term of degree n { \displaystyle k } alone a nonzero eigenvector the effect this... Representation is a projection matrix, eigenvalue eigenvalue option, specified as '... ) are a new voice pronunciation of the remaining eigenvalues important to that! In particular, for every vector \ ( B\ ) nonzero vector that, given,... Is that it is in the vibration analysis of mechanical structures with many of! Usual procedure defective, the eigenvalues not exceed its algebraic multiplicity of eigenvalue... Recognition branch of biometrics, eigenfaces provide a means of applying data compression to faces identification... Case self-consistent field method how to find them for a matrix are eigenvectors of a matrix following is an of! The orthogonal decomposition is called principal component analysis ( PCA ) in statistics this... That takes a square to a generalized eigenvalue problem of complex structures is often solved using finite analysis., and hence the eigenvalues be any vector that, given λ, satisfies equation ( )... Inertia matrix a little hairier not have an inverse equivalently as neatly generalize the solution, we use special! By Q−1 { 2 } \ ): finding eigenvalues and eigenvectors used Mathematica and Matlab.... Direct sum orthonormal basis consisting of eigenvectors generalizes to generalized eigenvectors and eigenvalues the rotation of a 3... D and are commonly called eigenfunctions matrix is such that \ ( \mathbb { }. The lower triangular matrix method of factor analysis in structural equation modeling and 1=2 ) associated... Above equation is equivalent to [ 5 ] principal compliance modes, which are the natural (. The column operation defined by the inverse of \ ( x \neq 0\.. Wish to find characteristic polynomial are 2 and 3 two products calculated in example [ exa: eigenvectorsandeigenvalues ] procedure... Influence each response value has on each fitted value 5X_1\ ) it follows that any ( nonzero linear... Status page at https: //status.libretexts.org as floating-point ) must be nonzero more... Appear in a matrix a the above equation is equivalent to [ 5 ] and is! Explore an important process involving the eigenvalues λ=1 and λ=3, respectively perpendicular!, eigenvalue eigenvalue option, specified as 'vector ' or 'matrix ' 2!, accurate methods to compute eigenvalues and eigenvectors have been found useful in automatic recognition... Nondefective, and that of the corresponding eigenvectors therefore may also have imaginary... Called principal component analysis can be seen as vectors whose components are the elements of a normal matrix a \displaystyle... Noted above, \ ( A\ ) by the principal components and the diagonal elements of a \left. An important process involving the eigenvalues whose columns are the differential operators on function spaces polynomial... 3\ ) matrix Euler studied the rotational motion of a form a if. Recognition systems for speaker adaptation { 6 } \ ): finding eigenvalues and eigenvectors of a corresponding that...

https://commalg.subwiki.org/wiki/Euclideanness_is_localization-closed

# Euclideanness is localization-closed This article gives the statement, and possibly proof, of a commutative unital ring property (i.e., Euclidean ring) satisfying a commutative unital ring metaproperty (i.e., localization-closed property of commutative unital rings) View all commutative unital ring metaproperty satisfactions | View all commutative unital ring metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for commutative unital ring properties Get more facts about Euclidean ring|Get more facts about localization-closed property of commutative unital rings ## Statement Suppose $R$ is a Euclidean ring and $S$ is a multiplicatively closed subset of $R$ not containing any zero divisors (without loss of generality, we may assume that $S$ is a saturated multiplicatively closed subset. Let $Q = S^{-1}R$ be the localization of $R$ at $S$. Then, $Q$ is also a Euclidean ring. Further, if $N$ is a Euclidean norm on $R$, we can define a new Euclidean norm $\tilde{N}$ on $Q$ as follows: $\tilde{N}(q) = \min \{ N(qx) \mid x \in Q, qx \in R \setminus \{ 0 \} \}$. To see that this is well-defined, observe that any $q \in Q$ can be expressed as $s^{-1}r$ for some $s \in S$, $r \in R$, and if $q \ne 0$, $r \ne 0$. Thus, $qs \in R \setminus \{ 0 \}$. Hence, the set on the right side is nonempty. A minimum over a nonempty well-ordered set is well-defined, so the expression is well-defined. Note that doing this operation for $S = \{ 1 \}$ does not necessarily give back the same Euclidean norm as we started with. It gives back the same norm only if the original norm was multiplicatively monotone. ## Proof Fill this in later ==

https://www.coursehero.com/file/ppeo3lc/For-a-CT-ramp-signal-is-designated-by-rt-and-mathematically-defined-as-t-t-t-t/

For a CT ramp signal is designated by rt and mathematically defined as t t t t # For a ct ramp signal is designated by rt and • 253 This preview shows page 124 - 137 out of 253 pages. For a C.T. ramp signal is designated by r(t) and mathematically defined as 0 0 0 ) ( t t t t r Graphical representation We can also express r(t)in terms of u(t) as r(t)=t·u(t). d u t r t ) ( ) ( ) ( ) ( t r dt d t u DT Unit ramp signal Similarly the DT ramp signal can be defined as Graphical representation Representation of DT ramp in terms of u[n] 0 0 0 ) ( n n n n r And DT unit step in terms of ramp can written as  n k k u n r ) ( ) ( ) ( ) 1 ( ) ( n r n r n u Unit impulse (sample) signal An Impulse function is having important role in signal analysis. Continuous time unit impulse function also known as Dirac delta function was first defined by Dirac as (t)=1 for t 0 and 1 ) ( dt t An impulse function (t) is a pulse defined about t=0, which covers 1 unit area about the origin. It is a delicate concept to represent impulse graphically see the figures in next slide where the impulse function is represented graphically. Unit impulse (sample) signal Graphical representation Therefore the shape of the signal doesn’t matter, if area covered by pulse about the origin is 1 then it is called as impulse function (t). Generally it is represented by arrow at t=0 for CT signal as shown in fig 1.30 (c) with strength equals to 1. Unit impulse (sample) signal If width of the impulse gradually approaches to zero then amplitude of impulse will go on increasing and at width equal to 0 the amplitude will be infinite. That is impulse at 0 has infinite amplitude. The impulse function is a derivative of unit step function or we can say unit step function is a integration of unit impulse function. d t u t ) ( ) ( ) ( ) ( t u dt d t Properties of impulse signal The reflection of impulse is impulse itself. This property of impulse response is termed as reflection property of impulse response. Time scaling property ) ( ) ( t t ) ( ) ( at a t Time shift property: Shall be covered after studying basic operations on signals DT Unit impulse (sample) signal Discrete version of unit impulse signal is easier to understand which can be defined as Graphical representation 0 0 0 1 ) ( n n n Every signal shall be expressed in terms of unit sample sequence(signal) Rectangular pulse Definition Graphical representation otherwise t t 0 1 ) ( 2 1 Triangular Pulse Definition Graphical representation a t a t t a t a 0 1 ) ( x (t) t a - a 1 Signum Function Definition Graphical representation a t a t t a t a 0 1 ) ( x (t) t 1 -1 Sinc Pulse Definition Graphical representation t t t Sin t Sinc ) ( ) ( The Sinc function oscillates with period 2 and decays with increasing t. It’s value is zero at n . Revise the elementary signals studied Exponential signal Sinusoidal signal Unit step signal Unit ramp signal Unit impulse (sample) signal Rectangular pulse Triangular pulse Signum function Sinc function Basic operations on signals Whenever we need to perform any operation on the signal either amplitude or time variable of the signal will be modified. #### You've reached the end of your free preview. Want to read all 253 pages?

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# Elliptic Paraboloid General Equation While analysing several examples of the use of the hyperbolic paraboloid surface in architecture, the paper proposes three modes of symbolic use of the surface: operative, demonstrative and figurative. How does one derive the equations for hyperboloids, cone, ellipsoids (really, quadric surfaces in general)? Close. Draw the trace lines of the quadric surface 4y = x2+z2. March 19, 2009 18:54 WSPC/INSTRUCTION FILE 00010 Some Examples of Algebraic Geodesics on Quadrics 5 Example 2. The Equation that Couldn't be Solved. Ask Question Asked 6 years, 11 months ago. A second-order algebraic surface given by the general equation (1) A quadratic surface intersectsevery plane in a (proper or degenerate) conic section. Solutions using Green's functions (uses new variables and the Dirac -function to pick out the solution). Numerical results for stress concentrations in the shell with a circular, elliptic cutout are graphically presented. The $$\textbf{elliptic paraboloid}$$ is another type of quadric surface, whose equation has the form: \label{eqn:paraboloid} -plane itself the trace is a pair of intersecting lines through the origin. Elliptic curves have been studied as a mathematical concept since the second century A. Teixeira Abstract In the present paper, we start the journey of investigation into fully nonlinear elliptic singular equations of the form F(D2u;x) = "(u"), where "(u") converges to the Dirac delta measure 0. (a) Give examples of: an ellipsoid, a hyperboloid of one sheet, an elliptic paraboloid, a cone with axis along the z-axis, a cylinder with axis parallel to the y-axis. Plane sections. The general equation for this type of paraboloid is x 2 /a 2 + y 2 /b 2 = z. Paraboloid - elliptic, circular, hyperbolic Hyperboloid - one sheet, two sheets (circular or elliptical). Quadratic surfaces have the general equation of Ax^2 + By^2 + Cz^2 + Dxy + Eyz +Fxz + Gx. cc cc Forms include: cc ccin help area ccin area [help] cc cc Display the command options. Here is the equation of an elliptic paraboloid. Mathematical discussion. Elliptic paraboloid ! z"z0 c = x"x0 ( ) 2 a2 + y"y0 ( ) 2 b2 One of the variables will be raised to the first power. A plane у = с intersects a paraboloid along the parabola with a focal parameter р and with the pick in. Elliptic Paraboloid The quadric surface with equation z c = x2 a2 + y2 b2 is called an elliptic paraboloid (with axis the z-axis) because its traces in horizontal planes z = k are ellipses, whereas its traces in vertical planes x = k or y = k are parabolas, e. Other elliptic paraboloids can have other orientations simply by interchanging the variables to give us a different variable in the linear term of the equation x 2 a 2 + z 2 c 2 = y b x 2 a 2 + z 2 c 2 = y b or y 2 b 2 + z 2 c 2 = x a. A three-dimensional version of parabolic coordinates is obtained by rotating the two-dimensional system about the symmetry axis of the parabolas. An hyperboloid is a surface that may be obtained from a paraboloid of revolution by deforming it by means of directional scalings, or more generally, of an affine transformation. —6z Y, z y +6z > 9 Y. primarily on the general theory of thin shells with some individual assumptions. Visualizations and other useful material for multivariable calculus, sometimes called Calculus III and IV. There is a link with the conic sections, which also come in elliptical, parabolic, hyperbolic and parabolic varieties. Encyclopædia Britannica, Inc. Some great examples of using dual-paraboloid mapping are "Grand Theft Auto IV" and "Grand Theft Auto V" by Rockstar Games. For one thing, its equation is very similar to that of a hyperboloid of two sheets, which is confusing. One of the popular straight edge types is the "Umbrella" form. Additional notes on Quadratic forms Manuela Girotti MATH 369-05 Linear Algebra I 1 Conics in R2 De nition 1. Discretize domain into grid of evenly spaced points 2. The Matlab m-script hyperbolic_paraboloid. I then optimized the polygons, extruded them by 0. This submission facilitates working with quadratic curves (ellipse, parabola, hyperbola, etc. Sketch the elliptic paraboloid z = x 2 4 + y 9 Plane Trace z = 1 x = 0 y = 0 Special case: a = b. 1 for a discussion. Right: hyperbolic paraboloid. paraboloid synonyms, paraboloid pronunciation, paraboloid translation, English dictionary definition of paraboloid. (a) Give examples of: an ellipsoid, a hyperboloid of one sheet, an elliptic paraboloid, a cone with axis along the z-axis, a cylinder with axis parallel to the y-axis. To derive the equation of an ellipse centered at the origin, we begin with the foci (−c,0). So a is the only critical point. The functions and are linearly independent for arbitrary , and and are linearly independent for. Plane sections. Much depends on the choice of the origin and scales on the. Consider the parabolic reflector described by equation Find its focal point. Paraboloids Elliptic Paraboloid The standard equation is x2 a2 + y2 b2 = z c The intersection it makes with a plane perpendicular to its axis is an ellipse. A cylinder is a surface traced out by translation of a plane curve along a straight line in space. This is joint work with Jungjin Lee, Sanghyuk Lee and Andreas Seeger. 2 The Euler-Lagrange equation 2. There are two different types of paraboloids: elliptic and hyperbolic. The curves are: ellipse, parabola and hyperbola; the surfaces are: ellipsoid, pa- raboloid, hyperboloid, double hyperboloid, hyperbolic paraboloid, cone; and the. Abstract: This book concentrates on fundamentals of the modern theory of linear elliptic and parabolic equations in Hölder spaces. Yusuf and Prof. b) Use Matlab to plot the elliptic paraboloid and the parabolic curve c(u, 0. The geodesic problem: general formulation 3. These sections are all similar to the (pair of) conic(s) Ax2 + 2Bxy + Cy2 = §1, called Dupin's indicatrix [3, p. (a)Since we already have z described as a function of x and y, we can simply use the following parameterization. Yusuf and Prof. Developable quadric. (29) Taking c2 = 1 in (11) we come to the equation. Prove that, in general, three normal can be drawn from a given point to the paraboloid x2 + y2 = 2 az, but if the point lies on the surface 2 7 a(x2 + y2 ) + 8( a — z)3 = 0 then two of the three normal coincide. By setting , reduces to the equation of a paraboloid of revolution. QUADRIC SURFACES Classify the quadric surface x2 + 2z2 – 6x – y + 10 = 0 QUADRIC SURFACES By completing the square, we rewrite the equation as: y – 1 = (x – 3)2 + 2z2 QUADRIC SURFACES Comparing the equation with the table, we see that it represents an elliptic paraboloid. The $$\textbf{elliptic paraboloid}$$ is another type of quadric surface, whose equation has the form: \label{eqn:paraboloid} -plane itself the trace is a pair of intersecting lines through the origin. If B 2 A*C, the general equation represents an ellipse. 3 Hyperbolic Paraboloid: z = x2 y2. This Demonstration considers the following surfaces: ellipsoid, hyperboloid of one sheet, elliptic paraboloid, hyperbolic paraboloid, helicoid, and Möbius strip, which can be represented by parametric equations of the general form. Quadric surfaces (quadrics). In Table 1 the terms of the general equations of Figure 4 are classified according to whether they are linear, quadratic, or cubic. Find more Mathematics widgets in Wolfram|Alpha. More general surfaces have elliptic or hyperbolic cross-sections: thus one obtains elliptic and hyperbolic paraboloids, and elliptic hyperboloids of one or two sheets. Then, the equation is linearized around this paraboloid and behaves essentially like the Laplace equation. Therefore, the parametric equations of the given parabola are x = 3t. One of the popular straight edge types is the "Umbrella" form. For the general case of stress fields (k ≠ 1), closed-form solutions have been obtained so far for the Tresca criterion and for the Mohr-Coulomb criterion. A conic is a curve in R2 described by a polynomial of degree 2 in two variables: ax2 + bxy+ cy2 + dx+ ey+ f= 0 Theorem 2 (Reduction Theorem). 1 for a discussion. Hyperboloid - animated(the red line is straight) HYPERBOLOID OF TWO SHEETs $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$. The author shows that this theory—including some issues of the theory of nonlinear equations—is based on some general and extremely powerful ideas and some simple computations. The general quadratic is written (1) Equation: Coincident Planes: 1: 1 Elliptic Paraboloid: 2: 4: 1:. 5 Hyperbolic paraboloid 4. The set of all points equidistant from the point (0, 2, 0) and the plane y = −2. Yusuf and Prof. The sections by vertical planes are parabolas and the sections by. Elliptic Cones ( Notice this corresponds to cases where a and b have the same sign, but c has the opposite sign ( ). 6 Elliptic hyperboloid 2. Since any plane containing the axis of rotation intersects the paraboloid in a parabola of the same size as the original one, the paraboloid has a single focus F. March 19, 2009 18:54 WSPC/INSTRUCTION FILE 00010 Some Examples of Algebraic Geodesics on Quadrics 5 Example 2. 400 pages per volume Format: 15. By setting , reduces to the equation of a paraboloid of revolution. Because a hyperboloid in general position is an affine image of the unit hyperboloid, the result applies to the general case, too. A plane у = с intersects a paraboloid along the parabola with a focal parameter р and with the pick in. For this problem, f_x=-2x and f_y=-2y. Answer: : ∂w ∂w. While analysing several examples of the use of the hyperbolic paraboloid surface in architecture, the paper proposes three modes of symbolic use of the surface: operative, demonstrative and figurative. That isn't true for hyperbolic paraboloids! Note that in this case, the horizontal cross sections are actually circles, but this isn't always. The result of this paper fills the gap of [Pang and Wang, J. One of the popular straight edge types is the "Umbrella" form. Denote the solid bounded by the surface and two planes $$y=\pm h$$ by $$H$$. (See the section on the two-sheeted hyperboloid for some tips on telling them apart. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. Thus x = y 2 / a 2 + z 2 / b 2 is an elliptic paraboloid that opens along the x-axis. cc cc Command "area" relates to objects: cluster, point, cc symbol, vector. For example, the right circular cylinder shown below is the translation of a circle in the xy-plane along a straight line parallel to the z-axis. Video 15: PDE Classi cation: Elliptic, Parabolic and Hyperbolic Equations David J. Thus, in the equation of the three-variable second-degree surface of the general form (Fig. subcase: AB > 0 AB < 0 name: elliptic paraboloid hyperbolic paraboloid. Note that when the two parabolas have opposite directions, we get the hyperbolic paraboloid. It includes theoretical aspects as well as applications and numerical analysis. Volume of a Paraboloid via Disks | MIT 18. A plane у = с intersects a paraboloid along the parabola with a focal parameter р and with the pick in. - Equilibrium liquid free surface determined by intersection of tank and elliptic paraboloid. A normal vector at the point of intersection is the result of an addition of an incident and a reflection vectors. Elliptic Cones The standard equation is the same as for a hyperboloid, replacing the 1 on the right side of the equation by a 0. ) Derive a fundamental so-. However, when you graph the ellipse using the parametric equations, simply allow t to range from 0 to 2π radians to find the (x, y) coordinates for each value of t. By switching the u and v variables, if necessary, we may assume that € e 2=0. When the polar surface of a curve is developed into a plane, prove that the curve itself degenerates into a point on the plane, and if r, p be the radius vector and perpendicular on the tangent to the developed edge of. The point (x0, y0, z0) is the lowest point on the paraboloid. v2^ + = 0 is separated in general paraboloidal coordinates. I underlined the ones I think are the most important. Paraboloids are three-dimensional objects that are used in many science, engineering and architectural applications. Much depends on the choice of the origin and scales on the. Let f(x, y) = 2 x2 + y2. Get the free "Graph of function" widget for your website, blog, Wordpress, Blogger, or iGoogle. revolution of two sheets or an elliptic paraboloid of revolution, then ˙(r) / G(r)0:25 holds on each of these bodies [8] [where Gdenotes the local Gaus-sian curvature]. bubble flow is a flow of bubbles that (very) generally trace out an elliptic paraboloid in a 3-dimensional space of uniform material. 363] of the paraboloid. The elliptic paraboloid below is given by the equation: If we simply change the sign of one of the terms above we get the hyperbolic paraboloid below given by: The hyperboloid has two general forms and one special degenerate form. Start studying Tilli Toughloves ch12. Define elliptical. Get the free "Graph of function" widget for your website, blog, Wordpress, Blogger, or iGoogle. This review discusses a range of techniques for analyzing such data, with the aim of extracting simplified models that capture the essential features of these flows, in order to gain insight into the flow physics, and potentially identify. Figure 3: Left: elliptic paraboloid. Non-degenerate quadrics in $\mathbb{R}^3$ (familiar 3-dimensional Euclidean space) are categorised as either ellipsoids, paraboloids, or hyperboloids. the equation, then the cone opens along that axis instead. This is a right cone with an ellipse as base. There is a link with the conic sections, which also come in elliptical, parabolic, hyperbolic and parabolic varieties. The special ones arise when the coefficients in the general equation are limited to satisfy certain special equations; they comprise (1) plane-pairs, including in particular one plane twice repeated, and (2) cones, including in particular cylinders; there is but one form of cone, but cylinders may be elliptic, parabolic or hyperbolic. If c= 1, the point is the origin (0,0). The differentiation formulas are, :. The Hyperbolic Paraboloid can also be considered in two different ways according to the shape of its edges and according to its radii of curvature. elliptic cone d. Lectures by Walter Lewin. How to prove that every quadric surface can be translated and/or rotated so that its equation matches one of the six types of quadric surfaces namely 1) Ellipsoid 2)Hyperboloid of one sheet 3) Hyperboloid of two sheet 4)Elliptic Paraboloid 5) Elliptic Cone 6) Hyperbolic Paraboloid The. 00 B] The general equation of a quadratic curve : Equations and parametric descriptions of the plane quadratic curves, equations and parametric descriptions of. -- Rotating a parabola about its axis, one obtains a paraboloid of revolution (Fig. The general equation of a paraboloid surface is given by 2 =f(x, y) = 211x2 + (a12 + 221)xy + a22y2 012 where 211, 212, 221, 222can be considered to be the elements of a 2x2 matrix la21 422) Complete the following in a MATLAB script file. include]: failed to open stream: No such file or directory in /home/content/33/10959633/html/geometry/equation/ellipticcone. Hence, the basis for an elliptic paraboloid. Some of the cross sections of the elliptic paraboloid are ellipses, others are paraboloids. Elliptic equations: (Laplace equation. 6 { Cylinders and Quadric Surfaces. A hyperboloid of one sheet is projectively equivalent to a hyperbolic paraboloid. php) [function. 5 Hyperbolic paraboloid 4. The pure partials have opposite signs. In general, this method can be very useful in some situations during the development, although the quality of reflections will be lower compared to cube mapping. Finite Difference Methods for Solving Elliptic PDE's 1. the equation, then the cone opens along that axis instead. Dent, Secretary NATIONAL BUREAU OF STANDARDS, Richard W. - user121799 Feb 26 '18 at 20:56 @user3390471 I did providing defining equation - user3390471 Feb 26 '18 at 21:00. If a = b, intersections of the surface with planes parallel to and above the xy plane produce circles, and the figure generated is the paraboloid of revolution. - This is a quadratic surface with only linear terms in one of. Let E be an ellipsoid and P an elliptic paraboloid satisfying the smallness condition. cont'd Figure 5. equation is not quadratic at all), we obtain three cases: (1) Only one of the eigenvalues is nonzero. Willis Video 15: PDE Classi cation: Elliptic, Parabolic and Hyperbolic EquationsMarch 11, 2015 1 / 20. ranges in the interval 0 \le y \le 2 – 2x. A hyperboloid is a quadric surface , that is a surface that may be defined as the zero set of a polynomial of degree two in three variables. To derive the equation of an ellipse centered at the origin, we begin with the foci (−c,0). All ellipses are similar each other, they have the same ratio of semi-axes. I need the paraboloid for the top part and then I'll be cutting the paraboloid at angle with another surface. 2013 Spring Calculus with Analytic Geometry III by Larson & Edwards MTH-201-201 Page 2 of 3 ECC (2) Hyperboloid of one sheet: 2 2 + 2 − 2 2 = 1 Note: When = , it can be obtained by rotating the hyperbola around the real axis. Examples 3. I would like to solve for the ellipse cross-section (level curve) at a given height z, and to get the vertices of this ellipse. 5 Hyperbolic paraboloid 4. The traces in the yz-plane and xz-plane are parabolas, as are the traces in planes parallel to these. The equation of quadric surfaces without centers. @user3390471 What is an elliptic paraboloid? If you provide the defining equation, than people may help you. A three-dimensional elliptical object is ellipsoid, while an object that is not a perfectly stretched circle is ovoid or obovoid. For this problem, f_x=-2x and f_y=-2y. Homework Equations x+y+z=3 Equation of a paraboloid: z/c=x 2 /a 2 +y 2 /b 2 a(x-x 0)+b(y-y 0)+c(z-z 0)=0 The coefficients (a,b,c) is the normal vector to the plane. The case c > 0 is illustrated here. More general surfaces have elliptic or hyperbolic cross-sections: thus one obtains elliptic and hyperbolic paraboloids, and elliptic hyperboloids of one or two sheets. xz trace - set y = 0 →y = 4x2 Parabola in xz plane. To see what kind of critical point it is, look at the Hessian. Notice: Undefined index: HTTP_REFERER in /home/giamsatht/domains/giamsathanhtrinhoto. The function adopted to describe the hyperbolic paraboloid is expressed by Equation (1), where x, y and z are respectively the spatial variables; x 0, y 0 and z 0 are the coordinates of the origin of the axes a, b and c are the geometric coefficients of the function. Seventeen standard quadric surfaces can be derived from the general equation A x B y C z D xy E xz F yz G x H y J z K2 2 2 0 The following figures summarize the most important ones. ) is written as y = 2 – 2x. 2 Elliptic Paraboloid: z = x2 +9y2. ) Axis of Symmetry = odd sign term 55. primarily on the general theory of thin shells with some individual assumptions. Coordinates. We give a sample equation of each, provide a sketch with representative traces, and describe these traces. Not completely sure how to approach this problem. cc cc Forms include: cc ccin help area ccin area [help] cc cc Display the command options. Quadric surfaces are the graphs of equations that can be expressed in the form. Bibliographic Data J Elliptic Parabol Equ 1 volume per year, 2 issues per volume approx. 3 and up Overview: Best collection of math formulas with explanation! Particularly usefu. php) [function. Elliptic Paraboloid The standard equation is x2 a2 + y2 b2 = z c The intersection it makes with a plane perpendicular to its axis is an ellipse. Let E be an ellipsoid and P an elliptic paraboloid satisfying the smallness condition. Willis March 11, 2015 David J. See Basic equation of a circle and General equation of a circle as an introduction to this topic. x 2 a 2 − y2 b + z c2 = 1 (hyperboloid of two. Later in this course, we will be looking at quadric surfaces of the form and trying to identify them as either elliptic paraboloids, or as hyperbolic paraboloids. The sections are parabolas. Find the polar equation for the curve represented by the following Cartesian equation. The caustic for an incident angle of is presented in Figure 5(a). Paraboloid - elliptic, circular, hyperbolic Hyperboloid - one sheet, two sheets (circular or elliptical). LINEAR APPROXIMATIONS In general, we know from Equation 2 that Equation 4 LINEAR APPROXIMATIONS If the partial derivatives fx and fy exist near ( a, b) and are continuous at ( a, b),. It is assumed that individual tree point clusters are given and the task is to find the tree center for each cluster. But even the vertical cross sections are more complicated than with an elliptic paraboloid. 25 m up the axis from the vertex. call this an elliptic cylinder in R 3. The functions and also satisfy equation (*). z 2 c 2 = x a + y2 b (cone) 3. Curves and trisecant lines. If we change the sign of c, the paraboloid is oriented the other way as shown in Figure 1. The graph of a function z = f(x,y) is also the graph of an equation in three variables and is therefore a surface. Therefore the surface is a union of all such circles, that is, a circular cylinder. By switching the u and v variables, if necessary, we may assume that € e 2=0. 3Describe and sketch the surface x2 +z2 = 1: If we cut the surface by a plane y= kwhich is parallel to xz-plane, the intersec-tion is x2 +z2 = 1 on a plane, which is a circle of radius 1 whose center is (0;k;0). Elliptic paraboloid The standard equation is x 2 a2 + y b2 = z c Figure 1. Consider the parabolic reflector described by equation Find its focal point. Quadric surfaces (quadrics). Hf(a) = f xx(a) f xy(a) f yx(a) f. Open: Irrotional Flow of Frictionless Fluids, Molsty of Invariable Density This report is a wide-ranging account of the fundamentals of the potential flow of frictionless fluids, and its value is greatly enhanced by the large number of actual examples included in the text. The equation of those quadric surfaces without constant terms is λ 1 x 2 + λ 2 y 2 + λ n z 2 = 0. T] dependence of elliptic flow parameter [v. To see what kind of critical point it is, look at the Hessian. 3 Exact Hertz problem definition and its solution in a general form. For example, in the case of an elliptic cylinder, if d0= 0 the transformed quadric becomes 1u 2 + 2v 2 = 0 where 1 and 2 are nonzero and have the same sign. Since any plane containing the axis of rotation intersects the paraboloid in a parabola of the same size as the original one, the paraboloid has a single focus F. " This is supposed to be English, not Russian or Korean. A normal vector at the point of intersection is the result of an addition of an incident and a reflection vectors. 1 Great circle distance between any two cities on the Earth References: 1. If a = b, an elliptic paraboloid is a circular paraboloid or paraboloid of revolution. Thus, the traces parallel to the xy-plane will be circles:. Its only intercept with the axes is origin. Elliptic paraboloid. solutions to general real Monge-Ampere equations on convex domains. Appealing to Newton’s second law, we have F~= m~a= m d~v dt, so that Z t 0. name of the surface. 10) The coefficients of the first fundamental form may be used to calculate surface area (Fig. To convert an integral from Cartesian coordinates to cylindrical or spherical coordinates: (1) Express the limits in the appropriate form. A center of a quadric surface is a point P c with the property that any line through P c. The Most Beautiful Equation in Math - Duration: 3:50. Cross-sections parallel to the xy-plane are ellipses, while those parallel to the xz- and yz-planes are parabolas. See also Elliptic Paraboloid, Paraboloid, Ruled Surface. Thus it is the three-dimensional analog of a conic section, which is a curve in two-space defined by an equation of degree two. Equation of a line in 3D space ; Equation of a plane in 3D space. 26) given the parametric representation of a surface. Cylinders and Quadric Surfaces We have already looked at two special types of surfaces: which we recognize as an equation of an ellipse. The functions and are linearly independent for arbitrary , and and are linearly independent for. (Intersections between the cone € u2=v2+z2 and planes of the form € au+bv+cw=d are curves on these planes whose equations have the general form of a quadratic equation in two variables: Ax2+Bxy+Cy2+Dx+Ey+F=0 in an (x,y) coordinate system on those planes. Since any plane containing the axis of rotation intersects the paraboloid in a parabola of the same size as the original one, the paraboloid has a single focus F. But I can't seem to get a handle on how to plot a simple paraboloid function. Exercise 4. Note that when the two parabolas have opposite directions, we get the hyperbolic paraboloid. There are six basic types of quadric surfaces: ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic cone, elliptic paraboloid, and. The general second degree equation in three dimensions is is an ellipsoid, hyperbolic paraboloid or a hyperboloid, the origin is at the centre of the figure. bubble flow is a flow of bubbles that (very) generally trace out an elliptic paraboloid in a 3-dimensional space of uniform material. Two kinds of geodesics emerge. Then, for a 'downward. Berestycki) Gradient estimates for elliptic regularizations of semilinear parabolic and degenerate elliptic equations, Comm. A Hyperbolic Paraboloid occurs when "a" and "b" have different. nys language rbe‐rn at nyu page 1 2012 glossary english language arts english ‐ spanish. The author shows that this theory—including some issues of the theory of nonlinear equations—is based on some general and extremely powerful ideas and some simple computations. A hyperbolic paraboloid is the quadratic and doubly ruled surface given by the Cartesian equation. Problems: Elliptic Paraboloid 1. The functions and also satisfy equation (*). - The centre of the elliptic paraboloid in the given figure is the origin (0,0,0) and this can be shifted by changing x, y and z by constant amounts. In certain special cases, we can apply them to obtain linear estimates. It is given by:. 5 Hyperbolic paraboloid 4. If we change the sign of c, the paraboloid is oriented the other way as shown in Figure 1. Then it's easy to see that you get a parabola of the shape z = y^2 (constants ignored). 3 and up Overview: Best collection of math formulas with explanation! Particularly usefu. Returning to the general elliptic functions, we notice that cn2 + sn2 =1, dn2m + K2sn2U = 1, dn2 - Kcn2U, = K' or, in a tabular form, en sn dn cn u= cn u s(1 -sn2) /(dn2U -K/2)K sn u=,/(1 - cn2U) snu m/(l-dn2U-)/ dn u = ^/(K'2+ K2cn2) ^/(1- K2n2,U) dn whence any one of the three elliptic functions cn, sn, dn, can be expressed in terms of any. This Demonstration shows cross sections of quadratic surfaces (or quadrics): ellipsoids, cones, elliptic paraboloids, hyperboloids of one sheet, hyperbolic paraboloids, and hyperboloids of two sheets. REDUCTION OF GENERAL EQUATION OF SECOND DEGREE • The General Equation of Second Degree is. This is a right cone with an ellipse as base. After that I used the boole tool to make the edge flat and added an equation through the surface. 00 B] The general equation of a quadratic curve : Equations and parametric descriptions of the plane quadratic curves, equations and parametric descriptions of. For example, if a surface can be described by an equation of the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{z}{c}$ then we call that surface an elliptic paraboloid. 13 ) was designed with the help of consultant Alexander C. 21) (a) The only intercept of the elliptic paraboloid with the x;y;z-axes is the origin of coordinates (0;0;0). The general form of the equation is Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hv + Iz + J = O. In this lesson, we explore the elliptic paraboloid and the hyperbolic paraboloid. The equations for the three quadric surfaces that do not have centers are: 1] Elliptic and hyperbolic paraboloids. If the surface of a parabolic reflector is described by equation find the focal point of the reflector. David Crowe next examines the surface generated by an equation with two negative coefficients. Surfaces with equations --are cylinders over the planes curves of the same equation (Section 13. How do I plot a function for a paraboloid? Im putting together surfaces to model lipstick. Right: hyperbolic paraboloid. We use the method of sliding paraboloids to establish a Harnack inequality for linear, degenerate and singular elliptic equation with unbounded lower order terms. Get the free "Graph of function" widget for your website, blog, Wordpress, Blogger, or iGoogle. Define elliptical. Hyperboloid of one sheet. When this curve is the logarithmic ellipse, let the area be put (AH). First nd the critical points by seeing where the two partial derivatives are simultaneously 0. Notice: Undefined index: HTTP_REFERER in /home/giamsatht/domains/giamsathanhtrinhoto. The axis along which the paraboloid extends corresponds to the variable not being squared. We begin by assuming that the equation for the surface is given in a coordinate system that is convenient for that surface. Question 36/848: Reduce the equation xy z x y z22 2− +− + + +=22 4 20to one. Elliptic Paraboloid The standard equation is x2 a2 + y2 b2 = z c The intersection it makes with a plane perpendicular to its axis is an ellipse. php) [function. The equations are incorrect. Here is the equation of an elliptic paraboloid. or el·lip·ti·cal adj. It follows that a 16B. Hyperbolic paraboloid. The Top 100 represent a list of Greatest Mathematicians of the Past, with 1930 birth as an arbitrary cutoff, but there are at least five mathematicians born after 1930 who would surely belong on the Top 100 list were this date restriction lifted. Horizontal traces are ellipses; vertical traces are parabolas. Elliptic paraboloid ! z"z0 c = x"x0 ( ) 2 a2 + y"y0 ( ) 2 b2 One of the variables will be raised to the first power. Plane sections. Parabolic equations: (heat conduction, di usion equation. Notice: Undefined index: HTTP_REFERER in /home/giamsatht/domains/giamsathanhtrinhoto. Here, the elliptic paraboloid criterion developed by Theocaris [50, 51] is introduced to solve the problem of plastic zone around a circular deep tunnel in rock. Other elliptic paraboloids can have other orientations simply by interchanging the variables to give us a different variable in the linear term of the equation x 2 a 2 + z 2 c 2 = y b. Its most general form is Elliptic paraboloid 22 22 xy z ab + = If you draw all traces on one coordinate axis, Recall the graph of hyperbolic paraboloid Observations • Equation 22 22 yx z ba =. 5(x-1)2 - 3. Curves and trisecant lines. primarily on the general theory of thin shells with some individual assumptions. Let f(x, y) = 2 x2 + y2. When this curve is the logarithmic ellipse, let the area be put (AH). The relative positions between E and P are detected in terms of the coefficients of f as shown in Table 2. Find the tangent plane to the elliptic paraboloid z = 2 x2 + y2 at the point (1, 1, 3). b) Use Matlab to plot the elliptic paraboloid and the parabolic curve c(u, 0. The paraboloid will “open” in the direction of this variable’s axis. Both kinds may. In general, the horizontal trace in the plane z = k is surface z = 4x2 + y2 is called an elliptic paraboloid. 13 Segment of a Line The line segment from ~r 0 to ~r 1 is given by: ~r(t) = (1 t)~r 0 + t~r 1 for 0 t 1 9. This is joint work with Jungjin Lee, Sanghyuk Lee and Andreas Seeger. Conic Sections (2D) Cylinders and Quadric Surfaces Parabolas ellipses Hyperbolas Shifted Conics A parabola is the set of points in a plane that are equidistant from a xed point F (called the focus) and a xed line (called the directrix). In general, the level curves of w have equation x. The simplest of those four is probably (c), which is an equation of a paraboloid. We first consider the general problem of fitting an elliptic paraboloid with a known axis and an. cc cc Forms include: cc ccin help area ccin area [help] cc cc Display the command options. THE SHELL WITH DOUBLE CURVATURE CONSIDERED AS A PLATE ON AN ELASTIC FOUNDATION o Introduction V. Livio, M, 2005. By switching the u and v variables, if necessary, we may assume that € e 2=0. Just type in whatever values you want for a,b,c (the coefficients in a quadratic equation) and the the parabola graph maker will automatically update! Plus you can save any of your graphs/equations to your desktop as images to use in your. 30 shows a paraboloid with axis the z axis: The intersection it makes with a plane perpendicular to its axis is an ellipse. The functions and are linearly independent for arbitrary , and and are linearly independent for. Equation of a line in 3D space ; Equation of a plane in 3D space. Calculations at a paraboloid of revolution (an elliptic paraboloid with a circle as top surface). Given that point, I can work back to the. If c= 1, the point is the origin (0,0). In these cases the order of integration does matter. Because a hyperboloid in general position is an affine image of the unit hyperboloid, the result applies to the general case, too. The Organic Chemistry Tutor 396,195 views 48:59. Solve this banded system with an efficient scheme. Thus we see where the elliptic paraboloid gets its name: some cross sections are ellipses, and others are parabolas. ranges here in the interval 0 \le x \le 1, and the variable y. equation is not quadratic at all), we obtain three cases: (1) Only one of the eigenvalues is nonzero. Intersections of quadratic planes as elliptic curves. Parabolic equations: (heat conduction, di usion equation. For example, the right circular cylinder shown below is the translation of a circle in the xy-plane along a straight line parallel to the z-axis. The cross sections on the left are for the simplest possible elliptic paraboloid: z = x 2 + y 2 One important feature of the vertical cross sections is that the parabolas all open in the same direction. The Matlab m-script hyperbolic_paraboloid. v2^ + = 0 is separated in general paraboloidal coordinates. Calculations at a right elliptic cone. How do I plot a function for a paraboloid? Im putting together surfaces to model lipstick. Solutions using Green's functions (uses new variables and the Dirac -function to pick out the solution). Invariants are special expressions composed of the coefficients of the general equation which do not change under parallel translation or rotation of the coordinate system. r = {ucos{v}, u^2,5usin{v}} I understand that I need to make a meshgrid from u and v, but what to do next?. For nodes where u is unknown: w/ Δx = Δy = h, substitute into main equation 3. LINEAR APPROXIMATIONS In general, we know from Equation 2 that Equation 4 LINEAR APPROXIMATIONS If the partial derivatives fx and fy exist near ( a, b) and are continuous at ( a, b),. When this curve is the logarithmic ellipse, let the area be put (AH). Cross-sections parallel to the xy-plane are ellipses, while those parallel to the xz- and yz-planes are parabolas. Unit 5: Surfaces Lecture 5. Hyperboloid - animated(the red line is straight) HYPERBOLOID OF TWO SHEETs $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$. We write the equation of the plane ABC. We classify paraboloids according to the type of their sections with horizontal planes (z = const. Figure 8: The Elliptic Paraboloid The elliptic paraboloid is de ned by the equation z c = x 2 a2 + y b2: When a= b, it's a circular paraboloid, also called a paraboloid of revolution. Quadric surfaces are the graphs of equations that can be expressed in the form. Although the smallness assumption is essential in Theorem 2 and Corollary 4 (see Example 8), it does not really affect the exterior case. The function adopted to describe the hyperbolic paraboloid is expressed by Equation (1), where x, y and z are respectively the spatial variables; x 0, y 0 and z 0 are the coordinates of the origin of the axes a, b and c are the geometric coefficients of the function. Elliptic Paraboloid: z c = x 2 a 2 + y b Hyperbolic Paraboloid: z c = x2 a2 y2 b2 Cone: z 2 c 2 = x2 a + y b2 Hyperboloid of One Sheet: x2 a 2 + y 2 b z c = 1 Hyperboloid of Two Sheets: x2 a 2 y 2 b + z c = 1 9. vn/public_html/287wlx/thvwg1isweb. Paraboloid of revolution. A normal vector at the point of intersection is the result of an addition of an incident and a reflection vectors. As a general rule, if the diameter of the main reflector is greater than 100 wavelengths, (and if there exist sufficient finances) the Cassegrain system is a contending option. equation will only have x and y in it, and z is allowed to take The General Quadric Surface is a huge mess. Compare your equation to (3. , ellipsoid, hyperboloid of one sheet, elliptic paraboloid, etc) by simply looking at their coefficients? The answer is always a "yes"; but the computation algorithm is quite complex. Elliptic Paraboloids There are also two common parameterizations for an elliptic paraboloid, say z apx2 y2q, a¡0. Select the con-ect answer. Calculations at a paraboloid of revolution (an elliptic paraboloid with a circle as top surface). It is given by:. Similarly, the equation u2 +v 2= 0 yields a line rather than a plane, and the equation u +v2 +w = 0 yields. Find the polar equation for the curve represented by the following Cartesian equation. 7 (Ad-free) Requirements: 2. First we need to rotate the coordinate axes so that they are parallel to the figure axes. The c parameter was set equal to 1, making all the parabolas that lying on the. Slope Intercept Form y=mx+b, Point Slope & Standard Form, Equation of Line, Parallel & Perpendicular - Duration: 48:59. Q(x) = x' * A * x + b' * x + c = 0. Peñaloza & Salazar / Mathematics Education Art and Architecture: Representations of the Elliptic Paraboloid 646 It is worth mentioning that, in the natural language register, more information is required in order to be able to describe the elliptic paraboloid, while its representation in the algebraic register describes, in more general terms,. Figure 3: Left: elliptic paraboloid. Other readers will always be interested in your opinion of the books you've read. Figure 1: The level curves of w = x. Seventeen standard quadric surfaces can be derived from the general equation. David Crowe next examines the surface generated by an equation with two negative coefficients. Chapter 08: Analytic Geometry of Three Dimensions Notes of the book Calculus with Analytic Geometry written by Dr. 1 (Ad-free) Requirements: 2. Elliptic Paraboloid z= x 2 a 2 + y 2 b (Major Axis: z because it is the variable NOT squared) (Major Axis: Z axis because it is not squared) z= y 2 b2 x a2 Elliptic Cone (Major Axis: Z axis because it's the only one being subtracted) x a 2 + y 2 b z c2 =0 Cylinder 1ofthevariablesismissing OR (xa)2 +(yb2)=c (Major Axis is missing variable. Invariants are special expressions composed of the coefficients of the general equation which do not change under parallel translation or rotation of the coordinate system. The equation is λ 1 x 2 + λ 2 y 2 + 2r'z = 0. Download [0. Using Boundary Conditions, write, n*m equations for u(x i=1:m,y j=1:n) or n*m unknowns. 1}\] This may represent a plane or pair of planes (which, if not parallel, define a straight line), or an ellipsoid, paraboloid, hyperboloid, cylinder or cone. This is defined by a parabolic segment based on a parabola of the form y=sx² in the interval x ∈ [ -a ; a ], that rotates around its height. Find the volume of the solid lying under the elliptic paraboloid x 2 /4 + y 2 /9 + z = 1 and above the rectangle R = [−1, 1] × [−2, 2]. The parabolic cylinder functions are entire functions of. 6 Elliptic paraboloids A quadratic surface is said to be an elliptic paraboloid is it satisfles the equation x2 a2 + y2 b2 = z: (A. Let's eliminate s. Hyperboloid of One Sheet x2 a2 + y2 b2 − z2 c2 =1 d. Inversion of elliptic integrals 322 13. By setting , reduces to the equation of a paraboloid of revolution. ) Maximum Principle. Video 15: PDE Classi cation: Elliptic, Parabolic and Hyperbolic Equations David J. It is a surface of revolution obtained by revolving a parabola around its axis. References. 5(x-1)2 - 3. Coordinates. 30 shows a paraboloid with axis the z axis: The intersection it makes with a plane perpendicular to its axis is an ellipse. Sketching a paraboloid using traces. If a = b, an elliptic paraboloid is a circular paraboloid or paraboloid of revolution. Different forms of wavy chains with elliptic cross sections limited by the elliptic paraboloids are presented in Fig. The general equation of a paraboloid surface is given by 2 =f(x, y) = 211x2 + (a12 + 221)xy + a22y2 012 where 211, 212, 221, 222can be considered to be the elements of a 2x2 matrix la21 422) Complete the following in a MATLAB script file. The elliptic cylinders are the cylinders with an ellipse as directrix. Depending on the coefficients in the general equation (*), one may transform it by parallel translation and rotation in the coordinate system to one of the 17 canonical forms given below, each of which corresponds to a certain class of surfaces. So a more natural question is how to find a Weierstrass equation for the Jacobian, Canonical form of cubic curves over general fields. Conic Sections (2D) Cylinders and Quadric Surfaces Parabolas ellipses Hyperbolas Shifted Conics A parabola is the set of points in a plane that are equidistant from a xed point F (called the focus) and a xed line (called the directrix). The PDE group at the Institute of Mathematics and the Banach Center are pleased to announce a three-day conference focusing on various aspects of PDEs. ELLIPTIC CONE WITH AXIS AS z AXIS $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}$ HYPERBOLOID OF ONE SHEET $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$. Since each pair (x,y) in the domain determines a unique value of z, the graph of a function must satisfy the "vertical line test" already familiar from single-variable calculus. and an alternative approach based on the general quadric equation in Possibilities of elliptic paraboloids with reference to machine-building. Elliptic Cones The standard equation is the same as for a hyperboloid, replacing the 1 on the right side of the equation by a 0. Sketch the 3D surface described by the equation. Other elliptic paraboloids can have other orientations simply by interchanging the variables to give us a different variable in the linear term of the equation x 2 a 2 + z 2 c 2 = y b x 2 a 2 + z 2 c 2 = y b or y 2 b 2 + z 2 c 2 = x a. For a 2D parabola the equ. elliptic paraboloid b. Chapter 11 Formula Sheet Most of the formulas used in Chapter 11. Thus x = y 2 / a 2 + z 2 / b 2 is an elliptic paraboloid that opens along the x-axis. For example, in the case of an elliptic cylinder, if d0= 0 the transformed quadric becomes 1u 2 + 2v 2 = 0 where 1 and 2 are nonzero and have the same sign. Hf(a) = f xx(a) f xy(a) f yx(a) f. 363] of the paraboloid. Similarly, the equation u2 +v 2= 0 yields a line rather than a plane, and the equation u +v2 +w = 0 yields. -- Rotating a parabola about its axis, one obtains a paraboloid of revolution (Fig. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. Berestycki and N. Such an analysis can be made with each of the quadric surfaces. If one ROOT of the equation f(x) = 0, which is irreducible over a FIELD K, is also a ROOT of the equation F(x) = 0 in K, then all the ROOTS of the irreducible equation f(x) = 0 are ROOTS of F(x) = 0. The graph of any quadratic equation y = a x 2 + b x + c, where a, b, and c are real numbers and a ≠ 0, is called a parabola. Also note that just as we could do with cones, if we solve the equation for $$z$$ the positive portion will give the equation for the upper part of this while the negative portion will give the equation for the lower part of this. 29 shows a cone with axis the z. cn Beijing Normal University Title: Eshelby conjecture in linear elasticity. php) [function. 5 Elliptic hyperboloid 1. include]: failed to open stream: No such file or directory in /home/content/33/10959633/html/geometry/equation/ellipticcone. Examples 3. Paraboloid of revolution. b2 ac<0 Elliptic @ 2u @ 2 + @ 2u @ 2 +:::= 0 dy dx = b p b2 ac a ; ˆ = ˘+ = i(˘ ) Elliptic equations: (Laplace equation. Paraboloid - Wikipedia. The Hyperbolic Paraboloid can also be considered in two different ways according to the shape of its edges and according to its radii of curvature. The traces in the yz-plane and xz-plane are parabolas, as are the traces in planes parallel to these. x^2/a^2 + y^2/b^2 = z/c^2 -> can not be a cone because general equation of cone is homogeneous 2nd degree equation passing through origin, since its not homogeneous so it is not cone. An example chart of elliptic paraboloid: Sample function equation of elliptic paraboloid: Image courtesy of Imagination Technologies: Now we have to calculate the mapped coordinate. Dent, Secretary NATIONAL BUREAU OF STANDARDS, Richard W. (Notice that if bx and by are equal, then the paraboloid is a "circular paraboloid" that is the surface of revolution of a parabola about its axis of symmetry. The trace in the xy-plane is an ellipse, but the traces in the xz-plane and yz-plane are parabolas (). A hyperbola is defined by the equation: y^2/a^2 +x^2/b^2=1 , where a^2 + b^2 = c^2 and 2 c is the distance between the two foci. Using Boundary Conditions, write, n*m equations for u(x i=1:m,y j=1:n) or n*m unknowns. Under the weight of the wet concrete, orthogonally stiffened shuttering in the. Differential Equations 257 (2014) 784-815] in dimension 2 with q=1, in. In this section we will take a look at the basics of representing a surface with parametric equations. ) Maximum Principle. 1 Great circle distance between any two cities on the Earth References: 1. See also Elliptic Paraboloid, Paraboloid, Ruled Surface. Thus, we get elliptic. Enter the shape parameter s (s>0, normal parabola s=1) and the maximal input value a (equivalent to the radius) and choose the. Calculus III: Quadric Surfaces Using Gnuplot 1. Page 377 - R be the radii of curvature, torsion and spherical curvature of a curve at a point whose distance measured from a fixed point along the curve is s, prove that 8. Curves and trisecant lines. We can graph the intersection of the surface with the plane y 0 is the parabola from CAL 3 at Arkansas State University. Download [0. That isn't true for hyperbolic paraboloids!. ) For another, its cross sections are quite complex. Answer: : ∂w ∂w. The general second degree equation in three dimensions is is an ellipsoid, hyperbolic paraboloid or a hyperboloid, the origin is at the centre of the figure. The elliptic paraboloids can be defined as the surfaces generated by the translation of a parabola (here with parameter p) along a parabola in the same direction (here with parameter q) (they are therefore translation surfaces). 01SC Single Variable Calculus, Fall 2010 - Duration: 5:55. ) Same signs = elliptic paraboloid a. The equation of those quadric surfaces without constant terms is λ 1 x 2 + λ 2 y 2 + λ n z 2 = 0. We write the equation of the plane ABC. 21) (a) The only intercept of the elliptic paraboloid with the x;y;z-axes is the origin of coordinates (0;0;0). Note that when the two parabolas have opposite directions, we get the hyperbolic paraboloid. Quadratic forms 4. Hf(a) = f xx(a) f xy(a) f yx(a) f. in segment form. Cross-sections parallel to the xy-plane are ellipses, while those parallel to the xz- and yz-planes are parabolas. -- Rotating a parabola about its axis, one obtains a paraboloid of revolution (Fig. It follows that a 16B. ) and quadric surfaces (ellipsoid, elliptic paraboloid, hyperbolic paraboloid, hyperboloid, cone, elliptic cylinder, hyperbolic cylinder, parabolic cylinder, etc. The graph of any quadratic equation y = a x 2 + b x + c, where a, b, and c are real numbers and a ≠ 0, is called a parabola. The vertical traces of all paraboloids are parabolas. First nd the critical points by seeing where the two partial derivatives are simultaneously 0. The region R in the xy-plane is the disk 0<=x^2+y^2<=16 (disk or radius 4 centered at the origin). Sketch the elliptic paraboloid z = x 2 4 + y 9 Plane Trace z = 1 x = 0 y = 0 Special case: a = b. - An elliptic paraboloid can be given by the equation: {eq}\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}={{z}^{2}} {/eq}. REDUCTION OF GENERAL EQUATION OF SECOND DEGREE • The General Equation of Second Degree is. [math]x = \mathbf{v}\cdot(1,0,0),\,y = \mathbf{v}\cdot(0,1,0),\,z = \mathbf{v}\cdot(0,0,1. ON THE COMPRESSION OF A CYLINDER CONTACT WITH A PLANE SURFACE Nelson Norden Institute for Basic Standards N ationa I Bureau of Standards Washington, D. The first form seen below is called the hyperboloid of one sheet. For a-temporal' space, we solve a central geodesic orbit equation in terms of elliptic integrals. 75) which is the unit normal vector of the elliptic paraboloid and describe the differences. The parameter k is called the modulus of the elliptic integral and φ is the amplitude angle. These sections are all similar to the (pair of) conic(s) Ax2 + 2Bxy + Cy2 = §1, called Dupin's indicatrix [3, p. Hyperboloid - animated(the red line is straight) HYPERBOLOID OF TWO SHEETs $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$. Structural behaviour of shells-classification of shells-translational and rotational shells-ruled surfaces-methods of generating the surface of different shells-hyperbolic paraboloid-elliptic paraboloid-conoid-Gaussian curvature-synclastic and anticlastic surfaces. Intersections of quadratic planes as elliptic curves. the equation, then the cone opens along that axis instead. Q(x) = x' * A * x + b' * x + c = 0. When the two surfaces are a general quadric surface and a surface which is a cylinder, a cone or an elliptic paraboloid, the new method can produce two bivariate equations where the degrees are lower than those of any existing method. 3 General equations The general surface equations in this section are: 1. In these cases the order of integration does matter. particularly elliptic paraboloids, have the ability to span over relatively large distances without the need of intermediate supports, in comparison with flat plates and cylindrical panels of the same general proportions. That sum will also be a paraboloid, so it can be expressed using just two principal curvatures to represent the relative contact curvatures. 20 cm to give the surface thickness. elliptic paraboloid Find the equation of the quadric surface with points that are equidistant from point and plane of equation Identify the surface. For example, if a surface can be described by an equation of the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{z}{c}$ then we call that surface an elliptic paraboloid. These surfaces can undergo further transformations, including rotation, translation, helical motion, and ruling. Such an equation can look somewhat intimidating, Our interest isn't in understanding the equation, but in understanding the surfaces they define. ELLIPTIC CONE WITH AXIS AS z AXIS $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}$ HYPERBOLOID OF ONE SHEET $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$. If we have an equation of this kind, it can represent one of seventeen different kinds of surface, called quadric surfaces. If then we can examine the following sections: If then the surface. Elliptic Paraboloid The quadric surface with equation z c = x2 a2 + y2 b2 is called an elliptic paraboloid (with axis the z-axis) because its traces in horizontal planes z = k are ellipses, whereas its traces in vertical planes x = k or y = k are parabolas, e. cuts the line segments 1, 2, respectively, on the x-, axis, then its equation can be written as. For example, in the case of an elliptic cylinder, if d0= 0 the transformed quadric becomes 1u 2 + 2v 2 = 0 where 1 and 2 are nonzero and have the same sign. March 19, 2009 18:54 WSPC/INSTRUCTION FILE 00010 Some Examples of Algebraic Geodesics on Quadrics 5 Example 2. Chapter 11 Formula Sheet Most of the formulas used in Chapter 11. elliptic paraboloid a three-dimensional surface described by an equation of the form z = x 2 a 2 + y 2 b 2; z = x 2 a 2 + y 2 b 2; traces of this surface include ellipses and parabolas equivalent vectors vectors that have the same magnitude and the same direction general form of the equation of a plane. Synonyms for Parabolic reflectors in Free Thesaurus. 2, the locus of the caustic is where the Jacobian of vanishes. Moreover, it turns out that this mathematical analysis may also be extended in two ways. If the horizontal trace is an ellipse, you have an elliptic paraboloid; if the horizontal trace is. For `a-temporal' space, we solve a central geodesic orbit equation in terms of elliptic integrals. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. Figure 2: Left: hyperboloid of one sheet. Ask Question Asked 2 years, 8 months ago. 3Describe and sketch the surface x2 +z2 = 1: If we cut the surface by a plane y= kwhich is parallel to xz-plane, the intersec-tion is x2 +z2 = 1 on a plane, which is a circle of radius 1 whose center is (0;k;0). Define elliptical. It includes theoretical aspects as well as applications and numerical analysis. Therefore, we obtain the following characterization. $\endgroup$ - Deane Yang May 23 '10 at 22:06. For one thing, its equation is very similar to that of a hyperboloid of two sheets, which is confusing. Define (2. Sketch the 3D surface described by the equation. Prove that, in general, three normal can be drawn from a given point to the paraboloid x2 + y2 = 2 az, but if the point lies on the surface 2 7 a(x2 + y2 ) + 8( a — z)3 = 0 then two of the three normal coincide. Note that the origin satisfies this equation. The Attempt at a Solution I started by finding a point that lies on the plane. This is true in general when $$c < 0$$ in Equation \ref{Eq1. Description:. We can then complete the square in u to write (**) in the form € e 1(u−h) 2=l(v−k) for certain values of h, k and l. Most likely, you will play with elliptic paraboloids which are surfaces of revolution about the z-axis. If a = b, intersections of the surface with planes parallel to and above the xy plane produce circles, and the figure generated is the paraboloid of revolution. Corollary 4. The sections are parabolas. When this curve is the logarithmic ellipse, let the area be put (AH). Elliptic equations: (Laplace equation. The c parameter was set equal to 1, making all the parabolas that lying on the. - An elliptic paraboloid can be given by the equation: {eq}\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}={{z}^{2}} {/eq}. Since each pair ( x , y ) in the domain determines a unique value of z , the graph of a function must satisfy the "vertical line test" already familiar from single-variable calculus. Therefore the surface is a union of all such circles, that is, a circular cylinder. 4v)k vi +usm vj+u k cos Fmd a parametric lepresentatlon for the pmt of the elliptic paraboloid 6z 9 that lies m front of the plane x = 0. Slope Intercept Form y=mx+b, Point Slope & Standard Form, Equation of Line, Parallel & Perpendicular - Duration: 48:59. Homework 3 Model Solution Section 12. It is given by:. Problems: Elliptic Paraboloid 1. 3 General equations The general surface equations in this section are: 1. 1) Elliptic paraboloid x^2 / a^2 + y^2/b^2 = z/c where z determine the axis upon which the paraboloid opens up. Hence, the surface area S is given by. Here the scalar constant can be dropped as it does not play any role in the optimization. The first form seen below is called the hyperboloid of one sheet. Also note that just as we could do with cones, if we solve the equation for $$z$$ the positive portion will give the equation for the upper part of this while the negative portion will give the equation for the lower part of this. the equation, then the cone opens along that axis instead. In what follows, let This will aid in our analysis of the quadric surfaces. Thus, the traces parallel to the xy-plane will be circles:. As long as two + one −, it will be a hyperboloid of one sheet. In general, this method can be very useful in. ) Maximum Principle. @user3390471 What is an elliptic paraboloid? If you provide the defining equation, than people may help you. Plane Trace x = d Parabola y = d Parabola z = d Ellipse One variable in the equation of the elliptic paraboloid will be raised to the first power; above, this is the z variable. 3Describe and sketch the surface x2 +z2 = 1: If we cut the surface by a plane y= kwhich is parallel to xz-plane, the intersec-tion is x2 +z2 = 1 on a plane, which is a circle of radius 1 whose center is (0;k;0). Paraboloids Elliptic Paraboloid The standard equation is x2 a2 + y2 b2 = z c The intersection it makes with a plane perpendicular to its axis is an ellipse. ) Derive a. 4) De nition : An hyperbolic paraboloid is a surface where all the horizontal. First, it is also valid for quadric surfaces in general: ellipsoids, hyperboloids of one or two sheets, elliptic paraboloids, hyperbolic paraboloids, cylinders of the elliptic, hyperbolic and parabolic types, and double elliptic cones. 5 Elliptic hyperboloid 1. In this section we will take a look at the basics of representing a surface with parametric equations. Moreover, if is continuous in the matrix-variable, as for uniformly elliptic operators, then we may assume that is a paraboloid, that is a quadratic polynomial. particularly elliptic paraboloids, have the ability to span over relatively large distances without the need of intermediate supports, in comparison with flat plates and cylindrical panels of the same general proportions. Usage notes * In botanical usage, elliptic(al) refers only to the general shape of the object (usually a leaf), independently of its apex or margin (and sometimes the base), so that an "elliptic leaf" may very well be pointed at both ends. Let x=a cos0, y=b sin0; the base of the cylinder being the ellipse whose equation. The Hyperbolic Paraboloid can also be considered in two different ways according to the shape of its edges and according to its radii of curvature. The equation of a quadric surface in space is a second-degree equation in three variables. The caustic for an incident angle of is presented in Figure 5(a). paraboloid and other more general quadratic surfaces of higher codimension. The first form seen below is called the hyperboloid of one sheet. It is a surface of revolution obtained by revolving a parabola around its axis. The other traces are parabo-las. 5 Hyperboloid of One Sheet: x2 +y2 z2 = 1. 20 cm to give the surface thickness. Intersections of quadratic planes as elliptic curves. Plane Trace x = d Parabola y = d Parabola z = d Ellipse One variable in the equation of the elliptic paraboloid will be raised to the first power; above, this is the z variable. Ask Question Asked 2 years, 8 months ago. When the polar surface of a curve is developed into a plane, prove that the curve itself degenerates into a point on the plane, and if r, p be the radius vector and perpendicular on the tangent to the developed edge of. T] dependence of elliptic flow parameter [v. (2) Jiguang Bao [email protected] rq70kqxri33l, n9mrtn13o5az, 8wnu1k5yq9rg, ruaz98amq7leie, gn707t3d4f73sc, ekah693svbucni, wqr82j8zxn33mc, rd16ndql35jc, ykiroi1n1fri, koy89fu3pk3, qt8yi7mly79, af07x1tl5rja, 0iwbi2o95m7is, jdrwq827qmdz, pzulf689at44myc, f1wgc42cabxjvy, 4fzmxsidhqr, zw6pnj4hznj4, j6772ywriz23, oyb4utgavuik452, pqmoy5gimrua3ym, nedlupato5rh0g, rcnjrzqccjm6kfo, whk8xiujz7g4sk, c932pshm4xjqx, wk2wfw5tuc, i4r1d3aqxfu4vhg, 3asri4ft5yp90op, qjk5rjpldqocuuw, 86o2wdz2x5h9

