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506475

Is the equation [imath]y=\sqrt x[/imath] a function? Is the equation [imath]y=\sqrt{x}[/imath] a function? A function has one only one [imath]y[/imath] value for a [imath]x[/imath] value. An equation is an expression. For the equation [imath]y=\sqrt{x}[/imath], [imath]y=\pm\sqrt{x}[/imath] Therefore, for a given [imath]x[/imath], there are 2 [imath]y[/imath] values. However when I plot this equation in the calculator it only shows me the top half. Please explain. Thanks!

492707

Why [imath]f(x) = \sqrt{x}[/imath] is a function? Why [imath]f(x) = \sqrt{x}[/imath] is a function (as I found in my textbook) since for example the square root of [imath]25[/imath] has two different outputs ([imath]-5,5[/imath]) and a function is defined as "A function from A to B is a rule of corre- spondence that assigns to each element in set A exactly one element in B.", so [imath]f(x) = \sqrt{x}[/imath] is not a function?

195932

Automorphism group of the quaternion group Let [imath]Q_8[/imath] be the quaternion group. How do we determine the automorphism group [imath]Aut(Q_8)[/imath] of [imath]Q_8[/imath] algebraically? I searched for this problem on internet. I found some geometric proofs that [imath]Aut(Q_8)[/imath] is isomorphic to the rotation group of a cube, hence it is isomorphic to the symmetric group [imath]S_4[/imath]. I would like to know an algebraic proof that [imath]Aut(Q_8)[/imath] is isomorphic to [imath]S_4[/imath].

519162

Automorphism of [imath]Q_8[/imath] Is there anyone could help me to prove that [imath]Aut(Q_8)=S_4[/imath]? Someone told me that there's an isomorphism between the rigid motions of cube and [imath]Aut(Q_8)[/imath], any ideas? Thank you!

507218

What do I need to know to prove this? [imath] \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi} [/imath] This is an identity put forward by Ramanujan (often used as "proof" of his genius): [imath] \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi} [/imath] How does one go about proving this? Alternatively, what does one need to know to be able to do so? Any help is appreciated.

14115

Motivation for Ramanujan's mysterious [imath]\pi[/imath] formula The following formula for [imath]\pi[/imath] was discovered by Ramanujan: [imath]\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\![/imath] Does anyone know how it works, or what the motivation for it is?

507230

Proving convergence Using the definition Use the definition of convergence to prove if [imath]x_n[/imath] converges to [imath]5[/imath], then [imath]\frac{x_n+1}{\sqrt{x_n-1}}[/imath] converges to [imath]3[/imath].

502858

Proof of convergence of sequences Problem: Show that if [imath]x_n\to 5[/imath] then [imath]\dfrac{x_n +1}{\sqrt{x_n -1}} \to 3[/imath]. So I know [imath]|x-5|<\epsilon[/imath] and I need to show that [imath]\left|\dfrac{x_n +1}{\sqrt{x_n -1}} -3\right| < \epsilon[/imath]. I tried applying the triangle inequality (and a bit of algebra) to get [imath]\left|\frac{x_n +1}{\sqrt{x_n -1}} -3\right| \leq |x_n -5| \left |\frac{1}{\sqrt{x_n -1}}\right| + \left|\frac{6}{\sqrt{x_n-1}} -3\right|.[/imath] From this I know that [imath]|x_n -5| <\epsilon[/imath] and can show that [imath]\left |\dfrac{1}{\sqrt{x_n -1}}\right|<1[/imath]. So my only problem is the second part of the sum. I have no idea what to do with it to get everything to be less than [imath]\epsilon[/imath].

14075

How do I tell if matrices are similar? I have two [imath]2\times 2[/imath] matrices, [imath]A[/imath] and [imath]B[/imath], with the same determinant. I want to know if they are similar or not. I solved this by using a matrix called [imath]S[/imath]: [imath]\left(\begin{array}{cc} a& b\\ c& d \end{array}\right)[/imath] and its inverse in terms of [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath], then showing that there was no solution to [imath]A = SBS^{-1}[/imath]. That worked fine, but what will I do if I have [imath]3\times 3[/imath] or [imath]9\times 9[/imath] matrices? I can't possibly make system that complex and solve it. How can I know if any two matrices represent the "same" linear transformation with different bases? That is, how can I find [imath]S[/imath] that change of basis matrix? I tried making [imath]A[/imath] and [imath]B[/imath] into linear transformations... but without the bases for the linear transformations I had no way of comparing them. (I have read that similar matrices will have the same eigenvalues... and the same "trace" --but my class has not studied these yet. Also, it may be the case that some matrices with the same trace and eigenvalues are not similar so this will not solve my problem.) I have one idea. Maybe if I look at the reduced col. and row echelon forms that will tell me something about the basis for the linear transformation? I'm not really certain how this would work though? Please help.

245749

How to tell if two matrices are similar? Two n-by-n matrices A and B are called similar if [imath] \! B = P^{-1} A P [/imath] for some invertible n-by-n matrix P. Similar matrices share many properties: Rank Determinant Trace Eigenvalues (though the eigenvectors will in general be different) Characteristic polynomial Minimal polynomial (among the other similarity invariants in the Smith normal form) Elementary divisors Given two square matrices A and B, how would you tell if they are similar? Constructing a [imath]P[/imath] in the definition seems difficult even if we know they are similar, does it? Not to mention, use this way to tell if they are similar. Are there some properties of similar matrices that can characterize similar matrices? Thanks!

508759

How do I know there are only 5 different groups of order 8? How many different groups are there in order 8? And how do I know which groups they are? I mean, is there anyone can teach me to calculate them? I want a proof, thank you! They are [imath]C_8[/imath], [imath]D_4[/imath], [imath]Q_8[/imath], [imath]C_{4h}=C_4 \times V_2[/imath], [imath]D_{2h}=D_2 \times V_2[/imath]

64406

How can you show there are only 2 nonabelian groups of order 8? It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations [imath]a^4=1[/imath], [imath]b^2=1[/imath] and [imath]bab^{-1}=a^3[/imath]. I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian group of order 8 is isomorphic to one of these?

508869

If [imath]f(x,y)\gt 0[/imath] for any [imath]x,y\in\mathbb R[/imath], then [imath]f(x,y)[/imath] has the minimum? Question : Let [imath]f(x,y)[/imath] be a polynomial with real coefficients having two variables. If [imath]f(x,y)\gt 0[/imath] for any real number [imath]x,y[/imath], then does [imath]f(x,y)[/imath] have the minimum? Motivation : I found the following question in a book without any proof: Let [imath]f(x)[/imath] be a polynomial with real coefficients having one variable. If [imath]f(x)\gt 0[/imath] for any real number [imath]x[/imath], then prove that [imath]f(x)[/imath] has the minimum. Proof : If [imath]\deg (f)=0[/imath], then it's obvious that [imath]f(x)[/imath] has the minimum. So, let [imath]\deg (f)=n\gt 0[/imath] and [imath]f(x)=a_nx^n+f_1(x)[/imath] where [imath]a_n\not=0, \deg(f_1)\lt n[/imath]. We get [imath]\lim_{x\to \pm\infty}f(x)=\lim_{x\to\pm\infty}|f(x)|=\lim_{x\to\pm\infty}|x|^n\left(\left|a_n+\frac{f_1(x)}{x^n}\right|\right)=\infty.[/imath] Hence, supposing [imath]G\gt 0[/imath] such that [imath]f(x)\gt f(0)[/imath] if [imath]|x|\gt G[/imath], let [imath]m=f(x_0)[/imath] be the minimum of the continuous function [imath]f(x)[/imath] in a closed-interval [imath][-G,G][/imath]. Then, since [imath]m\le f(0),[/imath] we know that [imath]f(x)[/imath] has the minimum [imath]m[/imath] at [imath]x=x_0[/imath]. Now the proof is completed. This got me interested the question about [imath]f(x,y)[/imath]. I can neither prove that [imath]f(x,y)[/imath] has the minimum nor find any counterexample. Can anyone help?

3820

Does a polynomial that's bounded below have a global minimum? Must a polynomial function [imath]f \in \mathbb{R}[x_1, \ldots, x_n][/imath] that's lower bounded by some [imath]\lambda \in \mathbb{R}[/imath] have a global minimum over [imath]\mathbb{R}^n[/imath]?

508987

How to Solve a Coupled Differential Equation? I came across the set of following coupled equations while studying cycloid motion in Griffiths' Intro to ED [imath]\ddot{y}=\omega \dot{z}[/imath] [imath]\ddot{z}=\omega (\frac{E}{B}-\dot{y})[/imath] I am at a loss as to how I may approach the problem to solve. Any help is appreciated.

508989

How to Solve the Coupled Differential Equations? I came across the set of following coupled equations while studying cycloid motion in Griffiths' Intro to ED [imath]\ddot{y}=\omega \dot{z}[/imath] [imath]\ddot{z}=\omega (\frac{E}{B}-\dot{y})[/imath] I am at a loss as to how I may approach the problem to solve. Any help is appreciated.

509034

Getting used to projective coordinates, need help describing (2) objects geometrically I'm trying to get an intuition for what things look like in projective coordinates. There are two curves that I have to work a problem with, but I'm not sure how to visualize them. They are [imath]V(u^2 X + v^2 Y + uvZ) \subset \mathbb{P}^2 \times \mathbb{P}^1 [/imath] [imath]V(u^2 X^2 + v^2 Y^2 + uvZ^2) \subset \mathbb{P}^2 \times \mathbb{P}^1[/imath] where [imath][u : v][/imath] are the projective coordinates of [imath]\mathbb{P}^1[/imath] and [imath][X : Y : Z][/imath] are the projective coordinates of [imath]\mathbb{P}^2[/imath]. Is there any way to describe these objects geometrically?

508575

Description of varieties in [imath]\mathbb{P}^2\times \mathbb{P}^1[/imath] If [imath][x:y][/imath] are coordinates of [imath]\mathbb{P}^1[/imath] and [imath][X:Y:Z][/imath] are coordinates of [imath]\mathbb{P}^2[/imath], what do the following varieties look like? [imath]V(x^2X+y^2Y+xyZ)\subset \mathbb{P}^2\times \mathbb{P}^1[/imath] [imath]V(x^2X^2+y^2Y^2+xyZ^2)\subset \mathbb{P}^2\times \mathbb{P}^1[/imath] As of now, I don't quite know how to visualize these varieties. A detailed geometric picture would be especially welcome.

509366

closed form [imath]f_n=\sqrt{2f_{n-1}}[/imath] ? I am trying to write up a proof for the convergence of this recursive function. I was wondering if there exists a closed form. Given first term in sequence is [imath]\sqrt{2}[/imath] and second is [imath]\sqrt{2\sqrt{2}}[/imath], and so on.

200416

How I can prove that the sequence [imath]\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}[/imath] converges to 2? Prove that the sequence [imath]$\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \ $[/imath] converges to [imath]$2$[/imath]. My attempt I proved that the sequence is increasing and bounded by [imath]$2$[/imath], can anyone help me show that the sequence converges to [imath]$2$[/imath]? Thanks for your help.

509403

Prove that when AC matrices are multiplied and equal the identity matrix the solution of Ax=b is consistent for all real numbers Continuing on my newbie dive into linear algebra I have this problem: Suppose that [imath]A = [a_{ij}][/imath] [imath]m×n[/imath] and [imath]C = [c_{ij}][/imath] [imath]n×m[/imath] and [imath]AC = I[/imath] . Prove that the system Ax = b is consistent for every [imath]b\in\mathbb{R}^{m}[/imath]

509380

Show that when [imath]BA = I[/imath], the solution of [imath]Ax=b[/imath] is unique I'm just getting back into having to do linear algebra and I am having some trouble with some elementary questions, any help is much appreciated. Suppose that [imath]A = [a_{ij}][/imath] is an [imath]m\times n[/imath] matrix and [imath]B = [b_{ij}][/imath] and is an [imath]n\times m[/imath] matrix and [imath]BA = I[/imath] the identity matrix that is [imath]n \times n[/imath]. Show that if for some [imath]b \in \Bbb{R}^m[/imath] the equation [imath]Ax = b[/imath] has a solution, then the solution is unique.

497205

convolve probit function with gaussian I want to prove the following, however, not sure where to start. [imath]\int\Phi(a)\mathcal{N}(a|\mu,\sigma^2)da=\Phi\left(\frac{\mu}{\sqrt{1+\sigma^2}}\right)[/imath] Where [imath]\Phi(\cdot)[/imath] is the probit function, defined as [imath]\Phi(a)=\int_{-\infty}^a\mathcal{N}(x|0,1)dx[/imath]

501952

Integral of an integral with variable limits I'd like to prove the following but not sure where to start: [imath]\int_{-\infty}^\infty\int_{-\infty}^a\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)dx\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(a-\mu)^2}{2\sigma^2}\right)da\\ =\int_{-\infty}^{\frac{\mu}{\sqrt{1+\sigma^2}}}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)dx[/imath] Note that the hard part is due to the fact that the [imath]a[/imath] is a variable limit, If it helps, the first term inside the integral is [imath]\mathcal{N}(x;0,1)[/imath] the second term is, [imath]\mathcal{N}(a;\mu,\sigma^2)[/imath]

507830

Pumping lemma [imath]L=\{a^ib^j | i \neq j ; i,j \ge 0\}[/imath] So, let's have language [imath]L=\{a^ib^j | i \neq j ; i,j \ge 0\}[/imath] I have to prove that it's not regular. \begin{align} \omega=a^nb^{n+1}=a^{n-1}ab^{n+1} \end{align} \begin{align} x&=a^k\\ y&=a\\ z&=a^{n-k-1}b^{n+1}\\ \\ i=2:xy^iz&=a^ka^ia^{n-k-1}b^{n+1}\\ &=a^ka^2a^{n-k-1}b^{n+1}\\ &=a^{n+1}b^{n+1}\\ \end{align} [imath]\#_a(\omega)=\#_b(\omega)\implies a^{n+1}b^{n+1} \not\in L [/imath] I'd like to ask if this is a correct proof?

289298

Pumping Lemma Excercise I am having a really tough time proving the following language is not regular using the pumping lemma. This whole time I have been working on pumping lemma problems, the variable of the power has been the same on all alphabet characters that make up the string for the language. [imath]L = \{0^i1^j | j \ne i\} [/imath] This is what I tried: Let [imath]p[/imath] be the pumping length of [imath]L[/imath]. Consider [imath]S= 0^p1^j[/imath], [imath]|s| > p[/imath], and [imath]|s| > j[/imath]. Thus [imath]s[/imath] must satisfy the pumping lemma. Consider splitting the string: I am assuming in order to prove this language is not regular via contradiction, we would have to find a case where the [imath]j[/imath] IS equal to [imath]i[/imath], even though they aren't supposed to be. I am not sure how to proceed from this point of the proof and would appreciate any suggestions. Many thanks in advance!

44522

(Theoretical) Multivariable Calculus Textbooks (Note that I have used bold text frequently simply to highlight the key points of my question for those who do not have the time to read through it thoroughly (it is not very long, however); I hope this is not considered offensive.) There are many textbooks on multivariable calculus. However, some textbooks on multivariable calculus do not focus very much on the theoretical foundations of the subject. For example, a textbook might state a result along the lines of "the order of partial differentiation is immaterial" without proof and ask the student to use this rule to solve problems. Similarly, theorems such as those due to Green and Stokes are often not proved in their full generality. Therefore, I ask the following question: What are some good theoretical multivariable calculus textbooks? Since "theoretical" is somewhat ambiguous, let me state the following criteria which I would like a "theoretical" textbook on multivariable calculus to satisfy: The textbook should be rigorous and it should not state a theorem without proof if the theorem is proved in at least one other multivariable calculus textbook. (Of course, the textbook may omit certain theorems; however, this criterion at least ensures that major theorems in multivariable calculus are not stated without proof and used purely for the sake of computations. Also, this criterion permits the textbook to state an interesting theorem if it is beyond the scope of all multivariable calculus textbooks.) The textbook should be primarily based on developing the theoretical foundations of multivariable calculus; therefore, applications such as learning how to compute the partial derivative of a function, learning how to solve extremum problems, learning how to compute etc. should be kept to a minimum. In particular, the textbook can assume that the reader has already seen at least an informal treatment of the subject where these aspects are emphasized. The textbook should have a rigorous treatment of differentiability in [imath]n[/imath]-dimensional Euclidean space (e.g., the inverse and implicit function theorems should be proven), Riemann integration in [imath]n[/imath]-space, and differential forms (e.g., Stokes theorem should be proven). It would also be a bonus if the book treated the general concept of a manifold. Textbooks with minimal prerequisites are preferred; however, please feel free to suggest books meeting the above criteria even if the prerequisites are quite demanding. Finally, it would also be preferable, but not essential, for the book to only treat multivariable calculus. Examples of books meeting the above criteria: "Analysis on Manifolds" by James Munkres, "Principles of Mathematical Analysis" by Walter Rudin, and "Calculus on Manifolds" by Michael Spivak. Although I have studied theoretical multivariable calculus already (four years ago), I could never find "the perfect book" (relative to myself, of course). Every book has its virtues; Rudin for its elegance, Munkres for its beautiful exposition, and Spivak for its "quick and dirty" approach. I am hoping that someone will be able to suggest a book that (relative to myself) is "perfect". Also, this question can be useful to other students who have not yet studied the subject and wish to learn it. Thank you very much for all answers! Please do feel free to suggest as many books as you can think of so we can form a big list. Also, please try to explain why a particular book is good or at least why you think it is good. I suppose it is fine to suggest a book that is already suggested provided you have a different view as to why the book is good.

