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330417 | Integral dependence and fraction fields
Consider [imath]\mathbb{Q}[x]\subset\mathbb{Q}(x)\subset\mathbb{Q}(x)[y]=:K[/imath], where [imath]y^2=x,[/imath] and let [imath]O_K[/imath] be the integral closure of [imath]\mathbb{Q}[x][/imath] in [imath]\mathbb{Q}(x)[y][/imath]. Show that [imath]\mathbb{Q}[x][y]=O_K[/imath]. I ask for some help/hints with this problem. Here is my work: [imath]\mathbb{Q}[x][y]\subset O_K[/imath] for, if [imath]a\in\mathbb{Q}[x][y][/imath] then [imath]a[/imath] is a root of [imath]f(y)=y-a(y)[/imath], which is a monic polynomial with coefficients in [imath]\mathbb{Q}[x][/imath]. Therefore [imath]a\in O_K[/imath]. Now I want to show [imath]O_K\subset\mathbb{Q}[x][y][/imath]. Let [imath]a\in O_K[/imath]. As [imath]a\in\mathbb{Q}(x)[y][/imath], I may write [imath] a=\sum \frac{r_j(x)}{s_j(x)}y^j \ , \quad r_j,s_j\in\mathbb{Q}[x] [/imath] The goal is to show that all [imath]s_j\in\mathbb{Q}[/imath]. Since [imath]a[/imath] is [imath]\mathbb{Q}[x][/imath]-integral we have an equation like [imath] a^n+q_1(x)a^{n-1}+\cdots+q_n(x)=0 \ , \quad q_j\in\mathbb{Q}[x] [/imath] but I am not sure how to use this now, plus the fact that [imath]y^2=x[/imath]. | 329284 | Integral closure of [imath]\mathbb{Q}[X][/imath] in [imath]\mathbb{Q}(X)[Y][/imath]
Consider the ring [imath]\mathbb{Q}[X][/imath] of polynomials in [imath]X[/imath] with coefficients in the field of rational numbers. Consider the quotient field [imath]\mathbb{Q}(X)[/imath] and let [imath]K[/imath] be the finite extension of [imath]\mathbb{Q}(X)[/imath] given by [imath]K:=\mathbb{Q}(X)[Y][/imath], where [imath]Y^2-X=0[/imath]. Let [imath]O_{K}[/imath] be the integral closure of [imath]\mathbb Q[X][/imath] in [imath]K[/imath]. Certainly [imath]O_{K}[/imath] contains both [imath]\mathbb{Q}[X][/imath] and [imath]Y[/imath], hence [imath] O_{K}\supseteq \mathbb{Q}[X][Y][/imath] My guess is that actually "=" holds. How can be proved this? |
28200 | A first order sentence such that the finite Spectrum of that sentence is the prime numbers
The finite spectrum of a theory [imath]T[/imath] is the set of natural numbers such that there exists a model of that size. That is [imath]Fs(T):= \{n \in \mathbb{N} | \exists \mathcal{M}\models T : |\mathcal{M}| =n\}[/imath] . What I am asking for is a finitely axiomatized [imath]T[/imath] such that [imath]Fs(T)[/imath] is the set of prime numbers. In other words in what specific language [imath]L[/imath], and what specific [imath]L[/imath]-sentence [imath]\phi[/imath] has the property that [imath]Fs(\{\phi\})[/imath] is the set of prime numbers? | 413247 | The set of primes as the spectrum of a first-order theory
In model theory, the finite spectrum of a first-order sentence [imath]\phi [/imath] (in a language with arbitrarily many constants, functions and relations is defined as the set of natural numbers [imath] n[/imath] such that [imath]\phi [/imath] has a model with exactly [imath] n [/imath] elements. How can I show that the set of primes is a spectrum for a sentence [imath]\phi [/imath]? My first idea was to take the theory of fields and augment it by some symbols and theorems excluding the fields of order [imath] p^n [/imath] with [imath] n > 1[/imath], but I have no idea how to do that. |
331429 | Is there a known distribution for this permutation with replacement problem?
Choose [imath]t[/imath] numbers from [imath]n[/imath] [imath](n>t)[/imath] distinct numbers with replacement and the order of the [imath]t[/imath] numbers matters. Say, [imath]P(X=1) = \dfrac{{numbers\ of\ unique\ t-set \ which\ has\ 1\ distinct\ number}}{n^t}[/imath] (i.e. [imath]n/n^t[/imath]), [imath]P(X=2)= \dfrac{{numbers\ of\ unique\ t-set \ which\ has\ 2\ distinct\ numbers}}{n^t}[/imath], [imath]...[/imath] [imath]P(X=t) = \dfrac{{numbers\ of\ unique\ t-set \ which\ has\ t\ distinct\ numbers}}{n^t}[/imath], [imath]P(X>t) = 0[/imath]. For example, [imath]n = 5[/imath] and [imath]t = 3[/imath], [imath]P(X=1) = \dfrac{5}{125}[/imath], [imath]P(X=2) = \dfrac{5 \times 4 \times 3}{125}= \dfrac{60}{125}[/imath], [imath]P(X=3) = \dfrac{5 \times 4 \times 3}{125}= \dfrac{60}{125}[/imath]. It seems to be very complicated for large [imath]t[/imath], is there a formula or even distribution function for that? Unique means different taking orders into consideration. e.g. (1,1,3) = (1,1,3) but not equal to (1,3,1) | 285820 | Throwing [imath]k[/imath] balls into [imath]n[/imath] bins.
I have the following question: Throwing [imath]k[/imath] balls into [imath]n[/imath] bins. What is the probability that exactly [imath]z[/imath] bins are not empty? I thought about something like: [imath]\Pr(z)=\frac{n! z^{k-z}}{n^k (n-z)!},[/imath] but this is not correct. Another idea is: [imath]\Pr(bin ~empty)=(1-1/n)^k[/imath], [imath]\Pr(bin ~Not ~empty)=1-(1-1/n)^k[/imath], so: [imath]\Pr(z)=(\Pr(bin ~empty))^{n-z}(\Pr(bin ~Not ~empty))^{z}\cdot A[/imath] So, how to obtain the [imath]A[/imath]? Thanks! |
331913 | Functional sequence
Let [imath](f_n)[/imath] be a sequence of functions [imath]\mathbb{R} \rightarrow \mathbb{R}[/imath]. Suppose that for any [imath](x_n)[/imath] convergent to [imath]x[/imath] we have [imath]f_n(x_n) \rightarrow f(x)[/imath]. Prove that [imath]f[/imath] in continuous, there is no assumption about [imath]f_n[/imath]'s continuity. Could you help me with that? | 310872 | If [imath]f_n(x_n) \to f(x)[/imath] whenever [imath]x_n \to x[/imath], show that [imath]f[/imath] is continuous
From Pugh's analysis book, prelim problem 57 from Chapter 4: Let [imath]f[/imath] and [imath]f_n[/imath] be functions from [imath]\Bbb R[/imath] to [imath]\Bbb R[/imath]. Assume that [imath]f_n(x_n)\to f(x)[/imath] as [imath]n\to\infty[/imath] whenever [imath]x_n\to x[/imath]. Prove that [imath]f[/imath] is continuous. (Note: the functions [imath]f_n[/imath] are not assumed to be continuous.) here's my attempt: assume [imath]x_n \to x[/imath]. we want to show that [imath]f(x_n) \to f(x)[/imath]. so [imath]|f(x_n) - f(x)| \leq |f(x_n)-f_n(x_n)| + |f_n(x_n)-f(x)|[/imath]. The second term can be made to be less than any [imath]\varepsilon > 0[/imath] for [imath]n[/imath] sufficiently large. i'm having trouble with the first term. can anyone help? thank you! |
332084 | Let [imath]G[/imath] a group and [imath]g,h∈G[/imath]. Prove that [imath]|gh|=|hg|[/imath].
Let [imath]G[/imath] a group and [imath]g,h∈G[/imath]. Prove that [imath]|gh|=|hg|[/imath]. A hint would be appreciated. | 238212 | Is it true that the order of [imath]ab[/imath] is always equal to the order of [imath]ba[/imath]?
How do I prove that if [imath]a[/imath], [imath]b[/imath] are elements of group, then [imath]o(ab) = o(ba)[/imath]? For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of [imath]ab[/imath] is [imath]2[/imath] and do the steps that lead me to conclude that [imath]ab=ba[/imath], so the orders must be the same. Is that the right way to do it? |
332171 | Prove that a non-zero, non-unit element [imath]a \in R[/imath] is irreducible
Prove that a non-zero, non-unit element [imath]a \in R[/imath] is irreducible iff its only divisors are units of [imath]R[/imath] and elements of [imath]R[/imath] which are associates of [imath]a[/imath] I have a proof below but i'm not sure if it is correct. I think I need to use [imath]c[/imath] in the second part but I'm not sure how. Here is my proof [imath]\Rightarrow[/imath] Assume [imath]a\ne 0[/imath], [imath]a[/imath] is not a unit [imath]\in R[/imath] is irreducible Then [imath]a=bc[/imath] where at least one of [imath]b[/imath] or [imath]c[/imath] is a unit Therefore if [imath]b[/imath] is a unit then [imath]c[/imath] is an associate of a and if [imath]c[/imath] is a unit then [imath]b[/imath] is an associate of a As [imath]a=bc[/imath] then [imath]b|a[/imath] and [imath]c|a[/imath] [imath]b[/imath] and [imath]c[/imath] are either units or associates of [imath]a[/imath] therefore the only divisors of [imath]a[/imath] are units of [imath]R[/imath] or associates of [imath]a\in R[/imath] [imath]\Leftarrow[/imath] Assume [imath]a\in R[/imath] can only be divided by units or associates Let [imath]b,c\in R[/imath] divide [imath]a[/imath], by assumption [imath]b[/imath] must either be a unit or an associate If [imath]b[/imath] is a unit then [imath]a[/imath] is irreducible else [imath]b[/imath] is an associate This means there exists a unit [imath]u\in R[/imath] where [imath]a=ub[/imath] Therefore [imath]u|a[/imath] so once again a is divided by a unit and is therefor irreducible | 327929 | Characterization of irreducible elements in integral domains.
Let [imath]R[/imath] be an integral domain. Show that a non zero, non unit element [imath]c[/imath] of [imath]R[/imath] is NOT reducible iff its only divisors are units of [imath]R[/imath] and elements of [imath]R[/imath] which are associates of [imath]c[/imath]. |
332130 | Prove that [imath]\mathcal{P}(A)⊆ \mathcal{P}(B)[/imath] if and only if [imath]A⊆B[/imath].
Here is my proof, I would appreciate it if someone could critique it for me: To prove this statement true, we must proof that the two conditional statements ("If [imath]\mathcal{P}(A)⊆ \mathcal{P}(B)[/imath], then [imath]A⊆B[/imath]," and, If [imath]A⊆B[/imath], then [imath]\mathcal{P}(A)⊆ \mathcal{P}(B)[/imath]) are true. Contrapositive of the first statement: If [imath]A \nsubseteq B[/imath], then [imath]\mathcal{P}(A) \nsubseteq \mathcal{P}(B)[/imath] If [imath]A \nsubseteq B[/imath], then there must be some element in [imath]A[/imath], call it [imath]x[/imath], that is not in [imath]B[/imath]: [imath]x \in A[/imath], and [imath]x \notin B[/imath]. Since [imath]x \in A[/imath], then [imath]\{x\} \in \mathcal{P}(A)[/imath]; moreover, since [imath]x \notin B[/imath], then [imath]\{x\} \notin \mathcal{P}(B)[/imath], which proves that, if [imath]A \nsubseteq B[/imath], then [imath]\mathcal{P}(A) \nsubseteq \mathcal{P}(B)[/imath]. By proving the contrapositive true, the original proposition must be true. To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement? | 685587 | P(A) [imath]\subset[/imath] P(B) implies A [imath]\subset [/imath] B proof or disproof.
P(A) [imath]\subset[/imath] P(B) implies A [imath]\subset [/imath] B proof or disproof. I have a strange feeling this is false but I do not know. Something to do with P(A) [imath]\subset[/imath] P(B) seems strange since P(B) is itself a powerself with P(A) being a subset. |
332611 | Group presentation where relations consist of inverse elements
Consider a group presentation [imath]G= \langle x_{1} , x_{2} , \ldots , x_{n} \mid r_{i} (x_{1} , x_{2} , \dots , x_{n} ) \rangle[/imath]. How does the presentation [imath]\langle x_{1} , x_{2} , \ldots , x_{n} \mid r_{i} (x_{1}^{-1} , x_{2}^{-1} , \ldots , x_{n}^{-1} ) \rangle[/imath] relate to [imath]G[/imath]? | 306722 | group presentation and the inverse elements of the generators
If [imath]G = \langle g_1 , g_2 , \ldots \mid r_i (g_1 , g_2 , \ldots ) = 1 \rangle [/imath] is a group presentation, then is it the case that [imath]\langle g_1^{-1} , g_2^{-1} , \ldots \mid r_i (g_1^{-1} , g_2^{-1} , \ldots ) = 1 \rangle [/imath] is also a group presentation for [imath]G[/imath]? |
332270 | Does π depends on the norm?
If we take the definition of π in the form: π is the ratio of a circle's circumference to its diameter. There implicitly assumed that the norm is Euclidian: \begin{equation} \|\boldsymbol{x}\|_{2} := \sqrt{x_1^2 + \cdots + x_n^2} \end{equation} And if we take the Chebyshev norm: \begin{equation} \|x\|_\infty=\max\{ |x_1|, \dots, |x_n| \} \end{equation} The circle would transform into this: And the π would obviously change it value into [imath]4[/imath]. Does this lead to any changes? Maybe on other definitions of π or anything? | 254620 | [imath]\pi[/imath] in arbitrary metric spaces
Whoever finds a norm for which [imath]\pi=42[/imath] is crowned nerd of the day! Can the principle of [imath]\pi[/imath] in euclidean space be generalized to 2-dimensional metric/normed spaces in a reasonable way? For Example, let [imath](X,||.||)[/imath] be a 2-dimensional normed vector space with a induced metric [imath]d(x,y):=\|x-y\|[/imath]. Define the unit circle as [imath]\mathbb{S}^1 := \{x\in X|\;\|x\|=1\}[/imath] And define the outer diameter of a set [imath]A\in X[/imath] as [imath]d(A):=\sup_{x,y\in A}\{d(x,y)\}=\sup_{x,y\in A}\{\|x-y\|\}[/imath] Now choose a continuous Path [imath]\gamma:[0,1]\rightarrow X[/imath] for which the image [imath]\gamma([0,1])=\mathbb{S}^1[/imath]. Using the standard definition of the length of a continuous (not necessarily rectificable) path given by [imath] L(\gamma):=\sup\bigg\{\sum_{i=1}^nd(\gamma(t_i),\gamma(t_{i+1}))|n\in\mathbb{N},0\le t_0\lt t_1\lt ... \lt t_n\le 1\bigg\}[/imath] we can finally define [imath]\pi[/imath] in [imath](X;\|.\|)[/imath] by [imath]\pi_{(X,\|.\|)}:=\frac{L(\gamma)}{d(\mathbb{S}^1)}[/imath] (This is way more well-defined than the old definition (check the rollbacks)) Examples: For the euclidean [imath]\mathbb{R}^2[/imath], [imath]\pi_{\mathbb{R}^2}=3.141592...[/imath] For taxicab/infinity norms, [imath]\pi_{(\mathbb{R}^2,\|.\|_1)}=\pi_{(\mathbb{R}^2,\|.\|_\infty)}=4[/imath] For a norm that has a n-gon as a unit circle, we have [imath]\pi_{(\mathbb{R}^2,\|.\|)}=??[/imath] (TODO: calculate) While trying to calculate values for [imath]\pi[/imath] for interesting unit circles, I have defined a norm induced by a unit circle: let [imath]\emptyset\neq D\subset X[/imath] be star-shaped around [imath]0\in D[/imath]. Define [imath]\lambda D:=\{\lambda d|d\in D\}[/imath]. Now the (quasi-)norm in [imath]X[/imath] is defined as [imath]\|x\|:=\min\{\lambda\mid x\in \lambda D\}[/imath]. In other words: the scaling factor required to make x a part of the border of D. This allows us to easily find norms for most geometric shapes, that have exactly that Geometric shape as a unit circle and have the property, that the choice of the radius for the definition of [imath]\pi[/imath] is insignificant (for example [imath]\pi=3[/imath] is calculated for regular triangles with (0,0) in the centroid). This allows for the following identity: let [imath]x\in\mathbb{R}^2[/imath], and [imath]S\in\partial D[/imath] be the Intersection of the Line [imath]\overline{x0}[/imath] with the border of D. Then we have [imath]\|x\|_D = \frac{\|x\|_p}{\|S\|_p}[/imath] where [imath]\|.\|p[/imath] is any (quasi)-norm in [imath]\mathbb{R}^2[/imath] (This follows out of the positive scalability of norms) EDIT: I have been thinking about this a bit more. However I found out that my induced norm defined earlier is not even a norm... It violates the subadditivity axiom: Let the unit circle be a equilateral triangle where the centroid marks the point (0,0). Here we find [imath]d(\mathbb{S}^1)=3[/imath] which violates the triangle equation, as [imath]d[/imath] is measured in between 2 points of the unit sphere; Therefore, we have [imath]\|a\|=1[/imath] and [imath]\|b\|=1[/imath], but [imath]\|a+b\|=3\nleq1+1=\|a\|+\|b\|[/imath]. Instead, we have a Quasinorm. This indeed allows for [imath]\pi=42[/imath], if the unit circle is, for example, graph of a function [imath]\varphi\mapsto(r(\varphi),\varphi)[/imath] in Polar coordinates where [imath]r(\varphi)=a*\sin(2\pi k)[/imath]. Questions Any other interesting norms? Is this definition reasonable, and is there any practical use to this? Feel free to share your thoughts. Mind me if I made some formal mistakes. And especially, how do I define a norm with [imath]\pi=42[/imath]? About the [imath]\pi=42[/imath] Prince Alis Answer below shows that there can be no such p-norm, for which [imath]\pi_p=42[/imath], In fact this holds true for every norm. However, you can easily define a quasi-norm that has any arbitrary [imath]\pi_{\|.\|}=\kappa\gt\pi[/imath]. For example, 1 + \frac{1}{a} \sin(b \theta)"> [imath]r(\theta)=1 + \frac{1}{a} \sin(b \theta)[/imath] Defines a quasinorm with [imath]\pi_{\|.\|_{a,b}}=\kappa[/imath] for every [imath]\kappa>\pi[/imath]. [imath]\kappa[/imath] can be increased by increasing a and b. The only thing left to do is find a and b for which we finally have [imath]\pi_{\|.\|_{a,b}}=42[/imath]. According to Mathematica, the Length of the boundary curve is 33.4581 (Took ~10 minutes to calculate) and the diameter is 4.5071, resulting in [imath]\pi=7.42342[/imath] for the norm given above ([imath]a=b=10[/imath]). I doubt I will be able to easily find a solution for [imath]\pi=42[/imath] using this method... (Testing manually, I got exemplary values [imath]a=9.95[/imath], [imath]b=175[/imath] with [imath]\pi=42.0649[/imath] which comes very close... On top of that, Prince Ali found a p-norm with [imath]p<1[/imath] for which [imath]\pi=42[/imath]. Thank you very much! |
332711 | Show that [imath]f[/imath] cannot have infinitely many zeroes in [imath][0, 1][/imath].
