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2311565 | If [imath]f[/imath] and [imath]g[/imath] are bounded and uniformly continuous prove that [imath]fg[/imath] is uniformly continuous
Please check my proof: Suppose [imath]f(x)[/imath] and [imath]g(x)[/imath] are bounded and uniformly continuous, then we obtain [imath]|x-y|<\delta \rightarrow |f(x)-f(y)|<\sqrt{M}\sqrt{|x-y|}[/imath] and [imath]|x-y|<\delta \rightarrow |g(x)-g(y)|<\sqrt{M}\sqrt{|x-y|}[/imath] added those we get [imath]|x-y|<\delta \rightarrow |f(x)-f(y)||g(x)-g(y)|<\sqrt{m}\sqrt{|x-y|}\sqrt{m}\sqrt{|x-y|}=M|x-y|[/imath] that satisfy Lipschitz condition, therefore [imath]f(x)g(x)[/imath] is uniformly continuous. | 470813 | two functions are uniformly continuous on some interval I and each is bounded on I then their product is also uniformly continuous on I .
prove that if two real valued functions are uniformly continuous on some interval [imath]I[/imath] and each is bounded on [imath]I[/imath] then their product [imath]f\cdot g=f(x)\cdot g(x)[/imath] is also uniformly continuous on [imath]I[/imath] . Is boundedness of each function on [imath]I[/imath] is necessary for the product |
2731151 | Probability that an [imath]m[/imath] digit number does not contain [imath]0[/imath], [imath]5[/imath], [imath]6[/imath] nor [imath]9[/imath]
Given an [imath]m[/imath]-digit number, what is the probability that it does not contain any of the following digits: [imath]\{0, 5, 6, 9\}[/imath]? | 2143621 | Probability that a k digit number does not contain certain digits
The probability that a [imath]k[/imath]-digit number does NOT contain the digits 0,5 or 9 [imath]0.3^k[/imath] [imath]0.6^k[/imath] [imath]0.7^k[/imath] [imath]0.9^k[/imath] I'm confused since the question has asked the number should not contain 0,5 OR 9. So, should the answer be [imath]0.9^k[/imath] or something else? |
2731359 | Trouble with a truth table for an inference in propositional logic
I'm beginning to study logic and was reading Graham Priest's Logic: a Very Short Introduction, and he includes this inference: [imath]q, \lnot q / p[/imath]. He then provides a truth table. But, for the life of me, I can't figure out what he's doing (he doesn't state it, which makes me feel even dumber) The final configuration of Ts and Fs for the final 'p' is (from top to bottom) TFTF, with the initial truth-value reference table for q and p being (naturally): q: TTFF p: TFTF Whence does he derive that particular configuration for the final 'p'? Is it that because the premise states 'q' and then negates it, it says nothing and therefore the truth values for p stay the same? I know I could have asked this more clearly, but I'm a little desperate | 2636126 | Proofs using theorems instead of axioms
I'm not sure how to prove these basic theorems in propositional calculus. Instead of using the standard axioms, we're supposed to use: Deduction Theorem (if [imath]\Phi, \alpha \vdash \beta[/imath] then [imath]\Phi \vdash \alpha \to \beta[/imath]), Reductio (if [imath]\Phi, \alpha \vdash \, [/imath], then [imath]\Phi \vdash \lnot \alpha[/imath]), Cut Rule (if [imath]\Phi \vdash \alpha[/imath] and [imath]\Psi, \alpha \vdash \beta[/imath] then [imath]\Phi \cup \Psi \vdash \beta[/imath]), Inconsistency Effect (if [imath]\Phi \vdash \, [/imath], then [imath]\Phi \vdash \beta[/imath] for every formula [imath]\beta[/imath]), and the Principle of Indirect Proof (if [imath]\Phi, \lnot \alpha \vdash \, [/imath], then [imath]\Phi \vdash \alpha[/imath]), as all the axioms can be deduced using these theorems. I don't really know how to start the proofs without using the axioms: i) prove that [imath]\lnot(\alpha \to \beta) ⊢ \alpha[/imath] ii) prove that [imath]\lnot\alpha \vdash \alpha \to \beta[/imath] Any suggestions on how to start these proofs or any insight at all would be greatly appreciated! Thanks! |
2730395 | Find [imath]\lim\limits_{n\to +\infty}\big(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\big)[/imath]?
Computing the limit: [imath] \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}} + \dfrac{1}{\sqrt{n^2 + 2}} + \cdots + \dfrac{1}{\sqrt{n^2 + n}}\right).[/imath] I've tried taking: We have [imath]\lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}}\right) = \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 2}}\right) = \ldots = \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + n}}\right) = 0.[/imath] Then, we obtain [imath] \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}} + \dfrac{1}{\sqrt{n^2 + 2}} + \cdots + \dfrac{1}{\sqrt{n^2 + n}}\right)\\ = \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}}\right) + \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 2}}\right) + \cdots + \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + n}}\right) = 0.[/imath] Any other approach to this is welcome. Thanks! | 2729507 | Compute the limit of the sequence [imath]{\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}[/imath]
I have to compute the limit of this sequence [imath]{\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}[/imath] as [imath]n\rightarrow\infty[/imath]. First I was thinking about some Riemann sum and and forced the [imath]n^{2}[/imath] outside the square root but the function was not so pleasant. |
2732345 | How can I find the subgroups of [imath]S_n[/imath]?
As the title says: how can I find the subgroups of [imath]S_n[/imath]? I know that [imath]S_n = <(1 2), (1 2 .. n)>[/imath] and [imath]S_n = <(1 i): i = 2, 3, .. n>[/imath]. But how do I find all subgroups? And how do I know their orders? | 76176 | Enumerating all subgroups of the symmetric group
Is there an efficient way to enumerate the unique subgroups of the symmetric group? Naïvely, for the symmetric group [imath]S_n[/imath] of order [imath]\left | S_n \right | = n![/imath], there are [imath]2^{n!}[/imath] subsets of the group members that could potentially form a subgroup. In addition, many of these subgroups are going to be isomorphic to each other. I feel that the question has an easy answer in terms of the conjugacy classes, but I don't see how. If the answer generalizes to all finite groups, please elaborate! |
2732559 | Expected value of the ratio
Suppose [imath]X_1, X_2, ..., X_n[/imath] are i.i.d.r.v., where [imath]X_i>0[/imath] and [imath]EX_i=\mu[/imath], [imath]Var(X_i)=\sigma^2[/imath]. Find [imath]E\left(\frac{X_1+X_2+...+X_m}{X_1+X_2+...+X_n}\right)[/imath], if [imath]m<n[/imath]. Any ideas how to even start this problem? | 2420091 | Find expectation of [imath]\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}[/imath] when [imath]X_1,\ldots,X_n[/imath] are i.i.d
Let [imath]X_1, \ldots, X_n[/imath] be i.i.d. random variables with expectation [imath]a[/imath] and variance [imath]\sigma^2[/imath], taking only positive values. Let [imath]m < n[/imath]. Find the expectiation of [imath]\displaystyle\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}[/imath]. My attemps to solve this probles are rather straightforward. Denote [imath]X = X_1 + \cdots + X_m[/imath] and [imath]Y = X_{m+1} + \dots + X_n[/imath]. So, [imath]X[/imath] has the expectation [imath]ma[/imath] and the variance [imath]m\sigma^2[/imath]. And [imath]Y[/imath] has the expectation [imath](n-m)a[/imath] and variance [imath](n-m)\sigma^2[/imath]. And also [imath]X[/imath] and [imath]Y[/imath] are independent. So we can compute the expectation by the definition [imath]\mathbb{E}\displaystyle\frac{X}{X+Y} = \int\limits_{\Omega^2}\frac{X(\omega_1)}{X(\omega_1) + Y(\omega_2)}\mathbb{P}(d\omega_1)\mathbb{P}(d\omega_2)[/imath]. But we do not know the distribution, so we do not have chance to calculate it. I would be glad to any help or ideas! |
2732575 | Solution of the equation [imath]x=\cos(x)[/imath]
I have to find the solution of the equation [imath]x=\cos(x)[/imath], [imath]x[/imath] is real. I solved it graphically but is there any other way to solve this except numerical approximation? I have seen the Dottie number question asked on this page but I am curious about other equations like [imath]x=\sin(x)[/imath]. [I know the solution is [imath]x=0[/imath]], [imath]x=\sec(x)[/imath] etc., and also when a function [imath]x-f(x)[/imath] has a zero? | 1208845 | How to compute Dottie number accurately?
Dottie number is root of this equation : [imath]cos \alpha = \alpha[/imath], [imath]\alpha \approx 0.73908513321516064165531208767\dots[/imath]. I wonder how can I compute it ? I have tried to do it with an approximating formula: [imath]\alpha = \frac{5\pi^2}{\alpha^2 + \pi^2} - 4[/imath] I have solved this equation and i got [imath]\alpha \approx 0.738305\dots[/imath]. So , how can i compute it accurately ? Can i use taylor series, etc. ? |
2730707 | Homotopy Groups of [imath]S^l \vee S^k[/imath]
Let [imath]S^l \vee S^k[/imath] wedge sum of two spheres [imath]S^l, S^k[/imath]. How can I simplify the calculation of homotopy groups [imath]\pi_n(S^l \vee S^k)[/imath]? I know that [imath]\pi_n(S^l \times S^k) = \pi_n(S^l) \oplus \pi_n(S^k)[/imath], as well [imath]H_n(S^l \vee S^k) = \pi_n(S^l) \oplus H_n(S^k)[/imath] holds, therefore we have [imath]\pi_n ^{ab}(S^l \times S^k) = \pi_n ^{ab}(S^l) \oplus \pi_n ^{ab}(S^k)[/imath], but do we have similar reduction rules for [imath]\pi_n(S^l \vee S^k)[/imath]? If, yes, why, but I suppose not generally. What other strategy can I use to simplify the calculation? | 913022 | Homotopy groups of a wedge sum
Let [imath]X_\alpha[/imath] be connected CW-complexes. Then from Hatcher's book, [imath]\pi_{n}(\prod_{\alpha} X_{\alpha})=\prod_{\alpha}\pi_{n}(X_{\alpha}).[/imath] Is it true in general that [imath]\pi_{n}(\bigvee_{\alpha} X_{\alpha})=\prod_{\alpha}\pi_{n}(X_{\alpha})?[/imath] |
2733126 | Why divide mean curvature by 2?
The mean curvature [imath]H[/imath] of a surface [imath]S[/imath] is given by [imath] H = \frac12(\kappa_1 + \kappa_2) [/imath] where [imath]\kappa_i[/imath] are the principal curvatures, which are the eigenvalues of the shape operator [imath]\text{S}[/imath] over our surface. My question is this: why divide by 2? As far as I can tell there is no reasonable explanation for why we define [imath]H = \frac12(\kappa_1 + \kappa_2)[/imath], and not simply [imath]H = \kappa_1 + \kappa_2[/imath]. At first I thought maybe it's the (arithmetic) "mean" of the principal curvatures, and the Gaussian curvature analogously is the geometric mean of the principal curvatures, but that's not correct because the Gaussin curvature [imath]K = \kappa_1\kappa_2[/imath], which isn't quite the geometric mean, so why should [imath]H[/imath] be the arithmetic mean? Every formula I see involving [imath]H[/imath] just involves an unnecessary [imath]2[/imath]. For example, [imath] H = \frac{EN - 2FM + GL}{2(EG - F^2)} [/imath] where [imath]E, F, G[/imath] are the coefficients of the first fundamental form, and [imath]L, M, N[/imath] are the coefficients of the second fundamental form. Why not just [imath] H = \frac{EN - 2FM + GL}{EG - F^2}? [/imath] Expressing it as the trace of the shape operator (or the Weingarten map) we see [imath] H = \frac12\text{trace}(S) [/imath] why not just [imath]H = \text{trace}(S)?[/imath] In the equation relating the three fundamental forms, we see [imath] \mathbf{III}-2H\mathbf{II}+K\mathbf{I} = 0 [/imath] why not just [imath] \mathbf{III}-H\mathbf{II}+K\mathbf{I} = 0? [/imath] Changing [imath]H[/imath] to [imath]\kappa_1 + \kappa_2[/imath] doesn't affect the theory of minimal surfaces where [imath]H = 0[/imath]. The only compelling evidence for why this might be a reasonable definition is that, for hypersurfaces: [imath] H = \frac1n\sum_{i=1}^n \kappa_i [/imath] according to Spivak, but I don't know much about hypersurfaces so I can't comment on how comparatively useful this definition is. So what's the deal? Was [imath]H = \kappa_1 + \kappa_2[/imath] originally, but when hypersurfaces were studied the definition [imath]H = \frac1n\sum\kappa_i[/imath] was more natural, and later adopted for the [imath]n = 2[/imath] case as well? | 1540919 | Mean curvature is the divergence of the normal
As a definition, I was told that for a surface in 3D, [imath] 2H = -\nabla \cdot \nu[/imath] where [imath]H[/imath] is the mean curvature and [imath]\nu[/imath] is the normal unit vector. In some results that I am studying, the factor 2 always disappears... Is this normal ? can we ignore the factor 2 and consider the definition "up to a constant" ? |
2732969 | Analytical solution of a convergent series
Which is the analytical solution of the infinite sum [imath]\frac{1}{1+1^{2}}+\frac{1}{1+2^{2}}+\frac{1}{1+3^{2}}+\frac{1}{1+4^{2}}+\cdots[/imath]? | 680825 | How to prove that [imath]\sum_{n=1}^{+\infty}\frac{1}{n^2+1}=\frac{-1+\pi \coth (\pi)}{2}[/imath]?