1514370

Rigorous book for several variable calculus I have lecture notes on several variable calculus. But it omits many proofs as the subject is so large, and for example the Taylor's formula was given only in a special case. Is there a rigorous book to fill the details, or should a student just fill the details as he or she sees something that is omitted? Lecture notes contains the euclidean space [imath]\mathbb R^n[/imath], real valued functions on [imath]\mathbb R^n[/imath], vector-valued functions, integrals in a plane and in higher dimensional spaces and integral formulas, like path integral, Green's formula in a plane, exact vector fields, surface integrals, Gauss's theorem, and Stokes's theorem.

510337

Euclid for polynomials I have a question bout euclid polynomials. If [imath]C(x) =x^4−1[/imath] and [imath]D(x) =x^3+x^2[/imath] How do I find a polynomials [imath]A(x)[/imath] and [imath]B(x)[/imath] such that [imath]A(x)C(x) +B(x)D(x) =x+1[/imath] for all [imath]x[/imath]?

509402

Polynomial Question Find polynomials [imath]A(x)[/imath] and [imath]B(x)[/imath] such that [imath]A(x)P(x) + B(x)Q(x) = x + 1[/imath] for all [imath]x[/imath] where [imath]P(x) = x^4 - 1[/imath] and [imath]Q(x) = x^3 + x^2[/imath]. I'm stumped on this question. I know that I'm supposed to apply the extended version of Euclid's algorithm for polynomials but I'm unsure of how to do that. I thought about trying to create some kind of linear system but guessing arbitrary coefficients but that wouldn't work as [imath]A[/imath] and [imath]B[/imath] don't have fixed degrees.

510465

A subset of [imath]Z(G)[/imath] If [imath]H[/imath] is a normal subgroup of [imath]G[/imath] and the [imath]|H|[/imath] is the smallest prime that divides [imath]|G|[/imath], then [imath]H \subset Z(G)[/imath]. First I note that (1) [imath]|G/C_G(H)|[/imath] divides the [imath]|Aut(H)|[/imath], how can i prove this using this (1)?

227971

[imath]G[/imath] group, [imath]H \trianglelefteq G[/imath], [imath]\vert H \vert[/imath] prime, then [imath]H \leq Z(G)[/imath] Let [imath]G[/imath] be a finite group. Let [imath]H \trianglelefteq G[/imath], with [imath]\vert H \vert = p[/imath], a prime, where [imath]p[/imath] is the smallest prime dividing [imath]\vert G \vert[/imath]. Prove that [imath]H \leq Z(G)[/imath]. (Hint: If [imath]a \in H[/imath], by normality, its conjugacy class lies inside [imath]H[/imath].) My approach so far: Let [imath]a \in H[/imath]. Then [imath]C(a) = \{gag^{-1} : g \in G\} \subseteq H[/imath]. Now, since [imath]\vert H \vert[/imath] is prime, it follows that [imath]H[/imath] is cyclic and moreover [imath]H[/imath] is abelian, hence [imath]C_H(a) = H[/imath] for all [imath]a \in H[/imath] and equivalently [imath][H: C_H(a)] = 1[/imath] for all [imath]a \in H[/imath]. Try to show [imath]\vert C(a) \vert = [G:C_G(a)] = [H:C_H(a)] = 1[/imath]. So, since [imath]\vert C(a) \vert = 1[/imath] for all [imath]a \in H[/imath], we have [imath]C(a) = \{a\}[/imath] for all [imath]a \in H[/imath] and equivalently [imath]a \in Z(G)[/imath] for all [imath]a \in H[/imath], so we can say [imath]H \subseteq Z(G)[/imath] and moreover [imath]H \leq Z(G)[/imath]. My main question is, how to show [imath][G:C_G(a)] = [H:C_H(a)][/imath]. But also, I haven't used the fact that [imath]p[/imath] is the smallest prime dividing [imath]\vert G \vert[/imath], so is my reasoning wrong anywhere?

509980

Systems of linear equations to calculate [imath]\alpha[/imath] and [imath]\beta[/imath] Point [imath]1[/imath]: When there is [imath]1[/imath] car passing the road, the average speed is [imath]50[/imath] km/h. Point [imath]2[/imath]: When there are [imath]5[/imath] cars passing the road, the average speed is [imath]45[/imath] km/h. Point [imath]3[/imath]: When there are [imath]12[/imath] cars passing the road, the average speed is [imath]38[/imath] km/h. A traffic engineering company decides to model the average speed (shown by [imath]u[/imath]) as a linear function of the number of cars (shown by [imath]n[/imath]). So we want to have [imath]u(n)=\alpha+\beta n[/imath] . Using Point [imath]1[/imath] , Point [imath]2[/imath] and Point [imath]3[/imath] information: write a system of linear equations to calculate [imath]\alpha[/imath] and [imath]\beta[/imath] using all three points; i.e. we will have three equations with two unknowns.

508663

Writing systems of linear equations and matlab help appreciated Does anyone know how I would go about answering the following question? A traffic engineering company has installed some trac cameras along a one lane road to find a relation between the number of cars passing the road and the average speed at which the cars move. The cameras show that: Point 1: When there is 1 car passing the road, the average speed is 50km/h. Point 2: When there are 5 cars passing the road, the average speed is 45km/h. Point 3: When there are 12 cars passing the road, the average speed is 38km/h. The traffic engineering company decides to model the average speed (shown by [imath]u[/imath]) as a linear function of the number of cars (shown by [imath]n[/imath]). So we want to have [imath]u(n) = \alpha + \beta n.[/imath] Using Point 1 and Point 2 information, write the system of linear equations to calculate [imath]\alpha[/imath] and [imath]\beta[/imath]. Write a MATLAB code to calculate [imath]\alpha[/imath] and [imath]\beta[/imath].

39079

A bijective function between a square and its side Is there a simple geometric proof that there exists a continuous bijective function between a square and its side? And is there some explicit continuous function or formula [imath]f^1(z)\mapsto (x,y)[/imath] and [imath]f(x,y)\mapsto z[/imath], with [imath](x,y) \in [0,1]\times[0,1][/imath] and [imath]z \in [0,1][/imath]? And is there a constructive proof that two sets of the same cardinality have a bijective function?

510573

Is there a continuous bijection between an interval [imath][0,1][/imath] and a square: [imath][0,1] \times [0,1][/imath]? Is there a continuous bijection from [imath][0,1][/imath] onto [imath][0,1] \times [0,1][/imath]? That is with [imath]I=[0,1][/imath] and [imath]S=[0,1] \times [0,1][/imath], is there a continuous bijection [imath] f: I \to S? [/imath] I know there is a continuous bijection [imath]g:C \to I[/imath] from the Cantor set [imath]C[/imath] to [imath][0,1][/imath]. The square [imath]S[/imath] is compact so there is a continuous function [imath] h: C \to S. [/imath] But this leads nowhere. Is there a way to construct such an [imath]f[/imath]? I ask because I have a continuous functional [imath]F:S \to \mathbb R[/imath]. For numerical reason, I would like to convert it into the functional [imath] G: I \to \mathbb R, \\ G = F \circ f , [/imath] so that [imath]G[/imath] is continuous.

511092

If [imath]a\equiv b\pmod m[/imath] and [imath]c+d\equiv 0\pmod m[/imath] then [imath]ac+bd\equiv 0\pmod m[/imath] If [imath]a\equiv b\pmod m[/imath] and [imath]c+d\equiv 0\pmod m[/imath] then [imath]ac+bd\equiv 0\pmod m[/imath]. The response, posted below is correct??

511046

If [imath]a+b\equiv0\pmod m[/imath] and [imath]c+d\equiv 0\pmod m[/imath] then [imath]ac\equiv bd\pmod m[/imath]. Demonstration. If [imath]a+b\equiv0\pmod m[/imath] and [imath]c+d\equiv 0\pmod m[/imath] then [imath]ac\equiv bd\pmod m[/imath] How to show?[imath]$$ I tried [/imath]a+b\equiv0\pmod m\Longrightarrow m\mid a+b\\c+d\equiv0\pmod m\Longrightarrow m\mid c+d\\a+b=mk\;\;\text{and}\;\;c+d=mj\\(a+b)(c+d)=m(mjk)\\m\mid ab+ad+bc+bd$$ But from here could solve anything.

511386

Divisible polynomial and first term How do I show that if [imath]b^2+ab+1[/imath] divides [imath]a^2+ab+1[/imath] for a, b are poaitive integers. Then [imath]a=b[/imath]?

429638

Divisibility by Quadratics $b^2+ba+1\mid a^2+ab+1\Rightarrow\ a=b$ The natural numbers [imath]a[/imath] and [imath]b[/imath] are such [imath]a^2+ab+1[/imath] is divisible by [imath]b^2+ba+1[/imath]. Prove that [imath]a = b[/imath]. I tried to algebraically manipulate it as follows: [imath](b^2 + ba + 1)k = a^2 + ab + 1[/imath] [imath][b(a + b) + 1]k = a(a + b) + 1[/imath] [imath]kb(a + b) + k = a(a + b) + 1[/imath] [imath]k - 1 = (a - kb)(a + b)[/imath] I'm stuck here. What should I do next? A case-by-case analysis of possible congruencies would be too tedious and inelegant.

511203

What is wrong with this proof of: [imath]2+2 = 5[/imath] I have seen this image and surprised that we can prove [imath]2 + 2 = 5[/imath]. can any one tell me what is wrong with this image. Prove that, [imath]2+2=5[/imath]. We know that, [imath]2+2=4[/imath] [imath]\begin{align}\Rightarrow2+2&=4-\dfrac92+\dfrac92\\\,\\ &=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\ &=\sqrt{16-2\cdot4\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\\ &=\sqrt{-20+\left(\dfrac92\right)^2}+\dfrac92\\ &=\sqrt{(5)^2-2\cdot5\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\qquad\qquad\qquad\\ &=\sqrt{\left(5-\dfrac92\right)^2}+\dfrac92\\ &=5-\dfrac92+\dfrac92\\ &=5\\\,\\&\therefore\,2+2=5\text{ (Proved)}\\ \end{align}[/imath]

457490

[imath]2+2 = 5[/imath]? error in proof [imath]\begin{align} 2+2 &= 4 - \frac92 +\frac92\\ &= \sqrt{\left(4-\frac92\right)^2} +\frac92\\ &= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\ &= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\ &= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\ &= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\ &= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\ &= \sqrt {\left(5-\frac92\right)^2} +\frac92\\ &= 5 + \frac92 - \frac92 \\ &= 5\end{align}[/imath] Where did I go wrong

511395

If [imath]\{f_i\}[/imath] generate the unit ideal in a ring, so do [imath]\{f_i^N\}[/imath] for any positive [imath]N[/imath] Let [imath]R[/imath] be a commutative ring, and let [imath]\{f_i\}[/imath] be a finite set of elements generating the unit ideal in R. Then [imath]\{f_i^N\}[/imath] also generate the unit ideal in [imath]R[/imath], for any positive [imath]N[/imath]. Why is this true?

243679

Finite number of elements generating the unit ideal of a commutative ring Let [imath]A[/imath] be a commutative ring with [imath]1[/imath]. Let [imath]f_1,\dots,f_r[/imath] be elements of [imath]A[/imath]. Suppose [imath]A = (f_1,\dots,f_r)[/imath]. Let [imath]n > 1[/imath] be an integer. Can we prove that [imath]A = (f_1^n,\dots,f_r^n)[/imath] without using axiom of choice? EDIT It is easy to prove [imath]A = (f_1^n,\dots,f_r^n)[/imath] if we use axiom of choice. Suppose [imath]A≠(f_1^n,…,f_r^n)[/imath]. By Zorn's lemma(which is equivalent to axiom of choice), there exists a prime ideal [imath]P[/imath] such that [imath](f_1^n,…,f_r^n) \subset P[/imath]. Since [imath]f_i^n \in P[/imath], [imath]f_i \in P[/imath] for all [imath]i[/imath]. Hence [imath]A = (f_1, \dots,f_n) \subset P[/imath]. This is a contradiction.

511645

Why can we get eigenvalue through solving [imath]det(A-\lambda I)=0[/imath]? In many notes, I see [imath]A\vec{v}=\lambda \vec{v}[/imath] [imath](A-\lambda)\vec{v}=\vec{0}[/imath] [imath]det(A-\lambda I)=0[/imath] How to get line 3 actually? Thanks.

501888

Why is [imath]\det(A - \lambda I)[/imath] zero? I was reading over some notes of mine from the Linear Algebra class I took a while ago in order to prepare myself for Computer Graphics course I'll be taking this Fall and I ran across a sentence that I can't quite grasp. It states that Since we have a non-zero vector [imath]x[/imath] in the nullspace of [imath]A - \lambda I[/imath], then [imath]\det(A - \lambda I) = 0[/imath]. From reading Wikipedia page on determinants I got that ... the system has a unique solution exactly when the determinant is nonzero; when the determinant is zero there are either no solutions or many solutions. So if we have vector [imath]x[/imath] in the nullspace of [imath]A-\lambda I[/imath], then we have either 0 solutions or many solutions. But I still don't understand why [imath]\det(A - \lambda I)[/imath] has to be equal zero?

511776

absolute value of a Root of a cubic equation Does the equation [imath]x^3+10x^2-100x+1729=0[/imath] fas atleast one complex root [imath]\alpha[/imath] such that [imath]|\alpha|[/imath] [imath]\gt 12[/imath]?? Since any cubic equation has atleast one real root, let it be say [imath]k[/imath]. Let the other two complex roots be [imath]\alpha+i\beta , \alpha-i\beta[/imath]. Then The product of the roots is [imath](\alpha^2+\beta^2)k=-1729.[/imath] Since [imath](\alpha^2+\beta^2) \ge 0[/imath], [imath]k\lt0[/imath]. From here I am unable to find any roots or deduce any further.

508416

How to show that the given equation has at least one complex root ,a s.t |a|>12 How do I show that the equation $x^3+10x^2-100x+1729[imath]=0$ has at least one complex root $a$ such that $|a|[/imath]>$[imath]12[/imath].

510176

Why is [imath]\forall x \in A:P(x)[/imath] equivalent to [imath]\forall x (x\in A \to P(x)) [/imath]? In the book that I'm studying from it defines [imath]\forall x \in A: P(x)[/imath] equivalent to [imath]\forall x (x\in A \to P(x))[/imath] without any explanation as to why it is that way. The same thing for the existential quantifier: [imath]\exists x \in A: P(x)[/imath] is equivalent to [imath]\exists x(x\in A \land P(x))[/imath]. I attempted to understand these definitions by applying to real world examples. Example for [imath]\forall x \in A: P(x)[/imath] equivalent to [imath]\forall x (x\in A \to P(x)) [/imath]: [Example One] All cars have wheels Let [imath]C[/imath] be a set of all cars and [imath]P(x)[/imath] stand for "[imath]x[/imath] has wheels." Then the English statement analyzed in logical form is [imath]\forall x \in C: P(x)[/imath] is equivalent to [imath]\forall x (x\in C \to P(x))[/imath]. Translating the statement means "If [imath]x[/imath] is a car, then [imath]x[/imath] has wheels." However, sometimes when I analyze it I get "For every [imath]x[/imath], [imath]x[/imath] is a car and [imath]x[/imath] has wheels." (In logical form, [imath]\forall x(x \in C \land P(x))[/imath]) which book says it's wrong. Again, I have the same trouble with the existential quantifier definition as well. [Example Two] Some cars have three wheels. Let [imath]C[/imath] be a set of all cars and [imath]P(x)[/imath] stand for "[imath]x[/imath] has three wheels." Then the English statement analyzed in logical form is [imath]\exists x \in C: P(x)[/imath] which is equivalent to [imath]\exists x(x \in C \land P(x))[/imath]. This translates to "There exists at least one [imath]x[/imath] such that [imath]x[/imath] is a car and [imath]x[/imath] has three wheels." But sometimes I translate these kinds of statements as [imath]\exists x(x \in C \to P(x))[/imath], which is wrong according to the book's definition. I'm asking how can I make it clear these two definitions are different. Also, why [imath]\forall x \in A: P(x)[/imath] is not equivalent to [imath]\forall x (x\in A \land P(x))[/imath] and that [imath]\forall x \in A: P(x)[/imath] is equivalent to [imath]\forall x (x\in A \to P(x))[/imath]. Similarly, why [imath]\exists x \in A: P(x)[/imath] is equivalent to [imath]\exists x(x\in A \land P(x))[/imath] and that [imath]\exists x \in A: P(x)[/imath] is not equivalent to [imath]\exists x(x\in A \to P(x))[/imath]? In short, can you help make it clear to me why these definitions are defined the way they are.

398492

Why is the universal quantifier [imath]\forall x \in A : P(x)[/imath] defined as [imath]\forall x (x \in A \implies P(x))[/imath] using an implication? And the same goes for the existential quantifier: [imath]\exists x \in A : P(x) \; \Leftrightarrow \; \exists x (x \in A \wedge P(x))[/imath]. Why couldn’t it be: [imath]\exists x \in A : P(x) \; \Leftrightarrow \; \exists x (x \in A \implies P(x))[/imath] and [imath]\forall x \in A : P(x) \; \Leftrightarrow \; \forall x (x \in A \wedge P(x))[/imath]?