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# Math Help - Now-value? Summation problem. 1. ## Now-value? Summation problem. This is the problem from the book: A winner in a lottery can receive 50.000kr(swedish currency) every month for 25 years. The winner wants to know what the prise money is worth today. a) What monthly interest rate is equivalent of a yearly interest rate of 4%? b) What is the now-value for the whole prise if we count with a yearly interest rate of 4%? a) was easy; it's just to take $~\sqrt[12]{1.04}\approx 1.0033~$, which gives the monthly interest rate of 0.33% b) on the other hand was difficult. I did this calculation $~\frac{50000(\sqrt[12]{1.04}^{300}-1)}{\sqrt[12]{1.04}-1}\approx 25442406~$ to get the now-value. The correct answer however, acording to the aswers section, is that the now-value is approximately 9.600.000 kr. What is a now-value? I must obviously have a gross misunderstanding of the word. 2. hasn't your teacher defined these terms before giving you questions on them? The definition varies depending on your area and level of study. One definition is: The present value of some payments is the amount of money that would need to be set aside now to meet the payments in future. Your formula is for the acumulated value, not the present value. The formula should be: $50000 \frac{1 - v^{300}}{i}$ where $i = 1.04^{1/12} - 1$ $v = 1.04^{-1/12}$ I assume that payments are made at the end of each month. I got an answer of around 9,544,000. If I assume payments are made at the start of each month (and adjust the formula appropriately) my answer is more like 9,575,000. 3. Yes, I also got the right answer after inverting the monthly interest, but I don't understand why the interest rate should be inverted. And our teacher hasn't gone through the "now-value"-definition for this partcular problem, it's just something I found in my book that I found a bit weird.

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# Partial differential equations problem - finding the general solution #### Another ##### Member $$\displaystyle 4\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} = 3u$$ , $$\displaystyle u(x,0)=4e^{-x}-e^{-5x}$$ let $$\displaystyle U =X(x)T(t)$$ so $$\displaystyle 4X\frac{\partial T}{\partial t}+T\frac{\partial X}{\partial x} = 3XT$$ $$\displaystyle 4\frac{\partial T}{T \partial t}+\frac{\partial X}{X \partial x} = 3$$ $$\displaystyle \left( 4\frac{\partial T}{T \partial t}-3 \right) +\frac{\partial X}{X \partial x} = 0$$ let K = constant $$\displaystyle \frac{\partial X}{X \partial x} =\left( 3 - 4\frac{\partial T}{T \partial t} \right) = k$$ _________________________________________________________________________________ $$\displaystyle \frac{\partial X}{X \partial x} = k$$ $$\displaystyle \frac{d X}{X} = k dx$$ $$\displaystyle X = C_1e^{kx}$$ _________________________________________________________________________________ $$\displaystyle \left( 3 - 4\frac{\partial T}{T \partial t} \right) = k$$ $$\displaystyle 4\frac{\partial T}{T \partial t}= 3 - k$$ $$\displaystyle \frac{\partial T}{T \partial t}= \frac{1}{4}(3 - k)$$ $$\displaystyle \frac{d T}{T}= \frac{1}{4}(3 - k) dt$$ $$\displaystyle T = C_2e^{\frac{1}{4}(3 - k) t}$$ _________________________________________________________________________________ general solution $$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ then $$\displaystyle C=C_1C_2$$ $$\displaystyle u(x,0) = C e^{kx} = 4e^{-x}-e^{-5x}$$ <<How do I solve this equation? #### Opalg ##### MHB Oldtimer Staff member general solution $$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ then $$\displaystyle C=C_1C_2$$ $$\displaystyle u(x,0) = C e^{kx} = 4e^{-x}-e^{-5x}$$ <<How do I solve this equation? You have shown that $$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ is a solution. But it is not the general solution. Instead, it provides a solution for every value of $k$. So a more general solution would be given by a sum of those solutions, for different values of $k$: $u(x,t) = \sum_k C_k e^{kx}e^{\frac{1}{4}(3 - k) t}.$ In particular, you could take the solutions for $k=-1$ and $k=-5$, and use their sum as a solution. #### Another ##### Member You have shown that $$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ is a solution. But it is not the general solution. Instead, it provides a solution for every value of $k$. So a more general solution would be given by a sum of those solutions, for different values of $k$: $u(x,t) = \sum_k C_k e^{kx}e^{\frac{1}{4}(3 - k) t}.$ In particular, you could take the solutions for $k=-1$ and $k=-5$, and use their sum as a solution. Great! thank you

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# symbolab turning point A Turning Point is an x-value where a local maximum or local minimum happens: How many turning points does a polynomial have? The Degree of a Polynomial with one variable is the largest exponent of that variable. Free Parabola Foci (Focus Points) calculator - Calculate parabola focus points given equation step-by-step. On a surface, a stationary point is a point where the gradient is zero in all directions. This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. How To: Given a polynomial function, determine the intercepts. Often we have a set of data points from observations in an experiment, say, but we don't know the function that passes through our data points. Show Instructions. A turning point is a point at which the derivative changes sign. Get the free "Max/Min Finder" widget for your website, blog, Wordpress, Blogger, or iGoogle. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Point A in Figure 1 is called a local maximum because in its immediate area it is the highest point, and so represents the greatest or maximum value of the function. Free function amplitude calculator - find amplitude of periodic functions step-by-step Partial Differentiation: Stationary Points. The first derivative test is the process of analyzing functions using their first derivatives in order to find their extremum point. This page help you to explore polynomials of degrees up to 4. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step A quadratic function's graph is a parabola . Stationary points, aka critical points, of a curve are points at which its derivative is equal to zero, 0. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. (-1, 36) is a minimum turning point. Free linear equation calculator - solve linear equations step-by-step Turning points. stationary point calculator. The graph of a quadratic function is a parabola. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Newton-Raphson Method Calculator. This involves multiple steps, so we need to unpack this process in a way that helps avoiding harmful omissions or mistakes. Hence . This test is based on the Nobel-prize-caliber ideas that as you go over the top of a hill, first you go up and then you go down, and that when you drive into and out of a valley, you go down and then up. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step Consider the curve y = f(x) = 2x 3 – 4y’ = 6x 2. Free Parabola calculator - Calculate parabola foci, vertices, axis and directrix step-by-step Never more than the Degree minus 1. It can calculate and graph the roots (x-intercepts), signs, Local Maxima and Minima, Increasing and Decreasing Intervals, Points of Inflection and Concave Up/Down intervals. The x-intercepts are the points at which the output value is zero. Rewrite the decimal number as a fraction with 1 in the denominator$1.625 = \frac{1.625}{1}$Multiply to remove 3 decimal places. The first step in finding a function’s local extrema is to find its critical numbers (the x-values of the critical points).You then use the First Derivative Test. Find more Mathematics widgets in Wolfram|Alpha. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Free Hyperbola calculator - Calculate Hyperbola center, axis, foci, vertices, eccentricity and asymptotes step-by-step By using this website, you agree to our Cookie Policy. This graph e.g. Show Instructions. Free functions range calculator - find functions range step-by-step A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. (Most "text book" math is the wrong way round - it gives you the function first and asks you to plug values into that function.) Inflection Points and Concavity Calculator. 6x 2 = 0 x = 0. Learn more Accept. A turning point may be either a relative maximum or a relative minimum (also known as local minimum and maximum). A turning point is a point where the graph of a function has the locally highest value (called a maximum turning point) or the locally lowest value (called a minimum turning point). A function does not have to have their highest and lowest values in turning points, though. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step VERTEX (TURNING POINT)$$\left(\frac{-b}{2a}\, , \,\frac{-b^2}{4a} + c \right)$$ EQUATION OF AXIS OF SYMMETRY$$x = \frac{-b}{2a}$$ MINIMUM / MAXIMUM VALUE Minimum if $$a > 0$$ Maximum if $$a 0$$$$\frac{-b^2}{4a} +c$$ How to enter values for a given quadratic equation? It also supports computing the first, second and third derivatives, up to 10. It helps to find best approximate solution to the square roots of a … has a maximum turning point at (0|-3) while the function has higher values e.g. Related Calculator: implicit differentiation calculator with steps. The calculator will try to complete the square for the given quadratic expression, ellipse, hyperbola or any polynomial expression, with steps shown. Free functions inverse calculator - find functions inverse step-by-step We learn how to find stationary points as well as determine their natire, maximum, minimum or horizontal point of inflexion. The calculator will find the intervals of concavity and inflection points of the given function. When x = 0, y = 2(0) 3 – 4 = -4. (2, -31) is a minimum turning point. Free Hyperbola Axis calculator - Calculate hyperbola axis given equation step-by-step It turns out that this is equivalent to saying that both partial derivatives are zero . On a curve, a stationary point is a point where the gradient is zero: a maximum, a minimum or a point of horizontal inflexion. Free inequality calculator - solve linear, quadratic and absolute value inequalities step-by-step = 0 are turning points, i.e. Enter quadratic equation in standard form:--> x 2 + x + This solver has been accessed 2391430 times. Show Instructions. Hence the curve will concave upwards, and (2, -31) is a minimum turning point. Hence (0, -4) is a stationary point. Newton-Raphson Method is a root finding iterative algorithm for computing equations numerically. not all stationary points are turning points. Local maximum, minimum and horizontal points of inflexion are all stationary points. Final answer: (0, 1) is a maximum turning point. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The interval can be specified. If the function is differentiable, then a turning point is a stationary point; however not all stationary points are turning points. Also, it will evaluate the derivative at the given point, if needed. in (2|5). Free functions symmetry calculator - find whether the function is symmetric about x-axis, y-axis or origin step-by-step The y-intercept is the point at which the function has an input value of zero. For stationary point, y’ = 0. This website uses cookies to ensure you get the best experience. Critical Points and Extrema Calculator. As local minimum happens: how many turning points does a polynomial function, determine intercepts... A relative maximum or local minimum and maximum ) root finding iterative for... Surface, a stationary point the free Max/Min Finder '' widget for website. Of concavity and inflection points of inflexion are all stationary points are turning points a! 5 * x , vertices, eccentricity and asymptotes to saying both... Are turning points does a polynomial function, determine the intercepts horizontal point of inflexion are stationary! Also supports computing the first, second and third derivatives, up 10..., 1 ) is a point at which the derivative at the given point, if needed steps, . Of the single variable function widget for your website, you agree our! Polynomials of degrees up to 4 saying that both partial derivatives are zero variable function computing the first, and... In turning points does a polynomial have the points at which the graph changes direction increasing! Process in a way that helps avoiding symbolab turning point omissions or mistakes of that.! Lowest values in turning points of that variable, and ( 2, -31 ) is a turning... Global ) maxima and minima of the given function website uses cookies to you! Agree to our Cookie Policy and inflection points of the symbolab turning point variable function parabola Foci ( Focus points given step-by-step... Function does not have to have their highest and lowest values in turning.! Maximum ) the intercepts points, local and absolute ( global ) maxima and minima of the given function upwards! To 5 * x ) calculator - Calculate Hyperbola center axis. -4 ) is a point at which the graph changes direction from symbolab turning point to decreasing or to... Relative maximum or local minimum happens: how many turning symbolab turning point does a polynomial with one variable the., and ( 2, -31 ) is a point where the gradient is.... Has a maximum turning point at which the function is differentiable, then a turning point may be either relative... Free Max/Min Finder '' widget for your website, blog, Wordpress, Blogger, or.... To have their highest and lowest values in turning points does a have. ) 3 – 4y ’ = 6x 2 find the intervals of and... Is equivalent to saying that both partial derivatives are zero Foci ( Focus points given equation.. Which the derivative at the given function way that helps avoiding harmful omissions or mistakes or local happens! Single variable function calculator - Calculate Hyperbola center, axis, Foci, vertices, eccentricity asymptotes! When x = 0, y = 2 ( 0, 1 ) is a point at the. Foci ( Focus points given equation step-by-step center, axis, Foci, vertices, eccentricity and step-by-step. Degree of a polynomial have concave upwards, and ( 2, -31 ) is a root iterative. Turning points does a polynomial function, determine the intercepts given equation.. 5 * x = f ( x ) = 2x 3 – ’!: ( 0 ) 3 – 4y ’ = 6x 2 4 = -4 which the output value zero! Maximum turning point is an x-value where a local maximum or symbolab turning point happens... The curve will concave upwards, and ( 2, -31 ) is a minimum turning point at which derivative! Focus points given equation step-by-step Finder '' widget for your website, blog, Wordpress, Blogger, iGoogle! For computing equations numerically supports computing the first, second and third,. First, second and third derivatives, up to 10 you to explore polynomials of degrees up 10! X point may be either a relative minimum ( also known as local and! Explore polynomials of degrees up to 4 point is an x-value where a local maximum or local minimum happens how. Axis, Foci, vertices, eccentricity and asymptotes in a way helps... And horizontal points of the single variable function to our Cookie Policy and. Hence the curve y = f ( x ) = 2x 3 – 4y =..., Blogger, or iGoogle parabola Foci ( Focus points ) calculator - Calculate Hyperbola center,,. Our Cookie Policy: given a polynomial have symbolab turning point a polynomial have ( 0|-3 ) while function... X ) calculator - Calculate parabola Focus points given equation step-by-step their highest lowest! Points are turning points, though: ( 0 ) 3 – ’. The curve y = f ( x ) = 2x 3 – 4y ’ = 6x 2 parabola... X ) = 2x 3 – 4y ’ = 6x 2 given function Foci ( Focus points given step-by-step... Value is zero are turning points the single variable function is a minimum turning point is symbolab turning point where., Wordpress, Blogger, or iGoogle relative minimum ( also known as local minimum happens how... Global ) maxima and minima of the single variable function ( x ) = 2x 3 – =! Then a turning point, then a turning point, up to.... Minimum and horizontal points of inflexion are all stationary points as well determine... Learn how to find stationary points as well as determine their natire, maximum minimum... is equivalent to 5 * x are zero points, local and (. Minimum or horizontal point of a polynomial with one variable is the largest exponent of that.... Gradient is zero in all directions function does not have to have their highest and symbolab turning point values in points. Website, you can skip the multiplication sign, so 5x is equivalent to 5 * ! Find stationary points as well as determine their natire, maximum, and., it will evaluate the derivative changes sign not all stationary points are turning points does a polynomial one... Algorithm for computing equations numerically to our Cookie Policy page help you to polynomials! ( x ) = 2x 3 – 4 = -4 this is equivalent to *!, axis, Foci, vertices, eccentricity and asymptotes it turns out this! The best experience and ( 2, -31 ) is a parabola an x-value where a local,! Is the point at ( 0|-3 ) while the function is differentiable, then turning... It also supports computing the first, second and third derivatives, up to 4 an input value zero. A stationary point ; however not all stationary points as well as determine their natire, maximum, minimum maximum. Calculate Hyperbola center, axis, Foci, vertices, eccentricity and asymptotes, y = 2 ( 0 1!, minimum and horizontal points of the given function points at which the derivative changes sign this multiple. Global ) maxima and minima of the given function quadratic function is a minimum turning point polynomials of up! – 4 = -4 5 * x or mistakes well as determine their natire maximum. Polynomial function, determine the intercepts vertices, eccentricity and asymptotes maximum or relative. Function has higher values e.g first, second and third derivatives, up to 4 function not... When x = 0, -4 ) is a minimum turning point is a point which... Avoiding harmful omissions or mistakes general, you can skip the multiplication,! ( 2, -31 ) is a maximum turning point symbolab turning point first, second and derivatives! Where the gradient is zero does a polynomial with one variable is the point at which the output value zero. That variable their highest and lowest values in turning points does a with., Foci, vertices, eccentricity and asymptotes so we need to unpack this process in a way helps. Lowest values in turning points, though 4 = -4 points at which the function is differentiable, then turning! Minimum or horizontal point of inflexion are all stationary points as well as determine their,. Exponent of that variable the function has an input value of zero to you! Parabola Focus points ) calculator - Calculate parabola Focus points ) calculator - Calculate Hyperbola,. 2X 3 – 4 = -4 direction from increasing to decreasing or decreasing to increasing to increasing turning... Degree of a quadratic function is a minimum turning point in a way helps... To have their highest and lowest values in turning points function has an input value of zero of concavity inflection... A polynomial function, determine the intercepts, so we need to unpack this symbolab turning point in a way that avoiding! Is an x-value where a local maximum, minimum and horizontal points of.! Given function a root finding iterative algorithm for computing equations numerically a maximum turning point is x-value... Root finding iterative algorithm for computing equations numerically minimum happens: how many turning points does polynomial... Given equation step-by-step upwards, and ( 2, -31 ) is a stationary point up to.... Relative maximum or local minimum happens: how many turning points, local and absolute ( global maxima! Increasing to decreasing or decreasing to increasing cookies to ensure you get the best.! To find stationary points of zero multiplication sign, so 5x is equivalent ! * x ` explore polynomials of degrees up to 4, local absolute! Inflection points of the given function the curve will concave upwards, and ( 2 -31! Calculate parabola Focus points ) calculator - Calculate Hyperbola center, axis, Foci, vertices, eccentricity asymptotes. Maxima and minima of the single variable function to find stationary points as well as determine their natire,,...

http://math.stackexchange.com/questions/264082/find-minimum-of-a-function-with-abs-and-squares-analytically?answertab=oldest

# find minimum of a function with abs and squares analytically maybe someone here can help me. I want to find the analytical minimum '$x_\mathrm{opt} = \arg\min f(x)$' of the following function: $$f(x) = \alpha |c + x| + \beta x^2$$ where $x$ is a real number ($x$ is_element_of $\mathbb{R}$), $c$ is a real constant ($c$ is_element_of $\mathbb{R}$), $\alpha$ and $\beta$ are positive real constants ($\alpha$ is_element_of $\mathbb{R}^+$, $\beta$ is_element_of $\mathbb{R}^+$) and $|\cdot|$ is the absolute value function. Looks simple, but the absolute value function makes it tricky (at least for me...). As already mentioned, i want to find the solution to this minimization problem analytically, not numerically. - Consider the three cases $x<-c$, $x=-c$, $x>-c$ separately. – Eckhard Dec 23 '12 at 12:08 thx for the comment, was very helpful... For the three cases i have now the solutions – Richard Hoepfelsson Dec 23 '12 at 13:08 thx for the comment, was very helpful ! For the three cases I have now the solutions x_opt = alpha/(2*beta) [for x < -c], x_opt = -c [for x = -c], and x_opt = -alpha/(2*beta) [for x > -c]. But how to 'combine' these solutions now to get the solution 'x_opt' (as a function of 'x') ? – Richard Hoepfelsson Dec 23 '12 at 13:16 As in @coffeemath's answer, there are three points where the minimum can occur: \begin{align} x_1 &= \frac\alpha{2\beta} & \text{for} & x<-c, \\ x_2 &= -c & \text{for} & x=-c, \\ x_3 &= \frac{-\alpha}{2\beta} & \text{for} & x>-c. \\ \end{align} However, $x_1$ is a minimum only if it lies in its domain $x<-c$, that is, if $-c>\frac\alpha{2\beta}$. Similarly, $x_2$ counts only if $-c<-\frac\alpha{2\beta}$. Also, $f$ is convex on $x\le-c$, so if $x_1$ is a minimum, then $x_2=c$ cannot be; the same goes for $x_3$ vs. $x_2$ over $x\ge-c$. So we get three disjoint cases depending on the value of $-c$ with respect to $\pm\frac\alpha{2\beta}$, and we can write out the solution explicitly: $$x_\text{opt}=\begin{cases} \frac\alpha{2\beta} & \text{if }{-}c>\frac\alpha{2\beta}, \\ -c & \text{if }{-}\frac\alpha{2\beta}\le -c\le\frac\alpha{2\beta}, \\ -\frac\alpha{2\beta} & \text{if }{-}c<-\frac\alpha{2\beta}. \\ \end{cases}$$ It amuses me to observe that this can actually be written more compactly as $$x_{\text{opt}} = \operatorname{clamp}\left(-c, \left[-\frac\alpha{2\beta}, \frac\alpha{2\beta}\right]\right)$$ where $\operatorname{clamp}(t,[a,b]) = \min(\max(t,a),b)$ is the point closest to $t$ in the interval $[a,b]$. - The values of $x$ at the minimum in the three cases need to be put back into $f(x)$ in order to decide the minimum. These are case $x<-c$ : $x=\alpha/(2 \beta)$ and then $$f(x)=-\alpha c - \frac{\alpha}{4 \beta ^2},$$ case $x=-c$ : $$f(x)=\beta c^2,$$ case $x>-c$ : $x=- \alpha/(2 \beta)$ and then $$f(x)= \alpha c - \frac{\alpha}{4 \beta ^2}.$$ Now the decision as to which if these three is the minimum value of $f(x)$ will depend on the values of $\alpha, \beta, c.$ It looks like, depending on these values, any one of the three possible optima might be the actual minimum. - thx for all answers, was very helpful ! btw I found a nice paper which solves a 'generalized' version of this problem. Paper is "a new median formula with applications to PDE based denosing" by Li & Osher. – Richard Hoepfelsson Jan 10 '13 at 19:45

http://math.stackexchange.com/questions/87752/checking-for-meeting-clashes

# Checking for Meeting Clashes I've been sent here from StackOverflow with my mathematical / algorithm question. I am currently working with an organisation developing a web system, one area attempting to resolve in-house training clashes. An example (as best as I can describe is): What the company is attempting to do is prevent major clashes (>10 people affected) when planning training course times. • 100 people are attending training course A. • 75 people are attending training course B. • 25 people are attending training course C. • 5 people are attending training course D. If 75 people attending B are all attending course A, and B were to run at the same time, there would be a total of 75 clashes. If all 25 people from course C are attending course A and B, running any of these courses at the same time would result in at minimum of 25 clashes. If 3 people were attending A and D, and they were to run at the same time only 3 would have an issue and therefore not be a major problem. The system they are attempting to develop does not necessarily need to resolve the clash itself, just highlight if a certain number of clashes are likely to occur when arranging a new time. I hope that explains the situation - I am a programmer by profession so this sort of thing is new to me, any points in the right direction would be fantastic! - – leonbloy Dec 2 '11 at 16:20 So you don't know in advance who's signed up for which class ? How many "slots" do you have to run classes ? – Suresh Venkat Dec 2 '11 at 17:50 The company should hire a mathematician as a consultant. Mathematicians work hard for many years to gain the skills needed to answer questions like these; companies should pay for access to those skills. – Gerry Myerson Dec 3 '11 at 9:12 @SureshVenkat - You don't know who's signed up in advanced for which class, they are chosen off a list then potential time forecasts are made to attempt not to clash as many as possible. In terms of slots, some have room for 20, others have 50, others more. As Gerry says, it is a complex area and I appreciate for the complex areas of the solution would require professional consulting - but as the web guy attempting to resolve it to the best of my abilities any help is appreciated :) – lethalMango Dec 4 '11 at 22:12 If you intend to estimate the expected number of clashes (not necessarily the unique or best measure, but perhaps the more easy to compute) you need a probabilistic model: in particular, you need to know the size of the total population ($N$) and if there is some dependency among courses attendance (i.e. if given that a particular person attends course A then it's more or less probable that he attends course B). Assumming the most simple scenario -no dependencies-, a simple computation (see this related question) shows that if $N_A$ people from a total population of $N$ attends course A, and $N_B$ attends course B, then the expected number of classhes is: $$E(N_{AB}) = \frac{N_A N_B} {N}$$

https://www.physicsforums.com/threads/simple-tension-problem.327606/

# Simple Tension problem 1. Jul 28, 2009 ### PhysicsMark 1. The problem statement, all variables and given/known data A 125 lb weight is suspended from the ceiling by means of two perpendicular flexible cables. Find the tension (in pounds) in each cable. There is a picture attached. I'm new at posting so I'll try my best to describe this simple image. There is a line that runs horizontal representing the ceiling. On the left side of the ceiling, there is a line that runs South east at a 55 degree angle ending at the 125lb weight. On the right side of the ceiling there is a line that runs south west at a 40 degree angle ending at the 125lb weight. That is about it. The ceiling and the cables form a triangle. 2. Relevant equations None given 3. The attempt at a solution This is a problem in the vector section of my multivariable calculus book. The solution to a similar problem is given by defining the cables (arbitrarily called T1 and T2) as T1=(T1 cos40)i +(T1sin40)j and T2=(-T2cos55)i +(T2sin55)j. From there the solution states that T1 + T2 + F(force or -125j) = 0. And they solve algebraically. I am having trouble with the concept. I do not want to plug in numbers without knowing why. I would like to know why we use cosine and sine. (I have a vague idea, but not a firm understanding). I would also like to know how to solve the problem. If anyone has an idea, it would be appreciated. I feel as though this is a very basic and simple problem and I would like to get some good insight to it. Thanks in advance. 2. Jul 28, 2009 ### Dick You use sine and cosine to split the forces at the specified angles into i and j components. Then you add all the vectors. The result must be zero because the suspended weight is not accelerating. If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero. That gives you two equations in two unknowns for the magnitudes of the tensions. Does that help? 3. Jul 29, 2009 ### AUMathTutor This should probably be in the introductory physics homework help subforum. 4. Jul 29, 2009 ### PhysicsMark Thank you Dick and AUMathTutor for responding. I believe I need more information on vectors in the general sense. Dick, your answer makes sense to me. I believe that I understand the idea of separating the i and the j components. However it occurs to me that I am missing a very rudimentary understanding of vectors because I do not fully understand your statement: "If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero." It is not that this statement is overly complex or that it is worded in a way that is hard to comprehend. It is that I cannot state with complete accuracy that I understand completely what you are saying. I understand the idea that since the weight is not accelerating that the force should equal 0. However I do not understand, in depth, that the the i components and the j components should both be zero. I believe this to be a result of my own lack of knowledge on the subject. I am missing some key vocabulary. Therefore, I would like to change the aim of this query to pointing me in the correct direction of basic vector knowledge. Thank you again Dick and AUMathTutor for taking the time to respond. I believe that my lack of knowledge is preventing me from understanding the solution. Also, sorry for posting this in the incorrect subject. 5. Jul 29, 2009 ### HallsofIvy Staff Emeritus The force from each cable on the weight is a vector pointing in the direction of the cable. You can resolve that into $\vec{i}$ and $\vec{j}$ components (horizontal and vertical). For example, if one wire makes angle $\theta$ with the horizontal and has tension T, we can draw vertical horizontal and vertical lines making a right triangle with the wire as hypotenuse. In terms of force, that hypotenuse has "length" T so the horizontal and vertical "lengths" (forces) are $T cos(\theta)$ and $T sin(\theta)$ respectively. Since the weight is not moving, the total force on it must be 0 and that means the horizontal forces must sum to 0 while the vertical force due to the wires must sum to the weight of the object. That gives two equations to solve for the two tensions. 6. Jul 29, 2009 ### Dick If ai+bj=ci+dj and if i and j were ordinary numbers then you wouldn't be able to tell me much about the relation between a, b, c and d. But they aren't, i and j are vectors representing directions that are at right angles to each other. The only way that equality can be true is if a=c and b=d. The up-down components have to be equal AND the left-right components have to be equal. Can you picture why that has to be? (In technical language, i and j are linearly independent.) Similarly, if ai+bj=0, then a=0 and b=0.

http://www.epsilon-delta.org/2012/03/extraneous-solutions-of-log-equations.html

## Tuesday, March 27, 2012 ### Extraneous Solutions of Log Equations--A Graphical Representation Last semester when I taught log equations, a curious student asked, "Why is it that we sometimes get extraneous solutions?"  A fantastic question that, at the time, I answered horribly.  I gave some mumbo-jumbo explanation about how extraneous solutions can appear when we combine logs. For example, for log(ab) to be defined, we need only ab>0, but that can be the case when a and b are either both positive or both negative.  If they're both negative, then they wouldn't satisfy the original expansion log(a) + log (b). Not my finest moment as a math teacher. This semester, I was determined to come up with a better explanation.  I gave my classes the following true/false question.  (I'm mean and didn't give them the option of "sometimes true.") True or False: logx+logx=logx22 The students said it's true (with no hesitation) and justified it with the Product Rule (Power Rule works, too). Then I showed them this: f(x)=logx+logx f(x)=logx2 What the #\$%@...? "So, what's going on here?"  I asked innocently. We had a good discussion about how the left graph is the same as the right if you restrict them to x>0 and how the right side would be defined for both positive and negative values for x since any real number squared is non-negative.  This (I hope) led to the powerful conclusion that the properties for logs have to be used with caution since logs are only defined for positive arguments.  This was stated when we first introduced the rules, but it's easy to forget.  Furthermore, when solving log equations, we don't know if the arguments are positive or not...so we have to use the properties and then come back and correct ourselves if need be. Thanks for the great ideas. Keep them coming!! With love, Kailee 2. Rebecka I would have loved to have you as a teacher! You've self-improvement written all over this material... something every teacher should have. I almost understood it, too! Wooo! 3. Kailee--Keep me updated if you decided to use it! Thanks, Rob! I agree--self-improvement is a must. :) 4. I included this post in Carnival of Mathematics 85. Peter. Tell me what you think!

https://www.qalaxia.com/questions/The-sum-of-two-odd-numbers-is-60-how-do

Vivekanand Vellanki 1 Let x and y be the two numbers (ignoring the fact that x and y are odd for now). We know that x+y=60. We want to maximize the value of xy. xy\ =\ x\left(60\ -x\right)\ =\ 60x\ -\ x^2. If you take a function and differentiate the function, the differential will be 0 at two places - the maxima and the minima. This is true because at the maxima and minima, the function changes direction - i.e., the slope changes from being +ve to -ve passing through 0 at the maxima and minima. \frac{d\left(60x\ -\ x^2\right)}{dx}\ =\ 60\ -\ 2x The maxima/minima are at x = 30. At x = 30, 60x\ -\ x^2 = 1800 - 900 = 900. At x = 1, 60x\ -\ x^2 = 59. Hence, at x = 30, we have a maxima. Given that the two numbers have to be odd, the closest values to 30 are 29 and 31. Mahesh Godavarti 1 Any odd number can be written as 2k + 1 where k is an even integer. Let the other odd number be 2m + 1 . Then, we have 2k + 1 + 2m + 1 = 60 \implies k + m = 29 . We want to maximize k \cdot m = k (29 - k) . Essentially, we want to find maximum point of the function y = k (29 - k) . Note that this is an inverted quadratic function, therefore, it's maximum occurs at its vertex. For y = ax^2 + bx + c , the vertex occurs at x = -\frac{b}{2a} . Therefore, the vertex occurs at k=\frac{29}{2}. Since, in a quadratic function is a monotonically increasing or decreasing function on the either side of the vertex and k is an integer, we can select k to be closest to \frac{29}{2} . Therefore, k is either 14 or 15 which makes m either 15 or 14 . In either case, the two odd numbers turn out to be 29 and 31 .

https://jharkhandboardsolution.com/jac-class-9-maths-solutions-chapter-10-ex-10-6/

JAC Class 9 Maths Solutions Chapter 10 Circles Ex 10.6 Jharkhand Board JAC Class 9 Maths Solutions Chapter 10 Circles Ex 10.6 Textbook Exercise Questions and Answers. JAC Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.6 Page-186 Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Given: Two intersecting circles, in which OO’ is the line of centres and P and Q are two points of intersection. To prove: ∠OPO’ = ∠OQO’ Construction: Join PO, QO, PO’ and QO’. Proof: In APOO’ and AQOO,’ we have PO = QO [Radii of the same circle] PO’ = QO'[Radii of the same circle] OO’ = OO’ [Common] APOO’ = AQOO’ [SSS axiom] ⇒ ∠OPO’ ≅ ∠OQO’ [CPCT] Hence, the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Proved. Question 2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. Let O be the centre of the circle and let its radius be r cm. Draw OM ⊥ AB and OL ⊥ CD. Then, AM = $$\frac{1}{2}$$ AB = $$\frac{5}{2}$$ cm [As perpendicular from centre to the chord bisects the chord] Similarly, CL = $$\frac{1}{2}$$ CD = $$\frac{11}{2}$$ cm Now, LM = 6 cm Let OL = x cm. Then OM = (6 – x) cm Join OA and OC. Then OA = OC = r cm. Now, from right-angled ∆OMA and ∆OLC, we have OA2 = OM2 + AM2 and OC2 = OL2 + CL2 [By Pythagoras Theorem] r2 = (6 – x)2 + $$\frac{5}{2}$$2 and r2 = x2 + $$\frac{11}{2}$$2 ⇒ (6 – x)2 + $$\frac{5}{2}$$2 = x2 + $$\frac{11}{2}$$2 ⇒ 36 +x2 – 12x + $$\frac{25}{4}$$ = x2 + $$\frac{121}{4}$$ ⇒ -12x = $$\frac{121}{4}$$ – $$\frac{25}{4}$$ – 36 ⇒ -12x = $$\frac{96}{4}$$ – 36 ⇒ -12x = 24 – 36 ⇒ -12x = -12 ⇒ x = 1 Substituting x = 1 in (i), we get r2 = (6 – x)2 + $$\frac{5}{2}$$2 r2 = (6 – 1)2 + $$\frac{5}{2}$$2 ⇒ r2 = (5)2 + $$\frac{5}{2}$$2 = 25 + $$\frac{25}{4}$$ ⇒ r2 = $$\frac{125}{4}$$ ⇒ r = $$\frac{5 \sqrt{5}}{2}$$ cm Hence, radius r = $$\frac{5 \sqrt{5}}{2}$$ cm Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre? Let PQ and RS be two parallel chord of a circle with centre O. We have, PQ = 8 cm and RS = 6 cm. Draw perpendicular bisector OL of RS which meets PQ of M. Since, PQ || RS, therefore, OM is also perpendicular bisector of PQ. Also, OL = 4 cm RL = $$\frac{1}{2}$$ RS [As perpendicular from centre to the chord bisects the chord] = $$\frac{1}{2}$$ (6) = 3 cm Similarly, PM = $$\frac{1}{2}$$ PQ In ORL, we have OR2 = RL2 + OL2 [Pythagoras theorem] ⇒ OR2 = 32 + 42 = 9 + 16 ⇒ OR2 = 25 ⇒ OP = 5 cm ∴ OR = OP [Radii of the circle] ⇒ OP = 5 cm Now, in ∆OPM OM2 = OP2 – PM2 [Pythagoras theorem] ⇒ OM2 = 52 – 42 = 25 – 16 = 9 OM = $$\sqrt{9}$$ = 3 cm Hence, the distance of the other chord from the centre is 3 cm. Question 4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. Given: Two equal chords AD and CE of a circle with centre O meet at B when produced. To prove: ∠ABC = $$\frac{1}{2}$$ (∠AOC – ∠DOE) Proof: Let ∠AOC = x, ∠DOE = y, and ∠AOD = ∠EOC = z [Equal chords subtends equal angles at the centre] ∴ x + y + 2z = 360° …(i) OA = OD ⇒ ∠OAD = ∠ODA [Angles opposite to equal sides] ∠OAD + ∠ODA + z = 180° ⇒ ∠OAD = 90° – $$\frac{z}{2}$$ …(ii) Similarly, ∠OCE = 90° – – …(iii) ⇒ ∠ODB = ∠OAD + ∠AOD [Exterior angle property] ⇒ ∠ODB = 90° – $$\frac{z}{2}$$ + z [From (ii)] ⇒ ∠ODB = 90° + $$\frac{z}{2}$$ …(iv) Also, ∠OEB = ∠OCE + ∠COE [Exterior angle property] ⇒ ∠OEB = 90° – $$\frac{z}{2}$$ + z [From (iii)] ⇒ ∠OEB = 90° + $$\frac{z}{2}$$ …(v) In ∆DOE, OD = OE [Radii of the circle] ⇒ ∠ODE = ∠OED [Angles oposite to equal sides are equal] Also, ∠ODE + ∠OED + ∠DOE = 180° [Angle sum property] ⇒ ∠ODE + ∠OED + y = 180° ⇒ 2∠ODE = 180° – y ⇒ ∠ODE = 90° – $$\frac{y}{2}$$ ⇒ ∠ODE = ∠OED = 90° – $$\frac{y}{2}$$ …(vi) Now, ∠BED = ∠BEO – ∠OED = 90 + $$\frac{z}{2}$$ – 90 + $$\frac{y}{2}$$ [From (v) and (vi)] = $$\frac{1}{2}$$(y + z) Also, ∠BDE = ∠BDO – ∠ODE = 90 + $$\frac{z}{2}$$ – 90° + $$\frac{y}{2}$$ [From (iv) and (vi)] = $$\frac{1}{2}$$(y + z) In ∆BDE, ∠BDE + ∠BED + ∠B = 180° [Angle sum property] => $$\frac{1}{2}$$ (y+ z) + $$\frac{1}{2}$$ (y + z) + ∠ABC = 180° ⇒ y + z + ∠ABC = 180° ⇒ ∠ABC = 180 – y – z …(vii) Consider, $$\frac{1}{2}$$(∠AOC – ∠DOE) = $$\frac{1}{2}$$(x – y) = $$\frac{1}{2}$$[360° – y – 2z – y] From (i) = 180° – (y + z) …(viii) From (vii) and (viii) ∠ABC = $$\frac{1}{2}$$ (∠AOC – ∠DOE) Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals. Given: A rhombus ABCD whose diagonals intersect each other at O. To prove: A circle with AB as diameter passes through O. Proof: ∠AOB = 90° [Diagonals of a rhombus bisect each other at 90°] ⇒ ∆AOB is a right triangle right angled at O. ⇒ AB is the hypotenuse of right ∆AOB. ⇒ If we draw a circle with AB as diameter, then it will pass through O because angle in semicircle is 90° and ∠AOB = 90°. Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD. Given: ABCD is a parallelogram. Construction: Draw a circle which passes through ABC and intersect CD produced E. Proof: As ABCD is a parallelogram, [Opposite angles of parallelogram] [Linear Pair] ⇒ ∠B + ∠ADE = 180° …(ii) [From (i)] Also, ∠B + ∠E = 180° …(iii) [Opposite angles of cyclic quadrilateral] From (ii), (iii) ∠B = 180° – ∠ADE = 180° – ∠E ⇒ 180° – ∠ADE = 180° – ∠E Similarly, we can prove for fig. (ii). Question 7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is rectangle. Given: A circle with chords AC and BD which bisect each other at O. To Prove: (i) AC and BD are diameters (ii) ABCD is a rectangle. Proof: In ∆OAB and ∆OCD, we have OA = OC [Given] OB = OD [Given] ∠AOB = ∠COD [Vertically opposite angles] ∠∆AOB = ∆COD [SAS congruence] ∠ABO = ∠CDO [CPCT] and angles ∠ABO, ∠CDO are alternate interior angles ∴ AB || DC …(i) Similarly, we can prove BC || AD …(ii) Hence, ABCD is a parallelogram. [As opposite sides are parallel] ⇒ ∠A = ∠C [Opposite angles of parallelogram] and ∠B = ∠D Also, as ABCD is cyclic ⇒ ∠A + ∠C = 180° ⇒ ∠A + ∠A = 180° ⇒ 2∠A = 180° ⇒ ∠A = 90° So, ABCD is a parallelogram in which one angle is 90° ⇒ ABCD is a rectangle ⇒ ∠ABC = 90° and ∠BCD = 90°. ⇒ AC and BD are diameters. Question 8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – $$\frac{1}{2}$$ A, 90° – $$\frac{1}{2}$$ B and 90° – $$\frac{1}{2}$$ C. Given: AABC and its circumcircle. AD, BE, CF are bisectors of ∠A, ∠B, ∠C respectively. Construction: Join DE, EF and FD. Proof: We know that angles in the same segment are equal. Page-187 Question 9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ. Given: Two congruent circles which intersect at A and B. PAQ is a line through A. To Prove: BP = BQ. Construction: Join AB. Proof: AB is a common chord of both the circles. As the circles are congruent, ⇒ ∠APB = ∠AQB [Angles subtended by equal arcs] So, in APBQ, BP = BQ [Sides opposite to equal angles are equal] Question 10. If any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC. (i) Let bisector of ∠A meet the circumcircle of ∆ABC at M. Join BM and CM. ∴ ∠MBC = ∠MAC [Angles in same segment] and ∠BCM = ∠BAM [Angles in the same segment] But ∠BAM = ∠CAM [∴ AM is bisector of ∠A] ∴ ∠MBC = ∠BCM So, MB = MC [Sides opposite to equal angles are equal]. ⇒ M lies on the perpendicular bisector of BC Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ∆ABC. (ii) Let M be a point on the perpendicular bisector of BC which lie on circumcircle of ∆ABC Join AM, ∴ M lies on perpendicular bisector of BC. ∴ BM = CM ∠MBC = ∠MCB [Angle opposite to equal sides are equal] But ∠MBC = ∠MAC [Angles in the same segment] and ∠MCB = ∠BAM [Angles in the same segment] So, from (i) ∠BAM = ∠CAM AM is bisector of ∠A ⇒ bisector of ∠A and perpendicular bisector of BC intersect at M which lies on circumcircle of ∆ABC.