I typed into my Mathematica:[imath]\sum _{n=1}^{\infty } \frac{1}{n^2+1}[/imath] , and the result was: [imath]\frac{-1+\pi \coth (\pi)}{2}[/imath] I know how to estimate the aforementioned sum , but I have no idea how to get its closed form. |
2732930 | Questions about existence of functions satisfying integral properties
Maybe it's elementary, but is it possible that a non-negative continuous function satisfies [imath]\int_{0}^{\infty}f(x)dx[/imath] exists (finite) and the limit [imath]\lim_{x\to\infty}f(x)[/imath] doesn't exist (neither finite nor infinite)? Also, is it possible to find a function such that the integral [imath]\int_{0}^{\infty}f(x)dx[/imath] exists (finite) but [imath]f(x)[/imath] is not bounded? If you could show me examples it would be appreciated. If these are not possible, please explain why. | 2304098 | Can an unbounded function have a finite integral?
I am wondering whether there exists a function such that: [imath]\lim_{x \rightarrow a}f(x)=\infty[/imath] at some point [imath]a[/imath] on the real axis but yet, [imath]\int_{-\infty}^{+\infty}\left|f(x)\right|\ dx<\infty[/imath] Does the fact that a function is unbounded imply that it has no finite integral? |
2733182 | Linear Independence of Power Vectors
Let [imath]p_{1},p_{2},\cdots,p_{K}[/imath] be a non-uniform probability distribution (i.e., not all [imath]p_{k}[/imath]s are the same) with [imath]K[/imath] being a fixed positive integer. Consider the following [imath]K\times K[/imath] matrix. [imath] A=\left( \begin{array}{cccc} 1&1&\cdots& 1 \\ p_{1}&p_{2}&\cdots & p_{K} \\ p_{1}^2&p_{2}^2&\cdots &p_{K}^2 \\ \vdots& \vdots & \vdots & \vdots \\ p_{1}^{K-1}&p_{2}^{-1}K&\cdots &p_{K}^{K-1} \end{array} \right) [/imath] Can [imath]\det(A)[/imath] be zero, in other words, can the the column vectors (or row vectors) of [imath]A[/imath] be linearly dependent? | 1526729 | How to show that the determinant of this matrix is in a nice product factorization,
Show that [imath]det \begin{bmatrix} 1 & 1 & \cdots &1 \\ \lambda_1 & \lambda_2 & \cdots &\lambda_n \\ \lambda^2_1 & \lambda^2_2 & \cdots &\lambda^2_n \\ \vdots & \vdots & \vdots & \vdots \\ \lambda^{n-1}_1 & \lambda^{n-1}_2 & \cdots &\lambda^{n-1}_n \\ \end{bmatrix} =\prod_{n\ge i>j\ge1}(λ_i−λ_j)[/imath] My work so far, I am trying to use induction on [imath]n[/imath]. For the base case n=2, the claim is true, since we have [imath](\lambda_2 - \lambda_1)[/imath]. I'm not sure how to carry out the inductive step. Any hints or suggestions are welcome. Thanks, |
2733785 | Computing Minima for degree 8 polynomial
Is there any better method than derivative or trial and error to calculate minima of [imath]f(x)=x^8-8x^6+19x^4-12x^3+14x^2-8x+9[/imath] The minima occurs at [imath]x=2[/imath] and [imath]f(2)=1[/imath] We were given option too [imath]-1,9,6,1[/imath] if that helps | 2198967 | The minimum value of [imath]x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9[/imath] is
The minimum value of [imath]f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9[/imath] is (a)-1 (b)9 (c)6 (d)1 Apart from trying to obtain [imath]1[/imath], which in this case is simple and [imath]f(2)=1[/imath] is there a standard method to approach such problems. Please keep in mind that this is an objective question in one of the competitive exam and you get around 5 mins to solve it. Also this is asked in elementary section, so only knowledge of basic calculus and polynomials is assumed. |
2734049 | For how many primes [imath]p[/imath] is [imath]p^2 + 2[/imath] is also prime?
For how many primes [imath]p[/imath] is [imath]p^2 + 2[/imath] also prime? I'm not sure but I think you can consider every prime number as [imath](3k+1)[/imath] or [imath](3k+2)[/imath] | 1248160 | Prove that if [imath]p[/imath] is a prime such that [imath]p^2+2[/imath] is a prime then [imath]p=3[/imath].
My problem in my solution is that I don't know if the operations I apply on congruence modulo n are admissible. I could really use some guiding: Attempt: Let there be [imath]p\ne 3[/imath] fulfilling the requirements aforementioned. Every prime number [imath]p\ne 2,3[/imath] satisfies: [imath]p\equiv1 \mod6[/imath] or [imath]p\equiv5 \mod6[/imath]. If [imath]p\equiv1 \mod6[/imath] then [imath]p^2\equiv1 \mod6[/imath] which means [imath]p^2+2\equiv3 \mod6[/imath] which is a clear contradiction. Here's the tricky part(of which I am not sure): [imath]p\equiv5 \mod6[/imath] [imath]\Rightarrow p^2\equiv5 \mod6[/imath] [imath]\Rightarrow p^2+2\equiv1 \mod6[/imath]. That means by definition: [imath]6|(p^2+2)-1[/imath] and [imath]6|p^2-5[/imath] and therefore(?): [imath]6|p^2-4[/imath] deriving [imath]2p^2\equiv 4 \mod 6[/imath] [imath]\Rightarrow p^2\equiv 2 \mod 6[/imath] which is a contradiction. |
1117412 | Proving [imath](\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})[/imath]
Problem: [imath](\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})[/imath] My work: Start by letting [imath]m=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}[/imath]. If [imath](a,p_i)=1[/imath] for some integer [imath]i[/imath], then we have that [imath] a^{\phi(p_i^{a_i})}\equiv 1\pmod{p_i^{a_i}} [/imath] by Euler's theorem. Since [imath]\phi(p_i^{a_i}) \mid \phi(m)[/imath], we have that [imath]p_i^{a_i}\mid (a^{\phi(m)}-1)[/imath] if [imath](a,p_i)=1[/imath]. Beyond this, I am stuck at the moment. | 861301 | Prove [imath]a^m\equiv a^{m-\phi(m)}\pmod m[/imath] for all positive integers
Prove that if [imath]a,m[/imath] are positive integers, then [imath]a^m\equiv a^{m-\phi(m)}\pmod m.\tag 1[/imath] If gcd[imath](a,m)=1[/imath] then this is Euler's theorem. Denote gcd[imath](a,m)=k[/imath] and [imath]a=xk,m=yk[/imath] then we need to prove [imath]a^{m-\phi(m)}(a^{\phi(m)}-1)\equiv 0\pmod m\tag2[/imath] If gcd[imath](k,y)=1[/imath] then [imath]k\mid a^{m-\phi(m)}[/imath] and [imath]y\mid a^{\phi(y)}-1\mid a^{\phi(m)}-1[/imath], hence [imath](2)[/imath] holds. However, how to prove it if gcd[imath](y,k)\gt1[/imath]? |
2734503 | Simple analysis problem that is giving me some grief.
I am given a function [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] that has the property that [imath]f(u+v)=f(u)+f(v)[/imath] for all [imath]u,v\in\mathbb{R}[/imath]. Then we define [imath]m=f(1)[/imath] and I am asked to prove that [imath]f(x)=mx[/imath] for all rational numbers [imath]x[/imath]. This should be pretty straightforward but I can't quite seem to nug this one out. What I wrote out so far is as follows: [imath]f(x)=f(\frac{p}{q})=f(\frac{1_1}{q}+...+\frac{1_p}{q})=f(\frac{1_1}{q})+...+f(\frac{1_p}{q})=pf(\frac{1}{q}).[/imath] This is where I am stuck. | 144559 | If [imath]f(x + y) = f(x) + f(y)[/imath] showing that [imath]f(cx) = cf(x)[/imath] holds for rational [imath]c[/imath]
For [imath]f:\mathbb{R}^n \to \mathbb{R}^m[/imath], if [imath]f(x + y) = f(x) + f(y)[/imath] for then for rational [imath]c[/imath], how would you show that [imath]f(cx) = cf(x)[/imath] holds? I tried that for [imath]c = \frac{a}{b}[/imath], [imath]a,b \in \mathbb{Z}[/imath] clearly [imath] f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right) [/imath] but I can't seem to finish it, any help? |
2734863 | Complete the square on cross-terms?
I am trying to complete the square on: [imath]xy - xz + yz = 1[/imath]. I tried plugging in: [imath]u = x + y[/imath] If you plug in [imath]u[/imath] you get: [imath]-y^2 + uy - uz + 2yz = 1[/imath] but I'm not sure where to go from here. | 2734782 | Complete the square on cross terms?
I am trying to complete the square on [imath]xy - xz + yz = 1[/imath]. I have tried plugging in [imath]u = x + y[/imath]. But I still end up with unsolvable cross terms. |
2735092 | Show that [imath]X_n[/imath] converge a.s
I want to solve the following exercise from Durrett: "Let [imath]X_n, n\geq0[/imath] be a submartingale with [imath]\sup X_n<\infty[/imath]. Let [imath]\xi_n = X_n - X_{n-1}[/imath] and suppose [imath]\mathbb{E}\left(\sup\xi^+\right)<\infty[/imath]. Show that [imath]X_n[/imath] converge a.s. " I am not sure to understand the hypothesis on the increments because I have the feeling that I can solve the exercise without it : [imath]\sup X_n<\infty[/imath] implies that we can find [imath]K[/imath] such that for all integer n [imath]X_n<K[/imath] thus we can consider the submartingale [imath]Y_n = X_n - K[/imath] and since [imath]\sup\mathbb{E}(Y_n^+)=0 < \infty[/imath] by the Martingale convergence theorem [imath]Y_n[/imath] converge a.s and thus [imath]X_n[/imath] too. Is this correct ? | 1578497 | Submartingale convergence (Durrett 5.3.1)
Exercise 5.3.1 in Durrett's "Probability Theory and Examples" states Let [imath]X_n[/imath], [imath]n\ge 0[/imath], be a submartingale with [imath]\sup X_n < \infty[/imath]. Let [imath]\xi_n=X_n-X_{n-1}[/imath], and suppose [imath]E(\sup \xi_n^+)<\infty[/imath]. Show that [imath]X_n[/imath] converges almost surely. Here are my thoughts: It looks like we need to use theorem 5.2.8 (Martingale convergence theorem), which states that [imath]X_n[/imath] converges almost surely if [imath]\sup EX_n^+<\infty[/imath]. I am trying to understand why we can just do the following (or at least I'm guessing we can't): Suppose [imath]X_n > 0[/imath] (edit: on some set with positive measure), then [imath]0< \sup X_n < \infty[/imath], and therefore [imath]EX_n^+ \le \sup X_n < \infty[/imath]. So we're done. Otherwise, if [imath]X_n \le 0[/imath] a.e., then [imath]\sup X_n^+ = 0[/imath] a.e. Once again, we're done. Am I missing something? Since I didn't use all the information, I probably am. Thanks for the help. |
2735257 | Rank of the sum of an all-ones matrix and an identity matrix
Let [imath]B[/imath] be an all-ones matrix of dimension [imath]n \times n[/imath], and [imath]I[/imath] be the identity matrix of dimension [imath]n \times n[/imath]. Define [imath]A = \frac{1}{n}B - I[/imath]. For [imath]n=3[/imath], [imath]A[/imath] would yield \begin{equation*} \begin{bmatrix} -2/3 & 1/3 & 1/3 \\ 1/3 & -2/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \end{bmatrix}. \end{equation*} How would one compute rank[imath](A)[/imath] for any [imath]n[/imath]? I have noted that any row [imath]r_i[/imath] of [imath]A[/imath] can be written as a linear combination of the other [imath]n-1[/imath] rows. Specifically, [imath]r_i = - \sum_{j\neq i} r_j[/imath]. This implies rank[imath](A) < n[/imath], but I don’t know how to compute the exact rank of [imath]A[/imath]. Any hint would be appreciated. | 2181367 | Rank of the [imath]n \times n[/imath] matrix with ones on the main diagonal and [imath]a[/imath] off the main diagonal
I want to find the rank of this [imath]n\times n[/imath] matrix \begin{pmatrix} 1 & a & a & \cdots & \cdots & a \\ a & 1 & a & \cdots & \cdots & a \\ a & a & 1 & a & \cdots & a \\ \vdots & \vdots & a& \ddots & & \vdots\\ \vdots & \vdots & \vdots & & \ddots & \vdots \\ a & a & a & \cdots &\cdots & 1 \end{pmatrix} that is, the matrix whose diagonals are [imath]1's[/imath] and [imath]a[/imath] otherwise, where [imath]a[/imath] is any real number. My first observation is when [imath]a=0[/imath] the rank is [imath]n[/imath] and when [imath]a=1[/imath] the rank is [imath]1.[/imath] Then I can assume [imath]a\neq 0, 1[/imath] and proceed row reduction to find its pivot rows. I obtain \begin{pmatrix} 1 & a & a & \cdots & a \\ 0 & 1+a & a & \cdots & a \\ 0 & a & 1+a & \cdots & a \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & a & a & \cdots & 1+a \end{pmatrix} by subtracting the first row multiplied [imath]a[/imath] for each row below the first, and then divides the factor [imath](1-a)[/imath], and stuck there. Any hints/helps? |
2733419 | How do I prove the following inequality (link included)?