511880

Boundary of union of two sets equals the union of their boundaries Plesae give some hint to solve the following problem: If [imath]A[/imath] and [imath]B[/imath] are two subsets of a topological space [imath]X[/imath] such that [imath]\overline{A}\cap \overline{B}=\emptyset[/imath], then [imath]\partial(A\cup B)=\partial A\cup \partial B[/imath], where [imath]\partial A[/imath] denotes the boundary of [imath]A[/imath].

218805

The boundary of union is the union of boundaries when the sets have disjoint closures Assume [imath]\bar A\cap\bar B=\emptyset[/imath]. Is [imath]\partial (A \cup B)=\partial A\cup\partial B[/imath], where [imath]\partial A[/imath] and [imath]\bar A[/imath] mean the boundary set and closure of set [imath]A[/imath]? I can prove that [imath]\partial (A \cup B)\subset \partial A\cup\partial B[/imath] but for proving [imath]\partial A\cup\partial B\subset \partial (A \cup B)[/imath] it seems not trivial. I tried to show that for [imath]x\in \partial A\cup\partial B[/imath] WLOG, [imath]x\in \partial A[/imath] so [imath]B(x)\cap A[/imath] and [imath]B(x)\cap A^c[/imath] not equal to [imath]\emptyset[/imath] but it seems not enough to show the result.

228443

Does radix 1 exists? Introduction This probably sounds a bit weird to ask, but the question is, is there such a base that is base 1. One thing we know is that this base will contain one symbol only, but will this symbol represent something i.e. a value, or nothing? If we try to see a connection between other bases, for instance base 10, 4, 3, 2, we see following: base 10 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} base 4 {0, 1, 2, 3} base 3 {0, 1, 2} base 2 {0, 1} So base 1 should have the symbol [imath]0[/imath], in order to fit this pattern. Number representation Let's say that we introduce a new symbol, which will be used in base 1, namely [imath]\alpha[/imath]. We will probably think that if we list the first numbers in base 1, we will get something like: [imath]\alpha,\alpha\alpha, \alpha\alpha\alpha, \alpha\alpha\alpha\alpha[/imath] Which is quite logical. But, if we observe other systems, such as base 10, we will see that 10 is actually built up of two symbols. Similarly, for radix 2, [imath]10_2[/imath] is equal to 2, which appears to be the base. So, would it not now be logical to claim that the definition of 1 (base 10) in base 1 is [imath]\alpha\alpha[/imath], or with other words, two symbols? Number conversion If we would, for example, convert [imath]\alpha\alpha\alpha\alpha[/imath] to radix 10, we would get following: [imath](\alpha\alpha\alpha\alpha)_1=\alpha*1^3+\alpha*1^2+\alpha*1^1 + \alpha*1^0 = (4\alpha)_{10}[/imath] Sure enough, [imath](\alpha\alpha)_1 = 1_{10} = (2\alpha)_{10}[/imath] So I suppose this statement is therefore true as well: [imath](\alpha\alpha\alpha\alpha)_1=(2\alpha)_{10} + (2\alpha)_{10} = 1 + 1 = 2[/imath] But the interesting part here is that our symbol [imath]\alpha[/imath] is nothing, i.e. does not have any value at all, which implies that all numbers do not exist as values. Conclusion I do not now how correct this is, so it would be great if someone could comment upon what is written here. The question is finally, is this plausible? Thank you, Artem

371972

What would base [imath]1[/imath] be? Base [imath]10[/imath] uses these digits: [imath]\{0,1,2,3,4,5,6,7,8,9\};\;[/imath] base [imath]2[/imath] uses: [imath]\{0,1\};\;[/imath] but what would base [imath]1[/imath] be? Let's say we define Base [imath]1[/imath] to use: [imath]\{0\}[/imath]. Because [imath]10_2[/imath] is equal to [imath]010_2[/imath], would all numbers be equal? The way I have thought Base 1 might be represented is tally marks, [imath]0_{10}[/imath] would be represented by nothing. So, [imath]5[/imath] in Base 1 would be represented by [imath]00000[/imath]? Or we could define Base 1 to use: [imath]\{[/imath]|[imath]\}[/imath] and [imath]5[/imath] would be |||||?

512526

Let [imath]n \in \Bbb N[/imath]. Let [imath]p>2[/imath] a prime number. Show that [imath]1^n+2^n+...+(p-1)^n \equiv 0 \pmod {p}[/imath] This is an exercise in my abstract algebra reader, in the chapter about polynomial rings. Let [imath]n \in \Bbb N[/imath]. Let [imath]p>2[/imath] a prime number. And let [imath]n[/imath] not divisble by [imath]p-1[/imath]. Show that [imath]1^n+2^n+...+(p-1)^n \equiv 0 \pmod {p}[/imath] Our teacher assistant gave a hint to solve this with induction. We have tried to solve this using induction, but after some hours tring, I don't believe anyomre that this is possible. At least, I don't think I'm able to do that with my current knowlegde about mathematics. If someone has an idea how to solve this with induction (or maybe an other way), then I'm all ears.

511444

[imath]1^n +2^n + \cdots +(p-1)^n \mod p =[/imath]? Calculate for every positive integer [imath]n[/imath] and for every prime [imath]p[/imath] the expression [imath]1^n +2^n + \cdots +(p-1)^n \mod p[/imath] I need your help for this. I don't know what to do, but I'll show you what I know. Wilson's theorem The identity [imath]X^p-X = \prod_{a \in \mathbb{F}_p-1}(X-a)[/imath] [imath]\exists a \in \mathbb{Z}, \ a^2 \equiv -1 \mod p \qquad \iff \qquad p \equiv 1 \mod 4[/imath] Now can you please provide me a hint?

263081

Sobolev space inequality If [imath]f\in H^2(\mathbb R^2)[/imath], I want to show that [imath]||f||_{L^\infty}\le c||f||_{H^1} [1+\ln(1+||f||_{H^2})][/imath] How can I get the "ln"? and how can I make it into a product of [imath]H^1[/imath] and [imath]H^2[/imath] norm? It is actually one of my Real variables's project. So how can I do this inequality in an relativly "elementary" way?

259336

Sobolev inequality If [imath]f\in H^2(\mathbb R^2)[/imath], I want to show that [imath]||f||_{L^\infty}\le c||f||_{H^2}[/imath] [imath]||f||_{L^\infty}\le c||f||_{H^1} [1+\ln(1+||f||_{H^2})][/imath] For 1, I use [imath]||f||_{L^\infty}\le \sup_{x\in \mathbb R^2} |f|\le\int_{\mathbb R^2} |\hat f(\xi)|d\xi = \int(1+|\xi|^2)^{-1}(1+|\xi|^2)|\hat f(\xi)|d\xi \le (\int (1+|\xi|^2)^{-2}d\xi)^2(\text {which is integrable in this case})||f||_{H^2}\to \text{is this correct?}[/imath] But for 2, I am stucked. How can I get the "ln"? and how can I make it into a product of [imath]H^1[/imath] and [imath]H^2[/imath] norm?

220079

Is there a simpler expression for [imath]f(f(f(...f(x)...)))[/imath] (total n '[imath]f[/imath]')? Is there a simpler expression for [imath]f(f(f(...f(x)...)))[/imath] (total n '[imath]f[/imath]') ? Thank you.

247710

Notation for repeated application of function If I have the function [imath]f(x)[/imath] and I want to apply it [imath]n[/imath] times, what is the notation to use? For example, would [imath]f(f(x))[/imath] be [imath]f_2(x)[/imath], [imath]f^2(x)[/imath], or anything less cumbersome than [imath]f(f(x))[/imath]? This is important especially since I am trying to couple this with a limit toward infinity.

513928

Coefficient of x^50 in the cube of a summation What is the coefficient of [imath]x^{50}[/imath] in [imath](\sum_{n=1}^{\inf} x^n)^3[/imath] ? Does there exist a combinatorial approach to this problem?

513174

Find coefficient of [imath]x^{50}[/imath] in [imath](\sum_{i=1}^{\infty}x^n)^3[/imath] How do you find the coefficient of [imath]x^{50}[/imath] in [imath](\sum_{i=1}^{\infty}x^n)^3[/imath]?

514505

A property of continuous functions A function [imath]f : [a,b] \to [a,b][/imath] is continuous for all [imath]x \in [a,b].[/imath] Prove that there exists a [imath]c\in [a,b][/imath] such that [imath]f(c) = c.[/imath]

211242

Fixed point in a continuous function Suppose that [imath]f[/imath] is a function defined in [imath][a;b][/imath] to [imath][a;b][/imath] and continuous on [imath][a;b][/imath]. The problem is I haven't the definition of the function, this is more abstract, but even if how can I prove that [imath]f[/imath] would have a fixed point?

143796

Number of [imath]\sigma[/imath] -Algebra on the finite set Let [imath]X[/imath] is a nonempty set with [imath]m[/imath] members . How many [imath]\sigma[/imath] -algebra can we make on this set?

671648

Is there a general way to count the number of sigma-algebras on a finite set? I was asked a high-school question today which was merely asking the number of sigma-algebras on a set. Let [imath]X[/imath] be a set Let [imath]S\triangleq \{\Sigma\subset P(X): \Sigma \text{ is a sigma-algebra on }X\}[/imath]. If [imath]X[/imath] is finite, what would be the cardinality of [imath]S[/imath]? The question i was asked was the case [imath]|X|=3[/imath]. In this case [imath]|S|=5[/imath]. What would be the cardinality of [imath]S[/imath] in general? Is it possible to find this value via elementary function?

496037

Prove that [imath]\sqrt 2 +\sqrt 3[/imath] is irrational. Please prove that [imath]\sqrt 2 + \sqrt 3[/imath] is irrational. One of the proofs I've seen goes: If [imath]\sqrt 2 +\sqrt 3[/imath] is rational, then consider [imath](\sqrt 3 +\sqrt 2)(\sqrt 3 -\sqrt 2)=1[/imath], which implies that [imath]\sqrt 3 − \sqrt 2[/imath] is rational. Hence, [imath]\sqrt 3[/imath] would be rational. It is impossible. So [imath]\sqrt 2 +\sqrt 3[/imath] is irrational. Now how do we know that if [imath]\sqrt 3 -\sqrt 2[/imath] is rational, then [imath]\sqrt 3[/imath] should be rational? Thank you.

457382

Can [imath]\sqrt{n} + \sqrt{m}[/imath] be rational if neither [imath]n,m[/imath] are perfect squares? Can the expression [imath]\sqrt{n} + \sqrt{m}[/imath] be rational if neither [imath]n,m \in \mathbb{N}[/imath] are perfect squares? It doesn't seem likely, the only way that could happen is if for example [imath]\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}[/imath], which I don't think is possible, but how to show it?

516297

Prove:[imath]A B[/imath] and [imath]B A[/imath] has the same characteristic polynomial. Suppose [imath]A B[/imath] are all [imath]n\times n[/imath] matrix. Prove:[imath]A B[/imath] and [imath]B A[/imath] has the same characteristic polynomial. If A is invertible \begin{align}P(A B)=|\lambda E-A B|=\left|\lambda A\cdot A^{-1}-A B\right|=\left|A\left(\lambda A^{-1}-B\right)\right|=\left|A\left\|\text{[imath]\lambda [/imath]A}^{-1}-B\right.\right|\end{align} \begin{align}P(B A)=|\lambda E-B A|=\left|\lambda A^{-1}\cdot A-B A\right|=\left|\left(\lambda A^{-1}-B\right)A\right|=\left|\left.\text{[imath]\lambda [/imath]A}^{-1}-B\right\|A\right|\end{align} Is it true? If A is not invertible How to prove? Edit: I found a proof by Block Matrix Multiplication, it's more straightforward, is it right? \begin{align}\left( \begin{array}{cc} E & 0 \\ -A & E \\ \end{array} \right).\left( \begin{array}{cc} \lambda E & B \\ \lambda A & \lambda E \\ \end{array} \right)=\left( \begin{array}{cc} \lambda E & B \\ 0 & \lambda E-A B \\ \end{array} \right)\end{align} \begin{align}\left( \begin{array}{cc} \lambda E & B \\ \lambda A & \lambda E \\ \end{array} \right).\left( \begin{array}{cc} E & 0 \\ -A & E \\ \end{array} \right)=\left( \begin{array}{cc} \lambda E-B A & B \\ 0 & \lambda E \\ \end{array} \right)\end{align} Another proof is by Sylvester determinant theorem which uses LU block decomposition.

311342

Do [imath] AB [/imath] and [imath] BA [/imath] have same minimal and characteristic polynomials? Let [imath] A, B [/imath] be two square matrices of order [imath]n[/imath]. Do [imath] AB [/imath] and [imath] BA [/imath] have same minimal and characteristic polynomials? I have a proof only if [imath] A[/imath] or [imath] B [/imath] is invertible. Is it true for all cases?

515488

category of linear maps Let [imath]V,W[/imath] be vector spaces. Let's define a category whose objects are linear map [imath]f:V\to W[/imath] and morphisms from [imath]f[/imath] to [imath]g[/imath] are pair of linear maps [imath](\alpha,\beta)[/imath] where [imath]\alpha:V\to V,\beta :W\to W[/imath] such that [imath]g \circ\alpha = \beta \circ f[/imath]. Does this category have some special name? If [imath]f_i[/imath] are indexed family of objects, is there product of them in this category?

518321

category of linear maps between fixed vector spaces Let [imath]V,W[/imath] be fixed [imath]k[/imath]-vector spaces. Let [imath]C[/imath] be the category whose objects are linear map [imath]f:V\to W[/imath] and morphisms from [imath]f[/imath] to [imath]g[/imath] are pair of linear maps [imath](\alpha,\beta)[/imath] where [imath]\alpha:V\to V,\beta :W\to W[/imath] such that [imath]g \circ\alpha = \beta \circ f[/imath]. I want to find the product in this category. Note that [imath]C[/imath] is the full subcategory of [imath]D[/imath], whose objects are linear maps between any [imath]k[/imath]-vector spaces, with morphisms [imath](\alpha,\beta)[/imath] such that [imath]g \circ\alpha = \beta \circ f[/imath]. I proved that in [imath]D[/imath], the product of linear maps [imath]f_i:V_i \to W_i[/imath], [imath] i \in I[/imath] is given by [imath]f: \prod_{i\in I} V_i \to \prod_{i\in I} W_i[/imath] where [imath]f((v_i))=(f_i(v_i))[/imath]. So given [imath]f_i:V \to W[/imath] in [imath]C[/imath], if there exists isomorphisms [imath]\prod V\simeq V[/imath] and [imath]\prod W\simeq W[/imath], we can also find the product in [imath]C[/imath]. But is it the necessary condition? In other words, if there exists product of [imath]f_i:V\to W[/imath] in [imath]C[/imath], then is it true that [imath]\prod V\simeq V[/imath] and [imath]\prod W\simeq W[/imath]?

517755

the division [imath]14^{256}[/imath] by [imath]17[/imath] What the rest of the division [imath]14^{256}[/imath] by [imath]17[/imath]? [imath]14^2\equiv9\pmod{17}\\14^4\equiv13\pmod{17}\\14^8\equiv16\equiv-1\pmod{17}\\(14^8)^{32}\equiv(-1)^{32}\equiv1\pmod{17}[/imath]The rest is [imath]1[/imath], correct?

511305

What is the remainder of dividing [imath]14^{256}[/imath] by [imath]17[/imath]? What is the remainder of dividing [imath]14^{256}[/imath] by [imath]17[/imath]? [imath]14^2\equiv 196\equiv 9 \pmod{17}\\14^{4}\equiv81\equiv13\pmod{17}\\14^8\equiv169\equiv16\pmod{17}\\14^{16}\equiv256\equiv1\pmod{17}\\14^{256}\equiv1^{16}\equiv1\pmod{17}[/imath] Soon rest is [imath]1[/imath], correct?

517835

How to determine whether critical points (of the lagrangian function) are minima or maxima? [imath]f(x,y) = 2x+y[/imath] subject to [imath]g(x,y)=x^2+y^2-1=0[/imath]. The Lagrangian function is given by [imath] \mathcal{L}(x,y,\lambda)=2x+y+\lambda(x^2+y^2-1), [/imath] with corresponding [imath] \nabla \mathcal{L}(x,y,\lambda)= \begin{bmatrix} 2 + 2\lambda x \\ 1+2\lambda y \\ x^2+y^2-1 \end{bmatrix}. [/imath] From the latter we can see that [imath]x=\frac{-1}{\lambda}[/imath] and [imath]y=\frac{-1}{2\lambda}[/imath], which we can substitute into [imath]x^2+y^2=1[/imath] to obtain [imath]\lambda = \pm\sqrt{\frac{5}{4}}[/imath]. Meaning that [imath]x = \pm \dfrac{2}{\sqrt{5}}[/imath] and [imath]y= \pm \dfrac{1}{\sqrt{5}}[/imath]. We find the critical points [imath](\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}})[/imath] and [imath](\frac{-2}{\sqrt{5}}, \frac{-1}{\sqrt{5}})[/imath]. I am confused on how I should proceed to check wether these points or minima or maxima? I know the Hessian is involved, but which one?

516746

Determine all the extrema of a function subject to a non-linear constraint. QUESTION Determine all extrema of the function [imath]f(x,y) = x+ 2y [/imath] subject to [imath]x^2 + y^2 - 80 = 0[/imath] ATTEMPT I don't think I understand what I'm supposed to do. This was in a test and I ended up trying to "graphically"or ïntuitively" find out how the [imath]f(x,y)[/imath] would behave in a circle of radius [imath]\sqrt80[/imath] Which left me with some pretty random numbers that turned out to also be wrong. What could I have done differently? (If possible could I get a bit of a detailed explanation or some links to that info.) Lagrange Multipliers: I tried that too on the paper and it also went horribly wrong (but I did get some points, though) but I couldn't quite figure it out.