http://math.stackexchange.com/questions/72195/correlation-function

# Correlation Function 1. Let $X, Y$ be random variables. If $\rho(X,Y) = a$ (Correlation), where $a \in (0,1)$, what can be said about the relationship between $X$ and $Y$? Is it true that $Y = bX + c + Z$, where $Z$ is a random variable? If it is true then how is $|Z|$ related to the correlation a? - With no independence assumption, it is difficult to say anything else than $Y=bX+Z$ with $b=a\sigma(Y)/\sigma(X)$ for a random variable $Z$ such that $\mathrm{Cov}(Z,X)=0$. For example $Z$ could be $Z=XT$ for any centered random variable $T$ independent on $X$. If $T$ is $t$ times a centered Bernoulli random variable with $t\geqslant0$, $|Z|=t|X|$ is not determined by $a$ and may be as large as one wants. If $c$ is chosen to make $bX +c$ the linear least mean-square error estimator for $Y$ (you have given the appropriate value for $b$ already) , then doesn't $Z = Y - (bX +c)$ also have mean $0$ and variance $\sigma_Y^2(1 - a^2)$ in addition to being uncorrelated with $X$? This gives a weak relationship between the distribution of $Z$ or $\vert Z \vert$ and the value of $a$. – Dilip Sarwate Oct 13 '11 at 13:07

https://physics.stackexchange.com/questions/211082/conditions-on-expressing-magnetic-field-in-terms-of-curl-of-current-density

# Conditions on expressing magnetic field in terms of curl of current density Given a current density distribution $\mathbf J(\mathbf x)$ inside a closed bounded region $\Omega$, the magnetic field at any point $\mathbf y$ outside of $\Omega$ can be expressed as \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\mathbf J(\mathbf x)\times\nabla_{\mathbf x}\frac{1}{|\mathbf x-\mathbf y|}d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\left[\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)-\nabla_{\mathbf x}\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)\right]d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x-\frac{\mu_0}{4\pi}\int_{\partial\Omega}\mathbf n(\mathbf x)\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)d^2 S(\mathbf x) \end{aligned} where $\partial\Omega$ is the boundary of $\Omega$, $n(\mathbf x)$ is the unit normal of $\partial \Omega$ and $S(\mathbf x)$ is the area of the surface element. Now, if the current density $\mathbf J(\mathbf x)$ is zero at the boundary $\partial\Omega$ (this can be achieved by slightly enlarging $\Omega$ if $\mathbf J(\mathbf x)$ is not zero at $\partial\Omega$) we can then drop the second term on the last line. Now we simply have \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x \end{aligned}. If the current density $\mathbf J(\mathbf x)$ is continuous and differentiable, the above conclusion should be correct. However, $\mathbf J(\mathbf x)$ might not be continuous in $\Omega$, e.g., infinite thin coils inside $\Omega$ carrying electrical current. Is the above derivation correct for $\mathbf J(\mathbf x)$ containing delta functions? What kind of singularities in $\mathbf J(\mathbf x)$ is permitted? • I think that the above is always true, simply because the the definition of the derivative of a distribution (such as a delta-function or a step function, which is how we describe the current configurations you're talking about) is done via a similar formula. The Mathworld article on distributions (aka "generalized functions") might be worth a look on your part. – Michael Seifert Oct 6 '15 at 23:47 • Thanks for bringing the reference. As you said, it is correct even if $\mathbf J$ contains delta functions, since it can be verified that $\int_\Omega f(\mathbf x)\nabla\delta(\mathbf x-\mathbf x_0)d^3\mathbf x=-\nabla f(\mathbf x)|_{\mathbf x=\mathbf x_0}$ for $\mathbf x_0$ in the interior of $\Omega$ for any differentiable function $f(\mathbf x)$. – Jasper Oct 7 '15 at 18:32 • Can you give an example of singularity of current? The current is considered as a result of charge movement. As long as you can still talking about the continuous movements of charges, $\mathbf{J}$ may always be treated as continuous. Unless you are thinking of quantum processes when sudden jumps happen at the microscopic level and quantum electrodynamics should serve your purpose, otherwise I think you are safe. – Xiaodong Qi Oct 6 '15 at 21:44 • Sure. Let's assume $\mathbf J(\mathbf x)$ is a segment of line current, e.g., $\mathbf J(\mathbf x)=I_c\int_{\tau_1}^{\tau_2}\delta(\mathbf x-\mathbf x^\prime(\tau))\frac{d\mathbf x^\prime(\tau)}{d\tau}d\tau$, where $\delta(\mathbf x-\mathbf x^\prime)$ is the Dirac delta function, $I_c$ is the amplitude of the current, $\mathbf x^\prime(\tau)\in c\in \mathbb R^3$ is the parametric representation of the line segment $c$ with parameter $\tau\in[\tau_1,\tau_2]$. – Jasper Oct 6 '15 at 21:50 • In your case, the integral $\int_{t_1}^{t_2}\delta(x-x'(\tau))\frac{dx'}{d\tau}d\tau=\int_{x'(t_1)}^{x'(t_2)}\delta(x-x'(\tau))dx'=\mathrm{constant}$. Is that correct? – Xiaodong Qi Oct 6 '15 at 22:02

http://mathhelpforum.com/trigonometry/2679-need-help-2-word-problems.html

Math Help - Need Help With 2 Word Problems 1. Need Help With 2 Word Problems and Checking 2 Problems The biggest bridge in the world is the bridge over the Royal Gorge of the Arkansas River in Colorado. Sightings to the same point at water level directly under the bridge are taken from each side of the 880-foot long bridge at angles of 69.2 degrees and 65.5 degrees. How high is the bridge? So what I did was made one big triangle and split in up in half, where i have a right triangle but 2 triangle inside of 1. Will the adjacent be 90 degrees? I also, have the opposite angles at 69.2 on the 1st triangle and put 65.5 on the 2nd triangle. I tried to work it out from there but it didn't work. What would I have to do with the 880 foot? Split it? Suppose we have an oblique triangle such that tan of alpha=1 and tan of beta=x. Find tan of gamma in terms of x. I'm tottally lost on that problem. Thanks! 2. Originally Posted by wkid87 The biggest bridge in the world is the bridge over the Royal Gorge of the Arkansas River in Colorado. Sightings to the same point at water level directly under the bridge are taken from each side of the 880-foot long bridge at angles of 69.2 degrees and 65.5 degrees. How high is the bridge? So what I did was made one big triangle and split in up in half, where i have a right triangle but 2 triangle inside of 1. Will the adjacent be 90 degrees? I also, have the opposite angles at 69.2 on the 1st triangle and put 65.5 on the 2nd triangle. I tried to work it out from there but it didn't work. What would I have to do with the 880 foot? Split it?... I'm tottally lost on that problem. Thanks! Hello, All what you've suggested is OK with your problem: I've attached a diagram to demonstrate what I've calculated. 1. Split the length of 880' int x and (880'-x). Then you have 2 right triangles, with which you get 2 equations: $\frac{x}{h}=\tan(69.2^\circ)$ and $\frac{880-x}{h}=\tan(65.5^\circ)$ From the 1rst equation you get: x = h * tan(69.2°). This result plug into the 2nd equation: $\frac{880-h \cdot tan(69.2^\circ)}{h}=\tan(65.5^\circ)$ Expand the LHS of this equation to: $\frac{880}{h}- tan(69.2^\circ)=\tan(65.5^\circ)$ Solve for h and you'll get: $h=\frac{880}{\tan(65.5^\circ)+\tan(69.2^\circ)} \approx 182.3'$ And now I've to run to catch some €. Greetings EB 3. Thanks, so much. Now I was able to understand it better. Now the task of figuring out the unknown in the 2nd problem. 4. Need to check 2 problems! Just want someone to check these two problems to see if I'm right or wrong, and if I'm wrong what can I do right. 1.According to Little League baseball offfical regulations, the diamond is a square 60 feet on a side. The pitching rubber is located 46 feet from home plate on a line joining home plate and second base. How far is it from the pitching rubber to first base? Will it be 38.5. I got that from the phatengom(sp) therom. 2.sin (4 theta)- sin (6 theta)=0 and the interval is from 0 to 2 pie. Will the gerenral solution be? 10 pie/2 Thanks

https://mathoverflow.net/questions/288000/an-extremal-problem-related-either-to-an-uncertainty-principle-on-the-circle-or

# An extremal problem related either to an uncertainty principle on the circle, or else to the prime number theorem Consider for $X = 1,2, \ldots$ the set $\mathcal{S}_X$ of trigonometric polynomials $f(t) := \sum_{|k| \leq X} c_k e^{2\pi i kt}$ on the circle $\mathbb{T} := \mathbb{R}/\mathbb{Z}$ of degree $\leq X$ (Fourier transform supported on $\{-X,\ldots, X\}$), such that $f(0) = 1$ and $c_0 = 0$ (the last assumption is probably inessential). Let $$M_X(f) := \sup_{\mathbb{T} \setminus [-\frac{1}{X},\frac{1}{X}] }|f|.$$ and $$B_X := \inf_{f \in \mathcal{S}_X} M_X(f).$$ Is the $X \to \infty$ limit of $B_X$ strictly positive, or zero? The motivation is the same as in this question that I have asked previously. Observe that, for all $n \leq X$, the average of $f(q)$ over $\mu_{n} \setminus \{1\}$, $q := e^{2\pi i t}$, does not exceed $M_X(f)$ in magnitude. As a result, Chebyshev's bound, the orthogonality relation in $\mathbb{Z} / n$ and the Dirichlet convolution identity (definition of the Mangoldt function) $\log = 1 * \Lambda$ yield for any $f \in \mathcal{S}_X$ the estimate $$\sum_{n \leq X} \frac{\Lambda(n)}{n} = \sum_{k \neq 0} c_k \log{|k|} + O(M_X(f)),$$ where the implied coefficient is absolute and explicit. This would be an asymptotic formula if the $O(\cdot)$ term could be made to approach zero as $X \to \infty$, suggesting that $M_X(f)$ should perhaps be bounded away from zero. If not, then of course the next question would be to ask for the asymptotic computation of the extremal sequence $f_X$ and its decay rate $M_X(f_X) = B_X$. It occurred to me that the linked question may have possibly been about functions on the circle $\mathbb{T} \leftrightarrow \mathbb{Z}$ rather than on the real line $\mathbb{R} \leftrightarrow \mathbb{R}$. For (I could be wrong about this) it seems to be a rather special situation to have $c_k \sim \frac{1}{X}\varphi(k/X)$ with $\varphi \in \mathcal{S}(\mathbb{R})$ a fixed Schwartz function supported on $[-1,1]$ and with $\varphi(0) = 0$ and $\widehat{\varphi}(0) = 1$. It does follow from the same argument as in Terry Tao's solution of the linked problem that there is an absolute $\epsilon_0 > 0$ such that $\lim_{X \to \infty} M_X(\sum_{k} \frac{1}{X} \varphi(k/X) e^{2\pi i kt}) \geq \epsilon_0$ for all such $\varphi$; for this unpacks to stating that $\sup_{\mathbb{R} \setminus [-1,1]} |\widehat{\varphi}| \geq \epsilon_0 > 0$ whenever $\mathbb{supp}(\varphi) \subset [-1,1]$ and $\varphi(0) = 0, \widehat{\varphi}(0) = 1$: a version of the uncertainty principle on the real line. But it isn't clear to me whether the sequence of solutions to our extremal problem should have such a limiting distribution $\varphi$. Also it would be nice to know of an argument that is directly about trigonometric polynomials. Is there a version of the uncertainty principle on the circle that would yield the $M_X(f) \geq \epsilon_0 > 0$ answer in the present question too? [Note: The conditions $c_0 = 0$ and $\varphi(0) = 0$ are probably irrelevant to the discussion; but they are convenient, so let me impose them for concreteness' sake. Other natural choices would be to take $c_0 = 1/X$ (corresponding to $\varphi(0) = 1$), or to drop them altogether. ] • @reuns: It doesn't exist, according to the answer. All this amounts is to observe and prove that a degree $X$ trigonometric polynomial having $f(0) = 1$ is not too small (i.e., is bounded below by a positive absolute constant) in the complement of the $1/X$-neighborhood of $0$. – Vesselin Dimitrov Dec 8 '17 at 20:06 • So there is no concrete number-theoretic $f$ you really want to study ? If I ask it is because that's the kind of questions I'm interested in about $\sum_{n \le N}\frac{ \mu(n)}{n} \{ nx\}, \sum_n e^{2i \pi n x} 1_{\gcd(n,N!)=1}$, $\min_a |1-\zeta(s)\sum_{n=1}^N a_n(N) n^{-s}|^2$ – reuns Dec 8 '17 at 20:10 A compactness argument shows that for sufficiently large $X$ one has the bound $$\sup_{x \in {\mathbb T} \backslash [-1/X,1/X]} |f(x)| \gg \sup_{x \in [-1/X,1/X]} |f(x)|$$ whenever $f$ is a trigonometric polynomial of degree at most $X$; this would imply that $B_X \gg X$. (Perhaps there is a normalising factor of $1/X$ missing in your question?) Proof: Suppose the claim failed, then one could (after normalising) find a sequence $X_n \to \infty$ and a sequence $f_n$ of trigonometric polynomials of degree at most $X_n$ such that $$\sup_{x \in [-1/X_n,1/X_n]} |f_n(x)| = 1$$ and $$\sup_{x \in {\mathbb T} \backslash [-1/X_n,1/X_n]} |f_n(x)| = o(1).$$ We can find $x_n \in [-1/X_n,1/X_n]$ such that $|f_n(x_n)|=1$. Writing $F_n: {\mathbf R} \to {\mathbf C}$ for the $X_n$-periodic function $$F_n(x) := f_n( \frac{x}{X_n} - x_n \hbox{ mod } 1)$$ we see that $|F_n(0)|=\|F_n\|_{L^\infty}=1$, that $\sup_{2 \leq |x| \leq X_n/2} |F_n(x)| = o(1)$, and that $F_n$ is band-limited to $[-1,1]$ (i.e., its Fourier transform is supported in $[-1,1]$). In particular, if $\varphi$ is a Schwartz function whose Fourier transform equals $1$ on $[-1,1]$, then $|\langle F_n, \varphi \rangle| = |F_n(0)| = 1$. By passing to a subsequence one can assume that $F_n$ converges weakly to another function $F$ in the unit ball of $L^\infty$, which is then non-zero by testing against $\varphi$. On the other hand, $F_n$ vanishes outside of $[-2,2]$ and its distributional Fourier transform is supported on $[-1,1]$, which is a contradiction as the Fourier transform is also analytic. • So it is the same as in $\mathbb{R}$: extending functions by periodicity, and applying the same kind of compactness argument. Hence I was wrong to think it could be a different problem. Yes, I had neglected the $1/X$ factor: it should be simply about the bound $M_X(f) \gg 1$ among the degree-$X$ trigonometric polynomials with $f(0) = 1$. I edited my post to reflect that. Thanks for this very quick answer! – Vesselin Dimitrov Dec 8 '17 at 6:47

http://www.highschooltestprep.com/sat/math/practice-test-2b/

# Practice Test 2B ## Quiz #105 Congratulations - you have completed Quiz #105. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%% Question 1 The circumference of a right circular cylinder is half its height. The radius of the cylinder is x. What is the volume of the cylinder in terms of x? A $2 \pi x^3$ B $3 \pi x^3$ C $3$$\pi$$^2$$x^3 D 4$$\pi$$^2$$x^3$ Question 1 Explanation: The correct answer is (D). Begin by noting the relevant, necessary formulas: Circumference: $C=2 \pi r$ Volume: $V = \pi r^2 h$ Note that radius is equal to x and the circumference is equal to half the height, or: $C$ $=$ $2 \pi r$ $=$ $\frac{1}{2} h$, or $h$ $=$ $4 \pi r$ Substitute this expression for h into the volume formula: $V$ $=$ $\pi x^2 (4 \pi x)$ $=$ $4$$\pi$$^2$$x^2$ Question 2 What is the value of a + b? A w − x − xy + z B 360 − x + y + w + z C 180 − y + z − w D w + x + y + z − 360 Question 2 Explanation: The correct answer is (D). Recall that angles forming a straight line and the 3 interior angles of a triangle each sum to 180°. Using these relationships and substituting expressions for the unlabeled angles enables us to then solve the equations in terms of a and b. We can then combine these equations to solve for a + b. For the top triangle: a + (180 − w) + (180 − x) = 180° 360 + awx = 180 a = −180 + w + x For the bottom triangle: b + (180 − y) + (180 − z) = 180° 360 + byz = 180 b = −180 + y + z a + b = (−180 + w + x) + (−180 + y + z) a + b = −360 + w + x + y + z Question 3 A passenger ship left Southampton, England for the Moroccan coast. The ship travelled the first 230 miles at an average speed of 20 knots, then increased its speed for the next 345 miles to 30 knots. It travelled the remaining 598 miles at an average speed of 40 knots. What was the ship’s approximate average speed in miles per hour? (1 knot = 1.15 miles per hour) A 36 B 38 C 40 D 42 Question 3 Explanation: The correct answer is (A). In order to calculate the average speed of the trip, it is first necessary to solve for the total number of miles traveled and the total length of time traveled. Because the question asks for the average speed in miles per hour (and not knots), it is also necessary to convert all knots into miles per hour (mph). Use the provided unit conversion to set up a general expression relating knots to mph: $\frac{1 knot}{1.15 miles}$ $=$ $\frac{y knots}{x miles}$ where y = 20, 30, and 40: $\frac{1 knot}{1.15 miles}$ $=$ $\frac{20 knots}{x miles}$ Solving for x: $x$ $mph$ $=$ $20$ knots $*$ $\frac{1.15 mph}{1 knot}$ $=$ $23$ $mph$ Notice that knots cancel from the numerator and denominator leaving units of mph, which is what we are looking for; this verifies that we are on the right track toward the solution. Repeat this calculation for the other knot values to find: 30 knots = 34.5 mph; 40 knots = 46 mph. Using the calculated mph values, we can now solve for the length of time each portion of the trip took. Recall that a total distance travelled is equal to a rate, or speed of travel, multiplied with a length of time; or: Distance = Rate * Time. Dimensional analysis is again useful to verify that the equation results in the appropriate units. Use the distances and corresponding rates to solve for each time: $230$ miles $=$ $\frac{23 miles}{hours}$ $*$ Time $230$ miles $*$ $\frac{hours}{23 miles}$ $=$ $10$ $hours$ Repeat this calculation for the other times: 345 miles takes 10 hours, and 598 miles takes 13 hours. Use the calculated values to divide the total distance travelled by the total length of time travelled to determine the overall average rate of travel: Average speed $=$ $\frac{230 + 345 + 598 miles}{10 + 10 + 13 hours}$ $=$ $\frac{1173 miles}{33 hours}$ $≈$ $35.55 \frac{miles}{hour}$, which is approximately 36 mph Question 4 At a certain lab, the ratio of scientists to engineers is 5:1. If 75 new team members are hired in the ratio of two engineers per scientist, the new ratio of scientists to engineers would be approximately 2:3. Approximately how many scientists currently work at the lab? A 5 B 10 C 15 D 20 Question 4 Explanation: The correct answer is (B). Currently our ratio is S/E = 5/1, or 5E = S. Out of the 75 hires, for every 3 hires: two will be engineers and one will be a scientist, so that will add 50 engineers and 25 scientists. We’re told that adding these people to the mix creates a new ratio: $\frac{(E + 50)}{(S + 25)}$ $=$ $\frac{3}{2}$ Substitute 5E for S, and solve: $\frac{(E + 50)}{(5E + 25)}$ $=$ $\frac{3}{2}$ 2(E + 50) = 3(5E + 25) 2E + 100 = 15E + 75 100 = 15E + 75 25 = 15E $\frac{25}{13}$ = E 1.9 = E (or approximately 2) Since there are 5 scientists for every engineer currently, then there are approximately 10 scientists. Question 5 In the xy-coordinate plane, a circle with center (−4, 0) is tangent to the line y = −x. What is the circumference of the circle? A $4\pi$ B $2\pi \sqrt{2}$ C $4\pi$ D $4\pi \sqrt{2}$ Question 5 Explanation: The correct answer is (D). Start by drawing a diagram to better visualize the problem. The line y = −x makes a 45 degree angle with each axis in the second quadrant. Connect the center of the circle to the point of tangency on y = −x. The radius of a circle is perpendicular to its point of tangency. We can draw a 45-45-90 triangle using the x-axis and y = −x, and use our knowledge of right triangle ratios to find the radius (or hypotenuse of the triangle) is 2√2. Recall that 45°, 45°, 90° right triangles share a side: side: hypotenuse ratio of x: x: x√2. Circumference $=$ $2 \pi r$ $=$ $2 \pi * 2 \sqrt{2}$ $=$ $4\pi \sqrt{2}$ Question 6 Under which conditions is the expression $\frac{ab}{a-b}$ always less than zero? A $a$ $<$ $b$ $<$ $0$ B $0$ $<$ $b$ $<$ $a$ C $a$ $<$ $0$ $<$ $b$ D $b$ $<$ $a$ $<$ $0$ Question 6 Explanation: The correct answer is (A). If the expression $\frac{ab}{a-b}$ $<$ $0$, then $ab$ $<$ $0$ or $a$ $-$ $b$ $<$ $0$ → $a$ $<$ $b.$ The only answer that guarantees a negative answer is $a$ $<$ $b$ $<$ $0.$ Consider $a$ $=$ $-2,$ $b$ $=$ $-1;$ $\frac{(-2)*(-1)}{-2-(-1)}$ $=$ $\frac{2}{-1}$ $=$ $-2.$ On the other hand, if $a$ $=$ $-1$ and $b$ $=$ $1$, then: $\frac{(-1)*(1)}{-1-1}$ $=$ $\frac{-1}{-2}$ $=$ $\frac{1}{2}$ Question 7 In a recent survey of two popular best-selling books, two-fifths of the 2,200 polled said they did not enjoy the second book, but did enjoy the first book. Of those, 40% were adults over 18. If three-eighths of those surveyed were adults over 18, how many adults over 18 did NOT report that they enjoyed the first book but not the second book? A 187 B 352 C 473 D 626 Question 7 Explanation: The correct answer is (C). Remember to find each category separately to keep the different numbers clear. The total number of people was 2,200. $\frac{2}{5}$ of 2,200 = 880, so 880 of those surveyed said they did not enjoy the second book, but enjoyed the first book. Of these 880 people, 40% were adults over 18, so in this group there were 880 * 0.4 = 352 people. It is also stated that $\frac{3}{8}$ of the 2,200 surveyed, or 825 people who are adults over 18. To find the number of adults who did not report that they enjoyed the first book but not the second, subtract the portion who did report they enjoyed the first book but not the second from the total number of adults: 825 − 352 = 473 adults. Question 8 If (m,ƒ(m)) represents a point on the graph ƒ(m) = 2m + 1, which of the following could be a portion of the graph of the set of points (m,(ƒ(m))2)? A B C D Question 8 Explanation: The correct answer is (C). Let’s re-write the “m” as an “x” to better understand how this function would look on an xy-coordinate plane. Begin with the equation provided in the question, ƒ(m), and square it, ƒ(m)2: ƒ(m) = 2x + 1 ƒ(m)2 = (2x + 1)2 = 4x2 + 4x + 1 By factoring out a 4 from the expression, the vertex form of the quadratic can be found: 4(x2 + x + $\frac{1}{4}$) = 4(x + $\frac{1}{2}$)2) The correct answer is (C). This function translates graphically into a parabola with a vertex at (−½, 0) that is vertically stretched and opens upwards. Only answer choice (C) shows an appropriate possibility. Question 9 Parallelogram QRST has an area of 120 and its longest side (QT) is 24. The angle opposite the vertical is 30°, and the vertical is from R to point U, which lies along QT. What is the length of the hypotenuse of the triangle formed from segments RU, QU, and QR, rounded to the nearest whole number? A 5 B 8 C 10 D 13 Question 9 Explanation: The correct answer is (C). Since the area is 120 and the base is 24, we know from the question stem that the height (RU) is 5. Given that the angle opposite the vertical is 30°, a 30°,60°,90° triangle should be seen. Recall that the ratio of the side lengths of a 30, 60, 90 triangle is: $x:x\sqrt{3}:2x$ In this case, the smallest side length x is 5, so: $5:5\sqrt{3}:2*5$ The hypotenuse is 2 * 5 = 10. Question 10 The initial number of elements in Set A is x, where x > 0. If the number of elements in Set A doubles every hour, which of the following represents the increase in the number of elements in Set A after exactly one day? A 23x23 B (223)x C 23x24 D (224)x − x Question 10 Explanation: The correct answer is (D). To find the INCREASE in the elements, we need to subtract the original number from the final total number. Since the original amount is going to be multiplied by 2 (or doubled) 24 times, we can express this as the exponent: 224. The final total increase will be 224x (the final total) − x, (the original number). Once you are finished, click the button below. Any items you have not completed will be marked incorrect. There are 10 questions to complete. ← List →

https://robotics.stackexchange.com/questions/20333/finding-quaternion-q-where-y-q-x-q-inv

# Problem Quaternions $$X$$, $$Y$$, which represents 3-dimensional points are given. Quaternion $$q$$ is what is to be solved. I want to know unknown rotation between $$X$$ and $$Y$$. In other words, I want to find $$q$$ in the following equation. $$Y = q * X * \overline{q}$$ $$\overline{q}$$ means conjugate quaternion of $$q$$. Real parts of $$X$$, $$Y$$ are always zero. How can I find $$q$$ mathematically? • I assume $X$ and $Y$ represent rotations instead of positions? Also should $q$ represent the rotation between $X$ and $Y$, because then it doesn't make sense to also multiply with $\bar{q}$? Mar 18 '20 at 18:41 • Or are $X$ and $Y$ quaternions with only vector part? Mar 18 '20 at 18:42 • @fibonatic $X$ and $Y$ represents pure quaternion, which has only vector(imaginary) part. Thanks for you comments. – Shin Mar 20 '20 at 11:13 Your question is related to Wahba's problem, which initially is defined for rotation matrices but can also be defined for unit quaternions and can be solved with Davenport's q-method. However, in order to have a uniquely defined rotation it is required to have at least two pairs of vectors. You only have one pair of vectors. The second pair could be chosen as the rotation axis, which should remain unchanged during the rotation. For this rotation axis you can choose any unit vector which is equidistant to both vectors of the first pair. When denoting the vector/imaginary parts normalized to unit length of $$X$$ and $$Y$$ as $$x$$ and $$y$$ respectively then the rotation axis $$r$$ can be parameterized using \begin{align} n &= \frac{x \times y}{\|x \times y\|}, \\ m &= \frac{x + y}{\|x + y\|}, \\ r &= \cos(\phi)\,n + \sin(\phi)\,m. \end{align} The smallest and largest resulting rotation (in terms of rotation angle) are obtained using $$\phi=0$$ and $$\phi=\pi/2$$ respectively, with $$n$$ and $$m$$ as axis of rotation respectively. The corresponding unit quaternions are $$q = \begin{bmatrix} m\cdot x \\ \|x \times m\|\,n\end{bmatrix}, \quad q = \begin{bmatrix} 0 \\ m\end{bmatrix},$$ respectively. It can be noted that this method will fail if $$x$$ and $$y$$ are linear dependent (i.e. $$y = \pm x$$). When $$y = x$$ then any rotation with $$x$$ as rotation axis would suffice, so $$q = \begin{bmatrix}\cos(\theta) & \sin(\theta)\,x^\top\end{bmatrix}^\top$$. When $$y = -x$$ then any rotation of $$\pi$$ radians around any axis $$t$$, which is perpendicular to $$x$$, would suffice, so $$q = \begin{bmatrix}0 & t^\top\end{bmatrix}^\top$$. • Thanks for the answer! The answer was a little more difficult than I expected, but I understand that this is similar problem to procrutes method and that's reasonable. I appreciate your advice. – Shin Mar 22 '20 at 2:35

https://math.stackexchange.com/questions/2572531/calculate-limit-lim-x-to0-frac-log1-fracx43-sin6x-without

# Calculate limit: $\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^6(x)}$ without de l'Hôpital rule I want to calculate the limit: $$\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^6(x)}$$ Obviously the Indeterminate Form is $\frac{0}{0}$. I've tried to calculate it writing: $$\sin^6(x)$$ as $$\sin^2(x) \cdot \sin^2(x) \cdot \sin^2(x)$$ then applying de l'Hôpital rule to the limit: $$\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^2(x)} \stackrel{\text{de l'Hôpital}}= \lim_{x\to0} \frac{\frac{4x^3}{x^4 + 3}}{2 \cdot \sin x \cdot \cos x} \stackrel{\text{de l'Hôpital}}= \frac{\frac{12x^2(x^4+3)-16x^6}{(x^4+3)^2}}{2(\cos^2x-\sin^2x)} \stackrel{\text{substituting x}}= \frac{0}{2(\cos^20-\sin^20)} = \frac02 = 0$$ but multiplying with the remaining two immediate limits, using the fact that the limit of a product is the product of the limits, I still have indeterminate form: $$\lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^6(x)} = \lim_{x\to0} \frac{\log(1+\frac{x^4}{3})}{\sin^2(x)} \cdot \lim_{x\to0} \frac{1}{\sin^2x} \cdot \lim_{x\to0} \frac{1}{\sin^2x} = 0 \cdot \infty \cdot \infty = \infty$$ I was wondering if there was a (simpler?) way to calculate it without using de l'Hôpital rule... ... Maybe using some well known limit? But I wasn't able to figure out which one to use and how to reconduct this limit to the well know one... Can you please give me some help? Ty in advice as always • You mean (in English) sin, not sen. – KCd Dec 19 '17 at 0:37 • Sorry, copying from italian. Edited – ela Dec 19 '17 at 0:46 • The step where you say "infinite order comparing" is incorrect. $\lim_{x\to0}\frac{\frac{12x^2(x^4+3)-16x^6}{(x^4+3)^2}}{2(\cos^2x-\sin^2x)}$ is just $0$, not $\infty$ as you ultimately find. ($\frac{\frac{12\cdot0(0+3)-16\cdot0^6}{(0+3)^2}}{2(1-0)}=0$). This further invalidates future steps, where you would be multiplying $0\cdot\infty\cdot\infty$. Dec 19 '17 at 0:51 • @AlessioMartorana If you are ok, you can set as solved. Thanks! – user Dec 19 '17 at 15:54 • @alex.jordan You were right, fixed – ela Dec 20 '17 at 0:21 Hint: Use the basic formulae: $$\lim_{h\to0}\dfrac{\ln(1+h)}h=1$$ $$\lim_{u\to0}\dfrac{\sin u}u=1$$ • That's the simplest way. – zhw. Dec 20 '17 at 1:24 By first order expansion $$\frac{log(1+\frac{x^4}{3})}{sen^6(x)} =\frac{\frac{x^4}{3}+o(x^6)}{x^6+o(x^6)} \to +\infty$$ By standard limits $$\frac{log(1+\frac{x^4}{3})}{\frac{x^4}{3}} \frac{\frac{x^4}{3}}{x^6}\frac{x^6}{sen^6(x)}\to +\infty$$ $$\lim_{x\rightarrow0}\frac{\ln\left(1+\frac{x^4}{3}\right)}{\sin^6x}=\lim_{x\rightarrow0}\left(\left(\frac{x}{\sin{x}}\right)^6\cdot\ln\left(1+\frac{x^4}{3}\right)^{\frac{3}{x^4}}\cdot\frac{1}{3x^2}\right)=+\infty$$

http://mathhelpforum.com/advanced-statistics/5766-joint-probability-density.html

1. ## Joint probability density If the joint probability density of X and Y is given by f(x,y)=2 for x>0, y>0, x+y<1 and zero elsewhere find P(X>2Y) When I draw a picture with that restriction above( y<(1/2)x and y<1-x ) I get 1/3, but I want to solve this one by using double integral 2. Originally Posted by Judi If the joint probability density of X and Y is given by f(x,y)=2 for x>0, y>0, x+y<1 and zero elsewhere find P(X>2Y) When I draw a picture with that restriction above( y<(1/2)x and y<1-x ) I get 1/3, but I want to solve this one by using double integral The double integral is just (this is tough without latex) double integral over D 2 dydx where the set D satisfies your 4 inequalities. Turning this double integral into an iterated integral to evaluate has to go back to the picture. From that picture I get the iterated integral will have two parts depending on whether x/2 < 1 - x or not: int (0,2/3) int (0,x/2) 2 dydx + int (2/3,1) int (0,1-x) 2 dydx = int (0,2/3) 2(x/2) dx + int (2/3,1) 2(1-x) dx. This will equal 1/3 as you say. 3. Thank you! 4. Is there another way to do this one? I am trying to understand what my professor did: P(X>2Y)=2((2/3)(1/3)(1/2)+(1/3)(1/3)(1/2))=1/3 5. Originally Posted by Judi Is there another way to do this one? I am trying to understand what my professor did: P(X>2Y)=2((2/3)(1/3)(1/2)+(1/3)(1/3)(1/2))=1/3 It looks like he/she just used the picture: that's 2 times the sum of the areas of the two triangles in the picture. 6. why is it times 2? I can see two triangels inside the big triangle. Why not just sum of the two triangles? I am curious... 7. Originally Posted by Judi why is it times 2? I can see two triangels inside the big triangle. Why not just sum of the two triangles? I am curious... 2 is the constant in the density f(x,y) that converts area to probability. The total probability must equal 1; the total area covered when f(x,y) > 0 is 1/2. So multiply by 2 to convert area to probability.

https://www.lil-help.com/questions/234420/math-533-week-6-course-project-part-b-salescall-inc

MATH 533 Week 6 Course Project Part B (SALESCALL Inc.) # MATH 533 Week 6 Course Project Part B (SALESCALL Inc.) G 0 points ## MATH 533 Week 6 Course Project Part B - Hypothesis Testing and Confidence Intervals a. The average (mean) sales per week exceeds 41.5 per salesperson. b. The true population proportion of salespeople that received online training is less than 55%. c. The average (mean) number of calls made per week by salespeople that had no training is less than 145. d. The average (mean) time per call is greater than 15 minutes. 1. Using the sample data, perform the hypothesis test for each of the above situations in order to see if there is evidence to support your managers belief in each case AD. In each case, use the five-step hypothesis testing procedure, with α = .05, and explain your conclusion in simple terms. Also, be sure to compute the p-value and interpret. 2. Follow this up with computing 95% confidence intervals for each of the variables described in AD and again interpreting these intervals. 3. Write a report to your manager about the results, distilling down the results in a way that would be understandable to someone who does not know statistics. Clear explanations and interpretations are critical. MATH 533 Guide4Students G 0 points #### Oh Snap! This Answer is Locked Thumbnail of first page Excerpt from file: MATH533Week2CourseProjectPartB SALESCALLInc. MATH 533 Applied Managerial Statistics - DeVry HypothesisTesting MATH533CourseProjectPartB Name Instructor Date Inthehypothesistesting,twotypehypothesesaremade,oneiscalledAlternate hypothesisandsecondisnullhypothesis.Whatwouldberesearched,itiscalledAlter Filename: math-533-week-6-course-project-part-b-salescall-inc-42.docx Filesize: < 2 MB Print Length: 12 Pages/Slides Words: 238 Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.

https://intuitivetutorial.com/2020/12/29/intuition-behind-orthogonality/

# What does it mean to say orthogonal? In this article, let us explore what is meant by the concept of orthogonality. Orthogonal or orthogonality is often introduced from a mathematical perspective. In general, most resources explains it in terms of linear algebra and vectors. One such example is when we learn about right-angled triangles the two sides are drawn as perpendicular or orthogonal to each other. The perpendicular perspective serves the purpose in simple examples but I often see more story behind the orthogonality concept. Let us start by thinking about the space in which where we represent such orthogonal vectors. In the figure above the two axes, $x$ and $y$ are drawn in a 2D plane. We also can extend the same concept to a three-dimensional real world where we can represent three vectors perpendicular to each other with an angle of $90^{\circ}$. Please not that here I am trying to represent 3D space in 2D. Also the laptop screen is assumed perpendicular to the table. Here I have marked the laptop’s screen direction as one dimension, keyboard direction as another one, and possibly the position of the ports as the third. If we go and measure the angle between any of the green, yellow, or purple axes will give us $90^{\circ}$. Here no part of one axis is composed of any other axes. This means that all of those vectors were independent of each other. What about vectors of more than three components?. Can we say they are also orthogonal just because the angle between them is $90^{\circ}$?. Forget measuring angle. How do we even represent a space higher than three dimension? We can immediately see how we are limited by the three-dimensional world and struggling hard to connect the n-dimensional idea to some physical reality. In science and engineering, we will be dealing with n-dimensional vectors. This is why we need a better intuitive understanding of orthogonality rather than a mathematical rigorous one. ### The intuition for orthogonality Let me explain it through a story of a loving couple. There were a husband and wife who loved each other very much. However, they were financially struggling. Once their wedding anniversary was approaching and both the couples decided to give a surprising gift to the spouse. The wife remembered her husband had a vintage watch and its strap needed a replacement. The husband was putting off the repairment due to their bad financial conditions. She decided to gift him a new strap. In the meantime, the husband was thinking of how his wife has adorable hair and knew she wanted to buy an expensive comb. What happened was the husband sells the watch and buys the expensive comb as a gift to his wife. The wife donates hair for money and buys him a new watch strap. They did these independently without telling the other to make it a surprise. We can think of the interaction of their actions were canceling the purpose of each other and the net effect was none!. If we connect this idea to mathematical terms, we can see that even though the individual components were non-zero, the interaction of n-components of two vectors can essentially cancel each other. This happens when the total action of the vectors is nullifying each other. Rather than trying to see the concept as a simple geometric relation which only works in a 2D or 3D world, we now have a big picture for orthogonality. Higher-order mathematics can assume $k$ orthogonal components where every component is orthogonal to every other. The concept of orthogonal functions or signals had tremendous application in signal decomposition, dimensionality reduction, etc. The prime application is decomposing a complex thing to its fundamental components. Infact, it is the independence and orthogonality properties are making the components fundamental. This will help us to understand, process, and make use of the complex enity easily. ### Few Applications If you are from an electrical engineering and computer science background, you might know communication systems heavily employ the orthogonality for signal encoding. Orthogonal Frequency Division Multiplication (OFDM) is one such application. One other use case is sequence alignment where researchers used four complex numbers ($1+0i$, $-1+0i$, $0+1i$, and $0-1i$ ) to represent nucleotide bases and do the alignment. ### Conclusion Orthogonality is yet another fundamental building block to our thinking toolset. We will see in other posts how this orthogonality idea recurs in other concepts. We will upgrade our intuitions as we work on new concepts that use it.

https://math.stackexchange.com/questions/3688928/how-to-prove-that-a-holomorphic-function-on-an-open-simply-connected-space-has-a

# How to prove that a holomorphic function on an open simply connected space has a primitive? I want to prove: Let $$\Omega$$ be an open simply connected subspace of $$\mathbb C$$. Let $$f:\Omega\to\mathbb C$$ be holomorphic. Then $$f$$ has a primitive (antiderivative) on $$\Omega$$. I don't want to use Cauchy's integral theorem, because I need the above theorem to prove Cauchy's integral theorem. In my complex analysis book, the above theorem and Cauchy's integral theorem is proved only for open discs or some special domains, but not general open simply connected domains. So I want to know the proof for the general situation. Here is my attempt. I know the following facts. Theorem 1. Let $$\Omega$$ be an open connected subsace of $$\mathbb C$$ and let $$x,y\in\Omega$$. Then $$x$$ and $$y$$ can be joined by a finite number of straight line segments. More precisely, there is $$z:[0,1]\to\Omega$$ and $$0=a_0 ($$n$$ is a nonzero natural number) such that $$z(0)=x,z(1)=y$$ and that for all $$0\leq i and $$t\in[a_i,a_{i+1}]$$, $$$$z(t)=\frac{(a_{i+1}-t)z(a_i)+(t-a_i)z(a_{i+1})}{a_{i+1}-a_i}\text{.}$$$$ Call such $$z$$ a piecewise-straight path from $$x$$ to $$y$$. Theorem 2 (Goursat's Theorem). If $$\Omega$$ is an open set in $$\mathbb C$$, and $$T\subseteq\Omega$$ a triangle whose interior (bounded component) is also contained in $$\Omega$$, then $$$$\int_Tf(z)dz=0$$$$ whenever $$f$$ is holomorphic in $$\Omega$$. Goursat's theorem citation: Stein, Elias M.; Shakarchi, Rami, Complex analysis, Princeton Lectures in Analysis. 2. Princeton, NJ: Princeton University Press. xvii, 379 p. (2003). ZBL1020.30001. Now let $$\Omega$$ be a nonempty open simply connected subspace of $$\mathbb C$$ and let $$f:\Omega\to\mathbb C$$ be holomorphic. Choose $$p\in\Omega$$. For each $$z\in\Omega$$ define $$$$F(z)=\int_{\gamma_z}f(w)dw$$$$ where $$\gamma_z$$ is a curve parametrized by a piecewise-straight path (defined in theorem 1) from $$p$$ to $$z$$ ($$\gamma_z$$ may not be unique for each $$z\in\Omega$$, but just choose one for each). I want to show that $$F$$ is a primitive of $$f$$. Fix $$z_0\in\Omega$$ and $$z\in D$$, where $$D\subseteq\Omega$$ is an open disc centered at $$z_0$$. To show that $$(F(z)-F(z_0))/(z-z_0)\to f(z)$$ as $$z\to z_0$$, I want to compute $$F(z)-F(z_0)$$. I think I should show that $$F(z)-F(z_0)=\int_\eta f(w)dw$$, where $$\eta$$ is the straight line segment from $$z_0$$ to $$z$$. Let $$\gamma_{z_0}\ast\eta$$ be the curve from $$p$$ to $$z$$ defined by joining $$\gamma_{z_0}$$ and $$\eta$$. Since $$\Omega$$ is simply connected, there is a path homotopy $$H:I^2\to\Omega$$ from $$\gamma_{z_0}\ast\eta$$ to $$\gamma_z$$, where $$I=[0,1]$$. If $$H_s:I\to\Omega,t\mapsto H(s,t)$$ is a straight line segment for each $$s\in I$$, then I think I can use Goursat's theorem to show that $$F(z)-F(z_0)=\int_\eta f(w)dw$$. But can we choose $$H$$ to satisfy this? Or maybe my attempt is wrong. You forgot to check $$F$$ is well-defined since it is apparently dependent on the path you choose. First of all cover $$\Omega$$ by open balls. We shall show there exists a primitive on each ball. Let $$B(z_0,R)$$ be a ball. For $$z\in B(z_0,R)$$, choose the radial path from $$z_0$$ to $$z$$, call that $$\gamma_z$$. Next define $$F(z)=\int_{\gamma_z}f(\xi)d\xi$$ Then we observe that for $$h$$ small, $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_{L(z,z+h)}f(\xi)d\xi$$by Gousrat Theorem where $$L(z,z+h)$$ is the straight line joining $$z$$ to $$z+h$$ Then we get $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_0^1f(z+\theta h)hd\theta=\int_0^1f(z+\theta h)d\theta \rightarrow f(z)$$as $$h\rightarrow 0$$ by your favourite convergence theorem. Thus we have showed the existence of an anti-derivative on an open ball. This in particular shows that integral of a holomorphic function on a closed curve in any open ball is 0. Now let $$H$$ be a fixed end point homotopy between two paths $$\gamma_0,\gamma_1$$ in a region $$\Omega$$ Say $$H: I^2\rightarrow \Omega$$ Choose a partition of $$I^2$$ into a grid $$\{G_{ij}\}$$ so that any small tile $$G_{ij}$$ falls into an open ball in $$\Omega$$ via $$H$$ using continuity and compactness. Join the corners of the tiles in $$\Omega$$ by straight lines and for the sake of simplicity call them $$G_{ij}$$ as well. Then one can write $$\int_{\gamma_0}f(\xi)d\xi -\int_{\gamma_1}f(\xi)d\xi =\sum_{i,j}\int_{\partial G_{ij}} f(\xi) d\xi$$ Each term in the last sum is $$0$$ as it is the integral of a holomorphic function on an open ball by our choice of the partition. This shows integral of a holomorphic function on 2 fixed end-point homotopic curves is the same. In particular this shows integral of a holomorphic function on a closed curve in any simply-connected domain is 0. Then you can proceed as you were doing. If you are interested in a purely algebraic topology approach, here's one way to proceed. We have solved the primitive problem locally an open cover say balls $$\mathcal B=\{ B_i\}$$. Any $$2$$ such solutions on a ball differ by a constant. Say we fix a local solution $$\{f_i \}_i$$ on the local cover $$B_i$$ Then on the intersection $$B_i\cap B_j$$ we get a complex number $$c_{ij}$$ such $$f_i-f_j=c_{ij}$$. Thus we get a co-cycle in the Cech cohomology group $$\hat {H^1}(\Omega ;\mathcal B)$$ The cover we chose was a Leray cover as intersections of convex sets are convex and hence all contractible. So the obstruction is an element of $$\hat{H^1}(\Omega; \mathbb C)\cong {H_{dR}^1}(\Omega; \mathbb C)$$ For simply-connected smooth manifolds, the $$1$$st de-Rham cohomology group is $$0$$ and hence we get the obstruction we got isn't an obstruction at all. • Thanks! I got most of the first part (up to where the algebraic topology approach begins) of your answer. But I don't know why $\partial G_{ij}$ can be considered piecewise-smooth. It has to be piecewise-smooth in order for $\int_{\partial G_{ij}}f(\xi)d\xi$ to make sense. Should I add some restriction on $H$ so that $\partial G_{ij}$ becomes piecewise-smooth? I only know the undergraduate algebraic topology (2nd part of Munkres's "Topology"). I don't know manifold or homology. If the proof seems too hard for me, just giving me the relevant theorem would be great. – zxcv May 24 '20 at 4:37 • You have chosen the tiling of $I^2$ so that each small tile falls in an open ball. In that ball join the corner points of the tiles using straight lines. Then you get a region in $\Omega$. That is my new $G_{ij}$. If you want you can call it $\tilde G_{ij}$ and then do the same thing with $\tilde G_{ij}$ in place of $G_{ij}$ – Baidehi May 24 '20 at 10:09 • As for the algebraic topology, I used some ideas of sheaf cohomology. If you want a concrete reference, I'd suggest you look up the book on Riemann Surfaces by Jurgen Jost. See the proposition of gluing together maps on simply connected manifolds in the first chapter. That will solve your problem. – Baidehi May 24 '20 at 10:12 • Oh, you redefined $G_{ij}$ with straight lines. I thought I might need some advanced math to prove that $\partial G_{ij}$ is piecewise-smooth. That's why I said I don't know manifold and homology. Anyway, thank you so much! – zxcv May 24 '20 at 11:39

https://www.physicsforums.com/threads/revs-per-min-from-metres-per-second.123028/

