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444325 | Show [imath]\lvert f\rvert\leq\lvert g\rvert\Rightarrow f=cg[/imath]
Let [imath]f[/imath] and [imath]g[/imath] be entire functions with [imath]\lvert f\rvert\leq\lvert g\rvert[/imath]. Show that then [imath] f=cg [/imath] for a constant [imath]c\in\mathbb{C}[/imath]. Can I just do it like this: [imath]\lvert f(z)\rvert\leq\sup\limits_{z\in\mathbb{C}}\lvert g(z)\rvert[/imath], so f is limited and therefore constant (Liouville), i.e. [imath]f(z)=c~\forall~z\in\mathbb{C}[/imath]. [imath]c=0[/imath]: Then [imath]0=f(z)=0\cdot g(z)[/imath] [imath]c\neq 0[/imath]: Then [imath]g(z)\neq 0[/imath] and therefore [imath]\frac{f(z)}{g(z)}[/imath] is defined. This is an entire function because [imath]f[/imath] and [imath]g[/imath] are entire function and it is [imath] \left\lvert\frac{f(z)}{g(z)}\right\rvert=\frac{\lvert f(z)\rvert}{\lvert g(z)\rvert}\leq 1, [/imath] so [imath]\frac{f(z)}{g(z)}[/imath] is a bounded entire function and therefore constant, lets say for a [imath]d\in\mathbb{C}[/imath] [imath] \frac{f(z)}{g(z)}=d~\forall~z\in\mathbb{C}\Leftrightarrow f(z)=d\cdot g(z). [/imath] This is my proof. Is it okay or does one need a special sentence for this proof? | 320576 | Entire function dominated by another entire function is a constant multiple
These two questions I didn't even find the way to solve So please if you can help Suppose [imath]f (z)[/imath] is entire with [imath]|f(z)| \le |\exp(z)|[/imath] for every [imath]z[/imath] I want to prove that [imath]f(z) = k\exp(z)[/imath] for some [imath]|k| \le 1[/imath] Can a non constant entire function be bounded in half a plane? Prove if yes , example if not. |
444413 | How to perform the integral [imath]\int_{-\infty}^\infty\frac{e^{-x^2}\sin(x) }{x }[/imath]?
Does anybody know how to perform the integral [imath] \int_{-\infty}^\infty\frac{e^{-x^2}\sin(x) }{x } [/imath] Thanks. | 208250 | Integral of sinc function multiplied by Gaussian
I am wondering whether the following integral [imath]\int_{-\infty}^{\infty} \exp( - a x^2 ) sin( bx ) / x[/imath] exists in closed form. I would like to use it for numerical calculation and find an efficient way to evaluate it. If analytical form does not exist, I really appreciate any alternative means for evaluating the integral. One method would be numerical quadrature including Gaussian quadrature, but it may be inefficient when the parameters a and b are very different in scale. thanks in advance |
444447 | bijection between [imath]\mathbb{N}[/imath] and [imath]\mathbb{N}\times\mathbb{N}[/imath]
I understand that both [imath]\mathbb{N}[/imath] and [imath]\mathbb{N}\times\mathbb{N}[/imath] are of the same cardinality by the Shroeder-Bernstein theorem, meaning there exists at least one bijection between them. But I can't figure out what such a bijection would be. The paper that I'm reading gives an example of an injective function [imath]f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}[/imath], [imath]f(n)=(0,n)[/imath], and an injective function [imath]g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}[/imath], [imath]g(a,b)=2^a3^b[/imath]. I was thinking perhaps if there were some way to combine these two, I could find a bijection, but I have no idea how to go about that or if it's even possible. What is an example of a bijection between these two sets, and please explain the process by which you found it? | 888730 | Bijection between natural numbers [imath]\mathbb{N}[/imath] and natural plane [imath]\mathbb{N} \times \mathbb{N}[/imath]
I know that is possible to build a bijection between the set of natural numbers [imath]\mathbb{N}[/imath] and the natural plane (the cartesian product of [imath]\mathbb{N}[/imath] by itself, [imath]\mathbb{N} \times \mathbb{N} = \mathbb{N}^2[/imath]. This is done by diagonally traversing the plane from zero upwards with triangles of growing size. Is there a simple algebraic form for that bijection? That is, is it possible to write explicitly some invertible [imath]f(i, j): \mathbb{N} \times \mathbb{N} \leftrightarrow \mathbb{N}[/imath]? I need to index in a simple way couples of natural numbers. |
444619 | Integral of the product of [imath]x^k[/imath] and the upper half circle of radius 2.
As I was browsing through the introduction of a paper, I came across the following equality: [imath]\displaystyle\frac1{2\pi}\int_{-2}^2x^k\sqrt{4-x^2}~dx=\begin{cases}\frac1{ k/2+1}\binom{k}{k/2}&\text{if [/imath]k[imath] is even;}\\0&\text{if [/imath]k[imath] is odd.}\end{cases}[/imath] Given that the equality in the case where [imath]k[/imath] is even was not intuitively obvious to me (the case where [imath]k[/imath] is odd is trivial), I decided to try to compute if myself. However, after trying to use most of the integral tricks I remembered, I am at a loss. I would appreciate any partial solution/hint of this if anyone knows it. | 380025 | Calculus Reduction Formula
For any integer [imath]k > 0[/imath], show the reduction formula [imath]\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx[/imath] for some constant [imath]C_{k}[/imath]. (original image) I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution? |
387330 | Big [imath]O[/imath] notation and little [imath]o[/imath] notation
I am a bit confused about the big [imath]O[/imath] and little [imath]o[/imath] notations. In other words, can any one show me with examples how these notations works? Explain in examples please? | 113095 | The big [imath]O[/imath] versus little [imath]o[/imath] notation.
I'm familiar to both the [imath]O[/imath] and little [imath]o[/imath] notation. I know they are of great use when studying limits, asymptotics and series. However, I'm not sure when to use one over another, particularly when solving problems. For example, you can put that, for [imath]x \to 0[/imath] [imath]\sin x = x + o(x)[/imath] since [imath]\frac{\sin x}{x}-1 \to 0[/imath] I've also recently read about a great definition for the derivative and strong derivative (Knuth), respectively, for [imath]\epsilon \to 0[/imath]: [imath]f(x+\epsilon)=f(x)+\epsilon f'(x)+o(\epsilon)[/imath] which clearly means that [imath]\frac{f(x+\epsilon)-f(x)}{\epsilon}- f'(x) \to 0[/imath] and the same for [imath]f(x+\epsilon)=f(x)+\epsilon f'(x)+o(\epsilon ^2)[/imath] Now consider the following [imath]\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}[/imath] I know that [imath]2x\left( {1 - \cos x} \right) = {x^3} + o\left( {{x^3}} \right)[/imath] [imath]\tan x = x + \frac{x^3}{3} + o\left( x^3 \right)[/imath] [imath]\int\limits_0^x {\frac{{\sin t}}{t}dt = x -\frac{x^3}{3\cdot 3!} +o\left( x^3 \right)} [/imath] So this gives [imath]\frac{{x - \frac{{{x^3}}}{{3 \cdot 3!}} + o\left( {{x^3}} \right) - x - \frac{{{x^3}}}{3} - o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}} = [/imath] [imath] \frac{{\left( { - \frac{1}{3} - \frac{1}{{18}}} \right){x^3} + o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}}[/imath] [imath]\frac{{\left( { - \frac{7}{{18}}} \right){x^3} + o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}} = - \frac{7}{{18}}[/imath] So how could I use the big [imath]O[/imath] to solve the same problem? And why should I choose big [imath]O[/imath] over little [imath]o[/imath], when [imath]o[/imath] seems more accurate than [imath]O[/imath]? NOTE: When I talk about the big [imath]O[/imath] I'm considering it for any [imath]x \to a[/imath], not only for asymptotic behaviour [imath]x \to \infty[/imath]. For example, for [imath]x \to 0[/imath] [imath]\sin x = O(x) [/imath] |
445401 | Prove that [imath]f(z)=z[/imath] on region [imath]D[/imath]
Given that [imath]D[/imath] be a bounded region containing [imath]0[/imath] and [imath]f:D\rightarrow D[/imath] be a holomorphic mapping such that [imath]f(0)=0,f'(0)=1[/imath] prove that f(z)=z for all [imath]z[/imath] in [imath]D[/imath] This problem reminded me of the Schwarz's lemma but the region is not [imath]\{z:|z|<1\}[/imath] so what theorem should I apply here? | 275270 | Holomorphic function [imath]\varphi[/imath] with fixed point [imath]z_0[/imath] such that [imath]\varphi'(z_o)=1[/imath] is linear?
This is an exercise in complex analysis: Let [imath]\Omega\subset{\Bbb C}[/imath] be open and bounded, and [imath]\varphi:\Omega\to\Omega[/imath] a holomorphic function. Prove that if there exists a point [imath]z_0\in\Omega[/imath] such that [imath] \varphi(z_0)=z_0\qquad\text{and }\qquad \varphi'(z_0)=1 [/imath] then [imath]\varphi[/imath] is linear. I'm trying work out the case [imath]z_0=0[/imath] first, in which [imath] \varphi(z)=z+\sum_{n=2}^{\infty}a_nz^2. [/imath] It suffices to show that [imath]a_n=0[/imath] for all [imath]n\geq 2[/imath]. If let [imath] \varphi(z)=z+a_2z^2+O(z^3) [/imath] then [imath] \varphi^k(0)=z+ka_2z^2+O(z^3), [/imath] and [imath] \varphi^k(0)=0,\quad (\varphi^k)'(0)=1. [/imath] If one can show that [imath]\{ka_2\}_{k=1}^{\infty}[/imath] is uniformly bounded, then one at least has [imath]a_2=0[/imath]. But I don't know how to go on. Any idea? |
445998 | Proving that a planar graph is bipartite
Let [imath]G[/imath] be a connected planar graph with a planar embedding where every face boundary is a cycle of even length. Prove that [imath]G[/imath] is bipartite. It is quite easy to prove the converse, but how to do this? My line of thought is to assume an odd cycle and show a contradiction, but I don't know where to head? This is homework, so please give hints only, not a full solution. | 445331 | Proving bipartition in a connected planar graph
Let [imath]G[/imath] be a connected planar graph with a planar embedding where every face boundary is a cycle of even length. Prove that [imath]G[/imath] is bipartite. Any hints/tips will be greatly appreciated. |
446039 | [imath]A[/imath] is similar to [imath]B[/imath] in [imath]E[/imath] if only if [imath]A[/imath] is similar to [imath]B[/imath] in [imath]F[/imath].
If [imath]E[/imath] is a field, [imath]F/E[/imath] is a field extension, let [imath]A[/imath] and [imath]B[/imath] be two matrices with entries in [imath]E[/imath] then [imath]A[/imath] is similar to [imath]B[/imath] in [imath]E[/imath] if only if [imath]A[/imath] is similar to [imath]B[/imath] in [imath]F[/imath]. I think it's true, but I don't know how to prove it. Any suggestion? Thanks. | 57242 | Similar matrices and field extensions
Given a field [imath]F[/imath] and a subfield [imath]K[/imath] of [imath]F[/imath]. Let [imath]A[/imath], [imath]B[/imath] be [imath]n\times n[/imath] matrices such that all the entries of [imath]A[/imath] and [imath]B[/imath] are in [imath]K[/imath]. Is it true that if [imath]A[/imath] is similar to [imath]B[/imath] in [imath]F^{n\times n}[/imath] then they are similar in [imath]K^{n\times n}[/imath]? Any help ... thanks! |
446119 | An equivalent norm
Is it possible to say that [imath]\|u\|_{H_{0}^{2}(\Omega)}[/imath] is equivalent to [imath]\|\Delta u\|_{2}[/imath]? | 301404 | eqiuvalent norms in [imath]H_0^2[/imath]
I have found that the [imath]H^2(D)[/imath] norm of a field with zero Cauchy data on [imath]\partial D[/imath] (i.e. in [imath]H_0^2(D)[/imath]) is equivalent to the [imath]L^2(D)[/imath] norm of its Laplacian, where D is simply connected with smooth boundary in [imath]\mathbb{R}^n[/imath]. How can i prove this using Poincare inequality; |
446096 | Does the series [imath]\sum\limits_{n=1}^\infty \sin \frac{1}{n}[/imath] converge?
Does the series [imath]\sum\limits_{n=1}^\infty \sin \dfrac{1}{n}[/imath] converge or diverge? | 9867 | Convergence/Divergence of [imath]\sum_{n=1}^{\infty} \sin(1/n)[/imath]
it is a question Convergence/Divergence of calculus II! Please give me a hand! Determine convergence or divergence using any method covered so far. [imath]\sum_{n = 1}^{\infty} \sin (1/n)[/imath] |
446472 | Finding [imath]i^{i^i}[/imath].
Express [imath]i^{i^i}[/imath] in the form [imath]a+bi[/imath] where [imath]a,b[/imath] are real. From euler's formula, I get [imath]\ln i=i\pi/2[/imath], which leads to [imath]i^i=e^{-\pi/2}[/imath]. Therefore, [imath]\ln i^{i^i}=i^i\ln i=i\pi e^{-\pi/2}/2,i^{i^i}=e^{i\pi e^{-\pi/2}/2}[/imath], but I don't know how to deal with this. Please help. Thanks. | 439820 | Find the value of [imath]\space\large i^{i^i}[/imath]?
Is [imath]\large i^{i^i}[/imath] real ? How to find it? Thank You! |
318606 | If [imath]\gcd(m,n)=1[/imath], then [imath]\mathbb{Z}_n \times \mathbb{Z}_m[/imath] is cyclic.
If [imath]\gcd(m,n)=1[/imath], then [imath]\mathbb{Z}_n \times \mathbb{Z}_m[/imath] is a cyclic group. Let's denote [imath]\mathbb{Z}_n=\langle1_n \rangle[/imath] and [imath]\mathbb{Z}_m=\langle1_m \rangle.[/imath] My proof goes as follows: since [imath]|\mathbb{Z}_n \times \mathbb{Z}_m|=mn[/imath] and [imath]\gcd(m,n)=1[/imath], [imath]|\langle 1_n,1_m\rangle|=\text{lcm}(|\langle 1_n\rangle|,|\langle 1_m\rangle|)=mn[/imath]. Hence the direct product is cyclic. Is my proof correct? | 445907 | The product of finitely many cyclic groups is cyclic
How to prove that the direct product of finitely many cyclic groups [imath]C_{n_1}\times C_{n_2}\times\cdots\times C_{n_m}[/imath] is cyclic if the [imath]n_i[/imath]'s are pairwise relatively prime? |
447022 | LU factorization of a 2X2 singular matrix
My book is asking me to show that every matrix of the form [imath]A=\begin{pmatrix}0&a\\0&b\end{pmatrix}[/imath] has a [imath]LU[/imath]-factorization and to show that even if [imath]L[/imath] is a unit lower triangular matrix that the [imath]LU[/imath] factorization is not unique. Doesn't this contradict the theorem for [imath]LU[/imath] factorization which says for a matrix [imath]A[/imath]: If the leading principal minors of [imath]A[/imath] are all nonsingular, then [imath]A[/imath] has an [imath]LU[/imath] decomposition. Clearly, the leading principal minors of [imath]A[/imath] are singular. What does this mean? What is the [imath]LU[/imath] factorization? | 445457 | Show that the LU decomposition of matrices of the form $\left[\begin{smallmatrix}0& x\\0 & y\end{smallmatrix}\right]$ is not unique
How can I show that every matrix of the form [imath]\begin{bmatrix}0& x\\0 & y\end{bmatrix}[/imath] has an [imath]LU[/imath] factorization and that even if [imath]L[/imath] is unit lower triangular there is not a unique factorization? I am completely stuck. Any ideas would be greatly appreciated |
447040 | Characterization of Hilbert spaces
Let [imath]X[/imath] be a Banach space for which there exists a constant [imath]\beta<\infty[/imath] such that for every finite-dimensional subspace [imath]B[/imath] of [imath]X[/imath] , [imath]d(B,\ell_2^n)\le\beta[/imath] (where [imath]\dim B=n[/imath]). Then [imath]X[/imath] is isomorphic to a Hilbert space. [imath]d(B,\ell_2^n):[/imath] Banach-Mazur distance [imath]\ell_2^n=(\mathbb{R}^n,||\cdot||_2)[/imath] Any hints would be appreciated. | 345857 | separable Banach space with Banach-Mazur distances to [imath]\ell_2^n[/imath] bounded must be isomorphic to [imath]\ell_2[/imath]?