517904

How to prove an equivalent definition of injective? Let [imath]A,B[/imath] be non-empty sets and [imath]f:A\to B[/imath] a function. Proof that [imath]f[/imath] is injective, iff [imath]f\circ g=f\circ h[/imath] implies that [imath]g=h[/imath] for all functions [imath]g,h:Y\to A[/imath], for every set [imath]Y[/imath]? I can see why this is. But how do I prove this? I get confused by the if and only if part.

481811

Prove that [imath]f[/imath] is one-to-one iff [imath]f \circ h = f \circ k[/imath] implies [imath]h = k[/imath]. Let [imath]f: A \to B[/imath] be a function. Prove that [imath]f[/imath] is one-to-one if and only if [imath]f \circ h = f \circ k[/imath] implies [imath]h = k[/imath], for every set [imath]C[/imath] and all choices of functions [imath]h: C \to A[/imath] and [imath]k: C \to A[/imath]. Here is my work: Assume that [imath]f[/imath] is one-to-one and let [imath]A = \{x_1, x_2\}[/imath] for two arbitrary [imath]x_1, x_2[/imath]. So we have [imath]f(x_1) = f(x_1)[/imath] and [imath]x_1 = x_2[/imath]. Assume [imath]h[/imath] and [imath]k[/imath] are one-to-one. (Is that okay assume that they are one-to-one?) And let [imath]C = \{z_1, z_2\}[/imath] so we have [imath]h(z_1)=h(z_2) = x_1[/imath] and [imath]k(z_1) = k(z_2) = x_2[/imath] since [imath]h: C \to A[/imath] and [imath]k: C \to A[/imath]. If we substitute [imath]h(z)[/imath] and [imath]k(z)[/imath] into [imath]x_1[/imath] and [imath]x_2[/imath] respectively, we have [imath]f(h(z)) = f(k(z))[/imath]. It follows [imath]f \circ h = f \circ k[/imath] implies [imath]h = k[/imath]. Conversely, we assume that [imath]f \circ h = f \circ k[/imath]. It follows [imath]f(h(z_1))=f(k(z_2))[/imath] follows [imath]z_1 = z_2[/imath]. Hence we can conclude that [imath]h[/imath] and [imath]k[/imath] are one-to-one. Since we know [imath]h: C \to A[/imath] and [imath]k: C \to A[/imath], ([imath]x_1[/imath] and [imath]x_2[/imath] are elements of [imath]A[/imath]) and [imath]h(z_1) = x_1[/imath] and [imath]h(z_2) = x_2[/imath]. Since [imath]h(z_1)=k(z_2)[/imath] implies [imath]x_1 = x_2[/imath] and since these two elements are in [imath]A[/imath], [imath]f[/imath] is onto. I know it is hard to read, but I really need your help to check my ambiguous proof. I know some part are wrong, so please let me know. Thank you very much.

441392

Prove that [imath]Ax = b[/imath] with complex solution, actually is rational solution Let [imath]Ax = b[/imath], a system of linear equations, [imath]A[/imath] is matrix of rational numbers, and this system has only one complex solution. Prove this solution is also rational. Can someone help me please? I prefer not only hints... thanks.

394223

System of linear equations having a real solution has also a rational solution. I saw this question Let [imath]A ∈ M_{m\times n}(\mathbb{Q})[/imath] and [imath]b ∈ \mathbb{Q}^m[/imath]. Suppose that the system of linear equations [imath]Ax = b[/imath] has a solution in [imath]\mathbb{R}^n[/imath]. Does it necessarily have a solution in [imath]\mathbb{Q}^n[/imath]? and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine. I considered the form of the equality as [imath] A^{(1)}x_1+\cdots+A^{(n)}x_n=b, [/imath] where [imath]A^{(i)}[/imath] is a column vector of [imath]A[/imath]. I then noticed that for [imath]x_i\in\mathbb{R\setminus Q}=\mathbb{T}[/imath] then, and this is where I think I'm doing something forbidden, each [imath]x[/imath] has the represenation [imath] x_1 = k_{11}\tau_1+\cdots+k_{1p}\tau_p \\ x_2= k_{21}\tau_1+\cdots+k_{2p}\tau_p \\ \vdots \hspace{4cm} \vdots \\ x_n=k_{n1}\tau_1+\cdots+k_{np}\tau_p,[/imath] where [imath]\tau_i[/imath] is a distinct irrational number, [imath]k_{ij}\in\mathbb{R}[/imath], and [imath]p[/imath] is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with [imath]p[/imath] and [imath]m[/imath]. I feel this method can lead me to the answer, but I'm not sure where to go from here. EDIT[imath]^1[/imath]: I end up getting something like this, I believe, after substitution: [imath] A(k^{(1)}\tau_1+\cdots+k^{(p)}\tau_p)=b[/imath] Here, [imath]k^{(i)}[/imath] is the vector [imath]\begin{pmatrix} k_{1i} \\ \vdots \\ k_{ni} \end{pmatrix}.[/imath] EDIT[imath]^2[/imath]: I think there is no discrepancy with [imath]p[/imath] and [imath]m[/imath] because [imath]A\in M_{m\times n}(\mathbb{Q})[/imath], [imath]K\in M_{n\times p}(\mathbb{R})[/imath], and [imath]\tau\in M_{p\times 1}(\mathbb{T})[/imath], so [imath] (m\times n)\cdot (n\times p)\cdot (p\times 1) = m\times 1. [/imath]

24397

Convex and bounded function is constant Let f be a convex and bounded function, meaning there is a constant [imath]C[/imath], such that [imath]f(x) < C[/imath] for every [imath]x[/imath]. I need to prove that [imath]f[/imath] is a constant function. Thanks!

518091

Show bounded and convex function on [imath]\mathbb R[/imath] is constant How can we show that a bounded and convex function on [imath]\mathbb R[/imath] is constant? Derivatives are of no use since the function does not have to differentiable. I saw an answer here I think a while ago but did not understand it at all. Since derivatives are useless, we would have to use the definition and somehow show that the function lies between two values which are equal to each other. But I am unable to progress any further.

518952

Rolle's theorem for a proof Consider the continuous functions [imath]f : \mathbb R \to \mathbb R, f(x) = 1 + e^x \cos(x)[/imath] and [imath]g : \mathbb R \to \mathbb R, g(x) = 1 + e^x \sin(x)[/imath]. Using Rolle's Theorem, prove that between any two roots of [imath]f[/imath] there exists at least one root of [imath]g[/imath]. A root of [imath]f[/imath] is a point [imath]x[/imath] in the domain of [imath]f[/imath] such that [imath]f(x) = 0[/imath] So how would you go about answering this one? Can anyone provide a brief proof? I have no idea and want to know before I tute tomorrow

509801

Root of continuous function using Rolle's Theorem [imath]f(x)= 1 + e^x\cos(x)[/imath] [imath]g(x)= 1 + e^x\sin(x)[/imath] Show that there is at least one root of [imath]g[/imath] between two roots of [imath]f[/imath] How do I start to answer this question and do I need to use Intermediate Value Theorem?

519058

Every nontrivial subgroup [imath]H[/imath] of [imath]S_9[/imath] containing some odd permutation contains a transposition. This is a true or false question. Apparently, it is false, but I don't follow. Clearly, if it contains an odd permutation, and an even/odd permutation is defined by the number of transpositions it could be broken into, then every permutation is a set of transpositions. At least, that's what I had gotten out of it. Am I wrong? I'm very confused with the answer.

515323

Does every non-trivial subgroup of [imath]S_9[/imath] containing an odd permutation necessarily contain a transposition? Does every non-trivial subgroup of [imath]S_9[/imath] containing an odd permutation necessarily contain a transposition? Here [imath]S_9[/imath] denotes the group of all permutations (i.e. bijections with itself) of the set [imath]\{1,2, \ldots, 9 \}[/imath] under the binary operation of composition of functions. By a transposition is meant a permutation that leaves all elements except two fixed, mapping either of these two onto the other. A permutation can only be expressed either as a product of an even number of transpositions or as a product of an odd number of transpositions (but not both) and as such is defined to be even or odd, respectively. Reference: Fraleigh p. 95 Question 23d. in A First Course in Abstract Algebra

247169

What are [imath]2222^{5555}+5555^{2222} \pmod 7[/imath] and [imath]9^{2n+1}+8^{n+2} \pmod{73}[/imath]? Tell me hint for solve : 1) [imath] 2222^{5555}+5555^{2222} \equiv \mathord? \pmod 7[/imath] 2) [imath] 9^{2n+1}+8^{n+2} \equiv \mathord ?\pmod{73}[/imath] thank you.

736488

Trouble finding remainder for this problem expression? [imath]\left[\frac{2222^{5555}}{7} + \frac{5555^{2222}}{7}\right][/imath] Please guide me through steps. Thanx..

519276

Zariski tangent space and [imath]K[\epsilon]/(\epsilon^2)[/imath] I want to prove that the Zariski tangent space at [imath]x\in X[/imath] ([imath]X[/imath] is an affine scheme) is isomorphic to [imath]Hom_K(X,K[\epsilon]/(\epsilon^2))[/imath] (K is the residue field at [imath]x[/imath]). I want to say that [imath]Hom_K(\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2,K)\simeq Hom_K(\mathcal{O}_{X,x}/\mathfrak{m}_{X,x}^2,K[\epsilon]/(\epsilon^2))[/imath] so I need splitting [imath]\mathcal{O}_{X,x}[/imath] into [imath]\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\oplus K[/imath], and I build this splitting using the exact sequence [imath]\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\rightarrow \mathcal{O}_{X,x}/\mathfrak{m}_{X,x}^2\rightarrow K[/imath] so I need a splitting map [imath]\sigma:K\rightarrow \mathcal{O}_{X,x}/\mathfrak{m}_{X,x}^2[/imath]. How can I do to find [imath]\sigma[/imath]?

339165

Zariski Tangent Space and [imath]k[\varepsilon]/\varepsilon^2[/imath] Let [imath]X[/imath] be a scheme over a field [imath]k[/imath], and let [imath]x\in X[/imath] be a rational point, that is, we have [imath]k(x):=\mathcal{O}_x/\mathfrak{m}_x\cong k[/imath]. Let [imath]\alpha:\mathfrak{m}_x/\mathfrak{m}_x^2\rightarrow k[/imath] be an element of [imath]T_x=(\mathfrak{m}_x/\mathfrak{m}_x^2)^*[/imath], the Zariski tangent space to [imath]X[/imath] at [imath]x[/imath]. Using [imath]x[/imath] and [imath]\alpha[/imath], I'd like to define a [imath]k[/imath]-morphism of schemes Spec [imath]k[\varepsilon]/\varepsilon^2\rightarrow X[/imath]. In the course of the construction, I need to define a local map of local rings [imath]\mathcal{O}_x\rightarrow k[\varepsilon]/\varepsilon^2[/imath]. Using [imath]\alpha[/imath], what is a natural way to define such a map? It might be helpful to note that I have an injection [imath]k\hookrightarrow\mathcal{O}_x[/imath] arising from the [imath]k[/imath]-scheme structure on [imath]X[/imath]. I think this leads to a decomposition [imath]\mathcal{O}_x/\mathfrak{m}_x^2=k\oplus(\mathfrak{m}_x/\mathfrak{m}_x^2)[/imath] as [imath]k[/imath] vector spaces, from which the desired map would follow, but I can't seem to explicitly see what this decomposition is inside of [imath]\mathcal{O}_x/\mathfrak{m}_x^2[/imath]. This is part of exercise II.2.8 in Hartshorne.

519755

Prove connectivity of graph with vertices of degree [imath]\geq \lfloor \frac n2 \rfloor[/imath] Claim: A graph with vertices of degree at least [imath]\lfloor \frac n2 \rfloor[/imath] where [imath]n = [/imath] number of vertices and [imath]n \geq 3[/imath] is connected. I tried to prove this by contradiction, but I didn't know what to make of the [imath]\lfloor \frac n2 \rfloor[/imath] part.

329087

Proving a simple graph is a connected graph Does any proof exist that a simple graph with [imath]n[/imath] vertices such that the least vertex degree is [imath]\geq \frac{n-1}{2}[/imath] is a connected graph? (i.e. does such a proof have a name?)

520295

Why isn't the graph of [imath]y=|x|[/imath] a smooth manifold? Consider the graph of [imath]y=|x|[/imath] from [imath]-1<x<1[/imath]. Equip it with a single chart, the projection onto the [imath]x[/imath]-axis. Is it now a smooth manifold? It seems like it shouldn't be smooth, but perhaps with only one chart, any manifold is smooth (because it trivially satisfies the compatibility condition)? What chart could we put on there that would not be smoothly compatible with this first one?

45673

Is [imath]M=\{(x,|x|): x \in (-1, 1)\}[/imath] not a differentiable manifold? Let [imath]M=\{(x,|x|): x \in (-1, 1)\}[/imath]. Then there is an atlas with only one coordinate chart [imath](M, (x, |x|) \mapsto x)[/imath] for [imath]M[/imath]. We don't need any coordinate transformation maps to worry about differentiablity. So I thought [imath]M[/imath] is a differentiable manifold. However my teacher says it is not. He says the sharp corner at [imath]x = 0[/imath] is a problem. I can't understand why it is a problem.

61714

Intuition for Smooth Manifolds Consider the graphs of the functions [imath]f_1(x) = |x|[/imath], and [imath]f_2(x) = x[/imath] under the subspace topology of [imath]\mathbb{R}^2[/imath]. Both of these graphs are smooth manifolds, just pick coordinate charts to be [imath](x, f_i(x)) \leftrightarrow x[/imath]. Moreover, they are diffeomorphic via the map [imath](x, f_1(x)) \rightarrow (x, f_2(x))[/imath]. This seems to clash with my intuition. For example, the graph of [imath]f_1[/imath] has a corner, so it "shouldn't" be smooth, much less diffeomorphic to [imath]f_2[/imath], which is just a straight line. Can someone explain what's going on here? In light of these examples, how should I visualize smooth manifolds and diffeomorphisms?

820880

Graph of a continuous function is a smooth manifold? Let [imath]f:(a,b)\to \mathbb{R}[/imath] be a continuous function and define [imath]\Gamma(f) = \{(x,f(x)):x\in (a,b)\}[/imath]. The two maps [imath]\Psi: \Gamma(f)\to (a,b)[/imath] given by [imath](x,f(x))\mapsto x[/imath] and [imath]\Phi: (a,b)\to \Gamma(f)[/imath] given by [imath]x\mapsto (x,f(x))[/imath] are both continuous and inverse to each other. Therefore [imath]\Gamma(f)[/imath] is homeomorphic to [imath](a,b)[/imath], and [imath]\{(\Gamma(f),\Psi)\}[/imath] is an atlas consisting of a single chart. This means [imath]\Gamma(f)[/imath] is a smooth manifold. This confuses me greatly. Somehow I feel [imath]\Gamma(f)[/imath] should be a smooth manifold only when [imath]f[/imath] is smooth, not merely continuous. What is wrong here?

500132

Find the number of distinct real roots of [imath](x-a)^3+(x-b)^3+(x-c)^3=0[/imath] Problem:Find the number of distinct real roots of [imath](x-a)^3+(x-b)^3+(x-c)^3=0[/imath] where [imath]a,b,c[/imath] are distinct real numbers Solution:[imath](x-a)^3+(x-b)^3+(x-c)^3=0[/imath] [imath]3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0[/imath] By Descartes rule of sign,number of positive real roots [imath]=3[/imath] But are they distinct [imath]?[/imath] Answer :- number of distinct real roots [imath] =1[/imath]

444828

How many real roots does [imath](x-a)^3+(x-b)^3+(x-c)^3[/imath] have? Let [imath]a,b,c[/imath] be distinct real numbers. What is the number of distinct real roots of the equation [imath](x-a)^3+(x-b)^3+(x-c)^3=0[/imath]? [imath]1[/imath], [imath]2[/imath], [imath]3[/imath], Depends on the value of [imath]a,b,c[/imath]. How can I solve this?

520342

Verify the identiy, (cos(x+h) - sin x)/h = cos x * ((cos h - 1)/h)- sin x * (sin h /h) Verify the identity: [imath]\frac{(\cos(x+h) - \cos x)}{h} = \cos x \left(\frac{\cos h - 1}{h}\right)- \sin x \left(\frac{\sin h }{h}\right)[/imath] =(Cosxcosh - sin x sin h -cos x)/h. I can't think of where to go from here. Thanks

402487

Intuition of Addition Formula for Sine and Cosine The proof of two angles for sine function is derived using [imath]\sin(A+B)=\sin A\cos B+\sin B\cos A[/imath] and [imath]\cos(A+B)=\cos A\cos B-\sin A\sin B[/imath] for cosine function. I know how to derive both of the proofs using acute angles which can be seen here http://en.wikibooks.org/wiki/Trigonometry/Addition_Formula_for_Cosines but pretty sure those who have taken trig know what I'm talking about. So I know how to derive and prove both of the two-angle functions using the acute angles, but what I am completely confused about is where those triangles came from. So for proving the two-angle cosine function, we look at two acute angles, [imath]A[/imath] and [imath]B[/imath], where [imath]A+B<90[/imath] and keep on expanding. So my question is, where did those two triangles come from and what is the intuition behind having two acute triangles on top of each other?