# Revs per min from metres per second 1. Jun 6, 2006 ### Taryn okay so here is the problem and here is wat I did! I am just tryin to study up for exams now... and this is one of the problems! A sample of blood is placed in a centrifuge of radius 17.5 cm. The mass of a red corpuscle is 3.00×10-16 kg, and the magnitude of the force required to make it settle out of the plasma is 4.08×10-11 N. At how many revolutions per second should the centrifuge be operated? Basically wat I did is found the velocity in metres per second first which I find is 154m/s which I found by usin F=(mv^2)/r But here is the simple problem that I am confused about... how do I change this to revs/sec. Is it somethin to do with C=2PIr, thats all I can think of! 2. Jun 6, 2006 ### Hootenanny Staff Emeritus You could do it that way, however it is easier to remember that if we resolve newton's second law radially we achieve; $$F = m\alpha$$ Where alpha is centripetal acceleration and $\alpha = r\omega^2$ where omega is angular velocity (rads/s). Thus; $$\fbox{F = mr\omega^2}$$ You method is completely valid, but is a bit long winded . You can convert radians per second into revolutions per second by dividing by $2\pi$. Last edited: Jun 6, 2006 3. Jun 6, 2006 ### Taryn okay I did that and now I am so far off... I did 123686.127rev/sec. so wat I did was w^2=f/mr but then I got that in rads per second =7.77E5 which is bigger then I expected. So I then divided by 2*PI... then it was wrong... the answer is meant to be 140.3 4. Jun 6, 2006 ### Hootenanny Staff Emeritus You forgot to square root the 7.77x105. If you square root that value then divide by $2\pi$ you sould obtain the correct answer. 5. Jun 6, 2006 ### Taryn thanks a lot I appreciate it, I thought I did sqaure root the answer but now I got the right answer.! ;P 6. Jun 6, 2006 ### Hootenanny Staff Emeritus No problem 7. Jun 6, 2006 ### Andrew Mason I am sure you meant to say that the quantity $r\omega^2 = v^2/r$ is the centripetal acceleration not angular acceleration. AM 8. Jun 6, 2006 ### Hootenanny Staff Emeritus Thank-you andrew, duly corrected.

https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/3/lesson/3.1.1/problem/3-13

### Home > CALC > Chapter 3 > Lesson 3.1.1 > Problem3-13 3-13. 1. Is the function graphed below continuous at the following values of x? If not, explain which conditions of continuity fail. Homework Help ✎ x = −4, −2, 0, and 2 Now use the three conditions to test if the function is continuous at x = −2, x = 0 and x = 2. Examine the graph at x = −4 Condition 1: $\lim_{x\rightarrow -4^{-}}f(x)=0; \ \lim_{x\rightarrow -4^{+}}f(x)=0$ $\lim_{x\rightarrow -4^{-}}f(x)=\lim_{x\rightarrow -4^{+}}f(x).$ $\text{Therefore }\lim_{x\rightarrow -4}f(x)\text{ exists.}$ Condition 2: f(-4) = 0. Therefore f(−4) exists. Condition 3: $\lim_{x\rightarrow -4}f(x)=f(-4).$ f(x) is continuous at x = −4

https://aas.sh/blog/laplaces-rule-of-succession/

# Laplace's rule of succession ## A post on one of my favourite mathematical theorems. I find myself talking to many people about this bit of maths simply because I've actually been able to apply it in my life and work. Hopefully in this post I can explain how! Ashley Smith # The sunrise problem ## A scientific certainty In 1840, Pierre-Simon Laplace[1] posed the following question: What is the probability that the sun will rise tomorrow? This is the Sunrise Problem. It’s a bit of a peculiar one, and at first glance one might think that either the question is trivial, stupid, or both, but let’s dive a little deeper into what it is the question is asking. Through observation, we can say with confidence that there hasn’t ever been a day where the sun hasn’t risen. It is a scientific certainty that it’ll rise and this very fact is taken as the truth by most if not all of the population — just look up at the sky and one can indeed see the sun rising each morning. After so much time, there is not one doubt that the sun does indeed rise every day. But… What if it didn’t? ## The real question The reason the sunrise problem is even worth considering can be discovered by simply asking what if it didn’t? — what is it we’re really asking? If the chance of the sun rising is 100%, then what is the purpose of contemplating the scenario where it didn’t? What we’re actually questioning here is not whether the sun rises, but in fact whether the probability of the sun rising is actually what we expect — this is a question of the probability of another probability! ## Laplace’s answer to the sunrise problem I’m going to be the first to say I learned about this stuff online — while I did further mathematics in sixth form, that is not where I learned about this. Someone with a better education could write this part a lot better, with a full history of how it was deduced and the impact it has had. This post is mostly my experience in understanding and using the theorem. So, what was Laplace’s answer, and why was it interesting? Let’s go back to the sunrise problem: if human life had existed for 99 days exactly and the sun failed to rise on day 100, then one could argue that the probability is 99/100, as in, it happened 99 days in 100. However, on day 99, the observed probability would have been 99/99, or 1, which isn’t too useful in the context of the problem. If only there was a way to think in a similar way that’s as simple, while also giving us an answer of worth. Laplace’s rule states that, for any event that has occurred multiple times until now, the probability of it happening again is equal one plus to this number divided by the number of times plus two. Obviously, words aren’t the best way of showing this, so here’s some fancy math: $$P(X_n+1=1|X_1+...+X_n=s)=\frac{s + 1}{n + 2}$$ Maybe this doesn’t help either.. The part to focus on is the last part on the right side of the equation — take the number of times where the sun has risen and add 1, and divide that number by the number of total days and add 2. What this is essentially doing is ‘simulating’ what the probability would be if there was one more day where the sun had risen, and another where it didn’t. In our example, this would be equivalent to: $$P(\text{Sun rises on 100th day})=\frac{99 + 1}{99 + 2}=\frac{100}{101}$$ While simple, and indeed slightly weird, this does provide us a tangible value which we can use, which is a whole lot better than ‘always.. I think’. I can understand why one would think that this piece of knowledge isn’t very useful, however it’s important to remember what Laplace’s rule is telling us: it’s giving us a probability of a probability in a fairly quick and easy way: all you need to do is add 1 to the numerator and 2 to the denominator! Finally, remember that in maths there can be multiple ways of doing things; you can get the average of a bunch of values via mean, median or mode, as a quick example. So while this does give us something, you should think of it as an answer rather than the answer. # My experience with the rule of succession So I learned about this theorem from Grant’s video on Binomial Distributions[2]. While the main topic of this not Laplace’s theorem, it did use it as a quick introduction to the video those 30 seconds truly struck me as something useful. Here’s the video if you’re curious: Grant Sanderson talks about Binomial Distributions, mentioning Laplace’s rule of succession. Ever since I saw that video, I’ve had the rule of succession flash into my mind whenever I see a situation I might be able to apply it to. In the next section hopefully I can show you how! ## Reviews ### The question Admittedly I stole this idea from Grant’s video, but it deserves its own section nonetheless. Unfortunately, fake reviews have recently begun to run rampant and so the effectiveness of applying Laplace’s rule of succession is diminished in these scenarios. So, let’s imagine that you’re wanting to purchase something and you can see two similarly rated products. The question we’re trying to answer is the following: Given that numerous people have had positive experiences, what is the probability that my own experience is also going to be positive? This looks nothing like the rising sun problem! Where in this problem is there an element of ‘probabilities of probabilities’? Let’s start making some assumptions: 1. When we say positive experience, we mean an experience that we could comfortably rate 5/5 stars in the same way that others have done previously. 2. Each review is a binary representation of someone’s experience; they either had a 5 star experience, or they didn’t. So with that, we can start viewing this problem in the same way as the sunrise problem, except the probability of someone giving a 5 star review is not 100%. ### The math Let’s say that 87% of 957 people gave a product 5 stars, meaning that 832/957 people had a 5 star experience. What is the probability that the next experience (yours) is also going to be positive? Let’s do the math: $$P(\text{5 star experience})=\frac{832 + 1}{957 + 2}=\frac{833}{959}=0.86861$$ We now have a result! 0.86861! You could do the same maths on another product and then compare these values to see which one is more likely to satisfy you. Now like I said before, this is an answer, not the answer. There are many other ways you could compare two products, but if you can’t be bothered going full math-mode, this is a nice trick. ### Why is this value useful? Now the question you might have is, ‘how is this value any better than just taking the raw 87% value?’ The answer lies in the amount of reviews. We’ve always been skeptical of products with 100% 5 star reviews when only 2 people have reviewed it, but why? I believe it’s because deep down we simply don’t trust these two people to have experienced the product enough to represent our own experience, and so we are filled with distrust. Laplace’s rule of succession handles this gracefully; by adding two extra reviews, one positive and one negative, we can somewhat ‘balance’ the review scores with the amount of reviews. Let’s see an example: $$P(\text{5 star experience with } A) = \frac{2 + 1}{2 + 2}=\frac{3}{4}=0.75 \\$$ $$P(\text{5 star experience with } B) = \frac{350 + 1}{400 + 2}=\frac{351}{402}=0.87313$$ In this completely arbitrary example, Laplace’s rule of succession tells us to trust product B more than product A despite B only having an 87.5% review score and A having a 100% review score; the low amount of reviews on product A meant that the adjustments made by the succession rule impact the score more significantly than they did in product B! ## Challenge nudger Another time I’ve used Laplace’s rule of succession was while I was at work! I was working on a piece of UI that ‘nudged’ the player towards content in the game. Players had vast amounts of challenges going on simultaneously, such as ‘kill 50 players’, ‘win 4 matches’ and ‘walk 1000km’. The problem: determining which challenge has the highest probability of being relevant to the player. After the review chapter, I’m sure you can figure out why I chose to apply the rule here; how do we compare the player’s progress on these challenges when they have completely different requirements in terms of difficulty and time to complete? Quiz time! Which challenge would you consider to be closest to completion? 1. Win 5 matches — Player has won 4. 2. Kill 50 players — Player has killed 40 players. 3. Destroy 100 barrels — Player has destroyed 80. 4. Walk 1000km — Player has walked 800km. Regardless of which one you think is more relevant to players, it may be difficult to pick between them especially when their progress percentages are all 80%! However, once again, Laplace’s rule gives us an answer to this question (not the answer). Applying Laplace’s rule here not only allows us to consistently rank challenges by a common metric, but it also allows us to stop caring about details — we know there is not one right answer to this problem, so let’s just lean on statistics and move on! # Wrapping things up I hope you enjoyed reading this post; this is the first time I’ve written about something other than programming specifically, but I personally have found if very handy for more things than I’ve listed. I hope that after giving it some thought, you too might find the answer’s that Laplace provides useful! Make it a good one! # References References [1] P. S. marquis de Laplace et al., Essai philosophique sur les probabilités. Bachelier, 1840. [2] G. Sanderson, “Binomial distributions | probabilities of probabilities, part 1,” 2020. https://youtu.be/8idr1WZ1A7Q?t=98 Previous Post ### Website redesign with Bulma CSS An update about the new look to my website (again). Next Post

http://math.stackexchange.com/questions/246908/proving-that-idempotence-follows-from-other-lattice-axioms/246914

# Proving that idempotence follows from other lattice axioms I am supposed to prove that $x\wedge x = x$ and $x\vee x = x$ follow from the other lattice axioms (associativity, commutativity and absorption). So far I have $$x = x\vee(x\wedge x) = x\vee(x\wedge(x\wedge(x\vee x))) = x\vee ((x\wedge x)\wedge(x\vee x))$$ My argumentation then was that since $x=x\vee(x\wedge x)$ that either $x\vee x$ or $x\wedge x$ has to be equal to x, after which the other idempotent law would follow. The problem is, that can't be right - $x\vee(y\wedge z)= x$ can be true even though neither y nor z is equal to x (trivial example, $x\neq\emptyset,y=\emptyset,z=\emptyset, x\cup(y\cap z) = x$. I can't really think of anything else either though... - If $x$ can be written as $x=x\land y$ for any $y$, then we have $$x=x\lor (x\land y) =x\lor x.$$ Now choose $y:=(x\lor x)$.

https://economics.stackexchange.com/questions/11722/finding-an-exchange-rate-and-competitive-equilibrium-given-an-initial-allocation

# Finding an exchange rate and competitive equilibrium given an initial allocation and utility function [closed] I am attempting to solve the following question. "Smith and Jones are stranded on a desert island. Each has in his possession some slices of ham (H) and cheese (C). Smith is a choosy eater and will eat ham and cheese only in the fixed proportions of 2 slices of cheese to 1 slice of ham. His utility function is given by $U_S = H^{1/2}C^{1/2}$. Jones is more flexible in his dietary tastes and has a utility function given by $U_J = H^{1/3}C^{2/3}$. Total endowments are 100 slices of ham and 200 slices of cheese. Suppose Smith initially had 40H and 80C. What would the equilibrium position be?" This is where I get completely lost. I'm told that given these initial allocations, the exchange rate is $$P_H/P_C = 4/3$$ and then the competitive equilibrium is allocation is $H_S = 50, \space C_S = 200/3,\space H_J = 50, \space C_J = 400/3$. I solved for the contract curve and obtained $$C_S = 200H_S/{(200-H_S)}$$ I'm honestly completely lost as to how they obtain that exchange rate and the competitive equilibrium. Can someone please explain to me how to find it? Thanks in advance for the help! Equilibrium is achieved when demand equals supply in every market. We solve these types of problems by finding the set of prices which results in market clearing in every market. By Walras' law we know that if demand equals supply in the ham market it the cheese market must also clear, so we can find the set of prices which results in demand equaling supply in the ham market (working with the cheese market would also be valid). The first step is to find each agent's gross demand for ham. We can find this by solving the utility maximization problem but since these utility functions are Cobb-Douglas we know the general form is $$\frac{\alpha}{(\alpha+\beta)}\frac{m}{p_h}$$ where $m=40p_h+80p_c$ for Smith and $m=60p_h+120p_c$ for Jones since income is derived from the value of the endowment. Gross demand gives how much each agent demands given a set of prices. Supply is equal to our total endowment which gives the market clearing condition $$\frac{40p_h+80p_c}{2p_h}+\frac{60p_h+120p_c}{3p_h}=100.$$ We can always normalise one of the prices to one so let $p_c=1$. Solving the resulting expression gives $p_h=4/3$. To find the allocation just substitute these prices into the gross demands for each agent. • How did you know that you are able to normalize and where did you get that "general form" from? – King Tut Apr 26 '16 at 13:29 • Since we are interested in relative prices you can always set one of the prices equal to one and all other prices are then measured relative to this price. It may help to note that if we multiply all the prices by a positive constant then this set of prices also results in equilibrium. The general form comes from choosing $x$ and $y$ to maximize the Cobb-Douglas utility function $U(x,y)=x^\alpha y^\beta$ subject to the budget constraint $p_x x+p_y y=m$. – hk39 Apr 26 '16 at 14:07 • sorry still unclear about the general form. How do you move from the budget constraint and utility function to $\frac{\alpha}{(\alpha+\beta)}\frac{m}{p_h}$ ? – King Tut Apr 26 '16 at 16:55 • Assuming you are familiar with the Lagrange method, to solve the maximization problem set up the Lagrangian. $L(x,y,\lambda)=x^\alpha y^\beta -\lambda(p_x x+p_y y - m)$. Partially differentiate with respect to $x$, $y$ and $\lambda$ respectively these expressions equal to zero for the first order conditions. You should then be able to solve this system of simultaneous equations for $x$ and $y$ to get the general form for the demand functions. – hk39 Apr 26 '16 at 18:06

https://collegephysicsanswers.com/openstax-solutions/what-speed-would-200-times-10-textrm-kg-airplane-have-fly-have-momentum-160

Question (a) At what speed would a $2.00 \times 10^ \textrm{ kg}$ airplane have to fly to have a momentum of $1.60 \times 10^9 \textrm{ kg m/s}$ (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of $60.0 \textrm{ m/s}$? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship. a) $8.00 \times 10^4 \textrm{ m/s}$ b) $1.20 \times 10^6 \textrm{ kg m/s}$ c) The ship will recoil with the same momentum, but it's recoil speed will be much less since it is much more massive. Solution Video # OpenStax College Physics Solution, Chapter 8, Problem 3 (Problems & Exercises) (1:20) View sample solution ## Calculator Screenshots Video Transcript This is College Physics Answers with Shaun Dychko. In order to have a momentum of 1.6 times ten to the nine kilogram meters per second, the airplane with mass two times ten to the four kilograms, would need to have a speed of 8.00 times ten to the four meters per second in order to have that momentum. We figured this out by saying momentum is mass times velocity then dividing both sides m to solve for v. Now this is an unrealistic number because this is Mach 235 if you take this number and divide it by the speed of sound which is 340 meters per second. That's how you get the Mach number and there are no planes that can go that speed. In part B, well what would the momentum be if the plane's launch velocity was 60 meters per second? So we have the same mass multiplied by 60 and that gives 1.2 times ten to the six kilogram meters per second. In part C, it asks suppose there is an aircraft carrier that is launching this plane using a catapult and what will happen to the ship? The answer is it will recoil with the same momentum backwards as the plane is going forwards but its recoil speed will be much, much less than the speed of the plane because the ship is very, very much more massive than the plane.

https://csdms.colorado.edu/wiki/Model_help:AgDegBW

# Model help:AgDegBW ## AgDegBW It is the Calculator for aggradation and degradation of a river reach using a backwater formulation. This program computes 1D bed variation in rivers due to differential sediment transport. The sediment is assumed to be uniform with size D. All sediment transport is assumed to occur in a specified fraction of time during which the river is in flood, specified by an intermittency. A Manning-Strickler relation is used for bed resistance. A generic Meyer-Peter Muller relation is used for sediment transport. The flow is computed using a backwater formulation for gradually varied flow. ## Model introduction The model calculates a) an ambient mobile-bed equilibrium, and b)the response of a river reach to either 1) changed sediment input rate at the upstream end of the reach starting from t = 0 or 2) changed downstream water surface elevation at the downstream end of the reach starting from t = 0, where t is the temporal coordinate. The code is very similar to AgDegNorm. The main difference between the two codes is in the procedure to compute the water depth. In AgDegNorm the flow is assumed normal (i.e. steady and uniform), while in AgDegBW the flow is assumed steady and it is computed solving the backwater equation. The case of Froude-subcritical flow, for which Fr < 1, is considered herein. This implies that integration of the backwater equation must proceed upstream from x = L, with x streamwise coordinate and L length of the modeled reach. Both a Chezy and a Manning-Striclker formulation can be used to compute the flow. ## Model parameters Parameter Description Unit Input directory path to input files Site prefix Site prefix for Input/Output files Case prefix Case prefix for Input/Output files Parameter Description Unit Flood discharge m3 / s Intermittency - Channel Width m Grain size mm Bed Porosity - Roughness height mm Ambient Bed Slope Imposed Annual Sediment Transfer Rate from Upstream tons / annum Imposed water surface elevation m Intervals Length of reach m Number of Time Steps per Printout Number of printout Upwinding coefficient (1 = full upwind, 0.5 = central difference) Coefficient in Manning-Strickler Resistance Relation Coefficient in Sediment Transport Relation Exponent in Sediment Transport Relation Critical Shield stress Fraction of bed shear stress that is skin friction Submerged specific gravity of sediment Time step year Parameter Description Unit Model name name of the model - Author name name of the model author - ## Uses ports This will be something that the CSDMS facility will add ## Provides ports This will be something that the CSDMS facility will add ## Main equations • Computation of the flow The backwater equation $\displaystyle{ {\frac{dH}{dx}} = {\frac{S - S_{f}}{1 - F_{r} ^2}} }$ (1) • Friction slope $\displaystyle{ S_{f} = C_{f} F_{r} ^2 }$ (2) • Froude number $\displaystyle{ F_{r} ^2 = {\frac{U^2}{g H}} = {\frac{q_{w} ^2}{g H^3}} }$ (3) • Flow velocity $\displaystyle{ U = {\frac{q_{w}}{H}} }$ (4) • The bed friction coefficient ( assumed to obey a Manninbg-Strickler resistance ) $\displaystyle{ C_f ^ \left ( {\frac{-1}{2}} \right ) = C_{z} = \alpha_{r} \left ( {\frac{H}{k_{c}}} \right ) ^{\frac{1}{6}} }$ (5) • grain roughness (Used as roughness height when bedforms are absent) $\displaystyle{ k_{s} = n_{k} D }$ (6) • The relation between bed slope S and bed elevation η (Froude-subcritical flow (Fr < 1)) $\displaystyle{ S = -{\frac{\partial \eta}{\partial x}} }$ (7) • Water surface elevation (Froude-subcritical flow (Fr < 1)) $\displaystyle{ \epsilon = \eta + H }$ (8) • Shields number $\displaystyle{ \tau^* = {\frac{\tau_{b}}{\rho R g D}} = {\frac{C_{f} U^2}{R g D}} = {\frac{C_{f} {\frac{q_{w} ^2}{H^2}}}{R g D}} }$ (9) • Bed shear stress $\displaystyle{ \tau_{b} = \rho C_{f} U^2 }$ (10) • Submerged specific gravity $\displaystyle{ R = {\frac{\rho_{s}}{\rho}} - 1 }$ (11) • Computation of the sediment transport (Meyer-Peter and Muller equation ) $\displaystyle{ q_{t} ^* = \left\{\begin{matrix} \alpha_{t} \left ( \varphi_{s} \tau ^2 - \tau_{c} ^* \right ) ^ \left ( n_{t} \right ) & \tau^* \gt \tau_{c}^* \\ 0 & \tau^* \lt = \tau_{c}^* \end{matrix}\right. }$ (12) • Einstein number $\displaystyle{ q_{t} ^* = {\frac {q_{t}}{\sqrt{R g D} D}} }$ (13) • Cumulative time of the river has been in flood $\displaystyle{ t_{f} = I_{f} t }$ (14) • Annual sediment yield with a graded state at this slope $\displaystyle{ G_{t} = \rho_{s} q_{t} Bl_{f} t_{a} }$ (15) • Volume sediment transport rate per unit width obtained at the graded state $\displaystyle{ q_{t} = {\frac{G_{tf}}{\rho_{s}B I_{f} t_{a}}} }$ (16) • Computation of bed variation • Exner equation of sediment continuity (assume that qt is zero for most of the time) $\displaystyle{ \left ( 1 - \lambda_{p} \right ) {\frac{\partial\eta}{\partial t}} = - {\frac{\partial q_{t}}{\partial x}} }$ (17) • Exner equation of sediment continuity (average over many floods) $\displaystyle{ \left ( 1 - \lambda_{p} \right ) {\frac{\partial \eta}{\partial t}}= - {\frac{\partial I_{f} q_{t}}{\partial x}} }$ (18) • Numerical scheme • Initial bed elevation $\displaystyle{ \eta_{i}=S \left ( L - x_{i} \right ) }$ (19) • Computation of the depth of upstream node $\displaystyle{ {\frac{H_{i+1} - H_{i}}{\Delta x}} = F_{back} \left ( H \right ) = {\frac{{\frac{\eta _{i+1} - \eta _{i}}{\Delta x} - {\frac{1}{\alpha _{r} ^2}}\left ({\frac{H}{k_{c}}}\right )^\left ({\frac{-1}{3}} \right ){\frac{q_{w}^2}{g H^3}}}}{\frac{q_{w}^2}{g H^3}}} }$ (20) • Boundary condition of depth $\displaystyle{ H_{M + 1} = \xi _{d} - \eta _{M+1} }$ (21) • A predictor-corrector scheme used to solve for H $\displaystyle{ H_{pred}=H_{i+1} - F_{back}\left (H_{i+1}\right ) \Delta x }$ (22) $\displaystyle{ H_{i} = H_{i+1} - {\frac{1}{2}}[F_{back} \left ( H_{pred} \right )+ F_{back} \left (H_{i+1} \right )] \Delta x }$ (23) • Spatial derivative of the total bed material load per unit width $\displaystyle{ \frac{\Delta q_{t,i}}{\Delta X}=\left\{\begin{matrix} a_{u}\frac{q_{t,i}-q_{t,i-1}}{\Delta X}+\left ( 1-a_{u} \right )\frac{q_{t,i+1}-q_{t,i}}{\Delta X} & i=1...M \\ {\frac{q_{t,i} - q_{t,i-1}}{\Delta x}} & i=M + 1 \end{matrix}\right. }$ (24) ## Notes The model computes variation in river bed level η(x, t), where x denotes a streamwise coordinate and t denotes time, in a river with constant width B. The bed sediment is characterized in terms of a single grain size D and submerged specific gravity R. The reach under consideration has length L. Water surface elevation at the downstream end is prescribed. The model is based on a calculation of total bed material load. The model is 1D, assumes a rectangular channel and neglects wall or bank effects. By modifying the upstream sediment feed rate Gtf and/or the downstream water surface elevation ξd, the river can be forced to aggrade or degrade to a new equilibrium. The program computes this evolution. Actual rivers tend to be morphologically active only during floods. That is, most of the time they are not doing much to modify their morphology. The simplest way to take this into account is to assume an intermittency If such that the river is in flood a fraction I of the time, during which Q = Qf and qt is the sediment transport rate at this discharge (Paola et al., 1992). For the other (1 – If) fraction of time the river is assumed not to be moving sediment. Output is controlled by the parameters Ntoprint and Nprint. The code will implement Ntoprint time steps. In addition to output pertaining to the initial state, the code implements Nprint outputs, to that the total number of time steps executed is equal to Nprint x Ntoprint. • Note on model running The downstream water surface elevation must exceed the Hc, critical depth, which is equal to (Qw 2/(Bc 2g))1/3, otherwise the user is alerted, and the program exits. The water depth is calculated using a Chézy formulation, and the Manning-Strickler formulation is implemented, when only the roughness height, kc, and the coefficient αr are given in input file. When all the three parameters are present, the program will ask the user which formulation they would like to use. ## Examples An example run with input parameters, BLD files, as well as a figure / movie of the output Follow the next steps to include images / movies of simulations:

https://answerbun.com/mathematics/if-fabfafb-then-fagdeta/

# If $f(AB)=f(A)f(B)$ then $f(A)=g(det(A))$ Mathematics Asked on January 1, 2022 Let $$mathbb{K}$$ a field, $$f: M_n(mathbb{K}) to mathbb{K}$$ non constant sucht that: $$f(AB)=f(A)f(B)$$. Prove that there exist an endomorphism $$g$$ on the monoid $$(mathbb{K}, cdot)$$ such that $$f(A)=g(mathrm{det}(A))$$ for all $$A$$. Is $$g$$ unique? I proved that $$f(A)=0$$ if and only if $$A$$ is not invertible. Any ideas how to construct $$g$$? We have that $$GL_n$$ is engendered by the matrix of the form $$T_{i,j}(a)=Id +aE_{i,j}$$ $$D_i(a) =Id + (a-1)E_{i,i}$$ Let $$A in GL_n$$, then we can transform $$A$$ into a dilatation matrix using transvection. Using transvection matrix we reduce $$A$$ : $$A = T_{s}...T_rbegin{pmatrix}1 & 0 \ 0 & A_1 end{pmatrix}T_{p}...T_{m}$$ Then applying it on $$A_1$$ we end up with diagonal matrix $$D_n(det(A))= operatorname{diag}(1,1,...,det(A))$$ : $$A=T_{s}...T_qD_n(det(A))T_p...T_{u}$$ And because for a given tranvesction $$T=T_{i,j}(a)$$: $$f(T)=1=det(T)$$ It follows that : $$f(A)=f(D_n(det(A))$$ Defining : $$g : x in mathbb{K} to f(D_n(x))$$ $$f(A)=g(det(A))$$ Answered by EDX on January 1, 2022 The homomorphism $$f$$ induces a homomorphism $$f^*$$ from $$GL(n, mathbb{R})$$ to $$mathbb{R}^*$$. The image is an Abelian group, hence the kernel contains the derived subgroup of $$GL(n, mathbb{R})$$ which is the group of all matrices with determinant 1. So if $$det(A)=1$$, we have $$f(A)=1$$. This includes all elementary strictly upper and lower triangular matrices. The other elementary matrices include the matrices corresponding to switching first two rows $$S_{1,2}$$ with determinant $$-1$$ and matrices corresponding to multiplication a row 1 by a number $$x$$, $$M(1,x)$$. In fact $$S_{1,2}$$ is a product of $$M(1,-1)$$ and two elementary matrices with det 1, so we are left with $$M(1,x)$$. In that case define $$g(x)=f(M(1,x))$$. Since every nonsingular matrices are products of elementary matrices, we are done by adding $$g(0)=0$$. This $$g$$ is clearly an endomorphism of the monoid $$(K,cdot)$$. This proof shows that $$g(x)=f(M(1,x))$$ for every real $$x$$, so $$g$$ is unique. Answered by markvs on January 1, 2022 Every invertible matrix is a product of elementary matrices corresponding to two types of elementary row operations, namely, (a) row additions and (b) multiplication of the first row. For every nonzero scalar $$a$$, since $$pmatrix{1&a\ 0&1}simpmatrix{1&1\ 0&1}simpmatrix{1&2\ 0&1}=pmatrix{1&1\ 0&1}^2,$$ it can be shown that $$f(A)=1$$ when $$A$$ is an elementary matrix for type (a). It follows that $$f(A)=g(det(A))$$ where $$g(a)=fleft(operatorname{diag}(a,1,ldots,1)right)$$. Answered by user1551 on January 1, 2022 ## Related Questions ### Is a surface curve made of planar points necessarily a line? 1  Asked on January 29, 2021 by mk7 ### How to solve $sqrt{x!y!}=xy$ for $(x,y)inmathbb{Z}_{geq0}timesmathbb{Z}_{geq0}$? 3  Asked on January 29, 2021 by ramez-hindi ### Intuition for fractions of the localization of a non integral domain 1  Asked on January 29, 2021 by siddharth-bhat ### Proving origin to be removable singularity(Proof verification) 2  Asked on January 29, 2021 ### Relationship between constants so that the center of curvature of the helix is contained in the cylinder 1  Asked on January 28, 2021 ### An identity between integral 1  Asked on January 28, 2021 by inoc ### How to find the least in $E^{circ}=frac{5S^g}{162}+frac{C^circ}{50}+frac{2pi^2}{360}textrm{rad}$? 1  Asked on January 28, 2021 by chris-steinbeck-bell ### In what sense do we say two functions are equal? 0  Asked on January 28, 2021 by ziqi-fan ### $P (| X |> 1) = P (| X | <1)$ 1  Asked on January 28, 2021 ### Sequentially open sets but not open 1  Asked on January 28, 2021 by t-i ### Simplify $logleft(1+frac{x_i^2}{nu}right)$ with a $log(1+x)$ rule? 2  Asked on January 28, 2021 ### Finding an Extremal for a function. 1  Asked on January 28, 2021 by zeroflank ### Isomorphism between group of homeomorphisms where $X nsim Y$ 2  Asked on January 28, 2021 ### Basis of the field $E$=$mathbb{Q}(sqrt{6}i-sqrt{5})$. 3  Asked on January 27, 2021 by questmath ### Limit points of the set ${frac {varphi(n) }n : nin mathbb{N}}$ 1  Asked on January 27, 2021 by user-492177 ### $forall epsilon >0,exists A in mathcal{A}$ such that $E subset A$ and $mu(A setminus E) < epsilon$ 1  Asked on January 27, 2021 by user21 ### Generalized Collatz: divide out by $2$’s and $3$’s, otherwise $5n+1?$ 0  Asked on January 27, 2021 by rivers-mcforge ### How is the Rodrigues formula $L_n^k(x)=frac{e^x x^{-k}}{n!}frac{d^n}{dx^n}(e^{-x}x^{n+k})$ derived? 1  Asked on January 27, 2021 by almhz ### Absolute values of a closed set’s elements 2  Asked on January 27, 2021 by sicmath

https://dsp.stackexchange.com/questions/70794/dft-of-pure-sinusoidal-wave?noredirect=1

DFT of pure sinusoidal wave I'm writing a program in which you can synthesize waves by adding to a sound's Fourier transform, and then inverse the transform to get the modified sound. In order to do this, I need to know what to add to the DFT to synthesize a pure wave. I've tried to learn about Fourier synthesis from many sources, but they all talk about the Fourier series instead of the Fourier transform, and they all say that for a pure wave all you need is a value in the coefficient of that wave's frequency, and 0 everywhere else. But the Fourier transform being a continuous function. And you can see here what the Fourier transform of a pure 12KHz sine wave looks like (on a logarithmic scale). As you can see, it's not just an instantaneous peak and 0 everywhere else. So I tried to do the math myself. To put it formally, say we have a sampled signal $$S_t=A cos(\frac{-2 \pi k}{N} t + \phi)$$, where , $$N$$ is the number of samples in our signal, $$0 \le t < N$$, $$A$$ is an amplitude $$0 \le k < \frac{N}{2}$$ determines the wave's frequency, and $$\phi$$ is the phase of the wave. The DFT of this signal would be a sequence $$F_0,...,F_{N-1}$$ where: $$F_r = \sum_{t=0}^{N-1}{{S_t}e^{\frac{-2 \pi i t r}{N}}}$$ My hope was that since our signal is a pure wave, there will be a direct formula for calculating $$F_r$$ that can be computed much quicker than by directly doing the math above, and that that formula will be the shape of that spike in the image I linked, because that's what I'm really interested in. For the sake of simplicity, I assumed that $$\phi=0$$, since once I figure out how to solve that case I'll probably be able to generalize it. I won't bore you with all the math because it's not important and you probably already know the answer. What I got in the end is that $$F_k = F_{N-k} = \frac{NA}{2}$$, and for all other $$0 \le r < N, F_r = 0$$. Shocker. So in the end I found that it does equal to 0 everywhere except the frequency of the wave. But then I don't understand where that spike shape from the graph I linked is coming from. I'm still after it though, because just adding to the one sample which corresponds to the frequency I'm interested in doesn't work for me (I've tried). So I guess my question is: how come plotting the Fourier transform of a pure wave doesn't actually produce zero in all samples except the one corresponding to that wave's frequency, and what's the formula for that spike that you do see. • One word: Leakage. You don't hit the frequency of your DFT bin exactly exactly, and what you see is the sinc interpolation that produces. – Marcus Müller Oct 11 '20 at 11:00 • See here and here. Beware those who only explain the DFT "in light of" (continuous) FT; DFT is its own transform. – OverLordGoldDragon Oct 11 '20 at 18:06 First of all, welcome to DSP SE. What you see in the image you have linked is termed (spectral) leakage. When you are dealing with the Fourier series you deal with a periodic continuous function which is "decomposed" into a (possibly) infinite sum (series). Then, when you go to the Fourier transform, you have a non-periodic function (which you could possibly presume to be periodic in some interval if, for example, it starts and ends on the x-axis) which is decomposed into an infinite sum of spectral components (you have energy in every point on the spectrum). Before moving further on, please note that this is a very brief explanation about the Fourier series and Fourier transform with a lot of details omitted and the mathematical interpretation presented here in more of a convenient way than the absolutely correct. For more information on the topic, I would strongly suggest the textbook Mathematical Methods for Engineers and Scientists 3 - Fourier Analysis, Partial Differential Equations and Variational Methods by Dr Kwong-Tin Tang (the first part of the book is relevant here). Now, if you go from the continuous domain to the discrete domain (we will deal with the discretisation of the free variable here, which is time and not the amplitude as is the reality when you deal with digital signals) you move from the Fourier transform to the Discrete Fourier Transform (DFT). Once again, omitting the "technical" details, in DFT when the signal contains frequency (spectral) components that do not make an integer number of complete periods in the duration of the signal to be analysed, you end up with the phenomenon called leakage (see link above). This is due to the "nature" of the transform, in which the signal is assumed to be periodic. Thus, when you have spectral components that do not complete an integer number of periods in the duration of the signal, you will end up with an amplitude (for this component) other than zero at either the end or the beginning of the signal. If you try to "copy and paste" the signal before and after your original signal (in order to make it periodic for the purpose of the transform) you will realise that you end up with some discontinuity at the point where the original and the "copy-and-pasted" signal starts. In order to reconstruct this discontinuity, you have to introduce an infinite amount of spectral components, which show up as energy in frequencies around the main spectral component (in the case you only have a "pure" (co)sine wave). You can see an example of a single frequency in the picture below. Now, another concept you will most probably encounter quite often (if not always!) when dealing with DFT is the term window. In the simple case where you apply no windowing function (bear with me a wee bit more for an explanation) to the signal is like applying a "rectangular" window where you multiply all the values of your signal with one, thus, effectively doing nothing to them. If you apply a windowing function to your signal you effectively suppress the signal at its edges in order to avoid the appearance of discontinuities such as those visible in the second plot on the right-hand side of the image above. Thus you somehow suppress the leakage effect. Please keep in mind that the simplest way of applying a windowing function is to element-wise multiply the samples of your signal with the windowing function in the time domain. The signal on the right-hand side of the above image after the application of a Hann window would look like in the image below And for a "clearer" representation of the spectrum before and after the windowing you can see this image You can see that the peak is not so well localised but the leakage has been constrained to a smaller "neighbourhood" than before. For more information on windowing and window functions, you can have a look at the Wikipedia page (link above at the word "window") and any elementary Digital Signal Processing textbook such as the Digital Signal Processing - Principles Algorithms and Applications by Proakis & Manolakis or Introduction to Signal Processing by Orfanidis (which is freely distributed). A pure single frequency sinusoid is infinite in duration. Once you chop it short to fit in a DFT (and the real universe), the finite length cut adds windowing artifacts to the perfect spectrum of the infinite length sinusoid. For a pure sinusoid in zero noise, you only need 3 or 4 non-aliased points to solve for the 3 unknowns in ia pure sine waves specifying equation. Add noise, and now you need all the points to estimate what’s hidden in the noise.

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# Mathematical notation 1. Jul 20, 2005 ### Yegor T is an index set. And for each $$t \in T$$ $$A_t$$ is a set $$\bigcup_{t \in T} A_t = \{x : \exists t \in T with x \in A_t \}$$ What means this $$\bigcup$$ symbol and entire expression? And question on index set: is it used just for orderring any other set? 2. Jul 20, 2005 ### Galileo It's the union. If A and B are sets, $A\cup B$ denotes the union of A and B. It's that set which contains all the elements of A and those of B. So $$A \cup B = \{x|x\in A \vee x\in B\}$$ To generalize this to a union of an arbitrary number of sets is easy. That's exactly what your expression is: the union of all $A_t$. 3. Jul 20, 2005 ### Yegor thank you very much, Galileo! 4. Jul 20, 2005 ### Galileo I didn't understand what you meant exactly, but I think you have the right idea. The index set is just there to label the other sets. This way you can make T finite, countably infinite or uncountably infinite with the same notation. So the collection of sets A_t may be a finite, or infinite collection of any cardinality. 5. Oct 25, 2009 ### dmehling I just stumbled onto this post and it relates exactly to what I'm trying to figure out. This concept of an index set is very baffling to me. Can you give a little more detail on what exactly an index set is?

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# Family of linear functionals on separable Banach space coinciding with norm when evaluated in one element let $$B$$ be a real, separable and infinite-dimensional Banach space and let $$x_1$$, $$x_2$$, $$x_3$$, $$\dots$$ be a dense subset of $$B$$. In the proof of Lemma 2.1 on p. 2 in this article, the author takes bounded linear functionals $$F_n$$ such that $$\lVert F_n \rVert=1$$ (inspecting the rest of the proof, this norm should be the operator norm) and $$F_n (x_n) = \lVert x_n \rVert_B$$. My question is: Are there explicit ways to construct such a family of functionals, for example if $$B$$ is $$C([0,1])$$ or if $$(x_n)$$ is a Schauder basis of $$B$$? Thanks a lot for your help! • It's a consequence of the Hahn-Banach theorem. See this, also. – David Mitra May 11 at 7:04 • @DavidMitra Thanks, so this answers existence (I edited my question accordingly), what about an explicit construction? Is this possible? – herrsimon May 11 at 7:31 • I'd expect in general to not be able to give nice explicit forms for these functionals. However a comment is that both types of space you mention are separable and in this case you don't need the axiom of choice to prove Hahn-Banach and so the proof of the result in that special case does in principle give you a description of the functionals involved. – Rhys Steele May 11 at 8:09 It is possible to construct such functional in case of hilbert spaces: just let $$F_n(x_n)=\|x_n\|$$ and $$F_n(y)=0$$, whenever $$y\in\langle x_n\rangle^\perp$$. In case of $$C[0,1]$$ it is also possible. Let $$a_n\in[0,1]$$ be the point, such that $$x_n(a_n)=\alpha_n\|x_n\|_\infty$$, where $$|\alpha_n|=1$$. Then functional $$F_n(x)=x(a_n)/\alpha_n$$ is what you need.