If [imath]X[/imath] is a separable infinite-dimensional Banach space and [imath]C\in\mathbb{R}^+[/imath] is an upper bound for the Banach-Mazur distance [imath]d(E,\ell_2^n)[/imath] for all [imath]n\in\mathbb{N}[/imath] and all [imath]n[/imath]-dimensional [imath]E\leq X[/imath], why must [imath]X[/imath] be isomorphic to [imath]\ell_2[/imath]? I've been thinking along the following lines: let [imath]\{x_n:n\in\mathbb{N}\}[/imath] be a countable, dense, linearly independent set in [imath]X[/imath] and let [imath]E_n=\langle x_1,\dots,x_n\rangle[/imath] for all [imath]n[/imath]. Now for each [imath]n[/imath] there's a linear isomorphism [imath]T_n:E_n\rightarrow\ell_2^n[/imath] such that [imath]\|T_n\|\|T_n^{-1}\|\leq C[/imath]. If we can somehow put together these [imath]T_n[/imath] to form a linear isomorphism [imath]T:\cup_nE_n\rightarrow\cup_n\ell_2^n[/imath] = {finite-length sequences in [imath]\ell^2[/imath]}, then this extends by continuity to a linear isomorphism [imath]\bar{T}:X\rightarrow\ell_2[/imath]. But how do we get [imath]T[/imath] from the [imath]T_n[/imath]? Each [imath]T_n[/imath] is only defined on finitely many of the [imath]x_i[/imath]. Maybe some sort of infinite series? Or can we choose each [imath]T_n[/imath] so as to agree with [imath]T_{n-1}[/imath], and still keep the [imath]\|T_n\|\|T_n^{-1}\|\leq C[/imath] property? Many thanks for any help with this! |
447162 | decoupling and integrating differential equations
I am having trouble with the process of decoupling. If I have [imath]\frac{dx}{dt}=-x+y[/imath] [imath]\frac{dy}{dt}=-x-y[/imath] I am trying to figure out how to solve for [imath]x(t)[/imath] and [imath]y(t)[/imath] by decoupling the system so that I only have one variable but I can't seem to get anywhere | 446061 | Getting equation from differential equations
I have: [imath]\dfrac {dx} {dt}[/imath]=[imath]-x+y[/imath] [imath]\dfrac {dy}{dt}[/imath]=[imath]-x-y[/imath] and I am trying to find [imath]x(t)[/imath] and [imath]y(t)[/imath] given that [imath]x(0)=0[/imath] and [imath]y(0)=1[/imath] I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate? |
447471 | Is this function differentiable?
Suppose [imath]f[/imath] is a function defined on [imath]\mathbb{R}^{2}[/imath] as follows: [imath] f(x,y) = \frac{x^{3}-y^{3}}{x^{2}+y^{2}} \text{ if } (x,y) \neq (0,0)[/imath] and [imath]0[/imath] if [imath](x,y) = (0,0)[/imath]. I am asked to prove that the function is not differentiable at [imath](0,0)[/imath]. I calculated the partial derivatives and they are [imath]1[/imath] and [imath]-1[/imath] respectively, in the [imath]x[/imath] and [imath]y[/imath] directions. I am unable to proceed from here. Could any shed some light? EDIT: Based on the comments below, this is what I did: We must show that [imath] \frac{\frac{x^{3} - y^{3}}{x^{2} + y^{2}} -x + y}{\sqrt{x^{2}+y^{2}}}[/imath] does not tend to zero as [imath](x,y) \to (0,0)[/imath]. Now, if we take [imath]y = \alpha x[/imath], we get that this expression is equal to some non-zero constant involving [imath]\alpha[/imath] and hence does not tend to zero. | 357433 | Let [imath]f(x,y)=\frac{x^3-y^3}{x^2+y^2}[/imath]. Is f differentiable in [imath](0,0)[/imath]?
Let [imath]f(x,y)=\frac{x^3-y^3}{x^2+y^2}[/imath] My solution manual says that this function is not diffb. in [imath](0,0)[/imath] because it is not linear. Well my problem is that I don't see why this function is linear, and I also don't see why that would imply that [imath]f[/imath] is not differentiable. |
447141 | Let [imath]A \subset B \subset C[/imath] be subrings. If [imath]C[/imath] is finitely generated as an [imath]A[/imath]-module, is [imath]B[/imath] finitely generated as an [imath]A[/imath]-module?
Let [imath]A \subset B \subset C[/imath] be commutative rings. Suppose [imath]C[/imath] is a finitely generated [imath]A[/imath]-module. Can we conclude that [imath]B[/imath] is a finitely generated [imath]A[/imath]-module? Thoughts: Since [imath]C[/imath] is a finite [imath]A[/imath]-algebra, it is an integral [imath]A[/imath]-algebra, hence [imath]B[/imath] is an integral [imath]A[/imath]-algebra. Thus, it remains to show that [imath]B[/imath] is finitely generated as an [imath]A[/imath]-algebra. If [imath]A[/imath] is Noetherian, then the main claim is true by Prop 7.8 in Atiyah-MacDonald. But what if [imath]A[/imath] is not Noetherian? Note: This is not a homework question. I'm simply trying to gain a better understanding of how finite generation works. | 126946 | [imath]A\subseteq B\subseteq C[/imath] ring extensions, [imath]A\subseteq C[/imath] finite/finitely-generated [imath]\Rightarrow[/imath] [imath]A\subseteq B[/imath] finite/finitely-generated?
Let [imath]A\subseteq B\subseteq C[/imath] be commutative unital rings. Recall that the extension [imath]A \subseteq B[/imath] is finite / of finite type / integral, when [imath]B[/imath] is a finitely generated [imath]R[/imath]-module / when [imath]B[/imath] is a finitely generated [imath]A[/imath]-algebra / when [imath]\forall b \in B[/imath] [imath]\exists[/imath] monic polynomial [imath]f \in A[x][/imath] with [imath]f(b)=0[/imath]. Notation [imath]_AB[/imath] means "[imath]A[/imath]-module [imath]B[/imath]". We know that [imath]A\subseteq C[/imath] is integral iff [imath]A\subseteq B[/imath] and [imath]B\subseteq C[/imath] are integral (Grillet, Abstract Algebra, 7.3.3). Do we also have the following: [imath]A\subseteq C[/imath] is finite [imath]\Leftrightarrow[/imath] [imath]A\subseteq B[/imath] and [imath]B\subseteq C[/imath] are finite. [imath]A\subseteq C[/imath] is of finite type [imath]\Leftrightarrow[/imath] [imath]A\subseteq B[/imath] and [imath]B\subseteq C[/imath] are of finite type. I'm having problems with ([imath]A\subseteq C[/imath] finite [imath]\Rightarrow[/imath] [imath]A\subseteq B[/imath] finite) and with ([imath]A\subseteq C[/imath] of finite type [imath]\Rightarrow[/imath] [imath]A\subseteq B[/imath] of finite type). If [imath]_AB[/imath] is a direct summand of [imath]_AC[/imath] (for example when [imath]A[/imath] is a field), i.e. [imath]_AC= _A B\oplus _A B'[/imath] for some submodule [imath]B'[/imath] of [imath]C[/imath], then [imath]C=Ac_1+\cdots+Ac_n[/imath] implies [imath]c_i=b_i+b'_i[/imath] for some [imath]b_i\in B[/imath] and [imath]b'_i \in B[/imath], hence [imath]B = Ab_1 + \cdots + Ab_n[/imath]. But what if [imath]_AB[/imath] is not a direct summand of [imath]_AC[/imath]? And what about the 'finite type' case? |
447874 | If [imath]Df(c)=0[/imath] [imath]\forall c\in V[/imath] then [imath]f[/imath] is constant on [imath]V[/imath]
Theorem: Suppose that [imath]V[/imath] is open and connected in [imath]\Bbb R^n[/imath] [imath]f:V\to \Bbb R^m[/imath] is differentiable on [imath]V[/imath] If [imath]Df(c)=0[/imath] [imath]\forall c\in V[/imath] then [imath]f[/imath] is constant on [imath]V[/imath] I want to prove this theorem with mean value theorem for real valued functions. How to prove this? | 447524 | Why does zero derivative imply a function is locally constant?
I've been trying to prove to myself that if [imath]\Omega[/imath] is an open connected set in [imath]\mathbb{R}^n[/imath], then if [imath]f\colon\Omega\to\mathbb{R}^m[/imath] is a differentiable function such that [imath]f'(x)=0[/imath] for all [imath]x\in\Omega[/imath], then [imath]f[/imath] is constant. I've reduced the problem to just showing [imath]f[/imath] is locally constant on [imath]\Omega[/imath]. Given [imath]x_0\in\Omega[/imath], I know that [imath] \lim_{x\to x_0}\frac{\|f(x)-f(x_0)-f'(x_0)(x-x_0)\|}{\|x-x_0\|}=\lim_{x\to x_0}\frac{\|f(x)-f(x_0)\|}{\|x-x_0\|}=0. [/imath] This implies [imath]\lim_{x\to x_0}\|f(x)-f(x_0)\|=0[/imath]. So for any [imath]\epsilon>0[/imath], there exists some open ball [imath]B(x_0,\delta)[/imath] around [imath]x_0[/imath] in [imath]\Omega[/imath] such that [imath]\|f(x)-f(x_0)\|<\epsilon[/imath] for [imath]x\in B(x_0,\delta)[/imath]. But since the choice of the open ball changes with [imath]\epsilon[/imath], I don't think I can conclude [imath]\|f(x)-f(x_0)\|=0[/imath] for [imath]x\in B[/imath]. So this doesn't convince me that there actually exists a neighborhood of [imath]x_0[/imath] on which [imath]f[/imath] is constant. How can you make the leap from zero derivative to locally constant? |
448168 | calculating this improper integral
I have problems when calculating this improper integral [imath]\int_{-\infty}^{\infty}\frac{e^{-x^2}}{1+x^2}e^{\frac{2x^2t}{1+t}}dx~= ~?~~~|t|<1[/imath] assistance would be helpful. | 447663 | improper integral calculation
I do not know how to solve this integral and then expand the result in powers of [imath]t[/imath] [imath](1-t)^{-1/2}\int_{-\infty}^{\infty}\frac{\exp\left(\frac{2x^2t}{1+t}\right)}{1+x^2}dx=?,~~~~|t|<1.[/imath] any idea? |
448621 | When is a cellular automaton "bidirectional"?
From this paper. "A (bi-directional, deterministic) cellular automaton is a triplet [imath]A = (S;N;\delta)[/imath], where [imath]S[/imath] is an non-empty state set, [imath]N[/imath] is the neighborhood system, and [imath]\delta[/imath] is the local transition function (rule). This function defines the rule of calculating the cell’s state at [imath]t +1[/imath] time step, given the states of the neighborhood cells at previous time step [imath]t[/imath]" I haven't found other "bidirectional" CA via Google, so I wonder what does the term say about this automaton's properties. Does it mean that the state [imath]t+1[/imath] can be used to find out the state [imath]t[/imath]? | 447927 | When is a cellular automaton "bidirectional"?
From this paper. "A (bi-directional, deterministic) cellular automaton is a triplet [imath]A = (S;N;\delta)[/imath], where [imath]S[/imath] is an non-empty state set, [imath]N[/imath] is the neighborhood system, and [imath]\delta[/imath] is the local transition function (rule). This function defines the rule of calculating the cell’s state at [imath]t +1[/imath] time step, given the states of the neighborhood cells at previous time step [imath]t[/imath]" I haven't found other "bidirectional" CA via Google, so I wonder what does the term say about this automaton's properties. Does it mean that the state [imath]t+1[/imath] can be used to find out the state t? |
448691 | in a finite ring, left inverse implies right inverse
Let [imath]R[/imath] be a finite ring with [imath]1\neq 0[/imath]. Suppose there are [imath]x,y\in R[/imath] such that [imath]xy=1[/imath]. I have to prove that this implies [imath]yx=1[/imath]. I have read the question Left inverse implies right inverse in a finite ring but I dont understand why [imath]xy=1[/imath] implies that left multiplication with [imath]y[/imath] is one to one. Could you explain? Thanks. | 138541 | Left inverse implies right inverse in a finite ring
Let [imath]R[/imath] be a finite ring with identity [imath]$1$[/imath], and assume [imath]\exists x,y\in R[/imath] such that [imath] xy=1[/imath]. How can I show it implies [imath]yx=1[/imath]? |
336678 | Prove that the integral operator is bounded
Prove that the following operator is bounded on [imath]L^{2}(0, \infty)[/imath]: [imath]Af(x)[/imath] = [imath]\frac{1}{\pi} \int_{0}^{\infty} \frac{f(y)}{x+y}dy[/imath] with [imath]||A|| \le 1[/imath]. Attempt at Solution It can be shown that: [imath]\int_{0}^{\infty} \frac{|f(y)|}{|x+y|}dy \le [\omega(x)\int_{0}^{\infty} \frac{|f(y)|^{2}}{|x+y|}\omega(y)^{-1}dy]^{\frac{1}{2}}[/imath] I am trying to show that: [imath]\int_{0}^{\infty}\frac{1}{\pi}\omega(x)\int_{0}^{\infty} \frac{|f(y)|^{2}}{|x+y|}\omega(y)^{-1}dy = \frac{1}{\pi}\int_{0}^{\infty} \frac{1}{(1+y)y^{\frac{1}{2}}}\int_{0}^{\infty}|f(y)|^{2} = \int_{0}^{\infty}|f(y)|^{2}[/imath] I can't seem to find a suitable choice of [imath]\omega(x)[/imath] such that result holds. | 384546 | Show that the linear operator [imath](Tf)(x)=\frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy[/imath] satisfies [imath]\|T\|\leq 1[/imath].
Show that the linear operator [imath]T[/imath] given by [imath](Tf)(x) = \frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy[/imath] is bounded on [imath]L^2(0, \infty)[/imath] with norm [imath]||T|| \leq 1[/imath]. The professor also wrote, [imath]``[/imath]In fact, you should observe that [imath]||T|| = 1[/imath], where [imath]||T|| = \sup\limits_{f\neq 0}\frac{||Tf||_2}{||f||_2}."[/imath] |
449151 | [imath]M\le \max_{t\in [0,1]}|at+b\sin t-t^2|\;\;,\;\;\forall \;a,b\in \mathbb{R}[/imath]
Calculate the best value of [imath]M[/imath] [imath]\large M\le \max_{t\in [0,1]}|at+b\sin t-t^2|\;\;,\;\;\forall \;a,b\in \mathbb{R}[/imath] Any hints would be appreciated. | 449078 | Distance from a point to a plane in normed spaces
How can we calculate the distance from a point to a plane in normed spaces ? where we are not inner product, for example: Calculate the distance in [imath](C[0,1],||\cdot||)[/imath] endowed with the supremum norm [imath]||f||=\sup_{t\in[0,1]}|f(t)|[/imath] from the function [imath]f(t)=t^2[/imath] to the linear hull of the functions [imath]g(t)=t[/imath] and [imath]h(t)=\sin(t)[/imath] Any hints would be appreciated. |
449255 | Show that [imath]f=g[/imath] a.e on [imath][a,b][/imath] implies thats [imath]f=g[/imath] on [imath][a,b][/imath].
I'm reading Real Analysis by Royden (4th edition). The problem I'm working on is in the title, were [imath]f[/imath] and [imath]g[/imath] are continuous functions on [imath][a,b][/imath]. What I know so far is that the set [imath]\{x\,:\,f(x)\neq g(x)\}[/imath] has measure zero. Also, I realized that [imath](f-g)^{-1}(R-\{0\})=\{x\,:\,f(x)\neq g(x)\}[/imath]. (R for the reals) What I'm thinking is to some how show that this set is empty, that way [imath]f=g[/imath] on all of [imath][a,b][/imath]. Thanks for any hints or feedback! | 231103 | Does [imath]f(x)[/imath] is continuous and [imath]f = 0[/imath] a.e. imply [imath]f=0[/imath] everywhere?
I wanna prove that "if [imath]f: \mathbb{R}^n \to \mathbb{R}[/imath] is continuous and satisfies [imath]f=0[/imath] almost everywhere (in the sense of Lebesgue measure), then, [imath]f=0[/imath] everywhere." I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain [imath]\mathbb{R}^n[/imath] is restricted to [imath]\Omega \subseteq \mathbb{R}^n[/imath] that contains a neighborhood of the origin "[imath]0[/imath]"? |
449358 | True Or not: Compact iff every continuous function is bounded
Let [imath]X[/imath] be a topological space. My question is: If [imath]f:X\to \mathbb{R}[/imath] is bounded for all such continuous [imath]f[/imath], then is [imath]X[/imath] compact. Is is really? If [imath]X[/imath] is the subset of [imath]\mathbb{R}^d[/imath], then it is clear, beacause with Heine-Borel we get what we want (closed and bounded (with the help of the norm)), but is it true in general? I really hope there exists a non-compact space with the property above. | 181367 | Pseudocompactness does not imply compactness
It is well known that compactness implies pseudocompactness; this follows from the Heine–Borel theorem. I know that the converse does not hold, but what is a counterexample? (A pseudocompact space is a topological space [imath]S = \langle X,{\mathfrak I}\rangle[/imath] such that every continuous function [imath]f:S\to\Bbb R[/imath] has bounded range.) |
449423 | Help with this challenge of the second degree equation
Mathematics course, and I love this science, I am very curious, and these days, studying a book of calculus found in (book from the library of the University) a sheet full of challenges, to share with a colleague, he gave me this challenge, which he also failed to respond ... I tried, I tried, I used the formulas of quadratic equation, and I've been trying since Friday (here in Brazil is Sunday at 22:33), and could not, and I ask your help, and please believe me, I struggled, tried to make substitutions, tried both, and I'm confused ... I must find the values of [imath] a, b [/imath] and [imath] c [/imath]? Consider a parabola whose roots, [imath]x'[/imath] and [imath]x''[/imath], are such that [imath]x'=-x''[/imath] where [imath]x'> 0[/imath]. Define [imath]A = (x', 0)[/imath] [imath]B = (x'', 0)[/imath] and let [imath]V[/imath] be the vertex of the parabola. If the equation of this parabola is [imath]y = ax^2 +bx+ c[/imath] calculate the value of the discriminant [imath]\Delta=b^2-4ac[/imath] so that the triangle AVB is equilateral. | 448746 | A challenge on the parabola
Good Night ... Here I am editing the question and explaining the purpose of the same ... Mathematics course, and I love this science, I am very curious, and these days, studying a book of calculus found in (book from the library of the University) a sheet full of challenges, to share with a colleague, he gave me this challenge, which he also failed to respond ... I tried, I tried, I used the formulas of quadratic equation, and I've been trying since Friday (here in Brazil is Sunday at 22:33), and could not, and I ask your help, and please believe me, I struggled, tried to make substitutions, tried both, and I'm confused ... I must find the values of [imath] a, b [/imath] and [imath] c [/imath]? Consider a parabola whose roots, other are [imath]x'[/imath] and [imath]x''[/imath], and such that [imath]x'=-x''[/imath] where [imath]x'> 0[/imath]. Let [imath]A = (x', 0)[/imath] [imath]B = (x'', 0)[/imath] and [imath]V[/imath] the vertex of the parabola. Suppose that the equation of this parabola is [imath]y = ax^2 +bx+ c[/imath]. Calculate the value of the discriminant [imath]\Delta=b^2-4ac[/imath], so that the triangle is equilateral AVB. |
449484 | A question about the Littlewood-Paley decomposition.