520709

Question about a nested radical identity of Ramanujan involving [imath]\sqrt 5[/imath] Ramanujan stated this radical in his lost notebook: [imath]\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}[/imath] Where the pattern of the signs is [imath]++-+++-++++-+++++-\cdots[/imath] I do not have access to his lost notebook. I wonder how we can prove this. For periodic sign changes this kind of question is easy but for nonperiodic Im clueless.

515253

How to prove [imath]\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}[/imath] Ramanujan stated this radical in his lost notebook: [imath]\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}[/imath] I don't have any idea on how to prove this. Any help appreciated. Thanks.

521088

When a sequential space has unique sequential limits? A topological space is called a sequential space if a set [imath]A \subset X[/imath] is closed if and only if together with any sequence it contains all its limits. so, according to definition, is the following correct? Sequential space has unique sequential limit iff each countably compact subset is closed.( and hence sequential).

467577

‎sequential ‎space A‎‎ ‎sequential ‎space ‎has ‎unique ‎sequential ‎limits ‎iff ‎each ‎countably ‎compact ‎subset ‎is ‎closed. ‎Proof: ‎If‎ ‎[imath] \{ x_n \} [/imath] ‎is a‎ ‎sequence ‎converging ‎to ‎two ‎distinc ‎‎[imath]‎x‎[/imath] ‎and ‎‎[imath]‎y‎[/imath]‎, then ‎[imath] ‎\{ x‎ ‎\}‎ ‎\cup ‎\{‎‎ ‎x_n :‎ n‎ ‎\in‎ ‎\omega ‎\} ‎‎[/imath] ‎is a‎ ‎non ‎compact ‎set. conversely: ‎if ‎‎[imath]‎X‎[/imath] ‎is ‎sequential ‎and ‎has ‎unique ‎sequential ‎limits, ‎‎[imath]‎X[/imath] ‎is ‎‎[imath]‎T_1‎[/imath] .‎ ‎Let‎ [imath] K‎ ‎‎\subseteq ‎X [/imath]‎ ‎be ‎countably ‎compact.If ‎‎[imath] ‎\{ ‎x_n ‎\} ‎‎[/imath] ‎is a‎ ‎sequence ‎in ‎‎[imath]‎K‎[/imath] ‎and‎ [imath] ‎\{‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎\longrightarrow ‎x‎_{0}‎[/imath] ‎then‎ ‎[imath] ‎\{ ‎x‎_{0} ‎\}‎ ‎\cup ‎\{‎‎ ‎x_n :‎ n‎ ‎\in‎ ‎\omega ‎\} ‎‎[/imath] ‎is ‎sequentially ‎closed ‎and ‎hence ‎closed. ‎Thus ‎‎[imath]‎x_{0}‎[/imath] ‎is ‎the ‎only ‎possible ‎accumulation ‎poin ‎of‎ [imath] ‎\{ ‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎[/imath]‎. if [imath] ‎\{ ‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎[/imath] ‎is ‎infinite‎, ‎[imath]‎x_0 ‎\in ‎K‎[/imath]‎. ‎If ‎not ‎for ‎some ‎‎[imath]‎n‎[/imath]‎, [imath] ‎x‎_{0} =‎ ‎x‎_{n}‎‎ ‎\in K ‎[/imath].Hence ‎‎[imath]‎K‎[/imath] ‎is ‎closed.‎ (1) Why" ‎if ‎‎[imath]‎X‎[/imath] ‎is ‎sequential ‎and ‎has ‎unique ‎sequential ‎limits, ‎‎[imath]‎X[/imath] ‎is ‎‎[imath]‎T_1‎[/imath]? (2) Why "if [imath] ‎\{ ‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎[/imath] ‎is ‎infinite‎"?

73060

Why are there no non-trivial regular maps [imath]\mathbb{P}^n \to \mathbb{P}^m[/imath] when [imath]n > m[/imath]? Question. Let [imath]k[/imath] be an algebraically closed field, an let [imath]\mathbb{P}^n[/imath] be projective [imath]n[/imath]-space over [imath]k[/imath]. Why is it true that every regular map [imath]\mathbb{P}^n \to \mathbb{P}^m[/imath] is constant, when [imath]n > m[/imath]? I can't see any obvious obstructions: there are certainly homomorphisms of function fields (giving rise to the dominant rational maps), and we're not demanding the map be injective or anything. While it is clear that [imath](F_0 : \cdots : F_m)[/imath] cannot define a regular map on its own unless [imath]F_0, \ldots, F_m[/imath] are all constants, I don't see why it should be impossible to extend [imath](F_0 : \cdots : F_m)[/imath] by choosing some other [imath](G_0 : \cdots G_m)[/imath] which agrees with [imath](F_0 : \cdots : F_m)[/imath] on the intersection of their domains. Is there something conceptual I'm missing?

1932969

Morphisms from [imath]\mathbb{P}^2[/imath] to [imath]\mathbb{P}^1[/imath] Let [imath]K[/imath] be algebraically closed field. Can we show there are no non-constant morphisms from [imath]\mathbb{P}^2 \to \mathbb{P}^1[/imath]? What is the most elementary way to show this (assuming its true)?

521095

Existence of an isomorphism [imath]\mathbb{P}^n\times\mathbb{P}^m \rightarrow \mathbb{P}^{n+m}[/imath] There exist an isomorphism of varieties? [imath]\mathbb{P}^n\times\mathbb{P}^m \rightarrow \mathbb{P}^{n+m}[/imath] I am considering [imath]\mathbb{P}^n\times\mathbb{P}^m[/imath] as the product in the category of varieties.

508588

Birational isomorphism [imath]\mathbb{P}^n\times \mathbb{P}^m\to \mathbb{P}^{n+m}[/imath] One can show that [imath]\mathbb{P}^n\times \mathbb{P}^m[/imath] is birational to [imath]\mathbb{P}^{n+m}[/imath] by making note of the Zariski topologies and the canonical isomorphism between affine spaces [imath]\mathbb{A}^n\times \mathbb{A}^m\to \mathbb{A}^{n+m}[/imath]. Question: Does it follow that there is isomorphism [imath]\mathbb{P}^n\times \mathbb{P}^m\to \mathbb{P}^{n+m}[/imath]? If so, what is an example of such an isomorphism? Unfortunately I am having a hard time coming up with one, and haven't seen too many examples thus far.

521447

[imath]L_2[/imath] is of first category in [imath]L_1[/imath] (Rudin Excercise 2.4b) We mean here [imath]L_2[/imath], and [imath]L_1[/imath] the usual Lebesgue spaces on the unit-interval. It is excercise 2.4 from Rudin. There's several ways to show that [imath]L_2[/imath] is nowhere dense in [imath]L_1[/imath]. But in (b) they ask to show that [imath]\Lambda_n(f)=\int fg_n \to 0 [/imath] where [imath]g_n = n[/imath] on [imath][0,n^{-3}][/imath] and 0 otherwise, holds for [imath]L_2[/imath] but not for all [imath]L_1[/imath]. Apparantly this implies that [imath]L_2[/imath] is of the first Category, but I dont know how. Second, I can show this holds for [imath]L_2[/imath] but I cant find a counterexample in [imath]L_1[/imath]. Theorem 2.7 in Rudin says: Let [imath]\Lambda_n:X\to Y[/imath] a sequence of continuous linear mappings ([imath]X,Y[/imath] topological vector spaces) If [imath]C[/imath] is the set of all [imath]x\in X[/imath] for which [imath]\{\Lambda_n x\}[/imath] is Cauchy in [imath]Y[/imath], and if [imath]C[/imath] is of the second Category, then [imath]C=X[/imath]. So if we find a [imath]f\in L_1[/imath] such that [imath]\Lambda_n(f)[/imath] is not Cauchy, then we proved that [imath]L_2\subset C \subset L_1[/imath] is of the first category. However I dont see why showing that [imath]\Lambda_n(f)[/imath] does not converge to 0 for some [imath]f\in L_1[/imath] is enough here. Am I missing something?

512225

Measure spaces are proper subsets I want to prove that [imath]L^2[/imath] is of the first category in [imath]L^1[/imath], thus I have to prove that [imath]L^2[/imath] is the countable union of nowhere dense subsets. The hint I get is: Take [imath]g_n(x)=n[/imath] for [imath]0\leq x\leq\frac{1}{n^3}[/imath] and show [imath]\int{fg_n}\rightarrow 0[/imath] for all [imath]f\in L^2[/imath] but not for every [imath]f\in L^1[/imath]. My question: What does this hint help for proving that [imath]L^2[/imath] is of the first category? I don't know how to use this hint. Can someone help me? :) Thanks

522020

A discontinuous function such that [imath]f(x + y) = f(x) + f(y)[/imath] Is it possible to construct a function [imath]f \colon \mathbb{R} \to \mathbb{R}[/imath] such that [imath]f(x + y) = f(x) + f(y)[/imath] and [imath]f[/imath] is not continuous?

423492

Overview of basic facts about Cauchy functional equation The Cauchy functional equation asks about functions [imath]f \colon \mathbb R \to \mathbb R[/imath] such that [imath]f(x+y)=f(x)+f(y).[/imath] It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called additive functions. Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.) Is there some overview of basic facts about Cauchy equation and related functional equations - preferably available online?

521809

Is this implication true? Suppose that a real sequence [imath]u_n[/imath] is such that [imath]u_{n+1}-u_n \rightarrow0[/imath] That is not enough to prove that [imath]u_n[/imath] is convergent (take [imath]u_n=ln(n)[/imath]) Now what if [imath]u_n[/imath] is bounded ? I guess it does converge, but how to prove this ? I tried to show that it had only one accumulation point...

359344

Give an example of a bounded, non-convergent real sequence [imath](a_n)[/imath] s.t. [imath]a_n-a_{n-1}\rightarrow 0[/imath] This was the last part of an exam question. I have spent a long time looking for such an example but have failed (sequences when one of the three conditions is dropped are easily found but I couldn't find one when all three are required). Could you also give the way you thought of the example in your answer please? Thanks!

522385

Determinant of a block matrix How prove this equality for a block matrix: [imath]\det\left[\begin{array}[cc]\\A&C\\ 0&B\end{array}\right]=\det(A)\det(B)[/imath] I tried to use a proof by induction but I'm stuck on it. Is there a simpler method? Thanks for help.

75293

Determinant of a block lower triangular matrix I'm trying to prove the following: Let [imath]A[/imath] be a [imath]k\times k[/imath] matrix, let [imath]D[/imath] have size [imath]n\times n[/imath], and [imath]C[/imath] have size [imath]n\times k[/imath]. Then, [imath]\det\left(\begin{array}{cc} A&0\\ C&D \end{array}\right) = \det(A)\det(D).[/imath] Can I just say that [imath]AD - 0C = AD[/imath], and I'm done?

522627

Show that [imath](a+b+c)^\alpha[/imath] For any positive real [imath] a, b, c[/imath], show that [imath](a+b+c)^\alpha<a^\alpha+b^\alpha+c^\alpha, 0<\alpha<1[/imath]. I can show that it works for special cases like [imath]\alpha=1/p, p\in\mathbb{N}[/imath]... but I don't know how to generalize further...

299120

Show that [imath]a^x+b^x+c^x>(a+b+c)^x[/imath] for all [imath]a,b,c>0[/imath] and [imath]0[/imath] Hy all! I'm having trouble finding a proof for the following problem: Show that [imath]a^x+b^x+c^x>(a+b+c)^x[/imath], if [imath]a,b,c>0[/imath] and [imath]0<x<1[/imath] (over the real numbers). This inequality has been torturing me for long hours the least. I couldn't find anything on Google. This kind of equation isn't made for search engines tbh. Anyway, I really thinked about it a lot but didn't make any significant progress. It would be great to see a proof without the really high-end Mathematics. Any ideas to start with?

522887

Show that if [imath]\int_0^1 f(x) v(x) dx = 0[/imath] for every function v for which [imath]\int_0^1 v(x) dx = 0[/imath], then f is constant. Show that if [imath]\int_0^1 f(x) v(x) dx = 0[/imath] for every function v for which [imath]\int_0^1 v(x) dx = 0[/imath], then f is constant. I do not know how to do it.

221551

How to prove the function is a constant Let [imath]f\in C[0,1][/imath] be such that [imath]\forall \phi\in C[0,1][/imath] with [imath]\int_0^1\phi \, dx=0[/imath],we have [imath]\int_0^1f\phi \, dx=0[/imath]. Then [imath]f[/imath] is constant. May be someone can help me proof this problem. Thanks!

522671

Open maping theorem. Completeness assumption are important The open maping theorem between banach spaces says. Let [imath]T:X\to Y[/imath] be a linear,continuous and surjective map between the banach spaces [imath]X,Y[/imath] then [imath]T[/imath] is an open map. I need examples to show that the completeness assumption are important even in the presence of completeness of the other space. Thanks!

291671

Open Mapping Theorem: counterexample The Open Mapping Theorem says that a linear continuous surjection between Banach spaces is an open mapping. I am writing some lecture notes on the Open Mapping Theorem. I guess it would be nice to have some counterexamples. After all, how can you appreciate it's meaning without a nice counterexample showing how the conclusion could fail and why the conclusion is not obvious at all. Let [imath]\ell^1 \subset \mathbb{R}^\infty[/imath] be the set of sequences [imath](a_1, a_2, \dotsc)[/imath], such that [imath]\sum |a_j| < \infty[/imath]. If we consider the [imath]\ell^1[/imath] norm [imath]\|\cdot\|_1[/imath] and the supremum norm [imath]\|\cdot\|_s[/imath], then, [imath](\ell^1, \|\cdot\|_1)[/imath] is complete, while [imath](\ell^1, \|\cdot\|_s)[/imath] is not complete. In this case, the identity [imath] \begin{array}{rrcl} \mathrm{id}:& (\ell^1, \|\cdot\|_1)& \to &(\ell^1, \|\cdot\|_s) \\ & x & \mapsto & x \end{array} [/imath] is a continuous bijection but it is not open. I want a counterexample in the opposite direction. That is, I want a linear continuous bijection [imath]T: E \to F[/imath] between normed spaces [imath]E[/imath] and [imath]F[/imath] such that [imath]F[/imath] is Banach but [imath]T[/imath] is not open. This is equivalent to finding a vector space [imath]E[/imath] with non-equivalent norms [imath]\|\cdot\|_c[/imath] and [imath]\|\cdot\|_n[/imath], such that [imath]E[/imath] is complete when considered the norm [imath]\|\cdot\|_c[/imath], and such that [imath] \|\cdot\|_c \leq \|\cdot\|_n. [/imath] The Open Mapping Theorem implies that [imath]\|\cdot\|_n[/imath] is not complete. So, is anyone aware of such a counterexample?

522992

Prove that [imath]\tau(n) \leq 2\sqrt{n}[/imath] I'm looking at the following problem: Prove that for a positive integer [imath]n[/imath], [imath]\tau(n) \leq 2 \sqrt{n}[/imath] where [imath]\tau(n)[/imath] is the number of divisors of [imath]n[/imath]. So my idea was to split the set of the number of divisors of [imath]n[/imath] into two subsets: One subset containing the divisors those less than [imath]\sqrt{n}[/imath] and the other subset containing the divisors larger than [imath]\sqrt{n}[/imath] and argue that there must be at most [imath]\sqrt{n}[/imath] in each subset. But I simply can't seem to find that argument! Can you help?

29544

Prove that [imath]d(n)\leq 2\sqrt{n}[/imath] In Waclaw Sierpinski's book Elementary Theory of Numbers on page 168 there is the following exercise: "Exercises. 1. Prove that for natural numbers [imath]n[/imath] we have [imath]d(n) \leq 2\sqrt{n}[/imath]," where [imath]d(n)[/imath] is the number of divisors of n. As a hint right below is given: "The proof follows from the fact that of two complementary divisors of a natural number [imath]n[/imath] one is always not greater than [imath]\sqrt{n}[/imath]. I understand the hint but I don't know how it can be used to prove [imath]d(n)\leq 2\sqrt{n}[/imath]. Complementary divisors are pairs of divisors that when multiplied gives the number that is to be divided. For example the number [imath]120[/imath] has the complementary divisors: \begin{align*} & 1, 120 \\ & 2, 60 \\ & 3, 40 \\ & 4, 30 \\ & 5, 24 \\ & 6, 20 \\ & 8, 15 \\ & 10, 12 \\ \end{align*} How do you prove that [imath]d(n) \leq 2\sqrt{n}[/imath]?