Let [imath]f : \mathbb{R}\to \mathbb{R}[/imath] be a differentiable function such that [imath]f[/imath] and its derivative have no common zero in the closed interval [imath][0, 1][/imath]. Show that f cannot have infinitely many zeroes in [imath][0, 1][/imath]. | 288677 | on [imath]\mathbb{R}[/imath] such that [imath]f[/imath] and [imath]f'[/imath] has no common [imath]0[/imath] in [imath][0,1][/imath],
[imath]f[/imath] is differentiable on [imath]\mathbb{R}[/imath] such that [imath]f[/imath] and [imath]f'[/imath] has no common [imath]0[/imath] in [imath][0,1][/imath], we need to show [imath]f[/imath] can not have infinitely many [imath]0[/imath] in [imath][0,1][/imath].could any one give me hint? |
333123 | Finitely many prime ideals lying over the same prime ideal
Let [imath]A \subseteq B[/imath] an extension of rings such that [imath]B[/imath] is an [imath]A[/imath]-module finitely generated. Show that for every prime ideal [imath]\mathfrak{p} \subseteq A[/imath] there is only a finite number of prime ideals [imath]\mathfrak{q} \subseteq B[/imath] such that [imath]\mathfrak{q} \cap A = \mathfrak{p}[/imath]. | 270428 | In a finite ring extension there are only finitely many prime ideals lying over a given prime ideal
I'm trying to solve the exercise 6.7 of Miles Reid's Undergraduate Commutative Algebra (pag 93). How can I prove that if [imath]B[/imath] is a finite ring extension of [imath]A[/imath], there are only finitely many prime ideals of [imath]B[/imath] whose intersection with [imath]A[/imath] is a given prime ideal of [imath]A[/imath]? Thank you! |
333348 | When are the product of cyclic groups also cyclic?
In general, products of cyclic groups are not cyclic. If [imath]C_n[/imath] is the cyclic group of order [imath]n[/imath], [imath]C_{22} \times C_{33}[/imath] is not cyclic. But is there an easy way to tell when such a product will be cyclic? | 310304 | [imath]\mathbb{Z}_m \times \mathbb Z_n[/imath] is cyclic if and only if [imath]\gcd(m,n)=1[/imath]
Let [imath]m,n[/imath] be positive integers bigger than [imath]1[/imath]. Show that [imath]\mathbb{Z}_m \times \mathbb Z_n[/imath] is cyclic if and only if [imath]\gcd(m,n)=1[/imath]. I have no idea on how to start. Anyone hints are much helpful. |
334015 | finding [imath]cov(X,Y)[/imath]
suppose [imath]X\sim N(0,1)[/imath] , [imath]Y = \left\{ \begin{array}{ll} X, & \hbox{if } |X|\geq a, \\ -X, & \hbox{if }|X|< a. \end{array} \right. [/imath] how can find [imath]cov(X,Y)[/imath]. [imath]\phi,\Phi[/imath] are density function and distribution function standard normal | 330486 | The correlation between two normal distribution
Let [imath]X[/imath] have the [imath]N(0,1)[/imath] distribution and let [imath]a>0[/imath], show that the random variable [imath]Y[/imath] given by [imath]Y=\begin{cases} X & \text{if }|X|<a\\[5pt] -X &\text{if }|X|\geq a\; \end{cases}[/imath] has the [imath]N(0,1)[/imath] distribution. What is cov[imath](X,Y)[/imath]? Comments: Does that mean [imath]Y[/imath] has pdf: [imath]f_Y(y)=\begin{cases} f_X(y) & \text{if }|y|<a\\[6pt] f_X(-y) & \text{if }|y|\geq a\; \end{cases}[/imath] |
152243 | Real numbers equipped with the metric [imath] d (x,y) = | \arctan(x) - \arctan(y)| [/imath] is an incomplete metric space
I have to show that the real numbers equipped with the metric [imath] d (x,y) = | \arctan(x) - \arctan(y)| [/imath] is an incomplete metric space. Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that. Can anybody help me with this. Thanks for helping me. | 717578 | Does there exist a metric under which [imath]\mathbb{R}[/imath] is incomplete?
Does there exist a metric under which [imath]\mathbb{R}[/imath] is incomplete? |
323079 | Determine general form of control function andthus show this coul have been achieved earlier
Determine the general form of [imath]u_0, u_1 ~\text{and} ~ u_2[/imath] if a system of difference equations of the form [imath]x_{n+1} = Ax_n + Bu_n,[/imath] where: [imath]A = \begin{pmatrix} 3 & 2 & 2 \\ -1 & 0 & -1 \\ 0 & 0 & 1 \end{pmatrix}[/imath] and: [imath]B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ 1 & 0 \end{pmatrix}[/imath] is to be controlled for [imath]x_0 = 0 ~ to ~ x_3 = [2, 1, 2]^T[/imath] . Show this target could have been achieved at [imath]x_2[/imath] Solution So far I have caculated the controlability matrix to be [imath] C =\begin{pmatrix} 0&0&2&2&6&6\\ 0&1&-1&0&-3&-2\\ 1&0&1&0&1&0 \end{pmatrix}. [/imath] Thus the system is controlable Now putting Cv=x3 i have the 3 equations [imath] 2c+2d+6e+6f=2\\ b-c-3e-2f=1\\ c+e+f=2\\ [/imath] which i have then put into augmented matrix row echleon form which i have found to be [imath] a-d-2e-3f=1\\ b+d+f=2\\ c+d+3e+3e=1\\ [/imath] How do I now solve with so many unknowns? also can you please check my working so far is correct. many thanks | 333798 | Find a general control and then show that this could have been achieved at x2
Determine the general form of [imath]u_0, u_1 ~\text{and} ~ u_2[/imath] if a system of difference equations of the form [imath]x_{n+1} = Ax_n + Bu_n,[/imath] where: [imath]A = \begin{pmatrix} 3 & 2 & 2 \\ -1 & 0 & -1 \\ 0 & 0 & 1 \end{pmatrix}[/imath] and: [imath]B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ 1 & 0 \end{pmatrix}[/imath] is to be controlled for [imath]x_0 = 0 ~ to ~ x_3 = [2, 1, 2]^T[/imath] . Show this target could have been achieved at [imath]x_2[/imath] Solution So far I have caculated the controlability matrix to be [imath] C =\begin{pmatrix} 0&0&2&2&6&6\\ 0&1&-1&0&-3&-2\\ 1&0&1&0&1&0 \end{pmatrix}. [/imath] Thus the system is controlable Now putting Cv=x3 i have the 3 equations [imath] 2c+2d+6e+6f=2\\ b-c-3e-2f=1\\ c+e+f=2\\ [/imath] which i have then put into augmented matrix row echleon form which i have found to be [imath] a-d-2e-3f=1\\ b+d+f=2\\ c+d+3e+3e=1\\ [/imath] How do I now solve with so many unknowns? also can you please check my working so far is correct. many thanks |
333854 | Given [imath]f[/imath] holomorphic on the unit disk, [imath]|f(z)|\leq M[/imath], prove [imath]|f'(z)|\leq M(1-|z|)^{-1}[/imath].
Problem: Let [imath]f[/imath] be a holomorphic function on the unit disk and [imath]|f(z)|\leq M[/imath] for all [imath]z[/imath] in the unit disk. Prove that [imath]|f'(z)|\leq M(1-|z|)^{-1}[/imath] for all [imath]z[/imath] in the unit disk. Using a generalized version of the Schwarz lemma, I can prove that [imath]|f'(z)|\leq M(1-|z|^2)^{-1}[/imath] on the unit disk, however I cannot see a way to prove the inequality that is asked for. Any suggestions? | 188457 | A Pick Lemma like problem
Let [imath]f[/imath] be analytic on the unit disc [imath]D[/imath] and bounded in modulus by [imath]M[/imath] there. I want to show that [imath]|f'(z)|\le \frac{M}{1-|z|}[/imath] for all [imath]z\in D[/imath]. I want to use Schwarz's lemma here after some suitable FLTs, as in the proof of Pick's lemma, but I haven't made progress. Does anyone have an idea? Edited: Forgot a factor of [imath]M[/imath] in the inequality originally. |
334669 | [imath]g:\mathbb{C}\rightarrow\mathbb{C}[/imath] be analytic and [imath]g(0)\neq 0[/imath] we need to calculate [imath]\frac{1}{2\pi i}\int_{|z|=r>0} f(z)g(z) dz[/imath]
[imath]f:\mathbb{C}^{*}\to\mathbb{C}[/imath] be a analytic with a simple pole of order [imath]1[/imath] at [imath]0[/imath] wit residue [imath]a_{-1}[/imath]. [imath]g:\mathbb{C}\rightarrow\mathbb{C}[/imath] be analytic and [imath]g(0)\neq 0[/imath] we need to calculate [imath]\frac{1}{2\pi i}\int_{|z|=r>0} f(z)g(z) dz[/imath] Please give me Hints. | 332496 | calculate for [imath]r>0[/imath] [imath]\frac{1}{2\pi i} \int_{|z|=r}{f(z)g(z)dz}[/imath]
Let [imath]f : \mathbb{C}\setminus[/imath] {[imath]0[/imath]} [imath]\to \mathbb{C}[/imath] be an analytic function with a simple pole of order [imath]1[/imath] at [imath]0[/imath] with residue [imath]a_1[/imath]. Let [imath]g : \mathbb{C} \to \mathbb{C}[/imath] be analytic with [imath]g(0)\neq 0[/imath].calculate for [imath]r>0[/imath] [imath]\frac{1}{2\pi i} \int_{|z|=r}{f(z)g(z)dz}[/imath] my thoughts: the answer will be [imath]Res(f(z)).g(0)[/imath] that is [imath]a_1g(0)[/imath]. am I right? |
334582 | burnside lemma cube
Having n colors, use the lemma to find a formula for the number of ways to color the edges of the cube. What I have so far: I got [imath]|A/G| = \dfrac{n^{12} + 6n^3 + 3n^6 + 8n^4 + 6n^7}{24}[/imath] but when I try [imath]n=2[/imath] or [imath]n=3[/imath], the results seem too large. Thanks for your help! | 334462 | Using Burnside's lemma on the cube.
Having [imath]n[/imath] colors, use the lemma to find a formula for the number of ways to color the edges of the cube. Here is what I have so far: The Burnside lemma says that [imath]\displaystyle |X/G| = \dfrac{1}{|G|} \sum_{g \in G} |X^g|[/imath] where [imath]G[/imath] is a finite group, [imath]X^g[/imath] is the set of elements fixed by [imath]g \in G[/imath], and [imath]|x/G|[/imath] is the number of orbits. Now, using [imath]n[/imath] colors, we have the following: 1 identity element leaving [imath]n^6[/imath] elements of [imath]X[/imath] unchanged. 6 90-degree face rotations each leaving [imath]n^3[/imath] elements unchanged. 3 180-degree face rotations each leaving [imath]n^4[/imath] elements unchanged. 8 120-degree vertex rotations each leaving [imath]n^2[/imath] elements unchanged. 6 180-degree edge rotations each leaving [imath]n^3[/imath] elements unchanged. Using the above results and the formula from Burnside's lemma, we obtain [imath]\dfrac{n^6+6 \cdot n^3 + 3 \cdot n^4 + 8 \cdot n^2 + 6 \cdot n^3}{24} = \dfrac{n^6 + 3n^4 + 12n^3 + 8n^2}{24}[/imath]. I think I did it right but wanted to check with you guys. Thank you! |
333901 | Prove that if [imath]f[/imath] is uniformly continuous then the one sided limit [imath]\lim_{x\to 0^+} f(x)[/imath] exists.
If [imath]f(x)[/imath] is a continuous function on [imath](0,1][/imath], prove that if [imath]f[/imath] is uniformly continuous, then the one sided limit [imath]\lim_{x\to 0^+} f(x)[/imath] exists. | 334779 | Show that [imath]f[/imath] is uniformly continuous if limit exists
Let [imath]f(x)[/imath] be continuous on [imath](0,1][/imath]. Show that [imath]f[/imath] is uniformly continuous IFF [imath]\displaystyle \lim_{x\to0^+} f(x)[/imath] exists. Thoughts: Backward Proof: Let another function [imath]\overline f(x)[/imath] be continuous on [imath][0,1][/imath] which is equal to [imath]f(x)[/imath] plus the limit point. Thus the limit exists and it is uniformly continuous. So, as [imath]f(x)[/imath]. Forward Proof: I Don't really have any idea...?? Please help guys |
330643 | Vector space is cyclic if and only if it has finitely many invariant subspaces
Please help me prove that a vector space [imath]V[/imath] is cyclic due to the endomorphism [imath]f[/imath] if and only if it has finitely many [imath]f[/imath]-invariant subspaces. What I've tried: I've tried the not direct prof, to show that it has infninitely invariant subspaces, then it's cyclic will be inifnite, but it's not going anywhere... | 55085 | Necessary and sufficient conditions for a module to have finitely many invariant subspaces
I have had trouble answering the following question which is from a study guide to a qualifying exam I will be taking later this summer. I am thinking this question has something to do with cyclic vectors but I have not been able to put the two definitions together. Definition: If [imath]\alpha[/imath] is any vector in [imath]V[/imath], the [imath]T[/imath]-cyclic subspace generated by [imath]\alpha[/imath] is the subspace [imath]Z(\alpha;T)[/imath] of all vectors of the form [imath]g(T) \alpha[/imath], [imath]g \in F[x][/imath]. If [imath]Z(\alpha; T) = V[/imath], then [imath]\alpha[/imath] is called a cyclic vector for [imath]T[/imath]. Let [imath]V[/imath] be a finite-dimensional vector space over an infinite field [imath]F[/imath] and let [imath]T:V\rightarrow V[/imath] be a linear operator. Give to each [imath]V[/imath] the structure of a module over the polynomial ring [imath]F[x][/imath] by defining [imath]x \alpha = T(\alpha)[/imath] for each [imath]\alpha \in V[/imath] In terms of the expression for [imath]V[/imath] as a direct sum of cyclic [imath]F[x][/imath]-modules, what are necessary and sufficient conditions in order that [imath]V[/imath] have only finitely many [imath]T[/imath]-invariant [imath]F[/imath]-subspaces? Every linear operator I have encountered has finitely many [imath]T[/imath]-invariant subspaces. Is there a good example of one that has infinitely [imath]T[/imath]-invariant [imath]F[/imath]-subspaces? I was thinking that one direction might require [imath]T[/imath] not to have any cyclic vectors but I dont think this is the only hypothesis we need in order to answer even one direction for part 1. |
335485 | Almost surely constant
Let [imath]x[/imath] and [imath]y[/imath] be two independent random variables on a probability space [imath](X,\Sigma,\mathbb{P})[/imath] such that [imath]x+y[/imath] is almost surely constant. Show that both [imath]x[/imath] and [imath]y[/imath] must be almost surely constant. | 215207 | How can [imath]P(X+Y=\alpha)=1[/imath] and [imath](X,Y)[/imath] independent imply that [imath]X[/imath] and [imath]Y[/imath] are constant?
This is an exercise in Jacod and Protter's Probability Essentials: Let [imath]X[/imath] and [imath]Y[/imath] be independent random variables and [imath]P(X+Y=\alpha)=1[/imath] where [imath]\alpha\in{\Bbb R}[/imath] is some constant. Show that both [imath]X[/imath] and [imath]Y[/imath] are constant random variables. What I think is that one might use Borel-Cantelli theorem here. Since [imath] \bigcup_{i=0}^{\infty}\{X=i,Y=\alpha-i\}\subset\{X+Y=\alpha\}=\bigcup_{\beta\in{\Bbb R}}\{X=\beta,Y=\alpha-\beta\}, [/imath] we have [imath] \begin{align} P\bigg(\bigcup_{i=0}^{\infty}\{X=i,Y=\alpha-i\}\bigg)=\sum_{i=0}^{\infty}P(X=i,Y=\alpha-i)<\infty \end{align} [/imath] But this seems to give nothing. Also, I'm surprised about the result is that [imath]X=\gamma[/imath] for some constant [imath]\gamma[/imath] instead of [imath]P(X=\gamma)=1[/imath]. Any idea about how I can go on? |
335674 | Prove that for any graph G, either G or its complement [imath]\bar{G}[/imath] is connected
Prove that for any graph G, either G or its complement [imath]\bar{G}[/imath] is connected. I understand what it means to be a graphs complement and I can see that the above is the case but I am stuck on how to actually prove it. Any suggestions would be great. | 122184 | Given a simple graph and its complement, prove that either of them is always connected.