Let [imath]\{f_k(x)\}_{k=0}^\infty[/imath] be a Littlewood-Paley decompositon, that is, [imath] f_k \in C_c^\infty [/imath] [imath] \sum_{k=0}^\infty f_k (x) = 1,[/imath] [imath] \text{supp} f_0 \subset \{ |x| \leq 2 \},[/imath] [imath] \exists f \in C_c^\infty \; \text{such that}\; \text{supp} f \subset \{ 2^{-1} \leq |x| \leq 2 \} \; \text{satisfying}\; f_k (x) = f(x/2^k).[/imath] Then I hope to show that there exist [imath]C, C'>0[/imath] such that [imath] C' \sum_{k=0}^\infty \| f_k(g) \|_{L^2}^2 \le \| g \|_{L^2}^2 \le C \sum_{k=0}^\infty \| f_k(g) \|_{L^2}^2[/imath] for [imath]g \in L^2 (\mathbb R)[/imath]. Here [imath]C_c^\infty[/imath] means [imath]C^\infty[/imath] functions with compact support and [imath]\text{supp}[/imath] means the support. | 449345 | A property of Littlewood-Paley decomposition
Let [imath]f_j \in C_c^\infty[/imath] and assume there exists [imath]M>0[/imath] such that [imath]\text{supp}f_0 \subset \{ |x| \le M \}[/imath] and [imath]\text{supp} f \subset \{ 1/M \le |x| \le M \}[/imath]. Define [imath]f_j[/imath] by [imath] f_j (y) = f(y/2^j) [/imath] for [imath]j =1, 2, \cdots[/imath]. Suppose that [imath] \sum_{j=0}^\infty f_j (x) = 1.[/imath] Then how can I show that there exist [imath]C, C'>0[/imath] such that[imath] C' \sum_{j=0}^\infty \| f_j (g)\|_{L^2}^2 \le\| g \|_{L^2}^2 \le C \sum_{j=0}^\infty \| f_j (g)\|_{L^2}^2?[/imath] [imath]C_c^\infty[/imath] means [imath]C^\infty[/imath] functions with compact support and [imath]\text{supp}[/imath] means the support. |
313489 | How many triangles are formed by [imath]n[/imath] chords of a circle?
This is a homework problem I have to solve, and I think I might be misunderstanding it. I'm translating it from Polish word for word. [imath]n[/imath] points are placed on a circle, and all the chords whose endpoints they are are drawn. We assume that no three chords intersect at one point. a) How many parts do the chords dissect the disk? b) How many triangles are formed whose sides are the chords or their fragments? I think the answer to a) is [imath]2^n[/imath]. But I couldn't find a way to approach b), so I calculated the values for small [imath]n[/imath] and asked OEIS about them. I got A006600. And it appears that there is no known formula for all [imath]n[/imath]. This page says that [imath]\text{triangles}(n) = P(n) - T(n)[/imath] where [imath]P(n)[/imath] is the number of triangles for a convex n-gon in general position. This means there are no three diagonal through one point (except on the boundary). (There are no degenarate corners.) This number is easy to calculate as: [imath]P(n) = {n\choose 3} + 4{n\choose 4} + 5{n\choose5} + {n\choose6} = {n(n-1)(n-2)(n^3 + 18 n^2 - 43 n + 60)\over720}[/imath] The four terms count the triangles in the following manner [CoE76]: [imath]n\choose3[/imath]: Number of trianges with 3 corners at the border. [imath]4{n\choose4}[/imath]: Number of trianges with 2 corners at the border. [imath]5{n\choose5}[/imath]: Number of trianges with 1 corners at the border. [imath]n\choose6[/imath]: Number of trianges with 0 corners at the border. [imath]T(n)[/imath] is the number of triple-crossings (this is the number of triples of diagonals which are concurrent) of the regular [imath]n[/imath]-gon. It turns out that such concurrences cannot occur for n odd, and, except for obvious cases, can only occur for [imath]n[/imath] divisible by [imath]6[/imath]. Among other interesting results, Bol [Bol36] finds values of n for which [imath]4[/imath], [imath]5[/imath], [imath]6[/imath], and [imath]7[/imath] diagonals are concurrent and shows that these are the only possibilities (with the center for exception). The function [imath]T(n)[/imath] for [imath]n[/imath] not divisible by [imath]6[/imath] is: [imath]T(n) = {1\over8}n(n-2)(n-7)[2|n] + {3\over4}n[4|n].[/imath] where [imath][k|n][/imath] is defined as [imath]1[/imath] if [imath]k[/imath] is a divisor of [imath]n[/imath] and otherwise [imath]0[/imath]. The intersection points need not lie an any of lines of symmetry of the [imath]2m[/imath]-gon, e. g. for [imath]n=16[/imath] the triple intersection of [imath](0,7),(1,12),(2,14)[/imath]. If I understand the text correctly, it doesn't give a general formula for [imath]T(n)[/imath]. Also I've found a statement somewhere else that some mathematician wasn't able to give a general formula solving this problem. I haven't found a statement that it is still an open problem, but it looks like it to me. So am I just misunderstanding the problem, or misunderstanding what I've found on the web, or maybe it is indeed a very hard problem? It's the beginning of the semester, our first homework, and it really scares me. | 439093 | [imath]n[/imath] Lines in the Plane
How am I to "[u]se induction to show that [imath]n[/imath] straight lines in the plane divide the plane into [imath]\frac{n^2+n+2}{2}[/imath] regions"? It is assumed here that no two lines are parallel and that no three lines have a common point. Further, this is not a non-Euclidean problem, but I wouldn't mind a discussion on the non-Euclidean nature of the problem. I was thinking it might be easier to show using the unit sphere. A GOOD REFERENCE I FOUND THAT WILL BE USEFUL TO ANYONE COMING ACROSS THIS PROBLEM IS Concrete Mathematics, Graham, Knuth, Patashnik pp. 4–8. |
449156 | Equation involved in generating function of divisor function
There is an identity between the divisor function of the odd numbers and the "odd" divisor function of power [imath]3[/imath](I don't know if there is a name for function for this type, if there is , sorry for my ignorance), that is [imath]\left(\sum_{n=1}^{\infty}\sigma_1(2n-1)x^{2n-1}\right)^2=\sum_{k=1}^{\infty}\left(\sum_{d|n,\frac{n}{d}\mbox{ is odd}}d^3\right)x^k[/imath] where [imath]\sigma_1(n)[/imath] is the sum of all positive divisors of [imath]n[/imath]. This is equivalent to proving [imath]\left(\sum_{n=1}^{\infty}\sigma_1(2n-1)x^{2n-1}\right)^2=\sum_{k=1}^{\infty}\frac{k^3 x^k}{1-x^{2k}}[/imath] I tried some calculation via Mathemtica, it seems to be true.This sequence seems to be OEIS A007331, but I cannot find some way in cracking this problem. The square seems annoying and I can't get rid of it. Thanks for your attention and helping hand. | 445680 | How prove this [imath]\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}[/imath]
Given the positive integer numbers [imath]n[/imath],prove that [imath]\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}[/imath] where [imath]\sigma(n)[/imath] is defined as [imath]\sigma{(N)}=\sum_{d|N}d[/imath] This problem I think is very interesting, and have nice methods? Thank you everyone. and I find some of this or can see:http://www.advancesindifferenceequations.com/content/pdf/1687-1847-2013-84.pdf |
449703 | Does [imath]2^{\mathfrak m}=2^{\mathfrak n}[/imath] imply [imath]\mathfrak m=\mathfrak n[/imath]?
Suppose [imath]\mathfrak m[/imath] and [imath]\mathfrak n[/imath] are infinite cardinals. Does [imath]2^{\mathfrak m}=2^{\mathfrak n}[/imath] imply [imath]\mathfrak m=\mathfrak n[/imath]? | 376509 | Bijection between power sets of sets implies bijection between sets?
Is it true that if [imath]X[/imath] and [imath]Y[/imath] are sets and there is a bijection between [imath]\mathcal{P}(X)[/imath] and [imath]\mathcal{P}(Y)[/imath] then there is a bijection from [imath]X[/imath] to [imath]Y[/imath] ?. I believe this should be obvious, but I can't see why this is so. A proof or a counter example would be highly appreciated. Thanks. |
514 | Conjectures that have been disproved with extremely large counterexamples?
I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture. I'm sure that everyone here is familiar with it; it describes an operation on a natural number – [imath]$n/2$[/imath] if it is even, [imath]$3n+1$[/imath] if it is odd. The conjecture states that if this operation is repeated, all numbers will eventually wind up at [imath]$1$[/imath] (or rather, in an infinite loop of [imath]$1-4-2-1-4-2-1$[/imath]). I fired up Python and ran a quick test on this for all numbers up to [imath]$5.76 \times 10^{18}$[/imath] (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at [imath]$1$[/imath]. Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.) I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?" To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!" And he said, "It is my conjecture that there are none! (and if any, they are rare)". Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples? | 2423879 | Conjectures that are false for very large numbers
Hi I would like to teach my students about false conjectures and computer approaches to them. I need references and/or direct examples of statements of the form [imath](\forall n \geq n_0)P (n)[/imath] where [imath]P[/imath] is a relatively simple property (e.g. of elementary number theory) such that the smallest number [imath]m[/imath] for which [imath]P (m) [/imath] is false is ridiculously bigger than [imath]n_0[/imath]. |
111440 | Examples of patterns that eventually fail
Often, when I try to describe mathematics to the layman, I find myself struggling to convince them of the importance and consequence of "proof". I receive responses like: "surely if Collatz is true up to [imath]20×2^{58}[/imath], then it must always be true?"; and "the sequence of number of edges on a complete graph starts [imath]0,1,3,6,10[/imath], so the next term must be 15 etc." Granted, this second statement is less logically unsound than the first since it's not difficult to see the reason why the sequence must continue as such; nevertheless, the statement was made on a premise that boils down to "interesting patterns must always continue". I try to counter this logic by creating a ridiculous argument like "the numbers [imath]1,2,3,4,5[/imath] are less than [imath]100[/imath], so surely all numbers are", but this usually fails to be convincing. So, are there any examples of non-trivial patterns that appear to be true for a large number of small cases, but then fail for some larger case? A good answer to this question should: be one which could be explained to the layman without having to subject them to a 24 lecture course of background material, and have as a minimal counterexample a case which cannot (feasibly) be checked without the use of a computer. I believe conditions 1. and 2. make my question specific enough to have in some sense a "right" (or at least a "not wrong") answer; but I'd be happy to clarify if this is not the case. I suppose I'm expecting an answer to come from number theory, but can see that areas like graph theory, combinatorics more generally and set theory could potentially offer suitable answers. | 809912 | Statements with rare counter-examples
This is a soft question. I'm searching for examples of mathmatical statements (preferably in number theory, but other topics are also fine), that seem to be true, but are actually not. Statements where observing some examples would let you think it is always true, but then there is a well hidden counter-example. If Riemann's [imath]\zeta[/imath] function had a zero beside the critical line, this would be such an example I'm looking for. Or if Fermat's Last Theorem would be false. Do you know any such surprising counter-exmaples? |
450420 | Dynamics of entire function
How can we conclude that the transcendental entire functions need not have any fixed points?For example consider [imath]f(z)=e^z+z,[/imath] why is it so? | 31953 | Zeros of exponential
I'm just want to be sure if the function [imath]f(z)=e^{-iz}, z\in \mathbb C[/imath], has no complex or real zeros?? |
450422 | What is the convergence speed of logistic sequence?
I am looking at the sequence [imath]x_{n+1}=r\, x_n(1-x_n)[/imath] where [imath]r=1[/imath]. Let's choose [imath]x_1=1/2[/imath] so as to make the sequence convergent to 0. My question is: precisely how quickly does this sequence approach zero? From my numerical experiments [imath]\lim_{n\to\infty}n\,x_n=1[/imath] seems likely. (That is [imath]\mathrm O(1/n)[/imath] convergence rate.) Do you have more precise result or some mathematical proof? | 49676 | How fast does the sequence [imath]y_t[/imath] defined by [imath]y_{t+1}=y_t(1-y_t)[/imath] decay to zero?
The question is in the title; I'm looking for the exact decay rate. Naturally, assume the starting point [imath]y_0[/imath] belongs to [imath](0,1)[/imath]. This is motivated by one of the answers to a previous question of mine. |
450634 | Cardinality of Solutions to an Inequality
Show that the number of solutions in nonneg. int. of the ineq. [imath]x_1+x_2+\cdots +x_n\leq M,[/imath] where [imath]M[/imath] is a nonneg. int., is [imath]C(M+n,n)[/imath]. | 450554 | [imath]x_1+x_2+\cdots+x_n\leq M[/imath]: Cardinality of Solution Set is [imath]C(M+n, n)[/imath]
Show that the number of solutions in nonnegative integers of the inequality [imath]x_1+x_2+\cdots+x_n\leq M,[/imath] where [imath]M[/imath] is a nonnegative integer, is [imath]C(M+n, n)[/imath]. |
428926 | Find [imath]\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{3^n} + {5^n}}}[/imath]
Find the limit [imath]\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{3^n} + {5^n}}}.[/imath] Thanks. | 80340 | Convergence of [imath]\sqrt[n]{x^n+y^n}[/imath] (for [imath]x, y > 0[/imath])
How can I prove the convergence of the sequence [imath]b_n=\sqrt[n]{x^n+y^n}[/imath] where [imath]x, y > 0[/imath]? Can I divide it in two cases? Case 1: [imath]x > y[/imath]. [imath] b_n=\sqrt[n]{x^n+y^n} < \sqrt[n]{x^n+x^n} = \sqrt[n]{2 \cdot x^n}=x \cdot \sqrt[n]{2} [/imath] Case 2: [imath]x < y[/imath]. [imath] b_n=\sqrt[n]{x^n+y^n} < \sqrt[n]{y^n+y^n} = \sqrt[n]{2 \cdot y^n}=y \cdot \sqrt[n]{2}[/imath] Result: Does the sequence converges to [imath]\max(x,y)[/imath]? |
451220 | "Question: Show that [imath]n^5 - n[/imath] is divisible by 30; for all natural n"
Show that [imath]n^5 - n[/imath] is divisible by [imath]30;[/imath] [imath]\forall n\in \mathbb{N}[/imath] I tried to solve this three-way. And all stopped at some point. I) By induction: testing for [imath]0[/imath], [imath]1[/imath] and [imath]2[/imath] It is clearly true. As a hypothesis, we have [imath]30|n^5-n\Rightarrow n^5-n=30k[/imath]. Therefore, the thesis would [imath]30|(n+1)^5-n-1\Rightarrow (n+1)^5-n-1=30j[/imath]. By theorem binomial [imath](n+1)^5-n-1=n^5+5n^4+10n^3+10n^2+5n+1-n-1[/imath] [imath]=30k+5n^4+10n^3+10n^2+5n.[/imath] It was a little messy and not given to proceed. II)I tried by Fermat's Little Theorem: [imath]30|n^5-n\Rightarrow 5\cdot3\cdot2|n^5-n [/imath] Analyzing each case, we have; clearly [imath]5|n^5-n[/imath] (Fermat's Little Theorem). Now cases [imath]3|n^5-n[/imath] and [imath]2|n^5-n[/imath] I could not develop. III)In this I wanted your help especially like to solve using this because this exercise on this handout (Properties of the Greatest Common Divisor). Would show that [imath]gcd(n^5-n,30)=30[/imath] Are detailers, please, because'm coursing Theory of the numbers the first time. In the third semester of the degree course in mathematics What would be the idea to solve this by using the Greatest Common? This problem in this handout GCD? | 132210 | How to prove [imath]n^5 - n[/imath] is divisible by 30 without reduction
How can I prove that prove [imath]n^5 - n[/imath] is divisible by 30? I took [imath]n^5 - n[/imath] and got [imath]n(n-1)(n+1)(n^2+1)[/imath] Now, [imath]n(n-1)(n+1)[/imath] is divisible by 6. Next I need to show that [imath]n(n-1)(n+1)(n^2+1)[/imath] is divisible by 5. My guess is using Fermat little theory but I don't know how.. |
451007 | Asymptotic Equivalence of Another Recurrence:
So I have the following recurrence relation: [imath]f(n) = f(n-1) + f(\lceil n/2\rceil)+ 1[/imath] I already know that: If: [imath]g(n) = g(n-1) + 1[/imath] [imath]g(n) = O(n)[/imath] If: [imath]g(n) = g(\lceil n/2\rceil) + 1[/imath] [imath]g(n) = O(\log(n))[/imath] If: [imath]g(n) = g(n-1) + g(\lceil n/2\rceil)[/imath] [imath]g(n) = O(nlog(n))$)[/imath] Thus can I definitely conclude that: f(n) = O(n \log(n) * n * \log(n)) = O(n^2 \log(n)^2)$$ | 450802 | Strange Recurrence: What is it asymptotic to?