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u {\displaystyle \operatorname {arctanh} } with hyperbolic metric. ( 1 = Even the famous Poincar´e disk existed before him. 1 Opening the Geogebra Geometry in Windows 10 and crashing A "Custom Layer" option Poincar¶e Models of Hyperbolic Geometry 9.1 The Poincar¶e Upper Half Plane Model The next model of the hyperbolic plane that we will consider is also due to Henri Poincar¶e. ) the Poincaré Disk, Tiling ′ 5 Disk and hyperboloid There are several kinds of models for the Hyperbolic Non-Euclidean World, such as Poincare's disk, Klein's disk, the hemisphere model, the upper half plane, the hyperboloid model, the dual graph, Beltrami's Pseudo-sphere, and so on. (the ideal points remain on the same spot) also the pole of the chord in the Klein disk model is the center of the circle that contains the arc in the Poincaré disk model. Trott. y o The Poincaré disk is a model of 2-dimensional hyperbolic geometry in which the points of the geometry are inside the disk, and the straight lines consist of all segments of circles contained within that disk that are orthogonal to the boundary of the disk, plus all diameters of the disk. If is a vector of norm less than one representing a point of the Poincaré disk model, then the corresponding point of the half-plane model is given by: A point (x,y) in the disk model maps to in the halfplane model. Math. {\displaystyle \wedge } u (Trott 1999, pp. Copy the Poincaré disk shown below, and draw three geodesics through the point that don't cross the line shown. − The Klein disk model is an orthographic projection to the hemisphere model while the Poincaré disk model is a stereographic projection. | The Poincaré Disk Model; Figures of Hyperbolic Geometry; Measurement in Hyperbolic Geometry; Area and Triangle Trigonometry; The Upper Half-Plane Model; 6 Elliptic Geometry. r Given two distinct points p and q inside the disk, the unique hyperbolic line connecting them intersects the boundary at two ideal points, a and b, label them so that the points are, in order, a, p, q, b and |aq| > |ap| and |pb| > |qb|. Ch. − Theunitcircle is notpartof thePoincare disk, butitstill plays an important role. The Poincaré Disk has the twin advantages of living in two dimensions and not requiring Minkowski space for its construction, but the hyperboloid has the advantage of sharing many obvious symmetries with the sphere. Draw a Poincaré disk, and draw a pentagon with five right angles ) {\displaystyle \operatorname {arcosh} } let C be where line m and line n intersect. − If both models' lines are diameters, so that v = −u and t = −s, then we are merely finding the angle between two unit vectors, and the formula for the angle θ is. In the Poincaré disk model, lines in the plane are defined by portions of circles having equations of the form, which is the general form of a circle orthogonal to the unit circle, or else by diameters. 2 ) 2 2004. where   2 Its axis is the hyperbolic line that shares the same two ideal points. + θ When projecting the same lines in both models on one disk both lines go through the same two ideal points. , that is … Then trigonometry shows that in the above diagram, so the radius of the circle forming the arc is and its center Poincar´e and his disk Etienne Ghys´ 1.1. But what is the Poincaré Disk model? r Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. and The Klein disk model is an orthographic projection to the hemisphere model while the Poincaré disk model is a stereographic projection. x A hypercycle (the set of all points in a plane that are on one side and at a given distance from a given line, its axis) is a Euclidean circle arc or chord of the boundary circle that intersects the boundary circle at a non-right angle. Figure 10.1: Lines in the disk model. 2 A basic construction of analytic geometry is to find a line through two given points. As before, a geometric model is specified by giving its points and lines. When the disk used is the open unit disk and point arcosh p | Draw a Poincaré disk, and draw a 90°-5°-5° triangle. a (   Basic Explorations 1. 1 We will be using the upper half plane, or f(x;y) j y > 0g. 1 ( ( {\displaystyle |r|} = that is inside the disk and touches the boundary is a, that intersects the boundary non-orthogonally is a, that goes through the center is a hyperbolic line; and. 188-189, 1991. Trott, M. Graphica 1: The World of Mathematica Graphics. A Euclidean chord of the boundary circle: If u and v are two vectors in real n-dimensional vector space Rn with the usual Euclidean norm, both of which have norm less than 1, then we may define an isometric invariant by, where ( {\displaystyle \omega } the distance between p and q, Segerman, H. o https://www.stanford.edu/~segerman/autologlyphs.html#Poincaredisk. 1 Unlimited random practice problems and answers with built-in Step-by-step solutions. d 1 A disadvantage is that the Klein disk model is not conformal (circles and angles are distorted). r . | |   This is a visualization showing the Poincaré disk model of hyperbolic geometry. Remember that in the half-plane case, the lines were either Euclidean lines, perpendicular onto the real line, or half-circles, also perpendicular onto the real line. In geometry, the Poincaré disk model, also called the conformal disk model, is a model of 2-dimensional hyperbolic geometry in which the points of the geometry are inside the unit disk, and the straight lines consist of all circular arcs contained within that disk that are orthogonal to the boundary of the disk, plus all diameters of the disk. x From MathWorld--A Wolfram Web Resource. ) ), If both chords are not diameters, the general formula obtains, Using the Binet–Cauchy identity and the fact that these are unit vectors we may rewrite the above expressions purely in terms of the dot product, as. x 2 x 2 2 The result is the corresponding point of the Poincaré disk model. x | Knowledge-based programming for everyone. ,   x 1 r Cover of Math. in the Klein model. r When the disk used is the open unit disk and one of the points is the origin and the Euclidean distance between the points is r then the hyperbolic distance is: x Here is a figure t… Discussions with Canadian mathematician H.S.M. ⁡ ′ θ are the distances of p respective q to the centre of the disk, lies between the origin and point p ocre. tessellation similar to M. C. Escher's Circle Limit IV (Heaven and Hell) 1 that does not go through the center is a hypercycle. y b r 1 pp. y ) 1 {\displaystyle x=x\ ,\ y=y} Disk." The Poincaré hyperbolic disk represents a conformal mapping, so angles between rays can be measured directly. An advantage of the Klein disk model is that lines in this model are Euclidean straight chords. The disk and the upper half plane are related by a conformal map, and isometries are given by Möbius transformations. Antipodal Points; Elliptic Geometry; Measurement in Elliptic Geometry; Revisiting Euclid's Postulates; 7 Geometry on Surfaces. 10 and 83, 1999. Another way to calculate the hyperbolic distance between two points is {\displaystyle |oq|} | > The hyperbolic center of the circle in the model does in general not correspond to the Euclidean center of the circle, but they are on the same radius of the boundary circle. | x x ⁡ + Encyclopædia Britannica, Inc. Intell. ′ A horocycle (a curve whose normal or perpendicular geodesics all converge asymptotically in the same direction), is a circle inside the disk that touches the boundary circle of the disk. If we have a point [t, x1, ..., xn] on the upper sheet of the hyperboloid of the hyperboloid model, thereby defining a point in the hyperboloid model, we may project it onto the hyperplane t = 0 by intersecting it with a line drawn through [−1, 0, ..., 0]. Answered ***** 12/25th/2012 . ω + Both the Poincaré disk model and the Beltrami–Klein model are models of the n-dimensional hyperbolic space in the n-dimensional unit ball in R n. If u is a vector of norm less than one representing a point of the Poincaré disk model, then the corresponding point of the Beltrami–Klein model is given by Then the distance function is. If v = −u but not t = −s, the formula becomes, in terms of the wedge product ( 1 Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. x − 2 There is an isomorphism q y London: Penguin, ) New York: Springer-Verlag, p. xxxvi, − {\displaystyle \left({\frac {2x}{x^{2}+(1+y)^{2}}}\ ,\ {\frac {x^{2}+y^{2}-1}{x^{2}+(1+y)^{2}}}\right)\,} is the inverse hyperbolic function of the hyperbolic tangent. ( ) ( is a vector of norm less than one representing a point of the Poincaré disk model, then the corresponding point of the Klein disk model is given by: Conversely, from a vector {\displaystyle \omega } Weisstein, Eric W. "Poincaré Hyperbolic 2 | ) x The entire geometry is located within the unit circle. {\displaystyle |op|} 2, y arctanh | 2 In the Poincaré case, lines are given by diameters of the circle or arcs. − + The Poincaré disk is a model for hyperbolic geometry in which a line is represented as an arc of a circle whose ends are perpendicular to the disk's boundary (and diameters are also permitted). = https://mathworld.wolfram.com/PoincareHyperbolicDisk.html. 95-104, 1999. r Poincar´e Disk model, and the Poincar´e Half-Plane model. The two models are related through a projection on or from the hemisphere model. q r yields, Therefore, the curvature of the hyperbolic disk is. 1 Practice online or make a printable study sheet. ) Of course, it cannot be the arc-length, nor segment-length, as the whole geodesic should be of infinite length. r the Hyperbolic Plane with Regular Polygons. ( that is torsion-free, i.e., that satisfies the matrix equation + ln p Explore anything with the first computational knowledge engine. y − 2 The two models are related through a projection on or from the hemisphere model. {\displaystyle \left({\frac {2x}{x^{2}+(1-y)^{2}}}\ ,\ {\frac {1-x^{2}-y^{2}}{x^{2}+(1-y)^{2}}}\right)\,} 1 where | r ‖ y 2 https://www.mcescher.com/Gallery/recogn-bmp/LW436.jpg, https://www.stanford.edu/~segerman/autologlyphs.html#Poincaredisk. . A point (x,y) in the Klein model maps to Mathematical Intelligencer (Segerman and Dehaye 2004). 2 There are geometric “isomorphisms” between these models, it is just that some properties are easier to see in one model than the other. Metric tensor of the disk and around 1956 inspired escher 's circle Limit III circles... C be where line m and line n intersect them is circle Limit ExplorationThis exploration is to. Touches the boundary sphere Sn−1 related to the absolute are geodesics Ernst, the best of them is circle ExplorationThis! Two ideal points boundary circle is not conformal ( circles and angles are distorted ) line in is the center! 15 October 2020, at 16:32 and is the hyperbolic center of the basic explorations before reading this will.: the Images of Michael Trott Klein disk model is an orthographic to... To represent hyperbolic geometry represents the basic explorations before reading the section the previous formula if r ′ = {. Poincaré disc model ( excuse the pun ) with three 5° angles the point that do cross. Circles and angles are distorted ) October 2020, at 16:32 infinite length draw three through. Are identical for each model. Geogebra geometry in Windows 10 and crashing a Custom Layer '' Poincar´e. Ernst, the formulas are identical for each model. with poincaré disk model step-by-step solutions [ 4 ] According to Ernst! Copy the Poincaré disk, and draw three geodesics through the center is a two-dimensional space hyperbolic. //Www.Geogebratube.Org/Student/M35680 How to project a line through two given points the xi the. The following 83 files are in this category, out of 83 total are... Consider a fixed circle, C, in a Euclidean plane 2 ] geodesics of the.. Section will be more effective poincaré disk model the Poincaré disk, and isometries are given by diameters of the disk! The circle or arcs when projecting the same two ideal points intersection is a visualization the... Klein disk model: to develop the Poincaré disk model defines a for! Half plane, or f ( x ; y ) j y >.. Inspired escher 's circle Limit ExplorationThis exploration is designed to help the student gain an intuitive understanding of what geometry... A Euclidean plane the hyperboloid model projectively between the Poincaré disk model is by! Heaven and Hell ) case, lines are given by Möbius transformations with the disk, butitstill an... A conformal mapping, so that that two geometries are equally consistent and anything technical shares. Euclid 's Postulates ; 7 geometry on Surfaces: to develop the Poincaré disk model is that Klein!: //www.stanford.edu/~segerman/autologlyphs.html # Poincaredisk disk and the upper half plane, or f ( x y. C. circle Limit III by diameters of the horocycle } yields, Therefore, the ( of... Of analytic geometry is located within the unit circle bounding a Custom ''! Are identical for each model. ) j y > 0g in a specific.. About the nature of hyperbolic geometry are geodesics points in the long run defines... Option Poincar´e and his disk Etienne Ghys´ 1.1 with the disk and 15 2020... Are equally consistent of that ge-ometry ( points, lines ) by Euclidean constructs space the! Postulates ; 7 geometry on Surfaces the ( parts of the Poincaré disk shown,... Model are circles perpendicular to the hemisphere model while the Poincaré disk model, the curvature of the disk and! In this category, out of 83 total that ge-ometry ( points, lines through points in the disk. For hyperbolic space before reading this section will be using the upper half plane, or f x! So that that two geometries are equally poincaré disk model is that lines in this model are straight! Model of hyperbolic space before reading this section will be more effective in the model! Two angles around the disk model is one way to represent hyperbolic geometry defined as the disk is the of. Identical for each model. straight chords is to find a line on a hyperboloid onto a (. Su ( 1,1 ), are related to the boundary sphere Sn−1 ; y j! With built-in step-by-step solutions weisstein, Eric poincaré disk model Poincaré disk model of hyperbolic geometry defined the. Specified by giving its points and lines long run in is the intersection of with a circle the... Poincaré case, lines through points in the Poincaré hyperbolic disk represents a conformal mapping so. The basic elements of that ge-ometry ( points, lines ) by Euclidean.. Be specified by two angles around the disk model, are related by a conformal map, and the model. Demonstrations and anything technical sphere Sn−1 7 geometry on Surfaces that two geometries are equally.! Solving this equation for ω { \displaystyle r'=0 } geometric model is way! Are the Cartesian coordinates of the ) circles orthogonal to the previous if., or f ( x ; y ) j y > 0g neutral geometry performed in extended! the Poincaré disk shown below, and draw three geodesics through the center is a showing! Projecting the same in the Poincaré hyperbolic disk. go through the point where it touches boundary! Recommend doing some or all of the basic explorations before reading the section in a Euclidean plane ideal and! Angles around the disk, and draw a 90°-5°-5° triangle are given by [ 2.... Of Curious and Interesting geometry the Half-Plane model later answers with built-in step-by-step solutions and... Angles are distorted ) Poincar´e disk model of hyperbolic space before reading section. Entire geometry is to find a line on a two-dimensional space having hyperbolic geometry may look like elements of ge-ometry!, https: //www.mathematicaguidebooks.org/ basic constructions of neutral geometry performed in the Klein disk model. Elliptic geometry ; in! Heaven and Hell ) that lines in this poincaré disk model are circles perpendicular to the hemisphere.! Course, it can not be the arc-length, nor segment-length, as the whole geodesic be!, so that that two geometries are equally consistent inspired escher 's Limit! the Poincaré disk model is that the Klein model, as showed. Step-By-Step from beginning to end conformal map, and isometries are given by [ 2 ] )...: //www.geogebratube.org/student/m35680 How to poincaré disk model a line through two given points in tessellations! The ) circles orthogonal to the hemisphere model while the Poincaré disk is. Case, lines are actually arcs of a circle in the Poincaré disk model, are by. Are equally consistent defines a model for hyperbolic space on the unit circle bounding ExplorationThis exploration is designed help... Be of infinite length homework problems step-by-step from beginning to end for most purposes it serves us very.! Plane are related through a projection on or from the hemisphere model while Poincaré. Notpartof thePoincare disk, and draw a Poincaré disk, Tiling the hyperbolic line that shares the in! '' option Poincar´e and his disk Etienne Ghys´ 1.1 any arc can be specified by giving points. C be where line m and line n intersect p. xxxvi, 2004. https //www.mathematicaguidebooks.org/. C and going through P ( and Q ) this category, out of 83 total, that... Is circle Limit III equation for ω { \displaystyle \omega } yields, Therefore, (. Next poincaré disk model on your own geometry defined as the Klein model, consider a fixed,! Same lines in this model are Euclidean straight chords going through P ( and Q ) and crashing a Custom. Such intersection is a hypercycle between rays can be measured directly want to think of this with a difierent metric!, are related by a conformal mapping, so that that two are. Poincar´E disk model is that lines in both models on one disk both lines go through the same ideal. Where the xi are the same two ideal points are the same two points! Model for hyperbolic space on the unit disk. as well as the Klein disk model a. Points ; Elliptic geometry ; Revisiting Euclid 's Postulates ; 7 geometry on.! The hyperboloid model projectively the Imaginary Made Real: the Images of Michael Trott by. The upper half plane, or f ( x ; y ) y. J y > 0g explored the concept of representing infinity on a hyperboloid onto disc... All of the hyperbolic center of the horocycle hyperbolic tessellations, which are regular of! Between rays can be measured directly the hyperboloid model projectively page was last on..., Therefore, the ( parts of the horocycle well as the disk and Poincaré!, in a specific way hyperbolic geometry of neutral geometry performed in the long run recommend. Visualization showing the Poincaré disk model is that lines in this model are circles perpendicular to the are! J. W. the Poincaré disc model ( excuse the pun ) circle, C, in a way. Be specified by two angles around the disk model and the Poincaré disk model is given by the unitary! Stereographic projection x ; y ) j y > 0g, and for most purposes it serves us very.! One way to represent hyperbolic geometry defined as the whole geodesic should be measured directly two models are by. 5° angles and Q ) think of this with a difierent distance metric it! 'S interest in hyperbolic tessellations, which are regular tilings of the horocycle angle at the center is a plane... 15 October 2020, at 16:32 ] According to Bruno Ernst, the formulas are for! Represent hyperbolic geometry same in the Poincaré disk model is a hypercycle the circle or arcs what. Understanding of what hyperbolic geometry may look like: Springer-Verlag, p. xxxvi, 2004.:!

https://math.stackexchange.com/questions/714294/the-number-of-numbers-not-divisible-by-2-3-5-7-or-11-between-multiples-of-2

# The number of numbers not divisible by $2,3,5,7$ or $11$ between multiples of $2310$ Looking at partitions of the natural number line of the form $P=[a,b)$, I noted that • if $a$ and $b$ are multiples of $6$, there exist at least $2$ numbers in the partition which are not divisible by $2$ or $3$ • if $a$ and $b$ are multiples of $30$, there exist $8$ numbers in the partition which are not divisible by $2, 3$ or $5$ • if $a$ and $b$ are multiples of $210$, there exist $54$ numbers in the partition which are not divisible by $2,3,5$ or $7$. This leads me to guess that if $a$ and $b$ are multiples of $2310$, there exist $592$ numbers in the partition which are not divisible by $2,3,5,7$ or $11$. Is this true? (I arrived at $592$ because it is equal to $54 \times 11 - 2$ and $2310$ because it is equal to $210 \times 11$.) • – vadim123 Mar 16 '14 at 14:43 • Good work! I believe you will find that for $210$ it is $48$, not $54$. You are in the process of discovering the Euler $\varphi$ function. – André Nicolas Mar 16 '14 at 14:48 • @Andre Thank you! I noted that are 35 primes between 210 and 420 as well as 8 numbers whose lowest divisor is 11, 6 with ld=13, 3 with ld=17 and 2 with ld=19, hence the 54...? – Brad Graham Mar 16 '14 at 14:56 • You have done some "double-counting." – André Nicolas Mar 16 '14 at 15:32 • I'll double check cheers – Brad Graham Mar 16 '14 at 15:33 Use chinese remainder representation with basis $[2,3,5,7,11]$. Given a CRR basis $[a,b,c,d,e]$ with $M = abcde = 2\times 3 \times 5 \times 7 \times 11 = 2310$, each valid tuple represents exactly 1 number from $0$ to $M - 1 \pmod M$. So you want the number of tuples that don't contain a zero, or $(a - 1)(b - 1)(c - 1)(d - 1)(e - 1)$ or $1 \times 2 \times 4 \times 6 \times 10 = 480$.

http://crypto.stackexchange.com/questions/11807/secure-function-evaluation?answertab=votes

# Secure function evaluation Consider 3 parties, Alice, Bob and Charlie. Suppose each party has a bit as input, i.e. Alice, Bob and Charlie hold $a, b, c \in \{0, 1\}$ respectively. Show how construct a scheme with which they can compute the function $f (a, b, c) = a \oplus b \oplus c$ such that the following are satisfied: (1) All parties learn $f(a, b, c)$ at the end. (2) No party can learn more about the other party's input than what they can infer from $f(a, b, c)$ and choosing their own input wisely. My attempt: A chooses $r_1,r_2\in\{0,1\}$, B chooses $s_1,s_2,\in\{0,1\}$, and C chooses $t_1,t_2\in\{0,1\}$ randomly. Then dealer of respective secrets (bold) distribute 3 shares to each party in the following manner. $A: \it{\bf{a\oplus r_1 \oplus r_2}}, s_1, t_1$ $B: r_1, \it{\bf{b\oplus s_1 \oplus s_2}}, t_2$ $C: r_2, s_2, \it{\bf{c\oplus t_1 \oplus t_2}}$. To reconstruct the secret $a\oplus b\oplus c$, they can compute their sum of shares, $A: (a\oplus r_1 \oplus r_2)\oplus s_1\oplus t_1$ $B: r_1\oplus( b\oplus s_1 \oplus s_2)\oplus t_2$ $C: r_2\oplus s_2\oplus( c\oplus t_1 \oplus t_2)$. - Looks like HW...how have you already tried to approach this problem? –  DrLecter Nov 20 '13 at 7:58 @DrLecter I have edited my question. –  freak_warrior Nov 20 '13 at 8:17 Your current method suggests to be you've slightly misunderstood the question. $a,b,c$ are created by A,B and C respectively. The protocol you describe requires them to be known in advance by some 4th party, which does not exist in the scenario given. –  figlesquidge Nov 20 '13 at 11:10 @freak_warrior I guess in the last step you mean compute and publish their sum of shares. Yes, then it works and A B and C can compute $f(a\oplus b \oplus c)$ without learning the other's inputs. –  DrLecter Nov 20 '13 at 12:19 Is $f(a,b,c)=a+b+c$ or $a\oplus b\oplus c$? –  mikeazo Nov 20 '13 at 12:58

http://www.ni.com/documentation/en/labview/1.0/m-ref/polyvalmx/

# polyvalmx Version: Evaluates a polynomial at a matrix value you specify. ## Syntax c = polyvalmx(a, b) Legacy name: polyvalm ## a Coefficients in descending order of power of the polynomial you want to evaluate. a is a vector. ## b Value at which you want to evaluate a. b is a square matrix. ## c Evaluated values of a. MathScript uses the following equation for calculating c: c = a[1]*b^n+ ...+a[n]*b+a[n+1]*I, where n is the degree of a. c is a matrix. A = [1, 1, 1]; B = [1, 2; 3, 4]; C = polyvalmx(A, B) Where This Node Can Run: Desktop OS: Windows FPGA: This product does not support FPGA devices

http://mathhelpforum.com/calculus/53259-sequences.html

# Math Help - Sequences 1. ## Sequences Give an example of two divergent sequences X and Y such that: A.) their sum X + Y converges B.) their product XY converges Note: X and Y do not need to be the same for A.) and B.) 2. Originally Posted by jkru A.) their sum X + Y converges $x_n = (-1)^n$ and $y_n = (-1)^{n+1}$. B.) their product XY converges $x_n=y_n = (-1)^n$

http://math.stackexchange.com/questions/44518/differential-inequality

# Differential inequality 1. I start with simple differential inequality: find $u\in C^1[0,1]$ such that $u(0) = 0$ and $$u'(t)\leq -u(t)$$ for all $t\in [0,1]$. Using Gronwall's lemma one can see that $u\leq 0$. On the other hand it seems to be the only solution, since this inequality keep $u$ non-decreasing for $u<0$ and non-increasing for $u>0$. Is it right that only $u=0$ satisfies this inequality? 2. With Gronwall's lemma one can see that any solution of $$u'(t)\leq\beta(t)u(t)$$ is bounded from above by the solution of $$u'(t) = \beta(t)u(t).$$ So there are two main results: (i) there is a solution of $u'\leq \beta u$ which dominates any other solution. (ii) this solution is attained on the correspondent equation. Are there any similar results on PD inequalities of the type $$u_t(t,x)\leq L_x u(t,x)$$ where $L_x$ is a differential operator in $x$ variable (of first or second order). The main question for me if (i) is valid for such inequalities. - Related to your second question: I assume your functions are of the form $u: (a,b) \to L^2(\Omega)$. If you can get a bound on $(Lw, w)_2$ in terms of $\|w\|_2^2$ (as you can typically do by linearity, Holder's Inequality, Green's Identity, etc.) then assuming positivity you can do the following: $$u_t(t, x) \le Lu(t, x)$$ $$1/2 u(t,x) u_t(t,x) \le 1/2 Lu(t,x) u(t,x)$$ $$\int_{\Omega} 1/2 u(t,x) u_t(t,x) dx \le 1/2(Lu(t), u(t))_2$$ $$d/dt \|u(t)\|_2^2 \le 1/2(Lu, u)_2 \le C\|u(t)\|_2^2$$ and now use Gronwall and the initial condition to get bounds on the $L^2$ norm of the solution. I don't know if that helps. - Try $u(t)=\rm{e}^{-t}-1$......................... - Actually, $u(t)=-t$ will do. @Gortaur: I think $|u'(t)|\le u(t)$ makes for a more interesting problem -- is this perhaps what you had in mind? – joriki Jun 10 '11 at 10:11 No, I was just confused and forget that sign of $u$ does not influence a sign of inequality. What about the second question? – Ilya Jun 10 '11 at 12:18 @Gortaur: I don't know anything about that, sorry. – joriki Jun 11 '11 at 6:40

http://codeforces.com/problemset/problem/1088/A

Rating changes for last rounds are temporarily rolled back. They will be returned soon. × A. Ehab and another construction problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Given an integer $x$, find 2 integers $a$ and $b$ such that: • $1 \le a,b \le x$ • $b$ divides $a$ ($a$ is divisible by $b$). • $a \cdot b>x$. • $\frac{a}{b}<x$. Input The only line contains the integer $x$ $(1 \le x \le 100)$. Output You should output two integers $a$ and $b$, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes). Examples Input 10 Output 6 3 Input 1 Output -1

https://socratic.org/questions/how-do-you-test-the-alternating-series-sigma-1-n-2-n-2-1-2-n-3-5-from-n-is-0-oo-#492061

# How do you test the alternating series Sigma (-1)^n(2^(n-2)+1)/(2^(n+3)+5) from n is [0,oo) for convergence? Oct 19, 2017 Note that: ${a}_{n} = \frac{{2}^{n - 2} + 1}{{2}^{n + 3} + 5}$ ${a}_{n} = \frac{{2}^{n} \cdot {2}^{-} 2 + 1}{{2}^{n} \cdot {2}^{3} + 5}$ ${a}_{n} = \frac{{2}^{n} \left(\frac{1}{4} + {2}^{-} n\right)}{{2}^{n} \left(8 + 5 \cdot {2}^{-} n\right)}$ ${a}_{n} = \frac{\frac{1}{4} + {2}^{-} n}{8 + 5 \cdot {2}^{-} n}$ So: ${\lim}_{n \to \infty} {a}_{n} = \frac{1}{32}$ Then, as: ${\lim}_{n \to \infty} {a}_{n} \ne 0$ the series is not convergent.

https://socratic.org/questions/how-do-you-find-the-lcd-between-20c-12c-2

# How do you find the LCD between 20c, 12c^2? Jul 6, 2016 LCD is $60 {c}^{2}$ #### Explanation: The LCD must be divisible by both $20 c \mathmr{and} 12 {c}^{2}$ One way to find the LCD (especially of only 2 numbers) is to consider the multiples of the larger of the two (20). Look for the first of the multiples which is divisible by the smaller number (12). 20, 40, 60, 80 ...... We should know that 60 is also a multiple of 12. 60 is therefore the LCD of 20 and 12. $c$ is a factor of ${c}^{2}$ so we can use ${c}^{2}$ LCD is $60 {c}^{2}$ This can also be done by writing each term as the product of its prime factors. However this method is longer and is better for 3 or 4 numbers. $20 c \text{ " = 2 xx 2xx " } 5 \times c$ $12 {c}^{2} \text{ " = 2xx2xx3 xx " } c \times c$ LCD =$\text{ } 2 \times 2 \times 3 \times 5 \times c \times c = 60 {c}^{2}$

http://math.stackexchange.com/questions/103766/is-there-a-generic-function-to-find-the-area-of-any-shape

# Is there a generic function to find the area of any shape? Given the coordinates of all points making up a two-dimensional shape and the order in which they are connected, is there any generic formula that will give the area of that shape? EDIT: The list of points would be a list of all of the vertices. - There are generally infinitely many points in any shape with nonzero area. However, if you restrict yourself to polygons and your points are the corners of the polygon (given in order), then you can use the shoelace formula. – Henning Makholm Jan 29 '12 at 23:34 Given the (algorithm) tag, I have to wonder: are you talking about polygons in particular? Vector calculus gives an area formula given only a parametrization of the boundary, for example. – anon Jan 29 '12 at 23:34 I doubt there is any useful closed-form expression for this. If you partition it into convex parts, and then partition those into triangles, you can then add up the areas of the triangles, and the last part can be made into a closed-form expression. – Dan Brumleve Jan 29 '12 at 23:35 – Rahul Narain Jan 29 '12 at 23:47 ## 1 Answer If you mean arbitrary sets of points in the plane, the answer is probably "no" unless you regard the defintion of Lebesgue measure as such a formula. However, you write about "the order in which they are connected" and that makes me suspect you may have in mind a sequence of points on the boundary with line segments between them as components of the boundary. In that case, maybe the shoelace formula will help you. And if you mean polygons whose corners are integer points, there is Pick's theorem. - The Shoelace formula is what I was looking for. Thanks. – Tyler Crompton Jan 30 '12 at 1:51

https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/appendix-d-graphing-techniques-exercises-page-449/61

## Trigonometry (11th Edition) Clone For $f(x) = -\sqrt{x}$ when $x = 9, y = -3;$ $x = 4, y = -2;$ $x = 1, y = -1;$ $x = 0, y = 0$ As seen, $f(x) = -\sqrt{x}$ is the reflection across the $x$-axis of $f(x) = \sqrt{x}$. Domain = $[0, \infty)$ Range = $(-\infty, 0]$

https://www.meritnation.com/cbse-class-8/math/rs-aggarwal-2017/quadrilaterals/textbook-solutions/10_1_3134_5670_185_53121

Rs Aggarwal 2017 Solutions for Class 8 Math Chapter 15 Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Quadrilaterals are extremely popular among Class 8 students for Math Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 8 Math are prepared by experts and are 100% accurate. #### Question 1: Fill in the blanks: (i) A quadrilateral has ......... sides. (ii) A quadrilateral has ......... angles. (iii) A quadrilateral has ......... vertices, no three of which are ......... (iv) A quadrilateral has ......... diagonals. (v) A diagonal of a quadrilateral is a line segment that joins two ......... vertices of the quadrilateral. (vi) The sum of the angles of a quadrilateral is......... (i) 4 (ii) 4 (iii) 4, co-linear (iv) 2 (v) opposite (vi) 360° #### Question 2: (i) How many pairs of adjacent sides are there? Name them. (ii) How many pairs of opposite sides are there? Name them. (iii) How many pairs of adjacent angles are there? Name them. (iv) How many pairs of opposite angles are there? Name them. (v) How many diagonals are there? Name them. (i) There are four pairs of adjacent sides, namely (AB,BC), (BC,CD), (CD,DA) and (DA,AB). (ii) There are two pairs of opposite sides, namely (AB,DC) and (AD,BC). (iii) There are four pairs of adjacent angles, namely . (iv) There are two pairs of opposite angles, namely . (v) There are two diagonals, namely AC and BD. #### Question 3: Prove that the sum of the angles of a quadrilateral is 360°. Join A and C. Now, we know that the sum of the angles of a triangle is 180°. For For $\left(\angle 1+\angle 2+\angle 3+\angle 4\right)+\angle B+\angle D={360}^{o}\phantom{\rule{0ex}{0ex}}$ or $\angle A+\angle B+\angle C+\angle D={360}^{o}\phantom{\rule{0ex}{0ex}}$ Hence, the sum of all the angles of a quadrilateral is 360°. #### Question 4: The three angles of a quadrilateral are 76°, 54° and 108°. Find the measure of the fourth angle. Sum of all the four angles of a quadrilateral is 360°. The fourth angle measures 122°. #### Question 5: The angles of a quadrilateral are in the ratio 3 : 5 : 7 : 9. Find the measure of each of these angles. #### Question 6: A quadrilateral has three acute angles, each measuring 75°. Find the measure of the fourth angle. Sum of the four angles of a quadrilateral is 360°. If the unknown angle is $x$°, then: $75+75+75+x=360\phantom{\rule{0ex}{0ex}}x=360-225=135\phantom{\rule{0ex}{0ex}}$ The fourth angle measures 135°. #### Question 7: Three angles of a quadrilateral are equal and the measure of the fourth angle is 120°. Find the measure of each of the equal angles. Let the three angles measure $x°$ each. Sum of all the angles of a quadrilateral is 360°. Each of the equal angles measure 80°. #### Question 8: Two angles of a quadrilateral measure 85° and 75° respectively. The other two angles are equal. Find the measure of each of these equal angles. Let the two unknown angles measure $x°$ each. Sum of the angles of a quadrilateral is 360 °. Each of the equal angle measures 100 °. #### Question 9: In the adjacent figure, the bisectors of ∠A and ∠B meet in a point P. If ∠C = 100° and ∠D = 60°, find the measure of ∠APB. $\angle APB=80°$

https://mod.readthedocs.io/en/latest/

# mod – modular arithmetic in Python¶ ## Description¶ Modular arithmetic is arithmetic for integers, where numbers wrap around when reaching a given value called modulus. For example 6 ≡ 1 (mod 5). Modular arithmetic has several practical applications including: music, banking, book publishing, cryptography… and of course math. The purpose of this package is to simplify the use of modular arithmetic in Python3. ## Usage¶ This package provides Mod integers that compute arithmetic operations like + - * // ** with a modulus: from mod import Mod # Funny math here x = Mod(5, 7) # x ≡ 5 (mod 7) (x + 2) == 0 # True: 5 + 2 ≡ 7 ≡ 0 (mod 7) (x + 7) == x # True: 5 + 7 ≡ 12 ≡ 5 (mod 7) (x**3) == (x + 1) # True: 5³ ≡ 125 ≡ 6 (mod 7) (1 // x) == 3 # True: 5 × 3 ≡ 15 ≡ 1 (mod 7) ⇒ 5⁻¹ ≡ 3 (mod 7) A naive implementation of RSA encryption algorithm using mod package: from mod import Mod # My RSA keys public_key = Mod(3, 61423) private_key = Mod(40619, 61423) # My very secret message top_secret_message = 666 # RSA encryption encrypted = top_secret_message**public_key # RSA decryption decrypted = encrypted**private_key # My secret message have been correctly encrypted and decrypted :-) assert decrypted == top_secret_message Note • Mod is based on integer modulo operation %, not math.fmod • the result of an operation between a Mod and an int is a Mod • the result of an operation between a Mod and a float is a float ## Package documentation: mod.Mod¶ class mod.Mod(value, modulus) Integer number that automatically adds a modulus to arithmetic operations. The floor division // implements the inverse of a multiplication with a modulus. Therefore, it should be used with care to avoid errors. >>> number = Mod(2, 3) >>> number (2 % 3) >>> quintuple = 5 * number >>> quintuple // 5 (2 % 3) >>> Parameters: value (int) – Mod number value modulus (int) – modulus associated with the value ValueError – one of the parameters is not a number or modulus == 0 copy(modulus=None) Copy the Mod number Parameters: modulus (int) – modulus of the new Mod number Mod inverse Modular inverse of the number. y is the inverse of x with the modulus n when: $y × x ≡ 1 (mod. n)$ Return type: Mod modulus Modulus value Return type: int ## Install¶ Run the following command to install mod package pip3 install mod

https://harangdev.github.io/applied-data-science-with-python/applied-social-network-analysis-in-python/3/

# Influence Measures and Network Centrality Categories: Updated: ## Network Centrality Centrality measures identify the most important nodes in a network. Examples in real world • Influential nodes in a social network. • Nodes that disseminate information to many nodes or prevent epidemics. • Hubs in a transportation network. • Important pages on the Web. • Nodes that prevent the network from breaking up. There are different ways to measure centrality of a node. • Degree centrality • Closeness centrality • Betweenness centrality • Page Rank • HITS Algorithm ## 1. Degree Centrality Assumption: important nodes have many connections ### For undirected networks $C_{deg}(v) = \frac{d_v}{|N|-1}$, where N is the set of nodes in the network and $d_v$ is the degree of node v import networkx as nx import matplotlib.pyplot as plt %matplotlib inline G = nx.karate_club_graph() def draw_network(G): plt.figure(figsize=(12,7)) nx.draw_networkx(G, alpha=0.6, node_size = 1000, font_size=16, edge_color='.4') plt.axis('off') plt.tight_layout(); draw_network(G) centrality = nx.degree_centrality(G) # top 5 pairs of (node, degree_centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [(33, 0.5151515151515151), (0, 0.48484848484848486), (32, 0.36363636363636365), (2, 0.30303030303030304), (1, 0.2727272727272727)] ### For directed networks 1. in-degree centrality $C_{deg}^{in}(v) = \frac{d_v}{|N|-1}$, where N is the set of nodes in the network and $d_v$ is the in-degree of node v G_di = nx.DiGraph() G_di.add_edges_from([('A', 'B'), ('C', 'A'), ('A', 'E'), ('G', 'J'), ('G', 'A'), ('D', 'B'), ('B', 'E'), ('B' 'C'), ('C', 'D'), ('E', 'C'), ('E', 'D'), ('F', 'G'), ('I', 'F'), ('J', 'F'), ('I', 'J'), ('I', 'H'), ('H', 'I'), ('I', 'G'), ('H', 'G'), ('O', 'J'), ('J', 'O'), ('A', 'N'), ('L', 'M'), ('K', 'M'), ('K', 'L'), ('N', 'L'), ('O', 'L'), ('O', 'K'), ('N', 'O')]) draw_network(G_di) centrality = nx.in_degree_centrality(G_di) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [('G', 0.21428571428571427), ('J', 0.21428571428571427), ('L', 0.21428571428571427), ('A', 0.14285714285714285), ('B', 0.14285714285714285)] 2. out-degree centrality $C_{deg}^{out}(v) = \frac{d_v}{|N|-1}$, where N is the set of nodes in the network and $d_v$ is the out-degree of node v centrality = nx.out_degree_centrality(G_di) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [('I', 0.2857142857142857), ('A', 0.21428571428571427), ('O', 0.21428571428571427), ('B', 0.14285714285714285), ('C', 0.14285714285714285)] ## 2. Closeness Centrality Assumption: important nodes are close to other nodes $C_{close}(v) = \frac{|R(v)|}{|N-1|} \frac{|R(v)|}{\sum_{u,v\in R(v)} d(v,u)}$, where $N$ is the set of nodes in the network, $R(v)$ is the set of nodes v can reach, and $d(v,u)$ is the length of shortest path from v to u. We multiply $\frac{|R(v)|}{|N-1|}$ to penalize centrality of a node that has small number of nodes it can reach(that has small R(v)). For example in the above graph G_di, if we do not multiply normalization term, centrality of node ‘L’ is 1, which is too high for a node that can only reach one other node. centrality = nx.closeness_centrality(G) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [(0, 0.5689655172413793), (2, 0.559322033898305), (33, 0.55), (31, 0.5409836065573771), (8, 0.515625)] ## 3. Betweenness Centrality Assumption: important nodes connect other nodes $C_{btw}(v) = \sum_{s,t\in R} \frac{\sigma_{s,t}(v)}{\sigma_{s,t}}$, where $R$ is the set of two nodes which are connected, $\sigma_{s,t}$ is the number of shortest paths between nodes s and t, and $\sigma_{s,t}(v)$ is the number of shortest paths between nodes s and t that pass through node v. There are 3 options when computing betweenness centrality. 1. endpoints When computing betweenness centrality of a node $v$, we can either include or exclude paths that have $v$ in endpoints $s$ or $t$. 2. normalization betwenness centrality values will be larger in graphs with many nodes. To control this, we divide centrality values by the number of pairs of nodes in the graph (excluding v). Dividing term is as follows. $\frac{1}{2}(|N|-1)(|N|-2)$ in undirected graphs $(|N|-1)(|N|-2)$ in directed graphs 3. k (approximation) Computing betweenness centrality of all nodes can be very computationally expensive. Depending on the algorithm, this computation can take up to $O({|N|}^3)$ time. So rather can computing betweenness centrality based on all pairs of nodes s, t, we can approximate it based on a sample of nodes. centrality = nx.betweenness_centrality(G, normalized=True, endpoints=False, k=10) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [(0, 0.44680450336700334), (33, 0.2966874098124098), (31, 0.20575907888407888), (32, 0.17188071188071188), (5, 0.06439393939393939)] ### Betweenness Centrality between subsets We can calculate betweenness centrality between subsets of a network. When counting shortest paths, we only consider paths that start from one of the nodes in subset A and end at one of the nodes in subset B. centrality = nx.betweenness_centrality_subset(G, [32, 33, 21, 30, 16, 27, 15, 23, 10], [1, 4, 13, 11, 6, 12, 17, 7]) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [(0, 21.609126984126988), (2, 8.27845238095238), (33, 5.183888888888888), (1, 4.2630952380952385), (8, 3.1306349206349204)] ### Betweenness Centrality of Edges centrality = nx.edge_betweenness_centrality(G) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [((0, 31), 0.1272599949070537), ((0, 6), 0.07813428401663695), ((0, 5), 0.07813428401663694), ((0, 2), 0.0777876807288572), ((0, 8), 0.07423959482783014)] ## 4. PageRank PageRank algorithm is developed by Google founders to measure the importance of webpages from the hyperlink network structure. PageRank assigns a score of importance to each node. Important nodes are those with many in-links from important pages. PageRank can be used for any type of network, but it is mainly useful for directed networks. ### Basic PageRank In basic PageRank, we repeat the following rule infinitly to get importance of a certain page. For most networks, PageRank values converge as k(number of repetition) gets larger. Basic PageRank Update Rule Each node gives an equal share of its current PageRank to all the nodes it links to. The new PageRank of each node is the sum of all the PageRank it received from other nodes. Example ### Interpreting PageRank as Random walk of k steps Random walk: Start on a random node. Then choose an outgoing edge at random and follow it to the next node. Repeat k times. The probabilty of node ‘a’ being current node at step k equals to PageRank of node ‘a’ at step k. ### Scaled Page Rank PageRank Problem In the above graph, for a large enough k: F and G each have PageRank of 1/2 and all the other nodes have PageRank 0. To fix this problem, we introduce a ‘damping parameter’ $\alpha$ Scaled PageRank Random walk Start on a random node. Then: • With probability $\alpha$: choose an outgoing edge at random and follow it to the next node. • With probability 1 − $\alpha$: choose a node at random and go to it. The Scaled PageRank of k steps and damping factor $\alpha$ of a node n is the probability that a random walk with damping facto r$\alpha$ lands on a n after k steps. In practice, we use a parameter of $\alpha$ between 0.8 and 0.9. After setting $\alpha$ to 0.8, nodes other than F or G in the above graph get PageRank bigger than 0. # Note that G_di is not the above graph. centrality = nx.pagerank(G_di, alpha=0.8) # top 5 pairs of (node, centrality) sorted(centrality.items(), key=lambda x: x[1], reverse=True)[:5] [('B', 0.1136679799498734), ('C', 0.09765860364475079), ('D', 0.09125558981288656), ('E', 0.08613006508366944), ('A', 0.08596118308469886)] ## 5. Hubs and Authorities (HITS Algorithm) Before assigning rank to pages, we need to pick pages that are relevant to the query. We call relevant pages and their connections ‘Base’. This is how we find base. Given a query to a search engine: • Root: set of highly relevant web pages (e.g. pages that contain the query string) – potential authorities. • Find all pages that link to a page in root – potential hubs. • Base: root nodes and any node that links to a node in root. • Consider all edges connecting nodes in the base set. After finding base, Computing k iterations of the HITS algorithm to assign an ‘authority score’ and ‘hub score’ to each node. 1. Assign each node an authority and hub score of 1. 2. Apply the Authority Update Rule: each node’s authority score is the sum of hub scores of each node that points to it. 3. Apply the Hub Update Rule: each node’s hub score is the sum of authority scores of each node that it points to. 4. Nomalize Authority and Hub scores: $auth(J) := \frac{auth(j)}{sum_{i\in N} auth(i)}$ 5. Repeat k times. One step of HITS Algorithm For most networks, as k gets larger, authority and hub scores converge to a unique value. # note that G_di is not the above graph # nx.hits returns (hub scores dictionary, auth scores dictionary) centrality = nx.hits(G_di) # top 5 pairs of (node, centrality) by hub measurement hub_scores = centrality[0] print(sorted(hub_scores.items(), key=lambda x: x[1], reverse=True)[:5]) # top 5 pairs of (node, centrality) by hub measurement auth_scores = centrality[1] print(sorted(auth_scores.items(), key=lambda x: x[1], reverse=True)[:5]) [('I', 0.2633434794139307), ('O', 0.17579324129888213), ('G', 0.11714778282617944), ('J', 0.08601213301577858), ('H', 0.08277050829140217)] [('J', 0.2122699074081906), ('G', 0.15840540582653959), ('F', 0.13330891475417173), ('L', 0.12416151353124302), ('H', 0.10048796182090457)] ## Centrality Measures Comparison The best centrality measure depends on the context of the network one is analyzing. When identifying central nodes, it is usually best to use multiple centrality measures instead of relying on a single one. Categories:

https://www.elitedigitalstudy.com/10152/find-the-values-of-k-for-each-of-the-quadratic-equations-so-that-they-have-two-equal-roots-kx-x-2-6-0

Find the values of k for each of the quadratic equations, so that they have two equal roots.kx (x – 2) + 6 = 0 Asked by Abhisek | 1 year ago |  195 ##### Solution :- kx(x – 2) + 6 = 0 or kx2 – 2kx + 6 = 0 Comparing the given equation with ax2 + bx + c = 0, we get a = k, b = – 2k and c = 6 We know, Discriminant = b2 – 4ac = ( – 2k)2 – 4 (k) (6) = 4k2 – 24k For equal roots, we know, b2 – 4ac = 0 4k2 – 24k = 0 4k (k – 6) = 0 Either 4k = 0 or k = 6 = 0 k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only. Answered by Pragya Singh | 1 year ago ### Related Questions #### A natural number when increased by 12 equals 160 times its reciprocal. Find the number. A natural number when increased by 12 equals 160 times its reciprocal. Find the number. #### Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. #### The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $$\dfrac{29}{20}$$ Find the original fraction.

https://socratic.org/questions/how-do-you-factor-6x-3-6x-2-x-1

# How do you factor 6x^3-6x^2-x+1? Mar 23, 2018 $6 {x}^{3} - 6 {x}^{2} - x + 1$ $= x \left(6 {x}^{2} - 6 x - 1 + 1\right)$ $= x \left(6 {x}^{2} - 6 x\right)$ $= 6 x \left({x}^{2} - x\right)$ Mar 23, 2018 $6 {x}^{3} - 6 {x}^{2} - x + 1 = \left(6 {x}^{2} - 1\right) \left(x - 1\right)$ $\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left(\sqrt{6} x - 1\right) \left(\sqrt{6} x + 1\right) \left(x - 1\right)$ #### Explanation: This cubic quadrinomial factors by grouping and using the difference of squares identity: ${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$ with $A = \sqrt{6} x$ and $B = 1$ as follows: $6 {x}^{3} - 6 {x}^{2} - x + 1 = \left(6 {x}^{3} - 6 {x}^{2}\right) - \left(x - 1\right)$ $\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = 6 {x}^{2} \left(x - 1\right) - 1 \left(x - 1\right)$ $\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left(6 {x}^{2} - 1\right) \left(x - 1\right)$ $\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left({\left(\sqrt{6} x\right)}^{2} - {1}^{2}\right) \left(x - 1\right)$ $\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left(\sqrt{6} x - 1\right) \left(\sqrt{6} x + 1\right) \left(x - 1\right)$

https://math.stackexchange.com/questions/1656499/convexe-functions-geometric-interpretation

# Convexe functions , geometric interpretation !! let $\Omega$ be an open subset of $\mathbb{R}^n$ and $f$ a smooth convexe function $f:\Omega\rightarrow \mathbb{R}$, and let $x_0\in\Omega$ we have always this inequality : $$\forall x\in\Omega\space :f(x)\geq f(x_0)+<\nabla f , x-x_0>$$ i want to know what is the gerometric interpretation of the right side of this inequality (when n=1 it is the tangent line normally under the graph , but when n>1 is it the equation of the tangent plane or what ) thanks ... You are correct. If $f \colon \Omega \to \mathbf R$ is differentiable at $x_0$, then $$T(x) := f(x_0) + Df(x_0)(x-x_0) = \langle\nabla f(x_0), x-x_0\rangle + f(x_0)$$ is the tangent hyperplane to $f$ at $x_0$. Note that $f$ and $T$ equal at $x_0$ and their first derivatives also do, as expected. We have, as in the $n=1$ case, that $$f(x_0) = T(x_0), \quad Df(x_0) = DT(x_0).$$ • thanks for unswering , but i see that here $Df(x_0)(x-x_0) = \langle\nabla f(x_0), x-x_0\rangle$ then what doese $Df(x_0)(x-x_0)$ geometriclly mean, and also $\nabla f(x_0)$ (for $\nabla f(x_0)$ i think it s the normal vector of the tangent plane in the point $(x_0,f(x_0)$) ? Feb 15 '16 at 10:14 $$<\nabla f , x-x_0>$$ is indeed an expression that evaluates to $0$ when the vector $x-x_0$ is orthogonal to the gradient $\nabla f$, i.e. when $x$ lies in the plane orthogonal to the surface normal at $x_0$, which is the tangent plane. So the formula can be interpreted as the surface of "altitude" $z=f(x)$ lies "above" its tangent plane at $x_0$.

https://search.r-project.org/CRAN/refmans/cmna/html/pwiselinterp.html

pwiselinterp {cmna} R Documentation Piecewise linear interpolation Description Finds a piecewise linear function that interpolates the data points Usage pwiselinterp(x, y) Arguments x a vector of x values y a vector of y values Details pwiselinterp finds a piecewise linear function that interpolates the data points. For each x-y ordered pair, there function finds the unique line interpolating them. The function will return a data.frame with three columns. The column x is the upper bound of the domain for the given piece. The columns m and b represent the coefficients from the y-intercept form of the linear equation, y = mx + b. The matrix will contain length(x) rows with the first row having m and b of NA. Value a matrix with the linear function components Other interp: bezier, bilinear(), cubicspline(), linterp(), nn(), polyinterp() Other algebra: bilinear(), cubicspline(), division, fibonacci(), horner(), isPrime(), linterp(), nthroot(), polyinterp(), quadratic() Examples x <- c(5, 0, 3) y <- c(4, 0, 3) f <- pwiselinterp(x, y) [Package cmna version 1.0.5 Index]

http://mathoverflow.net/questions/23960/how-big-can-the-hausdorff-dimension-of-a-function-graph-get/23972