I was tasked to prove that when given 2 graphs [imath]G[/imath] and [imath]\bar{G}[/imath] (complement), at least one of them is a always a connected graph. Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail. So I really hope someone could give me a hint on how to approach this problem. |
335715 | diagonalise a symmetric bilinear form
I have a math question that I got stuck on and would like to ask you about: I have a symmetric bilinear form on [imath]\mathbb{Q}^4[/imath] which is described w.r.t the standard basis. [imath]g(v,w)=\mathbf{v}^\top Aw[/imath] where [imath]A= \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 5\\ 3 & 4 & 5 & 6\\ 4 & 5 & 6 & 7\end{pmatrix} [/imath] a) I need to find a basis on whch g is diagonal. - I see that the rank is 2, so two eigenvalues must be [imath]0[/imath], but for the basis, I do need to find all 4 eigenvectors right? Or is there perhaps an easier way to see a basis. b)I need to find one ( or a product of ) matrix (matrices) B (or B=[imath]B_1\cdot B_2 \cdot\cdot[/imath]) such that [imath]\mathbf{B}^\top AB[/imath]= diagonal for this one I would suggest [imath]B[/imath] to be orthogonal, and I would use the orthogonal eigenvector from a) and normalise them, right? c)Consider the same g on [imath]\mathbb{R}^4[/imath], I have to give a basis for which g is diagonal and has only [imath]1,-1,0[/imath] on the diagonal. I dont know to connect this with my given information. Any help would be greatly appreciated! Thanks in advance | 329304 | Bilinear Form Diagonalisation
We have the following symmetric bilinear form on [imath]\\Q^4[/imath] (vector space over rational numbers) with respect to standard basis [imath]\{e_1,e_2,e_3,e_4\}[/imath] [imath]g(v,w)=v^tAw[/imath] [imath]A=\begin{bmatrix}1 & 2 &3&4\\2 & 3 & 4 & 5\\3&4&5&6\\4&5&6&7\end{bmatrix}[/imath] How to find a basis on which g is diagonal. I suppose using Gram-Schmidt algorithm we find orthogonal basis, but could someone explain this explicitly, please? |
335266 | Hint for proving [imath]{\rm SL}_n(\mathbb{R})[/imath] is generated by matrices that are off-diagonal entries added to identity.
Let [imath]S = \{ I_n + a\cdot e_{i,j} \mid a\in\mathbb{R},\ i,j= 1,\ldots,n,\ i\neq j\}[/imath], where [imath]e_{i,j}[/imath] is the matrix with [imath]1[/imath] at entry [imath](i,j)[/imath] and zero elsewhere. I need a hint to help prove that [imath]\langle S\rangle ={\rm SL}_n(\mathbb{R})[/imath]. That is, such matrices generate the multiplicative group of matrices with determinant one. | 333496 | What are the generators for [imath]SL_n(\mathbb{R})[/imath] (Michael Artin's Algebra book)
The book asks you to prove that [imath]SL_n(\mathbb{R})[/imath] is generated by elementary (row operation) matrices in which one nonzero off-diagonal entry is added to the identity matrix. For example, [imath] \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} [/imath] acts by left multiplication on [imath]2\times2[/imath] matrices by adding [imath]a[/imath] times (row 2) to (row 1). Considering a simple example: [imath] M= \begin{bmatrix} a & 0 \\ 0 & 1/a \end{bmatrix}, a \neq 0[/imath] you can see that the matrix [imath]M[/imath] does belong to [imath]SL_n(\mathbb{R})[/imath]. However, the elementary matrices composing [imath]M[/imath] are of the below type and not of the first type (nonzero off-diagonal entry). [imath] \begin{bmatrix} c & 0 \\ 0 & 1 \end{bmatrix} [/imath] i.e. one nonzero diagonal entry added to the identity matrix. So [imath]M = \begin{bmatrix} a & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1/a \end{bmatrix} [/imath] So [imath]M[/imath] clearly isn't generated by the first type. What is going wrong? |
335866 | Trying to show that operator [imath]T((x_n))=(2^{-n}x_n)[/imath] is compact.
Consider [imath]T\colon\ell^2\to\ell^2[/imath] an operator such that [imath]T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 [/imath] Does anyone know how to prove that it is compact? I understand that a linear operator [imath]T:E_1\to E_2[/imath] is considered compact if for every bounded sequence [imath](a_n)[/imath] in [imath]E_1[/imath]; [imath](Ta_n)[/imath] has a converging semisequence in [imath]E_2[/imath]. So I understand that I have to find a converging semisequence in [imath]\ell^2[/imath] and show that [imath](X_n)[/imath] is bounded. I got some hints but Im having trouble in understanding the whole picture here. Thank you for your help. I do not understand how to sum up the hints that I got in the previous question or how to comment there that I dont fully understand the answers. I think that in this question I added some data that is important for focusing on what Im trying to figure out. can someone give me a push on that? Thanks, Jeremy | 320751 | Trying to prove that operator is compact
Consider [imath]T\colon\ell^2\to\ell^2[/imath] an operator such that [imath]T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 [/imath] Does anyone know how to prove that it is compact? I understand that I have to find a converging semisequence in [imath]\ell^2[/imath]. Thank you for your help. |
336204 | Math question help probability?
We have the events [imath]D,F[/imath] and [imath]G[/imath]. Their probabilities are [imath]P(D)=\cfrac{1}{6},\quad P(F)=\cfrac{1}{8} \;\;\mbox{ and }\;\; P(G)=\cfrac{1}{4}.[/imath] Find the probability that any of these events happens( It does not specify which in my book). My opinion: Wouldnt that be [imath]1[/imath]? THE EVENTS ARE INDEPENDENT | 336195 | Finding the probability that one of the given independent events happens
We have the events [imath]D[/imath], [imath]F[/imath] and [imath]G[/imath]. Their probabilities are [imath]P(D)=1/5[/imath], [imath]P(F)=1/6[/imath] and [imath]P(G) = 1/8[/imath]. Find the probability that one of these events happens. (It does not specify which.) Thank you dear fellows. Wouldnt that be 1? The events are independent. |
336219 | Numbers ending with 0
[imath]2n[/imath] ends with [imath]0[/imath] if [imath]n[/imath] ends with [imath]0[/imath]. So how can we know if 2n ended with 0 in the first place? | 327885 | [imath]n^2[/imath] for bases to prove that the last digit of base b ends with 0
Consider numbers written in base [imath]b[/imath], where [imath]x\leq q\leq z[/imath]. For which bases [imath]q[/imath] [imath]n^2[/imath] ends with if [imath]n[/imath] ends with [imath]0[/imath]. |
313991 | Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space?
I have a non homework related question from a text and require a nice clear proof/disproof please Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space? my ideas: if [imath]H[/imath] is the subset of the topological space [imath]X[/imath] if the subset is closed in the closed subspace, the complement is open in the subspace, which means the complement is of form [imath]U\cap H[/imath] for some [imath]U[/imath] open in [imath]X[/imath] if the subspace is closed the complement is open which means complement of [imath]H=U[/imath] for some open [imath]U[/imath] in [imath]X[/imath] kind thanks | 518223 | Prove that a subset is closed
Let [imath]Y[/imath] be a subspace of a topological space [imath](X, T)[/imath]. Prove that if [imath]A[/imath] is a closed subset of [imath]Y[/imath] and [imath]Y[/imath] is a closed subset in [imath]X[/imath] then [imath]A[/imath] is a closed subset of [imath]X[/imath] My attempt: Since [imath]Y[/imath] is a subspace of [imath]X[/imath], the open sets are of the form [imath]Y\cap U[/imath] with [imath]U\in T[/imath] and [imath]Y[/imath] is closed in [imath]X[/imath] implies that [imath]X-Y[/imath] is open and [imath]A[/imath] is closed in [imath]Y[/imath] implies [imath]Y-A[/imath] is open. I am really stuck at this point. Can someone please help. |
336554 | Abstract algebra isomorphism map help
Let [imath]D=\{m+n\sqrt{2} \mid (m,n)\in\Bbb{Z}^2\}[/imath], which is an integral domain. Let [imath]Q[/imath] be its field of fractions and [imath]\phi:D\to Q[/imath] be the usual map. Find an isomorphism: [imath]\alpha:\{a+b\sqrt{2} \mid (a,b)\in\Bbb{Q}^2\}\to Q[/imath] whose restriction to [imath]D[/imath] is [imath]\phi[/imath]. You should write the map explicitly, prove that it's injective and surjective. what is field of fractions. Have no idea about this question. someone help | 331757 | Integral domain isomorphism.
The question is as follows. Let [imath]D= \{m+n\sqrt 2\text{ s.t. }(m,n) \in \mathbb Z^2\}[/imath] which is an integral domain. Let [imath]Q[/imath] be its field of fractions and [imath]\phi: D \to Q[/imath] the usual map. Find an isomorphism [imath]\alpha : \{a+b\sqrt 2\text{ s.t. }(a,b) \in \Bbb Q^2\} \to Q[/imath] whose restriction to [imath]D[/imath] is [imath]\phi[/imath]. You should write the map explicitly, prove that it's injective and surjective. EDIT So i have gone over what is written below and some other things in my textbook. I have no idea what the usual map [imath]\phi: D \to Q[/imath] is i understand that this is an integral domain going to a field of fractions Q and i believe that the below post has defined what Q must look like as a field to have this integral domain embedded in it. my confusion arises in constructing the map from the above integral domain to the below Q and proving its a bijection i know that all the pieces are here i just can't quite fit them together. i have the operation so to speak on either side of [imath]\phi[/imath] defined by above and below just that the map is eluding me. |
336855 | How to see that [imath]St(\omega, \mathcal U)=\Psi(\mathcal E)[/imath] for any open cover [imath]\mathcal U[/imath]?
How to see that [imath]St(\omega, \mathcal U)=\Psi(\mathcal E)[/imath] for any open cover [imath]\mathcal U[/imath]? Thanks ahead. | 311739 | A question on star countable space
Here is a proposition from the paper of L.P. Aiken Star-covering properties. I cannot understand why (the last line) the space is star countable. Could somebody help me? Thanks ahead. The picture is not clear. You could see the Proposition 11 of the paper. Proposition 11. If [imath]\aleph_0\leq\kappa\leq\mathfrak{c}[/imath], there exists a m.a.d.f. [imath]\mathcal{E}\subseteq [\kappa]^\omega[/imath] such that [imath]\Psi(\mathcal{E})[/imath] is star-countable. Proof. Let [imath]\mathcal{C}\subseteq[\omega]^\omega[/imath] and [imath]\mathcal{D}\subseteq[\kappa\setminus \omega]^\omega[/imath] be maximal almost disjoint families such that [imath]|\mathcal{C}|=|\mathcal{D}|=\mathfrak{c}[/imath]. This is possible because [imath]\kappa\leq\mathfrak{c}[/imath]. Choose a bijection [imath]f:\mathcal{C}\to\mathcal{D}[/imath]. Define [imath]\mathcal{E}=\{C\cup f(C):C\in\mathcal{C}\}[/imath]. If [imath]A\in[\kappa]^\omega[/imath], [imath]|A\cap\omega|=\aleph_0[/imath] or [imath]|A\cap (\kappa\setminus\omega)|=\aleph_0[/imath]. In either case, [imath]A[/imath] has infinite intersection with some element of [imath]\mathcal{C}\cup\mathcal{D}[/imath], thus [imath]\mathcal{E}[/imath] is maximal. Then [imath]\Psi(\mathcal{E})[/imath] is star-countable because [imath]\mathsf{St}(\omega,\mathscr{U})=\Psi(\mathcal{E})[/imath], for any open cover [imath]\mathscr{U}[/imath] of [imath]\Psi(\mathcal{E})[/imath]. [imath]\;\;\Box[/imath] (original image) |
337283 | Weak Derivative is [imath]0[/imath]
If the weak derivative of some function [imath]u[/imath] on a open connected set is 0. Does it mean that [imath]u[/imath] is constant almost everywhere on the given set? | 78142 | The constant distribution
If [imath]u[/imath] is a distribution in open set [imath]\Omega\subset \mathbb R^n[/imath] such that [imath]{\partial ^i}u = 0[/imath] for all [imath]i=1,2,\ldots,n[/imath]. Then is it necessarily that [imath]u[/imath] is a constant function? |
336702 | Infinite powering by [imath]i[/imath]
Find the value of: [imath]i^{i^{i^{i^{i^{i^{....\infty}}}}}}[/imath] Simply infinite powering by i's and the limiting value. Thank you for the help. | 180936 | Complex towers: [imath]i^{i^{i^{...}}}[/imath]
If [imath]w = z^{z^{z^{...}}}[/imath] converges, we can determine its value by solving [imath]w = z^{w}[/imath], which leads to [imath]w = -W(-\log z))/\log z[/imath]. To be specific here, let's use [imath]u^v = \exp(v \log u)[/imath] for complex [imath]u[/imath] and [imath]v[/imath]. Two questions: How do we determine analytically if the tower converges? (I have seen the interval of convergence for real towers.) Both the logarithm and Lambert W functions are multivalued. How do we know which branch to use? In particular [imath]i^{i^{i^{...}}}[/imath] numerically seems to converge to one value of [imath]i2W(-i\pi/2)/\pi[/imath]. How do we establish this convergence analytically? (Yes, I have searched the 'net, including the tetration forum. I haven't been able to locate the answer to this readily.) |
337717 | Why is this fact about the totient function true?
[imath] \sum_{k<n} {gcd(k,n)=1}k = \frac{1}{2} n \phi(n)[/imath] This is a homework problem. I would ideally like to get to the final proof on my own. But at the moment I can't even decide how to begin. The only relationship I can see between the sum and the right side of the equation is that the left adds all the things relatively prime with [imath]n[/imath] less than [imath]n[/imath], i.e., the things elements of the group [imath]\Phi(n)[/imath]. The order of this group is [imath]\phi(n)[/imath]. But that doesn't seem to help me in looking for a way about solving this. | 336593 | Prove: [imath]\sum_{k[/imath]
Prove: [imath]\sum_{k<n, (k,n)=1}k = \frac{1}{2}n \varphi (n)[/imath] I have had strep throat and missed the lecture discussing properties of the Euler function. Any help in solving this is appreciated. Thank you! |
337872 | Find the necessary and sufficient conditions for all [imath] 41 \mid \underbrace{11\ldots 1}_{n}[/imath], [imath]n\in N[/imath].
Find the necessary and sufficient conditions for all [imath] 41 \mid \underbrace{11\ldots 1}_{n}[/imath], [imath]n\in N[/imath]. And, if [imath]\underbrace{11\ldots 1}_{n}\equiv 41\times p[/imath], then [imath]p[/imath] is a prime number. Find all of the possible values of [imath]n[/imath] to satisfy the condition. | 330975 | How to find the values of [imath]n[/imath] for which [imath]\displaystyle\underbrace{111\cdots1111}_{n}\equiv0 \pmod {41}[/imath]?
What are the values of [imath]n[/imath] satisfying [imath]\displaystyle\underbrace{111\cdots1111}_{n}\equiv0 \pmod {41}[/imath]? I think [imath]n=5k[/imath], with [imath]k=1,2,\cdots[/imath], but I can't prove it. |
337867 | Question about two different definitions of neighborhood in the topological context
I read two different definitions of neighborhood: The definition in Flegg's book "From Geometry to Topology": If [imath]X[/imath] is a set which, together with some topology [imath]T[/imath] on [imath]X[/imath], is a topological space, then any subset [imath]N[/imath] of [imath]X[/imath] is a neighborhood of an element [imath]x∈X[/imath] if it includes an open set containing [imath]x[/imath]. The definition from the textbook by O.Y.Viro et al.: By a neighborhood of a point one means any open set containing this point. The two definitions are different. And textbook in 2 also points out that, as with definition in 1, analysts and French mathematicians (following N. Bourbaki) prefer a wider notion of neighborhood: they use "neighborhood" for any set containing a neighborhood defined in 2. That means, neighborhood are defined ambiguously in pure topological space. Why neighborhood is defined differently? Is the ambiguity is allowed in its definition, because "neighborhood", which matters in metric space, doesn't matters anymore in topological space, but rather give its place to "open set"? | 80025 | A question about the definition of a neighborhood in topology
Let [imath]X[/imath] be a topological space, and [imath]x \in X[/imath] be a point. There are two prevalent conventions on how to define a neighborhood of [imath]x[/imath]: Alternative Definitions (Neighborhood): 1) A neighborhood of [imath]x[/imath] is any open subset [imath]W \subset X[/imath] such that [imath]x \in W[/imath]. (This convention is used in Munkres's book for example.) 2) A neighborhood of [imath]x[/imath] is a subset [imath]W \subset X[/imath] such that there exists an open set [imath]A[/imath] such that [imath]x \in A \subset W[/imath]. (For example this is the definition in Bourbaki's or Willard's "General Topology") Thus, every neighborhood in the sense of (1) is a neighborhood in the sense of (2), but not vice-versa. One often needs to show that a neighborhood of a point [imath]x[/imath] in the sense of (2) is actually open, and often this is a non-trivial verification from the given context. An example that I can come up with now (and this example was a motivation for asking this question) is the following: Let [imath]G[/imath] be a topological abelian group, and [imath]H[/imath] a subgroup of [imath]G[/imath] which is also a neighborhood of [imath]0 \in G[/imath] in the sense of 2). Then one can show that [imath]H[/imath] is in fact an open set. [The trick is to observe that for a given [imath]g \in G[/imath], the map [imath]\phi_{g} :G \rightarrow G[/imath] given by for [imath]x \in G, \phi_{g}(x) = g + x[/imath] is a homeomorphism, and so any neighborhood of a point [imath]g \in G[/imath] is of the form [imath]g + U[/imath], where [imath]U[/imath] is a neighborhood of [imath]0[/imath]. Thus, for an [imath]h \in H[/imath], [imath]h + H[/imath] is a neighborhood of [imath]h[/imath], and moreover, [imath]h + H \subset H[/imath] because [imath]H[/imath] is a subgroup.] The above proof shows that sometimes proving that a neighborhood in the sense of (2) is open is not completely trivial, while a neighborhood in the sense of (1) is always open. At times like these, I wonder why the second definition of a neighborhood is used at all. But I have learned topology primarily from Munkres, and so I might be ignorant of the advantages of definition (2). So, what do you think are some of the advantages of using (2) as a definition for a neighborhood of a point [imath]x \in X[/imath], where [imath]X[/imath] is a topological space? (This might be a duplicate question. But I searched a little bit and could not find a question that was exactly similar to this one. So, excuse me if I have asked something that was already asked.) |
338047 | How to effectively calculate [imath](1/\sqrt1 + \sqrt2) + (1/\sqrt2 + \sqrt3) +\cdots + (1/\sqrt{99} + \sqrt{100})[/imath]
I have this series: [imath]\frac{1}{\sqrt1 + \sqrt2} +\frac{1}{\sqrt2 + \sqrt3} +\frac{1}{\sqrt3 + \sqrt4} +\cdots+\frac{1}{\sqrt{99} + \sqrt{100}} [/imath] My question is, what approach would you use to calculate this problem effectively? | 310962 | Value of [imath] \sum \limits_{k=1}^{81} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{1}{\sqrt{1} + \sqrt{2}} + \cdots + \frac{1}{\sqrt{80} + \sqrt{81}} [/imath]?