So I have the following recurrence relation for the growth rate of an algorithm: [imath]T(n)[/imath] = time taken by algorithm to solve problem of size n: [imath]T(n) = T(n-1) + T(\lceil(n/2)\rceil)[/imath] Clearly this does not obey any polynomial law but now my question is what does it obey? I merely want an asymptotic equivalence. My guess is subexponential since this is faster than polynomial but clearly slower than any exponential form. What though is the question.... |
451149 | Showing a isomorphism exists between a quotient group and another group
We have [imath]G_1 = GL(2,\mathbb{R}) \times GL(2,\mathbb{R})[/imath] and [imath]G_2 = \{(C,D) \in G_1 \mid \det(C) = \det(D)\}[/imath]. We want to show that [imath]G_1/G_2 \cong {(\mathbb{R^*},\times})[/imath]. So my approach for this is that we know [imath]G_1/G_2 = \{(A,B)G_2:(A,B) \in G_1\}[/imath] and construct the map [imath]f:G_1 \rightarrow \mathbb{R^*}[/imath] given by [imath]f(A,B) = \det(AB)[/imath]. But I am having trouble showing that this map is isomorphic, particularly showing that its one - one. | 450659 | Factor groups of matrices
Let [imath]G=GL(2,\mathbb R)\oplus GL(2,\mathbb R)[/imath] and let [imath]H=\{(A,B)\in G\mid \det(A)=\det(B)\}[/imath]. Prove that [imath]G/H \simeq (\mathbb R^*,\times)[/imath]. I'm guessing I should use: "Let [imath]G[/imath] be a group and let [imath]H[/imath] be a normal subgroup of [imath]G[/imath]. The set [imath]G/H = \{aH\mid a \in G\}[/imath] is a group under the operation [imath](aH)(bH)=abH[/imath]." |
452137 | n is+ve integer, how many solutions [imath](x,y)[/imath] exist for [imath]\frac{1}{x} + \frac{1}{y} = \frac{1}{n}[/imath] with [imath]x[/imath], [imath]y[/imath] being positive integers and [imath](x \neq y)[/imath]
I wanted to know, how can i solve this. For a given positive integer n, how many solutions [imath](x,y)[/imath] exist for [imath]\frac{1}{x} + \frac{1}{y} = \frac{1}{n}[/imath] with [imath]x[/imath] and [imath]y[/imath] being positive integers and [imath](x \neq y)[/imath]. | 403036 | Natural number solutions to [imath]\frac{xy}{x+y}=n[/imath] (equivalent to [imath]\frac 1x+\frac 1y=\frac 1n[/imath])
I have a question about the following problem from a Putnam review: Let [imath]n\in \mathbb{N}[/imath]. Find how many pairs of natural numbers [imath](x, y)\in \mathbb{N}\times \mathbb{N}[/imath] solve [imath] \frac{xy}{x+y}=n. [/imath] I have found some solutions for all [imath]n[/imath], such as [imath](2n, 2n)[/imath] and [imath](n+1, n(n+1))[/imath], but I feel as though this is the wrong approach, as the question is only asking for the number of solutions to the equation. I don't really want a complete solution, but any hints would be greatly appreciated. |
452261 | Commutativity of ring of order [imath]p^2[/imath] with unity [imath]e[/imath] and characteristic [imath]p[/imath]
Let [imath]R[/imath] be a finite ring of order [imath]p^2[/imath] with unity [imath]e[/imath] and characteristic [imath]p[/imath]. This ring is commutative but I cannot get why it is. I know that this ring looks as [imath]\mathbb Z /p\mathbb Z \times Z /p\mathbb Z[/imath]. Could anyone help me to show the commutativity of this ring? | 305512 | Ring of order [imath]p^2[/imath] is commutative.
I would like to show that ring of order [imath]p^2[/imath] is commutative. Taking [imath]G=(R, +)[/imath] as group, we have two possible isomorphism classes [imath]\mathbb Z /p^2\mathbb Z[/imath] and [imath]\mathbb Z/ p\mathbb Z \times \mathbb Z /p\mathbb Z[/imath]. Since characterstic must divide the size of the group then we have two possibilities [imath]p[/imath] and [imath]p^2[/imath]. Now IU don't understand how can I reason to say that the multiplication is commutative and how can I conclude for the case when characterstic is [imath]p[/imath]? |
452337 | Suppose [imath]H \le G[/imath] and [imath]g^2\in H[/imath] for all [imath]g\in G[/imath]. Show [imath]H[/imath] is a normal subgroup of [imath]G[/imath]
Let [imath]G[/imath] be a group and [imath]H[/imath] a subgroup of [imath]G[/imath]. Suppose [imath]g^2\in H[/imath] for all [imath]g\in G[/imath]. Show [imath]H[/imath] is a normal subgroup of [imath]G[/imath]. I tried lots of methods, but failed. Any suggestion? Thanks. | 398422 | Let [imath]H[/imath] be a subgroup of a group [imath]G[/imath] such that [imath]x^2 \in H[/imath] , [imath]\forall x\in G[/imath] . Prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]
Let [imath]H[/imath] be a subgroup of a group [imath]G[/imath] such that [imath]x^2 \in H[/imath], [imath]\forall x\in G[/imath]. Prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]. I have tried to using the definition but failed. Can someone help me please. |
452410 | How prove this [imath]f(0)=f(1)[/imath],
let [imath]f:[0,1]\longrightarrow [0,1][/imath] be a continuous function such that [imath]f(0)=f(1)[/imath], show that : if the number [imath]l[/imath] is not of the form [imath]\dfrac{1}{n}[/imath] there exsits a function of this form on whose graph one cannot inscrible a horizontal chord of length [imath]l[/imath] This problem is from $Mathematical Analysis (Zorich) P170,problem 7(2),Thank you evryone | 36765 | Constructing a continuous function whose graph seems 'special'
I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise. Let [imath]f: [0,1]\to \mathbb R[/imath] be a continuous function such that [imath]f(0)=f(1)[/imath]. Show that for any [imath]n\in \mathbb N[/imath] there exists a horizontal closed interval of length [imath]\frac 1n[/imath] with endpoints on the graph of this function; if the number [imath]\ell[/imath] is not of the form [imath]\frac 1n[/imath] there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length [imath]\ell[/imath]. The first part can be proven like this: Consider [imath]g: [0,(n-1)/n] \to \mathbb R[/imath] given by [imath]g(x) = f(x) - f(x+1/n)[/imath]. Then [imath]\sum_{k=0}^{n-1} g(k/n) = 0[/imath] and therefore either all of these points are zero or there exists both a point where [imath]g[/imath] is positive and a point where [imath]g[/imath] is negative. By continuity, there must then also be a point where [imath]g = 0[/imath]. So we are done. Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for [imath]\ell < 1/2[/imath]. (For [imath]\ell > 1/2[/imath] the function [imath]f(x) = \sin(2\pi x)[/imath] will do.) Can anyone help me out? Cheers, |
332763 | Associativity of symmetric difference of sets
By [imath]A\oplus B[/imath] we denote the symmectric difference of two sets. The definition is [imath]A\oplus B =(A\setminus B) \cup (B\setminus A)[/imath]. Now I hope to show that [imath]A\oplus (B\oplus C) = (A\oplus B)\oplus C[/imath]. I remember that there's an elegant proof, but I forget its detail. (The first step is to show [imath]A\oplus (A\oplus B) = B[/imath], and maybe applying this formula, we can obtain associativity.) | 919969 | Prove that [imath]A \Delta (B \Delta C) = (A \Delta B) \Delta C[/imath].
Prove that [imath]A \Delta (B \Delta C) = (A \Delta B) \Delta C[/imath]. [imath]A \Delta (B \Delta C) = A \Delta ((B - C) \cup (C - B))[/imath] [imath]= [A - ((B-C) \cup (C-B))] \cup [((B-C)\cup(C-B)) - A][/imath] [imath]= [A \cap \sim((B-C)\cup(C-B))] \cup [((B-C)\cup(C-B)) \cap \sim A][/imath] From here I am unsure how to proceed and get to what I want. |
453224 | Why write [imath]\mathrm dx, \mathrm dt[/imath] etc. at the beginning of an integral?
I've noticed that many people here (on Math.SE) as well as elsewhere write integrals out like this: [imath]\int^a_b \mathrm dt \; f(t)[/imath] instead of the more common (at least from what I've seen): [imath]\int^a_b f(t) \; \mathrm dt[/imath] Some quantum mechanics books I've seen also do the same (put the [imath]\mathrm dt[/imath] in front.) Are there any practical reasons for this? | 387572 | Notation: Why write the differential first?
From reading answers here, I've noticed that some people write integrals as [imath]\int dx \; f(x)[/imath], while other people write them as [imath]\int f(x)\;dx[/imath]. I realize that there is no mathematical difference between the two notation forms, but was wondering why some people choose the first method over the second. Is there some place in higher maths that it becomes beneficial to write the differential first? (I, personally, have always used the second method, just because I was taught that way...) |
453163 | Proving Every open set in [imath]\Bbb R[/imath] is a countable union of open intervals.
This question is from William R. Wade's Introduction to Analysis book: Prove that every open set in [imath]\Bbb R[/imath] is a countable union of open intervals. I have no ideas honestly. Thank you. | 318299 | Any open subset of [imath]\Bbb R[/imath] is a at most countable union of disjoint open intervals. [Collecting Proofs]
This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome. |
146387 | Fourier transform of [imath] 1/|x|^{k}[/imath]
is ist possible to find the Fourier transform (direct and inverse ) of [imath] f(x)= \frac{1}{|x|^{k}} [/imath] for [imath] k=1,2,3,......[/imath] this function has a severe singularity at [imath]x=0[/imath] so i think this will exists only in the sense of distributions :) | 48430 | ([imath]n[/imath]-dimensional) Inverse Fourier transform of [imath]\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}[/imath]
I'd like to find the [imath]n[/imath]-dimensional inverse Fourier transform of [imath]\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}[/imath] i.e. [imath] \int_{\mathbb{R}^n} \frac{1}{ \| \mathbf{\omega} \|^{2\alpha}} e^{2 \pi i \mathbf{\omega}\cdot \mathbf{x} } d \mathbf{\omega} [/imath] where [imath]\mathbf{x} = ( x_0 , x_1 , \cdots , x_n )[/imath] is a spatial parameter in [imath]\mathbb{R}^n[/imath], [imath]\mathbf{\omega} = ( \omega_0 , \omega_1 , \cdots , \omega_n )[/imath], and [imath] \| \omega\| = \omega_0^2 + \omega_1^2 + \cdots + \omega_n^2 [/imath] All I've been able to come up with in the one-dimensional case is that the integral [imath] \int_{-\infty}^{+\infty} \frac{1}{ \| \omega \|^{2\alpha}} e^{2 \pi i \omega x } d \mathbf{\omega} [/imath] diverges because the lower power terms [imath]\omega^p[/imath] terms, for which [imath]p < 2\alpha[/imath], in expansion of the exponential [imath] e^{2 \pi i \omega x } = \sum_{p = 0}^{\infty} \frac{(2 \pi i \omega x)^p}{p!} [/imath] do not prevent [imath]\frac{1}{\| \omega \|^{2\alpha}}[/imath] from blowing up at the origin. I know that one possible way of regularizing this integral is to include a test function and consider the limit of the resulting integral, but I don't quite know how to do so. I've tried reading Gelfand and Shilov's Gneralized Functions vol 1 and while I understand bits of it on the whole its a bit heavy for me. Based on the papers that I've read I know that there are two cases (the latter of which appears to me more general) and two solutions in each. Case 1: 2[imath]\alpha[/imath] is an odd/even integer Case 2: 2[imath]\alpha[/imath] is integer or otherwise I'd appreciate help, if possible, coming up with both solutions. |
453304 | How to show that [imath]x^6-72[/imath] is irreducible over the rationals?
An exercise states to (try to) prove that the polynomial [imath]x^6-72[/imath] is irreducible over the rationals. As [imath]72=2^33^2[/imath], we cannot apply Eisenstein criterion without any additional trick. I was wondering if there is a neat way to show that this polynomial is irreducible? The only thing that comes to my mind is to deal with undefined coefficients and then solve polynomial systems, but maybe there is a more conceptual way? [That there are no linear factors is obvious. Reductions modulo [imath]2[/imath] and [imath]3[/imath] give [imath]X^6=0[/imath]. Reduction modulo [imath]p=73[/imath] gives (partial) factorization [imath]X^6+1=(X^2+1)(X^4-X^2+1)[/imath], but I am not sure if this information is of any help here.] | 443612 | Proving irreducibility of [imath]x^6-72[/imath]
I have the following question: Is there an easy way to prove that [imath]x^6-72[/imath] is irreducible over [imath]\mathbb{Q}\ [/imath]? I am trying to avoid reducing mod p and then having to calculate with some things like [imath](x^3+ax^2+bx+c)\cdot (x^3+dx^2+ex+f)[/imath] and so on... Thank you very much. |
453380 | Prove a set with this associative operation is a group
Suppose we have a non-empty set [imath]X[/imath] with an associative binary operation, so that for all [imath]x \in X[/imath] there exists a unique [imath]x'[/imath] such that [imath]xx'x=x[/imath]. Prove the set under this operation is a group. Of course, associativity is already given, and unique inverses are obvious, but I have no idea how to go about showing an identity element exists, nor how to show the set is closed with respect to the operation. How can one go about this? | 253514 | A semigroup [imath]X[/imath] is a group iff for every [imath]g\in X[/imath], [imath]\exists! x\in X[/imath] such that [imath]gxg = g[/imath]
The following could have shown up as an exercise in a basic Abstract Algebra text, and if anyone can give me a reference, I will be most grateful. Consider a set [imath]X[/imath] with an associative law of composition, not known to have an identity or inverses. Suppose that for every [imath]g\in X[/imath], there is a unique [imath]x\in X[/imath] with [imath]gxg=g[/imath]. Show that [imath]X[/imath] is a group. Note that I’m not asking for a proof (though a really short one would please me!), just some place where this has been published. |
453603 | If [imath]k[/imath] divides [imath]n[/imath] then there's only one subgroup of [imath]G[/imath] of order [imath]k[/imath]
Let [imath]G[/imath] be a cyclic group of order [imath]n[/imath] generated by, say, [imath]g[/imath]. Show that if [imath]k[/imath] divides [imath]n[/imath], then there is one, and only one, subgroup of [imath]G[/imath] of order [imath]k[/imath], and that this is generated by [imath]g^{n/k}[/imath]. I can see it being true if we take [imath]n[/imath] to be prime. The [imath]k[/imath] is either [imath]1[/imath] or [imath]n[/imath], thus if [imath]k=1[/imath], then the subgroup of order [imath]1[/imath] will be generated by [imath]e[/imath] and if [imath]k=n[/imath], then the subgroup will be [imath]G[/imath] itself generated by [imath]g[/imath] which is true since [imath]n[/imath] is prime. How about in a case where [imath]n[/imath] is not prime? | 410389 | Subgroups of a cyclic group and their order.
Lemma [imath]1.92[/imath] in Rotman's textbook (Advanced Modern Algebra, second edition) states, Let [imath]G = \langle a \rangle[/imath] be a cyclic group. (i) Every subgroup [imath]S[/imath] of [imath]G[/imath] is cyclic. (ii) If [imath]|G|=n[/imath], then [imath]G[/imath] has a unique subgroup of order [imath]d[/imath] for each divisor [imath]d[/imath] of [imath]n[/imath]. I understand how every subgroup must be cyclic and that there must be a subgroup for each divisor of [imath]d[/imath]. But how is that subgroup unique? I'm having trouble understanding this intuitively. For example, if we look at the cyclic subgroup [imath]\Bbb{7}[/imath], we know that there are [imath]6[/imath] elements of order [imath]7[/imath]. So we have six different cyclic subgroups of order [imath]7[/imath], right? Thanks in advance. |
453608 | Show that [imath]\sum\limits_{k=0}^{y-1}(-1)^k\frac{\binom{y-1}{k}}{k+x}=\frac{(x-1)!(y-1)!}{(x+y-1)!}[/imath].
Prove that for [imath]x,y[/imath] positive integers, [imath]\sum_{k=0}^{y-1}(-1)^k\frac{\binom{y-1}{k}}{k+x}=\frac{(x-1)!(y-1)!}{(x+y-1)!}[/imath] One way is to use the beta-gamma functions relation: http://en.wikipedia.org/wiki/Beta_function Is there a direct way to calculate the sum? | 72067 | Is there a combinatorial way to see the link between the beta and gamma functions?
The Wikipedia page on the beta function gives a simple formula for it in terms of the gamma function. Using that and the fact that [imath]\Gamma(n+1)=n![/imath], I can prove the following formula: [imath] \begin{eqnarray*} \frac{a!b!}{(a+b+1)!} & = & \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+1+b+1)}\\ & = & B(a+1,b+1)\\ & = & \int_{0}^{1}t^{a}(1-t)^{b}dt\\ & = & \int_{0}^{1}t^{a}\sum_{i=0}^{b}\binom{b}{i}(-t)^{i}dt\\ & = & \int_{0}^{1}\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}t^{a+i}dt\\ & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\int_{0}^{1}t^{a+i}dt\\ & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\left[\frac{t^{a+i+1}}{a+i+1}\right]_{t=0}^{1}\\ & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\\ b! & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{a!(a+i+1)} \end{eqnarray*} [/imath] This last formula involves only natural numbers and operations familiar in combinatorics, and it feels very much as if there should be a combinatoric proof, but I've been trying for a while and can't see it. I can prove it in the case [imath]a=0[/imath]: [imath] \begin{eqnarray*} & & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(b+1)!}{0!(i+1)}\\ & = & \sum_{i=0}^{b}(-1)^{i}\frac{b!(b+1)!}{i!(b-i)!(i+1)}\\ & = & b!\sum_{i=0}^{b}(-1)^{i}\frac{(b+1)!}{(i+1)!(b-i)!}\\ & = & b!\sum_{i=0}^{b}(-1)^{i}\binom{b+\text{1}}{i+1}\\ & = & b!\left(1-\sum_{i=0}^{b+1}(-1)^{i}\binom{b+\text{1}}{i}\right)\\ & = & b! \end{eqnarray*} [/imath] Can anyone see how to prove it for arbitrary [imath]a[/imath]? Thanks! |
452836 | Why is the 2 norm "special"?