For any two events [imath]A[/imath] and [imath]B[/imath], show that [imath](P(A\cap B))^2+(P(A\cap B^c))^2+(P(A^c\cap B))^2+(P(A^c\cap B^c))^2\ge\frac 1 4[/imath] I tried solving the question by writing [imath]A \cap B^c = A - (A \cap B)[/imath] and similiarly expanding the other brackets. But I am not able to proceed any forward. I also tried applying the Boole's inequality and then squaring the respective terms, but still didn't get the required answer. Can somebody help me with this question, please? I would be extremely grateful. | 2114667 | Proving [imath](P(AB))^2+(P(AB^c))^2+(P(A^cB))^2+(P(A^cB^c))^2\ge\frac{1}{4}[/imath] for any events [imath]A[/imath] and [imath]B[/imath]
Show that for any two events [imath]A[/imath] and [imath]B[/imath], [imath](P(AB))^2+(P(AB^c))^2+(P(A^cB))^2+(P(A^cB^c))^2\ge\frac{1}{4}[/imath] Using the Inclusion-Exclusion formula, De Morgan's law and [imath]P(AB^c)=P(A)-P(AB)[/imath], I was able to expand the [imath]\rm{l.h.s.}[/imath] as [imath](x+y)(x+y-4z)+2z(2z+1)+x(x-2)+y(y-2)+1\quad...\quad(*),[/imath] where [imath]x=P(A),y=P(B)[/imath] and [imath]z=P(AB)[/imath]. However I'm not being able to group the terms involved in the [imath]\rm{l.h.s.}[/imath] to form a 'suitable' expression and probably the way I have grouped them is not correct. The expression [imath](*)[/imath] is subject to the condition [imath]\max\{0,x+y-1\}\le z\le \min\{x,y\}[/imath] and obviously [imath]0\le x,y,z\le 1[/imath], but I am unable to use this. Is there an elegant/slick way to show the inequality? It probably can be shown without a lot of work and it's possible I'm missing something trivial here. |
2735506 | I am finding it difficult to understand this adjacency matrix property- graph theory
The following properties I am having difficulty with Let G be a graph with vertices v 1, v 2 , ..., vn and let A = ai,j be the adjacency matrix of G. The diagonal entries of A are all 0; that is, a i i = 0 for i = 1,...,n. The adjacency matrix is symmetric, that is a i j = a j i for all i, j. degv i i is the number of 1’s in row i; this is also the number of 1’s in column i (row i and column i are the same) I do not get these properties: The [imath](i,j)[/imath] entry of [imath]A^2[/imath] is the number of different walks of length [imath]2[/imath] from [imath]v_i[/imath] to [imath]v_j[/imath], in particular, the degree of [imath]v_i[/imath] is the ith main diagonal entry of [imath]A^2[/imath]. In general, for any [imath]k\gt 1[/imath], the [imath](i,j)[/imath] entry of [imath]A^k[/imath] is the number of walks of length [imath]k[/imath] from [imath]v_i[/imath] to [imath]v_j[/imath]. so.. If I have a triangle graph with vertices = v1, v2 , v3 And Adjacency matrix [imath]A=\begin{pmatrix}0 & 1 & 1\\1 & 0 & 1\\ 1 & 1 & 0\end{pmatrix}\text{ then }A^2=\begin{pmatrix}2 & 1 & 1\\1 & 2 & 1 \\ 1 & 1 & 2\end{pmatrix}[/imath] then the property [imath](i,j)[/imath] entry of [imath]A^2[/imath] is the number of different walks of length [imath]2[/imath] from [imath]v_i[/imath] to [imath]v_j[/imath], in particular, the degree of [imath]v_i[/imath] is the ith main diagonal entry of [imath]A^2[/imath]. means that from v1 to v1 the number of different walks with length two is 2? Any help is appreciated thanks | 2434064 | Proof - raising adjacency matrix to [imath]n[/imath]-th power gives [imath]n[/imath]-length walks between two vertices
I came across the formula to find the number of walks of length [imath]n[/imath] between two vertices by raising the adjacency matrix of their graph to the [imath]n[/imath]-th power. I took me quite some time to understand why it actually works. I thought it would be useful to write the proof by induction for this in my own words. Theorem: Raising an adjacency matrix [imath]A[/imath] of simple graph [imath]G[/imath] to the [imath]n[/imath]-th power gives the number of [imath]n[/imath]-length walks between two vertices [imath]v_i[/imath], [imath]v_j[/imath] of [imath]G[/imath] in the resulting matrix. Proof by induction: Let [imath]P(n)[/imath] be the predicate that the theorem is true for [imath]n[/imath]. We let [imath]F^{(n)}_{ij}[/imath] be the number of [imath]n[/imath]-length walks between vertex [imath]v_i[/imath] and [imath]v_j[/imath]. [imath]P(n)[/imath] is then the predicate that [imath]F^{(n)}_{ij} = A^n_{ij}[/imath]. We proceed by induction on [imath]n[/imath]. Base case: [imath]P(1)[/imath] Case 1: [imath]F^{(1)}_{ij} = A^{(1)}_{ij} = 1[/imath] if [imath]\{v_i, v_j\} \in E[/imath], so there is is a walk of length [imath]1[/imath] between [imath]v_i[/imath], [imath]v_j[/imath]. Case 2: [imath]F^{(1)}_{ij} = A^{(1)}_{ij} = 0[/imath] if [imath]\{v_i, v_j\} \notin E[/imath], so there can't be any walk of length [imath]1[/imath] between [imath]v_i[/imath] and [imath]v_j[/imath]. In both cases [imath]F^{(1)}_{i j} = A^{(1)}_{ij}[/imath] holds, hence [imath]P(1)[/imath] is true. Inductive step: [imath]P(n+1)[/imath] For purpose of induction, we assume [imath]P(n)[/imath] is true, that is [imath]F^{(n)}_{i j} = A^{(n)}_{ij}[/imath] holds for [imath]n[/imath]. We can express a walk of length [imath]n+1[/imath] from [imath]v_i[/imath] to [imath]v_j[/imath] as a [imath]n[/imath]-length walk from [imath]v_i[/imath] to [imath]v_k[/imath] and a walk of length 1 from [imath]v_k[/imath] to [imath]v_j[/imath]. That means, the number of [imath]n+1[/imath]-length walks from [imath]v_i[/imath] to [imath]v_j[/imath] is the sum over all walks from [imath]v_i[/imath] to [imath]v_k[/imath] times the number of ways to walk in one step from [imath]v_k[/imath] to [imath]v_j[/imath]. Thus: [imath]F^{(n+1)}_{ij} = \sum_{k=1}^{|V|} A_{kj}F^{(n)}_{ik} = \sum_{k=1}^{|V|} A_{kj}A^{(n)}_{ik}[/imath] Which is the formula for the dot-product, used in matrix multplications. Any feedback appreciated. |
2736053 | A binomial probabilistic inequality
Let [imath]n[/imath] and [imath]1\leq k \leq n[/imath] be natural numbers. Prove the inequality [imath]\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} \leq 1 - \frac{1}{e} [/imath] Equivalently, if [imath]X\sim[/imath] Bin([imath]n[/imath],[imath]\frac{k}{n+1}[/imath]), prove that [imath]\mathbb{P}[X\geq k] \leq 1 - \frac{1}{e}[/imath]. My attempt: It may be helpful to show that the LHS tends to a limit less than [imath]1-\frac{1}{e}[/imath] as [imath]n \to \infty[/imath] (already did that) and that the LHS is an increasing function on [imath]n[/imath] (have not done that). | 2714505 | Inequality with Binomial distribution
Let [imath]n[/imath] and [imath]1\leq k \leq n[/imath] be natural numbers. Prove the inequality [imath]\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} \leq 1 - \frac{1}{e} [/imath] Equivalently, if [imath]X\sim[/imath] Bin([imath]n[/imath],[imath]\frac{k}{n+1}[/imath]), prove that [imath]\mathbb{P}[X\geq k] \leq 1 - \frac{1}{e}[/imath]. My attempt: It may be helpful to show that the LHS tends to [imath]1-\frac{1}{e}[/imath] as [imath]n \to \infty[/imath] (already did that) and that the LHS is an increasing function on [imath]n[/imath] (have not done that). |
2736174 | Graph Theory - Triangles
This is a question taken from some notes, and I do not know how to solve it, nor I know how to start. Can anyone help me? Prove that [imath]|E| \geq \lfloor \frac{|V|}{2} \rfloor ^2 [/imath] implies the graph [imath]G[/imath] has at least one triangle. | 628877 | Mantel's Theorem proof verification
I found the following proof for Mantel's theorem in Lecture 1 of David Conlon's "Extremal graph theory" course. I cannot understand the equality that I have highlighted in the image was arrived at. I would appreciate some assistance. Theorem 1 (Mantel's theorem) *If a graph [imath]G[/imath] on [imath]n[/imath] vertices contains no triangle then it contains at most [imath]\frac {n^2}{4}[/imath] edges. First proof Suppose that [imath]G[/imath] has [imath]m[/imath] edges. Let [imath]x[/imath] and [imath]y[/imath] be two vertices in [imath]G[/imath] which are joined by an edge. If [imath]d(v)[/imath] is the degree of a vertex [imath]v[/imath], we see that [imath]d(x)+d(y)\leq n[/imath]. This is because every vertex in the graph [imath]G[/imath] is connected to at most one of [imath]x[/imath] and [imath]y[/imath]. Note now that [imath]\bbox[5px,border:2px solid red]{ \sum_x d^2(x)=\sum_{x,y\in E} \big( d(x)+d(y)\big) } \leq mn.[/imath] On the other hand, since [imath]\sum_x d(x)=2m[/imath], the Cauchy-Schwarz inequality implies that [imath]\sum_x d^2(x)\geq\frac{\big(\sum_x d(x)\big)^2}{n} \geq \frac{4m^2}{n}.[/imath] Therfore [imath]\frac{4m^2}{n} \leq mn,[/imath] and the result follows. [imath]\tag*{[/imath]\square[imath]}[/imath] For people with visulal limitations, the highlighted part is: [imath]\sum_x d^2(x)=\sum_{x,y\in E} \big( d(x)+d(y)\big) \leq mn.[/imath] |
2736546 | Fundamental group of infinite row of tori
Consider a family [imath] \lbrace T_i \rbrace_{i\in\mathbb{Z}} [/imath] of tori indexed on the integers such that every torus is attached to the preceding one and to the subsequent by a single point. How can we compute the fundamental group of this infinite row of tori? I want to stress the fact that «every torus is attached to the preceding one and to the subsequent by a single point» therefore, this is not a duplicate of the question Fundamental Group is free on infinite generators. | 1082073 | Fundamental Group of surface with infinite genus is free on infinite generators
This is question 16 of section 1.2 in Hatcher's Algebraic Topology. I have to show that the fundamental group of the space [imath]X[/imath] is free on an infinite number of generators. So here is my approach. Using van-Kampen, I can see that [imath]\pi_1(X)=\pi_1(U)\star\pi_1(V)[/imath]. Also [imath]U[/imath] and [imath]V[/imath] are homeomorphic. So I've to calculate [imath]\pi_1(U)[/imath]. I have that the fundamental group of [imath]U_g[/imath] (torus of genus [imath]g[/imath] with two points removed) is free on 3g [imath]2g+1[/imath] generators. Now in some sense, [imath]U=\lim_{g\to\infty}{U_g}[/imath] and so [imath]\pi_1(U)[/imath] is free on infinite generators. But how do I justify this? Is it even true? My second approach is, using van-Kampen, [imath]\pi_1(U)=\pi_1(U_1)\star\pi_1(U)[/imath]. This is an infinite recursion. But again I cannot proceed further. Any help is appreciated. Thanks! |
2736713 | Evaluate [imath]\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}[/imath]
A user just posted a similar question (What is the summation of [imath]\sum_{k=0}^{\infty}\frac{k}{2^k}[/imath]?) with [imath]n[/imath] instead of [imath]n^2[/imath], and this question got some original answers, so I would like to propose the question: Find the sum [imath]\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}.[/imath] I am really interested to see what original answers are given to this question. There are number of different approaches, some I havent seen or though of before. Thus although the question may be unoriginal, some of the solutions are not. Thus it is inappropriate to close this question. | 1211834 | Find limit for infinite sum
I'm trying to determine the limit of the sum [imath]\lim_{n\to\infty} \sum\limits_{k=1}^n k^2/2^k[/imath] Doing the convergence test shows the sum converges [imath]\lim_{n\to\infty} \frac{(k+1)^2/2^{k+1}}{k^2/2^k} = 1/2[/imath] so there must be a solution, but the normal tricks used to solve geometric series or telescoping series don't seem to work. |
2737141 | How can I calculate [imath]\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n}}\cdot(n/2)!\cdot(n/2)!}{n!} [/imath]?
How can I calculate this limit? [imath]\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n}}\cdot(n/2)!\cdot(n/2)!}{n!} [/imath] I thought to try the ratio test, but I don't know how to do it because that I get [imath](\frac{n+1}{2})![/imath] and [imath](\frac{n}{2})![/imath] such that it's not define. | 2737097 | How can I calculate this limit? ideas?