I tried my best, but I am totally clueless about it. Worse thing is we were supposed to arrive at the answer in approximately [imath] 2 [/imath] minutes. The correct answer is [imath] 8 [/imath], right? Can you kindly explain how to arrive at it? I hope it won’t be too much bother. Thank you. |
338178 | Prove that [imath]+[/imath] is associative where [imath]A+B=( A \setminus B )\cup( B \setminus A ) [/imath].
If [imath]A[/imath] and [imath]B[/imath] are sets, then [imath]A+B=( A \setminus B )\cup( B \setminus A ) [/imath]. Prove that [imath]+[/imath] is an associative operation. How do I prove these? | 325724 | Proof that the [imath](\mathcal P (\mathbb N),\triangle)[/imath] is an abelian group?
I'm stuck trying to figure out how to prove that [imath](\mathcal P (\mathbb N) \space , \space \triangle)[/imath] is an abelian group? I know the definition, but I'm confused how to incorporate the fact that I'm dealing with a powerset instead of just the integers or natural numbers, so I'm seeking help to understand how proofs on sets works. Also, I'm using [imath]\triangle[/imath] to mean symmetric difference. |
339354 | the last decimal digit of every even perfect number is always 6 or 28.
how to Prove that the last decimal digit of every even perfect number is always [imath]6[/imath] or [imath]28[/imath]. | 293443 | even perfect numbers and primes
Thank in advance to m.s.e site. I am looking for discussions/proof of the 1) If $p ([imath]2^r[/imath]/2$) is an even perfect number then [imath]p[/imath] should be in the form of [imath]2^r[/imath] - 1 2) Every even perfect number ends with 6 or 28. why/justify 3) If [imath]a^k[/imath] - 1 is prime for [imath]a > 0[/imath] and [imath]k\ge 2[/imath]. If we fix [imath]a = 2[/imath], then [imath]k[/imath] becomes prime. Also, in what cases of a, we get primes always? |
339317 | show that if [imath]g \cdot b \equiv 1 \pmod n[/imath], then [imath]b[/imath] is also a primitive root of [imath]U_n[/imath]
Show that if [imath]g[/imath] is a primitive root of [imath]U_n[/imath] and [imath]g \cdot b \equiv 1 \pmod n[/imath], then [imath]b[/imath] is also a primitive root of [imath]U_n[/imath]. What property of primitive root should I use? How about [imath]g \in U_n[/imath] is a primitive root if [imath]g^{\frac{\phi(n)}{p}} \not\equiv{1} \pmod n[/imath]?? because [imath]n[/imath] isnt necessary a prime.. thats not much I can do.. Any hints?? | 339304 | Prove that if [imath]g[/imath] is a primitive root of [imath]n[/imath] and [imath]g*b \equiv 1 \pmod n[/imath], then [imath]b[/imath] is also a primitive root of [imath]n[/imath].
Some useful facts I am trying to use: If the multiplicative group [imath]U_n[/imath] modulo [imath]n[/imath] is a cyclic group, a generator [imath]g[/imath] of [imath]U_n[/imath] is called a primitive root of [imath]n[/imath]. if [imath]g[/imath] in [imath]U_n[/imath] is a primitive root, then [imath]|g|= \phi(n)[/imath] where [imath]\phi[/imath] is the euler phi function. An element [imath]g[/imath] in [imath]U_n[/imath] is a primitive root if and only if [imath]g^{\phi(n)/p}[/imath] not congruent to [imath]1\pmod n[/imath] for each prime dividing [imath]\phi(n)[/imath] |
325491 | Proof of Wolstenholme's theorem
According to the theorem, if [imath]1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}[/imath] then we have to prove that [imath]r\equiv0 \pmod{p^2}[/imath]. (Given [imath]p>3[/imath], otherwise [imath]1+\dfrac{1}{2}=\dfrac{3}{2}[/imath], [imath]3 \not\equiv 0 \pmod 9[/imath].) I guess there's a [imath](\bmod p)[/imath] solution for this, but I don't really get how to start it. | 1318510 | If for a prime p [imath]1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{a}{b}[/imath] then show that p divides a. Moreover if [imath]p>3[/imath] then [imath]p^2[/imath] divides a.
Here is a problem from Topics in algebra - Herstein. If for a prime p [imath]1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{a}{b}[/imath] then show that p divides a. Moreover if [imath]p>3[/imath] then [imath]p^2[/imath] divides a. I am able to prove the first part by writing [imath]1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{(p-1)!+ \frac{(p-1)!}{2}+\frac{(p-1)!}{3}+\ldots + 1 }{(p-1)!}[/imath] we can show the numerator to be [imath]\equiv 0 (\bmod p)[/imath]. I cant seem to crack the second part. Thanks in advance for any help. |
339703 | Proving that there are relatively prime elements (Posa problem)
How to prove that for each [imath]n+1[/imath] an element from the set [imath]\{1,2,3,4,\ldots,2n-1,2n\}[/imath] there are two relatively prime elements between them | 49080 | To prove there exist two relatively prime numbers in a finite set
Prove that in any set of [imath]n+1[/imath] positive numbers not exceeding [imath]2n[/imath] there must be two that are relatively prime. |
339871 | Galois automorphisms and Field extensions
Let [imath]\alpha[/imath] be a root of [imath]x^3+3x-1[/imath] and let [imath]\beta[/imath] be a root of [imath]x^3-x+2[/imath]. I want to show that [imath]\alpha^2+\beta[/imath] has degree [imath]9[/imath]. There are many ways to do this, but I wish to solve the problem by the following line of argument: using the automorphisms of the normal closure of [imath]\mathbb{Q}(\alpha,\beta)[/imath] might allows us to show that the latter contains at most two cubic subfields, which thus have to be [imath]\mathbb{Q}(\alpha)[/imath] and [imath]\mathbb{Q}(\beta)[/imath]. Can somebody help? Edit: The answer by Jyrki Lahtonen in this other post goes exactely in that direction, but it is not completely clear to me, and I believe/hope the argument could me made cleaner. | 334597 | degree of a field extension
Let [imath]\alpha[/imath] be a root of [imath]x^3+3x-1[/imath] and [imath]\beta[/imath] be a root of [imath]x^3-x+2[/imath]. What is the degree of [imath]\mathbb{Q}(\alpha^2+\beta)[/imath] over [imath]\mathbb{Q}[/imath]? My guess is 9, because i found a monic polynomial of degree 9 with integer coefficient, irreducible in [imath]\mathbb{Q}[x][/imath] with [imath]\alpha^2+\beta[/imath] as a root. But i don't know if this suffices. Any help? |
340358 | Solving Chinese Remainder Theorem
If [imath]x = am + k = bn - k[/imath] for known [imath]a,b,k[/imath] and [imath]\gcd(a,b)[/imath], how to solve for [imath]x[/imath] using Chinese Remainder Theorem? | 340319 | Solving for minimum [imath]x[/imath] given divisibility constraints
If [imath]x-k[/imath] is divisible by [imath]a[/imath] and [imath]x+k[/imath] is divisible by [imath]b[/imath] with [imath]a, b, k[/imath] known, how do I solve for [imath]x[/imath]? All numbers here are positive integers. |
340768 | prove this integral
[imath]\int_{0}^{\frac{\pi}{2}}x^2\sqrt{\tan{x}}\sin{2x}dx=\frac{\pi\sqrt{2}}{32}\left[\frac{5\pi^2}{6}+2\pi-2\pi\log{2}-4+4\log{2}-2\log^2{2}\right] [/imath] I think this integral is very nice. and I think this integral have nice methods? my methods:let [imath]\sqrt{\tan{x}}=t[/imath] | 284933 | A definite integral with trigonometric functions: [imath]\int_{0}^{\pi/2} x^{2} \sqrt{\tan x} \sin(2x) \, \mathrm{d}x[/imath]
How could we get a closed form for the following integral: [imath]\int_{0}^{\frac{\pi }{2}}{{{x}^{2}}\sqrt{\tan x}\sin \left( 2x \right) \, \mathrm{d}x} \tag{1}[/imath] While the antiderivative of [imath]\sqrt{\tan x} \sin(2x)[/imath] can be expressed in terms of elementary functions according to Wolfram Alpha, the antiderivative of [imath]x^{2} \sqrt{\tan x} \sin(2x)[/imath] seeminly cannot be expressed in closed form. To evaluate [imath](1)[/imath], would introducing a parameter and differentating under the integral sign be helpful? |
339989 | strictly positive elements in [imath]C^*[/imath]-algebra
Let [imath]A[/imath] be a [imath]C^*\text{-algebra}[/imath] and [imath]A_+[/imath] denote the positive elements. An element [imath]a\in A_+[/imath] is called strictly positive if [imath]\overline{aAa}=A[/imath]. Want to find the following:a)What are the strictly positive elements of [imath]C_0(\Omega)[/imath] where [imath]\Omega[/imath] is locally compact Hausdorff space, and [imath]C_0(\Omega)[/imath] is the space of continuous complex valued functions that vanishes at [imath]\infty[/imath].b)If [imath]A[/imath] is unital, then [imath]a\in A_+[/imath] is strictly positive iff [imath]a[/imath] is invertible.c)if [imath](e_n)[/imath] is an approximate identity of [imath]A[/imath], then [imath]a:=\sum_{n=}^{\infty}\frac{1}{2^n}e_n[/imath]is strictly positive.What I tried and know:a) I know that positive elements in this space are functions with positive images, and thus, strictly positive elements are functions with positive images and now real root. is this correct? and how can I show that the positive elements are these kind of functions.b) I figures out that if [imath]a[/imath] in invertible then [imath]a[/imath] is strictly positive, but no idea how to do the other direction.c)I did this but using another characterization of strictly positive elements but want to do it with this one, and I don't know how.Thank you for your help. | 342020 | How to prove this element is strictly positive?
Let [imath]A[/imath] be a [imath]C^*\text{-algebra}[/imath] and [imath]A_+[/imath] denote the positive elements. An element [imath]a\in A_+[/imath] is called strictly positive if [imath]\overline{aAa}=A[/imath]. Want to prove: if [imath](e_n)[/imath] is an approximate identity of [imath]A[/imath] , then [imath]a:=\sum_{n=1}^{\infty}\frac{1}{2^n}e_n[/imath]is strictly positive. i.e, I want to prove that if [imath]b\in A[/imath], then I can find a sequence [imath]x_n\in A[/imath] such that [imath]b=\lim(ax_na)[/imath]. By an approximate identity of [imath]A[/imath] we mean that : [imath](e_n)[/imath] is increasing, positive,[imath]\|e_n\|<1[/imath] and for all [imath]a\in A[/imath] : [imath]\|(ae_n)-a\|[/imath] converges to [imath]0[/imath] . Any help or suggestion is really appreciated. Thank you! |
265483 | Evaluating the infinite product [imath]\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)[/imath]
Evaluate the infinite product [imath]\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).[/imath] I can't see anything in this limit , so help me please. | 575105 | Evaluating product [imath]\prod_{n=2}^\infty\left(1-\frac{1}{n^2}\right)[/imath]
I'm reading about infinite products in complex analysis, where there is a theorem like The product [imath]\prod_{n=1}^\infty\left(1+a_n\right)[/imath] converges absolutely iff the series [imath]\sum_{n=1}^\infty|a_n|[/imath] converges. Then an exercise is to show that [imath]\prod_{n=2}^\infty\left(1-\dfrac{1}{n^2}\right)=\dfrac{1}{2}[/imath] The theorem above guarantees that the product converges, but what is the method to evaluate its value? |
341100 | Help with graph induction question?
Given a graph [imath]G[/imath] with [imath]n[/imath] vertices, where [imath]n[/imath] is even, prove by induction that if every vertex has degree [imath]n/2 + 1[/imath], then [imath]G[/imath] must contain a 3-cycle. A 3-cycle is a set of 3 vertices, [imath]a; b; c[/imath] such that [imath]ab[/imath] is an edge, [imath]bc[/imath] is an edge and [imath]ac[/imath] is an edge. | 341847 | Proving a Graph contains a [imath]3[/imath]-cycle.
Given a graph [imath]G[/imath] with [imath]n[/imath] vertices, where [imath]n[/imath] is even, prove that if every vertex has degree [imath]\frac{n}{2}+1[/imath], then [imath]G[/imath] must contain a [imath]3[/imath]-cycle. (A [imath]3[/imath]-cycle is a set of [imath]3[/imath] vertices, [imath]a[/imath]; [imath]b[/imath]; [imath]c[/imath] such that [imath]ab[/imath] is an edge, [imath]bc[/imath] is an edge and [imath]ac[/imath] is an edge.) |
341269 | Prove by induction help?
I'm trying to study for a test and one of the practise questions is very confusing and not sure what to do: Prove by induction that [imath]\sum_{i=1}^n \frac{3}{4^i} < 1[/imath] for all [imath]n \geq 2[/imath] The furthest I'm able to get is getting rid of the summation. So I get [imath](3/4)^i n< 1[/imath] for all [imath]n \geq2[/imath]. Any help would be greatly appreciated. Thanks. | 340087 | How would you prove [imath]\sum_{i=1}^{n} (3/4^i) < 1[/imath] by induction?