Out of all the vector norms, the [imath]2[/imath] norm, or the Euclidean norm, seems to be "special". Primarily, I say this because we use the 2 norm as a means of determining the distance from one point to another. But what I don't understand is, why do we use the [imath]2[/imath] norm in particular for this? Why not some other norm? I realize that the unit disk of the [imath]2[/imath] norm forms a figure whose outer boundary is equidistant from the origin, but the only way (that I can think of) that we make that determination is by using the [imath]2[/imath] norm to measure the distance from the boundary of the unit disk to the origin. So what is it about the [imath]2[/imath] norm that makes us choose it as the measure of distance from one point to another? What special properties does it have that other norms don't have? | 221367 | Why do we use the Euclidean metric on [imath]\mathbb{R}^2[/imath]?
On the train home, I thought I would try to prove [imath]\pi[/imath] is irrational. I needed a definition, so I used: [imath]\pi[/imath] is the area of the unit circle. But what is a circle? A circle is the set of tuples [imath](x,y)[/imath] satisfying "[imath]x^2 + y^2 \leq 1[/imath]". That seems a little awkward - is there a more natural way to say it? The standard distance metric on [imath]\mathbb{R}^2[/imath] is "[imath]\sqrt{(x1-x2)^2 + (y1-y2)^2}[/imath]". A circle is the unit ball of this metric, and [imath]\pi[/imath] is its area. But why do we use this metric in the first place? Why not use the Taxicab metric - it's linear so it'd seem to satisfy even more nice mathematical properties. Because of the Pythagorean theorem. And what property of [imath]\mathbb{R}^2[/imath] lets us prove the Pythagorean theorem? This one took a little more thought. I think it comes about because we want to associate lengths to lines that are equivalent under translations and rotations. Translations seem pretty natural - but I'm having trouble defining rotations without being circular: If you multiply a coordinate tuple by the appropriate matrix, you get a rotation - but the matrix's entries involve [imath]\sin[/imath] and [imath]\cos[/imath], which I would define as the coordinates on the unit circle(which is circular). I can define the set of all infinite lines that go through the origin. Except for the vertical line, you might assume they all have the form [imath]L_m = \{(x,y)|y = mx\}[/imath] for some [imath]m \in \mathbb{R}[/imath]. Given an [imath]m \in \mathbb{R}[/imath], I can't find an obvious way to get the point on the unit circle corresponding to it. The complex number route just gives [imath]\sin[/imath] and [imath]\cos[/imath]. Circular, again. We can determine whether two vectors are orthogonal using the dot product. That reduces defining the entire unit circle to defining just a single quadrant(if we also assume the length of [imath]-1 * x[/imath] is the same as [imath]x[/imath]). A quick Google search shows circles were a primitive notion in Euclid's Elements. There is another question/answer given here, but I was still left confused as to what a rotation actually is in [imath]\mathbb{R}^2[/imath]: Why is the Euclidean metric the natural choice? So my questions are: Out of all the possible metrics, why choose the Euclidean metric as the natural one? I think it's rotation that's the root cause, but that might end up being a red herring. How do we define a natural equivalence of points up to rotation from first principles? This is related to #1. If we define the set of infinite lines through the origin, you can choose a representative element from each one and form a perimeter curve(assuming a continuous choice function). What properties does the circle have that would make it a natural choice? |
453961 | proof of [imath]\sum_{n=1}^\infty n \cdot x^n= \frac{x}{(x-1)^2}[/imath]
I know that the Series [imath]\sum_{n=1}^\infty n \cdot x^n [/imath] converges to [imath]\frac{x}{(x-1)^2}[/imath] but I'm not sure how to show it. I'm pretty sure that has been asked before, but I wasn't able to find anything... | 2108999 | problem with a series
I have a problem with series [imath]\sum_{k=1}^ \infty k*(\frac{1}{2})^{k+2}[/imath] How can I solve it? |
454426 | Calculating number of functions from a set of size [imath]m[/imath] to a set of size [imath]n[/imath]
In set theory and combinatorics, the cardinal number [imath]n^m[/imath] is the size of the set of functions from a set of size m into a set of size [imath]n[/imath]. I read this from this Wikipedia page. I don't understand, however, why this is true. I reason with this example in which [imath]M[/imath] is a set of size [imath]5[/imath], and [imath]N[/imath] is a set of size [imath]3[/imath]. For each element in set [imath]M[/imath], there are three functions to map the element from the set of size [imath]5[/imath] to an element in the set of size [imath]3[/imath]. By my reasoning, that means the total number of functions is just [imath]3*5[/imath], i.e. [imath]3[/imath] functions for each of the [imath]5[/imath] elements in the set. Why is it actually [imath]3^5[/imath]? I saw on this thread that the number of functions from a set of size [imath]n[/imath] to a set of size [imath]m[/imath] is equivalent to "How many [imath]m[/imath]-digit numbers can I form using the digits [imath]1,2,...,n[/imath] and allowing repetition?" I know how to answer that question, but I don't know why it's the same thing as finding the number of functions from the size [imath]n[/imath] set to the size [imath]m[/imath] set. | 84162 | Counting functions between two sets
We got this question in homework (excuse my poor translation): This question deals with counting functions between two sets: A. How many functions exist between the set [imath][1,2,...,n][/imath] and the set [imath]\{1,2\}[/imath]? How many of them are onto? B. How many functions exist between the set [imath]\{1,2\}[/imath] and [imath][1,2,...,n][/imath]? How many of them are injective? I don't really know where to start. I tried summing the Binomial coefficient, but it repeats sets. Any ideas to get me going? Thanks! |
454613 | A minimization problem
Define [imath]L(w,u)=\frac{1}{2}\|w-u\|^2+\beta \|\frac{w}{x}\|,~w,u\in R^n[/imath] where [imath]\frac{w}{x}=(\frac{w_1}{x_1},\dots, \frac{w_n}{x_n})[/imath] [imath]\|x\|=\sqrt{x_1^2+\cdots+x_n^2}[/imath] Given [imath]u[/imath], [imath]x[/imath] and [imath]\beta[/imath], how to get [imath]\arg\min_{w\in R^n}L(w,u)[/imath] | 454519 | Least Squares with [imath] {L}_{2} [/imath] (L2) Linear Norm Regularization (Not Squared)
Define [imath]L(w,u)=\frac{1}{2}\|w-u\|^2+\beta \left\|\frac{w}{x}\right\|,~w,u\in \Bbb{R}^n[/imath] where [imath]\frac{w}{x}=\left(\frac{w_1}{x_1},\ldots, \frac{w_n}{x_n}\right)[/imath] [imath]\|x\|=\sqrt{x_1^2+\cdots+x_n^2}[/imath] Given [imath]u[/imath], [imath]x[/imath] and [imath]\beta[/imath], how to get [imath]\arg\min_{w\in \Bbb{R}^n}L(w,u)[/imath] |
454570 | Order of [imath]\mathbb{Z}[i]/(1+i)[/imath]
I have to calculate the order of the ring [imath]\mathbb{Z}[i]/(1+i)[/imath]. This is how far I am: If [imath]a+bi\in \mathbb{Z}[i]/(1+i)[/imath] then there are [imath]n,m\in \mathbb{Z}[/imath] such that [imath]a+bi\equiv 0+ni[/imath] or [imath]a+bi\equiv m+0i [/imath]. This means that [imath]\mathbb{Z}[i]/(1+i)=\{x\in \mathbb{Z}[i]: Re(x)=0 \text{ or }Im(x)=0\}[/imath]. So the order of this ring has to be infinite. I have a feeling I have something wrong, but cant figure what. Is there something wrong in this calculation? Thanks. | 1787409 | Structure of Quotient Ring
I'm struggling to see what the structure of a quotient ring such as [imath]\mathbb Z[i] / (1+i)[/imath] is. I think it's supposed to be isomorphic to [imath]\mathbb Z / 2\mathbb Z[/imath] but I don't see how. Can someone give me a hand? |
454656 | area under a curve with integration
How do I find the exact area under the curve of the function [imath]f(x)=4+3x-x^2[/imath] on the interval [imath][-1,3][/imath] using integration. I'm super lost, and not quite sure how to start the problem. | 454621 | Exact area using limits and Riemann sum
the i need to find the exact area under tha curve of the function [imath]f(x)=4+3x-x^2[/imath] on the interval [imath][-1,3][/imath] using limits and a Riemann Sum. I have nothing started, because I am confused on where to start, and where I would need to go from there, I need help with the entire problem. I hope someone can help me! Thanks |
454670 | Why does [imath]16^0 = 1[/imath] and not [imath]0[/imath]
If [imath]16^2 = 16 \times 16[/imath] and if [imath]16^1 = 16[/imath] why doesn't [imath]16^0 = 0[/imath]? I don't understand the rationale. Does [imath]16^0[/imath] express the idea that [imath]16[/imath] is multiplied by nothing and therefore is [imath]1[/imath]? | 235081 | Numbers to the Power of Zero
I have been a witness to many a discussion about numbers to the power of zero, but I have never really been sold on any claims or explanations. This is a three part question, the parts are as follows... 1) Why does [imath]n^{0}=1[/imath] when [imath]n\neq 0[/imath]? How does that get defined? 2) What is [imath]0^{0}[/imath]? Is it undefined? If so, why does it not equal 1? 3) What is the equation that defines exponents? I can easily write a small program to do it (see below), but what about in equation format? I just want a little discussion about numbers to the power of zero, for some clarification. Code for Exponents: (pseudo-code/Ruby) def int find_exp (int x, int n){ int total = 1; n.times{total*=x} return total; } |
453609 | EDIT variational formulation-exercice
i have this problem. Let [imath]\Omega[/imath] a bounded domain, connexe and regular, and let [imath]f \in L^2(\Omega).[/imath] Let the variational problem: Find [imath]u \in H^1(\Omega)[/imath] such [imath]\int_{\Omega} \nabla u \nabla v dx + (\int_{\Omega} u dx)(\int_{\Omega} v dx) = \int_{\Omega} f v dx, \forall v \in H^1(\Omega)[/imath] Deduce the boundary problem associate to this variational problem ( study the two cases [imath]u \in H^2(\Omega)[/imath] and [imath]u \notin H^2(\Omega)[/imath]. in the case 1: [imath]u\in H^2(\Omega)[/imath] In this case, we have that [imath]\tag{1}\int_\Omega \nabla u\nabla v=-\int_\Omega v\Delta u+\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v,\ \forall\ v\in H^1(\Omega)[/imath] From [imath](1)[/imath], we conclude that [imath]\tag{2}-\int_\Omega v\Delta u+\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v+\int_\Omega (\int_\Omega u)v=\int f v,\ \forall\ v\in H^1(\Omega)[/imath] If we take [imath]v\in C_c^\infty(\Omega)[/imath] in [imath](2)[/imath], we can conclude by the Fundamental Lemma of Calculus of Variation that [imath]\tag{3}-\Delta u(x)+\int_\Omega u=f(x),\ a.e.\ x\in\Omega[/imath] My question is: how we use the Fundametal Lemme Of Calculus Of Variation to deduce this? and why we choose [imath]v\in C^{\infty} (\Omega)[/imath]? why we can't work with [imath]v \in H^1(\Omega)[/imath]? My second question is: we put [imath]a(u,v)=\int_{\Omega}A \nabla u \cdot \nabla v dx + (\int_{\Omega} udx)(\int_{\Omega}vdx), \forall v \in H^1(\Omega)[/imath] How we prouve that [imath]a[/imath] is coercitive. for this, i process by absurde, so we have to prouve that [imath]\forall v \in V, \exists \nu > 0; a(v,v)< \nu ||v||^2_V[/imath] we choise [imath]\nu=\dfrac{1}{n}[/imath] , so we have [imath]\forall n \in \mathbb{N}, \exists v \in H^1(\Omega):a(v_n,v_n)<\dfrac{1}{n}||v_n||^2_V[/imath] I don't understand how we use the Rellich theorem to conclude the coercitivity. | 410550 | variational question
Let [imath]\Omega[/imath] a bounded domain, connexe and regular, and let [imath]f \in L^2(\Omega).[/imath] Let the variational problem: Find [imath]u \in H^1(\Omega)[/imath] such [imath]\int_{\Omega} \nabla u \nabla v dx + (\int_{\Omega} u dx)(\int_{\Omega} v dx) = \int_{\Omega} f v dx, \forall v \in H^1(\Omega)[/imath] 1- Prouve that this variational problem admits a unique solution in [imath]H^1(\Omega).[/imath] 2- Deduce the boundary problem associate to this variational problem ( study the two cases [imath]u \in H^2(\Omega)[/imath] and [imath]u \notin H^2(\Omega)[/imath]. I dont't understand in the question 2, why we mus study the two cases [imath]u \in H^2[/imath] and [imath]u \notin H^2[/imath]? |
238000 | using stokes theorem to calculate a line integral
Use stokes theorem to show that: [imath]\int_c ydx + zdy +xdz = -\sqrt{3} \pi a^2[/imath] Where c is the suitably oriented intersection of the surfaces [imath]x^2 + y^2 +z^2=a^2[/imath] and the plane [imath]x+y+z=0[/imath]. | 172089 | Use Stokes's Theorem to show [imath]\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2[/imath]
I am a little stuck on the following problem: Use Stokes's Theorem to show that [imath]\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2,[/imath] where [imath]C[/imath] is the suitably oriented intersection of the surfaces [imath]x^2 + y^2 + z^2 = a^2[/imath] and [imath]x + y + z = 0[/imath]. OK, so Stokes's Theorem tells me that: [imath]\oint_{C}\vec{F} \cdot d\vec{r} = \iint_{S}\operatorname{curl} \vec{F} \cdot \vec{N} ~dS[/imath] I have calculated: [imath]\operatorname{curl} \vec{F} = -\vec{i} - \vec{j} - \vec{k}.[/imath] I then figured that on the surface [imath]S[/imath] we must have: [imath]\vec{N}dS = \vec{i} + \vec{j} + \vec{k}dxdy[/imath] since this follows from the equation of the given plane. However this will then give me: [imath]\operatorname{curl} \vec{F} \cdot \vec{N} = -1 -1 -1 = -3[/imath] And thus I would get, if project this onto the [imath]xy[/imath]-plane: [imath]\iint_{S} \operatorname{curl} \vec{F} \cdot \vec{N} ~dS = -3 \iint_{A} dA = -3 \pi a^2[/imath] which is obviously not correct. I would greatly appreciate it if someone could help me with this. I actually had multivariable calculus a few years ago, and I know that I knew this stuff then. However, now that I need it again I notice that I've become quite rusty. Thanks in advance :) |
453663 | Prove the inverse of the Hilbert matrix has integer entries
[imath]1 \frac{1}{2} ... \frac{1}{n}[/imath] [imath]\frac{1}{2} \frac{1}{3} ... \frac{1}{n+1}[/imath] [imath].[/imath] [imath].[/imath] [imath].[/imath] [imath]\frac{1}{n} \frac{1}{n+1} ... \frac{1}{2n-1}[/imath] Does the inverse of this matrix have integer entries? Prove your statement. I thought of multiplying every line [imath]i[/imath] by [imath]i![/imath], but it leads to a very complicated solution (if it leads to a solution at all). Source: Linear Algebra, Kenneth Hoffman and Ray Kunze. Section 1.6, exercise 12. | 430060 | Why does the inverse of the Hilbert matrix have integer entries?
Let [imath]A[/imath] be the [imath]n\times n[/imath] matrix given by [imath]A_{ij}=\frac{1}{i + j - 1}[/imath] Show that [imath]A[/imath] is invertible and that the inverse has integer entries. I was able to show that [imath]A[/imath] is invertible. How do I show that [imath]A^{-1}[/imath] has integer entries? This matrix is called the Hilbert matrix. The problem appears as exercise 12 in section 1.6 of Hoffman and Kunze's Linear Algebra (2nd edition). |
455560 | Limit of a summable sequence
If [imath]x=(x_1.x_2,x_3,...,x_n,...)[/imath] and [imath]\sum^\infty_{k=1} |x_k|^p<\infty[/imath],[imath]p\geq 1[/imath] can you help me to show that [imath]\lim_{k\to\infty}|x_k|^p=0[/imath]? | 107961 | If a series converges, then the sequence of terms converges to [imath]0[/imath].
Following the guidelines suggested in this meta discussion, I am going to post a proposed proof as an answer to the theorem below. I believe the proof works, but would appreciate any needed corrections. Theorem If a series [imath]\sum_{n=1}^{\infty}a_n[/imath] of real numbers converges then [imath]\lim_{n \to \infty}a_n = 0[/imath] |
455223 | An example of a simple module which does not occur in the regular module?
Let [imath]K[/imath] be a field and [imath]A[/imath] be a [imath]K[/imath]-algebra. I know, if [imath]A[/imath] is artinain algebra, then by Krull-Schmidt Theorem [imath]A[/imath] , as a left regular module, can be written as a direct sum of indecomposable [imath]A[/imath]-modules, that is [imath]A=\oplus_{i=1}^n S_i[/imath] where each [imath]S_i[/imath] is indecomposable [imath]A[/imath]-module Moreover, each [imath]S_i[/imath] contains only one maximal submodule, which is given by [imath]J_i= J(A)S_i[/imath], and every simple [imath]A[/imath]-module is isomorphic to some [imath]A/J_i[/imath]. My question is that, can you, please, tell me an example of a non simisimple algebra, or a ring, such that it has a simple module which does not occur in the regular module. By occur I mean it has to be isomorphic to a simple submodule of a regular module | 455093 | Example of a simple module which does not occur in the regular module?