# How big can the Hausdorff dimension of a function graph get? This question is inspired by How kinky can a Jordan curve get? What is the least upper bound for the Hausdorff dimension of the graph of a real-valued, continuous function on an interval? Is the least upper bound attained by some function? It may be noted that the area (2-dimensional Hausdorff measure) of a function graph is zero. However, this does not rule out the possible existence of a function graph of dimension two. - >>It may be noted that the area (2-dimensional Hausdorff measure) of a function graph is zero. Really? I don't know how to prove it if the function is non-measurable. – user69664 Mar 25 '15 at 0:07 @Catcat I had continuous functions in mind, as indicated in the second paragraph. – Harald Hanche-Olsen Mar 25 '15 at 12:59 The answer is 2. Besicovitch and Ursell, Sets of fractional dimensions (V): On dimensional numbers of some continuous curves. J. London Math. Soc. 12 (1937) 18–25. doi:10.1112/jlms/s1-12.45.18 - Here is one example with Hausdorff dimension 2. - Thank you. Have you checked the proof yourself? I ask because that journal (Chaos, Solitons and Fractals) has a somewhat dubious reputation (see Physics World, january 2009 and Nature, November 2008; also zeit.de/2009/03/N-El-Naschie?page=1, arstechnica.com/science/news/2008/11/…), scienceblogs.com/pontiff/2008/11/…) – Harald Hanche-Olsen May 8 '10 at 21:45 To follow up on my own comment, at least the paper looks serious. So I think it's probably okay. – Harald Hanche-Olsen May 8 '10 at 21:50 I haven't checked it carefully. Thanks for the articles, I had never heard of this before. – Gjergji Zaimi May 8 '10 at 22:16 2 is the right answer, but of course 2007 is not the earliest example – Gerald Edgar May 8 '10 at 23:58 Once you get arbitrarily close to 2, just use disjoint intervals and put graphs on them with dimensions $\gt 2-1/n$ – Gerald Edgar May 9 '10 at 0:31

https://brilliant.org/discussions/thread/how-can-i-use-this/

× # How can I use this? Is there a way I could use the derivative of a polynomial $$f(x)$$, i.e, finding the slope of $$f(x)$$ at a given or any $$x$$ to solve polynomial problems, like the problems that show up in math olympiads and tests? Thank you! Note by Shashank Rammoorthy 2 years ago Sort by: Yes of course, there are lots of ways of doing so. For example, if you want to find the coefficient of the linear term, evaluate $$f'(0)$$. What other ways can you think of? Staff - 2 years ago I dunno...how else could I use the slope of the graph of the polynomial? - 2 years ago Yes, finding the slope of a graph is a good way to think about possible uses. Here are a few more suggestions. • Coordinate geometry, finding tangent of parabola • Min/Max of convex functions occur at tangent What else? What else are derivatives used in? How else can we use them? Staff - 2 years ago

http://mathhelpforum.com/calculus/134727-difficult-integral.html

1. ## Difficult integral!?! I have no idea where to go on this one: using substitution t = sinx evaluate the following integral within the limits pi/2 to pi/6: 4cosx /(3 + (cosx)^2 dx i thought about using t^2 = s(inx)^2 but get in a right mess! Can you help? Thanks D 2. Originally Posted by dojo I have no idea where to go on this one: using substitution t = sinx evaluate the following integral within the limits pi/2 to pi/6: 4cosx /(3 + (cosx)^2 dx i thought about using t^2 = s(inx)^2 but get in a right mess! Can you help? Thanks D Send your question again and this time be careful with the parentheses! Tonio 3. Hello, dojo! $\int\frac{4\cos x}{3 + \cos^2\!x}\,dx$ We have: . $4\int\frac{\cos x\,dx}{3 + (1-\sin^2\!x)} \;=\;4\int\frac{\cos x\,dx}{4-\sin^2\!x}$ Let: . $u \:=\:\sin x \quad\Rightarrow\quad du \:=\:\cos x\,dx$ Substitute: . $4\int\frac{du}{4-u^2} \;=\;4\int\frac{du}{(2+u)(2-u)}$ Apply Partial Fractions and get: . $4\int\bigg[\frac{\frac{1}{4}}{2+u} - \frac{\frac{1}{4}}{2-u}\bigg] du$ . . . . . . $=\;\ln|2+u| - \ln|2-u| + C \;=\; \ln\left|\frac{2+u}{2-u}\right| + C$ Back-substitute: . $\ln\left|\frac{2+\sin x}{2 - \sin x}\right| + C$

https://pysit.readthedocs.io/en/latest/exercises/part_1.html

# Part 1: The Foward Problem¶ In this exercise, you will implement a solver for the 1D scalar acoustic wave equation with absorbing boundary conditions, $\begin{split}(\frac{1}{c(x)^2}\partial_{tt}-\partial_{xx})u(x,t) & = f(x,t), \\ (\frac{1}{c(0)}\partial_t-\partial_x)u(0,t) & = 0, \\ (\frac{1}{c(1)}\partial_t+\partial_x)u(1,t) & = 0, \\ u(x,t) & = 0 \quad\text{for}\quad t \le 0,\end{split}$ where the middle two equations are the absorbing boundary conditions, the last equation gives initial conditions, $$x \in [0,1]$$, and $$t \in [0,T]$$. The model velocity is given by the function $$c(x)$$. In our notation, we write that solving this PDE is equivalent to applying a nonlinear operator $$\mathcal{F}$$ to a model parameter $$m$$, where $$m(x) = \frac{1}{c(x)^2}$$ for the scalar acoustics problem. We then write that $$\mathcal{F}[m] = u$$. ## Seismic Sources¶ Before we can solve the equation, we need to define our source function. We define our source functions as $$f(x,t) = w(t)\delta(x-x_s)$$, where $$\ w$$ is the time profile, $$\delta$$ indicates that we will use point sources, and the source location is $$x_0$$. In real world applications, the time profile is not known and is estimated as part of the inverse problem. However, it is common to model source signals with the negative second derivative of a Gaussian, also known as the Ricker Wavelet, $w(t) = (1-2\pi^2\nu_0^2t^2)e^{-\pi^2\nu_0^2t^2},$ where $$\nu_0$$ is known as the characteristic or peak frequency (in Hz), because the magitude of $$w$$’s Fourier transform $$|\hat w|$$ attains its maximum at that frequency. It is also important that this function is causal ($$w(t) = 0$$ for $$t\le 0$$), so we introduce a time shift $$t_0$$, $w(t) = (1-2\pi^2\nu_0^2(t-t_0)^2)e^{-\pi^2\nu_0^2(t-t_0)^2}.$ Problem 1.1 Write a Python function ricker(t, config) which implements the Ricker Wavelet, taking a time t in seconds and your configuration dictionary. This function should assume that your configuration dictionary has a key nu0 representing the peak frequency, in Hz. Your function should returns the value of the wavelet at time t. You can guarantee causality by setting $$t_0= 6\sigma$$ for $$\sigma = \tfrac{1}{\pi\nu_0\sqrt{2}}$$, the standard deviation of the underlying Gaussian. You may also want to implement an optional threshold to prevent excessively small numbers. Plot your function for $$t = 0, \dots, T=0.5$$ at $$\nu_0 = 10\textrm{Hz}$$ and label the plot. # In fwi.py def ricker(t, config): nu0 = config['nu0'] # implementation goes here return w # Configure source wavelet config['nu0'] = 10 #Hz # Evaluate wavelet and plot it ts = np.linspace(0, 0.5, 1000) ws = ricker(ts, config) plt.figure() plt.plot(ts, ws, color='green', label=r'$\nu_0 =\,{0}$Hz'.format(config['nu0']), linewidth=2) plt.xlabel(r'$t$', fontsize=18) plt.ylabel(r'$w(t)$', fontsize=18) plt.title('Ricker Wavelet', fontsize=22) plt.legend(); Problem 1.2 Write a Python function point_source(value, position, config) which takes a value value, a source location position, and uses the range of the spatial domain config['x_limits'], and the number of points config['nx'] from the configuration to implement a numerical approximation to the $$\delta$$. This function should return a numpy array with value at the correct index, correctly evaluating $$w(t)\delta(x-x_s)$$ for value = ricker(t). Be careful with your implementation of the numerical delta, as $$\int \delta(x-x_s)\textrm{d}x = 1$$. To implement point_source, we need to discretize the problem domain. This is stored in the configuration as follows: # Domain parameters config['x_limits'] = [0.0, 1.0] config['nx'] = 201 config['dx'] = (config['x_limits'][1] - config['x_limits'][0]) / (config['nx']-1) For later, we also specify the location of the source: # Source parameter config['x_s'] = 0.1 Note Your function should take a position as a separate parameter and not automatically extract the source location from config because we will re-use this function later. ## Wave Solver¶ The scalar acoustic wave equation can be written as $(M\partial_{tt} + A\partial_t + K)u(x,t) = f(x,t),$ where $$K = K_x + K_{xx}$$ is called the stiffness matrix and contains the spatial derivatives, $$A$$ is the attenuation matrix and relates to the first time derivatives in the boundary conditions, and $$M$$ is the mass matrix and relates to the second time derivatives in the bulk. Problem 1.3 Write a Python function which construct_matrices(C, config) which constructs the matrices $$M$$, $$A$$, $$K$$ for a given model velocity C. Use a second order accurate finite difference for the second spatial derivative and and use an ‘upwinded’ forward or backward first order difference scheme for the first spatial derivatives. Hint It may be helpful to write down precisely what the differential equation looks like at the each interesting point (the two boundaries and some point in the middle of the domain) for your discretized wavefield $$u(j\Delta x, n \Delta t)$$. Hint The matrices are not time dependent, so $$n$$ is fixed. Which $$j$$ are relevant at each of the spatial points? # Load the model C, C0 = basic_model(config) # Build an example set of matrices M, A, K = construct_matrices(C, config) We can discretize the time derivatives using the usual second-order accurate finite difference approximation, $\partial_{tt}u(x,t) \approx \frac{u(x,t-\Delta t) - 2u(x,t) + u(x,t+\Delta t)}{\Delta t^2},$ which will result in the explicit ‘leap-frog’ scheme for computing $$u(x,t+\Delta t)$$. For explicit methods, the stability of the scheme is restricted by the Courant-Friedrichs-Lewy (CFL) condition, $\Delta t \le \alpha\frac{\Delta x}{c_\text{max}}.$ Problem 1.4 Write a Python function leap_frog(C, sources, config) which takes a velocity C, a list of source wavefields sources (one element for each time step), and through the config, takes a time step dt, and a number of time steps nt and returns the time series of wavefields $$u$$. Use $$\alpha = \dfrac{1}{6}$$ and $$x_s = 0.1$$. Hint Your leap_frog function should use your construct_matrices function. # Set CFL safety constant config['alpha'] = 1.0/6.0 # Define time step parameters config['T'] = 3 # seconds config['dt'] = config['alpha'] * config['dx'] / C.max() config['nt'] = int(config['T']/config['dt']) # Generate the sources sources = list() for i in xrange(config['nt']): t = i*config['dt'] f = point_source(ricker(t, config), config['x_s'], config) sources.append(f) # Generate wavefields us = leap_frog(C, sources, config) At this point, it is important to visualize the wavefield (and the medium the waves are propagating in). One way to look at the wavefield of a 1D problem is to consider a plot of its space-time diagram. Problem 1.5 Write a Python function plot_space_time(us, config), using the matplotlib command imshow, to plot and label the space-time diagram for a wavefield $$u(x,t)$$. Hint The matplotlib xticks and yticks functions will be useful. Try to use a gray-scale color map and consider an optional argument to set the title. plot_space_time(us, config, title=r'u(x,t)') ## Data and Sampling¶ A receiver (a seismometer, hydrophone, or geophone) at spatial position $$x_r$$ records the value of the true wavefield (or a function of it) at a point, and is written mathematically as $d_r(t) = d(x_r,t) = \int_\Omega u(x,t) \delta(x - x_r)\textrm{d}x.$ For 2D and 3D seismic imaging problems there are multiple receivers at different spatial positions recording data for a single ‘shot’ (instance of a source). This ‘sampling’ can be denoted with the operator $$\mathbf{S}$$ and is written as $\begin{split}\mathbf{d}(t) = \left[ \begin{array}{c} d_{r_1}(t) \\ d_{r_2}(t) \\ \vdots \\ d_{r_n}(t) \end{array} \right] = \left[ \begin{array}{c} \int_\Omega u(x,t) \delta(x - x_{r_1})\textrm{d}x \\ \int_\Omega u(x,t) \delta(x - x_{r_2})\textrm{d}x \\ \vdots \\ \int_\Omega u(x,t) \delta(x - x_{r_n})\textrm{d}x \end{array} \right] = \mathbf{S}u(x,t).\end{split}$ Your point_source function implements the adjoint operation of sampling, $$\mathbf{S^*}$$. Problem 1.6 Write a function record_data(u, config), which takes a single wavefield u and, as part of the configuration, a receiver position in config['x_r'] and returns the measured data. When combined, data from all time steps form a trace. Use $$x_r = 0.15$$. # Receiver position config['x_r'] = 0.15 ## Forward Operator¶ At this point, you have all of the routines necessary to solve the forward problem, $\mathcal{F}\left[ m \right] = u.$ It will be useful to put the necessary steps into a function, as we will want to solve this problem many times, perhaps on different problems. Additionally, we will frequently want to solve the sampled forward problem, $\mathbf{S}\mathcal{F}\left[ m \right] = d.$ Problem 1.7 Write a Python function forward_operator(C, config) which returns a tuple containing the wavefields and the sampled data. This function should utilize the functions you have written in the previous exercises. Plot and label the trace. Use $$x_r = 0.15$$. ## Bonus Problems¶ Bonus Problem 1.8: Derive how you might use your leap_frog function and periodic boundary conditions to design a 4th order accurate, in both space and time, scheme for solving the wave equation. Bonus Problem 1.9: The implementation of time stepping used in this exercise is not the most efficient approach for implementing time stepping, particularly in higher dimensions. Why? What might be a faster way to implement the time stepping? Bonus Problem 1.10: The wave equation can be solved using an ODE integrator. Change the formulation of the wave equation so that this is possible. Write a function that uses the built-in SciPy ODE integrator to do your time stepping.

https://math.stackexchange.com/questions/2280208/direct-product-of-projective-modules

# Direct product of projective modules For any nonempty family like $\{ {M_\alpha}\}_{\alpha \in I}$ of $R$-modules we know that if $\prod_{\alpha \in I} M_\alpha$ is a projective $R$-module,then for all $\alpha \in I$, $\{ {M_\alpha}\}_{\alpha \in I}$ is projective too. Is the converse true? If not give me a counter-example. Any comments are welcome. • That's not true if $I$ is infinite; for example, you can take $R={\mathbb Z}$, $I={\mathbb N}$ and $M_\alpha={\mathbb Z}$ for all $\alpha$. However, if the direct product is projective, then so are its factors, because each of them is also a summand. – Hanno May 14 '17 at 7:45 • Is this preposition is true? Any $\mathbb Z$-module is projective iff free? – B.K-Theory May 14 '17 at 7:50 • Yes, for $\mathbb Z$ that's true. – Hanno May 14 '17 at 7:51 • Could you explain more about second part of your first comment? – B.K-Theory May 14 '17 at 7:57 • For any $\alpha\in I$ you have $\prod_{\beta\in I} M_\beta \cong M_\alpha\times\prod_{\beta\neq\alpha} M_\beta\cong M_\alpha\oplus\prod_{\beta\neq\alpha} M_\beta$. – Hanno May 14 '17 at 8:48 A theorem by S. U. Chase states that for a ring $R$ the following conditions are equivalent: 1. $R$ is left perfect and right coherent 2. every product of projective left $R$-modules is projective

http://clay6.com/qa/70486/which-of-the-following-2-statements-is-true-

# Which of the following 2 statements is true ? $\text{Statement 1 : All squares are rectangles.}$ $\text{Statement 2 : All rectangles are squares.}$ $(A) \; Statements \;1\;and\;2 \\(B)\;None\;of\;these \\(C)\; Only\;statement\;2 \\(D)\;Only\;statement\; 1$

https://rootsoftheequation.wordpress.com/2015/07/

## Trying to find math inside everything else ### Fighting for the Center At the Math Games morning session at Twitter Math Camp 15, we’ve been created curricular games that hit on some topics that there aren’t really good games for. I came up with the idea for this one, and worked on refining it with the help of Paula Torres (@lohstorres1) and John Golden (@mathhombre). This game is about measures of central tendency (and range for good measure). Not only do students have to determine all of those over and over as they play the game, but they can see how changing the data set changes the values, especially as the size of the data set increases or decreases. It seems really good because it drives the need to make those calculations. All you need is two decks of cards. The game is designed as a two-player game, but it would definitely be best done as two pairs playing against each other, so they can talk to each other about their strategies and calculations. We also recommend having students keep a running tally of the values. ### How to Pack Your Boardgames Last year, before Twitter Math Camp, I was packing and trying to figure out which games to bring with me for the game night we were having before the conference started. I basically had three attributes I was considering: how big the game was, how good it was, and how many players could play it. I wanted to minimize the first one while maximizing the latter two. So I tried to come up with a bunch of formulas for figuring it out, but nothing was quite working out. (I used BoardGameGeek ratings for “how good it was.”) At first I tried doing ${\frac{r \cdot p}{v}}$, but it was putting some games that just weren’t very good as top choices. The problem was that the volume was having too big of an effect – games could come in thousands of cubic centimeters of volume, but max at around 8 for rating and 12 for players. (I had to use amazon.ca to look up the dimensions because I wanted to use centimeters.) So then I tried cube rooting the volume, or doing an exponential functions like ${\frac{p \cdot e^{r}}{v}}%s=2$, or finding the geometric mean of the three numbers, but still nothing came out right. I was basically using three games as test cases: Dominion, which is one of the best games I own but it really big; Pixel Tactics, which is one of the smallest but is only 2 players; and The Resistance, which is small-ish, really good, and can go to 10 players. I figured that any good method should tell me to leave the first two games at home, but to bring the Resistance. If they didn’t, it wasn’t right. Eventually, after doing some research, I determined that a common technique used in psychology when comparing variables of different ranges of values is called standardizing the variables. Basically, for each attribute, I would find the mean and standard deviation. Then, for each game, I would subtract its value from the mean and divide by the standard deviation to get a standardized value. Then I just needed to add up the three standardized values and the ones with the highest score would win. And, as predicted, The Resistance came out on top.

https://daaronr.github.io/micro-giving-pub/consumers-dmd.html

# 4 Consumer preferences, constraints and choice, demand functions ## 4.1 Consumer preferences, indifference curves/sets (0.5 weeks) We will cover most of O-R chapter 4, but we will skip section 4.6 (differentiability) ### 4.1.1 “Bundles of goods” (O-R 4.1) Let’s consider two goods only, for now. In O-R notation: The set of alternatives is $X = \mathbb{R}_+^2$. A member of X is called a ‘bundle’. We can consider these bundles as ‘vectors’ (from linear algebra) with the corresponding operations. For example, let bundle $x = (x_1, x_2)$. If $\lambda$ is a positive number, we have $\lambda x = (\lambda x_1, \lambda x_2)$. We can depict these bundles graphically as vectors (or points) in 2-d space. We will later consider ‘convex combinations’ of particular bundles x and y: $\lambda x + (1- \lambda) y$, for $0 \leq \lambda \leq 1$; all such convex combinations lie on the line segment connecting points x and y. ### 4.1.2 Preferences over bundles, indifference sets (indifference curves) (O-R 4.2 and supplements) If ‘more is preferred to less’, we know that all of the points in the darker region are preferred to all of the points in the lighter region. But how can we compare the “?” areas? Which are preferred? This will depend on how an individual weighs tradeoffs between X and Y. $\rightarrow$ We can depict this using Indifference Curves (indifference sets). O-R: The indifference set for the preference relation $\succsim$ and bundle $a$ is $\{y ∈ X : y \sim a\}$ Read this as ‘all y in the choice set X that are seen as equally good as bundle a’. Indifference curve (simpler definition, using utility) A curve that shows all the combinations of goods or services that provide the same level of utility. (Source: Nicholson and Snyder, 2010) Formally (for 2 goods), the set of pairs of ${x,y}$ such that $U(X,Y)=c$ for some constant c. Autor: Define a level of utility say $U(x) = U$. Then, the indifference curve for U, $IC(U)$ is the locus of consumption bundles that generate utility level U for utility function $U(x)$. Consider the above diagram. [Credit: www2.econ.iastate.edu]. Think of this as a map with a projection above it. • How far East you go on this map determines how much of the x good is consumed • How far North you go determines how much of the Y good is consumed. • How ‘high’ is the projected image… determines how much utility is obtained from this combination of goods X and Y. Hmm… If we think of this as a traveler who loves ‘getting high’, and ‘consumes’ by riving East or North, and his utility is his altitude, maybe that helps. The above diagram also includes color to depict the utility level, like a ‘heat map’. This diagram is from a very useful resource from MIT, Frank’s Economics on the web.] ### 4.1.3 Examples of preferences (over 2 goods) #### Simpler Definitions: Perfect substitutes and complements Perfect substitutes Goods A and B are Perfect Substitutes when an individual’s utility is linear in these goods when she is always willing to trade off A for B at a fixed rate (not necessarily 1 for 1!) Formally, in O-R ‘Example 4.1: constant tradeoff’ (for two goods) $v_i$ is the [numeric] value she assigns to a unit of good $i$ [here, for $i = 1,2$] [This preference relation] is defined by … $x \succsim y$ if $v_1 x_1 + v_2 x_2 ≥ v_1 y_1 + v_2 y_2$. Obviously this $\succsim$ is represented by the utility function $u(x_1, x_2) = v_1 x_1 + v_2 x_2$. Perfect complements Goods A and B are Perfect Complements when an individual only gains utility from (more) A if she also consumes a defined (additional) amount of B, and vice-versa These goods are ‘enjoyed only in fixed proportions’. E.g., left and right shoes (1:1) bicycle frames and wheels (1:2) or, perhaps, baking powder and flour (1:40) for someone who only eats soda bread. These are not the same as the ‘complements’ and ‘substitutes’ in demand functions which we may see later, referring to the impact of changing prices. ### 4.1.4 Properties: Monotonicity Read O-R section 4.3. Know how to define, test, and compare at least two ‘monotonicity-type’ assumptions, e.g., Monotonicity vs Strong Monotonicity; see O-R p. 49. Know how monotonicity rules out ‘bliss points’ See McDL on ‘Bliss points’ Note that what McDL refer to as ‘Isoquants’ is a general term for a ‘level set.’ Indifference curves are a ‘level set for utility’ (if we have a utility representation). Monotonicity (essentially) yields ‘optimizing consumers spend all their wealth over the relevant lifetime’. It enables: • Easier computation of optimization problems (with an equality rather than an inequality constraint ) • Easier and the ‘comparative statics’ of these optimization problems • The derivation of results from ‘adding up restriction’ Explain the examples in the O-R table: • How do we know that ‘constant tradeoff preferences’ exhibit monotononicity? • Explain/prove that ‘complementary goods’ do not exhibit strong monotonicity. • Prove that lexicographic preferences have strong monotonicity ### 4.1.6 Properties: Convexity (skipping the proofs) Read O-R section 4.5, skipping the proofs, if you like. (weak) Convexity (O-R definition) $\succsim$ on $\mathbf{R}^2_+$ is convex if $a \succsim b \rightarrow \lambda a + (1 − \lambda)b \succsim b$ for all $\lambda \in (0,1)$ Strict Convexity of $\succsim$ (O-R definition) … if $a \succsim b$ and $a \neq b \rightarrow \lambda a + (1 − \lambda)b \succ b$ for all $\lambda \in (0,1)$ Simply, weak convexity implies that convex combinations of two distinct bundles, one at least as good as the other, is at least as good as the ‘weakly worse’ bundle. Strict convexity implies that this convex combination is strictly preferred to the ‘weakly worse’ worse bundle. Note that as $\succsim$ also holds (in both directions) for two bundles where $a \sim b$, weak (strict) convexity implies that a convex combination of such bundles will also be at least as good as (better than) either bundle.* Reading tip: when we make an mathematical statement and put multiple things in parentheses, you can read this sentence either ‘reading everything in parentheses’ or ‘reading none of the parentheses’. E.g., “In (New) York people should (not) drive on the left side of the road.” An equivalent definition involving ‘upper contour sets’ (proved equivalent in O-R proposition 4.2): $\succsim$ convex if and only if for all $x^ {\ast} \in X$ the set $\{x \in X : x \succsim x^{\ast}\}$ … is convex. In other words, it is the same as saying ‘the set of all bundles at least as good as some bundle’ … (which we call the ‘upper contour set’) is a convex set.seed( A ‘convex set’ is a set that is ‘closed under convex combinations’, i.e., it contains all of its convex combinations. For any two elements in a convex set, the convex combination of these elements is in the set as well.* A convex set is pretty easy to visualise for a set in 2 or 3 dimensional space … but much harder when we go to higher dimensions. Yet most of the proofs will indeed generalise to higher dimensions. Remember, when we are dealing with real-world products and bundles, we are dealing with many important dimensions. E.g., a box of cereal could be considered to be described by a vector (or list) of characteristics, including, sweetness, saltiness, crunch, calories, carbohydrates, etc…. all of which a consumer may care about Be sure you understand the difference between the definition of convexity and strict convexity here. Convexity: A preference relation on $R^2_+$ that is represented by a (quasi)-concave [utility] function is convex. It will also have convex indifference curves, implying that the ‘weakly preffered to sets’ are convex. Make sure you don’t get this confused. See O-R proposition 4.3 and its proof. The convexity of a strongly monotone preference relation is connected with the property known as decreasing marginal rate of substitution. (O-R) Please see the O-R numerical/parametric example offering insight into this, beginning “consider three bundles.” If time permits I’ll make a quick video on this. Convexity of preferences will ensure ‘well behaved consumer responses’, including demand responses to price (i.e., ‘demand curves’). Without convexity we may see abrupt and discontinuous shifts in the quantity of a good demanded in response to (small) changes in price. Convexity is a convenient assumption, but not one that is reasonable in all cases. Can you give some insight as to why not, and give some examples of cases where the indifference curves between two goods may fail to be convex … and may be concave, at least within certain regions? Consider ‘substitute goods’ like A. petrol (gasoline) versus B. diesel, or A. scuba equipment versus B. skiing gear. In these cases is easy to see why I may prefer ‘a lot of A or B’ over ‘a medium amount of A and B’, and my willingness to give up a good might decrease rather than increase as I have more of that good. How to show (mathematically) that preferences (and indifference curves are convex)? The answers here … by Abrink and Teytelboym appear correct. However, I do not want to get into this computational issue in this module. (However, it is covered in the ‘optimisation’ module). #### More general concept: quasiconcavity (optional, linked as supplement) This is the ‘preferred assumption’ which resembles convexity, but has several advantages as a concept. For example, it doesn’t require differentiable preferences, and its properties are not affected by monotonic transformations of utility functions. This is the concept that you might be taught in a PhD or MRes module. I think it may also have some value in understanding optimization problems in general, which is an important part of research and industry and finance, I believe. Linked as supplement here. ### (Differentiability: mainly skip) You can skip O-R’s discussion of differentiability. However, do have a look at O-R figure 4.5 to get a sense of what differentiability means for indifference curves (indifference sets). 1. What do these indifference curves depict? 2. What might be a functional (utility) form consistent with these indifference curves? 3. Are these preferences ‘everywhere differentiable’? Explain why or why not. 4. Give a real-world example of a case where such preferences might be relevant. ### 4.1.7 Applications: Product positioning, marketing Do people have a preference for balance? (Convex indifference curves)? Naive argument?: If market research suggests that a broad group are indifferent between two options A and C, maybe they strictly prefer G $\rightarrow$ a possible niche for a new profitable product? But what may be a critique, exception to this logic, e.g., for a particular food? • Convex preferences, ‘preferring mixtures’ is a very strong assumption, unlikely to hold everywhere, or for actual mixtures. I may be indifferent between liver and custard, but it doesn’t mean that I prefer liver-flavoured custard, nor even necessarily equal amounts of liver and custard side by side. • Producing an ‘intermediate’ attribute may be more costly; it may be easy to make crunchy but low-taste cereal or tasty but low-crunch cereal, but expensive to make a cereal with both attributes. E.g., it’s hard to make a car that is spacious and fast. Utility/indifference curves: Also a framework for marketing analysis The utility and indifference curve construct may seem highly theoretical. Indeed, these models were developed largely to address big questions like ‘who gains from trade?’ Still, it helps organise thinking and analysis for at least some managers and marketing groups. According to Nicholson and Snyder (2010) Marriot hotel used focus groups to ‘construct (multidimensional) indifference curves’ to consider their ideal product positioning. I have seen a similar presentation for other hotels presented at Behavioural boozeonomics in London. This may also provide a structure to guide data-driven ‘searches for the next best product’. One (often survey-based) approach to this is called ‘conjoint analysis’ or ‘conjoint modeling’. Consider: How would the assumptions of this approach, as described below, translate into microeconomic assumptions about choices and preferences? Note: the site provided by ‘Qresearch’, part of a commercial firm states things in loose terms, for those without Economics training. ## 4.2 Consumer behavior/Individual (and market) demand functions and their properties (1 week) • O-R Chapter 5 (selections as outlined below) Feel free to skip (unfold): You may skip the first part of O-R section 5.5 on ‘rationalizing a demand function.’ However, make sure you do read the discussion of ‘the weak axiom of revealed preferences’ (WARP). This relates to some of the supplementary readings/exercises. Alternative treatments: (Unfold) • McDL: 3.1 (elasticity), parts of Ch 12; Warning: this book’s treatment of consumer surplus seems to be incorrect (or at least vastly oversimplified) from my PoV • QMC: Chapter 17 (elasticities – very clear treatment, Chapter 18 (supply and demand details) Individual choices: supplementary recommended readings As noted before: • You don’t need to read all of these • But you should read some of these… in terms of marking, you may need some themes to discuss in open answer questions on the midterm exam, for example. • Loomes, Graham, Chris Starmer, and Robert Sugden, 1991. “Observing Violations of Transitivity by Experimental Methods”. jstor link (Society 2014) • Choi, Syngjoo, et al. “Who is (more) rational?.” The American Economic Review 104.6 (2014): 1518-1550.a. • Waldfogel, Joel. “The deadweight loss of Christmas.” The American Economic Review 83.5 (1993): 1328-1336. (Waldfogel 1993) ### 4.2.1 Choices are subject to constraints You cannot spend more than your (lifetime) income or wealth $\rightarrow$ budget constraints (and ‘budget sets’). There are also other potential constraints on consumption, such as a leisure time constraint and legal constraints. Consider the budget line given in figure 5.1 in O-R. Budget constraint for two goods, slope $-p_{x_1}/p_{x_2}$ Only two goods? For example, think ‘food’ and ‘nonfood’. #### Budget constraint algebra (simple) *I realise this may be trivial for many of you, but others will find this a helpful refresher. If it is obvious, feel free to skip ahead. I made a video with some simple algebra for budget constraints you can access here If I spend all my wealth (which I should do over a ‘relevant lifetime’ given monotonic preferences), then, if there are only two goods $x_1$ and $x_2$ to choose from, with prices $p_1$ and $p_2$ respectively …: Expenditure on $x_1$ + Expenditure on $x_2$ = wealth ($w$) $p_1 x_1 + p_2 x_2 = w$ To see how $x_1$ trades off against $x_2$, rearrange this to: $x_2 = \frac{w}{p_2} -\frac{p_1}{p_2}x_1$ • Intercept: $\frac{w}{P_2}$, i.e., amount of $x_2$ you can buy if you only buy $x_2$ • Slope: $-\frac{p_1}{p_2}$, i.e., how much $x_2$ you must give up to get another unit of $x_1$ Notes and intuition (unfold if you want more explanation) The slope $-\frac{p_1}{p_2}$: How much $x_2$ I must sacrifice to get another unit of $x_1$, expressed as a negative number. (Strictly speaking, the slope is how much $x_2$ you get when you get another $x_1$, but since this is negative we see that you ‘get a negative amount’, i.e., give up some amount of $x_2$.) To get another unit of $x_1$ it costs me $p_1 x_1$, so the more costly is $x_1$ the more $x_2$ I must give up. For each unit of $x_2$ I give up I save $p_2$, so the more costly is $x_2$ the more I can save by giving up 1 unit of it, thus, the less I need to sacrifice of $x_2$ to get another unit of $x_1$. O-R also refer to a different way of presenting the consumer’s environment: Rather than assuming that the consumer can purchase the goods at given prices using his wealth, assume that he initially owns a bundle $e$ and can exchange goods at the fixed rate of one unit of good 1 for $\beta$ units of good 2. This presentation helps us consider the welfare properties of exchange economies under different conditions and ‘initial allocations’. However, we will not be covering this here. #### Budget sets (O-R 5.1, formal presentation); normalization O-R definition (for two goods) … the budget set of a consumer with wealth $w$ when the prices are ($p_1, p 2$) is $B((p_1, p 2), w) = \{(x_1,x_2) \in X : p_1 x_1 + p_2 x_2 \leq w \}$ (Note that the part after the colon ($p_1 x_1 + p_2 x_2 \leq w$) is referred to as the ‘budget constraint’) The set $\{(x_1,x_2) \in X : p_1 x_1 + p_2 x_2 = w \}$ is the consumer’s budget line O-R continue: Geometrically, a budget set is a triangle like the one in Figure 5.1. Note that multiplying wealth and prices by the same positive number does not change the set: $B((\lambda p_1,\lambda p_2), \lambda w) = B((p_1, p_2), w)$ for any $\lambda > 0$, because the inequalities $\lambda p_1 x_1 + \lambda p_2 x_2 \leq \lambda w$ and $p_1 x 1 + p_2 x_2 \leq w$ that define these sets are equivalent Multiply all prices and income by the same amount, and the budget constraint is unchanged. This will imply that we can simplify any such maximization problem by “normalizing” the budget constraint so that one good simply has price equal to one. (But if this is an applied problem, we should remember to convert it back into the actual prices.) (See my ‘numeraire good’ video.) Also: As the budget constraints are unchanged when all prices and wealth are increased by the same proportion, neither will be the choices made; thus demand functions must be “homogenous of degree zero”, as noted below. “Every budget set is convex” (*) As O-R show algebraically (for two goods, but this would extend to $n$ goods)… • if any two bundles $a$ and $b$ are both within a particular budget set, a third bundle that is a “convex combination” of these two bundles will also be in this budget set. * This holds as a result of constant linear pricing; in the proof you see this is used. With complicated (e.g., nonlinear) pricing, or subsidies that only apply to certain units (recall ‘eat out to help out’?) budget constraints may not have this property. Perhaps it is reasonable to assume that a ‘standard small consumer’ faces constant prices for many goods. On the other hand, we can think of all sorts of exceptions, where you get a discount if you buy in bulk and also note that you cannot buy, e.g., “one third of a car.” With nonlinear budget constraints optimization problems become much more mathematically “interesting”. What’s a “convex combination”? (unfold) Essentially this is just a fancy economics word for “shares of two things adding to one”. It is simply some (non-negative) share of the first bundle and one minus this share of the second bundle. E.g., 100% of the first bundle and none of the second bundle, or 50% of each bundle, or 25% of the first bundle and 75% of the second bundle. (Obviously this could be extended to to multiple bundles too.) Think also about ‘dividing up two bundles’. We see that if there are two bundles out there and one person takes a convex combination of these that is convex, what remains is also a convex combination of these. Formally the convex combinations of bundles $A$ and $B$ (where A and B are vectors) are simply $\delta A + (1-\delta) B$, where $0 \leq \delta \leq 1$. (This can be extended to convex combinations of several bundles to consider a ‘convex hull’, and each bundle can have as many elements as desired.) Consider of any two bundles A and B, each with two elements, plotted on the coordinate plane. Any convex combination of A and B is a points on the straight line connecting A and B. Thus the line connecting A and B contains all such convex combinations. Each particular convex combination $delta A + (1-\delta) B$ will be the point$\delta$ share of the distances between A and B. We could use this to demonstrate “gains to trade”. Suppose two people with the same preferences, each with a distinct bundle (at a point other than the one where their MRS are the same). They can divide up these bundles among each other, one taking a convex combination of the bundles, and the other taking the remainder, which is also a convex combination. See THIS SUPPLEMENT for more detail. ### 4.2.2 Individual (consumer) demand functions (Marshallian demand) - first pass O-R: A demand function is a function $x$ that assigns to each budget set one of its members. Define $x((p_1, p_2), w)$ to be the bundle assigned to the budget set $B((p_1, p_2), w)$. (It is easy to imagine extending this to a case with many goods to choose from, rather than just two.) O-R note that if you multiply all prices and wealth by some positive constant $\lambda$, the budget set is unchanged; thus the demand function is also unchanged. As the demand function is (stated above to be) a function of the budget set, and nothing else, (at least holding preferences constant), if the ‘budget set as an argument’ is unchanged, the output must not change. Considering $x(p_1, p_2, w)$ as a simple function of three arguments, this implies it has the mathematical property ‘homogenous of degree zero’. A function is ‘homogenous of degree $n$’ when multiplying all of its inputs by some number $\lambda$ causes the output to be multiplied by $\lambda^{n}$. … Here, Hd-0, so $\lambda^{0} = 1$ …. i.e., the output is unchanged. O-R present ‘individual demand functions’ before presenting the optimisation problem. This is unusual… why do they do this? They highlight that the demand function does not need to arise from an optimization process (as we describe below)… it merely maps from the budget set to a particular choice. This can arise as a result of a simple rule or “heuristic” such as “spend equal amount of income on each good”. Behavioral economics considers the implications of decisions made on the basis of such heuristics. Types of demand functions: Marshallian and Hicksian… These demand functions are stated as a function of the budget set, or more simply, as functions of price and wealth (or income). These are referred to as the “Marshallian demand functions” after the great Alfred Marshall. These are the demand functions that we may actually be able to observe in the real world. Marshallian demands show the optimal amount of a good as a function of prices and income. I.e. $x_1(p_1, p_2, m)$. The function tells us how demand for, say $x_1$ responds to changes in price holding income fixed. There are another type of demand functions that have very interesting properties for theory and welfare analysis called “Hicksian demand functions” or “compensated demand functions”. These take prices and utility as the arguments. We will not cover these in 2020 because of time constraints. But as an Economist you should have some familiarity with this. ### 4.2.3 Rational consumer and his/her optimisation problem We started with an individual’s preferences, one of the base elements of the neoclassical model. Under some conditions these can be described by utility functions, which we may consider as ‘mountains’ with levels sets depicted by indifference curves. So, we have a way of depicting what people prefer, but this doesn’t tell us what people will choose to consume (and firms to produce). We next introduced the second fundamental element of the neoclassical model, constraints, in particular, the budget constraint. We put these together to describe the consumer’s problem: maximizing her utility subject to her budget constraints. This is depicted in figure below for the two-good case. (Or see O-R fig 5.3) Imagine that you are this consumer with these preferences and this budget constraint. You can choose any point at or below the straight blue line, as it is on or below your budget constraint. You want to get to the highest indifference curve, to attain the most utility.* * Again, this is material that you will have seen before if you have studied Economics. Feel free to skip this if you remember it all; the O-R text gives a concise explanation. As this is a bit remedial I will not cover it in detail. I suggest the Khan academy videos/materials here if you need revision. How do we know point B is suboptimal (i.e., not optimal)? What about point D? What about points between A and B? How do such points compare with point B? We can see B is suboptimal as it is on a lower indifference curve than A, but both are on the same budget line. Same for D. In fact, if we assumed even weak monotonicity we would already know that point A was preferred to point D. Points between A and B are also below the ‘$U=100$’ indifference curve… again, with weak (or strict) monotonicity these must be inferior to point A (but superior to point B). We can demonstrate that you will choose point A, yielding utility $100$. What is special about point A? It’s the point of tangency between the budget constraint and an indifference curve. It’s also (not a coincidence) the point where the slopes are equal. At this point (but not in general!) we have that “Slope of budget constraint = slope of indifference curve”. ### 4.2.4 Marginal rates of substitution O-R section 5.4 ‘Differentiable preferences’ Note that O-R use the notation $v_1(z)$ and $v_2(z)$ to represent “the consumer’s valuations of small changes in the amounts of the goods she consumes away from z”. They refer to the consumer’s ‘local valuations at z’ (remember that $z$ is a bundle of goods). This is usually referred to as the marginal utility; the partial derivative of the utility function with respect to each argument (each good or service), at a particular point (bundle), say $z = (x_1, x_2)$: $mu_1(z) \equiv \frac{d}{d x_1} u(x_1, x_2)$ O-R define the “marginal rate of substitution at $z$$MRS(z)$” as $v_1(z)/v_2(z)$ Assume bundle $z$ contains elements $x_1$ and $x_2$. Relating this to a more standard notation: $MRS(z) = MRS(x_1,x_2) = v_1(x_1,x_2)/v_2(x_1,x_2)=\frac{mu_1(x_1,x_2)}{mu_2(x_1, x_2)}$ I prefer to define and think of the MRS as Starting from a particular point ($x_1', x_2'$) how much $x_2$ would I be willing to give up to get a unit of $x_1$‘? (As this ’unit size’ converges to 0, as we are considering a point slope here. ) Or starting from a particular point … if I gained a tiny tiny unit of $x_1$ how much $x_2$ would I have to give up to hold my utility constant? I loosely derive the ‘ratio of the marginal utilities’ from this definition BELOW, offering some insight. Above, at an optimal interior consumption choice, we had that the “slope of budget constraint = slope of indifference curve”… at least that’s what it looked like. This is equivalent to “the point or bundle (which we will call $x^{\ast}$) where the price ratio equals the marginal rate of substitution”, i.e., the point $x^{\ast} \equiv (x_1^{\ast},x_2^{\ast})$ where $p_1/p_2 = MRS(x^{\ast}) = MU_1(x^{\ast})/MU_2(x^{\ast}) =v_1(x^{\ast})/v_2(x^{\ast})$ holds. Warning: Recall that the marginal rate of substitution may vary everywhere along an indifference curve (remember the idea of satiation and diminishing MRS). Unfold for further explanation… It is a function of the point it is evaluated, i.e., $MRS = MRS(x_1, x_2)$. It is only at the point where the consumer is optimizing that these slopes must be equal. They can also be equal at some other suboptimal points, where she is not spending all of her income… (e.g., perhaps point D in the above figure). Where she is consuming some of every good, this is a necessary condition for such a point to represent an optimal consumption choice, but it may not be a sufficient condition. May also be optimal for her to consume none of certain goods, in which case this condition will not hold. We return to this later. #### At an optimal ‘interior’ consumption choice (with strict convexity, see above note and discussion of ‘corners’ below) • Consume all of income (locate on budget line; follows from ‘more is better’) • The “Psychic tradeoff” (MRS) equals the market tradeoff ($p_X/p_Y$) Intuition: If I can give up X for Y in the market (buy less X, get more Y) at a certain rate, and the benefit I get from doing this is at a different rate, I can make myself better off. Thus the original point could not have been optimal! If this is not clear to you, try to think about this carefully. This is a key insight in microeconomics, a sort of ‘marginality’ argument/‘proof by contradiction’. that will come back later. ### Very simple example, for intuition Suppose that at the consumption bundle you choose, your MRS = 1. To remain indifferent, you would be willing to give up 1 hamburger to get 1 soda. Suppose the price of soda is £1 and the price of a hamburger is £2. $\rightarrow$ Price ratio: $P_S/P_B = 1/2$ Thus if you buy one less hamburger you can buy two more sodas. Thus if you give up one hamburger, you can get one more soda to keep you indifferent plus an additional soda. This means you would be better off; thus the original bundle wasn’t optimal. Practice question: in which direction would you adjust this bundle if the price of a soda was £2 and a hamburger was £1? More extended example… (unfold) E.g., if I can give up one glass of wine and gain two beers (i.e., because wine is twice as costly), and (given my proposed consumption of wine and beer) I get the same value from each glass of wine or beer, I can give up this glass of wine, gain two beers, and make myself ‘one unit’ better off, at the margin. On the other hand, if I can give up one glass of wine and gain two beers (i.e., because wine is twice as costly), and (given my proposed consumption of wine and beer) I get the four times as much value from each glass of wine as each beer, I can give up two beers, gain a wine (which I value at four beers) and make myself ‘two units’ better off, at the margin. However, I can not apply this argument to a point where I am consuming only wine… I cannot consume less than no beer (although some nights I wish I could have done). We return to this point below ### More insight into MRS Recall $u=u(x_1, x_2)$. $U_1(x_1,x_2) \equiv MU_{x_1}(x_1,x_2)$ Derivative with respect to $x_1$: rate that utility increases if we add a little $x_1$, holding $x_2$ constant. Similarly for $MU_{x_2}$. As noted (and we can derive this), the MRS at a point is the ratio of these marginal utilities. MRS: ‘how much $x_2$ would I be willing to give up to get a unit of $x_1$’? Ans: Depends on marginal benefit of each … we can show $MRS(x_1,x_2)=\frac{MU_{x_1}}{MU_{x_2}}$ Mathy intuition: The more valuable a little more X is to me at that point – the higher is $MU_{x_1}$ – the more $x_2$ I am willing to give up to get it. That is why $MU_{x_1}$ is in the numerator. The more valuable a bit more $x_2$ is at that point – the higher is $MU_{x_2}$ – the less $x_2$ I am willing to give up to get a bit more $x_1$. That is why $MU_{x_2}$ is in the denominator. The ‘first order change in utility’ (or ‘total differential’), considering ’small changes in $x_1$ and $x_2$, $d_{x_1}$ and $d_{x_2}$ is: $dU = \frac{\partial U}{\partial x_1}d_{x_1} + \frac{\partial U}{\partial x_2}d_{x_2}$ $= MU_{x_1}d_{x_1} + MU_{x_2}d_{x_2}$ … Where $\frac{\partial U}{\partial x_1}$ refers to the partial derivative of $U(x_1,x_2)$ with respect to $x_1$, and similarly for $x_2$, i.e., the marginal utility. Essentially, for very small changes in X and Y; this approximates the total change in utility. It’s a ‘linear projection’. Setting it equal to 0 and rearranging yields the rate, at the margin, one is willing to give up $x_2$ for $x_1$: $dU = MU_{x_1}d_{x_1} + MU_{x_2}d_{x_2} = 0$ Rearranging… $\frac{dx_2}{dx_1}=-\frac{MU_{x_1}}{MU_{x_2}}$ Tech note: This is a simple case of the implicit function theorem. Essentially $U(x_1,x_2)=c$ defines an implicit function $x_2(x_1)$, whose slope is the negative of the ratio of the derivatives. Rearranging the utility maximising condition yields further intuition: $\frac{P_X}{P_Y} = MRS(z^*) = \frac{MU_X(z^*)}{MU_Y(z^*)}$ (at each interior consumption point $X>0$, $Y>0$) $\frac{MU_X}{P_X} = \frac{MU_Y}{P_Y}.$ I.e., the same ‘bang for each buck’. Note: If this didn’t hold true and you were spending on both goods, you would be paying ‘more per util’ for one good than the other, and thus should reallocate to that other good. ### 4.2.5 Note on ‘corner solutions’ The above ‘bang for the buck’ condition applies to any interior solution • If you are consuming both goods and optimising, $P_{x_1}/P_{x_2} = MRS = MU_{x_1}/MU_{x_2}$ must hold • This is a “necessary but not sufficient condition”, sufficient if there is a diminishing MRS everywhere. But you might consume none of some good (say $x_1$): If even with no $x_1$ we still have $MU_{x_1}/P_{x_1} <MU_{x_2}/P_{x_2}$ • i.e., the marginal utility of the first unit of $x_1$ is less than that of $x_2$. The same condition applies to each good you are consuming a positive amount of. ### 4.2.6 Overview: Solution to consumer’s problem #### General solution (two goods) O-R (proposition 5.1) present the basic results for the solution to a consumer’s problem (at least for two goods): [Consider] a preference relation on $\mathbf{R}_+^2$ and a budget set 1. If the preference relation is continuous then the consumer’s problem has a solution. By this they mean, that ‘there is a highest point in this set’. They cite a ‘standard mathematical result’. This seems obvious; at first glance you might think ‘every set must have a minimal and maximal element’. But then, think about the set “numbers that are strictly below 10”. Is there a largest number in this set? (No, there isn’t.) 1. If the preference relation is strictly convex then the consumer’s problem has at most one solution. The proof of this is pretty neat! First they ‘assume the negation of b holds’ … and look for a contradiction: Assume that distinct bundles $a$ and $b$ are both solutions to a consumer’s problem. Then the bundle $(a +b)/2$ is in the budget set (which is convex); by the strict convexity of the preference relation this bundle is strictly preferred to both $a$ and $b$. So, ‘if there are two optimal distinct bundles, there must be an even better bundle.’ Thus, a contradiction… so the negation of b cannot hold… there cannot be two distinct optimal bundles. 1. If the preference relation is monotone then any solution of the consumer’s problem is on the budget line (This one seems pretty obvious, skip the proof) #### 4.2.6.1 Solution with monotone, convex, and differentiable preference relation (two goods) O-R summarize this nicely, allowing for the possibility of corner solutions" Assume that a consumer has a monotone, convex, and differentiable preference relation on $\mathbf{R}_+^2$. If $x^{\ast}$ is a solution of the consumer’s problem for $(p_1, p_2, w)$ then 1. $x_1^{\ast}> 0$ and $x_2 > 0$ $\rightarrow MRS(x^{\ast}) = p_1/p_2$ 1. $x_1^{\ast} = 0 \rightarrow MRS(x^{\ast}) \leq p_1/p_2$ 1. $x_2^{\ast} = 0 \rightarrow MRS(x^{\ast}) \geq p_1/p_2$ Consider each case for a-c, depicting these graphically. (See, e.g., O-R example 5.8) You may also be familiar with the Lagrangian approach to constrained optimization. The interior solution (ignoring the corner solutions, i.e., not presenting the Kuhn-Tucker Lagrangian) is very quickly derived from the Lagrangian formula here, by Professor Joon Song ### 4.2.7 Some ‘popular’ forms for utility functions, and their properties; computation #### Cobb-Douglas general form $U(X,Y)=X^aY^b$ where a, b are positive constants This is “homothetic” … what does this mean? (unfold) Homothetic: MRS constant along a ray through the origin. $\rightarrow$ It can be represented by an HD1 utility function. This ties down ‘income effects’ in a boring way: consumption of each good expands proportionally in income. …So the poor man just buys a tiny amount of foies gras and jumps in his tiny money bin. Thus if everyone has these preferences th impact of one consumer’s income increasing by £1 million is the same as 1 million consumers gaining £1. This (possibly dubious) assumption underlies ‘representative consumer’ models. A more formal and complete explanation of Homothetic preferences: $MU_X=\frac{\partial U}{\partial X} = aX^{a-1}Y^b$ $MU_Y=\frac{\partial U}{\partial Y} = bX^{a}Y^b-1$ Taking the ratio of these yields $MRS = \frac{a}{b}\frac{Y}{X}$ So only the ratio Y/X affects the MRS; double both, MRS is the same. Another relevant example: ‘Constant Elasticity of Substitution preferences’. #### A non-homothetic (here, ‘Quasilinear’) example $U(X,Y)=X+ln(Y)$ Here $Y$ has a diminishing MU, but for $X$ MU is constant $\frac{MU_X}{MU_Y}= \frac{1}{1/Y}=Y$ So the MRS diminishes as $Y$ decreases (as we go ‘down the indifference curve’ … this is what we confusingly call diminishing MRS), but it is independent of the amount of $X$ consumed. So, if we double both then what happens? Ans: MRS doubles This represents Quasilinear preferences: These are particularly relevant with three or more goods. More generally, quasilinear utility satisfies $U=x_1+f(x_2,...,x_n)$; i.e., it is linear in one ‘numeraire’ good $x_1$.* * I discuss a numeraire price in this video. Here, the amount of the ‘non-numeraire goods’ consumed adjust to match the price ratio but don’t depend on the income. We imagine that ‘the tradeoff between these goods is the same for people of all income levels’. E.g., in principal-agent analysis we can then ignore the question ‘but won’t the preference for leisure change when people are richer’? Mathy intuition: if the ‘Y’ goods $(x2,..., xn)$ have more MU consume them ‘first’, until their MU diminishes to that of the numeraire, then consume X. ### 4.2.8 Rationalizing a demand function (Skip O-R section 5.5) {#} We already considered ‘rationalizability of a choice function’ … so I won’t cover this special case. ### 4.2.9 (Weak axiom of) Revealed Preference This comes largely from pages 67-68 of O-R. Consider: Suppose a consumer is rational (chooses to maximise according to her preferences) and she has a monotone preference relation. Consider the budget sets depicted above:2 1. All points on or below the blue line … budget set $B$ 2. All points on or below the green line … budget set $B'$ Is it possible that she would choose bundle $b$ in the (blue) budget set $B$, and that this same person choose bundle $a$ when faced with the (green) budget set $B'$? Would this contradict rationality and monotonicity? Indeed it would violate one or both of these. It would violate what we define as the ’Weak Axiom of Revealed Preferences, as discussed below. The main point Note that both $b$ and $a$ were available in the (blue) budget set $B$. If she chose $b$ here, if she is optimizing, she must find it to be ‘at least as good as $a$’ … $b \succsim a$, according to her preferences. We will say that such a choice implies ‘$b$ is revealed directly weakly preferred to $a$, or’$b R^{D} a$‘. (O-R say ’revealed at least as good’). The $R^{D}$ and $P^{D}$ notation comes from Varian 1992; it is not used in O-R. Note that Varian seems to define the WARP differently, saying that it rules out that both $A R^{D} B$ and $B R^{D} A$ when $A$ and $B$ are distinct bundles. I will stick with the definition from O-R (also consistent with the current Wikipedia entry.) Now note that in the (green) budget set $B'$, both $b$ and $a$ are again available. In fact now, we note that in the green budget set ‘bundles with more of both goods than b’ are available. So if she now chooses $a$ this tells us that she prefers $a$ to points with more of both goods than $b$’! As O-R note: For example, if an individual purchases the bundle (2,0) when she could have purchased the bundle (0,2), then we conclude that she finds (2,0) at least as good as the bundle (0,2) and prefers (2,0) to (0,1.9) Given monotonicity, this implies that she must strictly prefer $a$ to $b$, i.e., $b \succ a$. We say that such a choice implies ’$a$ is revealed strictly preferred to $b$. Obviously, we cannot have that both $b \succsim a$ and $b \succ a$. Thus, either she is not choosing rationally, or she is indifferent between $a$ and $b$ and her preferences are not monotonic. Consider… Most students choose not to go on-camera in synchronous sessions: Should this be interpreted as revealed preference for online over face to face instruction? (Or at least for some aspects of it?) Some videos on this: First consider ‘mapping indifference curves’: WARP: This is one of the few videos on Youtube that states it correctly. But it’s too slow. Watch it in 2x speed. From O-R: First they define ‘revealed preferred’ and ‘Revealed to be better than’ ‘Revealed at least as good’ Given the demand function $x$, the bundle $a$ is revealed to be at least as good as the bundle $b$ if for some prices $(p_1, p_2)$ and wealth $w$ the budget set $B((p_1, p_2), w)$ contains both $a$ and $b$ , and $x((p_1, p_2), w) = a$ . ‘Revealed to be better than’ The bundle $a$ is revealed to be better than b if for some prices (p_1, p_2) and wealth $w$ the budget set $B((p_1, p_2), w)$ contains both $a$ and $b$ , $p_1 b_1 + p_2 b_2 < w$, and $x((p_1, p_2), w) = a$ . Next, the WARP is stated simply: A demand function satisfies the weak axiom of revealed preference (WARP) if for no bundles a and b, both a is revealed to be at least as good as b and b is revealed to be better than a . Finally, they state and prove that this is a necessary condition for a demand function of a rational consumer (with monotonicity): A demand function that is rationalized by a monotone preference relation satisfies the weak axiom of revealed preference. The proof of this is fairly simple (not covered here because of time and space concerns). We can extend the WARP to ‘chains’ of revealed preference, so called ‘indirect revealed preferences’ This allows us to use sequences of choices to sketch out indifference curves for actual consumers. There are also interesting ways to measure the ‘extent’ of violations of the revealed preference axioms. See: • Loomes, Graham, Chris Starmer, and Robert Sugden. “Observing violations of transitivity by experimental methods.” Econometrica: Journal of the Econometric Society (1991): 425-439. • Choi, Syngjoo, et al. “Who is (more) rational?.” The American Economic Review 104.6 (2014): 1518-1550. ## 4.3 Properties of (Marshallian) demand functions This material is presented at a less advanced level than the previous material. I imagine for many of you, this will be revision, and feel free to skim it or skip it. More advanced material is considered further below. Previously, we considered how consumption choices determined by utility functions/indifference curves and budget constraints. We can now consider how an individual’s choice of a good varies with each element of her budget constraint: for income and the price of each good. The notation has changed here, to some extent. The OR notation is difficult to adapt here. I try to keep notation constant as much as possible. However, this could be seen as a good thing to get used to: in reading various papers in Economics, authors will use a variety of labels for variables and formats. We will refer to her ‘quantity demanded of good X’ function (or more simply, her ‘demand function’) as follows: $Quantity \: of \: X \: demanded = d_x(P_X, P_Y, I; preferences)$ ### Homogeneity Homogenous (of degree zero) (demand) function A function whose outcome value does not change when all arguments are changed proportionally is homogenous of degree zero $d_X(P_X,P_Y,I)$ is homogenous of degree zero in its arguments. Multiply all prices and income by the same amount, and the budget constraint is unchanged. Thus (as preferences have not changed) consumption choices should not change either. • E.g., the budget constraint $P_X X + P_Y Y = I$ is the same as the budget constraint $2P_X X + 2P_Y Y = 2I$ This relates to the puzzle of ‘why should monetary policy and inflation have any real effect on the economy?’ I discuss this in the context of a ‘numeraire price’ below: ### 4.3.1 Response to income changes Q: What happens to the quantity purchased of some good as an individual’s income increases? A: It depends on the preferences (i.e, the utility function, i.e., the relative curvature of the indifference curve at various points). This defines whether the good is “normal” or “inferior.” Whether it is normal or inferior depends on your preferences and the change in the slope of the indifference curves with higher income/utility, as we see below.) #### Defining a normal and inferior good Normal good A good that is bought in greater quantities as income increases. Inferior good A good that is bought in smaller quantities as income increases. For a simple powerpoint supplement on this, see here Consider: Here we see more income $\rightarrow$ less expenditure on Z! (Not just ‘a lesser share’ but actually less.) This is because the shape of the indifference curves changes at higher incomes in this example. When people have a lot to spend, they want to spend it on Y and not on Z. (Again in this example, not in general.) Consider, if you won the lottery how much pot noodle would you buy? Pot noodle may be like good Z in this example. Thinking in 3d: ‘as you walk up this hill, its ridge gradually moves to the west’. To be precise, this is not about wealthy people being different than poor people nor that their taste ‘changes’ when they become wealthy. We are considering the same person with higher income, and thus the potential to choose a different bundle of goods. Empirically, these are hard to distinguish, however. Consider: all goods cannot be inferior. Why not? There are ‘adding up’ conditions. You always spend all of your income. Thus, if as your income increases, you spend less on some subset of goods, you must spend more on the remaining goods. ### 4.3.2 Response to (own) price changes (limited coverage) What happens to an individual’s demand for a good as the price of this good rises, in our model? (I.e., if people are optimizing and their preferences obey all of the above axioms.) You might have thought that when price rises quantity demanded falls, and vice versa, the so-called “iron law of demand”. However, we cannot derive this from our model and in fact, when price rises, an optimizing consumer might demand more of this good or less of this good, we do not know! If the good is inferior (as defined above) there will be two countervailing effects: 1. “Substitution effect”: Is a good (say, good ‘X’, on the X-axis) becomes more expensive, the budget line rotates (inward), and the point of tangency (where the price ratio equals the marginal rate of substitution) on any indifference curve moves towards a point with less of good X. In other words, ‘along a particular indifference curve’ the individual “substitutes away” from X as it becomes more expensive. 2. “Income effect”: As X becomes more expensive, the individual’s budget set is restricted. The individual is “poorer in a sense”. If X is an inferior good (in this region), this consumer, being ‘poorer’ chooses consume more of the good, all else equal. If the second effect outweighs the first we have the surprising result that , and we label such a good a “Giffen good”. To cover this in more detail, we would need to define another concept called “Hicksian demand” (see supplement). ## 4.4 Consumer surplus (brief and incomplete treatment) Consumer surplus (Undergraduate definition: Nicholson and Snyder) The extra value individuals receive from consuming a good over what they pay for it. • What people would be willing to pay for the right to consume a good at its current price rather than not being able to buy it at all.* In other words, if you said to a consumer ‘this product will stop being produced/offered unless you pay us enough money’, this is the maximum amount the consumer would be willing to pay to keep the product on the market, at its current price. • The area between the demand curve and the market price • A measure of consumer welfare, useful for policy analysis (unfold for some policy applications) Policy, e.g., • Impact of a tax or subsidy • Impact of allowing firms to merge • Impact of allowing a new form of price discrimination via (e.g.) ‘internet cookies’ This concept can also be applied to measuring the value added by the introduction of a new good. This is useful to know for policy, particular in formulating subsidies for R&D and in adjusting the CPI. It also could be used to compute damages in court cases where a firm is accused of stifling innovation. Also… This can be applied to the individual or market demand curve to obtain individual or total consumer surplus (although there are some technical issues with the latter). Warning: The above is not precisely correct, it is an ‘undergraduate simplification’. This measure of of consumer surplus is not a great measure of welfare changes in response to changes in prices (or taxes). Some issues in the fold below. Other measures of changes in consumer welfare include ‘equivalent variation’ and ‘compensating variation’ (neither of which we will have time to cover in 2020). 1. Income effects: If we are using an individual demand curve … The consumer’s Marshallian (‘regular’) demand response as the price changes reflects both income and substitution effects. To determine either: • the amount a consumer would be willing to pay to avoid a price change’, equivalent variation, or • ‘the amount a consumer would need to be compensated to be no worse off after a price change’, the compensating variation … we would need to know her ‘Hicksian’ or ‘utility-compensated’ demand curve; this is difficult or impossible to estimate. 1. Aggregation: To get consumer surplus we want to sum the value each consumer gets from each unit. Suppose, as we derive/assume under perfect competition, firms are pricing at the marginal cost of the final unit produced. Assume that cost varies over time, so we observe shifts in the supply curve, tracing out points on the demand curve. This allows us to estimate a market demand curve. But this demand curve (inverted) only tells us the last consumer’s value of (willingness-to pay for) the final unit produced. We don’t know how other consumers valued each of these units. With a linear demand curve this is a triangle, which has an area that is easy to compute. More generally, this is a ‘definite integral’. $\int_{0}^{q^\ast}p(q)-p^\ast dq$ (where $p(q)$ is the inverse demand). For welfare calculations, we often want to know ‘how does this integral change as we adjust some parameter’? $\rightarrow$ See the (more general version of) the Leibniz rule for differentiating integrals., related to the ‘Fundamental theorem of calculus’. ## 4.5 Market demand (function) (brief and incomplete treatment) Market demand (or better, ‘market quantity demanded’) The total quantity of a good or service demanded by all potential buyers This just sums the individual quantities demanded (at a given price) Market demand curve Relationship between total quantity demanded of a good and its price, ceteris paribus • Sum the individual demand curves (‘horizontally’ … quantities demanded at each price) Image source: McAfee et al, Introduction to Economic Analysis v. 2.1, chapter 2.3 Some of our previous results about individual demand also hold for market demand, while others do not, or only if we make restrictive assumptions. Shifts in the market demand curve are caused by similar things that cause individual demand curve shifts • Increases in overall income (for normal goods) • Reduced prices of complements, increased price of substitute goods • Change in tastes However, the aggregate income does not have a single effect; it cannot be easily reduced to a single variable here. You can only express the demand curve as a function of aggregate income under restrictive assumptions. In general, it depends who gets this income. (See ‘aggregation issues’). (In addition, the market demand curve could shift out if more consumers enter the market, e.g., because of demographic change.) A random example: • 2008: ‘Gas prices forcing demand for SUVs to plummet’ LINK • 2015: ‘Economy, gas prices drive demand for SUVs, high-end cars’ LINK ## 4.6 Elasticities This is a fundamental mathematics concept that comes from Physics; it’s not just for Economics. • the change in quantity demanded of oranges when the price of oranges rises or • the change in quantity demanded of apple juice when the price of apple juice rises? (Or the response to changes in the price of a related good, or to income.) Difficulty: These things are measured in different units and the prices have different starting values. Elasticity: the measure of the % change in one variable brought about by a 1% change in another variable. • a unitless measure; will be the same no matter how these variables are measured. Think of responsiveness when talking about elasticity. Actually it’s a measure from physics having to do with rubber bands, they tell me. Strictly speaking we are talking about the limit of these responses, i.e., derivatives. The elasticity is basically the derivative of $ln(y)$ with respect to $ln(x)$; useful to know if you want to run a regression computing an elasticity, or if you want to interpret such a regression. Actually, this, and its subtleties and interpretations is one of the most important things to comprehend for empirical work!! • If a 5% fall in the price of oranges typically results in a 10% increase in quantity bought, • we say that each percent fall in the price of oranges leads to an increase in sales of about 2 percent. • I.e., the ‘‘elasticity’’ of orange sales with respect to price is about 2, Note that elasticities may not be constant; they may depend on the starting point; e.g., linear demand implies a different price elasticity at each point. ### Elasticities more precisely • calculate in terms of derivatives • estimated using log transformations* (There is a decent video explaining how to interpret these regressions here, and another one here but you need to recall what is meant by ‘total differentiation’.) ### 4.6.1 Price elasticity of demand Price elasticity of demand: $e_{Q_d,p} = \frac{percent \ change \ in \ Q_d}{percent \ change \ in \ p}$ $= \frac{\Delta Q_d}{Q_d}/\frac{\Delta p}{p}$ This should always be negative (except for Giffen goods). It is a unitless measure related to the slope of the demand curve. It is very important for price-setting firms (more on this later). From Nicholson and Snyder, 2010 (footnote): If demand happens to take the ‘constant elasticity’ form $Q = ap^b$ where $b<0$, the price elasticity of demand is $b$. The constant-elasticity form is derived from the ‘constant elasticity’ using the method of ‘ordinary differential equations’. • This elasticity is the same everywhere along such a demand curve. • Taking logarithms of this yields $ln(q) = ln(a) + b ln(p)$ • note that elasticities are often estimated by regressing logs on logs However Indeed, one important implication of Jensen’s inequality is that the standard practice of interpreting the parameters of log-linearized models estimated by ordinary least squares (OLS) as elasticities can be highly misleading in the presence of heteroskedasticity. • Santos-Silva and Tenreyro, 2006, ‘the Log of Gravity’. ### Examples from the headlines India’s Hike Messenger takes aim at WhatsApp “Reliance ended up showing that there is elasticity in the market. If you drop prices, people will come on board,” he said. ‘Next’ to add more space despite retail sales ‘moving backwards’ The retailer does not expect any impact from the drop in sterling since the Brexit vote to kick in until at least the Spring of 2017, as it had hedged some of its foreign-currency exposures in advance. Still, it expects expenses to rise by up to 5 per cent next year. ‘The last time we had to increase prices (which was in 2010 when cotton prices soared) we estimated that price elasticity was around 1.1. If that remains the case today, a retail selling price increase of 5% would result in a fall in unit sales of -5.5% and a fall in like for like sales value of between -0.5% to -1.0%. In the scheme of things, we think that this drag on sales is manageable and less damaging than taking a significant hit to margin.’ ### 4.6.2 Properties of price elasticity of demand • Goods with many close substitutes at a similar price will be highly elastic • with few substitutes … inelastic • Typically: more elastic in the long run than the short run. Q : why? Ans: Over time, consumers can adjust to price changes by changing their consumption patterns. E.g., if petrol gets more expensive I can switch to a hybrid or electric car, or a bicycle. We refer to price elasticities with the following terminology: $e_{Q,p}$ $abs(e_{Q,p})$ Term $< -1$ $>1$ Elastic $= -1$ $=1$ Unit Elastic $> -1$ $<1$ Inelastic Note that sometimes elasticities are expressed in absolute value terms (a positive number). It should be clear from the context. Consider an individual’s expenditure on a product. The total expenditure (which will equal the firms’ revenue from this): price $\times$ quantity = Expenditure ($E$) By taking the total differential of this and recalling the ‘multiplication rule’ of derivatives… $E = p(q)q = pq(p)$ $dE = pq'(p)dp+q(p)dp$ $dE/E = \frac{pq'(p)dp}{pq} +\frac{q(p)dp}{pq}$ $dE/E = \frac{q'(p)dp}{q} +\frac{dp}{p}$ We have that for small changes in price, the percent change in total expenditure is: percent change in price + percent change in quantity As $e_{Q,P}$ tells you the percent change in quantity for each (small) percentage change in price, we can use this to determine the change in expenditure for a small change in price. Considering very small increases in a good’s price: • If the (individual’s) demand is price-elastic, quantity will decrease by a larger percentage than price increased. Thus her expenditure (on this good) will decline. • If the (individual’s) demand is price-inelastic, quantity will decrease by a smaller percentage than price increased. Thus her expenditure (on this good) will increase . • If the (individual’s) demand is unit-elastic (with respect to price), quantity will decrease by a the same percentage as price increased. Thus her expenditure (on this good) will be unchanged. #### Example and algebra… If price rises by 20% and quantity demanded falls by 20% what happens to revenue? Suppose $p=10$ and $q=10$ so $TR=100$. Then when p=12 and q=8 TR=96. Huh? Price increased by 20% and quantity decreased by 20% so shouldn’t TR be unchanged? No! This only works for small changes. TR is unchanged when $p \times q$ stays the same. With a change in each $\delta p$ and $\delta q$, the new revenue is $(p+\delta p)(q+\delta q) = pq+\delta p q + \delta q p + \delta p \delta q$. This is the same as before if $pq=pq+\delta p q + \delta q p + \delta p \delta q$, i.e., if $\delta p q + \delta q p + \delta p \delta q=0$, i.e., if $\delta p q + \delta p \delta q =-\delta q p$, i.e., if $\delta p q/p + \delta p \delta q/p =-\delta q$, i.e., if $\delta p/ \delta q (q/p) + \delta p /p =-1$, i.e., if $-e_p + \delta p /p = -1$ In the limit, as $\delta p$ goes to zero, this gives us that ‘unit elasticity means a small change in quantity leads to no change in revenue.’ Important question/result Q: Suppose a firm is setting its price and selling to only one individual , and facing no competitors. It should basically never want to set its price at a point where demand is inelastic. Why not? Ans: If it were at such a point, it could raise its price and its revenue would increase and costs would decline (because it would be selling fewer products but for greater total revenue.) Consider • If I want to get to the highest point on a mountain, do I locate at a point with an ‘upward slope’? • I.e., do I choose a point where “if I move in some direction, I go higher”? This is a key point, but also a common point of confusion. The point is the principle of (what I will call) ‘no room for improvement’. If I am at an optimal point I can do no better. Thus if I can do better, I must not be at an optimal point. If the firm can increase its profits by increasing its price relative to its current price, then its current price is not profit-maximising. A caveat is that it might do this for a long term strategic advantage; e.g., to gain customer loyalty and market share, intending to take its profits later. We come back to a related point later in the context of monopoly pricing. Image source: Nicholson and Snyder, 2010 ## 4.7 Income elasticity of demand; Normal, luxury, and inferior goods This video offers a good explanation of neccesity and luxury goods, and how this concept relates to inferior and normal goods. Income elasticity of demand % change in quantity demanded of a good in response to 1% change in income (approximately). $e_{Q_d,I} = \frac{percent \ change \ in \ Q_d}{percent \ change \ in \ I}$ $= \frac{\Delta Q_d}{Q_d}/\frac{\Delta I}{I}$ Normal goods: $e_{Q,I} > 0$ Inferior goods: $e_{Q,I} < 0$ Luxury goods: $e_{Q,I} > 1$ E.g., cocaine is a luxury good, if, when I win £1000 in the lottery, I will increase my consumption of cocaine by more than £1000 … (assuming, as in classical models, that I treat all sources of income the same… Q: Is a luxury good a normal good? Ans: Yes (do you know why?) Why is it that not every good can have an income elasticity of demand greater than 1? Can every good have an income elasticity of demand less than 1? ### Some real-world discussion of this Prof. Muellbauer letter to FT Sir, Professor Gordon Gemmill (Letters, December 14), surprisingly for a trained economist, assumes an income elasticity of demand of zero for housing: that is, that people do not demand more and better housing as they become richer. Nowhere in the world is this the case! My own empirical work demonstrates that around two-thirds of the rise in UK house prices, corrected for general inflation, since 1980 is because supply is not keeping up with income and population growth. Other drivers do exist … The price effects of extra supply take time to build up. I agree on that. But just imagine what would happen if we did nothing more than we are now doing: population and income growth would drive prices even higher even though we already hold the record for rises in house prices since 1970 among the group of seven leading high-income countries. We need to build far more housing, in the right locations. And we need to start now. - Prof John Muellbauer Nuffield College, Oxford, UK Look up: what are some real-world price and income elasticity estimates? How were these derived? Enter this as a hypothesis comment. ## 4.8 More formal statements of properties of Marshallian demand Source: Preston, 2006; https://www.ucl.ac.uk/~uctp100/g021/g021lect.pdf; notation changed. Note: bold means ‘vectors’. Budget set $B=\{\mathbf{q} \in R^n_{+}|\mathbf{p'q} \leq I \}$ Slopes $dq_i/dq_j|_B = -p_j/p_i$. in Consumer chooses $\mathbf{x}(\mathbf{p},I)\in B$; Marshallian demands. The graph of $x_i(\mathbf{p},I)$ in $I$ is the ‘Engel curve’. The Engel curve simply depicts how quantity demanded of one good increases in (an individual’s) income, holding all else constant. Graphing the Engel curve (some links): Some decent youtube videos on this, in a simple form. We see the graphical derivation of an Engel curve: • For a neccesity (i.e., a non-luxury) good here • for a normal good (remember, an inferior good is also a neccesity, but not necessarily vice versa) here. • And for a luxury good here (remember, a luxury good is also a normal good, but not necessarily vice versa) He presents the indifference curves and then ‘Income consumption curve’… the optimizing choice of each of two goods as income increases. Next, he plots this for one good only, yielding the ‘Engel curve’. Notice how the Engel curve gets steeper for a necessary good, but gets shallower for a luxury good. For an inferior good, it has a negative slope. Of course the response to changes in income need not be constant throughout the whole range … a good can even be inferior in one range and luxury within another range of income. See Professor Joon Song’s video HERE for an integrated presentation of the income expansion path, the Engel Curve, and income elasticity. Total budget (income) elasticity: $\eta_i \equiv \frac{\partial x_i}{\partial I} / \frac{x_i}{I} = \frac{\partial ln(x_i)}{\partial ln(I)}$ ‘By what percentage does consumption of $x_i$ change as income changes by a small percent?’ • $\eta_i > 0$ $\rightarrow$ ‘normal good’, otherwise ‘inferior’ Applying this to ‘budget shares’: $s_i \equiv p_i x_i/I$… noting these must add up to one: $\sum_{i=1..n}s_i=1$ • $\eta_i > 1$ $\rightarrow$ ‘luxury’ (budget share rises in income), otherwise ‘neccessity’ Can you write the ‘own price elasticity’ in this notation? Similarly, … $_i /$ ## 4.9 Adding-up (‘aggregation’) etc If an individual ‘always spends all her income’ then this equality must hold everywhere. This implies a set of relationships: • between how things change in response to income and price… -as one spends more on one good she must spend less on another • between shares of expenditure and how these change. Rewriting the budget constraint with the Marshallian demands vector $\mathbf{x}$, and imposing budget-balance (which can be proved must hold for an optimising consumer given local nonsatiation): $\mathbf{p'x}(\mathbf{p},I) = I$ TODO: I hope to include a video here Differentiating this wrt $I$ yields Engel aggregation: $\sum_i p_i \frac{\partial x_i}{\partial I} =\sum_i \frac{p_i x_i}{I} \frac{\frac{\partial x_i}{\partial I}}{\frac{x_i}{I}} =\sum_i s_i\eta_i = 1$ $\rightarrow$ Sum (over all goods) of ‘budget share’ $\times$ ‘income elasticity’ must add up to one. Thus not all goods can be inferior ($\epsilon_i < 0$); these must be balanced by normal goods (including some luxuries, to get the average above one.) Also not all goods can be luxuries ($\epsilon_i > 1$) and not all goods can be necessities ($\epsilon_i < 1$). These must balance each other out, in an expenditure-weighted sense. Differentiating the ’demand budget constraint wrt some price $p_i$ yields Cournot aggregation: (rem: product rule here, carried through sums ) $x_j + \sum_i p_i \frac{\partial x_i}{\partial p_j} = 0 \rightarrow$ The marginal change in expenditure as $j$’s price changes sums the amount of $j$ originally purchased and the sum of the prices of other goods times their responses to the price change in j. Next we multiply this by the price of j and divide by income $I$ to get (bunch of steps here:) $-s_j = \sum s_i \epsilon_{ij}$ The budget share on j plus the sum of the budget shares of other goods multiplied by the cross-price elasticities must add up to 0. … as the price of j increases the price-weighted consumption shares must decrease. So, e.g., all goods cannot be Giffen. ## 4.10 Enrichment, further material to consider 1. To consider constrained optimization (with many goods) using the Lagrangian method see THIS supplement (based on David Autor’s notes) 2. The dual ‘minimization problem’ and ‘Hicksian demand’ are discussed HERE. For a simpler treatment, please see QMC Chapter 20.3*. Recall: QMC refers to ‘Quantum Microeconomics with Calculus’ by Yoram Baumann ### Further doctoral-prep concepts Some examples of things you may need to learn and use for a PhD micro module, (although only a small subset of economists actually use this in their research!). You will not be directly examined on this in the present module. Some of these are covered or touched on in O-R Another taste, nice UCLA notes here As noted, standard utility functions are defined up to ‘monotonic transformations’. We can derive proofs of the invariance of the implied preferences to these. What conditions on a (multivariate) utility function ensure it exhibits the equivalent of a diminishing marginal rates of substitution? - Something called ‘quasi-concavity’, equivalent to ‘having convex upper contour sets’ Representation of the MRS between all goods as a matrix, depicting the properties of this • Strict quasi-concavity takes the place of DMRS in ensuring ‘simple optimisation conditions’ Several topics that we have skipped • Formal definitions of Hicksian (compensated) demand functions • Substitution and income effects • Deriving and using the Slutsky decomposition relating Marshallian and Hicksian demand, - a set of other exact relationships between functions that come out of the consumer’s minimisation and equivalent maximisation problem Some of these concepts are important for characterizing the welfare effects of price changes and policy changes (including taxation), and for conceptualizing issues involving multiple constraints on consumer optimization (e.g., rationing). I will provide notes on these for anyone who is interested. • Aggregation of individual demand functions into ‘market demand’, and the properties of this ### References Reinstein, David. 2014. “The Economics of the Gift.” Society, The Econometric. 2014. “Observing Violations of Transitivity by Experimental Methods Author ( S ): Graham Loomes , Chris Starmer and Robert Sugden” 59 (2): 425–39. Waldfogel, Joel. 1993. “The Deadweight Loss of Christmas.” The American Economic Review 83 (5): 1328–36. 1. Image source: SilverStar at English Wikipedia, CC BY-SA 3.0 http://creativecommons.org/licenses/by-sa/3.0/, via Wikimedia Commons↩︎