How would you prove this by induction? [imath]\sum_{i=1}^{n} \frac{3}{4^i} < 1 \quad \quad \forall n \geq 2[/imath] I can do the base case but don't know how to to finish it. |
341598 | riesz frechet for a operator
Let [imath]H[/imath] be a Hilbert space with orthonormal basis [imath](e_{n})_{n\in\mathbb{N}}[/imath]. Furthermore, let [imath]T\colon H\rightarrow C[a,b][/imath] be a bounded operator. a) Let [imath]x\in [a,b][/imath]. Show that there is a unique [imath]g_{x}\in H[/imath] with [imath]\langle f,g_x\rangle=(Tf)(x)[/imath] and all [imath]f \in H[/imath]. My solution: Let [imath]x\in[a,b][/imath], define the linear continuous map [imath]L_{x}:H\rightarrow \mathbb{K}[/imath], with [imath]\mathbb{K}=\mathbb{C}[/imath] or [imath]\mathbb{K}=\mathbb{R}[/imath] by [imath] L_{x}(f):=T(f)(x).[/imath] Since [imath]H[/imath] is an Hilbert space and [imath]L_{x}:H\rightarrow \mathbb{K}[/imath] is a bounded linear functional on [imath]H[/imath] we can apply the Riesz-Frechet theorem. According to the Riesz-Frechet theorem there exists a unique [imath]g_{x}\in H[/imath] such that for [imath]x\in[a,b][/imath] and all [imath]f\in H[/imath] [imath]L_{x}(f)=T(f)(x)=(Tf)(x)=<f,g_{x}>.[/imath] Question 1: Is this correct? Or am I missing something? | 340780 | Bounded operator and Compactness problem
Let [imath]H[/imath] be a Hilbert space with orthonormal basis [imath](e_{n})_{n\in\mathbb{N}}[/imath]. Furthermore, let [imath]T\colon H\rightarrow C[a,b][/imath] be a bounded operator. a) Let [imath]x\in [a,b][/imath]. Show that there is a unique [imath]g_{x}\in H[/imath] with [imath]\langle f,g_x\rangle=(Tf)(x)[/imath] and all [imath]f \in H[/imath]. Updated version 1. So far I have: Let [imath]x\in[a,b][/imath], define the linear continuous map [imath]L_{x}:H\rightarrow \mathbb{K}[/imath], with [imath]\mathbb{K}=\mathbb{C}[/imath] or [imath]\mathbb{K}=\mathbb{R}[/imath] by [imath] L_{x}(f):=T(f)(x).[/imath] Since [imath]H[/imath] is an Hilbert space and [imath]L_{x}:H\rightarrow \mathbb{K}[/imath] is a bounded linear functional on [imath]H[/imath] we can apply the Riesz-Frechet theorem. According to the Riesz-Frechet theorem there exists a unique [imath]g_{x}\in H[/imath] such that for [imath]x\in[a,b][/imath] and all [imath]f\in H[/imath] [imath]L_{x}(f)=T(f)(x)=(Tf)(x)=\langle f,g_{x}\rangle.[/imath] Question 1: have I proven it correctly? Or am I missing some important details? b) Show that [imath]\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|(Te_{j})(x)|^{2}<\infty.[/imath] A hint is that [imath]||g_{x}||\leq ||T||_{H \rightarrow C[a,b]}[/imath]. Updated version 1. Question 2: how do I prove this hint? I want to prove it before I make use of it. For the rest of the problem I have this so far: Using part a), we have that: [imath]\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|(Te_{j})(x)|^{2}=\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}.[/imath] Since the [imath]e_{j}[/imath] form an orthonormal basis in [imath]H[/imath], Bessel's inequality yields [imath]\sup_{x\in [a,b]}\sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}\leq \sup_{x\in [a,b]}||g_{x}||_{2}^{2}.[/imath] Now by definition of the operator norm we have: [imath]\sup_{x\in [a,b]}\sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}\leq \sup_{x\in [a,b]}||g_{x}||_{2}^{2}\leq ||T||\cdot ||g_{x}||.[/imath] Using the hint and the fact that operator [imath]T[/imath] is bounded we get: [imath]\sup_{x\in [a,b]}\sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}\leq \sup_{x\in [a,b]}||g_{x}||_{2}^{2}\leq ||T||\cdot ||g_{x}||\leq ||T||\cdot ||T||_{{H \rightarrow C[a,b]}} <\infty.[/imath] Question 3: Is the proof now complete or am I missing a detail/making a mistake? In class for example for Bessel's inequality we usually had this form [imath]\sum_{j=1}^{\infty}|\langle f,e_{j}\rangle|^{2}\leq ||f||_{2}^{2}<\infty[/imath]. c) Show that [imath]\sum_{j=1}^{\infty} ||Te_{j}||_{L^{2}}^{2}< \infty[/imath]. Updated version 1. What I have so far: We define the function [imath]x\mapsto\sum_{j=1}^{\infty}|Te_{j}(x)|^{2}.[/imath] We have from part b): [imath]\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|(Te_{j})(x)|^{2}<\infty.[/imath] Integrating the function over [imath](a,b)[/imath] yields [imath]\int_{a}^{b}\sum_{j=1}^{\infty}|Te_{j}(x)|^{2}dx<\infty.[/imath] Now making use of the Fubini-Tonelli theorem we can interchange limit and integral and get: [imath]\int_{a}^{b}\sum_{j=1}^{\infty}|Te_{j}(x)|^{2}dx=\sum_{j=1}^{\infty}\int_{a}^{b}|Te_{j}(x)|^{2}dx=\sum_{j=1}^{\infty}||Te_{j}||_{L^{2}}^{2}<\infty.[/imath] Question 4: Is the proof now complete or am I missing a detail/making a mistake? In fact the interchange is not clear to me. We did treat Fubini in class but only shorty to change integrals not integral and sum. Usually when changing integral and summation we used monotone convergence. d) Show that [imath]T\colon H\rightarrow L^{2}(a,b)[/imath] is compact. We have to get the result by making use of estimates. Updated version 1. I could prove it without estimating and had: We note that [imath]H[/imath] and [imath] L^{2}(a,b)[/imath] are Hilbert spaces and the operator [imath]T\colon H\rightarrow L^{2}(a,b)[/imath] is a bounded linear operator as given before. Furthermore in part c) we obtained an inequality. Thus we have that [imath]T\colon H\rightarrow L^{2}(a,b)[/imath] is an abstract Hilbert-Schmidt operator. Now we can apply a certain theorem that states that every abstract Hilbert-Schmidt operator is compact. Now for the proof making use of estimates: To show that [imath]T[/imath] is compact, we have to approximate [imath]T[/imath] in the operator norm by finite rank operators. For [imath]N\in\mathbb{N}[/imath] define the linear operator [imath]T_{N}\colon H\rightarrow L^{2}(a,b)[/imath]. Question 5: I don't know how to show that [imath]T_{N}[/imath] is of finite-rank? Normally you would have something along the lines that [imath]T_{N}[/imath] has range within [imath]span\{f_{1},\ldots,f_{N}\}[/imath] and hence is of finite-rank. But for this problem I don't see it. I think you can say that [imath]ran(T)\subseteq C[a,b][/imath], but [imath]C[a,b][/imath] is infinite diminesional so this confuses me. Question 6: How do I show that [imath]||T-T_{N}||\rightarrow 0[/imath], so I can conclude that [imath]T[/imath] is a compact operator. |
341462 | How do I determine the intersection of span A and span B?
Consider the following 2 sets of vectors in [imath]\mathbb R^4[/imath]: [imath]A = \{v_1, v_2, v_3\}, B = \{w_1, w_2, w_3\}[/imath]. You are given that [imath]A[/imath] is a set of linearly independent vectors and that [imath]B[/imath] is a set of linearly independent vectors. Let [imath]v_1 = (3,1,4,1), v_2 = (5,9,2,6), v_3 = (5,3,5,8), w_1 = (9,7,9,3), w_2 = (2,3,6,4), w_3 = (6,2,8,4)[/imath]. (Wrote them sideways when it should be top to bottom in brackets but left to right shown) Determine the intersection of span[imath]\,A[/imath] and span[imath]\,B[/imath] and write your answer as the span of a set of linearly independent vectors. I'm really lost in class. Please show steps and answers that I can learn. Please help... Thank you | 341373 | Based on 2 sets of vectors in [imath]\mathbb{R}^4[/imath], how do I determine a system of equations? (Linear Algebra) Please help?
Consider the following 2 sets of vectors in [imath]\mathbb{R}^4[/imath]: [imath]A = \{v_1, v_2, v_3\}, B = \{w_1, w_2, v_3\}[/imath]. You are given that [imath]A[/imath] is a set of linearly independent vectors and that [imath]B[/imath] is a set of linearly independent vectors. The intersection of 2 sets is the set of elements that are common to both sets. Suppose [imath]u[/imath] is in the intersection of span [imath]A[/imath] and span [imath]B[/imath]. Determine a system of equations that could be used to determine all such [imath]u[/imath]. Please show steps and answers so that I can learn. Thank you so much. |
341884 | Green's Theorem.
I have a question I can't really solve using Green's theorem for some reason. There's a path C that's closed, it's a piecewise smooth curve formed by the traveling in straight lines between (-2, 1), (-2, -3), (1, -1), (1, 5) and back to (-2, 1). We're supposed to use Green's theorem to evaluate: [imath] \int_C (2xy)dx + (xy^2)dy [/imath] So I applied Green's theorem and got: [imath] \int\int_D (y^2 - 2x)dxdy [/imath] But the problem is, I can't find the region to evaluate the integral I got. The region doesn't look like a square/rectangle, and it's sort of uneven so I can't really find an equation that would best define it. Any sort of hints? | 338576 | Green's Theorem
Let [imath]C[/imath] be the closed,piecewise curve figured by traveling in straight lines between the points [imath](-2,1),(-2,-3),(1,-1),(1,5)[/imath] and back to [imath](-2,1)[/imath], in that order. Use Green's Theorem to evaluate the integral: [imath]\int_C (2 x y)dx + (x y^2)dy[/imath] So far I have used Green's Theorem and calculated the vector form of Green's Theorem. Since the region enclosed by C is y-simple, we can set up the double integral to first integrate in terms of [imath]y(x)[/imath] then [imath]x[/imath]. However, I am not getting the same answer as the book. |
279055 | What is the solution to the equation below?
Solve the equation below. [imath]x^2+\frac{81x^2}{(9+x)^2}=40[/imath] I couldn't solve it after trying many time. | 233131 | How one should solve [imath]x^2+\frac{81x^2}{(x+9)^2}=40[/imath]
As in the title, Please help me solve [imath]x^2+\frac{81x^2}{(x+9)^2}=40[/imath] Thanks. |
343973 | If [imath]f(x) \le g(x) \le h(x)[/imath] for all [imath]x\in D[/imath] and [imath]f[/imath], [imath]h[/imath] are Riemann integrable, then so is [imath]g[/imath]. True or False? (Check my work)
If [imath]f(x) \le g(x) \le h(x)[/imath] for all [imath]x\in [a,b][/imath], and f and h are Riemann integrable on [a,b], then so is g. True or false? Explain. A: True Proof: Since [imath]f\in R[a,b][/imath], [imath]\bar \int_a^b{f} = \int_a^b\bar{f}[/imath] (That's supposed to be an underbar), so [imath]\int_a^b{f} \le U(P,f)[/imath]. Since [imath]h\in R[a,b][/imath], [imath]\bar \int_a^b{h} = \int_a^b\bar{h}[/imath], so [imath]\int_a^b{h} \ge L(P,h)[/imath]. Therefore, since [imath]f(x)\le g(x) \le h(x)[/imath], and [imath]\int_a^b{f} \le U(P,f)[/imath] and [imath]\int_a^b{h}\ge L(P,f)[/imath], [imath]U(P,f) \le \int_a^b{g} \le L(P,h)[/imath] which implies [imath]U(P,f) \le L(P,g) \le U(P,g) \le L(P,h)[/imath] So [imath]g(x)\in R[a,b][/imath]. Found it: If [imath]f(x)[/imath] and [imath]g(x)[/imath] are Riemann integrable and [imath]f(x)\leq h(x)\leq g(x)[/imath], must [imath]h(x)[/imath] be Riemann integrable? | 85839 | If [imath]f(x)[/imath] and [imath]g(x)[/imath] are Riemann integrable and [imath]f(x)\leq h(x)\leq g(x)[/imath], must [imath]h(x)[/imath] be Riemann integrable?
Let [imath]f[/imath] and [imath]g[/imath] be Riemann integrable (real) functions and [imath]f(x)\leq h(x)\leq g(x).[/imath] Is it true that [imath]h(x)[/imath] is Riemann integrable? Can someone post a proof (if there is)? Thanks. |
336167 | Every number less than [imath]n![/imath] can be written as a sum of at most [imath]n[/imath] divisors.
Prove that, for every [imath]n > 1[/imath], any number that is less than [imath]n![/imath] can be written as a sum of at most [imath]n[/imath] distinct divisors of [imath]n![/imath]. [Hint: Use induction and Euclidean division with remainder]. | 178122 | Prove that for any [imath]x \in \mathbb N[/imath] such that [imath]x is the sum of at most n distinct divisors of n![/imath]
Prove that for any [imath]x \in \mathbb N[/imath] such that [imath]x<n![/imath] is the sum of at most [imath]n[/imath] distinct divisors of [imath]n![/imath]. |
342637 | Does [imath]\sum\frac{\sin n}{n}[/imath] converge?
Does [imath]\sum\frac{\sin n}{n}[/imath] converge? I have tried the comparison test, root test and ratio test but still can't prove it is convergent or divergent. | 15984 | Does [imath]\sum \limits_{n=1}^\infty\frac{\sin n}{n}(1+\frac{1}{2}+\cdots+\frac{1}{n})[/imath] converge (absolutely)?
I've had no luck with this one. None of the convergence tests pop into mind. I tried looking at it in this form [imath]\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}[/imath] and apply Dirichlets test. I know that [imath]\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n} \to 0[/imath] but not sure if it's decreasing. Regarding absolute convergence, I tried: [imath]|\sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}|\geq \sin^2 n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=[/imath] [imath]=\frac{1}{2}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}-\frac{1}{2}\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}[/imath] But again I'm stuck with [imath]\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}[/imath]. Assuming it converges then I've shown that [imath]\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}[/imath] doesn't converge absolutely. |
344536 | Curious [imath]\sum _{n=1}^{\infty} \frac{1}{n^2 - x^2}[/imath] identity
Let [imath]F(x) = \sum _{n=1}^{\infty} \frac{1}{n^2 - x^2}[/imath] It seems that for odd integer [imath]k[/imath] [imath]F\left(\frac{k}{2}\right) = \frac{2}{k^2}[/imath] My evidence is strictly computational, and I have no idea how to approach a proper proof. So standard questions are: is it a known fact is it a fact is it curious what is a proof strategy Please advise. | 141470 | Find the sum of [imath]\sum 1/(k^2 - a^2)[/imath] when [imath]0[/imath]
So I have been trying for a few days to figure out the sum of [imath] S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} [/imath] where [imath]a \in (0,1)[/imath]. So far from my nummerical analysis and CAS that this sum equals [imath] S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] [/imath] But I have not been able to prove this yet. Anyone know how? My guess is that the sum of this series is related to fourier-series but nothing particalr comes to mind. For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers. |
344785 | Can we really understand [imath]R[/imath] by studying [imath]R[/imath]-modules?
According to Algebra: Chapter 0, the category [imath]R\operatorname{-Mod}[/imath] reveals a lot about [imath]R[/imath]. However after completing the first eight chapters, I still found no examples where this happens. Can someone give some honest examples where the category of modules do tell us a lot about the underlying ring? Better still, such kind of information is very difficult if one looks at the ring directly. Thanks very much! | 124077 | What does [imath]R-\mathrm{Mod}[/imath] tell us about [imath]R[/imath]?
I have read the phrase "a good way to study a ring R is to study its modules" many times - though I cannot really see why this should be a general phenomenon. Is this phenomenon an analogue to (or an instance of?) the fact that studying field extensions of [imath]\mathbb{Q}[/imath] can tell us something about [imath]\mathbb{Q}[/imath] itself? For example, by studying factorizations in [imath]\mathbb{Z}[i][/imath], one can draw conclusions about factorizations in [imath]\mathbb{Z}[/imath]. Another example is factor rings (If [imath]R[/imath] is a ring and [imath]I[/imath] is in ideal, then [imath]R/I[/imath] as an [imath]R[/imath]-module). Yet another example is localizations. For example, if [imath]Y[/imath] is a variety (or a manifold), then the local ring at a point P tells us how functions on [imath]Y[/imath] behave near [imath]P[/imath]. But all these examples seem more like lucky coincidences (because so many objects are R-modules), than an instance of a general phenomenon (that "a good way to study a ring [imath]R[/imath] is to study its modules"). So is there more to the phrase "a good way to study a ring [imath]R[/imath] is to study its modules", or is it just a shorthand way of saying "to study a ring [imath]R[/imath], study its quotients, its localizations and its extension..."? |
344983 | Differentiation help required!
Let [imath]f_k:(a,b)\to \mathbb{R}[/imath] be differentiable and non-zero [imath](f_k(x)\neq0[/imath] for all [imath]a<x<b),\ k=1,\cdots,n[/imath] Find [imath]\frac{d}{dx}f_1(x)f_2(x)\cdots f_n(x)[/imath]. I know that the problem should be solved by using induction and [imath](f_1f_2)' = f_1'f_2 + f_1f_2'[/imath]. But how am I supposed to start? Help appreciated! | 344861 | Find [imath]\frac{d}{dx}\left(f_1(x)f_2(x)\cdots f_n(x)\right)[/imath].
Let [imath]f_k:(a,b)\to \mathbb{R}[/imath] be differentiable and non-zero [imath](f_k(x)\neq0[/imath] for all [imath]a<x<b),\ k=1,\cdots,n[/imath] Find [imath]\frac{d}{dx}f_1(x)f_2(x)\cdots f_n(x)[/imath]. Help appreciated! |
344686 | Solve this equation [imath] (x^3+100)^2 = (x^4-100)^3[/imath]
Solve this equation [imath](x^3+100)^2 = (x^4-100)^3[/imath] | 339106 | Algebraic equation problem - finding [imath]x[/imath]
[imath](x^2 +100)^2 =(x^3 -100)^3[/imath] How to solve it? |
345130 | A question on finite [imath]p[/imath]-groups.
Is true that if [imath]G[/imath] is a [imath]p[/imath]-group finite, say, [imath]\mid G \mid = p^d[/imath], then [imath]G[/imath] is [imath]d[/imath]-generated? | 344988 | A question about finitely generated [imath]p[/imath]-groups
My question is about finitely generated [imath]p[/imath]-groups. In general, a subgroup of a finitely generated group is not necessarily finitely generated. But, my question is about finite and finitely generated [imath]p[/imath]-groups. More specifically: if [imath]G[/imath] is a finitely generated [imath]p[/imath]-group, say, [imath]m[/imath]-generated and [imath]U[/imath] is a finitely generated subgroup of [imath]G[/imath], then is [imath]U[/imath] at most [imath]m[/imath]-generated? If not, can [imath]U[/imath] be generated by a number of elements that depends only on [imath]m[/imath]? |
331829 | How to calculate the sum of floors
The problem is to calculate the value [imath] \sum_{i = 1}^{1000} \left\lfloor \frac{1}{\sqrt{i} - \lfloor \sqrt{i} \rfloor} \right\rfloor [/imath] Where [imath] i [/imath] is not perfect square number. My thought is, if I can calculate the sum [imath] \sum_{m = 1}^{M} \sum_{k = 1}^{2m} \left\lfloor \frac{\sqrt{m^2+k} + m}{k} \right\rfloor [/imath] then I can solve the original problem. And I also know that [imath] \left\lfloor \frac{x + m}{n} \right\rfloor = \left\lfloor \frac{\lfloor x \rfloor + m}{n} \right\rfloor [/imath] So I just need to calculate [imath] \sum_{m = 1}^{M} \sum_{k = 1}^{2m} \left\lfloor \frac{2m}{k} \right\rfloor [/imath] Then, it's difficult for me. I have tried to exchange the sum-order of [imath] m [/imath] and [imath] k [/imath], that is [imath] \sum_{k = 1}^{2M} \sum_{m = 1}^{M} \left\lfloor \frac{2m}{k} \right\rfloor [/imath] While I could not calculate the sum of [imath] \frac{2m}{k} [/imath], in fact it is easier if it is [imath] \frac{m}{k} [/imath] So I ask some advices for you : ) | 326283 | Finding the sum [imath]\sum\limits_{n=1,n\neq m^2}^{1000}\left[\frac{1}{\{\sqrt{n}\}}\right][/imath],
Find the value [imath]\displaystyle\sum_{n=2,n\neq m^2}^{1000}\left[\dfrac{1}{\{\sqrt{n}\}}\right][/imath], by [imath]\{x\}=x-[x][/imath], [imath][x][/imath]was bracket function,for example:[imath][5.4]=5, [2.9]=2,[-1.1]=-2 [/imath]and so on. |
345603 | Show that closed unit ball in [imath] l^2 [/imath] is not compact.