How can I calculate this limit? [imath]\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}[/imath] I don't have idea and I will be happy for help. |
2737524 | Integral limit of function on unit interval
Let [imath]f[/imath] be a real valued continuous function defined on [imath][0,1][/imath] so that [imath]f(0) = 1,[/imath] [imath]f(1/2) = 2,[/imath] [imath]f(1) = 3.[/imath] Show the limit, [imath]\lim_{n \rightarrow \infty} \int_0^1 f(x^n)\,dx[/imath] exists and compute its limit. It seems as if we can use continuity to say if [imath]x_n[/imath] goes to [imath]x,[/imath] then [imath]f(x_n) \rightarrow f(x) = f(0) = 1[/imath], but I don't know why they give [imath]f(1/2)[/imath] and [imath]f(1)[/imath], plus [imath]x^n[/imath] is not uniformly continuous on [imath][0,1][/imath] so I'm a little unsure. | 2594866 | Evaluate [imath]\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx[/imath]
Let [imath]f[/imath] be a real, continuous function defined for all [imath]0\leq x\leq 1[/imath] such that [imath]f(0)=1[/imath], [imath]f(1/2)=2[/imath], and [imath]f(1)=3[/imath]. Show that [imath]\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx[/imath] exists and compute the limit. Attempt: Since [imath]f[/imath] is real-valued continuous on [imath]0\leq x\leq 1[/imath] and the boundaries [imath]f(0)=1[/imath] and [imath]f(1)=3[/imath], [imath]f[/imath] is bounded on the interval and the integral [imath]\displaystyle\int_0^1 f(x^n)dx[/imath] exists for any positive [imath]n[/imath]. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &=\int_0^1 f(0)dx=1. \end{align*} Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where [imath]f(x^n)[/imath] is dominated by a function [imath]g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)[/imath] and [imath]f_n(x^n)[/imath], on [imath]0\leq x\leq 1[/imath], converges pointwise to a function [imath]f[/imath] that takes value 1 for [imath]x\neq 1[/imath] and 3 for [imath]x=1[/imath]. My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified. |
726324 | An exercise in Gathmann's lecture notes about Projective spaces
Exercise 3.5.1 in Gathmann's lecture notes: Let [imath]L_1[/imath] and [imath]L_2[/imath] be two disjoint lines in [imath]\mathbb{P}^3[/imath], and let [imath]P\in \mathbb{P}^3 \setminus (L_1 \cup L_2)[/imath] be a point. Show that there is a unique line L meeting [imath]L_1[/imath], [imath]L_2[/imath], and [imath]P[/imath] (i. e. such that [imath]P \in L[/imath] and [imath]L \cap L_i \neq \emptyset[/imath] for [imath]i=1,2[/imath]. And is it true that a line in [imath]\mathbb{P}^3[/imath] is given by a system of two homegeneous linear equations? | 115949 | Unique line in [imath]\mathbb{P}^3[/imath] through a point [imath]p[/imath] and intersecting two disjoint lines
I'm a bit stuck with this exercise from a script I'm reading, and I'm not very familiar with projective [imath]n[/imath]-space yet. The problem: Let [imath]L_1[/imath] and [imath]L_2[/imath] be two disjoint lines in [imath]\mathbb{P}^3[/imath], and let [imath]p\in\mathbb{P}^3\smallsetminus(L_1\cup L_2)[/imath]. Show that there is a unique line [imath]L\subseteq\mathbb{P}^3[/imath] meeting [imath]L_1[/imath], [imath]L_2[/imath], and [imath]p[/imath] (i.e. such that [imath]p\in L[/imath] and [imath]L\cap L_i\neq\varnothing[/imath] for [imath]i=1,2[/imath]). To be honest, I already have a problem with the term 'line'. As I take it, a line in [imath]\mathbb{P}^3[/imath] should be something cut out by two degree-1 polynomials (homogeneous, since it wouldn't be well defined otherwise, right?). But what are disjoint lines in projective space? As far as I understood it, two distinct lines should always intersect in exactly one point there, so how can they be disjoint? Can it be that these two statements mean a different kind of 'line'? Any explanation of this, hints, or even a solution would be very appreciated. Thanks in advance! |
2732391 | Derivative of an analytic one-to-one map has no zeros
Let [imath]G[/imath] be a region in a complex plane. Suppose [imath]f:G\to\mathbb C[/imath] is analytic and one-one. Show that [imath]f'(z)[/imath] is not zero for any [imath]z[/imath] in [imath]G[/imath]. I was trying to do the contradiction method. But I can't conclude anything. Please help | 690499 | Analytic function not injective in any neighborhood of a point if the derivative at that point is zero?
Let [imath]f[/imath] be analytic in some neighborhood [imath]U[/imath] with [imath]z_0 \in U[/imath]. I've read that if [imath]f'(z_0) = 0[/imath], then [imath]f[/imath] is not one-to-one in any neighborhood of [imath]z_0[/imath]. Why is this the case? |
2737756 | Players A and B dice game
Player A and B are playing a game where they roll a dice and the first player who rolls [imath]6[/imath] wins. What is the probability that player A wins if they goes first? I said [imath]\dfrac{1}{6}[/imath] but my teacher said that was wrong. Not really the best at probability, sorry! | 1973508 | Alternate moves, first one to roll a 6 wins: what's the probability of winning?
Question: Players A and B play a sequence of independent games. Player A starts and throws a fair die and wins on a "six". If she fails then player B throws the same die and wins on a "six" or fails otherwise. If she fails then player A gets the die and throws again, and so on. The first player to throw a "six" wins. Find the probability that player A wins. Here is what I have done: A wins right away: [imath]\dfrac{1}{6}[/imath] A fails B wins: [imath]\dfrac{5}{6}\times \dfrac{1}{6}[/imath] A fails B fails A wins: [imath]\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}[/imath] Unlike another similar question which the player B can win on "five" and "six" after player A fails on the first throw, in this case, the games do not seem to have an endpoint. How should I calculate the probability that player A wins then? |
2738363 | Why is the possibility [imath]x-y = 0[/imath] rejected?
I'll state the question from my textbook below: If [imath]x (1+y)^{1/2} + y(1+x)^{1/2} = 0[/imath], for, [imath]-1<x<1[/imath], prove that [imath]\frac{dy}{dx} = - \frac{1}{(1+x)^2}[/imath] Here's how I tried proving it: [imath]x (1+y)^{1/2} + y(1+x)^{1/2} = 0[/imath] Rearranging and squaring we get: [imath]x^2(1+y) = y^2(1+x)[/imath] Simplifying and factorizing we get: [imath](x+y+xy)(x-y) = 0[/imath] Case I: [imath]x+y+xy = 0[/imath] From here we get the desired equation. Since the proof is irrelevant to my question, I'll skip it. But I note that if I replace the value of [imath]y[/imath] I get from this equation is in harmony with the original equation given in the question. Case II: [imath]x-y = 0[/imath] Clearly in this case, [imath]\frac{dy}{dx} = 1[/imath] This is where the problem arises. Isn't this a valid answer too? Probably not, because the value of [imath]y[/imath] I get from this equation satisfies the original equation only for [imath]x = -1,0[/imath]. Why is this? Here are the questions I'm looking answer for: 1) Is something wrong with squaring the equation? 2) Is there a mathematical reason to reject [imath]x-y = 0[/imath] as a possibility? I am also interested in alternative methods to prove the required equation. So if you have any, feel free to post them as an answer. Any help would be appreciated. | 2727358 | If [imath]x\sqrt{1+y}+y\sqrt{1+x}=0[/imath] find [imath]y'[/imath]
Find [imath]\frac{dy}{dx}[/imath] if [imath]x\sqrt{1+y}+y\sqrt{1+x}=0[/imath] for [imath]-1\leq x\leq 1[/imath] My Attempt [imath] x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\ 2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\ \frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\ \frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy} [/imath] How do I proceed further and find the derivative ? |
2737465 | Why does no minimal surface in [imath]\mathbb{R}^3[/imath] exist that is diffeomorphic to the [imath]2[/imath]-sphere?
I stumpled upon the following question in one of my exercise-sheets: Justify that there are no minimal surfaces in [imath]\mathbb{R}^3[/imath] that are diffeomorphic to the 2-sphere [imath]S^2[/imath] I have no idea though. Can someone elaborate? What's the approach i am supposed to take on this? I highly appreciate any hints! Thank you very much. | 544134 | There are no compact minimal surfaces
This is one of the exercises of 'Do Carmo' (Section 3.5, 12) How do you prove that there are no compact (i.e., bounded and closed in [imath]\mathbb{R}^3[/imath]) minimal surfaces? Thanks! |
2738527 | [imath]f[/imath] and [imath]|f|[/imath] are analytic.
If [imath]f[/imath] and [imath]|f|[/imath] are analytic on a domain(open and connected) [imath]D \subset \mathbb{C}[/imath] , then show [imath]f[/imath] is constant on [imath]D[/imath]. What I did: Since [imath]|f|[/imath] is analytic and since it is a real valued function then its derivative is zero in the domain and hence [imath]|f|[/imath] is constant. Thus we must have the derivative of [imath]f[/imath] is [imath]0[/imath] ( which I know how to prove). Thus [imath]f[/imath] must be constant. Is my logic correct? Thanks | 2454269 | If [imath]f[/imath] is analytic then [imath]|f|[/imath] is not constant unless [imath]f[/imath] is constant
Let [imath]f[/imath] be analytic in a domain (open connected set of the domain of definition) (that is, every point of that domain is such that [imath]f'[/imath] exists in that point and also in a neighborhood of radius [imath]r>0[/imath] of that point). Show that his absolute value [imath]|f|[/imath] can't be constant unless [imath]f[/imath] is constant too. Firstly, I'm a little confused about the statement. I must prove [imath]|f|[/imath] is constant [imath]\Rightarrow f[/imath] is constant? Assuming that's what I must show, here's my attempt: if [imath]f[/imath] and [imath]\overline{f}[/imath] are both analytic then [imath]f[/imath] is constant so if [imath]f[/imath] is not constant and [imath]f[/imath] is analytic we have that [imath]\overline{f}[/imath] is not analytic at some point [imath]p[/imath]. But [imath]|f| = \sqrt{f\overline{f}}[/imath] so [imath]|f|[/imath] is also not analytic at the point [imath]p[/imath] so it can't be constant because constant functions are analytics at all points of their domain. Is my attempt correct? Thanks in advance. |
2739543 | What is the radius of convergence of the Taylor series?
What is the radius of convergence of the taylor series of[imath] \frac{2\tan x}{1+4x^2}[/imath] around [imath]x = 0[/imath] ? my attempsts:The function [imath]f(x) := \frac{2\tan x}{1+4x^2}[/imath] is odd and the coefficient of [imath]x^{2n+1}[/imath] of its Taylor expansion at [imath]0[/imath] is equal to [imath]\frac{2\tan x}{1+4x^2}=2\sum_{n=0}^{\infty}(-1)^{2n+1}\frac{ x^{2n+1}}{2n+1}\cdot \sum_{k=0}^{\infty}(-4x^2)^{k}[/imath] Now i can not able proceed further ...pliz help as i have found this solution but i did not understand this...solution me...What is the radius of convergence of the Taylor series of [imath]\frac{2\tan x}{1+4x^2}[/imath] around [imath]x=0?[/imath] | 2662932 | What is the radius of convergence of the Taylor series of [imath]\frac{2\tan x}{1+4x^2}[/imath] around [imath]x=0?[/imath]
What is the radius of convergence of the Taylor series of [imath]\frac{2\tan x}{1+4x^2}[/imath] around [imath]x=0?[/imath] I have to find out [imath]a_n=\frac {f^{(n)}(0)}{n!}[/imath], but the [imath]n-[/imath]th derivative of the given function is not easy to handle. Please help. Thanks, |
2739383 | Calculate the limit [imath]\frac{1}{2}[/imath],[imath]\frac{\frac{1}{2}}{\frac{3}{4}}[/imath],...
[imath]a_1=\frac{1}{2},a_2=\frac{\frac{1}{2}}{\frac{3}{4}},a_3=\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}[/imath]... Prove that [imath]\lim\limits_{n\rightarrow \infty}{a_n}=\frac{\sqrt{2}}{2}[/imath] Notice that if [imath]\frac{i}{i+1}[/imath] is a numerator, then [imath]\frac{i+2^n}{i+1+2^n}[/imath] is a denominator. Each pair of number which is numerator is different from the pair of number that minus [imath]2^n[/imath]. For example, we notice that [imath]a_3=\frac{1}{2}\frac{4}{3}\frac{6}{5}\frac{7}{8}[/imath] and [imath]\frac{1}{2}\rightarrow 00,\frac{3}{4}\rightarrow 01,\frac{5}{6}\rightarrow 10,\frac{7}{8}\rightarrow 11[/imath] I check the result by MATLAB. | 2555815 | Find the limit of a series of fractions starting with [imath]\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}[/imath]
Problem Let [imath]a_{0}(n) = \frac{2n-1}{2n}[/imath] and [imath]a_{k+1}(n) = \frac{a_{k}(n)}{a_{k}(n+2^k)}[/imath] for [imath]k \geq 0.[/imath] The first several terms in the series [imath]a_k(1)[/imath] for [imath]k \geq 0[/imath] are: [imath]\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}, \, \frac{\frac{1/2}{3/4}/\frac{5/6}{7/8}}{\frac{9/10}{11/12}/\frac{13/14}{15/16}}, \, \ldots[/imath] What limit do the values of these fractions approach? My idea I have calculated the series using recursion in C programming, and it turns out that for [imath]k \geq 8[/imath], the first several digits of [imath]a_k(1)[/imath] are [imath] 0.7071067811 \ldots,[/imath] so I guess that the limit exists and would be [imath]\frac{1}{\sqrt{2}}[/imath]. |
2739560 | [imath]\sin(2\pi q)[/imath] is algebraic over [imath]\mathbb{Q}[/imath]
Let [imath]q[/imath] be an element of [imath]\mathbb{Q}[/imath] (rational numbers). How can I prove that [imath]\sin (2\pi q)[/imath] is algebraic over [imath]\mathbb{Q}[/imath] for any [imath]q[/imath]? I am trying the method: Euler formula: [imath]e^{i\theta} =\cos \theta+i\sin \theta[/imath] so [imath]2\sin\theta = ie^{i\theta}-ie^{-i\theta}[/imath]. [imath]i[/imath] is root of [imath]x^2+1[/imath]. So if I prove that [imath]e^{i\theta}[/imath] and [imath]e^{-i\theta}[/imath] is algebraic over [imath]\mathbb{Q}[/imath]f or [imath]\theta=2\pi q[/imath], it will be ok. | 2188739 | Is [imath]\sin(p \pi/q)[/imath] necessarily an algebraic number?