523497

No. of positive eigen values of [imath]3\times 3[/imath] real matrix Suppose [imath]A[/imath] is a [imath]3\times 3[/imath] symmetric matrix such that : [imath]\begin{pmatrix} a & b & 1 \end{pmatrix} A \begin{pmatrix} a \\ b\\ 1\end{pmatrix} = ab -1[/imath] for all [imath]a,b\in \mathbb{R}[/imath] Question is to find no. of positive eigen values of [imath]A[/imath]. and rank of [imath]A[/imath]. What i have observed so far is : As [imath]A[/imath] is symmetric matrix, it should have real eigen values. item As [imath]x^TAx[/imath] is not positive, for [imath]x= \begin{pmatrix} a \\ b\\ 1\end{pmatrix}[/imath], ([imath]a=b=\frac{1}{2}[/imath]) we see that not all eigen values of [imath]A[/imath] are positive As [imath]x^TAx[/imath] is not negative, for [imath]x= \begin{pmatrix} a \\ b\\ 1\end{pmatrix}[/imath], ([imath]a=2,b=1[/imath]) we see that not all eigen values of [imath]A[/imath] are Negative So, [imath]A[/imath] does have positive eigen values and also negative eigen values. Now, I have little less clarity particularly here : If i take [imath]a=b=1[/imath] the, I have [imath]\begin{pmatrix} 1 & 1 & 1 \end{pmatrix} A \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix} = 1 -1=0[/imath] I want to conlcude from this that [imath] A \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix} =0\begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}[/imath] somehow converting row marrix in one side to column matrix in other side. If this is true, then i would have [imath]0[/imath] as one eigen value. So, I have one positive, one negative, one [imath]0[/imath] eigen value. So, Rank of [imath]A[/imath] should be [imath]2[/imath] and no. of positive eigen values is [imath]1[/imath]. Please help me to sort out some mistakes (if there are any) and help me to make this answer a little messier than what i have written. This problem is already asked some time ago but, the answer for that was "to try with a brute force" which i thought is not the only way to go. I had some start up but that is not entire answer, so i can not post that as an answer there.. So, I thought i should ask here at the risk of getting negative votes. P.S : Here is the link for a copy of this what is no. of positive eigen value of symmetric matrix A with some given relationship

426719

what is no. of positive eigen value of symmetric matrix A with some given relationship Suppose A is a 3*3 symmetric matrix s.t. [imath]\begin{pmatrix} x & y & 1 \\ \end{pmatrix} A \begin{pmatrix} x \\ y\\ 1\end{pmatrix} = xy -1 [/imath] let p be the no. of positive eigen value of A and q = rank(A)-p, then p=? q=? what matrix A should be I can't find??

523780

Prove by induction that [imath]n < 2 ^n [/imath] where [imath]n \in \mathbb{N}[/imath] Example question in a textbook that I don't understand. Proof works for n = 1 Setting for k makes [imath]k < 2^k [/imath] Setting for k + 1 makes [imath]k+1 < 2^{k+1} [/imath]. Here, I would be stuck, the book takes the equation to: [imath]k+1<2^k +1\leq 2^k+2^k = 2 \cdot 2^k=2^{k+1}[/imath]. NB: [imath]<2^k +1[/imath] is not a typo. There doesn't seem to be a good explanation for this in the book (although it does mention adding 1to both sides of the equation), could I have some advice on how the method used works?

449672

Prove that [imath] n < 2^{n}[/imath] for all natural numbers [imath]n[/imath]. Prove that [imath] n < 2^{n} [/imath] for all natural numbers [imath]n[/imath]. I tried this with induction: Inequality clearly holds when [imath]n=1[/imath]. Supposing that when [imath]n=k[/imath], [imath]k<2^{k}[/imath]. Considering [imath]k+1 <2^{k}+1[/imath], but where do I go from here? Any other methods maybe?

523965

Prove Divisibility test for 11 Prove Divisibility test for 11 "If you repeatedly subtract the ones digit and get 0, the number is divisible by 11" Example: 11825 -> 1182 - 5 = 1177 1177 -> 117 - 7 = 110 110 -> 11 - 0 = 11 11 -> 1-1 = 0 Therefore 11825 is divisible by 11. Note 11825 = 1075*11 I was thinking that we let x = [imath]a_ka_{k-1}.....a_1a_0[/imath] where the following [imath]a_i[/imath]'s are the digits.

71638

[imath]11[/imath] divisibility We know that [imath]1331[/imath] is divisible by [imath]11[/imath]. As per the [imath]11[/imath] divisibility test, we can say [imath]1331[/imath] is divisible by [imath]11[/imath]. However we cannot get any quotient. If we subtract each unit digit in the following way, we can see the quotient when [imath]1331[/imath] is divided by [imath]11[/imath]. [imath]1331 \implies 133 -1= 132[/imath] [imath]132 \implies 13 - 2= 11[/imath] [imath]11 \implies 1-1 = 0[/imath], which is divisible by [imath]11[/imath]. Also the quotient is, arrange, all the subtracted unit digits (in bold and italic) from bottom to top, we get [imath]121[/imath]. Which is quotient. I want to know how this method is working? Please write a proof.

524205

Evaluate the integral [imath]\int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{(1-e^{ax})(1-e^{bx})}dx[/imath] Evaluate the integral [imath]\int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{(1-e^{ax})(1-e^{bx})}dx[/imath] for constants [imath]a,b[/imath]. I tried separation by partial fractions, but couldn't do it.

522157

Evaluate [imath]\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx[/imath] Given the integral [imath]\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx[/imath] How can I evaluate it? Thanks!

524559

Prove [imath]\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3[/imath] Prove [imath]\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3[/imath] I tried to express it as a recursive sequence, but I failed.

165671

Convergence of the sequence [imath]\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots[/imath] I recently came across this problem Q1 Show that [imath]\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }} = 3[/imath] After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists. So my question is Q2 Prove that the sequence[imath]\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}[/imath] converges. Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help . If the solution from that book is required I can post it but it is not complete as I mentioned. Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.

524638

Prove that [imath]\sin(\sin...(\sin(x))..)[/imath] converges asymptotically to zero I'm not able to mathematically prove that the equation [imath]x(k+1)=\sin(x(k))[/imath] converges asymptotically to zero. By a simple thought it can be concluded that for any [imath]x(0) \in \mathbb{R}[/imath] it applies [imath]x(1)=\sin(x(0)) \qquad \qquad \quad x(1) \in [-1,1][/imath] [imath]x(2)=\sin(\sin(x(0))) \qquad \, \, \, \, \, x(2) \in [-0.84,0.84][/imath] [imath]x(3)=\sin(\sin(\sin(x(0)))) \, \, \, \, x(3) \in [-0.74,0.74][/imath] ... Which means that with every step the solution is closer and closer to zero, however I'm not able to prove it. Does anyone know how to solve this problem please?

45283

Compute [imath] \lim\limits_{n \to \infty }\sin \sin \dots\sin n[/imath] I need your help with evaluating this limit: [imath] \lim_{n \to \infty }\underbrace{\sin \sin \dots\sin}_{\text{$n$ compositions}}\,n,[/imath] i.e. we apply the [imath]\sin[/imath] function [imath]n[/imath] times. Thank you.

522365

Idempotents in a ring of fractions of the tensor product of Gaussian integers Let [imath]S=\{x^0,x^1,x^2,...\}\subset \mathbb{Z}[/imath] be the multiplicatively closed subset generated by [imath]x[/imath]. What are the nontrivial idempotents in the total quotient ring [imath]S^{-1}(\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{Z}[i])[/imath]? I came across this question reading about the Chinese remainder theorem, and would like to see concrete examples of how to find the idempotents. Presumably one would find them using the isomorphism given by CRT. Could you show me the cases for [imath]x=2[/imath] and [imath]x=3[/imath]?

517210

Tensor Product of the fraction field of the Gaussian Integers Let [imath]\mathbb{Z}[i]=\left\{a+bi:a,b \in \mathbb{Z}\right\}[/imath] be the ring of Gaussian integers [imath](i^{2}=-1)[/imath] and let [imath]\mathbb{Q}[i]=\left\{a+bi:a,b \in \mathbb{Q}\right\}[/imath] be its fraction field. 1) Show that the tensor product [imath]\mathbb{Q}[i] \otimes_{\mathbb{Q}} \mathbb{Q}[i][/imath] is isomorphic to the product [imath]\mathbb{Q}[i] \times \mathbb{Q}[i][/imath] of two copies of the field. 2) Let [imath]S=\left\{1,2,2^2,2^3,...\right\} \subseteq \mathbb{Z}[/imath] be the multiplicatively closed subset generated by 2. Find two distinct idempotents in the algebra [imath]S^{-1}(\mathbb{Z}[i] \otimes_{\mathbb{Z}} \mathbb{Z}[i])[/imath] other than [imath]0[/imath] and [imath]1[/imath]. For the first part, I thought I would try to use the presentation [imath]\mathbb{Q}[i] \simeq \mathbb{Q}[T]/(T^2+1)[/imath] and see what comes from that. Any help would be much appreciated!

525039

How to find $\ \lim_{n \to \infty} \frac{\log n}{\sqrt n}$? Find the limit of the following expression: [imath]\lim_{n \to \infty} \frac{\log n}{\sqrt n}[/imath] As I plot this function, it seems that this goes to [imath]1[/imath]. Is it correct?

126099

How do I take the limit as [imath]n[/imath] goes to [imath]\infty[/imath] of $\frac{\sqrt{n}}{\log(n)}$? How do take this limit: [imath] \lim_{n\to\infty} \frac{\sqrt{n}}{\log(n)}[/imath] I have a feeling that it is infinity, but I'm not sure how to prove it. Should I use L'Hopitals Rule?

525794

Is [imath]||f||_p[/imath] continuous in [imath]p[/imath] I just started learning about [imath]L^p[/imath] spaces today and I have this question: Let [imath](X,\scr{M},\mu)[/imath] be a measure space. Let [imath]f:X\rightarrow \mathbb{C}[/imath] be measurable. Consider [imath]\eta:]0,\infty[\rightarrow[0,\infty][/imath] that sends [imath]p[/imath] to [imath]||f||_p[/imath]. Is [imath]\eta[/imath] continuous ? My approach was to show that the function [imath]p\rightarrow \int_X |f|^P d\mu[/imath] is continuous by showing that for every sequence [imath]\{p_n\}[/imath] that converges to [imath]p[/imath] we have [imath]\int_X |f|^{p_n}d\mu[/imath] converges to [imath]\int_X |f|^p d\mu[/imath] using Lebesgue's dominated convergence theorem, but I couldn't as I wasn't able to find a dominating function. I also think that the claim of my approach is stronger than the claim "[imath]\eta[/imath] is continuous". Thank you

133773

Is [imath]p\mapsto \|f\|_p[/imath] continuous? Suppose that [imath]\|f\|_p < \infty[/imath] for all [imath]1\leq p < p'[/imath], I want to know if the the following is true and in that case how to show it [imath]p \mapsto \|f\|_p[/imath] is continuous on [imath][1,p')[/imath] Or maybe we need to impose some more constraints such as finite measure space. In case of finite measure space, I tried to use Egoroff's theorem.

525945

Showing that complement of [imath]\mathbb{Q}^2[/imath] in [imath]\mathbb{R}^2[/imath] is connected The complement of [imath]\mathbb{Q}^2[/imath] in [imath]\mathbb{R}^2[/imath] is connected. I have to solve this problem. At first, I tried to solve as following: For any 2 disjoint neighborhoods of [imath]\mathbb{R}^2[/imath] \ [imath]\mathbb{Q}^2[/imath], whose union is [imath]\mathbb{R}^2[/imath] \ [imath]\mathbb{Q}^2[/imath], one of them must be empty. But I couldn't deduce the last goal. How can this problem be solved?

356444

Topological proof to show a complement set of a set is polygonally connected I started learning topology recently, so I'm very new to this. The question is as follows: Given the set [imath]A = \{ (x,y)\;;\;x, y\in\mathbb{Q} \}=\mathbb{Q}^2[/imath], show that the complement set [imath]A^c=\mathbb{R}^2\setminus A[/imath] is polygonally connected. As an attempt on this question, I plan to show that the set [imath]A^c[/imath] is connected, thus using a theorem, I can say [imath]A^c[/imath] is polygonally connected. But here is the part that I got confused: By definition, the set [imath]A^c[/imath] is connected iff it cannot be written as the union of [imath]2[/imath] nonempty separated sets. In this case, I believe [imath]A^c[/imath] is the set of all [imath](x,y)[/imath] where [imath]x,y[/imath] belong to [imath]\mathbb R- \mathbb Q[/imath]. But how can I express the fact that [imath]A^c[/imath] is not the union of [imath]2[/imath] nonempty separated sets ? Based on my knowledge, I know both sets [imath]\mathbb Q[/imath] and [imath]\mathbb R-\mathbb Q[/imath] are dense, but how is this helpful to solving this problem? I think a reason is because for any collection of [imath](x,y)[/imath] where [imath]x,y [/imath]are in [imath]\mathbb Q[/imath], I can always find a point in [imath]\mathbb R- \mathbb Q[/imath] which is a boundary point of that certain collection. But is my thought ok ? Would someone please help me on this? I'm kinda lost >_< Thanks ^^

526431

Graph divisibility problem For [imath]p[/imath] and [imath]q[/imath] distinct primes. Two conditions are given: vertices of graph [imath]G[/imath] are integers in the set [imath]\{0, 1, 2, \ldots, pq-2, pq-1\}[/imath]; there is an edge of graph [imath]G[/imath] between [imath]a[/imath] and [imath]b[/imath] if [imath]ab[/imath] either [imath]p∣a−b[/imath] or [imath]q∣a−b[/imath]. In this case, how many edges of [imath]G[/imath] are there?

520954

Counting edges in a specially defined graph [imath]G[/imath] has vertices [imath]\{0, \ldots, pq - 1\}[/imath], where [imath]p, q[/imath] are different primes. There is an edge between [imath]x[/imath] and [imath]y[/imath] if [imath]p \mid x - y[/imath] or [imath]q \mid x - y[/imath]. How many edges does [imath]G[/imath] have? I looked at the cases where [imath]p=2[/imath] and [imath]q=3, 5, 7, 11, 13, 17, 23[/imath] and the formula I got was [imath]|E|=q^2[/imath], but it doesn't apply to [imath]p=3[/imath]. For [imath]p = 3, q = 5[/imath], there are [imath]45[/imath] edges. How do I derive a closed form formula for this?

525947

Suppose that [imath]5\leq q\leq p[/imath] are both prime. Prove that [imath]24|(p^2-q^2)[/imath]. Suppose that [imath]5\leq q\leq p[/imath] are both prime. Prove that [imath]24|(p^2-q^2)[/imath]. This is what I got so far. I figured that since [imath]p,q[/imath] are bigger than [imath]5[/imath], there are only odd primes for this conjecture. Then I can rename [imath]p=2m+1,q=2n+1,m,n∈Z[/imath]. We want to show [imath](p^2-q^2 )=24k,k∈Z[/imath]. By substituting [imath]2m+1[/imath] for [imath]p[/imath] and [imath]2n+1[/imath] for [imath]q[/imath], we get [imath]((2m+1)^2-(2n+1)^2)[/imath]. If we expand it, we get [imath]4(m^2+n^2+m-n)[/imath] Q. I think I can make [imath]4(m^2+n^2+m-n)[/imath] look like [imath]24k[/imath]. How should I approach ? And my prof. commented that "I would take this proof in small steps. Rather than trying initially to prove that [imath]24 | p^2-q^2[/imath], look at the progress you already have. You proved [imath]4 | p^2-q^2[/imath]. Can you prove 8 is a factor? What about 3? Then see if you can put together what you have to get 24 as a factor." I am stuck here. Any helps?

507451

Suppose that [imath]p[/imath] ≥ [imath]q[/imath] ≥ [imath]5[/imath] are both prime numbers. Prove that 24 divides ([imath]p^2 − q^2[/imath]) I suppose I need to use prime factorization. I want to show [imath]p^2-q^2=24k[/imath] for some integer [imath]k[/imath] . How can I start this proof?

496760

If [imath]f(z)g(z) = 0[/imath] for every [imath]z[/imath], then [imath]f(z) = 0[/imath] or [imath]g(z) = 0[/imath] for every [imath]z[/imath]. This is for homework, and I would really appreciate a hint. The question states "If [imath]f[/imath] and [imath]g[/imath] are holomorphic on some domain [imath]\Omega[/imath] and [imath]f(z)g(z) = 0[/imath] for every [imath]z \in \Omega[/imath], then [imath]f(z) = 0[/imath] or [imath]g(z) = 0[/imath] for every [imath]z \in \Omega[/imath]." I tried contrapositive first. So suppose there exists points [imath]z_0[/imath] and [imath]z_1[/imath] in [imath]\Omega[/imath] such that [imath]f(z_0) \neq 0[/imath] and [imath]g(z_1) \neq 0[/imath]. But I immediately get stuck here because I can't say anything about [imath]f(z_0)g(z_0)[/imath] or [imath]f(z_1)g(z_1)[/imath]. Maybe a direct proof would be better? So by the Cauchy integral formula, I can say [imath] \int_{\partial \Omega} \frac{f(w)}{w - z} dw \int_{\partial \Omega} \frac{g(w)}{w - z} dw = f(z)g(z) = \int_{\partial \Omega} \frac{f(w)g(w)}{w - z} dw = 0 [/imath] for all [imath]z \in \Omega[/imath]. Again I get stuck here, because I don't see how this implies [imath]f(z) = 0[/imath] or [imath]g(z) = 0[/imath] for all [imath]z \in \Omega[/imath].

1048256

If [imath]fg=0[/imath] on a connected region , where [imath]f,g[/imath] are analytic on the connected region , then is either of [imath]f,g[/imath] identically [imath]0[/imath]? If [imath]f,g[/imath] are analytic on a connected region [imath]G[/imath] , and [imath]fg \equiv0[/imath] , then is it true that either [imath]f[/imath] or [imath]g[/imath] is identically [imath]0[/imath] ?

526630

If [imath]|f(z)| \leq C_1 + C_2 |z|^k[/imath] then [imath]f[/imath] is a polynomial of degree at most [imath]k[/imath] This is a weird problem that I could use some hints on how to solve: If [imath]f:\mathbb{C}\to \mathbb{C}[/imath] is analytic and for some [imath]k \in \mathbb{N}[/imath] there exist constants [imath]C_1,C_2>0[/imath] such that [imath]|f(z)|\leq C_1 + C_2|z|^k[/imath] then [imath]f[/imath] is a polynomial of degree at most [imath]k[/imath]. I am pretty much stumped here. Can you guide me/provide some hints?