I have a prove using the defination of compact set.But I want to prove it using sequential criteria of compactness.How is it possibe? | 115344 | Proving that the unit ball in [imath]\ell^2(\mathbb{N})[/imath] is non-compact
So on my homework it says that to prove the unit ball in [imath]\ell^2(\mathbb{N})[/imath] is non-compact, it suffices to find countably many elements [imath]x_n[/imath] of [imath]\ell^2(\mathbb{N})[/imath] with [imath]\lVert x_n\rVert \leq \frac{1}{2}[/imath] such that [imath]\lVert x_n - x_m\rVert \geq \delta[/imath] for some [imath]\delta \in (0, \frac{1}{2})[/imath] and [imath]n \neq m[/imath]. Why is that? Thanks |
328261 | Trying to define [imath]\mathbb{R}^{0.5}[/imath] topologically
A few days ago, I was trying to generalize the defintion of Euclidean spaces by trying to define [imath]\mathbb{R}^{0.5}[/imath]. Question: Is there a metric space [imath]A[/imath] such that [imath]A\times A[/imath] is homeomorphic to [imath]\mathbb{R}[/imath]? I am interested also in seeing examples of [imath]A[/imath] which are only topological spaces Edit: If there exists a topological space [imath]A[/imath] such that [imath]A\times A\cong \Bbb R[/imath], then [imath]A\times \{a\}[/imath] is a subspace of [imath]A\times A[/imath] ([imath]a\in A[/imath]). Hence [imath]A\times\{a\}[/imath] can be embedded in [imath]\mathbb{R}[/imath], since [imath]A\cong A\times \{a\}[/imath]. Thus [imath]A[/imath] can be embedded in [imath]\mathbb{R}[/imath]. Therefore [imath]A[/imath] is metrizable. Thank you | 57375 | [imath]\mathbb R = X^2[/imath] as a Cartesian product
I wonder if it is possible to consider [imath]\mathbb R[/imath] as a Cartesian product [imath]X\times X[/imath] for some set [imath]X[/imath]. From the point of view of the dimensionality, there are spaces with a Hausdorff dimension [imath]1/2[/imath] (sort of Cantor sets), but I guess there are other problems in this construction unrelated to the dimension. Edited: replying on the comment by Asaf. I want that for all [imath]r\in\mathbb R[/imath] there exists a unique representation [imath]r=\langle x,y\rangle[/imath] where [imath]x\in X[/imath] and [imath]y\in X[/imath]. Also, what if we want to have [imath]X[/imath] to be a topological space and [imath]h:X\times X\to\mathbb R[/imath] to be a homeomorphism? |
337565 | A question comparing [imath]\pi^e[/imath] to [imath]e^\pi[/imath]
I was doing an algebra problem set following a chapter on logarithms and exponentiation, and it presented this "bonus question": Without using your calculator, determine which is larger: [imath]e^\pi[/imath] or [imath]\pi^e[/imath]. I wasn't able to come up with anything, and I'm just curious how you might tackle this, keeping in mind it came out of a college algebra textbook, less than halfway through, so I don't imagine the author had any super-advanced tactics in mind. | 7892 | Comparing [imath]\pi^e[/imath] and [imath]e^\pi[/imath] without calculating them
How can I calculate, without calculator or similar device, the values of [imath]\pi^e[/imath] and [imath]e^\pi[/imath] in order to compare them? |
346088 | Find [imath]\lim\limits_{(x,y) \to(0,0)} \frac{xy^2}{ x^2 + y^4} [/imath]
I am doing some exercises for my Calculus 3 exam and I get stuck in this exercise: (I need to find the limit or say that it does not exist) [imath]\lim_{(x,y) \to(0,0)} \frac{xy^2}{ x^2 + y^4} [/imath] I tried to change it polar coordinates (so that I have only one variable) but it is messy and I dont understand it... Can someone explain me how to do it? Thanks! | 174190 | Prove that [imath]\lim\limits_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0[/imath]
[imath]\lim_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0[/imath] Please, Anyone could suggest me some way for this?. Thanks. |
342573 | integrable functions and real-value on (X,M,μ)
If f and g are integrable functions and real-value on (X,M,μ) , which assertion is correct? 1-[imath]fg\in L^ 1 (μ)[/imath] 2-[imath]fg\in L^ 2 (μ)[/imath] 3-[imath](f^2+g^2)^{(1/2)}\in L^ 1 (μ)[/imath] 4-No. | 346460 | Combinations of integrable functions
If [imath]f[/imath] and [imath]g[/imath] are integrable functions and real-valued on [imath](X,M,\mu)[/imath] , which assertion is correct? [imath]fg\in L^1 (\mu)[/imath] [imath]fg\in L^2 (\mu)[/imath] [imath]\sqrt{f^2 +g^2}\in L^1 (\mu)[/imath] None of the above. |
346466 | How to correctly write a binomial distribution for a [imath]50[/imath] questions exam
Using binomial distribution I want to know what is the chance of getting [imath]70\%[/imath] or greater in a [imath]50[/imath] question exam, each question having a true/false option to select from. What is the correct formula and can you show what the exact figures are in condition for [imath]70\%[/imath] or greater and show what each feature in the condition represents so I can use it to work out of scores (for example [imath]60\%[/imath] or greater, [imath]50\%[/imath] or greater). I want the answer to be represented as a percentage so that I know know that the chances of getting [imath]70\%[/imath] or greater is [imath]...\%[/imath]. Thanks. | 345831 | Probability of getting >70% in exam with 50 yes/no questions
In a paper containing 50 yes/no questions, I am trying to find the probability of getting 70%. Using binomial distribution, [imath]P(X\ge70\%)=\sum_{k=25}^{50} \binom{50}{k}\left(\frac{1}{2}\right)^{50}[/imath] The following result was obtained: 70% (Grade A in university) in Assessment: 0.199913% chance I am not sure I have followed the formula correctly so looking for approval and guidance. |
345978 | Prove that if [imath]\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)[/imath] then either [imath]A \subseteq B[/imath] or [imath]B \subseteq A[/imath].
Prove that for any sets [imath]A[/imath] or [imath]B[/imath], if [imath]\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)[/imath] then either [imath]A \subseteq B[/imath] or [imath]B \subseteq A[/imath]. ([imath]\mathcal P[/imath] is the power set.) I'm having trouble making any progress with this proof at all. I've assumed that [imath]\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)[/imath], and am trying to figure out some cases I can use to help me prove that either [imath]A \subseteq B[/imath] or [imath]B \subseteq A[/imath]. The statement that [imath]\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)[/imath] seems to be somewhat useless though. I can't seem to make any inferences with it that yield any new information about any of the sets it applies to or elements therein. The only "progress" I seem to be able to make is that I can conclude that [imath]A \subseteq A \cup B[/imath], or that [imath]B \subseteq A \cup B[/imath], but I don't think this gives me anything I don't already know. I've tried going down the contradiction path as well but I haven't been able to find anything there either. I feel like I am missing something obvious here though... | 246491 | Stuck with proof for [imath]\forall A\forall B(\mathcal{P}(A)\cup\mathcal{P}(B)=\mathcal{P}(A\cup B)\rightarrow A\subseteq B \vee B\subseteq A)[/imath]
I came to point where I suppose for case 1 that [imath]A\subseteq B[/imath] and conclusion is trivial. For case 2 I suppose that [imath]A\not\subseteq B[/imath] and try to prove [imath]B\subseteq A[/imath], but that gets me nowhere. Any pointers here are most welcome. |
346561 | Show [imath]\tau=\tau^*[/imath] if [imath]\tau^*\subset \tau[/imath]
Let [imath](X,\tau)[/imath] be compact and [imath](X,\tau^*)[/imath] be a Hausdorff space. How can we show that [imath]\tau=\tau^*[/imath] if [imath]\tau^*\subset \tau[/imath]? | 153734 | Point set topology question: compact Hausdorff topologies
[imath]\tau_1,\tau_2,\tau_3[/imath] are topologies on a set such that [imath]\tau_1\subset \tau_2\subset \tau_3[/imath] and [imath](X,\tau_2)[/imath] is a compact Hausdorff space. Could any one tell me which of the following are correct? [imath]\tau_1=\tau_2[/imath] if [imath](X,\tau_1)[/imath] is compact Hausdorff. [imath]\tau_1=\tau_2[/imath] if [imath](X,\tau_1)[/imath] is compact. [imath]\tau_2=\tau_3[/imath] if [imath](X,\tau_3)[/imath] is Hausdorff. [imath]\tau_2=\tau_3[/imath] if [imath](X,\tau_3)[/imath] is compact. |
346619 | Integral between [imath]-\pi[/imath] and [imath]\pi[/imath]
How can I show that [imath]\int_{-\pi}^{\pi}\sin (mt) \sin (nt) {dt}=\begin{cases}\ 0 \mbox{ if } m \neq n\\ \pi \mbox{ if } m=n \end{cases}[/imath]. I want to prove the above property by expressing sinAsinB as a sum of complex exponentials using Euler's formulas. Thank you | 327110 | Show that [imath]\int_{-\pi}^\pi\sin mx\sin nx d x[/imath] is 0 [imath]m\neq n[/imath] and [imath]\pi[/imath] if [imath]m=n[/imath] using integration by parts
Show that [imath]\int_{-\pi}^{\pi}\sin{mx}\,\sin{nx}\, d x =\begin{cases} 0&\text{if }m\neq n,\\ \pi&\text{if }m=n. \end{cases}[/imath] by using integration by parts. I've done the following, but I'm not sure if I went the wrong direction, if I messed up some calculation, or if I'm almost there and just can't see what to do next... [imath]\int_{-\pi}^{\pi}\sin{mx}\,\sin{nx} \, d x=-\left(\frac{n}{n^2-m}\right)\sin{mx}\cos{nx}+\left(\frac{m}{n^2-m}\right)\cos{mx}\sin{nx}+C[/imath] [imath]=-2\left(\frac{n}{n^2-m}\right)\sin{m\pi}\cos{n\pi}+2\left(\frac{m}{n^2-m}\right)\cos{m\pi}\sin{n\pi}[/imath] Now ... I figure that if [imath]n=m[/imath], then I can just as well replace them all with a 3rd variable... say [imath]z[/imath]... [imath]=-2\left(\frac{z}{z^2-z}\right)\sin{z\pi}\cos{z\pi}+2\left(\frac{z}{z^2-z}\right)\cos{z\pi}\sin{z\pi}[/imath] Wouldn't that equal 0? Or am I completely mistaken? Addition: By following the suggestions below and using the product-to-sum forumulas, I got the following: [imath]\frac{1}{2}\int\cos{((n-m)x)}\ dx-\frac{1}{2}\int\cos{((n+m)x)}\ dx=\frac{\sin{((n-m)x)}}{2(n-m)}-\frac{\sin{((n+m)x)}}{2(n+m)}[/imath] So now if [imath]n=m[/imath], then the first quotient will end up dividing by 0... |
346634 | Show [imath]z\to\frac{2}{1-z}[/imath] sends unit semicircle to [imath]x=1[/imath] and line [imath]x=1[/imath] to [imath]x=0[/imath]
How can we show that [imath]z\to\frac{2}{1-z}[/imath] sends the unit semicircle to [imath]x=1[/imath] and the line [imath]x=1[/imath] to [imath]x=0[/imath]? [imath]x=1[/imath] implies [imath]z=1+iy[/imath]. Then the transformation sends it to i*(2/y). | 341274 | Möbius Transformation help
Hey guys I need help on these 2 questions that I am having trouble on. 1) Show that the Möbius transformation [imath]z \rightarrow \frac{2}{1-z}[/imath] sends the unit circle and the line [imath]x = 1[/imath] to the lines [imath]x = 1[/imath] and [imath]x = 0[/imath], respectively. 2) Now deduce from this that the non-Euclidean distance between the unit circle and the line [imath]x = 1[/imath] tends to zero as these non-Euclidean lines approach the x-axis. I know that the non-euclidean distance between the lines [imath]x=0[/imath] and [imath]x=1[/imath] goes to zero as [imath]y[/imath] approachs [imath]\infty[/imath]. But I dont know how to deduce that the unit circle would go to [imath]x=1[/imath] and [imath]x=1[/imath] to [imath]x=0[/imath] and how would I deduce from this. Its from my old past tests and I am trying to practice but I got stuck on this question. Please help out thank you |
346922 | A weird open in the real numbers
A have the next problem in a topology homework titled "A weird open". We take [imath]\mathbb R[/imath] with the usual metric. Let [imath]f[/imath] be a a bijection from the natural numbers onto the rational numbers in [imath][0,1][/imath]. Let [imath]\{r_n\}_{n\in \mathbb N}[/imath] be a positive decreasing sequence such that [imath]r_1< 1/2[/imath]. And we define the open [imath]\mathcal O(f,r):=\displaystyle\bigcup_{n\in\mathbb N} B(f(n),r_{n})[/imath]. Where [imath]B(f(n),r_{n})[/imath] is the open interval with center [imath]f(n)[/imath] and radius [imath]r_{n}[/imath]. Now if [imath]r_n \rightarrow 0[/imath]. Is there a function [imath]f[/imath] such that [imath]\mathcal O(f,r)\neq [0,1][/imath]? (Note that [imath]r_n[/imath] is an arbitrary sequence which holds certain conditions, I need to determine [imath]f[/imath]). | 346854 | A union of balls centered at all the rationals on [imath][0,1][/imath] with decreasing radius [imath]r_n \to 0[/imath] relation with [imath][0,1][/imath]
Given a strictly decreasing sequence [imath]r_n[/imath] such that [imath]r_n\in [0,1][/imath], [imath]r_1 < \frac{1}{2}[/imath], and [imath]\lim\limits_{n\to \infty} r_n=0[/imath]. Question: Does there exist a bijective sequence [imath]x:\mathbb N\to \mathbb Q\cap[0,1][/imath] such that [imath]\bigcup_{n\in \mathbb N}J(x_n,r_n)\ne [0,1][/imath] where [imath]J(a,b)=(a-b,a+b)\cap[0,1][/imath] and [imath]x_n[/imath] denotes [imath]x(n)[/imath]? |
347378 | Finite Union of proper subspaces of [imath]\mathbb C^2[/imath] can equal to [imath]\mathbb C^2[/imath]?
My instructor for Linear Algebra gave us a problem to think about but am quite unsure on how to approach it: Let [imath]V_1, V_2, ... V_{100}[/imath] be [imath]100[/imath] proper subspaces of the complex vector space [imath]V=\mathbb C^2[/imath]. Can it be possible that [imath]\bigcup _{i=1}^{100} V_i = \mathbb C^2[/imath]? | 145869 | A finite-dimensional vector space cannot be covered by finitely many proper subspaces?
Let [imath]V[/imath] be a finite-dimensional vector space, [imath]V_i[/imath] is a proper subspace of [imath]V[/imath] for every [imath]1\leq i\leq m[/imath] for some integer [imath]m[/imath]. In my linear algebra text, I've seen a result that [imath]V[/imath] can never be covered by [imath]\{V_i\}[/imath], but I don't know how to prove it correctly. I've written down my false proof below: First we may prove the result when [imath]V_i[/imath] is a codimension-1 subspace. Since [imath]codim(V_i)=1[/imath], we can pick a vector [imath]e_i\in V[/imath] s.t. [imath]V_i\oplus\mathcal{L}(e_i)=V[/imath], where [imath]\mathcal{L}(v)[/imath] is the linear subspace span by [imath]v[/imath]. Then we choose [imath]e=e_1+\cdots+e_m[/imath], I want to show that none of [imath]V_i[/imath] contains [imath]e[/imath] but I failed. Could you tell me a simple and corrected proof to this result? Ideas of proof are also welcome~ Remark: As @Jim Conant mentioned that this is possible for finite field, I assume the base field of [imath]V[/imath] to be a number field. |
347571 | How to show that this is a complete metric space
Let [imath](X;d)[/imath] be a metric space and [imath]\mathrm{C_b}(X,\mathbb{R})[/imath] denote the set of all continuous bounded real valued functions defined on [imath]X[/imath], equipped with the uniform metric: [imath] \rho(f,g) = \sup\{\, \lvert f(x) - g(x)\lvert : x \in X \,\}. [/imath] Show that [imath]\mathrm{C_b}(X,\mathbb{R})[/imath] is a complete metric space with respect to the metric [imath]\rho[/imath]. | 71121 | Space of bounded continuous functions is complete
I have lecture notes with the claim [imath](C_b(X), \|\cdot\|_\infty)[/imath], the space of bounded continuous functions with the sup norm is complete. The lecturer then proved two things, (i) that [imath]f(x) = \lim f_n (x)[/imath] is bounded and (ii) that [imath]\lim f_n \in \mathbb{R}[/imath]. I don't understand why it's not enough that [imath]f[/imath] is bounded. I think the limit of a sequence of continuous functions is continuous and then if [imath]f[/imath] is bounded, it's in [imath]C_b(X)[/imath]. So what is this [imath]\lim f_n \in \mathbb{R}[/imath] about? Many thanks for your help. |
28329 | Nice proofs of [imath]\zeta(4) = \frac{\pi^4}{90}[/imath]?