Let [imath]K[/imath] be a field and [imath]A[/imath] be a [imath]K[/imath]-algebra. I know, if [imath]A[/imath] is artinain algebra, then by Krull-Schmidt Theorem [imath]A[/imath] , as a left regular module, can be written as a direct sum of indecomposable [imath]A[/imath]-modules, that is [imath]A=\oplus_{i=1}^n S_i[/imath] where each [imath]S_i[/imath] is indecomposable [imath]A[/imath]-module Moreover, each [imath]S_i[/imath] contains only one maximal submodule, which is given by [imath]J_i= J(A)S_i[/imath], and every simple [imath]A[/imath]-module is isomorphic to some [imath]A/J_i[/imath]. My question is that, can you please tell me an example of a non simisimple algebra, or a ring, such that it has a simple module which does not occur in the regular module. By occur I mean it has to be isomorphic to a simple submodule of a regular module |
455799 | Correct Demonstration?
If [imath]\frac{a}{(a,b)}\mid c \;\ \Rightarrow \;\ a\mid b\cdot c[/imath] [imath]\frac{a}{(a,b)}\mid c\Rightarrow c=\frac{a}{(a,b)}\cdot k\Rightarrow b\cdot c=\frac{a \cdot b}{(a,b)}\cdot k\Rightarrow b\cdot c=a(\frac{b}{(a,b)}\cdot k)\Rightarrow[/imath][imath]a\mid b\cdot c[/imath] This is correct? | 20093 | How can I prove that [imath]a|bc[/imath] if and only if [imath]\frac{a}{(a,b)}|c[/imath]?
I tried to start by showing that [imath] \frac{a}{\gcd(a,b)} [/imath] is always an integer, let's call it [imath]d[/imath], because [imath]a[/imath] is always a multiple of [imath](a,b)[/imath] based on the definition of a g.c.d. I then tried to show that if [imath]a | b[/imath] then [imath]a|bc[/imath] from the theorem that states that [imath]a|b[/imath] implies that [imath]a|bc[/imath] for any integer [imath]c[/imath], but I cant seem to prove that [imath]a | b[/imath] or that [imath]a|bc[/imath] only if [imath]d \mid c[/imath] (where [imath]d = \frac{a}{\gcd(a,b)}[/imath] ). Any help would be much appreciated. |
455772 | If [imath]MM^{'}[/imath] is positive definite, [imath]M[/imath] is invertible?
Supposing I have a square real matrix M. If [imath]MM^{T}[/imath] is positive definite, is [imath]M[/imath] invertible? I came up with the proof [imath]MM^{T}=M\times I \times M^{T}[/imath], that is equal to say [imath]MM^{T}[/imath] is congruent to an identity matrix, where [imath]M[/imath] is the transformation matrix. However, I couldn't prove why [imath]M[/imath] should be non-degenerating. Can anyone gives me a hint on this? Thanks a lot. | 55996 | [imath]A^{T}A[/imath] positive definite then A is invertible?
Say if [imath]A[/imath] is an [imath]n \times n[/imath] matrix, why is it that if [imath]A^{T}A[/imath] is positive definite, the matrix [imath]A[/imath] is then invertible? All I know is [imath]A^{T}A[/imath] gives a symmetric matrix but what does [imath]A^{T}A[/imath] is positive definite tell or imply or hint about the matrix [imath]A[/imath] itself that leads to the fact that it will be invertible? |
455931 | German tank problem, simple derivation
I was reading the recent question on the German tank problem, and had trouble with one of the derivations in this section. [imath]\sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} = \frac{k(N+1)}{k+1}[/imath] I understand where the left-hand side comes from, in the context of the problem. I'm just having trouble arriving at the right-hand side. \begin{align*} \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} &= \sum_{m=k}^N m \frac{(m-1)!}{(k-1)!(m-k)!} \frac{k! (N-k)!}{N!}\\ &= k \sum_{m=k}^N \frac{m!}{(m-k)!} \frac{(N-k)!}{N!} \end{align*} This looks promising, since I have found the [imath]k[/imath] that appears in the right-hand side, but I don't know what to do with the sum. | 65398 | Why does this expected value simplify as shown?
I was reading about the german tank problem and they say that in a sample of size [imath]k[/imath], from a population of integers from [imath]1,\ldots,N[/imath] the probability that the sample maximum equals [imath]m[/imath] is: [imath]\frac{\binom{m-1}{k-1}}{\binom{N}{k}}[/imath] This make sense. But then they take expected value of the sample maximum and claim: [imath]\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}} = \frac{k(N+1)}{k+1}[/imath] And I don't quite see how to simplify that summation. I can pull out the denominator and a [imath](k-1)![/imath] term out and get: [imath]\mu = \frac{(k-1)!}{\binom{N}{k}} \sum_{m=k}^N m(m-1) \ldots (m-k+1)[/imath] But I get stuck there... |
455665 | Difficult problem in the analysis of sequences
Let [imath]a_n[/imath] be a sequence of real numbers and let: [imath]\lim_{n \to \infty}(a_{n+1}-a_n)=0[/imath] Prove that every [imath]a \in (\lim\text{inf}\,a_n, \lim\text{sup}\,a_n)[/imath] is a limit point of [imath]a_n[/imath]. | 195646 | If a sequence satisfies [imath]\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0[/imath] then the set of its limit points is connected
Prove that if a sequence satisfies [imath]\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0[/imath] then the set of its limit points is connected. My professor once mentioned a proposition likewise but I cannot find the exact version of it. Can anyone help? |
456512 | expected value in Poisson distribution
E(X) in Poisson Dist, \begin{align} \mathrm{E}(X) &= \sum_{k=0}^{\infty} \frac{k \lambda^k e^{-\lambda}}{k!}= \\ &= e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k+1}}{k!}=\\ &= \lambda \end{align} then, [imath]\text{E}(X(X-1)) = \sum_{k=0}^{\infty} \frac{k(k-1) \lambda^{k(k-1)} e^{-\lambda}}{k(k-1)!}[/imath] how can I simplify this ? | 456235 | Poisson random variable with parameter [imath]\lambda>0[/imath]
I am currently studying Poisson distribution. I could understand \begin{align} \mathrm{E}(X) &= \sum_{k=0}^{\infty} \frac{k \lambda^k e^{-\lambda}}{k!}= \\ &= e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k+1}}{k!}=\\ &= \lambda \end{align} but [imath]\text{E}(X(X-1)) = \lambda^2[/imath] why is this true ? How can I check it ? I have no idea to check [imath]E(X-1)[/imath] and [imath]E(X(X-1))[/imath] instead of just [imath]X[/imath]. |
456475 | Existence of a limit of a function
Let let [imath]f(x,y)=\frac{x^3-y^3}{x^2+y^2}[/imath] and [imath](x,y)\not=(0,0)[/imath] How to prove that the function has a limit as [imath](x,y)\to (0,0)[/imath] Should I look at firstly on the x axis and then on y-axis? | 153134 | What is $\lim_{(x,y)\to(0,0)} \frac{(x^3+y^3)}{(x^2-y^2)}$?
In class, we were simply given that this limit is undefined since along the paths [imath]y=\pm x[/imath], the function is undefined. Am I right to think that this should be the case for any function, where the denominator is [imath]x^2-y^2[/imath], regardless of what the numerator is? Just wanted to see if this is a quick way to identify limits of this form. Thanks for the discussion and help! |
456595 | Prove: [imath]\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2[/imath]
While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove [imath]\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2[/imath] by induction. I have seen it many times and proved it before but can't remember what it was I did. I see that for the first two terms [imath]n = 1, n=2[/imath] I get: for [imath]n = 1[/imath], [imath]\frac{1}{1^2} = 1 < 2[/imath] for [imath]n = 2[/imath], [imath]\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2[/imath] Now I am stumped, I know I want to show this works for the [imath]n+1[/imath] term and am thinking, let the series [imath]\sum_{n=1}^\infty \frac{1}{n^2} = A(n)[/imath] Then look to show the series holds for [imath]A(n+1)[/imath] But [imath]A(n+1) = A(n) + \frac{1}{(n+1)^2}[/imath] But now what? If I tried [imath]A(n+1) - A(n) = \frac{1}{(n+1)^2}[/imath] , but would have to show that this is less than [imath]2 - A(n)[/imath]. I am stuck. Thanks for your thoughts, Brian | 333417 | Need to prove the sequence [imath]a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}[/imath] converges
I need to prove that the sequence [imath]a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}[/imath] converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble: Monotonic: The sequence seems to be monotone and increasing. This can be proved by induction: Claim that [imath]a_n\leq a_{n+1}[/imath] [imath]a_1=1\leq 1+\frac{1}{2^2}=a_2[/imath] Need to show that [imath]a_{n+1}\leq a_{n+2}[/imath] [imath]a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}[/imath] Thus the sequence is monotone and increasing. Boundedness: Since the sequence is increasing it is bounded below by [imath]a_1=1[/imath]. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound. Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence? Thanks so much in advance! |
456563 | Prove that [imath](a,b)R(c,d) \longleftrightarrow ad=bc [/imath] is equivalence relation on [imath]A=R^2-\{(0,0)\}[/imath]
I am trying to prove that [imath](a,b)R(c,d) \longleftrightarrow ad=bc [/imath] is equivalence relation on [imath]A=\mathbb{R}^2-\{(0,0)\}[/imath] [imath]A[/imath] is all points on the plane. If I want to show that is reflexivity so I need to take [imath]a[/imath] and [imath]c[/imath], set [imath](a,a)\in R[/imath] and [imath](c,c)\in R[/imath] how I can show that [imath]a^2 = c^2[/imath]? how to show transitivity? Thanks! | 255224 | Equivalence Relation problem
Let [imath]S[/imath] be the relation on [imath]\mathbb{N} \times \mathbb{N}[/imath] defined by [imath](a,b)S(c,d)[/imath] if and only if [imath]ad=bc[/imath]. Prove this is an equivalence relation on [imath]\mathbb{N} \times \mathbb{N}[/imath]. I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity? Reflexive: Let [imath](x,y)S(x,y)[/imath]. Then [imath]xy = yx[/imath]. So [imath]S[/imath] is reflexive. Symmetric: Suppose [imath](a,b)S(c,d)[/imath]. Then, [imath]ad = bc[/imath]. Therefore, [imath]da = cb[/imath] and [imath]cb = da[/imath]. Therefore, [imath](c,d)S(a,b)[/imath]. Thus, [imath]S[/imath] is symmetric. Transitive: Suppose [imath](a,b)S(b,c)[/imath]. Then [imath]ac = bb[/imath]. This is where I'm stuck. Any ideas? |
77227 | Proving this relation is transitive
Let [imath]r[/imath] be a relation on [imath]A \times A[/imath] such that [imath](a,b) r (c,d) \iff ad = bc.[/imath] How can I show that this relation is transitive, ie. [imath](a, b)r(c,d)[/imath] and [imath](c,d)r(e, f) \implies (a,b)r(e,f)[/imath]? I tried to say that [imath](a,b) r (c,d)[/imath] means that [imath]c=ka[/imath] and [imath]d=kb,[/imath] [imath]k[/imath] a coefficient, such that [imath]ad = bc \iff akb = bka,[/imath] and going on from there, but I'm not sure this is valid for all values of [imath]a, b, c,[/imath] and [imath]d[/imath]. How can I show the transitivity of the relation? | 513447 | [imath](a,b)R(x,y) \iff ay=bx[/imath] is an equivalence on [imath]\Bbb Z \times (\Bbb Z\setminus\{0\})[/imath]
Define a relation [imath]R[/imath] on [imath]\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})[/imath] by [imath](a,b)R(x,y)\iff ay=bx.[/imath] [imath]a)[/imath] Prove that [imath]R[/imath] is an equivalence relation. [imath]b)[/imath] Describe the equivalence classes corresponding to [imath]R[/imath]. I know that an equivalence relation satisfies transitivity, reflexivity, and symmetry. Also, I know that the Cartesian Product is [imath]A\times B=\{(a,b):a\in A[/imath] and [imath]b\in B\}. [/imath]If I could get a hint for [imath]a)[/imath], that would be appreciated. For part [imath]b)[/imath], could someone explain what an equivalence class is? Please refrain from complete solutions. Thank you. |
338890 | Prove that the closed unit ball of [imath]L^2[a,b][/imath] is closed in [imath]L^1[a,b][/imath]
Prove that the closed unit ball of [imath]L^2[a,b][/imath] is closed in [imath]L^1[a,b][/imath]. My friends and I have literally been pouring over this problem for days now without success. We've been using Hölder's inequality, and especially its special case the Cauchy-Schwarz Inequality. We've also been using Minkowski's (the triangle) Inequality, and also the inequality: [imath]||f||_{p_1}\leq(b-a)^{\frac{p_2-p_1}{p_1p_2}}\;||f||_{p_2}[/imath] for [imath]1\leq p_1\leq p_2\leq\infty[/imath] The other inequalities being: [imath]||fg||_1\leq||f||_p\cdot ||f||_q[/imath] for [imath]\frac{1}{p}+\frac{1}{q}=1[/imath] and if [imath]p=1[/imath] then [imath]q=\infty[/imath] (Hölder's) [imath]||f+g||_p\leq ||f||_p+||g||_p[/imath] (Triangle) [imath]||fg||_1\leq ||f||_2\cdot ||g||_2[/imath] (Cauchy-Schwarz) I'll just say that we've been a lot of places with these inequalities, but nowhere we want to go. We defined [imath]A[/imath] to be the closed unit ball in [imath]L^2[a,b][/imath]: [imath]A:=\{\;f\in L^2[a,b] : ||f||_2\leq 1\}.[/imath] And now we want to consider functions in [imath]L^1[a,b][/imath] (and thus with the [imath]L^1[/imath] metric), and prove that if every open ball around a function contains a point from [imath]A[/imath], then that function is actually in [imath]A[/imath]. Can anyone help us with this? Thanks. | 457275 | Is unit ball in [imath]L_2(0,1)[/imath] closed in [imath]L_1(0,1)[/imath]?