86772

Show that if [imath]|f(z)| \leq M |z|^n[/imath] then [imath]f[/imath] is a polynomial max degree n I can't prove this statement, can anybody show me how to prove it? [imath]f:\mathbb{C}\rightarrow \mathbb{C} \in \mathcal{O}(\mathbb{C}), \exists n\in \mathbb{N}, R >0 , M>0 : |f(z)| \le M|z|^{n} \ \ \forall |z|>R \Rightarrow \deg(f)\le n [/imath] To show is that if there exists such an [imath]M[/imath], that then [imath]f[/imath] is a polynomial of max degree [imath]n[/imath]. I started like this: [imath]f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n[/imath] So if I put this into the inequality: [imath]|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .[/imath]

527183

What is the integral of this gaussian I want to know what is the following integral [imath]\int e^{-(y-\mu)^T \Lambda(y-\mu) } dy[/imath] I am trying to see the properties of gaussian integral but I couldn't find anything for this one. Any help guys? I want to know how given where Z(x) is the partition function This is the paper If the integration that you guys have given is correct there should be [imath]|\Lambda|^{1/2}[/imath] at the end instead of just [imath]|\Lambda|[/imath] in [imath]\frac{1}{Z(x)}[/imath]. I doubt that there is a type in the paper. So I must be missing something Can anyone explain why that guy didn't have [imath]|\Lambda|^{1/2}[/imath] but [imath]|\Lambda|[/imath] instead???

525295

Confusion related to gaussian distribution I was reading this paper where it had a gaussian distribution model. I mean gaussian is given by [imath]P(y) = \frac{e^{-\frac{1}{2}(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n/2}|\Sigma|^{1/2}}[/imath] But is [imath]\frac{e^{-(y -\mu)^T \Sigma^{-1}(y -\mu)}}{2\pi^{n}|\Sigma|}[/imath] Gaussian as well? The paper used their model saying the gaussian had the above form. It is valid?

527497

Can there be a real solution to the square root of -1? Basically what the title says: Can there be a real solution to the square root of [imath]-1[/imath] (or any negative number in fact) or is it defined to be unreal? Because of this: [imath] \begin{align} \sqrt{-1} & = (-1)^\frac 12 \\[6pt] & = (-1)^\frac 24 \\[6pt] & = \sqrt[4]{(-1)^2} \\[6pt] &= \sqrt[4]{1} \\[6pt] &= 1 \end{align}[/imath]

472227

math fallacy problem: [imath]-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1[/imath]? I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: [imath]-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?[/imath]

527651

Order of a conjugate permutation From Finan: Let [imath]\sigma\in{S_n}[/imath], define the [imath]\mathbf{order}[/imath] of [imath]\sigma[/imath] to be the smallest positive integer such that [imath]\sigma^m=(1)[/imath]. Prove that if [imath]\sigma[/imath] has order [imath]m[/imath] then [imath]\tau\sigma\tau^{-1}[/imath] has order [imath]m[/imath] for all [imath]\tau\in{S_n}[/imath]. So [imath]\tau[/imath] can have its own order, so let [imath]order(\tau)=p[/imath]. Thus [imath]\tau^p=(1)[/imath]. If [imath](m,p)=1[/imath], then [imath]order(\tau\sigma)=mp[/imath]. But doesn't [imath]order(\tau^{-1})=p[/imath]? Thus [imath]order(\tau\sigma\tau^{-1})=pmp[/imath]. So them this implies [imath](\tau\sigma\tau^{-1})^{pmp}[/imath]... Not sure where to go with this.

418761

Order of Permutation : If [imath]\tau \in S_n[/imath] has order [imath]m[/imath], then [imath]\sigma \tau\sigma^{-1}[/imath] has also order [imath]m[/imath]. I dont understand the following very simple statement: If [imath]\tau \in S_n[/imath] has order [imath]m[/imath], then [imath]\sigma \tau\sigma^{-1}[/imath] has also order [imath]m[/imath]. The proof is: Suppose [imath]\tau[/imath] has order [imath]m[/imath]. [imath](\sigma \tau \sigma^{-1})^{m} = \sigma \tau^{m}\sigma^{-1}= (1)[/imath] Then (using http://www.math.niu.edu/~beachy/abstract_algebra_2ed/guide/soln23.pdf exercise 17) they say that the order of [imath]\sigma\tau\sigma^{-1}[/imath] cannot be less than [imath]n[/imath] (remember we are in [imath]S_n[/imath]), since [imath](\sigma\tau\sigma^{-1})^k=1 [/imath] implies [imath]\sigma\tau^k\sigma^{-1} =(1)[/imath], and then [imath]\tau^k=\sigma^{-1} \sigma = (1)[/imath] I don't understand the last part.... Who can help me?

527847

College number theory problem - need a pointer! [imath]n[/imath] divides [imath]2^{2^n+1}+1[/imath] [imath]\implies n[/imath] divides [imath]2^{2^{2^n+1}+1}+1[/imath]? There are two ways to try to prove this. One is above, the other is its de Morgan counterpart: [imath]n[/imath] doesn't divide [imath]2^{2^{2^n+1}+1}+1 \implies[/imath] [imath]n[/imath] doesn't divide [imath]2^{2^n+1}+1[/imath]. Disproving it requires only one example of course. Tried using [imath]\gcd(2^a+1, 2^b+1) = 2^{\gcd(a, b)}+1[/imath] (where [imath]a[/imath] and [imath]b[/imath] are odd positive integers), stuck on both ends. I figured out that if [imath]n[/imath] divides [imath]2^n+1[/imath] then n divides both [imath]2^{2^n+1}+1[/imath] and [imath]2^{2^{2^n+1}+1}+1[/imath] but this implication doesn't work backwards (e.g. [imath]n=57[/imath]). Would appreciate some help.

519568

Does [imath]n \mid 2^{2^n+1}+1[/imath] imply [imath]n \mid 2^{2^{2^n+1}+1}+1[/imath]? There are two ways to try to prove this. One is in the title, the other is its de Morgan counterpart: [imath]n \nmid 2^{2^{2^n+1}+1}+1 \implies n \nmid 2^{2^n+1}+1[/imath]. Disproving it requires only one example of course. Tried using [imath]\gcd(2^a+1, 2^b+1) = 2^{\gcd(a, b)}+1[/imath] (where [imath]a[/imath] and [imath]b[/imath] are odd positive integers), stuck on both ends. I figured out that if [imath]n[/imath] divides [imath]2^n+1[/imath] then n divides both [imath]2^{2^n+1}+1[/imath] and [imath]2^{2^{2^n+1}+1}+1[/imath] but this implication doesn't work backwards (e.g. [imath]n=57[/imath]). Would appreciate some help. EDIT1 Eric's pointer wasn't enough for me. Trying to bump by editing instead of reposting (sorry, not sure how to). EDIT2 This is not much but might save some time for someone. Using user101140's notation and the [imath]a^n+b^n[/imath] identity [imath]f(n) = 2^n+1 = 3 \sum\limits_{k=0}^{n-1} (-2)^k[/imath] [imath]f(f(n)) = 3 \sum\limits_{k=0}^{2^n} (-2)^k[/imath] [imath]f(f(f(n))) = 3 \sum\limits_{k=0}^{2^{2^n+1}} (-2)^k[/imath] Also, [imath]n \mid f(n) \implies n \mid f(f(n))[/imath] is due to [imath]n \mid f(n) \implies f(n) \mid f(f(n))[/imath] so the proof might be something along the lines of [imath]n \mid f(f(n)) \implies f(f(n)) \mid f(f(f(n)))[/imath]. (Please don't bash me if that's stupid.)

527857

Rational algebraic numbers A polynomial with integer coefficients is an expression of the form: [imath]f(x)=a_nx^n+a_{n−1}x^{n−1}+\cdots+a_1x+a_0,[/imath] where [imath]a_n, a_{n−1},\ldots,a_1,a_0[/imath] are integers and [imath]a_n\ne 0[/imath]. A zero of the polynomial is a [imath]c\in\Bbb R[/imath] such that [imath]f(c)=0[/imath]. A real number is said to be algebraic if it is a zero of some polynomial with integer coefficients 1) Show that every rational number is algebraic. 2) Show that if [imath]a[/imath], [imath]b[/imath] and [imath]k[/imath] are positive integers, then the [imath]k[/imath]th root of [imath]a/b[/imath] is algebraic. I have no idea what to do, this is an advanced problem, I'm just curious to the solution.

527508

Showing rational numbers are algebraic A polynomial with integer coefficients is an expression of the form: [imath]f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0[/imath] where [imath]a_n[/imath], [imath]a_{n-1}, \ldots, a_1, a_0[/imath] are integers and [imath]a_n[/imath] is not equal to [imath]0[/imath]. a zero of the polynomial is a [imath]c \in \mathbb{R}[/imath] such that [imath]f(c)=0[/imath] A real number is said to be algebraic if it is a zero polynomial with integer coefficients 1) Show that every rational number is algebraic 2) Show that if [imath]a[/imath], [imath]b[/imath] and [imath]k[/imath] are positive integers, then the [imath]k[/imath]-th root of [imath]a/b[/imath] is algebraic I don't even know where to start on this. What is a zero of a polynomial with integer coefficients?

528055

If [imath]A\ B[/imath] are symmetric positive definite matrices then [imath]A>B[/imath] iff [imath]\sqrt{A}>\sqrt{B}[/imath] My guess is, it only holds one way i.e. [imath]A>B[/imath] then [imath]\sqrt{A}>\sqrt{B}[/imath] but not otherwise. Any proof or counterexample would be appreciated.

510895

If [imath]A^2\succ B^2[/imath], then necessarily [imath]A\succ B[/imath] I remember reading somewhere about the following properties of non-negative definite matrix. But I don't know how to prove it now. Let [imath]A[/imath] and [imath]B[/imath] be two non-negative definite matrices. If [imath]A^2\succ B^2[/imath], then it necessarily follows that [imath]A\succ B[/imath], but [imath]A\succ B[/imath] doesn't necessarily leads to [imath]A^2\succ B^2[/imath]. How can you prove it? Thanks!

11412

Any two points in a Stone space can be disconnected by clopen sets Let [imath]B[/imath] be a Stone space (compact, Hausdorff, and totally disconnected). Then I am basically certain (because of Stone's representation theorem) that if [imath]a, b \in B[/imath] are two distinct points in [imath]B[/imath], then [imath]B[/imath] can be written as a disjoint union [imath]U \cup V[/imath] of open sets where [imath]a \in U, b \in V[/imath]. However, I can't seem to prove this directly. The proof should be fairly straightforward, so I am sure I'm missing something obvious. (As an exercise for myself, I'm trying to prove Stone's representation theorem, and I need this as a lemma.)

1573116

Splitting of Connected Components We say [imath]U,V[/imath] is a splitting of a Topological space [imath]X[/imath] if , [imath] X = U \cup V [/imath] where [imath]U,V[/imath] are Open, disjoint , non empty subsets of [imath]X[/imath] In a book I'm reading it says that for any Compact Haussdorf topological space and given two distinct connected components, a splitting can be chosen such that they lie on different sides of the split. My question is, how do we show this? I realize compact haussdorf implies Normal but I'm not able to make an explicit splitting using that.

528104

decomposing a group into a direct sum of cyclic groups This is a question that I asked 2 days ago. But it was not solved..... Let [imath]\mathbb{Z}_n ^*[/imath] be the set of all units in [imath]\mathbb{Z}_n[/imath] and [imath](\mathbb{Z}_n ^*)^2[/imath] = [imath]\{ a^2 | a \in \mathbb{Z}_n ^*\} [/imath] Then, decompose the factor group [imath]\mathbb{Z}_n ^* / (\mathbb{Z}_n ^*)^2[/imath] when [imath]n[/imath] is an odd prime when [imath]n[/imath] is a product of two distinct odd primes [imath]p, q[/imath] I solved the first as following: In [imath]\mathbb{Z}_p[/imath], since [imath]x^2=(p-x)^2[/imath] holds, the order of [imath](\mathbb{Z}_n ^*)^2[/imath] is exactly the half of that of [imath]\mathbb{Z}_n ^*[/imath]. Thus, the answer is [imath]\mathbb{Z}_2[/imath]. However the second is more complicated. I didn't know that. Please help me to solve this.

525893

decomposing a factor group into a direct sum of cyclic groups Let [imath]\mathbb{Z}_n ^*[/imath] be the set of all units in [imath]\mathbb{Z}_n[/imath] and [imath](\mathbb{Z}_n ^*)^2[/imath] = [imath]\{ a^2 | a \in \mathbb{Z}_n ^*\} [/imath] Then, decompose the factor group [imath]\mathbb{Z}_n ^* / (\mathbb{Z}_n ^*)^2[/imath] 1.when [imath]n[/imath] is an odd prime 2.when [imath]n[/imath] is a product of two distinct odd primes [imath]p, q[/imath] I solved the first as following: In [imath]\mathbb{Z}_p[/imath], since [imath]x^2=(p-x)^2[/imath] holds, the order of [imath](\mathbb{Z}_n ^*)^2[/imath] is exactly the half of that of [imath]\mathbb{Z}_n ^*[/imath]. Thus, the answer is [imath]\mathbb{Z}_2[/imath]. However the second is more complicated. I didn't know that. Please help me to solve this.

529260

Finding the associated matrix of a linear operator Let [imath]V[/imath] be a complex vector space of dimension [imath]n[/imath] with a scalar product, and let [imath]u[/imath] be an unitary vector in [imath]V[/imath]. Let [imath]H_u: V \to V[/imath] be defined as [imath]H_u(v) = v - 2 \langle v,u \rangle u[/imath] for all [imath]v \in V[/imath]. I need to find the minimal polynomial and the characteristic polynomial of this linear operator, but the only way I know to find the charactestic polynomial is using the associated matrix of the operator. I don't know how to find this matrix because I don't know how to deal with the scalar product. Is there some other way to find the characteristic polynomial? If not, how can I find the associated matrix of this linear operator? Thanks in advance.

528783

What's the associated matrix of this linear operator? Let [imath]V[/imath] be a complex vector space of dimension [imath]n[/imath] with a scalar product, and let [imath]u[/imath] be an unitary vector in [imath]V[/imath]. Let [imath]H_u: V \to V[/imath] be defined as [imath]H_u(v) = v - 2 \langle v,u \rangle u[/imath] for all [imath]v \in V[/imath]. I need to find the characteristic polynomial of this linear operator, but the only way to find it that I know of is using the associated matrix of the operator. I don't know how to find this matrix because I don't know how to deal with the scalar product. Is there some other way to find the characteristic polynomial? If not, how can I find the associated matrix of this linear operator? Thanks in advance.

528683

A question about bounded operators on banach space Let [imath]L(X)[/imath] denotes the Banach algebra of all bounded linear operators acting on a Banach space [imath]X[/imath]. And [imath]T[/imath] is not invertible. Can we find a invertilbe bounded operator series [imath]\{T_{n}\}[/imath] such that [imath]T_{n}\rightarrow T[/imath], [imath]n\rightarrow \infty[/imath], here [imath]T_{n}[/imath] convergence in the operator norm.

92079

Closure of the invertible operators on a Banach space Let [imath]E[/imath] be a Banach space, [imath]\mathcal B(E)[/imath] the Banach space of linear bounded operators and [imath]\mathcal I[/imath] the set of all invertible linear bounded operators from [imath]E[/imath] to [imath]E[/imath]. We know that [imath]\mathcal I[/imath] is an open set, and if [imath]E[/imath] is finite dimensional then [imath]\mathcal I[/imath] is dense in [imath]\mathcal B(E)[/imath]. It's not true that [imath]\mathcal I[/imath] is dense if we can find [imath]T\in\mathcal B(E)[/imath] injective, non surjective with [imath]T(E)[/imath] closed in [imath]E[/imath], since such an operator cannot be approximated in the norm on [imath]\mathcal B(E)[/imath] by elements of [imath]\mathcal I[/imath] (in particular [imath]E[/imath] has to be infinite dimensional). So the question is (maybe a little vague): is there a nice characterization of [imath]\overline{\mathcal I}^{\mathcal B(E)}[/imath] when [imath]E[/imath] is infinite dimensional? Is the case of Hilbert space simpler?

529687

the closed interval [0,1] ( with the usual topology) A topological space is called a US-space provided that each convergent sequence has a unique limit. A topological space is called [imath]T_B[/imath] if each compact subset is closed. The bellow example show that [imath]US[/imath] space is not [imath]T_B[/imath] space. Suppose X with the closed interval [0,1] ( with the usual topology) and attach a new point [imath]z[/imath] whose neighborhoods are open dense subsets of [0,1].[0,1] is a compact non closed subspace of X and thus X is not [imath]T_B[/imath] space. However no sequence in [0,1] converges to [imath]z[/imath] and in particular sequences in X have unique limits. Why is it right "no sequence in [0,1] converges to [imath]z[/imath] and in particular sequences in X have unique limits"?

461809

[imath] KC [/imath] spaces imply [imath] US [/imath] spaces , but vise versa is false. In the [imath] US [/imath] space , each convergent sequence has unique limit. In the [imath] KC [/imath] space , every compact subset is closed. It easy to show that [imath] KC [/imath] spaces imply [imath] US [/imath] spaces. The example below show that [imath] US [/imath] spaces do not imply [imath] KC [/imath] spaces. Let [imath] X = [ 0, 1 ] [/imath] with the the usual topology and attach a new point z whose neighborhood are open dense subsets of [imath] [ 0,1 ][/imath]. Observe [imath] [ 0,1 ][/imath] is a compact non-closed of [imath]X[/imath] and then [imath]X[/imath] is not a [imath]KC[/imath] space. However no sequence in [imath] [ 0,1 ][/imath] converge to [imath]z[/imath] and in particular all convergent in [imath]X[/imath] have unique limits. can you give me more details? Why [imath]X[/imath] is not closed? Why no sequence in [imath] [ 0,1 ][/imath] converges to [imath]z[/imath] and in particular all convergent sequences in [imath]X[/imath] have unique limits ?