Here [imath]p[/imath] and [imath]q[/imath] are co-prime integers. Is it necessarily an algebraic number? |
2739613 | Showing that 2 elements of a vectorspace are equal
I don't know how to solve the following problem. Suppose that dim(V) = n and W = [imath]\mathbb{R}[/imath]. The vectorspace V* = Hom(V, [imath]\mathbb{R}[/imath]) is called the dual space of V. An element [imath]\phi[/imath] [imath]\in[/imath] V* is called a (linear) functional. Now, suppose that v, w [imath]\in[/imath] V such that [imath]\phi[/imath](v) = [imath]\phi[/imath](w) for all [imath]\phi[/imath] [imath]\in[/imath] V*. Can we conclude that v = w? Thanks for your help. | 1138758 | Linear functional f(v)=0 imply v=0
Good day, I want to prove the following theorem: 1) For any nonzero vector [imath] v \in V [/imath], there exists a linear funtional [imath] f \in V^*[/imath] for wich [imath]f(v) \neq 0 [/imath] I know that if [imath]f[/imath] is a lineal functional then we have 2 posibilities 1)[imath]\dim \ker(f)=\dim V[/imath] 2)[imath]\dim \ker(f)=\dim V-1[/imath] I've tried to suppose that, for all [imath]v \neq 0 [/imath] and [imath]f \in V^*[/imath] We have [imath]f(v)=0[/imath], Then [imath]Im(f)=0[/imath] for all [imath]f \in V^*[/imath] and [imath] v \in V[/imath] but then the only linear functional is [imath]f=0[/imath] ([imath]V^*=\{0:V \to F\}[/imath]) and from this follows: [imath]\dim \ker(f)=\dim V[/imath]. But 2) suggest the existence of a linear functional which [imath]\dim \ker(f)=\dim V-1[/imath], so ¿is that the contradiction? |
2739213 | How can I find the maximum curvature of [imath]y = \ln x[/imath]?
Title says it all. I know that you can use the following equation to find curvature (kappa) for a parametric curve, [imath]\frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|} [/imath] but I don't know how to do this for regular functions. Answer is at the point [imath]\left(\frac{1}{\sqrt2},\ln\big(\frac{1}{\sqrt2}\big)\right)[/imath]. | 316328 | At what point is the curvature of [imath]y=-\ln(x)[/imath] maximal?
I used the equation [imath]k = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}.[/imath] Just by looking at its would think the max is at [imath]x = 1, y = 0[/imath]. What is the correct answer? |
2737414 | How to factor [imath]5[/imath] in [imath]\mathbb{Q}[\sqrt[3]{2}][/imath]?
I'm learning to use PARI/GP and would like to factor the number [imath]5[/imath] in the number field [imath]K=\mathbb{Q}[\sqrt[3]{2}][/imath] Can someone help me interpret the output? ? K = nfinit(x^3-2); idealfactor(K,5) %4 = [[5, [ 2, 1, 0]~, 1, 1, [-1, 2, -4; -2, -1, 2; 1, -2, -1]], 1; [5, [-1, -2, 1]~, 1, 2, [ 2, 0, 2; 1, 2, 0; 0, 1, 2]], 1] This is supposed to be a factorization of [imath](5)[/imath] into ideals, but also I'd like to see a factorization into numbers in [imath]\mathcal{O}_K[/imath] if possible. I can say just a little bit more ? K = nfinit(x^3-2); ? P = idealprimedec(K,5) %1 = [[5, [2, 1, 0]~, 1, 1, [-1, 2, -4; -2, -1, 2; 1, -2, -1]], [5, [-1, -2, 1]~, 1, 2, [2, 0, 2; 1, 2, 0; 0, 1, 2]]] ? P = idealprimedec(K,5); ? [p1,p2]=P; ? p1 %2 = [5, [2, 1, 0]~, 1, 1, [-1, 2, -4; -2, -1, 2; 1, -2, -1]] ? p1.e %3 = 1 ? p1.f %4 = 1 ? p1.gen %5 = [5, [2, 1, 0]~] and some notations (from the PARI/GP docs) The result is a vector of prid structures, each representing one of the prime ideals above p in the number field nf. The representation pr = [p,a,e,f,mb] of a prime ideal means the following: a is an algebraic integer in the maximal order [imath]\mathbb{Z}_K[/imath] and the prime ideal is equal to [imath]\mathfrak{p} = p \mathbb{Z}_K + a \mathbb{Z}_K[/imath]; e is the ramification index; f is the residual index; finally, mb is the multiplication table attached to the algebraic integer [imath]b[/imath] is such that [imath]\mathfrak{p}^{-1} = \mathbb{Z}_K + b/p\mathbb{Z}_K [/imath] which is used internally to compute valuations. In other words if p is inert, then mb is the integer 1, and otherwise it is a square t_MAT whose j-th column is b.nf.zk[j]. This say something like, as ideals in [imath]\mathbb{Q}(\sqrt[3]{2})[/imath]: [imath] (5) = \big(2 + \sqrt[3]{2}+{\color{#CCCCCC}0}\sqrt[3]{4},5\big)\times\big(-1 - 2\sqrt[3]{2}+\sqrt[3]{4},5\big) [/imath] This is a different set of generators than the linked question so there might be a chance to re-open. these are basically the same generators as in the previous question. | 1924378 | How to factorize [imath]5[/imath] in [imath]\mathbb{Z}[\root 3 \of 2][/imath]?
Since [imath]5[/imath] has a norm of [imath]125[/imath] in this domain, and [imath]N(1 + (\root 3 \of 2)^2) = 5[/imath], it seems like a sensible proposition that [imath]5 = (1 + (\root 3 \of 2)^2) \pi_2 \pi_3[/imath], where [imath]\pi_2, \pi_3[/imath] are two other numbers in this domain having norms of [imath]5[/imath] or [imath]-5[/imath]. This is supposed to be a unique factorization domain, right? I am encouraged by the fact that [imath]N\left(\frac{5}{1 + (\root 3 \of 2)^2}\right) = N(1 + 2 \root 3 \of 2 - (\root 3 \of 2)^2) = 25.[/imath] But I am discouraged by the fact that [imath]\frac{5}{(-1 - (\root 3 \of 2)^2)(1 + (\root 3 \of 2)^2)} = \frac{7 - 6 \root 3 \of 2 - 2 (\root 3 \of 2)^2}{5}[/imath] is an algebraic number but not an algebraic integer. I have found a couple of other numbers with norms of [imath]5[/imath] or [imath]-5[/imath] but can't get them to multiply to [imath]5[/imath] in any of the combinations of three of them that I have tried. I feel like I'm going around in circles. |
2740311 | Proof by induction for [imath]\left(1−\frac14\right)\cdot\left(1−\frac19\right)\cdots\left(1−\frac1{n^2}\right)[/imath] for all natural numbers [imath]n[/imath] with [imath]n \ge 2[/imath]
Make a conjecture about a formula for the product: [imath]\left(1−\frac14\right)\cdot\left(1−\frac19\right)\cdot\cdots\cdot\left(1−\frac1{n^2}\right)[/imath] for all natural numbers [imath]n[/imath] with [imath]n \ge 2[/imath]. Then, state a theorem about the formula and use mathematical induction to prove your theorem. Any help appreciated! I don't know what the conjecture is, thats what I need help with. All I found was the result is always between 0 and 1. I don't know if there should be something more to that or not. | 1464207 | Proving [imath]\prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}[/imath] by induction
So I have to prove the following using induction. [imath]{\displaystyle \prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right)} = \frac{n+1}{2n}[/imath] I showed the basis step that if [imath]n=i=2[/imath], then the two functions are equal [imath]\frac{3}{4}[/imath], and I know that the induction step involves simplifying the function where [imath]n=n+1[/imath]. But I'm not sure how to do it with the following algorithm. I am quite confused. Can someone show me how I can prove this in the induction step? |
2727367 | A problem of limit: [imath]\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{(1/n)}[/imath]
I need to find the limit of [imath]\lim\limits_{n\to \infty}\left\{\left(\frac{2}1\right) \left(\frac{3}2\right)^{2} \left(\frac{4}3\right)^{3} ... \left(\frac{n+1}n\right)^{n} \right\}^{(1/n)}[/imath] Cancelling out the same quantities in numerators and denominators, I have reached the step: [imath]\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{\frac1n}[/imath] Now I am stuck here. I have searched google for help, but found only the result for [imath]\lim\limits_{n\to\infty}\left\{\frac{1}{n!} \right\}^{(\frac1n)}[/imath]. Which formulae/properties should I use now to proceed from this stage? Edit: The options for the answer are: a) [imath]e[/imath], b) [imath]1/e[/imath], c) [imath]\pi[/imath], d) [imath]1/\pi[/imath] | 2734257 | Why does [imath]\;\lim_{n\to \infty }\frac{n}{n!^{1/n}}=e[/imath]?
I have tried to show that this limit : [imath]\lim\limits_{n\to \infty }\frac{n}{n!^{\frac 1 n}}=e[/imath] using [imath] \lim (1+\frac 1 n)^{\frac 1 n} , n \to \infty [/imath] , I don't find any equivalence , however wolfram alpha says that is [imath]e[/imath] as shown here, then how do I evaluate it ? |
2740976 | Find the value of [imath]\lim_{n\to\infty}\frac{1+2^{1/2}+3^{1/3}+\cdots+n^{1/n}}{n}[/imath]
The given problem is a problem of calculus. I have tried this problem only to find that the numerator is a divergent series. No idea how to do this problem [imath]\lim_{n\to\infty}\frac{1+2^{1/2}+3^{1/3}+\cdots+n^{1/n}}{n}[/imath] | 240585 | Computing [imath]\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}[/imath].
I have a problem with the calculation of the following limit. \begin{equation} \lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n} \end{equation} I do not know where to start! Thank you very much |
2740785 | Evaluating the power series of arctan at x=1.
I have that [imath]\tan^{-1}(x)=\sum_{n=1}^\infty \frac{(-1)^nx^{2n+1}}{2n+1}[/imath] for [imath]|x|<1[/imath] I also know that the series [imath]\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}[/imath] converges since it is decreasing in absolute value and alternating. I am wondering why we can evaluate the power series at [imath]x=1.[/imath] Is it because the series is a uniformly convergent sum of continuous functions defined on [imath]x\in[0,1][/imath], and therefore the sum function, or [imath]\tan^{-1}[/imath], is continuous at [imath]x=1[/imath]? So for a sequence of points [imath]\{x_n\}\in[0,1][/imath] with [imath]\lim_{n\to\infty} x_n=1[/imath], we must have that [imath]\lim_{n\to\infty} \sum_{n=1}^\infty \frac{(-1)^n(x_n)^{2n+1}}{2n+1}= \lim_{n\to\infty} \tan^{-1}(x_n)=\tan^{-1}(1)[/imath]? General question: if you can know that a power series of continuous functions converges on some open interval, say, [imath](-1,1),[/imath] but you can show the convergence at an extremity, say [imath]x=1,[/imath] then is it fair to say the continuous "sum function" known to converge on [imath](-1,1)[/imath] is also continuous on an interval containing an extremity, like [imath][0,1][/imath]? Is continuity of the function the sole reason this is possible? | 2740376 | Evaluating power series on the radius of convergence for [imath]\tan^{-1}(x)[/imath]
Suppose that I know the series for [imath](\tan^{-1})'=\sum_{n=0}^{\infty} (-1)^nx^{2n}[/imath] converges when [imath]\lvert x\rvert \lt 1[/imath]. Then I know that [imath]\tan^{-1}(x)=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{2n+1}[/imath] only for [imath]\lvert x \rvert \lt 1[/imath]. My professor asks me: How do I know that [imath]\lim_{x\to\ 1}\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \lim_{x\to\ 1}\frac{(-1)^nx^{2n+1}}{2n+1}\;?[/imath] In general, once the radius of convergence has been found for a power series, I think I can rest assured the series obtained by differentiating or integrating term by term has the same radius of convergence. What more do I need to consider to talk about the points on the radius? |
2740675 | Using gronwall's lemma to prove inequalities involving max[imath]x(t)[/imath].
Given [imath]\frac{\partial x}{\partial t}+x^3=yx^2[/imath], [imath]x(0)=x_{0}[/imath], where [imath]x[/imath] : [imath]\mathbb{R}^+ \mapsto \mathbb{R}[/imath] and [imath]y[/imath] : [imath]\mathbb{R}^+ \mapsto \mathbb{R} [/imath] with [imath]\int_{0}^{t} y^2(s) ds < \infty[/imath] for all [imath]t[/imath] if [imath]x(t)[/imath] is continuous function satisfying the above equation in [imath][0,T][/imath] [imath] \max_{t\in[0,T]}|x(t)|^2 \le |x(0)|^2e^{\int_0^T|y|^2\,\mathrm ds}.[/imath] [imath] \int_0^T|x|^4\,\mathrm ds <|x(0)|^2+\max_{s\in[0,T]}|x(s)|^2\int_0^Ty^2\,\mathrm ds.[/imath] It seems like it has something to do with Gronwalls lemma. My first approach was to set [imath]x(t)'<y(t)x(t)^2[/imath] and find bound for [imath]x(t)[/imath], but it did not work nicely. Can anyone help me? | 2740581 | [imath]\max_{t\in[0,T]}|x(t)|^2 \le |x(0)|^2e^{\int_0^T|y|^2\,\mathrm ds}.[/imath] using gronwall's lemma?