More exactly, let [imath]L_1(0,1)[/imath] be the space of integrable functions [imath]f : (0,1)->R \to \mathbb{R}[/imath] with norm [imath]\|f\|_1 = \int_0^1 |f(x)|dx|dx[/imath]. Let B be subset of [imath]L_1(0,1)[/imath] such that [imath]f \in B[/imath] if and only if [imath]\int_0^1 |f(x)|^2 dx <= 1 \leq 1[/imath]. Is B a closed set in [imath]L_1(0,1)[/imath]? In other words, if [imath]f_n[/imath] converges to f in [imath]L_1(0,1)[/imath] and [imath]\int |f_n(x)|^2 dx <= 1 \leq 1[/imath] for all n, does it imply that [imath]\int |f(x)|^2 dx <= 1 \leq 1[/imath]? (Note that convegrence in [imath]L_1(0,1)[/imath] does not imply convegrence in [imath]L_2(0,1)[/imath]). |
457366 | The number of solutions for [imath]x+y+z=n[/imath]
How do I approach this problem? I know the formula but do not how it had come. Could you please explain to me the procedure, with examples if possible. stars and bars theorem | 382935 | Combinatorics Distribution - Number of integer solutions Concept Explanation
I reading my textbook and I don't understand the concept of distributions or number of solutions to an equation. It's explained that this problem is 1/4 types of sampling/distributions problems. An example is provided to illustrate: In how many ways can 4 identical jobs (indistinguishable balls) be distributed among 26 members (urns) without exclusion (since one member can do multiple jobs)? A sample outcome might be: [imath]\text{_____________}[/imath] | A | B | C |...| Z | [imath]\text{--------------------}[/imath] [imath]\text{_____________}[/imath] | o | oo| | | ||| o | [imath]\text{_____________}[/imath] | A | B | C |... | Z | [imath]\text{--------------------}[/imath] Thus, the question is reduced to, "How many [imath](26-1+4)[/imath] letter words are there consisting of four circles and [imath](26-1)[/imath] vertical lines?" Therefore, the solution is: [imath]\binom{26-1+4}{4}[/imath] I really don't understand why its [imath](26-1+4)[/imath]. There's only 26 different spots or "urns" to place the 4 jobs. Can someone please explain? I looked through another text to try and understand and I found it explained as such: There are [imath]\binom{n+r-1}{r-1}[/imath] distinct nonnegative integer valued vectors [imath](x_1,...,x_r[/imath] satisfying the equation [imath](x_1 + ... + x_r = x_n[/imath] for [imath]x\ge0[/imath]. [imath]\spadesuit[/imath] How in the world are they deriving this? For distinct positive integers I understand: Assume I have 8 balls (n=8) and I have 3 urns (r=3): o^o^o^o^o^o^o^o, where o represents a ball and ^ represents a place holder where an urn could be placed. For this scenario: There are [imath]\binom{n-1}{r-1}[/imath] distinct positive integer valued vectors [imath](x_1,...,x_n)[/imath] satisfying the equation: [imath]x_1 + ... + x_n = n, x_i>0, i=1,..,r[/imath] [imath]\clubsuit[/imath] It's clear that I could have this specific case ooo|ooo|oo. Here the bar represents a divide for the urn and you see I have 3 sections. So that case is clear. Can anyone please explain this problem to me? I don't understand the nonnegative integer case. Also, people who post tend to be crazy smart and explain things very in a complicated manner. I'd appreciate it if it could be explained in layman's terms as much as possible. Thank you!!! |
447923 | For what natural numbers is [imath]n^3 < 2^n[/imath]? Prove by induction
Problem For what natural numbers is [imath]n^3 < 2^n[/imath]? Attempt @ Solution For [imath]n=1[/imath], [imath]1 < 2[/imath] Suppose [imath]n^3 < 2^n[/imath] for some [imath]n = k \ge 1[/imath] It looks like the inequality is true for [imath]n = 0[/imath], [imath]n = 1[/imath] and [imath]n\ge10[/imath] But, how can I prove this through induction? | 70661 | Prove [imath]2^n > n^3[/imath]
Let [imath]P(n)[/imath] be the property: [imath]2^n > n^3[/imath]. Let's use mathematical induction to prove that [imath]P(n)[/imath] is true for [imath]n\geq10[/imath]. Basis: [imath]P(10): 2^{10} > 10^3 \Leftrightarrow 1024 > 1000[/imath] which is true. Hypothesis: [imath]P(k): 2^k > k^3[/imath] Inductive step: we have to prove [imath]P(k) \Rightarrow P(k+1)[/imath]: [imath]2^{k+1} = 2\cdot 2^k > 2\cdot k^3 = k^3 + k^3[/imath] Everything is clear up to here, but then it goes on like this: [imath]k^3 + k^3 \geq k^3 + 7\cdot k^2[/imath] Question here: How to prove that [imath]k^3 > 7\cdot k^2[/imath] ? I know it's because [imath]n\geq 10[/imath], but how to prove it? Besides empirically - which shows it's true. Then the proof goes on: [imath]k^3 + k^3 \geq k^3 + 3\cdot k^2 + 3\cdot k + 1[/imath] Question here: I suppose it's the same "technique" like above, isn't it? If not, how to prove that: [imath]k^3 \geq 3\cdot k^2 + 3\cdot k + 1[/imath] ? |
457911 | Whether [imath](0,1)[/imath] and [imath](0,1][/imath] are homeomorphic
In connection with the question continuous onto map from [imath](0,1)\to (0,1][/imath] I would like to know whether [imath](0,1)[/imath] and [imath](0,1][/imath] are homeomorphic. The map mentioned in the above question is onto but not a bijection. So does such a continuous bijection exist? | 240414 | Is there exist a homemoorphism between either pair of [imath](0,1),(0,1],[0,1][/imath]
As the topic is there exist a homeomorphism between either pair of [imath](0, 1),(0,1],[0,1][/imath] |
458474 | [imath]u(x,y)[/imath] harmonic and bounded in punctured disc; show [imath]0[/imath] is a removable singularity
I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis: If [imath]u[/imath] is harmonic and bounded in [imath]0 < |z| < \rho[/imath], show that the origin is a removable singularity in the sense that [imath]u[/imath] becomes harmonic in [imath]|z|<\rho[/imath] when [imath]u(0)[/imath] is properly defined. My Solution So Far: This problem is screaming out to use the big Picard theorem...I would like to define a harmonic conjugate function [imath]v:\{0<|z|<\rho\} \to \mathbb{C}[/imath], also bounded, and such that [imath]u,v[/imath] satisfy the Cauchy-Riemann equations. Then we would have that [imath]f(z):=u(x,y) + i v(x,y)[/imath] is analytic and bounded in [imath]\{0<|z|<\rho\}[/imath], so [imath]0[/imath] is a removable singularity, and thus we can fill it in for [imath]u[/imath]. I need to show that there exists such a [imath]v[/imath] and that it is also bounded. If the domain were simply connected, we could define [imath]v(x,y) = \int_{(x_0,y_0)}^{(x,y)} -u_2(s,t)ds + u_1(s,t)dt, [/imath] which would be well-defined because the integrand is closed, owing to the harmonicity of [imath]u[/imath], so the integral is path-independent. But obviously it is not simply connected...how can we overcome this problem? As to [imath]v[/imath] bounded, even if we could define it as above, I'm not sure how we could bound it on the open punctured disk. (I think we could bound it on any compact subset...) | 184490 | Can any harmonic function on [imath]\{z:0<|z|<1\}[/imath] be extended to [imath]z=0[/imath]?
After looking at this question for quite some time, I've asked a couple of other students, and they also couldn't seem to come up with an answer. This is from an old qualifying exam at our university. Let [imath]u[/imath] be a harmonic function bounded on the set [imath]\{z:0 < |z| < 1\}[/imath]. Can it always be defined at the point [imath] z= 0[/imath] to become harmonic on the whole unit disk? The standard argument with the logarithm doesn't work here, as it must be bounded on the punctured disk, and the logarithm blows up at [imath]0[/imath]. Also, we can't use any kind of harmonic conjugate argument because our domain is not simply connected. Thus, I think it's probably true, but I haven't been able to come up with a proof. Thanks in advance for any help! |
458746 | Show that [imath]\sum\limits_{n=1}^\infty\frac{\sqrt{a_n}}{n^p}[/imath] converges for p > [imath]\frac{1}{2}[/imath]
Let [imath]\sum_{n=1}^\infty{a_n}{}[/imath] be a convergent series of positive terms. Show that [imath]\sum_{n=1}^\infty\frac{\sqrt{a_n}}{n^p}[/imath] converges for p > [imath]\frac{1}{2}[/imath] | 418982 | Convergence of a series [imath]\sum\limits_{n=1}^\infty\left(\frac{a_n}{n^p}\right)^\frac{1}{2}[/imath]
I've got a question about the convergence of a series during studying analysis. If I know that a series of positive real numbers [imath]\sum_{n=1}^\infty a_n[/imath] converge, why does [imath]\sum_{n=1}^\infty\left(\frac{a_n}{n^p}\right)^\frac{1}{2}[/imath] also converge for [imath]p>1[/imath]? Although I know about many convergence tests, I don't know how to apply those tests for this case. Since this problem is the form of "series A converge → series B converge", I've been thinking that it must be verified by using some "comparison" tests. Is this thinking correct? All advice is welcome^_^ Thanks. |
105433 | Does every set have a group structure?
I know that there is no vector space having precisely [imath]6[/imath] elements. Does every set have a group structure? | 1344256 | Can we always make a group?
Can we always find an operation on non-empty set, which create a group[imath]?[/imath] I cann't imagine, how not, but is it proof for that? |
459504 | Find x inequality
Find [imath]x[/imath]([imath]x[/imath] belongs to [imath]\mathbb{Z}[/imath]) from this inequality: [imath]-\frac1{28}\leq\frac1{3^n+1}\leq\frac1{28} [/imath] I tried something,but I think it's wrong.. [imath]-3^n-1\leq28\leq3^n+1[/imath] [imath]-3^n\leq29\leq3^n+2[/imath] | 459426 | Finding [imath]x[/imath] from inequality: [imath]\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}[/imath]
Find [imath]x[/imath] in [imath]\mathbb{Z}[/imath] satisfying this inequality: [imath]\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}.[/imath] I tried something, but I don't think it's correct. [imath]-\frac{1}{28} \le \frac{3^n + 2}{3^n + 1} - 1 \le \frac{1}{28}[/imath] I arrived at: [imath]-3^n - 1 \le 28 \le 3^n + 1.[/imath] I don't know if it's ok or not. |
460240 | Proving that [imath]8\mid n(n^{2}-1)(3n+2)[/imath]
I was trying and could not, as it shows that [imath]8\mid n(n^{2}-1)(3n+2);\forall n \in \text{N}[/imath] Induction; looking eight consecutive numbers, what to do and how to do?[imath]$[/imath]$Sorry, forgot to add a detail: I understand the statement without using modular arithmetic (content not yet studied). | 455043 | [imath]24\mid n(n^{2}-1)(3n+2)[/imath] for all [imath]n[/imath] natural problems in the statement.
"Prove that for every [imath] n [/imath] natural, [imath]24\mid n(n^2-1)(3n+2)[/imath]" Resolution: [imath]24\mid n(n^2-1)(3n+2)[/imath]if[imath]3\cdot8\mid n(n^2-1)(3n+2)[/imath]since[imath]n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)[/imath]and[imath]8\mid n(n^{2}-1)(3n+2)?[/imath][imath]$[/imath]$ Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you. |
457787 | Smallest [imath]n[/imath] such that [imath]G[/imath] is a subgroup of the symmetric group [imath]S_n[/imath]
A well-known result, Cayley's theorem, says that any group is isomorphic to [imath]S_n[/imath] for some [imath]n[/imath]. Given a (finite) group [imath]G[/imath], is there a standard name for the smallest such [imath]n[/imath]? This seems like a very basic concept which has surely been studied and named, but my searches have found nothing so far. Is it easy to calculate [imath]n[/imath]? An obvious lower bound is the Kempner number A002034[imath](|G|)[/imath]. Note: there is a related question but it doesn't address my issue. | 191446 | "Efficient version" of Cayley's Theorem in Group Theory
I'm considering finite groups only. Cayley's theorem says the a group [imath]G[/imath] is isomorphic to a subgroup of [imath]S_{|G|}[/imath]. I think it's interesting to ask for smaller values of [imath]n[/imath] for which [imath]G[/imath] is a subgroup of [imath]S_n[/imath]. Obviously, it's not always possible to do better than Cayley's theorem. But sometimes it is possible (for example, [imath]\mathbb{Z}_6[/imath] as a subgroup of [imath]S_5[/imath]). So I'm asking: Given a finite group [imath]G[/imath], is there an algorithmic way to find or approximate the minimal [imath]n[/imath] for which [imath]G[/imath] is isomorphic to a subgroup of [imath]S_n[/imath]? If the answer to [imath](1)[/imath] is not known, is it known for specific classes of groups? In particular, for finite abelian groups, is it true that for a prime [imath]p[/imath], the minimal [imath]n[/imath] for [imath]\mathbb{Z}_{p^{t_1}} \times \mathbb{Z}_{p^{t_2}}[/imath] is [imath]p^{t_1}+p^{t_2}[/imath] (I can prove that is is true for different primes [imath]p_1[/imath] and [imath]p_2[/imath], but have problems when it's the same prime in both factors). Thanks! |
460288 | Find the remainder when [imath]11^{2013}[/imath] is divided by [imath] 61[/imath]
How do I find the remainder when [imath]11^{2013}[/imath] is divided by [imath]61[/imath]? Brute force? Without a calculator? How did people do that? | 419542 | Finding the remainder of [imath]11^{2013}[/imath] divided by [imath]61[/imath]
How am I suppose to find the remainder when [imath]11^{2013}[/imath] is divided by [imath]61[/imath]? |
426981 | Need help in proving that [imath]\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}[/imath]
We need to prove that [imath]\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}[/imath] I have tried and it gets confusing. | 1698487 | Prove [imath]\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1+\sin\theta}{\cos\theta}[/imath]
Prove by writing: [imath]\theta = 2A[/imath] that: [imath]\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1+\sin\theta}{\cos\theta}[/imath] First I subbed in the [imath]2A[/imath] such that: [imath]\frac{\sin 2A-\cos 2A+1}{\sin 2A+\cos 2A-1}=\frac{1+\sin 2A}{\cos 2A}[/imath] Then I considered the following trigonometric formulae: [imath]\begin{align} \sin 2A &= 2\sin A\cos A \\ \cos 2A &= \cos^{2}A - \sin^{2}A \\ \cos 2A &= 2\cos^{2}A-1 \\ \cos 2A &= 1-2\sin^{2}A \end{align}[/imath] I have taken various approaches but keep ending up with: [imath]\frac{\cos A+\sin A}{\cos A-\sin A}[/imath] on the left hand side. But I can't seem to get the right hand side to agree with this. I end up with something like: [imath]\frac{1+2\sin A\cos A}{1-2\sin^{2}A}[/imath] I've spent a few hours on this and can't find the solution. Help appreciated. |
460931 | Prove [imath]f=x^p-a[/imath] either irreducible or has a root. (arbitrary characteristic) (without using the field norm)
Let [imath]K[/imath] be an arbitrary field, [imath]p[/imath] a prime and [imath]a\in K[/imath]. Show [imath]f=x^p-a[/imath] is either irreducible in [imath]K[x][/imath] or has a root in [imath]K[/imath]. My strategy was to split this up into a case for each characteristic. The characteristic [imath]p[/imath] case is easy. Assume [imath]f[/imath] is reducible [imath]=gh[/imath] both smaller degree [imath]g[/imath] irreducible, with [imath]\alpha[/imath] a root of [imath]g[/imath]. Then in [imath]K(\alpha)[x][/imath], [imath]f=(x-\alpha)^p[/imath]. So [imath]g=(x-\alpha)^k[/imath], and all the other factors of [imath]f[/imath] are of the same form. Hence they are all equal to the [imath]minpoly(\alpha)[/imath] and the degree of [imath]minpoly(\alpha)|p[/imath] so it is [imath]1[/imath] or [imath]p[/imath] and we are done. The characteristic [imath]0[/imath] case and characteristic [imath]q\ne p[/imath] I get stuck on. If [imath]K[/imath] contains a primitive [imath]p[/imath]th root of unity [imath]\zeta[/imath], then over [imath]K(\alpha)[/imath], [imath]f[/imath] splits into linear factors. So the degree of the field extension for any root of [imath]f[/imath] is the same, so the degree of such a root is either [imath]1[/imath] or [imath]p[/imath]. If [imath]K[/imath] does not have a primitive [imath]p[/imath]th root of unity, then simply adjoin one. We know [imath][K(\zeta):K]|p-1[/imath] and by the previous step either [imath]f[/imath] is irreducible over [imath]K(\zeta)[/imath] or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form [imath]x-\zeta^ia^{1/p}[/imath], but I am unsure what to do from here. The characteristic [imath]q[/imath] case seems similar to the characteristic [imath]0[/imath] case but again I am stuck trying to finish the proof. Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did. | 266171 | Irreducibility of a polynomial if it has no root (Capelli)
Let [imath]F[/imath] be a field of arbitrary characteristic, [imath]a\in F[/imath], and [imath]p[/imath] a prime number. Show that [imath]f(X)=X^p-a[/imath] is irreducible in [imath]F[X][/imath] if it has no root in [imath]F[/imath]. This answer to a related question mentions the result is due to Capelli. I can prove the result if [imath]F[/imath] has characteristic [imath]p[/imath] as follows. Suppose [imath]f[/imath] is reducible: [imath]f(X)=g(X)h(X)[/imath] with [imath]g(X)[/imath] an irreducible factor of degree [imath]m[/imath], [imath]1\le m<p[/imath]. Then if [imath]\alpha[/imath] is a root of [imath]g[/imath] in some extension field [imath]K[/imath] of [imath]F[/imath], we have [imath]f(X)=X^p-\alpha^p=(X-\alpha)^p[/imath] so its divisor [imath]g(X)[/imath] must be of the form [imath](X-\alpha)^m[/imath]. Since the coefficient of [imath]X^{m-1}[/imath] in [imath]g[/imath] is in [imath]F[/imath], we have [imath]m\alpha\in F[/imath]. So [imath]\alpha\in F[/imath] because [imath]m[/imath] is invertible modulo [imath]p[/imath]. How would you show the result in other characteristics? |
461070 | show that [imath]\int_{0}^{\infty}e^{-x}\ln(x)dx=-\gamma=\Gamma'(1) [/imath]
show that [imath]\int_{0}^{\infty}e^{-x}\ln(x)dx= -\gamma=\Gamma'(1)[/imath] where [imath]\gamma[/imath] is Euler–Mascheroni constant I started with [imath]\Gamma(z)=\left[ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)e^{-z/k}\right]^{-1}.[/imath] by take log then derivative with respect to z and put z=1 I got [imath]\Gamma'(1)=-1-\gamma+\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})[/imath] but I dont know how to compute [imath]\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})[/imath] I know the answer is 1 but how ? this is my way . can any one show different way ? or give me how to compute [imath]\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})[/imath] | 419026 | Computing [imath] \int_0^\infty \frac{\log x}{\exp x} \ dx [/imath]
I know that [imath] \int_0^\infty \frac{\log x}{\exp x} = -\gamma [/imath] where [imath] \gamma [/imath] is the Euler-Mascheroni constant, but I have no idea how to prove this. The series definition of [imath] \gamma [/imath] leads me to believe that I should break the integrand into a series and interchange the summation and integration, but I can't think of a good series. The Maclaurin series of [imath] \ln x [/imath] isn't applicable as the domain of [imath] x [/imath] is not correct and I can't seem to manipulate the integrand so that such a Maclaurin series will work. Another thing I thought of was using [imath] x \mapsto \log u [/imath] to get [imath] \int\limits_{-\infty}^\infty \frac{\log \log u}{u^2} \ du [/imath] and use some sort of contour integration, but I can't see how that would work out either. |
446796 | Theory of Numbers Properties of GCD (Greatest Common Divisor)
I am a student in Undergraduate Mathematics, and I'm struggling to number theory ... I have this problem gcd, and do not know how to do it, and still do not study, congruences, Diophantine equations or, among other matters more advanced ... I'm used to divisibility, and some properties and/or theorems gcd ... Help me, please ... question a) Show that if [imath](a, b) = 1[/imath], [imath]\Longrightarrow[/imath] [imath](a · c, b) = (c, b)[/imath]. | 342092 | number theory [imath]\gcd(a,bc)=\gcd(a,c)[/imath]
Suppose [imath]\forall (a;b;c;d) \in \mathbb{Z}^4[/imath] as [imath]\gcd(a,b)=\gcd(c,d)=1[/imath]: How can I prove that [imath]\gcd(a,bc)=\gcd(a,c)[/imath]? Also, how to prove that [imath]\gcd(ac,bd) = (\gcd(a,d))*(\gcd(b,c))[/imath]? |
460856 | components of a system
Suppose a system has [imath]10[/imath] components and that at a particular time the [imath]j[/imath]th component is working with probability [imath]1=j[/imath] for [imath]j = 1,2,\ldots,10[/imath]. How many components do you expect to be working at that particular time? Can I get a start on this? Thanks | 449946 | the number of components working at a particular time
Suppose a system has [imath]10[/imath] components and that a particular time the [imath]j[/imath]'th component is working with probability [imath]1/j[/imath] for [imath]j=1,2,\dots,10[/imath]. How many components do you expect to be working at that particular time? I have only found the probability of the components and I know that the sum of the possible probabilities is [imath]1[/imath]. How do I go about solving this question? |
461625 | Prove that [imath]||x|-|y|| \leq |x-y|[/imath]
[imath]||x|-|y|| \leq |x-y|[/imath] when [imath](x,y \in R^k)[/imath] In Principles of MA(Rudin), the author said one sees easily that [imath]||x|-|y|| \leq |x-y|[/imath] when [imath](x,y \in R^k)[/imath] (p.88, Rudin) from the triangle inequality. But I'm not sure how to use the triangle inequality to show this. Can you help me show this? | 406443 | How to prove [imath]\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert[/imath]?