529884

Prove that [imath]\{ \sin(x), \sin^2(x), \sin^3(x)\}[/imath] is linearly independent in [imath]F(\Bbb R)[/imath] Prove that [imath]\{ \sin(x), \sin^2(x), \sin^3(x)\}[/imath] is linearly independent in [imath]F(\Bbb R)[/imath]. I tried plugging in [imath]\left\{ 0, \frac {\pi} {2}, \pi, \frac {3\pi}{2}\right\}[/imath] but that doesn't work. How should I prove this?

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Are the functions [imath]\sin^n(x)[/imath] linearly independent? The following problem is from Golan's linear algebra book. I have posted a proposed solution in the answers. Problem: For [imath]n\in \mathbb{N}[/imath], consider the function [imath]f_n(x)=\sin^n(x)[/imath] as an element of the vector space [imath]\mathbb{R}^\mathbb{R}[/imath] over [imath]\mathbb{R}[/imath]. Is the subset [imath]\{f_n:\ n\in\mathbb{N}\}[/imath] linearly independent?

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Let [imath]R[/imath] be an infinite commmutative ring which contains a zero divisor. Show that there exists infinite many zero divisors. I'm making exercises to prepare for my ring theory exam. Let [imath]R[/imath] be an infinite commmutative ring which contains a zero divisor. Show that there exists infinite many zero divisors. Let [imath]a\in R[/imath] a zero divisor. Then [imath]a⋅b=0[/imath] for some element [imath]b≠0[/imath] in [imath]R[/imath]. If [imath]a[/imath] or [imath]b[/imath] has infinite order, then I can find infinite zero divisors. But otherwise, I don't see why this should be true. I think I need to do something with the fact that [imath]R[/imath] is commutative. A hint or a detailed solution are both appreciated.

315735

If [imath]R[/imath] is an infinite ring, then [imath]R[/imath] has either infinitely many zero divisors, or no zero divisors Please help me to prove that if [imath]R[/imath] is an infinite ring, then [imath]R[/imath] has either an infinite number of zero divisors, or it has no zero divisors.

530004

Why is [imath]ℤ[X]/(X^2-1)[/imath] not isomorphic with [imath]ℤ×ℤ[/imath]? Why is [imath]ℤ[X]/(X^2-1)[/imath] not isomorphic with [imath]ℤ ×ℤ[/imath] ? I understand why this is true in the case of [imath]\mathbb{Q}[/imath]. But [imath](X-1)+(X+1) ≠ℤ[X][/imath], so therefore I can't use the same reasoning. I don't see how I can proof that there can't be an isomorphism.

454493

[imath]\mathbb{Z}[X]/(X^2-1)[/imath] is not isomorphic with [imath]\mathbb{Z}\times \mathbb{Z}[/imath] I have to show that the ring [imath]\mathbb{Z}[X]/(X^2-1)[/imath] is not isomorphic with [imath]\mathbb{Z}\times \mathbb{Z}[/imath]. I know that [imath](\mathbb{Z}\times\mathbb{Z})^*=\{(\pm1,\pm1)\}[/imath], so I thought I should be looking for elements which have inverses in [imath]\mathbb{Z}[X]/(X^2-1)[/imath] and hopefully find more or less than 4. But I didnt succeed, so I need hints. Thanks.

530043

Orthogonal completion in nonhilbert spaces Let [imath]X[/imath] be some Hilbert space. There is the widely known theorem in functional analysis which states that for each closed subspace [imath]H\subset X[/imath] we have [imath]H\bigoplus H^{\perp}=X[/imath]. Now we do not suppose [imath]X[/imath] to be Hilbert space (but it is endowed with a scalar product). Is it true that there always exists such closed subspace [imath]H\subset X[/imath] such that [imath]H\bigoplus H^{\perp}\ne X[/imath]? So, I know that the completness of [imath]X[/imath] is important in the proves given in books, but how to build such closed subspace [imath]H[/imath]?

82499

A counterexample to theorem about orthogonal projection Can someone give me an example of noncomplete inner product space [imath]H[/imath], its closed linear subspace of [imath]H_0[/imath] and element [imath]x\in H[/imath] such that there is no orthogonal projection of [imath]x[/imath] on [imath]H_0[/imath]. In other words I need to construct a counterexample to theorem about orthogonal projection when inner product space is not complete.

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[imath]f:S^1 \rightarrow R[/imath] continuous means there exists a point [imath]x[/imath] such that [imath]f(x) = f(-x)[/imath]. Let [imath]S^1[/imath] be the unit circle in [imath]\mathbb{R}^2[/imath], then for a continuous map from [imath]S^1[/imath] into [imath]\mathbb{R}[/imath], there exists [imath]x \in S^1[/imath] such that [imath]f(x) = f(-x)[/imath]. I know that [imath]S^1[/imath] is path connected and so connected. And the intermediate value theorem. For any choice of [imath]-x, x[/imath], and any [imath]r[/imath] between [imath]f(-x)[/imath] and [imath]f(x)[/imath] there is [imath]c \in S^1[/imath] such that [imath]f(c) = r[/imath]. Let [imath]\psi : [0,2\pi] \rightarrow S^1, \ x \mapsto e^{i(a + x)}[/imath] for any [imath]a\in \mathbb{R}[/imath] be a circular path in [imath]S^1[/imath] identified with the circle group as a subset of [imath]\mathbb{C}[/imath]. Let [imath]g(x) = f(x) - f(-x)[/imath]. Let [imath]a\in S^1[/imath] be any point such that there exist points in the image of [imath]g[/imath] to the right of [imath]g(a)[/imath] for at least a run of [imath]\epsilon[/imath]. Now translate everything to near the origin with [imath]h(x) = g(x) - g(a) - \epsilon/2[/imath] Can't figure this one out.

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[imath]f : S^1 \to\mathbb R[/imath] is continuous then [imath]f(x)=f(-x)[/imath] for some [imath]x\in S^1[/imath] Question is to prove : [imath]f : S^1 \to \mathbb R[/imath] is continuous then [imath]f(x)=f(-x)[/imath] for some [imath]x\in S^1[/imath] I guess it would be helpful to use intermediate value theorem Assuming [imath]f(x)\neq f(-x)[/imath] then given any [imath]p\in (f(-x),f(x))[/imath] (assuming [imath]f(-x)<f(x)[/imath]) there exists [imath]y\in (-x, x)[/imath] such that [imath]f(y)=p[/imath] I am not very sure of how to use this.. It would be helpful if someone can give some hint which would help me to solve this.. Thank you.

530730

A matrix problem 2 Find a [imath]3 \times 3[/imath]-matrix [imath]A[/imath] such that [imath]A^2=\begin{pmatrix} 13 & 9 & -9 \\ 0 & 4 & 0 \\ 12 & 12 & -8 \\ \end{pmatrix}[/imath] and [imath]a_{11}, a_{22}>0[/imath] I noticed that [imath]A^2-4I[/imath] is somewhat simpler form but don't know how to find [imath]A[/imath]. How can I find it?

530600

Given [imath]A^2[/imath] where A is matrix, how find A? Problem is simple. Given [imath]A^2=\begin{bmatrix}13 & 9 &-9 \\ 0 & 4 & 0 \\ 12 & 12 & -8 \end{bmatrix}[/imath] How find [imath]A[/imath]? I think a method using eigenvalues and I find them. But I can't find an actual [imath]A[/imath]. Is it right to use eigenvalues?

529745

Why least square problem always has solution for arbitrary b? We know from linear algebra, the least square solution of linear equation system : [imath]Ax=b[/imath] always exists. That is, the equation [imath]A^TAx=A^Tb[/imath] always has at least one solution. Most of us explain it existence by interpreting least square problem as a matter of orthogonal projection. It's correct, I know. But can we prove this by showing that the rank of matrix [imath]A^TA[/imath] is equal to the rank of the augmented matrix:[imath][A^TA,A^Tb][/imath]for any given vector b? I've worked for hours but failed to show that. Although there is a similar thread in mathematics.SX, but I personally think there is no good answer presented there. So I address this problem again. Thanks at first! [update] Now I come up with a proof for this proposition: suppose that rank [imath]A^TA[/imath]=k, and the matrix [imath][A^TA,A^Tb][/imath] has one more column than [imath][A^TA][/imath], so rank [imath][A^TA,A^Tb]\ge k[/imath]. But on the other hand, we have [imath][A^TA,A^Tb]=A^T[A,b][/imath], using the rank inequality, we have [imath]rank [A^TA,A^Tb]\le rank A^T[/imath]. Since (we can prove that) [imath] rank A^TA= rank A= rank A^T[/imath] for any given matrix A. So the inequality is actually [imath]rank [A^TA,A^Tb]\le k[/imath]. Combining the two inequalities, we have [imath]rank [A^TA,A^Tb]=rank A^TA[/imath]. Q.E.D.

72222

Existence of least squares solution to [imath]Ax=b[/imath] Does a least squares solution to [imath]Ax=b[/imath] always exist?

333941

How to computer [imath]\int_0^1\log(\sin \pi x)\text{d}x[/imath] under complex methods. the below integral is coming from M.Stein's complex analysis. chapter 3 exercise 9. Show that [imath]\int_0^1\log(\sin \pi x)\text{d}x=-\log 2[/imath] use the contour shown in the above figure Then I recall the definition of [imath]\log(z)[/imath],it must be defined on a simply connected domain [imath]\Omega[/imath] with [imath]1\in \Omega[/imath] and [imath]0 \not\in \Omega[/imath]. In order to use the above contour to computer the integral. maybe consider [imath]\int_{\gamma}\log(\sin \pi z)\text{dz}[/imath],where [imath]\gamma[/imath] is the contour. but for [imath]z=0[/imath], there is no definition. I think the author arrange the exercise here must expect the reader use residue formula to computer it. but I don't know how to. thanks very much

341643

Showing that [imath]\int_0^1 \log(\sin \pi x)dx=-\log2[/imath] I need help with a textbook exercise (Stein's Complex Analysis, Chapter 3, Exercises 9). This exercise requires me to show that [imath]\int_0^1 \log(\sin \pi x)dx=-\log2[/imath] A hint is given as "Use the contour shown in Figure 9." Since this is an exercise from Chapter 3, I think I should use the residue formula or something like that. But the function [imath]f(x)=\log(\sin \pi x)[/imath] becomes singular on [imath]x=0[/imath] and [imath]x=1[/imath], which makes the contour illegal for the residue theorem. Can anyone give me a further hint on this problem? Many thanks in advance! P.S. This is my first time on Math Stack Exchange. If you find my post ambiguous, let me know.

519338

How do I find where a curve hits the [imath]xy[/imath] plane with vectors? I need to find [imath]r'(t)[/imath] and [imath]||r'(t)||[/imath] of [imath]r(t)=<t,t,t^2>[/imath], and tell where the curve hits the [imath]xy[/imath] plane (if it does). Also, I need to say something about how the curve looks like (do I just plot some points and figure out how it looks like?). I understand how to find [imath]r'(t)[/imath] and [imath]||r'(t)||[/imath], but how do I tell where the curve hits the [imath]xy[/imath] plane, and how do I describe the curve? Here are my solutions for [imath]r'(t)[/imath] and [imath]||r'(t)||[/imath]: [imath]r(t)=<t,t,t^2>[/imath] [imath]r'(t) = <1,1,2t>[/imath] [imath]||r'(t)||=\sqrt{2+4t^2}[/imath]

531052

Find where [imath]r(t)=[/imath] hits the [imath]x-y[/imath] plane I have to find [imath]r'(t)[/imath] and [imath]||r'(t)||[/imath] for [imath]r(t)=<t,t,t^2>[/imath], which I know how to do. [imath]r'(t)=<1,1,2t>[/imath] [imath]||r'(t)||=\sqrt{2+4t^2}[/imath] The problem is that my professor didn't explain how to find where each curve hits the [imath]x-y[/imath] plane (if it hits). I'm guessing that I have to make [imath]z=0[/imath], but where?

63774

What is the limit [imath]\lim\limits_{n \to \infty} \frac{n^{\sqrt{n}}}{2^n}[/imath]? The question is rather: Prove the following: [imath] \lim_{x \to \infty} \frac{x ^ {\sqrt x} }{ 2 ^ x} = 0. [/imath] I was thinking of using the Squeeze Theorem (might not be the right way to go), but finding an upper-bound function proved to be quite tricky.

1009296

What is [imath]\lim_{n\rightarrow\infty}\frac{n^{\sqrt{n}}}{2^{n}}=? [/imath] I'm trying to calculate the following limit: [imath]\lim_{n\rightarrow\infty}\frac{n^{\sqrt{n}}}{2^{n}}[/imath] I assume it's equal to [imath]0[/imath], but can't find a way to prove it. I don't even know where to start..

531382

question about contractive and hereditary property A topological space is [imath]T_B[/imath] if each compact subset is closed. A topological space is called a US-space provided that each convergent sequence has a unique limit. A topological property [imath]\mathcal{R}[/imath] is contractive if [imath](X,\tau)[/imath] has property [imath]\mathcal{R}[/imath] and if [imath]\tau^{\prime} \subset \tau[/imath], then (A, [imath]\tau^{\prime}[/imath]) has property [imath]\mathcal{R}[/imath]. A topological space is called a C- C-space provided that the compact subsets are precisely the closed sets. Is property C-C , US contractive, hereditary? Is property [imath]T_B[/imath] contractive?

530719

contractive property A topological property [imath]\mathcal{R}[/imath] is contractive if [imath](X,\tau)[/imath] has property [imath]\mathcal{R}[/imath] and if [imath]\tau^{\prime} \subset \tau[/imath], then (A, [imath]\tau^{\prime}[/imath]) has property [imath]\mathcal{R}[/imath]. A topological space is [imath]T_B[/imath] if each compact subset is closed. A topological space is called a US-space provided that each convergent sequence has a unique limit. A topological space is called a C- C-space provided that the compact subsets are precisely the closed sets. Is property C-C , US contractive, hereditary? Is property [imath]T_B[/imath] contractive?

531981

Hartshorne, Prop II.5.2(d,e): direct and inverse images of sheaves associated to modules Hartshorne, Algebraic Geometry, Proposition II.5.2, reads (in part): Let [imath]A[/imath] be a ring and let [imath]X = \operatorname{Spec} A[/imath]. Also let [imath]A \to B[/imath] be a ring homomorphism and [imath]f : \operatorname{Spec} B \to \operatorname{Spec} A[/imath] be the corresponding map of spectra. Then: [...] (d) for any [imath]B[/imath]-module [imath]N[/imath] we have [imath]f_\ast(\widetilde{N}) \cong \widetilde{_{A}N}[/imath], where [imath]_A N[/imath] means [imath]N[/imath] considered as an [imath]A[/imath]-module; (e) for any [imath]A[/imath]-module [imath]M[/imath] we have [imath]f^\ast(\widetilde{M}) \cong \widetilde{M \otimes_A B}[/imath]. The corresponding proof in the text reads: The last statements... follow directly from the definitions. I don't get it. For instance, [imath](f_\ast\widetilde{N})(U) = \left\{s : f^{-1}(U) \to \bigsqcup_{f(\mathfrak{p}) \in U} N_\mathfrak{p} \; \middle\vert \; s(\mathfrak{p}) \in N_\mathfrak{p}, s = \frac{n}{b} \text{ locally}\right\}[/imath] while [imath]\widetilde{_{A}N}(U) = \left\{t : U \to \bigsqcup_{\mathfrak{q} \in U} ({}_AN)_\mathfrak{q} \; \middle\vert \; s(\mathfrak{q}) \in ({}_AN)_\mathfrak{q}, s = \frac{n}{a} \text{ locally}\right\}[/imath] I can't see the connection. What am I missing?

309389

Inverse image of the sheaf associated to a module In Hartshorne, Algebraic geometry it's written, that for every scheme morphism [imath]f: Spec B \to Spec A[/imath] and [imath]A[/imath]-module [imath]M[/imath] [imath]f^*(\tilde M) = \tilde {(M \otimes_A B)}[/imath]. And that it immediately follows from the definition. But I don't know how to prove it in simple way. Could you help me?

532257

Prove if prime can be written as [imath]2^n+1[/imath], [imath]n = 2^k[/imath] Prove that if prime can be written as [imath]\ 2^n + 1[/imath] then [imath]n = 2^k[/imath], [imath]\;\;n, k \in \mathbb N[/imath]. I am pretty new in this part of math.

140804

If [imath]2^n+1[/imath] is prime, why must [imath]n[/imath] be a power of [imath]2[/imath]? A little bird told me that if [imath]2^n+1[/imath] is prime, then [imath]n[/imath] is a power of [imath]2[/imath]. I tend not to trust talking birds, so I'm trying to verify that statement independently. Suppose [imath]n[/imath] is not a power of [imath]2[/imath]. Then [imath]n = a \cdot 2^m[/imath] for some [imath]a[/imath] not a power of [imath]2[/imath] and some integer [imath]m[/imath]. This gives [imath]2^n+1 = 2^{a \cdot 2^m}+1[/imath]. Now I suspect there's a way to factor that, but I don't see how. Can someone give me a hint?