Given [imath]\frac{\partial x}{\partial t}+x^3=yx^2[/imath], [imath]x(0)=x_{0}[/imath], where [imath]x[/imath] : [imath]\mathbb{R}^+ \mapsto \mathbb{R}[/imath] and [imath]y[/imath] : [imath]\mathbb{R}^+ \mapsto \mathbb{R} [/imath] with [imath]\int_{0}^{t} y^2(s) ds < \infty[/imath] for all [imath]t[/imath] if [imath]x(t)[/imath] is continuous function satisfying the above equation in [imath][0,T][/imath] [imath]\Large \max_{t\in[0,T]}|x(t)|^2 \le |x(0)|^2e^{\int_0^T|y|^2\,\mathrm ds}.[/imath] It seems like it has something to do with Gronwalls lemma. My first approach was to set [imath]x(t)'<y(t)x(t)^2[/imath] and find bound for [imath]x(t)[/imath], but it did not work nicely. Can anyone help me? |
2741513 | Choosing eigenvectors for SVD (or EVD)
I'd like to find the SVD for the following matrix: [imath]\left(\begin{array}{cc} 5 & 5\\ -1 & 7 \end{array}\right)[/imath] I've calculated the eigenvalues and eigenvectors of [imath]XX^{T}[/imath] and the eigenvectors of [imath]X^{T}X[/imath], arriving at [imath]X=\left(\begin{array}{cc} 5 & 5\\ -1 & 7 \end{array}\right)=U\Sigma V^{T}=\left(\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{array}\right)\left(\begin{array}{cc} \sqrt{80} & 0\\ 0 & \sqrt{20} \end{array}\right)\left(\begin{array}{cc} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}}\\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right)^{T}[/imath] This does not work. However if I multiply the second column of [imath]U[/imath] by -1: [imath]\left(\begin{array}{c} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{array}\right)\mapsto\left(\begin{array}{c} -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{array}\right)[/imath] Everything works out. The question is why? Both are normalized orthogonal eigencectors, shouldn't either one work? How do I choose? | 2069085 | Problem with Singular Value Decomposition
I have a very trivial SVD Example, but I'm not sure what's going wrong. The typical way to get an SVD for a matrix [imath]A = UDV^T[/imath] is to compute the eigenvectors of [imath]A^TA[/imath] and [imath]AA^T[/imath]. The eigenvectors of [imath]A^TA[/imath] make up the columns of [imath]U[/imath] and the eigenvectors of [imath]AA^T[/imath] make up the column of [imath]V[/imath]. From what I've read, the singular values in [imath]D[/imath] are square roots of eigenvalues from [imath]AA^T[/imath] or [imath]A^TA[/imath], and must be non-negative. However, for the simple example [imath]A = \left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right][/imath], [imath]A^TA[/imath] and [imath]AA^T[/imath] are both the identity, and thus have eigenvectors [imath]\bigg\lbrace \left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\ ,\ \left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg\rbrace[/imath]. Clearly, the eigenvalues are 1 and 1, so our decomposition ought to be: \begin{align} \left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right] = \left[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right]\left[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right]\left[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right] \end{align} What has gone wrong? |
2740848 | Evaluating a triple integral by converting using change of variables
I have the double integral: [imath] \int_0^1 \int_0^{1-x} e^{\frac {y}{(x+y)}} \;dy\; dx [/imath] Ignore any difficulty this integral may have at the origin. I was given a hint to simplify the exponent by taking it as a chunk. I thus set [imath]v=\frac{y}{(x+y)}[/imath], but am not sure what to do for you and go from there? Can someone please help? Thanks! | 1270924 | Double integral with transformation; possible error in limits
I'm asked to show that the following integral is true with the transformation [imath]u = x+y[/imath] and [imath]y=uv[/imath]: [imath]\int_0^1 \int_0^{1-x} e^{y/(x+y)} dx dy = \frac{e-1}2 [/imath] I found the determinant of the jacobian of the transformation is [imath]-u[/imath], and the exponent is just [imath]v[/imath] so the integral becomes [imath]\iint_R |-u|e^v du dv = \frac {u^2}2 e^v|_R[/imath] [imath]=\frac{(x+y)^2}2e^{y/(x+y)}|_{x=0}^{1-x}\;|_{y=0}^1[/imath] And plugging it all in doesn't work (AFAICT). For one, the x's don't all cancel. I've looked at it 100 times. It's definitely written dxdy not dydx even though there is definitely an x in the integral. Maybe this is a typo and why my calculation isn't working out? I tried reversing the order to dydx and it didn't work. I tried changing the limit of integration to [imath]1-y[/imath] and it didn't work out. I'm not even sure how to figure out the domain of integration in the uv-plane. Flipping things around yields y=uv and x=u-uv. But if my domain is bounded by x=0 and y=0 and y=1 (ignoring the last limit for now), then I have uv=o=u and uv=1. !?! Am I just making an algebra mistake plugging everything in? Is there a way to handle a variable in it's own integral that I'm unaware of? |
2741456 | Prove if [imath]a>b>0[/imath] and [imath]n\in\mathbb{N}[/imath] then [imath]a^{1/n}-b^{1/n}<(a-b)^{1/n}[/imath]
If [imath]a>b>0[/imath] and [imath]n\in\mathbb{N}[/imath] then [imath]a^{1/n}-b^{1/n}<(a-b)^{1/n}[/imath] I know [imath]f(x)=x^{1/n}-(x-1)^{1/n}[/imath] is strictly decreasing if [imath]x>1[/imath]. Since after taking the derivative I get that [imath]f'(x)=\frac{1}{n}x^{\frac{1-n}{n}}-\frac{1}{n}(x-1)^{\frac{1-n}{n}}[/imath] which is decreasing for all [imath]x>1[/imath] Now evaluating [imath]f(1)[/imath] I get just [imath]1[/imath] and evaluating [imath]f\left(\frac{a}{b}\right)[/imath] I get [imath]\frac{a}{b}^{\frac{1}{n}}-\left(\frac{a-b}{b}\right)^{\frac{1}{n}}[/imath]. Now I'm not sure how to then show that [imath]a^{1/n}-b^{1/n}<(a-b)^{1/n}[/imath] | 1030311 | An inequality for powers of reals: [imath]a^{1/n}-b^{1/n}<(a-b)^{1/n}[/imath]
Let [imath]a>b>0[/imath] and let [imath]n \in N[/imath] satisfy [imath]n \geq 2[/imath]. Prove that [imath]a^{1/n}-b^{1/n}<(a-b)^{1/n}.[/imath] If we let [imath]a=x[/imath] and [imath]b=x-1[/imath], then we need to show that [imath]f(x):=x^{1/n}-(x-1)^{1/n}[/imath] is decreasing for [imath]1\leq x[/imath] (which means that [imath]f'(x)\leq 0[/imath] and [imath]f[/imath] is differentiable), and evaluate [imath]f[/imath] at [imath]1[/imath] and [imath]a/b[/imath]. |
2742217 | Determinant of an [imath]n \times n[/imath] matrix
In my linear algebra class, my professor gave us this determinant for practice: [imath]det\pmatrix {1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ \vdots & \vdots & \vdots & \dots & \vdots \\ n & 1 & 2 & \dots & n-1}[/imath] Where the [imath]i[/imath]-th row or column is the set [imath]\{1,2,\dots,n\}[/imath] shifted [imath]i-1[/imath] times to the left. He recommended we add rows [imath]2[/imath] through [imath]n[/imath] to row [imath]1[/imath]. And then factor out the constant. This yields: [imath]\frac{n(n+1)}{2}det\pmatrix {1 & 1 & 1 & \dots & 1 \\ 2 & 3 & 4 & \dots & 1 \\ \vdots & \vdots & \vdots & \dots & \vdots \\ n & 1 & 2 & \dots & n-1}[/imath] However, I cannot for the life of me find a useful way to simplify this further. I have tried adding the first column to each subsequent column, subtracting the first row from each subsequent row, but nothing seems to simplify the matrix. Perhaps, there is an inductive solution, but I haven't been able to find one simply from the [imath]2\times2[/imath] and [imath]3\times3[/imath] cases. Any help would be appreciated. | 2629389 | Solve [imath]n[/imath]th order determinant
Solve [imath]n[/imath]th order determinant: [imath] \begin{pmatrix} 1 & 2 & 3 & \cdots & n-2 & n-1 &n\\ 2 & 3 & 4 & \cdots & n-1& n& 1\\ 3 & 4 & 5 & \cdots & n &1 &2\\ \vdots& \vdots& \vdots& \ddots& \vdots& \vdots&\vdots\\ n-2 &n-1&n &\cdots &n-5 &n-4 &n-3\\ n-1 &n&1 &\cdots &n-4 &n-3 &n-2\\ n &1&2 &\cdots &n-3 &n-2 &n-1\\ \end{pmatrix} [/imath] I tried adding [imath]2[/imath]nd to [imath]n[/imath]th row to the [imath]1[/imath]st row, then the first row is: [imath] \begin{pmatrix} \frac{n(n+1)}{2} &\frac{n(n+1)}{2} \cdots \frac{n(n+1)}{2}\\ \end{pmatrix} [/imath] then I took out [imath]\frac{n(n+1)}{2}[/imath] from the first row to get [imath] \begin{pmatrix} 1 & 1 & \cdots& 1\\ \end{pmatrix} [/imath] and then tried to work with that but I don't think this is the correct approach because anything I tried doesn't look nice. |
2742609 | Is the union of countable sets (S)U(S x S)U(S x S x S)U..... countable?
Consider a set X = $\bigcup_{n=1}^\infty[imath]S^n$, where $S^n$ = S x S x S x S ... (n times)[/imath] Question: Is X countable? My attempt: For two countable sets S and T, I found it easy to prove that their Cartesian product is countable. I defined a function f: S x T -> $\Bbb N$, f(m,n) = $2^m3^n$ and proved it was an injection, then defined a function g: [imath]\Bbb N[/imath] -> S x T, g(n) = (0,n) and showed it was also an injection. Therefore by Cantor-Bernstein, there exists a bijection. So | S x T | = |[imath]\Bbb N[/imath]| and S x T is countable. Is there some way to extend this idea to the infinite union of Cartesian products of this countable set S ? My first thought was to define some function f([imath]s_{1}[/imath], ..., ) and in some way show that the injection that existed for S x T will always exist for any amount of Cartesian products. I also know that the for some countable sets A and B, C = B-A --> AUB = AUC and its easy to show that AUB is countable (considering both cases: C is finite or C is infinite and countable) Completely stuck on marrying the two ideas though! Any help appreciated | 563671 | [imath]Seq (\mathbb{N})[/imath] of all finite sequences of elements of [imath]\mathbb{N}[/imath] is countable.
I have few questions to the proof of this claim that is illustrated in a textbook. [imath]Seq (\mathbb{N})[/imath] of all finite sequences of elements of [imath]\mathbb{N}[/imath] is countable. Proof: Since [imath]Seq(\mathbb{N}) = \bigcup^{\infty} _{n=0} \mathbb{N}^n,[/imath] it suffices to produce a sequence [imath]\left \langle {a_n | n \in \mathbb{N}}\right \rangle [/imath] of enumerations of [imath]\mathbb{N}^n,[/imath] i.e. for each [imath]n \in \mathbb{N}, a_n =\left \langle {a_n(k)| k \in \mathbb{N}}\right \rangle [/imath] is an infinite sequence, and [imath]\mathbb{N}^n = \{a_n(k) | k \in \mathbb{N}\}[/imath] Let [imath]g[/imath] be a bijection from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N} \times \mathbb{N}.[/imath] Define [imath]a_1(i) = \left \langle {i}\right \rangle, \forall i \in \mathbb{N}; a_{n+1}(i) = \left \langle {b_0,...,b_{n-1},i_2}\right \rangle, [/imath] where [imath]g(i) = (i_1,i_2)[/imath] and [imath]\left \langle {b_0,...,b_{n-1}}\right \rangle = a_n(i_1).[/imath] Hence [imath]a_n[/imath] is onto [imath]\mathbb{N}^n.[/imath] My questions: This means [imath]a_2(i) = \left \langle {i_1, i_2}\right \rangle,[/imath] where [imath]g(i) =(i_1,i_2).[/imath] Then [imath]a_3(i) = \left \langle {i_1,i_2,i_2}\right \rangle [/imath] [imath],a_4(i) =\left \langle {i_1,i_2,i_2,i_2}\right \rangle [/imath] and [imath]a_n(i) =\left \langle {i_1,i_2,...,i_2}\right \rangle[/imath] etc? If so, given [imath]f = \left \langle {f(0),...,f(n-1)}\right \rangle \in \mathbb{N}^n,[/imath] I don't understand how could [imath]f [/imath] be expressed as [imath]a_n(i),[/imath] for some [imath]i \in \mathbb{N}?[/imath] Appreciate any clear instruction. Thank you. |
2743088 | Let [imath]f: \Bbb R \to \Bbb R[/imath] be such that [imath]|f(x) -f(y)| \le |x - y |^2 \ \forall \ x,\ y \in \Bbb R[/imath]. How can I show that [imath]f[/imath] is constant?