I know some nice ways to prove that [imath]\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6[/imath]. For example, see Robin Chapman's list or the answers to the question "Different methods to compute [imath]\sum_{n=1}^{\infty} \frac{1}{n^2}[/imath]?" Are there any nice ways to prove that [imath]\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?[/imath] I already know some proofs that give all values of [imath]\zeta(n)[/imath] for positive even integers [imath]n[/imath] (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for [imath]\zeta(4)[/imath]. I would be particularly interested in a proof that isn't an adaption of one that [imath]\zeta(2) = \pi^2/6[/imath]. | 630044 | [imath]\zeta(4)=\sum_ {k=1}^{\infty}{\frac{1}{k^4}}[/imath]
How to Find [imath]\zeta(4)=\sum_ {k=1}^{\infty}{\frac{1}{k^4}}[/imath] the most basic way possible? I know it's [imath]\pi^4/90[/imath] but to arrive at this figure? Curious, because I need it to solve the integral [imath]\int_{0}^{\infty}x^3\cdot(e^x-1)^{-1}\;dx[/imath]. |
348048 | Chessboard and Catalan numbers
Let's consider labelling the square of a [imath]2\times n[/imath] chessboard from [imath]1[/imath] to [imath]2n[/imath] such that the numbers increase from left-to-right and bottom-to-top. Prove that the number of such labellings equals the [imath]n[/imath]-th Catalan number [imath]C_n[/imath] I still have no clue how to even start this problem after all the "research" I've done. Extremely confused. | 347902 | Catalan number interpretation
I have a [imath]2 \times n[/imath] chessboard where numbers are increasing from left to right and top to bottom. I want to show that the number of arrangements is the [imath]nth[/imath] catalan number. for example one such arrangement with n = 3 could be: [imath]\pmatrix{1 3 4\\ 256}[/imath] I think it would be easiest to use: [imath]C_n = \frac{1}{n+1} {2n\choose n}[/imath] What I see is that the first entry must be [imath]1[/imath] and the last entry must be [imath]2n[/imath] Also, if one row is complete (and valid) it completely determines the other row. So it suffices to satisfy one row. Thus we must do [imath]2n\choose n[/imath] to fill a row. However, a bunch of these combinations would not satisfy these conditions. What is the reason to divide by [imath]n+1[/imath]? I tried to use the sequence of [imath](n+1)[/imath] [imath]1[/imath]'s and [imath](n-1)[/imath] [imath]-1[/imath]'s trick (matching parenthesis example that is counted by catalan) but could not establish the proper bijection. All help is greatly appreciated. |
348018 | Geometric Interpretation for Definite Integrals with [imath]\pi[/imath] in the result
What is the geometric interpretation for the following integral? What is a nice geometric interpretation for the following integral (possibly in relationship to a circle) that emphasizes why we get the result of π in the right hand side? [imath]\int^\infty_{-\infty} \frac{1}{1+x^2}dx = \pi[/imath] I do know, of course that the indefinite integral for the integrand is [imath]\tan^{-1} x[/imath]. In the answers, I'd also appreciate examples of other integrals with a geometric interpretation. | 2899 | Can this standard calculus result be explained "intuitively"
Recently I stumbled upon someone who said he wanted to understand why [imath]\arctan x = \int\dfrac{dx}{1+x^2}[/imath] At first I was confused. This is an easy result in any integral calculus course. But then he explained that although he understood the proof, he wanted to understand it "intuitively". He wanted to see why it was in terms of arclength and addition and subtraction. My question is: Is there an "intuitive" way to explain this identity? |
348730 | A proof that [imath]1=2[/imath]. May I know why it’s false?
im not good at formatting. May I know why it’s false [imath]x^2= x+x+x+\ldots(x\; \text{times})[/imath] apply derivative on both sides [imath]=> \frac{d}{dx}(x^2)=\frac{d}{dx}(x+x+x+\ldots(x\; \text{times}))[/imath] [imath]=> 2x=1+1+1+\ldots(x\; \text{times})[/imath] [imath]=> 2x=x[/imath] [imath]=> 2=1[/imath] | 758798 | What's wrong with these equations?
My friend Boris (Boryan) gave me a task, and completely refuses to give the answer what's wrong here. [imath]x^2=\overbrace{x+\cdots+x} ^{x\text{ times}}[/imath] [imath](x^2)'=(x+\cdots+x)'[/imath] [imath]2x=1+\cdots+1[/imath] [imath]2x=x[/imath] [imath]2=1[/imath] Yeah! I've succesfully copypasted latex formulas! I think the problem is in non-formal symbols. It brings me to question, what is the result for [imath](\sum_{i=1}^{x}x)' = ?[/imath] That's usually an obstacle for those who memorised many things without clear understanding of definitions. So, I'm interested in fundamental mistake of this equations, because I want to get out of this mess) |
349376 | Nilradical and Jacobson's radical.
Let A be a commutative ring with 1. 1) Prove that a sum of a nilpotent element and an invertible element is invertible. 2) Prove that if [imath]f=a_0+a_1x+\dots+a_nx^n \in A[x][/imath] a) [imath]\exists f^{-1}\in A[x] \Leftrightarrow a_0[/imath] is invertible and the other coefficients are nilpotent. b) f is nilpotent [imath]\Leftrightarrow [/imath] all its coefficients are nilpotent. p.s. Those are the first two in a series of problems. The rest easily follow from each other. I'm only struggling with the first two. | 19132 | Characterizing units in polynomial rings
I am trying to prove a result, for which I have got one part, but I am not able to get the converse part. Theorem. Let [imath]R[/imath] be a commutative ring with [imath]1[/imath]. Then [imath]f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}[/imath] is a unit in [imath]R[X][/imath] if and only if [imath]a_{0}[/imath] is a unit in [imath]R[/imath] and [imath]a_{1},a_{2},\dots,a_{n}[/imath] are all nilpotent in [imath]R[/imath]. Proof. Suppose [imath]f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}[/imath] is such that [imath]a_{0}[/imath] is a unit in [imath]R[/imath] and [imath]a_{1},a_{2}, \dots,a_{r}[/imath] are all nilpotent in [imath]R[/imath]. Since [imath]R[/imath] is commutative, we get that [imath]a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}[/imath] are all nilpotent and hence also their sum is nilpotent. Let [imath]z = \sum a_{i}X^{i}[/imath] then [imath]a_{0}^{-1}z[/imath] is nilpotent and so [imath]1+a_{0}^{-1}z[/imath] is a unit. Thus [imath]f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)[/imath] is a unit since product of two units in [imath]R[X][/imath] is a unit. I have not been able to get the converse part and would like to see the proof for the converse part. |
349386 | recursive sequence of nested intervals: how to find the limit value
I came across the following interesting exercise: Let [imath]b_{n+1}=\frac{1}{2}(a_n+b_n)[/imath] and [imath]a_{n+1}=\sqrt{a_nb_n}[/imath] with [imath]a_1>0[/imath], [imath]b_1>0[/imath] and [imath]b_1>a_1[/imath]. Show that [imath]I_n=[a_n, b_n][/imath] is a sequence of nested intervals. My question is not about the fact that [imath]I_n[/imath] is a sequence of nested intervals. This is relatively straightforward to show. But I have not been able to calculate the limiting value [imath]\lim\limits_{n\to\infty} a_n = \lim\limits_{n\to\infty} b_n[/imath] as a function of [imath]a_1[/imath] and [imath]b_1[/imath] (even though this is not part of the exercise). Any ideas are greatly appreciated! | 347283 | Limit of [imath]a_{k+1}=\dfrac{a_k+b_k}{2}[/imath], [imath]b_{k+1}=\sqrt{a_kb_k}[/imath]?
Let [imath]a,b>0[/imath] and let [imath]a_0=a[/imath], [imath]b_0=b [/imath], [imath]a_{k+1}=\dfrac{a_k+b_k} 2[/imath],[imath]b_{k+1}=\sqrt{a_kb_k}[/imath] [imath]\quad k\geq0[/imath]. This converges to a number between a and b. Also [imath]a_k>b_k[/imath] for [imath]k\geq1[/imath] (AM-GM inequality). Can we find the limit explicitly in terms of [imath]a[/imath] and [imath]b[/imath]? |
345013 | Evaluating [imath]\int_0^{2\pi} \sqrt{1+\cos^2(x)}\ dx[/imath]
I want to evaluate this integral [imath]\int_0^{2\pi} \sqrt{1+\cos^2(x)}\ dx[/imath] But I cannot find a useful substitution/strategy. Could you please give me a hint? I was thinking proving that this is equal to [imath]\int_0^{2\pi} \sqrt{1+\sin^2(x)}\ dx[/imath] but don't know how to proceed. | 45089 | What is the length of a sine wave from [imath]0[/imath] to [imath]2\pi[/imath]?
What is the length of a sine wave from [imath]0[/imath] to [imath]2\pi[/imath]? Physically I would plot [imath]y=\sin(x),\quad 0\le x\le {2\pi}[/imath] and measure line length. I think part of the answer is to integrate this: [imath] \int_0^{2\pi} \sqrt{ 1 + (\sin(x))^2} \ \rm{dx} [/imath] Any ideas? |
349728 | Direct product of finite cyclic groups of coprime orders
The Question is this: How many generators are there of the group [imath]G\times H[/imath], if [imath]G[/imath] and [imath]H[/imath] are cyclic groups of order [imath]m[/imath] and [imath]n[/imath], which are coprime? Let's say that [imath]G[/imath] is generated by [imath]g[/imath], and [imath]H[/imath] by [imath]h[/imath]. Here I already proved that [imath](g,h)[/imath] is generator of [imath]G\times H[/imath] but I can't come up with another one. So does [imath]G\times H[/imath] really have any other generator? Thanks! | 272371 | Another point of view that [imath]\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}[/imath] is cyclic.
I was thinking that the product of groups [imath]\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}[/imath] is not cyclic, but [imath]\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}[/imath] is cyclic if p is an odd prime. Then I found logic that [imath]\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}[/imath] is certainly cyclic if [imath]m[/imath] and [imath]n[/imath] are coprimes. I searched on google a proof of that, to compare with mine and the answer is that [imath]\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}[/imath] is isomorphic to [imath]\mathbb{Z}/mn\mathbb{Z}[/imath] (which is clearly cyclic) [imath]\iff[/imath] [imath]m[/imath] and [imath]n[/imath] are coprimes. But I got another proof and I want to know what you think about it : I claim that [imath]([1]_m,[1]_n)[/imath] generates [imath]\mathbb{Z}/m\mathbb{Z}[/imath] [imath]\times\mathbb{Z}/n\mathbb{Z}[/imath], where [imath][a]_d[/imath] is the canonical projection of [imath]a\in \mathbb{Z}[/imath] into [imath]\mathbb{Z}/d\mathbb{Z}[/imath]. Let's take multiples of [imath]([1]_m,[1]_n)[/imath] until we get [imath]m \cdot[/imath] [imath]([1]_m,[1]_n) = ([0]_m,[m]_n)[/imath] (where I suppose [imath]m=\min(m,n)[/imath]... otherwise, consider [imath]\mathbb{Z}/n\mathbb{Z}[/imath] [imath]\times\mathbb{Z}/m\mathbb{Z}[/imath] and [imath]\min(m,n)=n[/imath]). The next step is [imath]([1]_m,[m+1]_n)[/imath]. This pair is not equal to [imath]([1]_m,[1]_n)[/imath], since [imath]m+1 \not\equiv1[/imath] mod [imath]n[/imath] ([imath]m[/imath] and [imath]n[/imath] are supposed to be coprimes). We can now continue adding [imath]([1]_m,[1]_n)[/imath] to the results. The last step is precisely reached when you got [imath]([mn]_m,[mn]_n)=([0]_m,[0]_n)[/imath]. At this point, the fact that [imath]m[/imath] and [imath]n[/imath] are coprimes is absolutely necessary because if they are not coprimes, [imath]lcm(m,n)\ne m\cdot n[/imath] and [imath]([lcm(m,n)]_m,[lcm(m,n)]_n)=([0]_m,[0]_n)[/imath] but all the [imath]m \cdot n[/imath] possible pairs are not reached, only the [imath]lcm(m,n)[/imath] are ! |
350165 | Recursive/Fibonacci Induction
1) Let [imath]F_n[/imath] denote the $n^t[imath]^h$ Fibonacci number. Prove by induction: [/imath] F_n = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}}{\sqrt{5}} $$ Clear explanation would be appreciated. | 94989 | Prove by induction Fibonacci equality
[question:] Prove by induction that the i th Fibonacci number satisfies the equality [imath]F_i=\frac{\phi^i-\hat{\phi^i}}{\sqrt5}[/imath]where [imath]\phi[/imath] is the golden ratio and [imath]\hat{\phi}[/imath] is its conjugate. [end] I've had multiple attempts at this, the most fruitful being what follows, though it is incorrect, and I cannot figure out where I am going wrong: [my answer:] What follows is an incorrect approach, after which comes the approach that I should have done Inductive Hypothesis: [imath]F_i=\frac{\phi^i-\hat{\phi^i}}{\sqrt5}, i\in\mathbb{N} [/imath] (I include 0) Base case: [imath]F_0=\frac{\phi^0-\hat{\phi^0}}{\sqrt5}=0[/imath] Proof: [imath]\begin{eqnarray*} F_{i+1}&=&\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ &=&\frac{\phi*\phi^i-\hat{\phi}*\hat{\phi^i}}{\sqrt5}\\ &=&\frac{\frac{1+\sqrt5}{2}*\phi^i-\frac{1-\sqrt5}{2}*\hat{\phi^i}}{\sqrt5}\\ &=&\frac{\frac{\phi^i+\sqrt5\phi^i}{2}-\frac{\hat{\phi^i}-\sqrt5\hat{\phi^i}}{2}}{\sqrt5}\\ &=&\frac{\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{2}}{\sqrt5}\\ &=&\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{2*\sqrt5}\\ &=&\frac{1}{2}\left(\frac{\phi^i+\sqrt5\phi^i-\hat{\phi^i}+\sqrt5\hat{\phi^i}}{\sqrt5}\right)\\ &=&\frac{1}{2}\left(\frac{\sqrt5\phi^i}{\sqrt5} + \frac{\sqrt5\hat{\phi^i}}{\sqrt5} +\frac{\phi^i-\hat{\phi^i}}{\sqrt5}\right)\\ &=&\frac{1}{2}\left(\phi^i+\hat{\phi^i}+F_{i}\right)\text{ by inductive hypothesis}\\ &=&\frac{1}{2}\left(\sqrt5*...\right)\\ \end{eqnarray*}[/imath] Actually I just saw my error in that line (I eventually multiplied by [imath]\frac{\sqrt5}{\sqrt5}[/imath] and substituted for [imath]F_i[/imath], but I see that the conjugates are added there, not subtracted). Correct approach: [imath]\begin{eqnarray*} F_{i+1}&=&F_{i} + F_{i-1}\\ &=&\frac{\phi^i-\hat{\phi^i}}{\sqrt5}+\frac{\phi^{i-1}-\hat{\phi^{i-1}}}{\sqrt5}\\ &=&\frac{\left(\phi+\hat{\phi}\right)\left(\phi^i-\hat{\phi^i}\right)-\phi\hat{\phi}\left(\phi^{i-1}-\hat{\phi^{i-1}}\right)}{\sqrt5}\text{ (see answer for why this works)}\\ &=&\frac{\phi^{i+1}-\phi\hat{\phi^i}+\hat{\phi}\phi^i-\hat{\phi^{i+1}}-\phi^i\hat{\phi}+\phi\hat{\phi^i}}{\sqrt5}\\ &=&\frac{\phi^{i+1}-\hat{\phi^{i+1}}-\phi\hat{\phi^i}+\phi\hat{\phi^i}-\hat{\phi}\phi^i+\hat{\phi}\phi^i}{\sqrt5}\\ &=&\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ \text{Q.E.D., punk problem.} \end{eqnarray*}[/imath] From what I searched on the web, there is no available source proving this of the Fibonacci sequence - all inductive hypothesis proofs prove the actual sequence with induction. Bill, it's always easy to forget about the inherent properties of special numbers... I didn't even realize [imath]\phi[/imath] + [imath]\hat{\phi}[/imath] = 1 in your first hint, as well as the multiplication, until you pointed it out. Alex, I'm sure that going through the problem that way will produce the result, also, but with Bill's, it's much faster. Thanks for the help. |
350292 | Prove if [imath]g[/imath] is an element of order [imath]d[/imath] and [imath]d[/imath] divides [imath]n[/imath] then [imath]gn = 1[/imath].
Prove if [imath]g[/imath] is an element of order [imath]d[/imath] and [imath]d[/imath] divides [imath]n[/imath], then [imath]gn = 1[/imath]. | 350277 | If [imath]g[/imath] is an element of order [imath]d[/imath] and [imath]d[/imath] divides [imath]n[/imath], then [imath]gn = 1[/imath]
Prove if [imath]g[/imath] is an element of order [imath]d[/imath] and [imath]d[/imath] divides [imath]n[/imath], then [imath]gn = 1[/imath]. I need help on how to prove the converse which comes from this theorem that I solved. Suppose [imath]gn = 1[/imath] in a group [imath]G[/imath] and let [imath]d[/imath] be the order of [imath]g[/imath]. Then [imath]d[/imath] divides [imath]n[/imath]. Proof: By the division algorithm, we may write [imath]n = dq + r[/imath] with [imath]0 < r < d[/imath]. Then [imath]g^r = g^{n-dq}=g^n=(g^d)^{-q}=1 \times 1=1.[/imath] If r > 0, this contradicts minimality of the order [imath]d[/imath]. Hence [imath]r = 0[/imath] and so [imath]d[/imath] divides [imath]n[/imath]. |
350390 | Is there any function [imath]f: \mathbb{R}^2 \to \mathbb{R}^2[/imath] which is [imath]C^1[/imath] and has an invertible derivative matrix at all points, but is not 1-1?
Is there any function [imath]f: \mathbb{R}^2 \to \mathbb{R}^2[/imath] which is [imath]C^1[/imath] and has an invertible derivative matrix at all points, but is not 1-1? Thanks! | 234229 | The implicit and the inverse function
This is a simple problem but I am confused about the results. Suppose the [imath]f:\mathbb{R}^2\longrightarrow\mathbb{R}^2[/imath] is a differentiable mapping in [imath]\mathbb{R}^2[/imath] such that [imath]\det(d_pf)\neq 0[/imath] for all [imath]p\in\mathbb{R}^2[/imath]. Has the mapping [imath]f[/imath] an inverse?. If it is not true, what conditions would lack for it to be?. |
350426 | Why [imath]r^{(p+1)/2} = -r \pmod{p}[/imath].