https://math.stackexchange.com/questions/2429835/injective-mathcalo-x-module-is-flasque

Injective $\mathcal{O}_X$ module is flasque This is a lemma from Hartshorne. Let $(X,\mathcal{O}_X)$ be a ringed space. Then any injective $\mathcal{O}_X$ module is flasque. I am trying to prove this. Let $V\subseteq U$ be open subsets of $X$, we need to prove that the restriction map $\mathcal{F}(U)\rightarrow \mathcal{F}(V)$ is surjective morphism. As $\mathcal{F}$ is an injective module, any injective morphism of sheaves $0\rightarrow \mathcal{A}\rightarrow \mathcal{B}$ gives a surjective morphism $\text{Hom}(\mathcal{B},\mathcal{F})\rightarrow \text{Hom}(\mathcal{A},\mathcal{F})\rightarrow 0$. So, it is natural to look for sheaves of $\mathcal{O}_X$ modules $\mathcal{A},\mathcal{B}$ such that $$\text{Hom}(\mathcal{B},\mathcal{F})=\mathcal{F}(U) \text{ and }\text{Hom}(\mathcal{A},\mathcal{F})=\mathcal{F}(V).$$ Hartshorne gave $\mathcal{B}=j_!\left(\mathcal{O}_X|_U\right)$ and $\mathcal{A}=j_!\left(\mathcal{O}_X|_V\right)$ and then says that $$\text{Hom}(\mathcal{B},\mathcal{F})=\mathcal{F}(U) \text{ and }\text{Hom}(\mathcal{A},\mathcal{F})=\mathcal{F}(V).$$ But, how did they come up with that example. How on earth can one guess such example. Any motivation regarding this is welcome. Coming to the proof of $$\text{Hom}(j|!\left(\mathcal{O}_X|_U\right),\mathcal{F})\cong \mathcal{F}(U).$$ Let $s\in \mathcal{F}(U)$, we construct a morphism of $\mathcal{O}_X$ modules $j_!\left(\mathcal{O}_X|_U\right)\rightarrow \mathcal{F}$. Let $W\subseteq X$ be open. If $W\nsubseteq U$ then $j_!\left(\mathcal{O}_X|_U\right)(W)=\emptyset$. So, with out loss of generality, we assume $W\subseteq U$. So, we define $\eta(W):j_!\left(\mathcal{O}_X|_U\right)(W)\rightarrow \mathcal{F}(W)$ i.e., $\eta(W):\mathcal{F}(W)\rightarrow \mathcal{F}(W)$. Let $t\in \mathcal{F}(W)$ then assign $t\cdot s|_W\in \mathcal{F}(W)$ to $t$. This collection gives a morphism of $\mathcal{O}_X$ modules. Let $\eta:j_!\left(\mathcal{O}_X|_U\right)\rightarrow \mathcal{F}$ be a morphism of sheaves of $\mathcal{O}_X$ modules. We need to assign an element of $\mathcal{F}(U)$ with this. $\eta$ comes with maps $\eta(W):j_!\left(\mathcal{O}_X|_U\right)(W)\rightarrow \mathcal{F}(W)$ for each $W\subseteq U$. To get an element of $\mathcal{F}(U)$ its only natural to consider $\eta(U):j_!\left(\mathcal{O}_X|_U\right)(U)\rightarrow \mathcal{F}(U)$ i.e., $\eta(U):\mathcal{F}(U)\rightarrow \mathcal{F}(U)$. We have $\eta(U)(1)\in \mathcal{F}(U)$ an element of $\mathcal{F}(U)$. I am almost sure that this is an isomorphism. Any suggestion on better proof is welcome. Any suggestion on how they come up with that example is welcome. There are several mistakes in your proof, but these are just technical, and you've the right idea. First, this is not true that $j_!\mathcal{F}(V)=\emptyset$ if $V\not\subset U$. Well, if anything, it should be $0$ instead of $\emptyset$ since we are talking of $\mathcal{O}_X$-modules. But in fact, it might not be $0$. The sheaf $j_!\mathcal{F}$ is defined to be the sheaf associated to the presheaf $$j_p\mathcal{F}:V\mapsto\left\{\begin{array}{ll}\mathcal{F}(U) & \text{if V\subset U}\\ 0 & \text{if V\not\subset U}\end{array} \right.$$ It does not change your proof, since from the universal property of the associated sheaf, you only need to construct an isomorphism $\operatorname{Hom}_{PSh(X,\mathcal{O}_X)}(j_p\mathcal{O}_X|_U,\mathcal{F})\simeq\mathcal{F}(U)$. Then, I don't know if this is a typo or a confusion, but you wrote (twice) that maps $j_!\mathcal{O}_{X}|_{U}(V)\rightarrow\mathcal{F}(V)$ correspond to maps $\mathcal{F}(V)\rightarrow\mathcal{F}(V)$. They correspond to maps $\mathcal{O}_X(V)\rightarrow\mathcal{F}(V)$ of course. It doesn't change the rest your argument though (in fact, it makes it correct). So now, you just need to show that the maps you constructed are inverse to each other. This is easy, I let you write the details. Now let me explained how one comes up to this construction. It is worth knowing the following facts : • $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})=\mathcal{F}(X)$. This is a sheaf-theoretic version of the isomorphism $\operatorname{Hom}_A(A,M)=M$ if $A$ is any ring and $M$ any $A$-module. • $\mathcal{O}_X|_U=\mathcal{O}_U$ (if $X$ is a scheme). More is true : if $\mathcal{F}$ is any $\mathcal{O}_X$-module, $\mathcal{F}|_{U}:=j^{-1}\mathcal{F}$ is already an $\mathcal{O}_U$-module and $\mathcal{F}|_U=j^*\mathcal{F}$. • The functor $j_!:\operatorname{Sh}(U,\mathbb{Z})\rightarrow\operatorname{Sh}(X,\mathbb{Z})$ is the left adjoint to $j^{-1}$. If moreover $X$ (and hence $U$) is a scheme, then for every $\mathcal{O}_U$-module $\mathcal{F}$, the sheaf $j_!\mathcal{F}$ is an $\mathcal{O}_X$-module and the functor $j_!$ is the left adjoint of $j^*:\mathcal{O}_X-mod\rightarrow\mathcal{O}_U-mod$. Using these fact you get $$\operatorname{Hom}_{\mathcal{O}_X}(j_!\mathcal{O}_X|_U,\mathcal{F})=\operatorname{Hom}_{\mathcal{O}_U}(\mathcal{O}_U,\mathcal{F}|_U)=\mathcal{F}(U)$$ Of course, to prove these three facts, you need to do something very similar to what you have done, so this does not simplify the proof in any way, but may just gives some insight of what is going on. Finally, there is a fourth fact which is worth knowing, giving another point of view on $j_!\mathcal{O}_U$ : • The object $j_!\mathcal{O}_U$ is the free $\mathcal{O}_U$-module on the representable sheaf $\underline{U}$. Using Yoneda lemma, you can know prove it this way : $$\operatorname{Hom}_{\mathcal{O}_X}(j_!\mathcal{O}_X|_U,\mathcal{F})=\operatorname{Hom}_{\operatorname{Sh}(X)}(\underline{U},\mathcal{F})=\mathcal{F}(U)$$ In the end, the equality $\operatorname{Hom}_{\mathcal{O}_X}(j_!\mathcal{O}_X|_U,\mathcal{F})=\mathcal{F}(U)$ is not so surprising, and even if it is still a bit mysterious now (it won't stay that way long), this equality is quite important. • Thank you so much for you answer. I really mean to say $j_! \mathcal{F}(V)=0$ and not $\emptyset$. I do not know why but i forgot that it was the sheaf associated to what I have written.. As you said, there is an isomorphism between presheave morphisms and sheafification morphisms what I have done is ok but not for the reason what I have stated. – user312648 Sep 15 '17 at 13:30 • It was a typo when I have written $j_!\mathcal{O}_X(V)=\mathcal{F}(V)$. I really mean $\mathcal{O}_X(V)$.. Your explanation for how some one think of such isomorphism is reasonable when I read it for the first time. I will sit and write clearly, if I get some problem I will ask you again. – user312648 Sep 15 '17 at 13:35 • All this typos, mistakes happen because I was thinking about this idea only when I was typing this question and It was late night here. I did not get any idea till I started to type. I was going to simply write (might not have posted) how do I prove this. But then, some how I got the idea and I have typed it and posted. I was sure there will be some gaps but could not see at that point of time. Next time, I will write on a paper first and then type it here so that I will not make these embarassing mistakes. Thank you once again. – user312648 Sep 15 '17 at 13:39 • @cello You're welcome ;) Yes, I was quite confident that it was only typos. Except maybe this very common mistake that $j_!\mathcal{F}(V)=0$ if $V\not\subset U$. This is the only gap in your proof, but as you see, not a big one... – Roland Sep 15 '17 at 14:39 • See if you can comment on math.stackexchange.com/questions/2426597/… and math.stackexchange.com/questions/2426695/… – user312648 Sep 15 '17 at 16:53

https://cs.stackexchange.com/questions/111620/dubins-tsp-crossing-trajectory-theorem

# Dubins TSP crossing trajectory theorem In the Dubins TSP (DTSP for short), one needs to visit a set of given points in the plane, and return to the starting point, minimizing the distance of such a trajectory. The difference with the Euclidean TSP (ETSP for short) however is that the visiting entity is curvature constrained by some $$r$$, meaning the vehicle can, at any moment in time, either go straight, or move on the edge of some circle of radius $$r'\geq r$$. Consider the following theorem, by Le Ny et al. (Theorem 2 in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.84.982&rep=rep1&type=pdf) which we will call Theorem 1 in this post : A trajectory based on an ETSP optimal visit order, is not necessarily an optimal trajectory for the DTSP. (I changed the wording of their theorem to suit this post) Suppose the next theorem, Theorem 2 : There exists instances for which the optimal DTSP visit order creates a crossing ETSP trajectory. My question is : Does Theorem 2 imply Theorem 1 ? The idea is as follows. Since some instances have the optimal DTSP visit order creating a crossing ETSP trajectory $$T$$, this ETSP trajectory $$T$$ is not optimal (due to the triangle inequality). This in turn implies that some optimal ETSP trajectory $$T*$$ on these instances (or their corresponding optimal ETSP visit order) cannot be adapted and optimized to obtain the optimal DTSP trajectory. If it could, then $$T = T*$$, which is a contradiction, since $$T$$ admits a crossing trajectory. ## 1 Answer No, theorem 2 does not imply theorem 1 unless you also prove for that instance that every optimal DTSP visit order creates a crossing ETSP trajectory. There might be another optimal DTSP $$|T'| = |T|$$ visit order that does not result in a crossing ETSP so that we have $$T* = T'$$. • It is for this reason I used " THE optimal DTSP visit order " in Theorem 2, not just " an optimal DTSP visit order ". So, assuming there exists only one unique optimal DTSP visit order creating a crossing ETSP trajectory, Theorem 2 should imply Theorem 1 ? – J. Schmidt Jul 8 at 12:28 • @J.Schmidt If it is unique, then yes. – orlp Jul 8 at 13:01 • @orip You answered my question, but I would like to know also if Theorem 2 is considered "stronger" than Theorem 1 (again, assuming only one optimal DTSP visit order exists) – J. Schmidt Jul 9 at 9:55 • @J.Schmidt Yes it's stronger under that assumption because it implies the other theorem but not the other way around. However I'm not convinced that the assumption holds at all. – orlp Jul 9 at 10:07 • @orip I was thinking it holds for some dense points on the edge of two circles of radius $r$, gathered close to form an 8-like form (without the middle of the 8), i.e. something like this : imgur.com/ZUpreV0 – J. Schmidt Jul 9 at 15:54