Let [imath]f: \Bbb R \to \Bbb R[/imath] be such that [imath]|f(x) -f(y)| \le |x - y |^2 \ \forall \ x,\ y \in \Bbb R[/imath]. How can I show that [imath]f[/imath] is constant? To me the statement looks an awful lot like something having to do with continuity/uniform continuity, but the fact that [imath]|x - y|[/imath] is being raised to the power of [imath]2[/imath] is confusing me.. I also recognize that the problem doesn't say anything about continuity, it's just what it looks like to me. Any tips? | 2369044 | [imath]f[/imath] Holder continuous with Holder exponent [imath]p>1\implies f \text{ is constant}[/imath]
Say I have a function on a an interval [imath]I[/imath] in [imath]\mathbb{R}[/imath] [imath]f: I \to Y [/imath] where [imath]Y[/imath] is any metric space.Say [imath]f[/imath] satisfies [imath]d_{Y}(f(y), f(x)) \leq C\cdot|y-x|^p[/imath], for all [imath]y,x \in I[/imath] where [imath]p \in (1,\infty)[/imath] , i.e. [imath]p[/imath] is bigger than [imath]1[/imath] (so this is much stronger than mere Holder Continuity). I want to show that [imath]f[/imath] is constant on [imath]I[/imath]. Here is what I got so far. [imath]f[/imath] is obviously continuous. Also, it makes intuitive sense for [imath]f[/imath] to be constant, since [imath]p >1[/imath] will make [imath]|y-x|^p[/imath] very small for for [imath]|y-x| \ll 1[/imath]. I also feel like I have to look at the expression [imath]\frac{d_{Y}(f(y), f(x))}{|y-x|} \leq C\cdot|y-x|^{p-1}[/imath] (Note: There is no notion of differentiability here). I'm also thinking that splitting up the interval [imath][x,y][/imath] (assuming [imath]x<y[/imath]) will help me like so: [imath]d_{Y}(f(y), f(x))\leq\sum_{k=1}^{k=n}d_{Y}(f(x_k), f(x_{k-1})) \leq C\cdot\sum_{k=1}^{n} |x_k-x_{k-1}|^p = \sum_{k=1}^{n} (\frac{1}{n})^p = n^{1-p} [/imath] [imath] (y = x_n, x = x_0)[/imath]. Is this the answer, that last equality chain? Edit: That last chain of equations, as suggested in the comments below should be [imath]d_{Y}(f(y), f(x))\leq\sum_{k=1}^{k=n}d_{Y}(f(x_k), f(x_{k-1})) \leq C\cdot\sum_{k=1}^{n} |x_k-x_{k-1}|^p [/imath] [imath]= C\sum_{k=1}^{n} (\frac{|y-x|}{n})^p =|y-x|^p n^{1-p} [/imath] |
2743507 | Isomorphisms between Alternating groups and modulo groups
I'm studying Group theory at the moment and am struggling to see how [imath]\mathbb{A}_3\cong Z/3Z[/imath] How can you form an isomorphism that sends say (ijk) [imath]\in \mathbb{A}_3[/imath] to either of {0,1,2} [imath]\in Z/3Z[/imath] ? | 1194504 | Prove any group with 3 elements is isomorphic to [imath]\mathbb{Z}_3[/imath]
Let [imath]G = \left\{e, a, b \right\}[/imath] Since [imath]G[/imath] has [imath]3[/imath] elements, [imath]a \not= e , \,b \not= e, \,a \not= b[/imath] [imath]G[/imath] is closed so [imath]ab \in G[/imath] If [imath]ab = a[/imath], this means that [imath]b = e[/imath] and if [imath]ab = b[/imath], [imath] a = e [/imath] So we are left with [imath]ab = e[/imath] Define a function [imath]\varphi (a) = [1][/imath] [imath]\varphi (b) = [2][/imath] To show that this function is an isomorphism, [imath]\varphi (e) = \varphi (ab) = \varphi (a) + \varphi (b) = [1] + [2] = [0][/imath] So [imath]\varphi (e) = [0][/imath] So this function takes the identity to the identity, so it's an isomorphism. |
2743620 | Solving Heat Equation with Transport Term
I'd like to solve this equation: [imath] u_t = u_{xx} + u_x [/imath] for [imath]t>0[/imath] with initial condition: [imath] u(0,x) = \cos (2 \pi x) [/imath] And [imath]1[/imath]-periodic, i.e we have that [imath]u(t,x) = u(t,x+1).[/imath] I was going to solve it using separation of variables, but I was thinking another way, namely solving it using a fourier series. I.e, first I assume that [imath]u[/imath] has fourier series expansion: [imath] u(t,x) = \sum_{k \in \mathbb{Z}} a_k(t)e^{2\pi ikx} [/imath] Now, I would like to differentiate the coefficients in the series and set them equal to each other. Something akin to: [imath] a_k'(t) = 2\pi ike^{2\pi i k x} ak(t) + (2 \pi ik )^2 ak(t) e^{2 \pi i k x} [/imath] And then solve for [imath]ak(t)[/imath]. Two questions: 1) is this method valid? I.e is it true that the series are equal if and only if the coefficients are? 2) If I can use this method, how do I solve the ODE for [imath]a_k(t)[/imath]? Apologies if this is particularly obvious, I have never taken an ODEs class | 2252366 | Solve [imath]u_t=u_{xx}+u_x[/imath]
For [imath]u:\mathbb{R}\times [0,1][/imath] with boundary conditions [imath]u(0,x)=\cos (2\pi x)[/imath] and [imath]u(t,0)=u(t,1)[/imath]. Solve [imath]u_t=u_{xx}+u_x[/imath]. I had this on an exam and tried to write [imath]u[/imath] as a product of two single variate functions and convert to an ode using the usual methods but things got messy and I couldn't finish. I also tried to write [imath]u[/imath] as a Fourier series in [imath]x[/imath] with coefficients [imath]a_k(t)[/imath] depending on [imath]t[/imath], but this also didn't seem to yield anything. Is there a more clever way to approach this, possibly with Fourier series? I heard from a friend there may be a Fourier transform method since the equation was "homogeneous in momentum space." Thanks! |
2743792 | Evaluate the integral of [imath]\int_0^\infty \frac{k}{(1+x)^4} \, dx[/imath]
Evaluating the integral: [imath]u=(1+x), \qquad du=1\,dx[/imath] [imath]\int_0^\infty \frac k {\frac{u^{-4+1}}{-4+1}} \, du \to \left. \frac{-3k}{(1+x)^{-3}} \right|_0^\infty[/imath] but the answer given has: [imath]\left.\frac{-1}{3(1+x)^3} \right|_0^\infty[/imath] What did I do wrong? | 2737688 | Evaluating a PDF to find an unkown constant
Evaluate the PDF to find K [imath]1 = \int_0^{\infty} \frac{k}{(1+x)^4}[/imath] using u substitution I got: [imath]u=1+x[/imath] and [imath]du= dx[/imath] [imath]\int_0^{\infty} \frac{k}{u^4} \to \frac{k}{\frac{-1}{3} u^3}[/imath] Since [imath]u^4 du = \frac{u^{-4+1}}{-4+1}=[/imath] [imath]\frac{u^{-3}}{-3}[/imath] which evaluating gives: [imath]1=\frac{k}{\frac{-1}{3(1+x)^3}} \vert_0^{\infty} \to \frac{-3k}{(1+x)^3} \vert_0^{\infty} = 0 -3k[/imath] and [imath]k= \frac{-1}{3}[/imath] somehow though they came up with [imath]\frac{-k}{3} \frac{1}{(1+x)^3}[/imath] after integrating [imath]\int_0^{\infty} \frac{k}{(1+x)^3}[/imath] Was my approach correct to use substitution? Where does my work go wrong? |
2742995 | Distribution of [imath]Y = X_{1}X_{2}, X_{i} \sim \operatorname{N}(0,1)[/imath]
I've got a question about the distribution of the random variable [imath]Y = X_{1}X_{2}[/imath], where [imath]X_i \sim \operatorname{N}(0,1)[/imath], and [imath]X_1[/imath], [imath]X_2[/imath] independent. Specifically, I need to find the MGF of [imath]Y[/imath]. Here's what I've tried: The [imath]r^\text{th}[/imath] moment of [imath]Y[/imath] is given by: [imath]\begin{align} \operatorname{E} (Y^r) & = \operatorname{E} (X_1^r X_2^r) \\ & = \operatorname{E} (X_1^r) \operatorname{E} (X_2^r) & \text{because } X_1 \text{ and } X_2 \text{ are independent} \\ \end{align}[/imath] Since [imath]X_i \sim \operatorname{N}(0,1)[/imath], I know that the [imath]r^\text{th}[/imath] central moment is given by: [imath] \operatorname{E}(X_i^r) = \begin{cases} 0 & r \text{ odd} \\ (2r-1)! & r \text{ even} \\ \end{cases}[/imath] Because [imath]\operatorname{E}(X_i^r)[/imath] does not depend on [imath]i[/imath], [imath]\operatorname{E} (X_1^r) = \operatorname{E} (X_2^r)[/imath] and so [imath]\operatorname{E} (X_1^r) \operatorname{E} (X_2^r) = ((2r-1)!)^2 \quad \text{for } r \text{ even}[/imath] But I can't really derive a formula for the [imath]r^{\text{th}}[/imath] integral, especially if I don't know how the formula for [imath]\operatorname{E}(X_i^r)[/imath] looks before plugging in [imath]t=0[/imath]. Are there other ways to derive the MGF of [imath]Y[/imath]? EDIT: I also need to show that [imath]Y[/imath] can be expressed as a difference of two independent, Gamma-distributed variables. | 2026175 | Finding the M.G.F of product of two random variables.
We are given two independent standard normal random variables [imath]X[/imath] and [imath]Y[/imath]. We need to find out the M.G.F of [imath]XY[/imath]. I tried as follows : \begin{align} M_{XY}(t)&=E\left(e^{(XY)t}\right)\\&=\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(XY)t}f_X(x)f_y(y)dxdy \\ &=\dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(xy)t}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &= \dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-\dfrac{(x-ty)^2}{2}}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dy \\ &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{-y^{2}(\frac{1}{2}+t^2)}dy \\ &= \dfrac{1}{\sqrt{1+2t^2}} \end{align} Is this correct ? |
2743545 | Baffled with [imath]\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}[/imath]
Calculate [imath]\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}[/imath] Personal work: [imath]\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^{\sin x}\cdot\cos x-\sin x}\cdot e^{\sin x}\over \sin x})=\cdots[/imath] This gets to nowhere. Also, I substituted [imath]t=e^{\sin x}[/imath] but I could not replace the [imath]e^x[/imath]. | 1851380 | Limit: [imath]\lim\limits_{x\to0}\frac{e^x-e^{\sin x}}{x-\sin x}[/imath]
Obviously, when substituted, expression gives [imath]\frac{0}{0}[/imath], which leads me to conclusion to use L'Hopital's rule. However, it is again [imath]\frac{0}{0}[/imath]. How can I find this limit? |
2741438 | Homotopy equivalence gives the mapping cone null homotopic?
Let [imath]\mathcal{A}[/imath] be an additive category. And let [imath]C(\mathcal{A})[/imath] be the category of the chain complexes. Now let [imath]f:X_* \rightarrow Y_*[/imath] be a chain map in [imath]C(\mathcal{A})[/imath] and [imath]C(f)[/imath] be its mapping cone. As we've already known, there's a proposition: If [imath]C(f)[/imath] is contractible, then [imath]f:X_* \rightarrow Y_*[/imath] is a homotopy equivalence. My question: Is the inverse problem true? To be explicitly, if [imath]f:X_* \rightarrow Y_*[/imath] is a homotopy equivalence, then does [imath]C(f)[/imath] necessarily to be contractible? I guess that this is not true in general for a part of the proof of the proposition doesn't holds conversely. Am I right? If I'm right, is there any counterexample? Thanks in advance. | 1917288 | explicit formula for contraction of conus of homotopy equivalence
Let [imath]f: K^\bullet \to L^\bullet[/imath] be homotopy equivalence. From theory of triangulated categories it follows that [imath]C(f)^\bullet[/imath] is contractible. But how can i produce explicit formula for homotopy of [imath]Id_{C(f)^\bullet}[/imath] and [imath]0_{C(f)^\bullet}[/imath]? |
2744797 | Nilpotent element [imath]a^2[/imath] in a polynomial quotient ring
[imath]\textbf{Problem Statement : }[/imath] Let [imath]R : = \mathbb{Z}/2\mathbb{Z}[x][/imath] and let [imath]I[/imath] be the ideal [imath]I = R \cdot(x^{17} -1)[/imath] Is there a non-zero element a of the quotient ring [imath]A = R/I[/imath] such that [imath]a^2 = 0[/imath]? I do not believe there is, as if there was, then [imath]a^2[/imath] would have to be contained within the ideal. However [imath]x^{17} -1[/imath] is reducible and is equal to [imath](x-1)(\sum_{i=0}^{16} x^i)[/imath], which are both irreducible so there can be no element [imath]a \notin I[/imath] but [imath]a^2 \in I[/imath]. Am I on the right track? | 109149 | Squares in Polynomial Quotient Ring [imath] (\mathbb{Z}/2)[x] /(x^{17} - 1) [/imath]
Let [imath]R = (\mathbb{Z}/2)[x] [/imath] and [imath]I[/imath] the ideal generated by the polynomial [imath]x^{17} - 1[/imath]. Is there a non-zero element of [imath]R/I[/imath] s.t. the square of that element is zero in [imath]R/I[/imath]? I know a quotient ring [imath]R/P[/imath] is an integral domain iff [imath]P[/imath] is a prime ideal, and since [imath]R[/imath] is a PID, the prime elements are the irreducible, but [imath]x^{17} - 1 [/imath] is reducible in [imath]R[/imath], hence [imath]R/I[/imath] is not an integral domain. So there are nonzero [imath]a,b \in R/I[/imath] s.t. [imath]ab = 0 [/imath], but [imath]b[/imath] might not be [imath]a[/imath], and that's where I got stuck. I've also tried finding [imath]f \in R[/imath] s.t. [imath] f^2 = g \cdot (x^{17} - 1)[/imath], but don't know how to attack this methodically. |
2744976 | Find the last [imath]3[/imath] digits of [imath]13^{1201}[/imath].