I'm trying to show that [imath]\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert[/imath]. A hint would be nice. |
461752 | Prove that [imath]f(x)[/imath] is irreducible in [imath]\mathbb{Q}[/imath] (rational fileld).
[imath]f(x)=\left(x-a_1\right)^2\left(x-a_2\right)^2\text{...}\left(x-a_n\right)^2+1[/imath], where [imath]a_1,a_2,\text{...},a_n[/imath] are different integers. Prove that [imath]f(x)[/imath] is irreducible in [imath]\mathbb{Q}[/imath] (rational fileld). One thought: [imath]f(x)[/imath] is a polynomial with integer coefficients, and assume [imath]p[/imath] is an root of [imath]f(x)[/imath], then [imath]f(p)\geq 1[/imath], there is no possible to make [imath]f(p)=0[/imath], so [imath]f(x)[/imath] is irreducible, I think my thought is too easy, and is wrong... | 317622 | [imath][(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1[/imath] is irreducible over [imath]\mathbb Q[/imath]
Suppose that [imath]a_1,a_2, \cdots, a_n[/imath] are [imath]n[/imath] different integers. Then [imath][(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1[/imath] is irreducible over [imath]\mathbb Q[/imath]. I've no idea why it is true. Thanks very much. |
461803 | How can I find the center of ellipse given an Arc and ellipse radii?
I have an arc with start and end points and also I have width and height of the ellipse. Using these can I find center of the ellipse? hi Stefan, how can i calculate [imath](x,y)[/imath] center , can u give idea? | 53093 | How to find the center of an ellipse?
I have the following data:- I have two points ([imath]P_1[/imath], [imath]P_2[/imath]) that lie somewhere on the ellipse's circumference. I know the angle ([imath]\alpha[/imath]) that the major-axis subtends on x-axis. I have both the radii ([imath]a[/imath] and [imath]b[/imath]) of the ellipse. I now need to find the center of this ellipse. It is known that we can get two possible ellipses using the above data. I have tried solving this myself but the equation becomes so complex that I always give up. This is what I have done till now:- I took the normal ellipse equation [imath]x^2/a^2 + y^2/b^2 = 1[/imath]. To compensate for the rotation and translation, I replaced [imath]x[/imath] and [imath]y[/imath] by [imath]x\cos\alpha+y\sin\alpha-h[/imath] and [imath]-x\sin\alpha+y\cos\alpha-k[/imath], respectively. [imath]h[/imath] and [imath]k[/imath] are x and y location of the ellipse's center. Using these information I ended up with the following eq:- [imath]a B_1\pm\sqrt{a^2 B_1^2 - C_1(b^2 h^2 - 2 A_1 b^2 h)} = a B_2\pm\sqrt{a^2 B_2^2 - C_2 (b^2 h^2 - 2 A_2 b^2 h)} \quad (1)[/imath] where [imath]A = x\cos\alpha +y\sin\alpha[/imath], [imath]B = -x\sin\alpha+y\cos\alpha[/imath] and [imath]C = a^2 B^2 + A^2 b^2 - a^2 b^2[/imath]. Now the only thing I need to get is [imath]h[/imath] from (1). All other values are known, but I am not able to single that out. Anyway if the above equations looks insane then please solve it yourself, your way. I could have drifted into some very complicated path. |
461828 | [imath]g(x+y)=g(x)g(y)[/imath] for all [imath]x,y \in\mathbb{R}[/imath]. If [imath]g[/imath] is continuous at [imath]0[/imath], prove that [imath]g[/imath] is continuous on [imath]\mathbb{R}[/imath]
[imath]g(x+y)=g(x)g(y)[/imath] for all [imath]x,y \in\mathbb{R}[/imath]. If [imath]g[/imath] is continuous at [imath]0[/imath], prove that [imath]g[/imath] is continuous on [imath]\mathbb{R}[/imath]. | 151032 | If [imath]f\colon \mathbb{R} \to \mathbb{R}[/imath] is such that [imath]f (x + y) = f (x) f (y)[/imath] and continuous at [imath]0[/imath], then continuous everywhere
Prove that if [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath] is such that [imath]f(x+y)=f(x)f(y)[/imath] for all [imath]x,y[/imath], and [imath]f[/imath] is continuous at [imath]0[/imath], then it is continuous everywhere. If there exists [imath]c \in \mathbb{R}[/imath] such that [imath]f(c) = 0[/imath], then [imath]f(x + c) = f(x)f(c) = 0.[/imath] As every real number [imath]y[/imath] can be written as [imath]y = x + c[/imath] for some real [imath]x[/imath], this function is either everywhere zero or nowhere zero. The latter case is the interesting one. So let's consider the case that [imath]f[/imath] is not the constant function [imath]f = 0[/imath]. To prove continuity in this case, note that for any [imath]x \in \mathbb{R}[/imath] [imath]f(x) = f(x + 0) = f(x)f(0) \implies f(0) = 1.[/imath] Continuity at [imath]0[/imath] tells us that given any [imath]\varepsilon_0 > 0[/imath], we can find [imath]\delta_0 > 0[/imath] such that [imath]|x| < \delta_0[/imath] implies [imath]|f(x) - 1| < \varepsilon_0.[/imath] Okay, so let [imath]c \in \mathbb{R}[/imath] be fixed arbitrarily (recall that [imath]f(c)[/imath] is nonzero). Let [imath]\varepsilon > 0[/imath]. By continuity of [imath]f[/imath] at [imath]0[/imath], we can choose [imath]\delta > 0[/imath] such that [imath]|x - c| < \delta\implies |f(x - c) - 1| < \frac{\varepsilon}{|f(c)|}.[/imath] Now notice that for all [imath]x[/imath] such that [imath]|x - c| < \delta[/imath], we have [imath]\begin{align*} |f(x) - f(c)| &= |f(x - c + c) - f(c)|\\ &= |f(x - c)f(c) - f(c)|\\ &= |f(c)| |f(x - c) - 1|\\ &\lt |f(c)| \frac{\varepsilon}{|f(c)|}\\ &= \varepsilon. \end{align*}[/imath] Hence [imath]f[/imath] is continuous at [imath]c[/imath]. Since [imath]c[/imath] was arbitrary, [imath]f[/imath] is continuous on all of [imath]\mathbb{R}[/imath]. Is my procedure correct? |
461970 | Number of idempotents and commutativity
Let [imath]R[/imath] be a finite ring (that does not necessarily contain the identity) such that more than three fourth of its elements are idempotents. Then show that [imath]R[/imath] must be commutative. If [imath]R[/imath] is a ring with 1 then this case I have proved, but without assuming the existence of it how can it be shown? | 445367 | Sufficiently many idempotents and commutativity
It is a well-known result that if a ring [imath]R[/imath] satisfies [imath]a^2=a[/imath] for each [imath]a\in R[/imath], then [imath]R[/imath] must be commutative. See here for proof. I am wondering whether the same result holds for finite rings if we only assume sufficiently many (but not necessarily all) elements of the ring are idempotents. Recall that an element [imath]a\in R[/imath] is called idempotent if [imath]a^2=a[/imath]. For example, suppose [imath]R[/imath] is a finite ring in which at least [imath]80[/imath]% elements are idempotents. Can we conclude that [imath]R[/imath] is commutative? More generally, Does there exist an absolute constant [imath]0<k<1[/imath], such that whenever a finite ring [imath]R[/imath] satisfies [imath]\frac{\textrm{Number of idempotents}}{|R|}\ge k[/imath] then [imath]R[/imath] is commutative. Motivation: We know that if every element of a group [imath]G[/imath] satisfies [imath]a^2=1[/imath], then [imath]G[/imath] must be abelian. This is a relatively easy exercise. However, it turns out that we only need 75% percent of elements to satisfy [imath]a^2=1[/imath] in order force [imath]G[/imath] to be abelian. See here. For reference, [imath]a\in G[/imath] is called involution if [imath]a^2=1[/imath]. |
462282 | The roots of [imath]X^2-1[/imath] in the ring [imath](\mathbb{Z}/n\mathbb{Z})[X][/imath]
I don't know how to prove the following. Let [imath]n[/imath] be an odd natural number with [imath]t[/imath] different prime divisors. Then the polynomial [imath]X^2-1[/imath] has exactly [imath]2^t[/imath] roots. Own tries First of all: [imath]X^2-1=(X-1)(X+1)[/imath]. We say that [imath]a \in \{1, 2, \cdots, n \}[/imath] is a root if and only if [imath](a-1)(a+1)=kn[/imath] for some integer [imath]k[/imath]. Now observe that [imath]p|(a-1) \iff p\nmid(a+1)[/imath], except when [imath]p=2[/imath]. If [imath]a[/imath] is odd, [imath]2[/imath] divides both [imath]a+1[/imath] and [imath]a-1[/imath]. If [imath]a[/imath] is even, [imath]2[/imath] divides neither. Luckely [imath]n[/imath] is odd. This means that a prime divisor [imath]p[/imath] from [imath]n[/imath] only can appear in maximal one of the numbers [imath]a+1, a-1[/imath]. It looks like we only need to prove that there exists [imath]k \in \mathbb{Z}[/imath] so that [imath]p[/imath] appears in at least one of these numbers: [imath]kn+a-1, kn+a+1[/imath], because, be previous observations, there are exactly [imath]2^t[/imath] was to "arrange" prime divisors in the factorisation of [imath]\bar{a}-\bar{1}[/imath] and [imath]\bar{a}+\bar{1}[/imath]. Is this correct? | 385857 | [imath]x^2 = 1[/imath] in the ring [imath]\mathbb{Z}/n \mathbb{Z}[/imath]
I need to solve this equation: [imath]x^2 = 1[/imath] in the ring [imath]\mathbb{Z}/n \mathbb{Z}[/imath]. I know that [imath](n - 1)(n - 1) \equiv 1 \pmod n[/imath], in general [imath](n-a)^2 \equiv a^2 \pmod n[/imath]. I also know that for [imath]p[/imath] prime all elements of [imath]\mathbb{Z}_p[/imath] are invertible and here [imath]1[/imath] and [imath]p-1[/imath] are their own inverses. Could you tell me what other solutions there are to this equation? |
462322 | Suppose [imath]p[/imath] is an odd prime. Show that [imath]1^{p-1} +2^{p-1}+ \ldots +(p-1)^{p-1}\equiv -1\pmod p[/imath]
Suppose [imath]p[/imath] is an odd prime. Show that [imath]1^{p-1} +2^{p-1}+ \ldots +(p-1)^{p-1}\equiv -1\pmod p[/imath]. I think I need to use Wilson's Theorem on this but I'm not sure how. I believe I am suppose to factor it somehow too but also I'm lost at this point. | 461336 | Show that [imath]1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p[/imath]
Show that [imath]1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p[/imath] So, I use Fermat's little theorem, that is if [imath]p[/imath] does not divide [imath]a[/imath], then [imath]a^{p-1}[/imath] is congruent to [imath]1[/imath] (mod [imath]p[/imath]). But I'm still confused. If [imath]1^{p-1}[/imath] is congruent to [imath]1[/imath] (mod [imath]p[/imath]), [imath]2^{p-1}[/imath] is congruent to [imath]1[/imath] (mod [imath]p[/imath]), and [imath]p[/imath] does not divide [imath]p-1[/imath], so [imath](p-1)^{p-1}[/imath] is congruent to [imath]1[/imath] (mod [imath]p[/imath]). How does that equal [imath]-1[/imath] (mod [imath]p[/imath])? |
461466 | What "double bracket" means?
What does this symbol mean? Please i need help: Its the symbol just before "[imath]H^{t-1}[/imath]". Thanks for help | 461816 | What is the double bracket notation used here?
Its kind of a bracket but I'm not sure what it means. I have two ideas about it: It means the number of times the expression in satisfied or it changes for [imath]1[/imath] or [imath]0[/imath] depending on the result every time the [imath]i[/imath] value changes. |
77656 | Is the power set of the natural numbers countable?
Some explanations: A set S is countable if there exists an injective function [imath]f[/imath] from [imath]S[/imath] to the natural numbers ([imath]f:S \rightarrow \mathbb{N}[/imath]). [imath]\{1,2,3,4\}, \mathbb{N},\mathbb{Z}, \mathbb{Q}[/imath] are all countable. [imath]\mathbb{R}[/imath] is not countable. The power set [imath]\mathcal P(A) [/imath] is defined as a set of all possible subsets of A, including the empty set and the whole set. [imath]\mathcal P (\{\})=\{\{\}\}, \mathcal P (\mathcal P(\{\}))=\{\{\}, \{\{\}\}\} [/imath] [imath]\mathcal P(\{1,2\})=\{\{\}, \{1\},\{2\},\{1,2\}\}[/imath] My question is: Is [imath]\mathcal P(\mathbb{N})[/imath] countable? How would an injective function [imath]f:S \rightarrow \mathbb{N}[/imath] look like? | 657549 | Is continuum [imath] 2^{\mathbb{R}}[/imath]?
I had questions if exist A [imath]\neq \emptyset [/imath] that |A|=|P(A)| And my doubt is if [imath] |2^{\mathbb{R}}|=|\mathbb{R}|[/imath]? |
462566 | Prove that [imath]n! \geq 2^{n-1}[/imath] for [imath] n\geq1[/imath]
Mathematical Induction:-Prove that [imath]n! \geq 2^{(n-1)}[/imath] for [imath]n\geq 1[/imath]. I tried mathematical induction but could not | 2668193 | Prove by induction: [imath]n! \ge 2^{(n-1)}[/imath] for any [imath]n \ge 1[/imath]
I need help with this exercise. What I've done so far is prove the exercise when [imath]n=1[/imath]. So: [imath]n=1[/imath] [imath]1!\ge2^{(1-1)}[/imath] [imath]1\ge2^0[/imath] [imath]1\ge1[/imath] Which is true Therefore, now that I assume that the assumption is correct, I want to prove that with [imath]n+1[/imath], it will also be true. So, what I've done now is: [imath]P(n) \implies P(n+1) [/imath] [imath]n!+n+1\ge 2^{(n-1)}...[/imath] And my problem is that I do not know how to add the [imath]n+1[/imath] in the right side of the equation, therefore I'm not been able to finish the exercise. Thank you so much for your help. If something's not very clear, please let me know. I'll try to be clearer next time :) |
462817 | Prove all base 10 positive integers are divisible by 3 if the sum of their digits is divisible by 3
I came across a strange property of [imath]3[/imath] in base [imath]10[/imath]; for all integers I have tried, the following rule seems to be true: The integer '[imath]n[/imath]' is divisible by [imath]3[/imath] if the sum of the digits of the integer are also divisible by three. Is there a proof to show that this in fact is true? Or a counter example to disprove? An example of the rule: Suppose '[imath]n[/imath]' = [imath]123456789[/imath]. Then [imath] 1+2+3+4+5+6+7+8+9 = 45 [/imath] [imath] 4+5 = 9 [/imath] [imath] 9 \mod 3 = 0 [/imath] Therefore [imath] 123456789 \mod 3 = 0 [/imath] Afterword: For those of you who are wondering, I realized this rule while working on a database where every third row had a value I was wanted. What I wanted was a way to determine if a given row number '[imath]n[/imath]' was divisible by [imath]3[/imath] without actually performing the function [imath]n/3 = 0?[/imath] and so far the rule I came up hasn't failed. | 341202 | How to prove the divisibility rule for $3\, $ [casting out threes]
The divisibility rule for [imath]3[/imath] is well-known: if you add up the digits of [imath]n[/imath] and the sum is divisible by [imath]3[/imath], then [imath]n[/imath] is divisible by three. This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule: [imath]1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439[/imath] [imath]124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524[/imath] This works for as many numbers as I've tried. However, I'm not sure how this may be proven. Thus, my question is: Given a positive integer [imath]n[/imath] and that [imath]3\mid\text{(the sum of the digits of [/imath]n[imath])}[/imath], how may we prove that [imath]3\mid n[/imath]? |
462852 | if [imath]f(x)[/imath] is summable square function, then...
I have a question. If a normed function, that is to say [imath] \int_{-\infty}^{\infty}|f(x)|^2dx<\infty~~~\text{(summable square function)}[/imath] then, [imath]\underset{\begin{array}{c} x\rightarrow+\infty\\ x\rightarrow-\infty \end{array}}{lim}f(x)=0[/imath] It is true, why? thanks | 462080 | summable square function implies...?
I have difficulty to demonstrate this: [imath] \int_{-\infty}^{\infty}|f(x)|^2dx<\infty~~~\text{(summable square function)}[/imath] then, [imath]\lim_{|x|\rightarrow\infty }f(x)=0[/imath] thank you. |