If [imath]p \equiv 3 \pmod 4[/imath], prove that [imath]−r[/imath] is not a primitive root modulo [imath]p[/imath]. If [imath]p \equiv 1 \pmod 4[/imath], prove that [imath]−r[/imath] is a primitive root modulo [imath]p[/imath]. The (start of the ) Proof: Recall that the order of an element a k modulo n is precisely ordn(a)/gcd(k,ordn(a)). Thus, the order of −r is precisely p − 1/gcd((p + 1)/2, p − 1). Thus we need to determine gcd((p + 1)/2, p − 1). it seems from the answer that r^(p+1)/2 = -r mod p. Why? | 346869 | Determining the order of [imath]-a[/imath] if [imath]a[/imath] is a primitive root
Let [imath]p[/imath] be a prime such that [imath]p\equiv1\pmod4[/imath]. Prove that [imath]a[/imath] is a primitive root modulo [imath]p[/imath] if and only if [imath]-a[/imath] is a primitive root mod [imath]p[/imath]. Let [imath]p[/imath] be a prime such that [imath]p\equiv3\pmod4[/imath]. Prove that [imath]a[/imath] is a primitive root modulo [imath]p[/imath] if and only if [imath]−a[/imath] has order [imath]\dfrac{p−1}{2}[/imath]. For the second one, there is a theorem that says that if the orders of [imath]c[/imath] and [imath]d[/imath] are relatively prime, then the order of [imath]cd[/imath] is the product of the orders of [imath]c[/imath] and [imath]d[/imath]. So, if the order of [imath]-1[/imath] is [imath]2[/imath] and the order of [imath]-a[/imath] is [imath]\dfrac{p−1}{2}[/imath], then the order of [imath]a[/imath] is [imath]p-1[/imath]. But, what about the other direction? |
350421 | Finite group cyclic
Let [imath]G[/imath] be a finite group. If for each m[imath]\in[/imath]N, [imath]x^m=e[/imath] has at most [imath]m[/imath] solutions in [imath]G[/imath], [imath]G[/imath] is cyclic. Can you give me a hint of this problem? I don't know. | 346936 | Finite group for which [imath]|\{x:x^m=e\}|\leq m[/imath] for all [imath]m[/imath] is cyclic.
Let [imath]G[/imath] be a finite group. For each positive integer [imath]m[/imath], if [imath]x^{m}=e[/imath] has at most [imath]m[/imath] solutions in [imath]G[/imath], [imath]G[/imath] is cyclic. What I have thought is that [imath]n=\sum_{d\mid n}\phi(d)[/imath] can be used to solve this and showing that [imath]|G|[/imath] order element in [imath]G[/imath] exists is enough. |
350513 | Find GCD of functions/polynomials
Let [imath]\mathbb F[/imath] be any field [imath]a \neq b[/imath] be two elements of [imath]\mathbb F[/imath] Find the GCD of [imath] f(x) = x + a [/imath] and [imath]g(x) = x + b[/imath]. Also find the polynomials [imath]s(x)[/imath] and [imath]t(x)[/imath] such that [imath]s(x)f(x) + t(x)g(x)[/imath] equals the GCD. My work so far: [imath]d(x) = s(x)f(x)+t(x)g(x)[/imath] [imath]d(x) = s(x)[x+a] + t(x)[x+b][/imath] [imath]d(x) = x[s(x) + t(x)] + s(x)*a +t(x)*a[/imath] Thoughts on how I can find the polynomials? Is there an explicit solution for the GCD [imath]d(x)[/imath]? | 348642 | How to compute the gcd of [imath]x+a[/imath] and [imath]x+b[/imath], where [imath]a\neq b[/imath]?
Having some trouble with a fairly basic field question. Let [imath]F[/imath] be any field and [imath]a \neq b[/imath] where [imath]a,b\in F[/imath]. Find the greatest common divisor of [imath]f(x)=x+a[/imath] and [imath]g(x)=x+b[/imath], and also find polynomials [imath]s(x)[/imath] and [imath]t(x)[/imath] so that [imath]s(x)f(x)+t(x)g(x)= \gcd[/imath] Oddly the [imath]s(x)f(x)+t(x)g(x)=\gcd[/imath] is probably the easier part of this question but I am having a lot of trouble swallowing this notation; I'm only somewhat sure as to why [imath]a=b[/imath] is bad (I think it would allow [imath]f(x)=g(x)[/imath]). Any help much appreciated. |
348775 | Proving the total number of subsets of S is equal to [imath]2^n[/imath]
Student here! Just reading Liebecks Introduction to pure mathematics for fun and I made an attempt at proving the total number of subsets of S is equal to [imath]2^n[/imath]. I realized that the total number of subsets of S is just: [imath] \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} + 1[/imath] And so I put together the cool looking formula [imath] \displaystyle \sum_{r=1}^n \frac{n!}{r!(n-r)!} = 2^n - 1[/imath] But I have no clue how to approach proving this formula analytically. Any help? So if I try a proof by induction: [imath]\displaystyle \sum_{r=1}^n \frac{n!}{r!(n-r)!} = 2^n - 1[/imath] [imath]\displaystyle \sum_{r=0}^n \frac{n!}{r!(n-r)!} = 2^n[/imath] obviously true for [imath]n = 1[/imath]: [imath]1 + \frac{1}{1!0!} = 2 = 2[/imath] assume true for [imath]n[/imath], I could multiply both sides by [imath]2[/imath]? [imath]\displaystyle 2\sum_{r=0}^n \frac{n!}{r!(n-r)!} = 2^{n+1}[/imath] How could I go about proving that [imath]\displaystyle 2\sum_{r=0}^n \frac{n!}{r!(n-r)!} = \sum_{r=0}^{n+1} \frac{(n+1)!}{r!((n+1) - r)!}[/imath] | 570801 | Prove the following combination?
I need a quick proof of the following combination [imath]\binom{n+1}1+\binom{n+1}2+\binom{n+1}3+\dots++\binom{n+1}{n+1}=2^{n+1}-1[/imath] |
350514 | Compact formula for [imath]\sum_k k![/imath]
Is there any compact formula for: [imath]\sum_{k=0}^n k![/imath] I've tried to find it using one method for summation, but I was able to receive only compact formula for [imath]\sum_k k! \cdot k = (n+1)!-1[/imath] I've typed it into wolfram, but answer is also pretty complicated. | 227551 | [imath]\sum k! = 1! +2! +3! + \cdots + n![/imath] ,is there a generic formula for this?
I came across a question where I needed to find the sum of the factorials of the first [imath]n[/imath] numbers. So I was wondering if there is any generic formula for this? Like there is a generic formula for the series: [imath] 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} [/imath] or [imath] 1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + n^{2} = \frac{n(n+1)(2n + 1)}{6} [/imath] Is there is any formula for: [imath] 1! +2! +3! + 4! + \cdots + n! [/imath] and [imath] {1!}^2 +{2!}^2 +{3!}^2 + \cdots + {n!}^2 [/imath]? Thanks in advance. If not, is there any research on making this type of formula? Because I am interested. |
350931 | Finite subgroups.
(Index of a subgroup [imath]H[/imath] in a group [imath]G[/imath] is the number of distinct left cosets of [imath]H[/imath] in [imath]G[/imath], it is denoted as [imath]|G : H|[/imath]) How to solve that problem. Let [imath]H[/imath] and [imath]K[/imath] be subgroups of a finite group [imath]G[/imath], satisfying [imath]H ≤ K ≤ G[/imath]. Prove that [imath]|G : H| = |G : K||K : H|[/imath]. | 7002 | Finite Index of Subgroup of Subgroup
Prove the following: If [imath]H[/imath] is a subgroup of finite index in a group [imath]G[/imath], and [imath]K[/imath] is a subgroup of [imath]G[/imath] containing [imath]H[/imath], then [imath]K[/imath] is of finite index in [imath]G[/imath] and [imath][G:H] = [G:K][K:H][/imath]. So this is basically a bijective proof? The number of cosets of [imath]H[/imath] in [imath]G[/imath] equals the number of cosets of [imath]K[/imath] in [imath]G[/imath] times the number of cosets of [imath]H[/imath] in [imath]K[/imath] by the multiplication principle? Reference: Fraleigh p. 103 Question 10.35 in A First Course in Abstract Algebra |
350875 | Difficulty involving Rings and Subrings Proving lub contained in fields?
Oe of the questions in my textbooks is as follows. Let [imath]R= \left\{ \frac{n}{10^{k}}: n \in\Bbb Z, k>0\right\}[/imath] Consider [imath]S[/imath] a subset of [imath]R[/imath] where [imath]S=\left\{ \frac{3}{10},\frac{33}{100},\frac{333}{1000},\dots\right\}[/imath] I am asked to show that this set has an upper bound in [imath]R[/imath]. Then prove whether or not it has a least upper bound in [imath]\Bbb Q[/imath] as well as if it has a least upper bound in [imath]R[/imath]. How does one prove something such as this? I am familiar with calculus and the reals but im having a hard time translating that knowledge into this weird realms of fields. if there some explanation that could help me borrow from that intuition? | 331726 | Weird Sub-ring/field? question
Let [imath]R=\{\frac{n}{10^{k}} \mid (n,k) \in Z, k>-1\}[/imath] which is the sub-ring of the Rational numbers ( assumed true) Consider S a subset of R [imath]S= \{(3/10),(33/100),(333/100),...\}[/imath] Show that this set has an upper bound in R. Does it have a least upper bound in the rationals? yes 1/3 Explanation: Lim as n goes to infinity of [imath](10^{n-1} *3)/10^{n}[/imath] = 1/3 [imath]1/3 \in \Bbb Q[/imath] Does it have a least upper bound in R? No Explanation: cant take the limit of a subring at infinity. |
351465 | The null space of [imath]A[/imath] is the plane [imath]x+y=0[/imath] and [imath]T(1,0,0)=(1,1,0)[/imath]
Find the standard matrix [imath]A[/imath]? So I started off like this: [imath]x+y=0[/imath] [imath]y=-x[/imath] [imath]\begin{pmatrix}x1 &-x1& z1\\x2& -x2& z2\\x3& -x3& z3\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}[/imath] Any hints would be appreciated thank you!:) | 349748 | The null space of A is the plane x+y=0
Find the standard matrix [imath]A[/imath] for the operator [imath]T[/imath], given that the null space of [imath]A[/imath] is the plane [imath]x+y=0[/imath] and [imath]T(1,0,0)=(1,1,0)[/imath] The biggest issue I am having with this question is that I can't figure out how to turn the equation for the plane into something I can use to find the null space. Any help with this part would be appreciated. Thanks. |
351653 | Let [imath]r[/imath], [imath]s[/imath], [imath]t[/imath] be the roots of the equation [imath]x^3 - 6x^2 + 5x + 1[/imath]. What is the value of [imath](2-r)(2-s)(2-t)[/imath]?
the question is mentioned in my math olympiad. please explain how to solve the problem. I have factorised the equation to [imath]-x^2+1[/imath], [imath]x-6x[/imath], [imath]-5x +1[/imath]. I am only in year 6.Help! | 351650 | Let [imath]r,s,t[/imath] be the roots of the equation [imath] x^3 - 6x^2 + 5x + 1[/imath]. What is the value of [imath](2-r)(2-s)(2-t)[/imath]?
Let [imath]r,s,t[/imath] be the roots of the equation [imath] x^3 - 6x^2 + 5x + 1[/imath]. What is the value of [imath](2-r)(2-s)(2-t)[/imath]? The question is mentioned in my math olympiad. Please explain how to solve the problem. I have factorised the equation to [imath]-x^2+1, x-6x, -5x +1.[/imath] I am only in year 6. |
351772 | Using [imath]\frac{\epsilon}{3}[/imath] to show a limit implies uniform continuity?
If [imath]f:[0,\infty)\to\mathbb{R}[/imath] is continuous and [imath]\lim_{x\to\infty}f(x)=L<\infty[/imath], then [imath]f[/imath] is uniformly continuous on [imath][0,\infty)[/imath]. Using the formal definition of a limit with [imath]\frac{\epsilon}{3}[/imath] to obtain a positive number [imath]M[/imath] such that [imath]f(x)[/imath] is within [imath]\frac{\epsilon}{3}[/imath] to [imath]L[/imath] for all [imath]x\geq M[/imath]. I need to know the case when [imath]x\in[0,M][/imath] and [imath]y\in(M,\infty)[/imath]. Show the definition of continuity holds in this case using [imath]\frac{\epsilon}{3}[/imath]. | 346292 | uniform continuity on [imath][0, +\infty)[/imath]
Let [imath]f:[0,\infty)\rightarrow R[/imath] be a continuous function. If [imath]\lim_{x\rightarrow+\infty} f(x)[/imath] is finite, show that [imath]f[/imath] is uniformly continuous. Also, can I change "[imath]\lim_{x\rightarrow+\infty} f(x)[/imath] is finite" to "[imath]f[/imath] is bounded" and get the same conclusion? Thank you! |
349011 | Find the Galois group of [imath]x^3 -2[/imath] over [imath]\mathbb{Q}[/imath].
Show that it is a non-abelian group of order 6, then under the Galois correspondence find the (fixed) subfield corresponding to the subgroup of G of order 3. I've found the splitting field which is [imath]\mathbb{Q}((-1)^{1/3} \ \sqrt[3]{2})[/imath], and the roots are [imath]\sqrt[3]{2}[/imath], [imath](-1)^{1/3} \ \sqrt[3]{2}[/imath], and [imath](-1)^{2/3}\sqrt[3]{2}[/imath]. I know the next step is to work out the permutations of the roots - presumably there will be a 3-cycle and a 2-cycle, and at a guess I'd say the Galois group will be isomorphic to [imath]D_{6}[/imath]. However, I'm not sure where to go from there. With regards to the subfield, again at a guess it seems likely it will be permutations of the cube root of unity. Is this correct? And if so how do I show it? Edited to correct the splitting field and roots. | 56804 | Galois group of [imath]x^3 - 2 [/imath] over [imath]\mathbb Q[/imath]
I know the Galois group is [imath]S_3[/imath]. And obviously we can swap the imaginary cube roots. I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots. I know that if you have a 3-cycle and a 2-cycle operating on three elements, you get [imath]S_3[/imath]. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle. But this doesn't feel very "constructive" to me. I wonder if I've made myself understood in terms of what kind of argument I'd like to see? |
352383 | Proving that if [imath]A^n = 0[/imath], then [imath]I - A[/imath] is invertible and [imath](I - A)^{-1} = I + A + \cdots + A^{n-1}[/imath]
Let [imath]A[/imath] be a squared matrix, and suppose there exists an [imath]n\in \Bbb N[/imath] in a way that [imath]A^n=0[/imath]. Show that [imath]I-A[/imath] is invertible and that [imath](I-A)^{-1}=I+A+\cdots+A^{n-1}[/imath] I don't have a clue where to start from. | 1181393 | If [imath]A^2=0[/imath], then [imath]I−A[/imath] is invertible
If [imath]A^2=0[/imath], then show that [imath]I−A[/imath] is invertible. I am getting nowhere that leads me to the hint: [imath]I+A[/imath]. |
352413 | Bijection between [imath]\mathbb R^\mathbb N[/imath] and [imath]\mathbb R[/imath]
[imath]\mathbb R^\mathbb N[/imath] is the set of all functions from the naturals to the reals. I have to prove that [imath]\mathbb R^\mathbb N[/imath] has the same cardinality as [imath]\mathbb R[/imath]. I found an injective function from [imath]\mathbb R[/imath] to [imath]\mathbb R^\mathbb N[/imath], but can't seem to find one for [imath]\mathbb R^\mathbb N[/imath] to [imath]\mathbb R[/imath]. | 243590 | Bijection from [imath]\mathbb R[/imath] to [imath]\mathbb {R^N}[/imath]
How does one create an explicit bijection from the reals to the set of all sequences of reals? I know how to make a bijection from [imath]\mathbb R[/imath] to [imath]\mathbb {R \times R}[/imath]. I have an idea but I am not sure if it will work. I will post it as my own answer because I don't want to anchor your answers and I want to see what other possible ways of doing this are. |
352440 | Let [imath](X,d)[/imath] be a metric space and assume that [imath]B_r^d(x)=B_s^d(y)[/imath]. is [imath]r=s[/imath] and [imath]x=y[/imath]
Let [imath](X,d)[/imath] be a metric space and assume that [imath]B_r^d(x)=B_s^d(y)[/imath] where: [imath]B_r^d=\{ a \in X | d(a,x) < r\}[/imath] Now, is it always true that (a) [imath]r=s[/imath] (b) [imath]x=y[/imath] I made an elaborate argument on this question why both these statements should be true. However, I later doubted this conclusion. Lets take the following metric: [imath]d(x,y)=1 \text{ if } x\neq y,\ 0 \text{ if } x=y[/imath] Then [imath]B_{0.5}^d(x)=B_{0.4}^d(x)=\{x\}[/imath] assuming for the moment that the balls have the same centre. Does this mean that [imath]r[/imath] does not necessarily equal [imath]s[/imath]? And how about if they do not have the same centre? | 351101 | If 2 open balls define the same space, is it true that x=y and r=s?
Let [imath](X,d)[/imath] be a non-empty metric sapce, [imath]r[/imath] and [imath]s[/imath] are postive radii, and [imath]b_r^{d}(x)=b^d_s(y)[/imath] for some [imath]x,y \in X[/imath]. Is it true that [imath]r=s[/imath] ? Is it true that [imath]x=y[/imath]? My answer would be something like: no, because for example consider the discrete metric space [imath](X,d_0)[/imath]. Then [imath]b_{35}(x)= X = b_{2}(y)[/imath], where [imath]2\neq 35[/imath]. Is this a good counter example? The same would be true for [imath]x \neq y[/imath]? |
333217 | Expansion of Lambert [imath]W[/imath] for negative values
What is a good approximation for the Lambert [imath]W(x)[/imath] function for values between [imath]\frac{-1}{e}[/imath] and [imath]0[/imath]? Is it simply [imath]x-x^2[/imath]? If so, what bounds are there on the error? | 53191 | Approximating Lambert W for input below 0
As a small part of a much bigger project, I need to be able to approximate the numerical output of the Lambert W function. I have found decent approximations (good up to at least 4 decimal places), for W for inputs on [imath][3,\infty)[/imath] and [imath][0,3)[/imath]. I thought this was going to be sufficient, but it turns out much of the input to the function will have to be below zero. What is a good (and not too complicated) approximation of W for negative input? Keep in mind I'm a programmer, not a mathematician. |