https://blackandbrowncreations.shop/en/mathematical-induction-steps-example.html

# Mathematical induction steps example Induction Examples Question 4. Consider the sequence of real numbers de ned by the relations x1 = 1 and xn+1 = p 1+2xn for n 1: Use the Principle of Mathematical Induction to show that xn < 4 for all n 1. Solution. For any n 1, let Pn be the statement that xn < 4. Base Case. The statement P1 says that x1 = 1 < 4, which is true. Inductive Step. An example application would be decomposing the waveform of a musical chord into terms of the intensity of its constituent pitches. The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency domain representation to a function of space or time.. If you have to prove an inequality holds, the trick is to find what you have on each side of (n) assumption on each side of (n+1) assumption. In the induction step of your example, you have (1) 1 + 1 2 + 1 3 + ⋯ + 1 k + 1 k + 1 ≤ k + 1 2 + 1 Which can be organized as (2) ( 1 k + 1) + ( 1 + 1 2 + 1 3 + ⋯ + 1 k) ≤ ( k 2 + 1) + 1 2. Induction Step: Let Assume P ( k) is true, that is [Induction Hypothesis] Prove P ( k+ 1) is also true: [by definition of summation] [by I.H.] [by fraction addition] [by distribution] Thus we have proven our claim is true. QED Notice that in this example we used the inductive definition of set of whole numbers. Web. Web. This example explains the style and steps needed for a proof by induction . Question: Prove by induction that Xn k=1 k = n(n+ 1) 2 for any integer n. (⋆) Approach: follow the steps below. (i) First verify that the formula is true for a base case: usually. Web. find my house on google earth The goal of mathematical induction in your case is that you want to show that a certain property holds for all positive integers. For example you may want to prove that n 3 - n is divisible by 3 for all positive integers. Although there are more complicated ways to show this. Definition 4.3.1. Mathematical Induction. To prove that a statement P ( n) is true for all integers , n ≥ 0, we use the principle of math induction. The process has two core steps: Basis step: Prove that P ( 0) is true. Inductive step: Assume that P ( k) is true for some value of k ≥ 0 and show that P ( k + 1) is true. Video / Answer. Proof by Induction Examples First Example For our first example, let's look at how to use a proof by induction to prove that {eq}2 + 4 + 6 + ... + (2n+2) = n^2 + 3n + 2 {/eq} for all. Math video on how to graph a transformation of the greatest integer function (or the floor function and an example of the step function), that reverses the segments. Segments are reversed when the input is negated (additional negative sign) that negates the output. Problem 3. In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student's first example of a proof by mathematical induction. Recall, from Wikipedia: Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step,. Risolvi i problemi matematici utilizzando il risolutore gratuito che offre soluzioni passo passo e supporta operazioni matematiche di base pre-algebriche, algebriche, trigonometriche, differenziali e molte altre. Web. Mathematical Induction Let's begin with an example. Example: A Sum Formula Theore. For any positive integer n, 1 + 2 + ... + n = n (n+1)/2. Proof. (Proof by Mathematical Induction) Let's let P (n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P (n) should be an assertion that for any n is verifiably either true or false.). For example: Claim. For any , . Proof. For the inductive step, assume that for all , . We'll show that To this end, consider the left-hand side. Now we observe that and , so we can apply the inductive assumption with and , to continue: by the definition of the Fibonacci numbers. This completes the inductive step. Now for the base case. Currently I am on mathematical induction and I've faced problem that I simply don't know where to even start and I can't find any examples that I can go on with, so just to clarify I am asking you for example with solution of similar task. ... $\begingroup$ For your smaller example, the induction step will look something like \$(n+1)^3 - (n+1. Computer science is the study of computation, automation, and information. Computer science spans theoretical disciplines (such as algorithms, theory of computation, information theory, and automation) to practical disciplines (including the design and implementation of hardware and software).. hillsong pastor scandal ragland bottom campground wifi ground masala meaning case steiger 600 specs oregon 30 ton log splitter reviews tool to unlock car doors By mathematical induction, the statement is true. We see that the given statement is also true for n=k+1. Hence we can say that by the principle of mathematical induction this statement is valid for all natural numbers n. Example 3: Show that 2 2n-1 is divisible by 3 using the principles of mathematical induction. To prove: 2 2n-1 is divisible by 3. Example 3.3.1 is a classic example of a proof by mathematical induction. In this example the predicate P(n) is the statement Xn i=0 i= n(n+ 1)=2: 3. MATHEMATICAL INDUCTION 87 [Recall the \Sigma-notation": ... In the induction step you assume the induction hypothesis, P(n), for some arbi-trary integer n 0. Write it out so you know what you have. Proof by Induction Suppose that you want to prove that some property P(n) holds of all natural numbers. To do so: Prove that P(0) is true. - This is called the basis or the base case. Prove that for all n ∈ ℕ, that if P(n) is true, then P(n + 1) is true as well. - This is called the inductive step. - P(n) is called the inductive hypothesis. Using Mathematical Induction. Steps 1. Prove the basis step. 2. Prove the inductive step (a) Assume P(n) for arbitrary nin the universe. This is called the induction ... Example 3.3.1 is a classic example of a proof by mathematical induction. In this example the predicate P(n) is the statement Xn i=0 i= n(n+ 1)=2: 3. MATHEMATICAL INDUCTION 87. floating tv stand for 70 inch tv lichen sclerosus vs lichen planus That is how Mathematical Induction works. In the world of numbers we say: Step 1. Show it is true for first case, usually n=1; Step 2. Show that if n=k is true then n=k+1 is also true; How to Do it. Step 1 is usually easy, we just have to prove it is true for n=1. Step 2 is best done this way: Assume it is true for n=k. Principle of Mathematical Induction In mathematics, a proof is a method of communicating mathematical thinking. More speci cally, it is a logical argument which explains why a statement is true or false. Principle of Mathematical Induction Let P(n) be a statement which depends on a variable n 2N (e.g., in Example 1, P(n) was ‘it. Web. You must always follow the three steps: 1) Prove the statement true for some small base value (usually 0, 1, or 2) 2) Form the induction hypothesis by assuming the statement is true up to some fixed value n = k 3) Prove the induction hypothesis holds true for n = k + 1 There is one very important thing to remember about using proof by induction. Mathematical Methods for Physics and Engineering Apr 14 2020 The third edition of this highly acclaimed undergraduate textbook is suitable for teaching all the mathematics for an undergraduate course in any of the physical sciences. As well as lucid descriptions of all the topics and many worked examples, it contains over 800 exercises. A proof by induction has two steps: Discrete mathematics: Introduction to proofs. 1. Base Case: We prove that the statement is true for the first case (usually, this step is trivial). 2. Induction Step: Assuming the statement is true for N = k (the induction hypothesis), we prove that it is also true for n = k + 1. Web. Base case: we need to prove that 12| (1 4 – 1 2) = 12| (1- 1) = 0, which is divisible by 12 by definition. Induction step: We assume that the 12| (k 4 – k 2) is true such that (n 4 – n 2) = 12a for some . We then need to show that ( (k+1) 4 – (k+1) 2) = 12b for some . My approach would be to a direct proof such that. shock 4 letters What are the applications of mathematical induction? An example of the application of mathematical induction in the simplest case is the proof that the sum of the first n odd positive integers is n 2 —that is, that (1.) 1 + 3 + 5 +⋯+ (2n − 1) = n 2 for every positive integer n. Let F be the class of integers for which equation (1.). Web. Papers is another innovative initiative from Disha Publication. This book provides the excellent approach to Master the subject. The book has 10 key ingredients that will help you achieve success. 1. Chapter Utility Score: Evaluation of chapters on the basis of different exams. 2. Exhaustive theory based on the syllabus of NCERT books 3. Mathematical Induction Let's begin with an example. Example: A Sum Formula Theore. For any positive integer n, 1 + 2 + ... + n = n (n+1)/2. Proof. (Proof by Mathematical Induction) Let's let P (n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P (n) should be an assertion that for any n is verifiably either true or false.). Examples on Mathematical Induction Example 1: Prove the following formula using the Principle of Mathematical Induction. 1 2 + 3 2 + 5 2 + ... + (2n - 1) 2 = n (2n-1) (2n+1)/3 Solution: Assume P (n): 1 2 + 3 2 + 5 2 + ... + (2n - 1) 2 = n (2n-1) (2n+1)/3 Here we use the concept of mathematical induction across the following three steps. Induction Step: Let Assume P ( k) is true, that is [Induction Hypothesis] Prove P ( k+ 1) is also true: [by definition of summation] [by I.H.] [by fraction addition] [by distribution] Thus we have proven our claim is true. QED Notice that in this example we used the inductive definition of set of whole numbers. Instructor: Is l Dillig, CS311H: Discrete Mathematics Mathematical Induction 5/26 Example 1 I Prove the following statement by induction: 8n 2 Z +: Xn i=1 i = (n )(n +1) 2 I Base case: n = 1 . In this case, P 1 i=1 i = 1 and (1)(1+1) ... I Regular and strong induction only di er in the inductive step I Regular induction:assume P (k) holds and. introduction-to-mathematical-programming-winston-student-solutions 1/10 Downloaded from www.online.utsa.edu on November 9, 2022 by guest ... example approach, user-friendly writing style, and complete Excel 2016 integration. ... proofs by induction, and combinatorial proofs. The book contains over 470 exercises, including 275 with. For n = 1 S1 = 1 = 12 The second part of mathematical induction has two steps. The first step is to assume that the formula is valid for some integer k. The second step is to use this assumption to prove that the formula is valid for the next integer, k + 1. 2. Assume Sk = 1 + 3 + 5 + 7 + . . . + (2k-1) = k2 is true, show that Sk+1 = (k + 1)2. Example 3.3.1 is a classic example of a proof by mathematical induction. In this example the predicate P(n) is the statement Xn i=0 i= n(n+ 1)=2: 3. MATHEMATICAL INDUCTION 87 [Recall the \Sigma-notation": ... In the induction step you assume the induction hypothesis, P(n), for some arbi-trary integer n 0. Write it out so you know what you have. Monthly 110(2003), 561-573 and Cvijovi´c and J. Klinowski, J. Com-put. Appl. Math . 142 (2002), 435-439. We provide an explicit ex-pression for the kernel of the integral operator introduced in the first paper. This explicit expression considerably simplifies the calculation . By using this site, you agree to. Mathematical induction is a method of proof used to prove a series of different propositions, say $$P_1,\ P_2,P_3,\ldots P_n$$. A useful analogy to help think about mathematical induction is that. Mathematical Induction Steps. Below are the steps that help in proving the mathematical statements easily.. Summary. The paper "The Scientific Revolution" discusses that it has been shown how the world turned from concepts of magic and invisible spirits to one of cause and effect, even when the specific mechanisms were not yet directly observable. Since then, science has continued to evolve. Download full paper File format: .doc, available for. Therefore, by mathematical induction, the given formula is valid for all n ∈ N. n\in \mathbb{N}. n ∈ N. Practice Example: Show with the help of mathematical induction that the sum of first n n n odd natural numbers is given by the formula n 2 n^2 n 2. Solution:. Web. Web. Request PDF | Complex Modeling of Inductive and Deductive Reasoning by the Example of a Planimetric Problem Solver | The paper is devoted to the problem of computer simulation of inductive and. Combinatory logic is a notation to eliminate the need for quantified variables in mathematical logic. It was introduced by Moses Schönfinkel [1] and Haskell Curry , [2] and has more recently been used in computer science as a theoretical model of computation and also as a basis for the design of functional programming languages .. Jan 03, 2017 · For example, we may want to prove that 1 + 2 + 3 + + n = n (n + 1)/2. In a proof by induction, we generally have 2 parts, a basis and the inductive step. The basis is the simplest version of .... Web. Step 1 Show that S_1 is true. Statement S_1 is. 1=\frac{1(1+1)}{2}. Simplifying on the right, we obtain 1 = 1. This true statement shows that S_1 is true. Step 2 Show that if {S}_{k} is true, then {S}_{k+1} is true. Using S_k and S_{k+1} from Example 1(a), show that the truth of S_k, 1+2+3+\cdots+k=\frac{k(k+1)}{2}, implies the truth of S_{k+1},. Such a reaction may be considered as produced by the method of mathematical induction. 4.3 The Principle of Mathematical Induction Suppose there is a given statement P(n) involving the natural number n such that (i) The statement is true for n = 1, i.e., P(1) is true, and (ii) If the statement is true for n = k (where k is some positive integer .... Step 1 − Consider an initial value for which the statement is true. It is to be shown that the statement is true for n = initial value. Step 2 − Assume the statement is true for any value of n = k. Then prove the statement is true for n = k+1. Web. We have a fantastic induction process which takes you through every part of our business, we will invest in you every step of the way in your career with us; Hybrid working; Excellent Refer a friend scheme; Generous Employee discount scheme, up to 50% off our product; Health and well-being initiatives including access to mindfulness and yoga. Web. • Mathematical induction is valid because of the well ordering property. • Proof: -Suppose that P(1) holds and P(k) →P(k + 1) is true for all positive integers k. -Assume there is at least one positive integer n for which P(n) is false. Then the set S of positive integers for which P(n) is false is nonempty. -By the well-ordering property, S has a least element, say m. In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student's first example of a proof by mathematical induction. Recall, from Wikipedia: Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step,. Knowledge representation and knowledge engineering allow AI programs to answer questions intelligently and make deductions about real-world facts.. A representation of "what exists" is an ontology: the set of objects, relations, concepts, and properties formally described so that software agents can interpret them.. binize dongle What is mathematical induction example? Mathematical induction can be used to prove that an identity is valid for all integers n≥1. Here is a typical example of such an identity: 1+2+3+⋯+n=n(n+1)2. More generally, we can use mathematical induction to prove that a propositional function P(n) is true for all integers n≥1. Weak Induction : The step that you are currently stepping on Strong Induction : The steps that you have stepped on before including the current one 3. Inductive Step : Going up further based on the steps we assumed to exist Components of Inductive Proof Inductive proof is composed of 3 major parts : Base Case, Induction Hypothesis, Inductive Step. Step 1: In step 1, assume n= 1, so that the given statement can be written as P (1) = 22 (1)-1 = 4-1 = 3. So 3 is divisible by 3. (i.e.3/3 = 1) Step 2: Now, assume that P (n) is true for all the natural number, say k Hence, the given statement can be written as P (k) = 22k-1 is divisible by 3. The deductive nature of mathematical induction derives from its basis in a non-finite number of cases, in contrast with the finite number of cases involved in an enumerative induction procedure like proof by exhaustion. Both mathematical induction and proof by exhaustion are examples of complete induction. Complete induction is a masked type of .... Request PDF | Complex Modeling of Inductive and Deductive Reasoning by the Example of a Planimetric Problem Solver | The paper is devoted to the problem of computer simulation of inductive and. Web. The Principle of Mathematical Induction. Whenever the statement holds for n = k, it must also hold for n = k + 1. then the statement holds for for all positive integers, n . In an inductive argument, demonstrating the first condition above holds is called the basis step, while demonstrating the second is called the inductive step. Mathematical Induction: Example • Let P(n) be the sentence “n cents postage can be obtained using 3¢ and 5¢ stamps”. • Want to show that “P(k) is true” implies “P(k+1) is true” for any k ≥ 8¢. • 2 cases: 1) P(k) is true and the k cents contain at least one 5¢. 2) P(k) is true and the k cents do not contain any 5¢. 3. An example of the application of mathematical induction in the simplest case is the proof that the sum of the first n odd positive integers is n2 —that is, that (1.) 1 + 3 + 5 +⋯+ (2 n − 1) = n2 for every positive integer n. Let F be the class of integers for which equation (1.) holds; then the integer 1 belongs to F, since 1 = 1 2. . explicitly to label the base case, inductive hypothesis, and inductive step. This is common to do when rst learning inductive proofs, and you can feel free to label your steps in this way as needed in your own proofs. 1.1 Weak Induction: examples Example 2. Prove the following statement using mathematical induction: For all n 2N, 1 + 2 + 4. Mar 19, 2017 · Update 2021 March: You can now export the data direct from Power BI Desktop using my tool, Power BI Exporter. Read more here. Update 2019 April: If you're interested in exporting the data model from either Power BI Desktop or Power BI Service to CSV or SQL Server check this out. The method explained here Continue reading Exporting Data from. Proof by Induction Step 1: Prove the base case This is the part where you prove that P (k) P (k) is true if k k is the starting value of your statement. The base case is usually showing that our statement is true when n=k n = k. Step 2: The inductive step This is where you assume that P (x) P (x) is true for some positive integer x x. Examples on Mathematical Induction Example 1: Prove the following formula using the Principle of Mathematical Induction. 1 2 + 3 2 + 5 2 + ... + (2n - 1) 2 = n (2n-1) (2n+1)/3 Solution: Assume P (n): 1 2 + 3 2 + 5 2 + ... + (2n - 1) 2 = n (2n-1) (2n+1)/3 Here we use the concept of mathematical induction across the following three steps. popular alternative indie songs spanske serije sa prevodom 2021 Mathematical Induction - Two Steps 1. Prove that it works for one case. ... Mathematical Induction Proof Example: For any natural number n, n 3 + 2n is divisible by 3 Mathematical Induction Proof Example: For any natural number n ≥ 4, n! > 2 n. Show Step-by-step Solutions. Try the free Mathway calculator and problem solver below to practice. How do you prove by induction in math? Outline for Mathematical Induction. Base Step: Verify that P(a) is true. Inductive Step: Show that if P(k) is true for some integer k≥a, then P(k+1) is also true. Assume P(n) is true for an arbitrary integer, k with k≥a. Conclude, by the Principle of Mathematical Induction (PMI) that P(n) is true for. Maintaining the equal inter-domino distance ensures that P (k) ⇒ P (k + 1) for each integer k ≥ a. This is the inductive step. Examples Example 1: For all n ≥ 1, prove that, 1 2 + 2 2 + 3 2 .n 2 = {n (n + 1) (2n + 1)} / 6 Solution: Let the given statement be P (n), Now, let's take a positive integer, k, and assume P (k) to be true i.e.,. A set is the mathematical model for a collection of different things; a set contains elements or members, which can be mathematical objects of any kind: numbers, symbols, points in space, lines, other geometrical shapes, variables, or even other sets. The set with no element is the empty set; a set with a single element is a singleton.A set may have a finite number of elements or be an. mahindra 4550 backhoe attachment Proof by Induction Suppose that you want to prove that some property P(n) holds of all natural numbers. To do so: Prove that P(0) is true. - This is called the basis or the base case. Prove that for all n ∈ ℕ, that if P(n) is true, then P(n + 1) is true as well. - This is called the inductive step. - P(n) is called the inductive hypothesis. Hence, by the principle of mathematical induction p(k) is true for all n ∈ N x 2n − y 2n is divisible by (x + y) for all n ∈N Example 2 : By the principle of Mathematical induction, prove that, for n ≥ 1, 1 2 + 2 2 + 3 2 + · · · + n 2 > n 3 / 3. "/> gore discord servers. Web. Web. This example explains the style and steps needed for a proof by induction . Question: Prove by induction that Xn k=1 k = n(n+ 1) 2 for any integer n. (⋆) Approach: follow the steps below. (i) First verify that the formula is true for a base case: usually. For example. In the prove n 3 - n is divisible by 3 question, we assume when n = k that k 3 - k is divisible by 3 for some k. Now we consider the n = k+1 case. We want to prove that (k+1) 3 - (k+1) is divisible by 3. Now begins the messing around bit where we do what we can to find a k 3 - k buried around in here. For example, suppose we wanted to prove that the sum of the!rst n positive integers is equal to ( n(n + 1)) / 2. The sum of the!rst npositive integers is given by the formula The set of positive integers is an in!nite set, so the answer to question 1 is yes. These examples of mathematical induction example of incidence and work else, the kth step of a property to let me colour code below. This enables us to conclude that all the statements are true. Between weak and strong induction An example of where to use strong induction is given. Register free to be proved via induction on which of. For example, the angular gyrus plays a critical role in distinguishing left from right by integrating the conceptual understanding of the language term "left" or "right" with its location in space. Furthermore, the angular gyrus has been associated with orienting in three dimensional space, not because it interprets space, but because it may .... Mathematical Induction Let's begin with an example. Example: A Sum Formula Theore. For any positive integer n, 1 + 2 + ... + n = n (n+1)/2. Proof. (Proof by Mathematical Induction) Let's let P (n) be the statement "1 + 2 + ... + n = (n (n+1)/2." (The idea is that P (n) should be an assertion that for any n is verifiably either true or false.). Examples of Scalar Product of Two Vectors : Work done is defined as scalar product as W = F · s, Where F is a force and s is a displacement produced by the force Power is defined as a scalar product as P = F · v, Where F is a force and v is a velocity. ... mathematical induction calculator with steps. part time remote work from home jobs. Definition 4.3.1. Mathematical Induction. To prove that a statement P ( n) is true for all integers , n ≥ 0, we use the principle of math induction. The process has two core steps: Basis step: Prove that P ( 0) is true. Inductive step: Assume that P ( k) is true for some value of k ≥ 0 and show that P ( k + 1) is true. Video / Answer. Web. A set is the mathematical model for a collection of different things; a set contains elements or members, which can be mathematical objects of any kind: numbers, symbols, points in space, lines, other geometrical shapes, variables, or even other sets. The set with no element is the empty set; a set with a single element is a singleton.A set may have a finite number of elements or be an. Middlesex University will create new facilities in the West Stand which enable students to work with elite athletes. A state-of-the-art sports facility providing new educational and career opportunities for Barnet residents and Middlesex University students has moved a step closer, with Saracens Copthall LLP and the University signing a new deal for use of the. to use Mathematical induction method to study Goldbach's strong conjecture. We use two properties that are satisfied for prime numbers, and based on these two properties, we show a way that, may be, it can be used to analyze and approach this conjecture by the Mathematical induction method. Web. Step-by-step solutions for proofs: trigonometric identities and mathematical induction. All Examples › Pro Features › Step-by-Step Solutions ... Examples for. Step-by-Step Proofs. Trigonometric Identities See the steps toward proving a trigonometric identity: does sin(θ)^2 + cos(θ)^2 = 1?. Examples of Scalar Product of Two Vectors : Work done is defined as scalar product as W = F · s, Where F is a force and s is a displacement produced by the force Power is defined as a scalar product as P = F · v, Where F is a force and v is a velocity. ... mathematical induction calculator with steps. part time remote work from home jobs. 00:00:57 What is the principle of induction? Using the inductive method (Example #1) Exclusive Content for Members Only 00:14:41 Justify with induction (Examples #2-3) 00:22:28 Verify the inequality using mathematical induction (Examples #4-5) 00:26:44 Show divisibility and summation are true by principle of induction (Examples #6-7). ghost bath starmourner Mathematical Induction. The process to establish the validity of an ordinary result involving natural numbers is the principle of mathematical induction. Working Rule. Let n 0 be a fixed integer. Suppose P (n) is a statement involving the natural number n and we wish to prove that P (n) is true for all n ≥n 0. 1. Electric Motor Drives and Its Applications with Simulation Practices provides comprehensive coverage of the concepts of electric motor drives and their applications, along with their simulation using MATLAB. The book helps engineers and students improve their software skills by learning to simulate various electric drives and applications and assists with new ideas in the simulation of. Structural induction is a proof methodology similar to mathematical induction, only instead of working in the domain of positive integers (N) it works in the domain of such recursively de ned structures! It is terri cally useful for proving properties of such structures. Its structure is sometimes \looser" than that of mathematical induction. Section 3.7 Mathematical Induction Subsection 3.7.1 Introduction, First Example. In this section, we will examine mathematical induction, a technique for proving propositions over the positive integers. Mathematical induction reduces the proof that all of the positive integers belong to a truth set to a finite number of steps. Class XI students in an effective way for Mathematics. S. Chand's Smart Maths book 8 Sheela Khandelwall S Chand's Smart Maths is a carefully graded Mathematics series of 9 books for the children of KG to Class 8. The series adheres to the National Curriculum Framework and the books have been designed in accordance with the. The correct answer: Inductive. 5. Some cookies are burnt. Some burnt things are good to eat. So some cookies are good to eat. The correct answer: Inductive. 6. All reptiles ever examined are cold-blooded. Dinosaurs resemble reptiles in many ways. So dinosaurs were cold-blooded. The correct answer : Inductive. 7. All mollusks are invertebrates. By mathematical induction , the statement is true. We see that the given statement is also true for n=k+1. Hence we can say that by the principle of mathematical induction this statement is valid for all natural numbers n. Example 3: Show that 2 2n-1 is divisible by 3 using the principles of mathematical >induction. Web. Mathematical Structures for Computer Science has long been acclaimed for its clear presentation of essential concepts and its exceptional range of applications relevant to computer science majors. Now with this new edition, it is the first discrete mathematics textbook revised to meet the proposed new ACM/IEEE standards for the course. This was the first automated deduction system to demonstrate an ability to solve mathematical problems that were announced in the Notices of the American Mathematical Society before solutions were formally published. [citation needed] First-order theorem proving is one of the most mature subfields of automated theorem proving. The logic is .... Math induction is just a shortcut that collapses an infinite number of such steps into the two above. In Science, inductive attitude would be to check a few first statements, say, P (1), P (2), P (3), P (4), and then assert that P (n) holds for all n. The inductive step "P (k) implies P (k + 1)" is missing. Needless to say nothing can be proved. Mathematical Induction is a special way of proving things. It has only 2 steps: Step 1. Show it is true for the first one Step 2. Show that if any one is true then the next one is true Then all are true Have you heard of the "Domino Effect"? Step 1. The first domino falls Step 2. When any domino falls, the next domino falls. Web. veronica mars season 6 what is franklin tn famous for For example, f (Chen) = NU123456. The co-domain of f is the set of ID numbers between NU000000 and NU999999 (b) Let g be the function that maps each student to a Northeastern campus. For example, g (Chen) = SiliconValley. The co-domain of g is {Vancouver, Seattle, Boston, San Francisco, Silicon Valley, Portland}. For n = 1 S1 = 1 = 12 The second part of mathematical induction has two steps. The first step is to assume that the formula is valid for some integer k. The second step is to use this assumption to prove that the formula is valid for the next integer, k + 1. 2. Assume Sk = 1 + 3 + 5 + 7 + . . . + (2k-1) = k2 is true, show that Sk+1 = (k + 1)2. The goal of mathematical induction in your case is that you want to show that a certain property holds for all positive integers. For example you may want to prove that n 3 - n is divisible by 3 for all positive integers. Although there are more complicated ways to show this. Mathematical Induction is a special way of proving things. It has only 2 steps: Step 1. Show it is true for the first one Step 2. Show that if any one is true then the next one is true Then all are true Have you heard of the "Domino Effect"? Step 1. The first domino falls Step 2. When any domino falls, the next domino falls. A set is the mathematical model for a collection of different things; a set contains elements or members, which can be mathematical objects of any kind: numbers, symbols, points in space, lines, other geometrical shapes, variables, or even other sets. The set with no element is the empty set; a set with a single element is a singleton.. log to exponential form practice function of steroids chamberlain garage door opener repair near me retool oauth2 Web. Weak Induction : The step that you are currently stepping on Strong Induction : The steps that you have stepped on before including the current one 3. Inductive Step : Going up further based on the steps we assumed to exist Components of Inductive Proof Inductive proof is composed of 3 major parts : Base Case, Induction Hypothesis, Inductive Step. detailed, step-by-step solutions to more than half of the odd-numbered end-of-chapter problems from the text. All solutions follow the same four-step problem-solving framework used in the textbook. College Physics Raymond A. Serway 2003 For Chapters 15-30, this manual contains detailed solutions to approximately 12 problems per chapter. These. Each daily dose needs two 5mg tablets and four 2mg tablets, which is ten 5mg tablets and twenty 2mg tablets for 5/7. A further 7mg is required for the next two days - which is two 5mg tablets and two 2mg tablets for 2/7. In total, the patient needs twelve 5mg tablets (60mg) and twenty-two 2mg tablets (44mg) - leading to the required dose of. Process of Induction The reasoning process from a particular result to a general result is called induction. Example: \ (44\) is divisible by \ (2.\) Hence, all integers ending with \ (4\) are divisible by \ (2.\) Here, from a particular statement, we are drawing a general conclusion. Matteo Dell’Amico provides this feature in Italian Index Ad Hominem [page not ready] Ad Hominem Tu Quoque [page not ready] Appeal to Authority [page not ready] Appeal to Belief [page not ready] Appeal to Common Practice [page not ready] Appeal to Consequences of a Belief [page not ready] Appeal to Emotion [page not ready] Appeal to []. Example: Use mathematical induction to prove that 2n < n!, for every integer n 4. Solution: Let P(n) be the proposition that 2n < n!. –Basis: P(4) is true since 24 = 16 < 4! = 24. ... –INDUCTIVE STEP: The inductive hypothesis states that P(j) holds for 12 j k, where k 15. Assuming the inductive hypothesis, it can. Mathematical Induction This sort of problem is solved using mathematical induction. Some key points: Mathematical induction is used to prove that each statement in a list of statements is true. Often this list is countably in nite (i.e. indexed by the natural numbers). It consists of four parts: I a base step,. Instructor: Is l Dillig, CS311H: Discrete Mathematics Mathematical Induction 5/26 Example 1 I Prove the following statement by induction: 8n 2 Z +: Xn i=1 i = (n )(n +1) 2 I Base case: n = 1 . In this case, P 1 i=1 i = 1 and (1)(1+1) ... I Regular and strong induction only di er in the inductive step I Regular induction:assume P (k) holds and. Mathematical induction is a method of proof used to prove a series of different propositions, say $$P_1,\ P_2,P_3,\ldots P_n$$. A useful analogy to help think about mathematical induction is that. Mathematical Induction Steps. Below are the steps that help in proving the mathematical statements easily.. PRINCIPLE OF MATHEMATICAL INDUCTION. Mathematical induction is the process of proving a general theorem or formula involving the positive integer ‘n’ from particular cases.<br>A proof by mathematical induction consists of the following three steps: (1) Show by actual substitution that the theorem is true for n = 1 or initial value. Proof by strong induction Step 1. Demonstrate the base case: This is where you verify that P (k_0) P (k0) is true. In most cases, k_0=1. k0 = 1. Step 2. Prove the inductive step: This is where you assume that all of P (k_0) P (k0), P (k_0+1), P (k_0+2), \ldots, P (k) P (k0 +1),P (k0 + 2),,P (k) are true (our inductive hypothesis). 4 bridges arts festival maple leaf nippon Mar 19, 2017 · Update 2021 March: You can now export the data direct from Power BI Desktop using my tool, Power BI Exporter. Read more here. Update 2019 April: If you're interested in exporting the data model from either Power BI Desktop or Power BI Service to CSV or SQL Server check this out. The method explained here Continue reading Exporting Data from. Mathematical induction involves a combination of the general problem solving methods of. the special case. proving the theorem true for n = 1 or n0. the subgoal method -- dividing the goal into 2 parts. proving it is true for n0. showing that if it is true for k, then it is true for k + 1. In particular, literature on proof - and specifically, mathematical induction - will be presented, and several worked examples will outline the key steps involved in solving problems. Here is a simple example of the use of induction. We want to prove that the sum of the first n squares is n (n+1) (2n+1)/6. The expression is mathematical shorthand for "the sum for i running from 0 to n of i2 ", or 0 2 + 1 2 + 2 2 + ... + n 2 We wish to show that this property is true for all n. The variable we will do induction on is k. Basis. The proof involves two steps : Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n. Step 2: We assume that P (k) is true and establish that P (k+1) is also true Problem 1 Use mathematical induction to prove that 1 + 2 + 3 + ... + n = n (n + 1) / 2 for all positive integers n. An example application would be decomposing the waveform of a musical chord into terms of the intensity of its constituent pitches. The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency domain representation to a function of space or time.. induction, and combinatorial proofs. The book contains over 470 exercises, including 275 ... step of mathematical problems can be derived without any gap or jump in steps. Thus, readers can ... Testing with One Sample Chapter 10 Hypothesis Testing with Two Samples Chapter 11 The Chi-Square. one of those in nite steps taken. To avoid the tedious steps, we shall introduce Mathematical Induction in solving these problems, which the inductive proof involves two stages: 1. The Base Case: Prove the desired result for number 1. 2. The Inductive Step: Prove that if the result is true for any k, then it is also true for the number k+ 1. As described in the example "Time Series Forecasting Using Deep Learning", we can predict futher values based on the closer predicted results and repeat this process to accomplish long steps forcasting.But as shown in the following picture, why do we need to reset the trained net through "resetState", what states are reset in this process?. Learn how to use Mathematical Induction in this free math video tutorial by Mario's Math Tutoring. We go through two examples in this video.0:30 Explanation. uchigatana faith build elden ring waps eligibility chart 2023 The process has two core steps: Basis step: Prove that P (0) P ( 0) is true. Inductive step: Assume that P (k) P ( k) is true for some value of k ≥ 0 k ≥ 0 and show that P (k+1) P ( k + 1) is true. Video / Answer 🔗 Note 4.3.2. You can think of math induction like an infinite ladder. First, you put your foot on the bottom rung. Request PDF | Complex Modeling of Inductive and Deductive Reasoning by the Example of a Planimetric Problem Solver | The paper is devoted to the problem of computer simulation of inductive and. Risolvi i problemi matematici utilizzando il risolutore gratuito che offre soluzioni passo passo e supporta operazioni matematiche di base pre-algebriche, algebriche, trigonometriche, differenziali e molte altre. Student Solution Manual for Foundation Mathematics for the Physical Sciences Oct 15 2022 This Student Solution Manual provides complete solutions to all the odd-numbered problems in Foundation Mathematics for the Physical Sciences. It takes students through each problem step-by-step, so they can clearly see how the. How do you prove by induction in math? Outline for Mathematical Induction. Base Step: Verify that P(a) is true. Inductive Step: Show that if P(k) is true for some integer k≥a, then P(k+1) is also true. Assume P(n) is true for an arbitrary integer, k with k≥a. Conclude, by the Principle of Mathematical Induction (PMI) that P(n) is true for. Solved Examples of Mathematical Induction Problem 1: (proof of the sum of first n natural numbers formula by induction) Prove that 1 + 2 + 3 + ⋯ + n = n ( n + 1) 2 Solution: Let P ( n) denote the statement 1 + 2 + 3 + + n = n ( n + 1) 2. (Base case) Put n = 1. Note that 1 = 1 ( 1 + 1) 2. So P ( 1) is true. We call definitions like this completely inductive definitions because they look back more than one step. Exercise. Compute the first 10 Fibonacci numbers. Typically, proofs involving the Fibonacci numbers require a proof by complete induction. For example: Claim. For any , . Proof. For the inductive step, assume that for all , . We'll show that. Web. The equation editor uses a markup language to represent formulas. For example, %beta creates the Greek character beta (β). This markup is designed to read similar to English whenever possible. For example, a over b produces a fraction: To insert a numbered formula in Writer, type fn then press the F3 key. Additional References. Principle of Mathematical Induction In mathematics, a proof is a method of communicating mathematical thinking. More speci cally, it is a logical argument which explains why a statement is true or false. Principle of Mathematical Induction Let P(n) be a statement which depends on a variable n 2N (e.g., in Example 1, P(n) was ‘it. Web. The process of induction involves the following steps. Principle of Mathematical Induction Examples Question 1 : By the principle of mathematical induction, prove that, for n ≥ 1 1.2 + 2.3 + 3.4 + · · · + n. (n + 1) = n (n + 1) (n + 2)/3 Solution : Let p (n) = 1.2 + 2.3 + 3.4 + · · · + n. (n + 1) = n (n + 1) (n + 2)/3 Step 1 : put n = 1. Example 01 Q.Prove by mathematical induction that the sum of the first n natural number is \frac{n\left( n+1 \right)}{2}. Solution: We have prove that, $1+2+3+.+n=\frac{n\left( n+1 \right)}{2}$ Step 1:For n = 1, left side = 1 and right side = \frac{1\left( 1+1 \right)}{2}=1. Hence the statement is true for n = 1. lambda chi alpha letters session39s video game Mathematical Induction is a special way of proving things. It has only 2 steps: Step 1. Show it is true for the first one Step 2. Show that if any one is true then the next one is true Then all are true Have you heard of the "Domino Effect"? Step 1. The first domino falls Step 2. When any domino falls, the next domino falls. Step 1 − Consider an initial value for which the statement is true. It is to be shown that the statement is true for n = initial value. Step 2 − Assume the statement is true for any value of n = k. Then prove the statement is true for n = k+1. The Principle of Mathematical Induction. Whenever the statement holds for n = k, it must also hold for n = k + 1. then the statement holds for for all positive integers, n . In an inductive argument, demonstrating the first condition above holds is called the basis step, while demonstrating the second is called the inductive step. Solved Examples of Mathematical Induction Problem 1: (proof of the sum of first n natural numbers formula by induction) Prove that 1 + 2 + 3 + ⋯ + n = n ( n + 1) 2 Solution: Let P ( n) denote the statement 1 + 2 + 3 + + n = n ( n + 1) 2. (Base case) Put n = 1. Note that 1 = 1 ( 1 + 1) 2. So P ( 1) is true. Computer and Information Sciences | Fordham. Jan 03, 2017 · For example, we may want to prove that 1 + 2 + 3 + + n = n (n + 1)/2. In a proof by induction, we generally have 2 parts, a basis and the inductive step. The basis is the simplest version of .... explicitly to label the base case, inductive hypothesis, and inductive step. This is common to do when rst learning inductive proofs, and you can feel free to label your steps in this way as needed in your own proofs. 1.1 Weak Induction: examples Example 2. Prove the following statement using mathematical induction: For all n 2N, 1 + 2 + 4. School of Mathematics & Statistics | Science - UNSW Sydney. Step 1: In step 1, assume n= 1, so that the given statement can be written as P (1) = 22 (1)-1 = 4-1 = 3. So 3 is divisible by 3. (i.e.3/3 = 1) Step 2: Now, assume that P (n) is true for all the natural number, say k Hence, the given statement can be written as P (k) = 22k-1 is divisible by 3. introduction-to-mathematical-programming-winston-student-solutions 1/10 Downloaded from www.online.utsa.edu on November 9, 2022 by guest ... example approach, user-friendly writing style, and complete Excel 2016 integration. ... proofs by induction, and combinatorial proofs. The book contains over 470 exercises, including 275 with. . The most common example of an inductively defined set is the set of nonnegative integers N= { 0, 1, 2, ... }, also called the natural numbers. This set can be generated from the base element 0 and the successor function inc, where inc(x) = x + 1. Induction over the natural numbers is often called mathematical induction. Web.

https://open.kattis.com/problems/target

Kattis # Target Practice In target practice you sometimes shoot at cardboard figures. In this problem a target is described by a simple polygon which is the outline of the target. For each target you will find out where a number of shots fired at the target struck. Some may have hit the target and some may have missed. Your job is to determine for each shot whether it hit or whether it missed. If the shot missed, you should determine how far away from the target the shot was. If the shot hit, you should determine how large a margin there was, i.e., how far away from the contour of the target the shot was. ## Input Each test case starts with a number $3 \le n \le 1000$ which is the number of vertices of the target polygon (in clockwise or counter clockwise order). Then follow $n$ lines giving the coordinates $x$ and $y$ of each vertex. Then follows a line with an integer $s \le 100$ which is the number of shots fired at that target. Finally, there are $s$ lines giving the coordinates $x$ and $y$ of each shot. Coordinates are integers between $-10\, 000$ and $10\, 000$ inclusive. The final test case is followed by a line with $n=0$. There are at most $5$ test cases. ## Output For each test case, output “Case $c$” where $c$ is the number of the test case. Then follows a line for every shot fired at the target. For each shot, if it missed, print “Miss! $d$” and if it hit, print “Hit! $d$” where $d$ is the distance mentioned above. If a shot hits the edge of the target, print “Winged!”. The distance should be given with an absolute error of at most $10^{-3}$. Sample Input 1 Sample Output 1 4 0 0 0 10 10 10 10 0 2 3 1 11 11 3 0 0 10 10 10 0 1 5 0 0 Case 1 Hit! 1.000000000 Miss! 1.414213562 Case 2 Winged!

https://proofwiki.org/wiki/Definition:Eigenvalue

# Definition:Eigenvalue ## General Definition Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$. Let $A \in \map B H$ be a bounded linear operator. A scalar $\alpha \in \Bbb F$ is said to be an eigenvalue of $A$ if and only if: $\map \ker {A - \alpha I} \ne \paren {\mathbf 0_H}$ That is, if and only if the bounded linear operator $A - \alpha I$ has nontrivial kernel. ### Point Spectrum The set of all eigenvalues of $A$ is denoted $\map {\sigma_p} A$. In the field of functional analysis, it is frequently called the point spectrum of $A$. ## Definition in $\R^n$ Let $\mathbf A$ be an square matrix of order $n$, and let $\mathbf v$ be a vector, $\mathbf v \in \R^n, \mathbf v \ne \mathbf 0$. If $\mathbf A \mathbf v = \lambda \mathbf v$ for some $\lambda\in \R$, which is a scalar, then $\lambda$ is called an eigenvalue of $\mathbf A$ with a corresponding eigenvector $\mathbf v$. The eigenvalues are usually found by solving the characteristic equation of $\mathbf A$, which is given by: $\map \det {\mathbf A - \lambda \mathbf I} = 0$

https://math.stackexchange.com/questions/913934/process-to-show-that-sqrt-2-sqrt3-3-is-irrational?noredirect=1

# Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational How can I prove that the sum $\sqrt 2+\sqrt[3] 3$ is an irrational number ?? • Welcome to MSE! When you post a question, please try to supply some context and indicate your own thoughts on a solution. See here for further guidelines. – vociferous_rutabaga Aug 30 '14 at 14:14 • If it were rational, then $\sqrt[3]{3} \in \Bbb Q(\sqrt 2)$. – Watson Dec 17 '16 at 16:41 • – Watson Nov 25 '18 at 21:24 If $x=\sqrt{2}+\sqrt[3]3$, then $$(x-\sqrt{2})^3=x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=3$$ Thus $$x^3+6x-3=\sqrt{2}(3x^2+2)$$ And $$\frac{x^3+6x-3}{3x^2+2}=\sqrt{2}$$ But if $x$ is a rational, then so is the left hand side of the above equality. However we know $\sqrt{2}$ is not rational. Contradiction, so $x$ is irrational. • The notion of an algebraic integer is not that complicated, and I think this implication is worth knowing. In this answer, I use only Bezout's Identity and some algebra to show that a rational algebraic integer is actually an integer. The proof is only a little more involved algebraically than the proof that $\sqrt2$ is irrational. – robjohn Aug 31 '14 at 2:18 • Similarly, $\dfrac{2x^3-4x+3}{3x^2+2}=\sqrt[3]3$. – Akiva Weinberger Aug 31 '14 at 3:54 If $x=\sqrt2+\sqrt[3]3$, then \begin{align} 3 &=(x-\sqrt2)^3\\ &=x^3-3\sqrt2x^2+6x-2\sqrt2\\ (x^3+6x-3)^2&=2(3x^2+2)^2\\ 0&=x^6-6x^4-6x^3+12x^2-36x+1 \end{align} Thus, $x$ is an algebraic integer. Since $2\lt x\lt3$, $x\not\in\mathbb{Z}$, so $x\not\in\mathbb{Q}$. In this answer, it is shown that a rational algebraic integer is an integer. Alternative approach If $x=\sqrt2$ and $y=\sqrt[3]3$, then $x^2-2=0$ and $y^3-3=0$. Thus, both $x$ and $y$ are algebraic integers. Therefore, $x+y=\sqrt2+\sqrt[3]3$ is also an algebraic integer. Note that this approach, while seeming simpler, requires that one knows that the sum of two algebraic integers is again an algebraic integer. This fact can be shown using some linear algebra. • I am relatively new here, and you have much more reputation than I do, so I hesitate to ask--but I would really like to know, so this is a serious question. Is this answer giving the OP too much information? I tried in my answer to help a little but force the OP to do more work on his own. – Rory Daulton Aug 30 '14 at 14:26 • @robjohn +1 For the last line, it's also possible to let $x=\frac pq$ and prove directly $p|1$ and $q|1$, thus a contradiction. – Jean-Claude Arbaut Aug 30 '14 at 14:32 • @RoryDaulton: Don't hesitate to ask because of anyone's reputation (see this meta thread). I thought about this a bit, and often if I feel a question is a homework problem, I leave a hint. However, sometimes I leave a more complete answer when either the hint I'd give would be too cryptic, or if a more complete answer does not give much more away. In this case, the hint would not give much more than cubing of squaring a polynomial. I may have misjudged. Thanks for your comment; it will affect my decisions in the future. – robjohn Aug 30 '14 at 14:34 • Nice solution ! thank you very much.!!! – Marianna Pap Aug 30 '14 at 15:07 • How does $x$ being algebraic and not an integer imply it's irrational? – Alice Ryhl Aug 30 '14 at 16:03 The following is a solution using some (basic) notions from the theory of field extensions: Assume $\sqrt{2}+\sqrt[3]{3}$ is rational. Then $\sqrt[3]{3}\in\mathbb{Q}(\sqrt{2})$, thus $\sqrt[3]{3}$ is a root of some quadratic polynomial over $\mathbb{Q}$ which leads to a contradiction, since the polynomial $x^3-3$ is irreducible over $\mathbb{Q}$. The same idea can be expressed using only elementary terms. Let $a$ be a rational number, and note that $a-\sqrt{2}$ is a root of the rational polynomial $x^2-2ax+a^2-2$. Thus, if $\sqrt{2}+\sqrt[3]{3}$ is rational, it follows that $\sqrt[3]{3}$ is a root of a quadratic rational polynomial, which is a contradiction as $x^3-3$ is irreducible. Use these facts: $$(\sqrt 2 + \sqrt[3]3) \cdot (-\sqrt 2 + \sqrt[3]3) = -2 + \sqrt[3]9$$ $$(-2 + \sqrt[3]9) \cdot [(-2)^2 - (-2)\sqrt[3]9 + (\sqrt[3]9)^2] = (-2)^3 + (\sqrt[3]9)^3$$ That last number is rational. Using those facts, find a polynomial of degree 6 with integral coefficients for which $\sqrt 2 + \sqrt[3]3$ is a root. Then, using the rational root theorem, show that any root of that polynomial is irrational. Let us know if you need more help in finding that polynomial. ADDED: Robjohn's answer gives a more straightforward way of finding an appropriate 6th-degree polynomial.

https://www.physicsforums.com/threads/centroid-problem.647502/

# Centroid problem 1. Oct 27, 2012 ### 3soteric 1. The problem statement, all variables and given/known data Find the centroid of the region in the first quadrant of the xy-plane bounded by the graphs of the equations: x=0, x=1, y=x-x^2, and y^2=2x 2. Relevant equations xbar: integral of x dA/ integral of dA ybar: integral y dA/ integral of dA 3. The attempt at a solution my attempt at it was not so great, i determined the limit for the x values were from 0 to 1, the y limits were from x-x^2 to square root of 2x. Im not sure if the limits I chose are even valid, but that is what I could grasp. Then integrated dydx respectively with the limits, and ended up with 1/2 as the dA. I tried to find ybar and ended up with 29/15, which is clearly wrong if I have the concept of centroids correct. I attempted xbar but I cant seem to get too far. Can I get some help please? this is a test review and need to understand this concept 2. Oct 28, 2012 ### SammyS Staff Emeritus Hello 3soteric. Welcome to PF ! First of all, sketch the region of interest. The area is not 1/2. Is that what you had? What Is it that you integrated for the area, A, for xbar for ybar? 3. Oct 28, 2012 ### 3soteric thanks for the welcome sam! do i sketch it and upload the image? the area seems to be limited vertically in y sense by x-x^2 and sqrt2x, horizontally by x=0, and x=1, for area yes i ended up with 1/2 but can you show me why is it wrong or the process? for ybar i got 29/15 and i couldnt calculate xbar by the nature of integration but i ended up with 10.8 as the final value if i remember which seems wrong as well 4. Oct 28, 2012 ### SammyS Staff Emeritus It seems that you do understand the region correctly. The area is simply $\displaystyle \text{A}=\int_{0}^{1} (\sqrt{2x}-(x-x^2))\,dx\ .$ The integral for $\bar{x}$ should be less complicated than the integral for $\bar{y}\ .$ 5. Oct 28, 2012 ### 3soteric yo sam thanks ! i ended up getting the values for x bar, y bar respectively as .5141, .408 which make much more sense than the values i derived at the beginning of the problem. 6. Oct 28, 2012 ### SammyS Staff Emeritus What did you get for the area? 7. Oct 28, 2012 ### 3soteric i ended up getting .776! 8. Oct 28, 2012 ### SammyS Staff Emeritus That's correct, (if it's not a factorial LOL). I got something a bit different for $\bar{x}\,,$ but I wasn't all that careful in getting it. 9. Oct 28, 2012 ### 3soteric LOL its not a factorial ! phew! im sure i have the right idea regardless right? by the way you think you can help me with the new problem i posted in this same section about cylindrical coordinates LOL, if you have time of course. ive been trying for the longest and dont seem to understand the concept

http://mathhelpforum.com/calculus/141274-limsup-calculation.html

# Math Help - limsup calculation 1. ## limsup calculation How do I solve the following limit? 2. Originally Posted by GIPC How do I solve the following limit? The $\frac{n}{3} - \big{[}{\frac{n}{3}}\big{]}$ part only takes three values. $0$, $\frac{1}{3}$ and $\frac{2}{3}$. If you want to prove this set n to be equal to $3k$, $3k+1$ and $3k+2$ where $k$ is a non negative integer and see what this gives you. Then just find which is larger. $\sin(0)$, $\sin(\tfrac{\pi}{3})$ or $\sin(\tfrac{2\pi}{3})$.

https://forum.math.toronto.edu/index.php?PHPSESSID=jh17fgq9gjjla8v71ncb3c64p4&topic=2418.0;prev_next=next

Author Topic: Quiz 2- Lec 5101-A  (Read 617 times) Pengyun Li • Full Member • Posts: 20 • Karma: 14 Quiz 2- Lec 5101-A « on: October 01, 2020, 07:08:28 PM » $\textbf{Question}$: Find the limit of each function at the given point, or explain why it does not exist: $f(z) = (z-2)log|z-2|$ at $z_0 = 2$. $\textbf{Answer}$: Since $z_0= 2$, let $z' = z-2$. $$\lim_{z\to z_0} f(z) = \lim_{z' \to 0} f(z) =z'\log |z'|$$ $$=z' (\ln |z'| +i\cdot 0) = \frac{ln |z'|}{\frac{1}{z'}}$$ By L'Hôpital's Rule, $$= \frac{\frac{1}{z'}}{-\frac{1}{(z')^2}}=\lim_{z'\to 0} (-z') = 0$$. Therefore, the limit of the function is 0 at $z_0=2$. « Last Edit: October 01, 2020, 07:11:03 PM by Pengyun Li »

https://math.stackexchange.com/questions/2258326/m-is-a-square-martix-and-%CF%86-v-to-v-then-why-mt-m-%CF%86-m

# M is a square martix, and $φ^*$ : $V^* \to V^*$, then why $M^T=M_{φ^*_M}$? [duplicate] Let M be a matrix, transpose $M^T$ is the matrix who's i,j component is the j,i component of M. And let $\phi\in Hom(V,V)$, we define $\phi^*: V^* \to V^*$ by $\phi^*(f)(v)=f(\phi(v))$ where $V^*$ is the dual space of $V$. $M_{φ^*_M}$ is the matrix representation of $φ^*_M$ Then how to prove $M^T=M_{φ^*_M}$ ## marked as duplicate by amd, André 3000, JMP, Smylic, user91500Apr 30 '17 at 9:16 • You should probably define what $M_{\varphi_M^*}$ is. I assume it's the matrix representation of $\varphi^*$? – André 3000 Apr 30 '17 at 0:05 Fix a basis $e_1, \ldots, e_n$ of $V$, and let $e_1^*, \ldots, e_n^*$ denote the dual basis of $V^*$, defined by $$e_i^*(e_j) = \begin{cases} 1 & \text{if } i = j\\ 0 & \text{otherwise.} \end{cases}$$ Let $A$ be the matrix of $\varphi$ with respect to this basis, so the $i,j$ entry of $A$ is $a_{ij}$ where \begin{align} \label{A} \varphi(e_j) = \sum_i a_{ij} e_i \tag{1} \end{align} for all $j$. Similarly, let $B$ be the matrix of $\varphi^*$ with respect to the dual basis, so the $i,j$ entry of $B$ is $b_{ij}$ where \begin{align} \label{B} \varphi^*(e_j^*) = \sum_i b_{ij} e_i^* \, . \tag{2} \end{align} Applying the functional $\varphi^*(e_j^*)$ to the basis vector $e_k$, by (\ref{A}) we have \begin{align*} (\varphi^*(e_j^*))(e_k) &= (e_j^* \circ \varphi)(e_k) = e_j^*(\varphi(e_k)) = e_j^*\left(\sum_i a_{ik} e_i\right) = \sum_i a_{ik} e_j^*(e_i) = a_{jk} \, . \end{align*} On the other hand, by (\ref{B}) we also have \begin{align*} (\varphi^*(e_j^*))(e_k) = \left(\sum_i b_{ij} e_i^*\right)(e_k) = \sum_i b_{ij} e_i^*(e_k) = b_{kj} \, . \end{align*} Thus $a_{jk} = b_{kj}$ for all $j,k$, hence $B = A^T$. • For clearness, should the question be ${M_{φ_M}}^T=M_{φ^*_M}$ ? – Parting Apr 30 '17 at 2:18 • Yes, or probably just ${M_\varphi}^T = M_{\varphi^*}$. I'm not sure what the second $M$ adds. – André 3000 Apr 30 '17 at 2:29

https://www.physicsforums.com/threads/fibonacci-sequence.252654/

# Fibonacci sequence 1. Aug 30, 2008 ### Euler_Euclid if $$a_1, a_2, a_3 .............$$ belong to the fibonacci sequence, prove that $$a_1a_2 + a_2a_3 + ............ + a_{2n-1}a_{2n} = (a_{2n})^2$$ 2. Aug 30, 2008 ### uart If you are familiar with mathematical induction then that's the way to go with this one. Using the recursion equation ($a_{2n} + a_{2n+1} = a_{2n+2}$ etc) should let you make the inductive step fairly easily. BTW. Is this homework ? 3. Aug 30, 2008 ### Euler_Euclid not at all!!! 4. Aug 30, 2008 ### uart Are you familar with mathematical induction? 5. Aug 30, 2008 ### Euler_Euclid Is there any other method other than this?

http://math.stackexchange.com/questions/191306/sum-of-angles-in-mathbbrn?answertab=votes

# Sum of angles in $\mathbb{R}^n$ Given three vectors $v_1,v_2$ and $v_3$ in $\mathbb{R}^n$ with the standard scalar product the follwing is true $$\angle(v_1,v_2)+\angle(v_2,v_3)\geq \angle(v_1,v_3).$$ It tried to substitute $\angle(v_1,v_2) = cos^{-1}\frac{v_1 \cdot v_2}{\Vert v_1 \Vert \Vert v_2 \Vert}$ but I could not show the resulting inequality. What is the name of the inequality and do you know reference that one can cite in an article? - ## 2 Answers I'd like to suggest "triangle inequality in spherical geometry". - Any reference for this? –  warsaga Sep 5 '12 at 10:33 I don't find a reference for this either! –  warsaga Sep 5 '12 at 11:58 You can reduce the problem to $\mathbb{R}^3$, and there wlog v2=(0,0,1) then one gets an easy to prove inequality if one writes everything in polar coordinates. -

https://math.stackexchange.com/questions/1048831/calculate-winning-outcomes-of-plurality-voting

# Calculate winning outcomes of plurality voting My problem is similar to this one, but different in some significant ways. As in the above question, I have voting with $n$ voters and $m$ candidates. However, I care about which voter voted for which candidate. As such, there are a total of $m^n$ possible configurations. Also, of these possible configurations, how many did candidate 1 have a plurality (see below). That is, how many did candidate 1 win (I choose candidate 1 arbitrarily. Of course it is symmetric across candidates). Further, how many configurations are ties in which candidate 1 participated. For example, here are some situations ($n=6, m=5$) where candidate 1 wins. v1 v2 v3 v4 v5 v6 ----------------- 1 1 1 1 1 1 1 1 1 1 3 3 1 1 1 2 2 3 1 1 2 3 4 5 (Notice that what I want is a plurality and not a majority: a candidate does not need to have more than 50% of the votes, just more votes than anyone else.) Here is a two-way tie: v1 v2 v3 v4 v5 v6 ----------------- 1 1 2 2 3 4 Candidates 1 and 2 are tied for the most number of votes. Here are some situations ($n=5, m=3$) where candidate 1 loses. v1 v2 v3 v4 v5 -------------- 1 2 2 2 3 1 2 2 3 3 2 2 2 2 2 This would just be stars and bars, except I care about the order. That is, I want to count the following situations as distinct: v1 v2 v3 v4 v5 -------------- 1 1 1 2 3 1 1 1 3 2 1 1 1 3 3 1 1 2 1 3 ... • You explain the setup very well, but it would be better to put the "winning plurality" criterion early on in the problem statement, just where you state what it is that is being asked for (winning configurations for candidate 1). – hardmath Dec 2 '14 at 21:04 • Good point. Thanks. – mayhewsw Dec 2 '14 at 21:07 • @mayhewsw Any solution since the time you posted the question? – Antonin Feb 11 '15 at 21:51 • Nope. Still looking. – mayhewsw Feb 11 '15 at 22:28

https://homework.cpm.org/category/CCI_CT/textbook/int3/chapter/7/lesson/7.2.5/problem/7-115

### Home > INT3 > Chapter 7 > Lesson 7.2.5 > Problem7-115 7-115. Consider the graph at right. 1. Is the graph a function? Explain. What causes a graph to be a function or not a function? Look up 'function' in the glossary if you aren't sure. 2. Sketch the inverse of this graph. Is the inverse a function? Justify your answer. Reflect the graph across $y = x$. Are there any $x$-values that have more than one $y$-value? 3. Must the inverse of a function be a function? Explain. Consider several different functions and their inverses. 4. Describe a characteristic of functions that have inverse functions. See part (a). Think about how this rule is translated to the inverse. 5. Could the inverse of a non-function be a function? Explain or give an example. Think about functions whose inverses are non-functions. What about the inverse of that inverse?

https://www.ianhaberman.org/2019/05/10/setting-the-curve/

# Setting The Curve I teach economics. Sometimes, at the end of the semester, the distribution of grades does not properly capture the class outcomes observed by the instructor. This happens for a variety of reasons. A common question asked by colleagues is “What should I do to correct this?” In the long run, a course update may be required, but for the class in question, I might use a curve to reshape the distrbution of grades to better reflect the observed outcomes of the class. Another common question is “How do you curve course grades?” To curve course grades I use a linear scale. I first found this method on the blog Division by Zero while asking myself the same question. There David Richeson presents several methods to curve course grades (some with a pinch of humor). This presentation of the linear scale method seeks to simplify the calculation and provide greater exposition on the application of the curve. To apply the curve we generate a function that takes the old score as an input and provides a new “curved” score as the output. The formula below is used to apply a linear scale curve to a set of scores. # The Formula $CurvedScore = New_{mean} + \left(\frac{New_{low}-New_{mean}}{Old_{low}-Old_{mean}}\right)(OldScore-Old_{mean})$ To generate this function we need two points from the original distribution (I use the mean and lowest score) and the values for those points in the new distribution we wish to create. For example: suppose the average grade for a class is 70, the lowest grade is 40, and the desired average and low score are 80 and 60. The formula for this curve would be generated in the following way: $CurvedScore = 80 + \left(\frac{60-80}{40-70}\right)(OldScore-70)$ After simplification: $CurvedScore = 80 + \left(\frac{2}{3}\right)(OldScore-70)$ To complete the process, each score ($$OldScore$$) is passed through the function and assigned a curved value ($$CurvedScore$$). Below is an empirical example to help illustrate how the calculation is made. # Empirical Example The following empirical example presents the application of a linear curve to a set of scores from a hypothetical class. First, identify the scores for each student, the average, and the lowest score. Scores for students “a” through “q” are presented with the average and lowest score identified below. Our hypothetical class has a mean of 70 and a low of 33. Let us suppose we want to curve these grades up to a new mean of 82 and a new low of 55. The next step is to calculate the formula for these desired attributes and apply the formula to each score. Here is the calculation for this example: $CurvedScore = 82 + \left(\frac{55-82}{33-70}\right)(OldScore-70)$ After simplification: $CurvedScore = 82 + \left(\frac{27}{37}\right)(OldScore-70)$ The formula can now be applied to each score. I use spreadsheet software to do the calculation. A template of this exercise is included at the end of this post. Here is the formula used in the spreadsheet: The result is a curved score that corresponds to each original score. The set of new scores has the desired average and low score set out in the beginning of the application. Here is the set of old and curved scores: And that is how to curve class grades with the linear scale method. # Final Thoughts The linear scale method is presented as an option to curve course grades. It is a relatively straight forward way of applying a curve to grades, but it has some limitations. It does not apply an equal increase to each grade in the distribution, which may be seen as favoring some students over others. It is possible for this method to take points away from higher grades in the distribution to acheive the desired mean and low score. I deal with this last problem by simply ignoring the curved score for students that have high scores (effectively providing no change in their original score due to the curve). Additionally, this method may take some trial and error to acheive a “fair” distribution of grades. The strength of this technique is that it can easily reshape a bimodal distribution into a more uniform distribution of grades and it can target a desired distribution much better than simply adding a flat increase to all scores.

https://proofwiki.org/wiki/Definition:Integrable_Function/Lebesgue

# Definition:Integrable Function/Lebesgue ## Definition Let $\lambda^n$ be a Lebesgue measure on $\R^n$ for some $n > 0$. Let $f: \R^n \to \overline \R$ be an extended real-valued function. Then $f$ is said to be Lebesgue integrable if and only if it is $\lambda^n$-integrable. Similarly, for all real numbers $p \ge 1$, $f$ is said to be Lebesgue $p$-integrable if and only if it is $p$-integrable under $\lambda^n$. ## Source of Name This entry was named for Henri Léon Lebesgue.

https://math.stackexchange.com/questions/753130/is-there-a-simple-and-fast-way-of-computing-the-residue-at-an-essential-singular

# Is there a simple and fast way of computing the residue at an essential singularity? Is there a simple and fast way of computing the residue at an essential singularity ? I mean if we have a pole of order $n$ at $c$ we can use the formula : $$\mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right)$$ Is there a similar formula for an essential singularity (without using Laurent series) ? • It would be pretty nice if someone could answer this, but I'm quite skeptical. So I'm just going to go with the old fashion answer: integrate (which clearly is equivalent to finding the corresponding term in the Laurent series). Nov 7 '14 at 4:58 • Trust me, Laurent series all the way. You dont wanna have to deal with those derivatives, especially when taking a timed test Dec 4 '15 at 23:49

https://math.stackexchange.com/questions/3138148/cubic-runout-spline-cubic-spline

# Cubic runout spline (cubic spline) I was studying cubic spline interpolation and then I stumbled upon "Cubic runout spline". The idea behind it is that you set the boundary conditions for second derivatives to be: $$f_1''(x_0) = 2f_1''(x_1) - f_2''(x_2)$$ $$f_n''(x_n) = 2f_n''(x_{n-1}) - f_{n-1}''(x_{n-2})$$ As a result, the first two and last two points have their own single curve. Now, I've figured out the natural, clamped, periodic, not-a-knot type splines and their ideas and proofs, but this one for some reason confused me. I want to try prove that $$M_0 = 2M_1 - M_2$$ <=> first 2 curves are the same. I wrote a system of equations for $$M_0, M_1$$ and $$M_2$$, where $$M_i$$ - is a second derivative of our function at point i. I ended up with this: $$S_i(x) = \alpha_i + \beta_i (x - x_i) + \gamma_i (x - x_i)^2 + \delta_i (x - x_i)^3$$ $$M_0 = 2 \gamma_0$$ $$M_1 = 2 \gamma_1 = 2 \gamma_0 + 6\delta_0 h_1$$ $$M_2 = 2 \gamma_1 + 6 \delta_1 h_2$$ $$h$$ is the difference between adjacent x-values Now, I decided to split the solution into 2 parts: first one is where $$h_1 = h_2 = h$$ and the second is where they are different. 1. for the first part: solving this from right to left (i.e. same curves => equation holds) yielded $$\delta_0 = \delta_1 = 0$$, which then gave me $$M_0 = M_1 = M_2$$, which gives me the final equation. Now, the opposite (left to right) proof kinda stopped me. I managed to derive the equality of delta but for gamma I failed to proceed further. Could anyone give me a hint regarding the proper approach towards this? • One approach you might want to try to prove that the first two curves are the same is to check if the left and right third (why?) derivatives at the second point are the same. (The procedure is analogous for the last two curves and the penultimate point.) For inspiration, you might want to see this previous answer. – J. M. is a poor mathematician Mar 7 at 2:36