I believe I am supposed to use a mod in this problem. I am unsure exactly how to do so. Help would be much appreciated and thanks in advance! | 290394 | Finding the last two digits [imath]123^{562}[/imath]
This is my question: Find the last 2 digits of [imath]123^{562}[/imath] I don't even understand where to begin on this problem. I know I'm supposed to use Euler's theorem but I have no idea how or why. Any help? Thanks |
2745977 | Integral of limit is limit of integral and is finite, so the integrals over subsets also match
Can someone please assist me with this question? Let [imath]f_n[/imath] be a sequence of non-negative measurable functions on [imath](\Omega,\mathcal{A},\mu)[/imath] that converge to [imath]f[/imath] pointwise. Suppose that [imath]\int_\Omega f \ \mathrm{d}\mu = \lim_{n\to\infty}\int_\Omega f_n \ \mathrm{d}\mu < \infty[/imath] Then show that for any [imath]A \in \mathcal{A}[/imath] [imath]\int_A f \ \mathrm{d}\mu = \lim_{n\to\infty}\int_A f_n \ \mathrm{d}\mu.[/imath] Trivially, I can use Fatou's lemma to show that [imath]\int_A f \ \mathrm{d}\mu \leq \lim_{n\to\infty}\int_A f_n \ \mathrm{d}\mu[/imath]. But as for the reverse inequality, I'm not sure. | 531011 | Integral of limit of pointwise convergent sequence of non-negative measurable functions
Suppose [imath]\{f_n\} \subset L^+[/imath] , [imath]f_n \rightarrow f[/imath] pointwise, and [imath]\int f = \lim \int f_n < \infty[/imath]. Then [imath]\int_{E} f = \lim \int_{E} f_n[/imath] for all [imath]E \in \mathcal{M}[/imath]. However, this needn't be true if [imath]\int f = \lim \int f_n = \infty[/imath]. This is the exercise 13 on page 52, Real Analysis, by Gerald B. Folland. [imath]L^+[/imath] is the space of all measurable functions from [imath](X,\mathcal{M}, \mu)[/imath] to [imath][0, \infty][/imath]. I can find a counterexample satisfying [imath]\int f = \lim \int f_n = \infty[/imath]. Let [imath]X = \Bbb N[/imath], [imath]\mathcal{M} = \mathcal{P}(\Bbb N)[/imath] and [imath]\mu[/imath] be the counting measure. [imath]f_n = \chi_{\{2i+1:i \leq n\} \cup \{2j:n \leq j \leq 2n \}}[/imath]. Then [imath]\int f = \lim \int f_n = \infty[/imath]. But when [imath]E = \{2k: k \in \Bbb N\}[/imath], Then [imath]\int_{E} f = 0[/imath] and [imath]\lim \int_{E} f_n = \infty[/imath]. But how to show the part "[imath]f_n \rightarrow f[/imath] pointwise, and [imath]\int f = \lim \int f_n < \infty[/imath]. Then [imath]\int_{E} f = \lim \int_{E} f_n[/imath] for all [imath]E \in \mathcal{M}[/imath]"? |
2745056 | Monotonicity in this argmin function
I have the following function: [imath]f(a)=\text{argmin}_{t\in\mathbb{R}} \left\{\sum_{i=1}^n\frac{1}{2}(1-a)(b_i-t)^2+a| b_i-t |\right\}\quad ,a\in[0,1][/imath] Is there a way to show that the function is monotone in [imath]a[/imath]? Edit: for [imath]a=0[/imath] we have the mean of [imath]\beta_i[/imath]s and for [imath]a=1[/imath] we have the median. I'm interested in the cases when [imath]a\in(0,1)[/imath], in particular that the function is bounded by the mean and median. | 2573211 | Monotonicity in argmin function
I have the following equation [imath]f(a)=\text{argmin}_{t\in\mathbb{R}} \sum_{i=1}^n\frac{1}{2}(1-a)(b_i-t)^2+a| b_i-t |, a\in[0,1][/imath] How does one prove that the function is monotone in a? |
1044429 | Integral over the unit ball in [imath]\mathbb{R}^n[/imath]
Let [imath]f(x)=|x|^r[/imath] on [imath]B_1(0)[/imath] real valued function.Where [imath]B_1(0)[/imath] is the unit ball in [imath]\mathbb{R}^n[/imath]. I am trying to show that if [imath]r>1-n[/imath] f has a weak derivative. ATTEMPT: I know from the definition of the weak derivative that I need to check whether there is a function [imath]g:B_1(0)\rightarrow \mathbb{R}[/imath] such that [imath]\int_{B_1(0)}f\phi'dx=-\int_{B_1(0)}g\phi dx,[/imath] for all test functions [imath]\phi\in C_{c}^{\infty}(B_1(0))[/imath]. But since [imath]x\in \mathbb{R}^n[/imath], I dont know how to proceed. Please help. | 593327 | First order weak derivatives of [imath]f(x)=|x|^r[/imath]
Let [imath]f(x)=|x|^r[/imath] for a given real number [imath]r[/imath]. Show that [imath]f[/imath] has first order weak derivatives on the unit ball [imath]B_1(0)\subset \mathbb{R}^n[/imath] provided that [imath]r > 1-n[/imath]. Does anyone have an idea on how to prove this? |
2747181 | Find the limit of [imath]a_{n + 1} = \frac{{1 + a_n + a_n b_n }}{{b_n }} [/imath]
Let [imath]a_n[/imath] and [imath]b_n[/imath] be two sequences defined by the recurrence relation \begin{align} a_{n + 1} = \frac{{1 + a_n + a_n b_n }}{{b_n }},\qquad b_{n + 1} = \frac{{1 + b_n + a_n b_n }}{{a_n }} \end{align} with [imath]a_1=1[/imath] and [imath]b_1=2[/imath]. Find [imath]\mathop {\lim }\limits_{n \to \infty } a_n[/imath]. Simple manipulations yield that \begin{align} a_n = \frac{{1 + b_n }}{{b_{n + 1} - b_n }},\qquad b_n = \frac{{1 + a_n }}{{a_{n + 1} - a_n }} \end{align} I tried to find the first few terms of [imath]a_n[/imath] and [imath]b_n[/imath] but it seem that they doesn't have a general form. Once can see that both sequences increasing fast and randomly. Also, I tried to insert [imath]b_n[/imath] in [imath]a_n[/imath], but I got a complicated result and it doesn't help or indicate to anything. Any help is appreciated. Thanks | 401637 | Calculate the limit of two interrelated sequences?
I'm given two sequences: [imath]a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}[/imath] as well as an initial condition [imath]a_1=1[/imath], [imath]b_1=2[/imath], and am told to find: [imath]\displaystyle \lim_{n\to\infty}{a_n}[/imath]. Given that I'm not even sure how to approach this problem, I tried anyway. I substituted [imath]b_{n-1}[/imath] for [imath]b_n[/imath] to begin the search for a pattern. This eventually reduced to: [imath]a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}[/imath] Seeing no pattern, I did the same once more: [imath]a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}[/imath] While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the [imath]a_n[/imath] equation, then take the limit of it. How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated! Edit: While there is a way to prove that [imath]\displaystyle\lim_{n\to\infty}{a_n}=5[/imath] using [imath]\displaystyle f(x)=\frac{1}{x-1}[/imath], I'm still looking for a way to find the absolute form of the limit, [imath]\displaystyle\frac{1+2a+ab}{b-a}[/imath]. |
2747652 | Why is the sum of outer products equal to the matrix product of a matrix and its transpose , so [imath]A^TA = \sum_{i=1}^n a_i a_i^T[/imath]?
Why is the sum of outer products equal to the matrix product of a matrix and its transpose? So [imath]A^TA = \sum_{i=1}^n a_i a_i^T[/imath], where [imath]A = [ a_0, a_1 , ... , a_n ] [/imath], [imath]a_i \in \mathbb{R}^k[/imath]. An answer or a link to the answer would be greatly appriciated. It seems so simple but I just cannot figure it out. It is essentially this question of which I do not understand the answer: Matrix [imath]A^T A[/imath] as sum of outer products | 2335457 | Matrix [imath]A^T A[/imath] as sum of outer products
I have recently read in a script about statistical methods in a chapter about linear regression that: Given an [imath]n \times k[/imath]-matrix [imath]A[/imath], we have [imath]A^T A = \sum_{i=1}^{n} a_i^T a_i[/imath] where [imath]a_i[/imath] denotes the [imath]i[/imath]-th row of [imath]A[/imath]. Unfortunately, the author doesn't give a proof of that and I can't figure out one myself. Maybe someone can help me. source: script on page 10 |
2044734 | Formal way to proof that [imath]k! > (k/2)^{(k/2)}[/imath] for any [imath]k \in \{1,2,3,4...\}[/imath]
I am looking for a formal way to proof that [imath]k! > (k/2)^{(k/2)}[/imath] for any [imath]k \in \{1,2,3,4...\}[/imath] I tried proof by induction but this didn't really work out. So I kind of know why [imath]k! > (k/2)^{(k/2)}[/imath], but can't come up with a good way to write it down. | 1615650 | How to show that for all k, [imath]k! \ge (k/2)^{k/2}[/imath]
I'm working on a homework problem that has me showing a "[imath]\Omega(n\log k)[/imath] lower bound on the number of comparisons needed to sort a sequence of [imath]n[/imath] elements, when the input sequence consists of [imath]\frac{n}{k}[/imath] subsequences each containing [imath]k[/imath] elements. Each element in a subsequence will be larger than its preceding subsequence and smaller than its succeeding subsequence." Essentially, I just need to sort [imath]k[/imath] elements in each of the [imath]\dfrac{n}{k}[/imath] subsequences. I realize that to do this I need to show that the maximum possible height of a binary tree [imath]2^h \geq (k!)^{n/k}[/imath]. I did this to try and solve it (assuming we are using [imath]\log_2x)[/imath]: [imath]\log(2^h) \geq \log((k!)^{n/k})[/imath] [imath]h \geq \dfrac{n}{k}\log(k!)[/imath] This is as far as I got. However, my professor told me that in order to solve the problem I would need to prove that for all [imath]k[/imath], [imath]k! \geq \left(\dfrac{k}{2}\right)^{k/2}[/imath] I can obviously see why this would be true, but I don't really know how to prove it. Looking it up online I found an answer that someone posted stating: (1)[imath]\qquad[/imath] [imath]k! = k(k-1)(k-2)\ldots(2)(1)[/imath] (2)[imath]\qquad[/imath] [imath]k! \geq k(k-1)(k-2)\dots\left(\dfrac{k}{2}\right)[/imath] (3)[imath]\qquad[/imath] [imath]k! \geq k(k-1)(k-2)\ldots\left(\dfrac{k}{2}\right) \geq \left(\dfrac{k}{2}\right)\left(\dfrac{k}{2}\right)\ldots\left(\dfrac{k}{2}\right)[/imath] (4)[imath]\qquad \geq \left(\dfrac{k}{2}\right)\left(\dfrac{k}{2}\right)\ldots\left(\dfrac{k}{2}\right) = \left(\dfrac{k}{2}\right)^{k/2}[/imath] For (2), half of terms are greater than [imath]\frac{k}{2}[/imath] and half are smaller drop all terms less than [imath]k[/imath] Since [imath]\dfrac{k}{2} \geq k[/imath] we can write (3) From this answer I know that by plugging in [imath]\dfrac{k}{2}[/imath] for [imath]k![/imath] in the original equation will give me the answer [imath]\dfrac{n}{2}\log\left(\dfrac{k}{2}\right)[/imath]. But I don't understand why we dropped half of [imath]k![/imath] and how we could know that [imath]\dfrac{k}{2} \geq k[/imath]. I understand part (4) but I don't really understand the transformations taking place in between (1) and (2), and (2) and (3). |
2747626 | Suppose [imath] x+y+z=0 [/imath]. Show that [imath] \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} [/imath].
How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation. In addition, this identity is supposed to be true: [imath] \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .[/imath] | 851985 | [imath] \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) [/imath]
I have a question. I tried so to solve it, but there is a problem. that is i don't have any idea to findout how can i work with degrees 4,5,7 ... this is the problem : let [imath] x , y [/imath] and [imath] z [/imath] three real numbers such [imath] x+y+z = 0 [/imath]. prove : [imath] \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) [/imath] Please think and write your solutions! ; ) |
2747586 | Question on Derivatives Real Analysis
Here is the question: Let [imath]f(x)[/imath] be a three times differentiable function on [imath][−1, 1][/imath] such that [imath]f(−1) = 0[/imath], [imath]f(0) = 0[/imath], [imath]f(1) = 1[/imath] and [imath]f′(0) = 0[/imath]. Prove that [imath]f′′′(x) ≥ 3[/imath] for some [imath]x ∈ (−1, 1)[/imath]. My attempt: Well I think I have to use either the mean-value theorem or Taylor's Theorem, or both. I have no idea. Can someone just give a hint in order for me to start on this problem? Thank you very much!! | 1904905 | Applying MVT and IVT (?) to show something about f'''? (edit: actually, Taylor's Theorem using Lagrange remainder)
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times. However, I can't see how to proceed. I went down what I thought were the typical paths and just reached dead ends. Does someone maybe have a hint? I think that with the appropriate nudge I should be able to solve this question. If not, I'll edit and ask if someone knows how to solve the whole thing. Here's the question: Let [imath]f: \mathbf R \to \mathbf R[/imath] be such that [imath]f[/imath], [imath]f'[/imath], [imath]f''[/imath], and [imath]f'''[/imath] exist and are continuous on [imath]\mathbf R[/imath] and satisfying [imath]f(-3)=-1[/imath], [imath]f(0)=0=f'(0)[/imath], and [imath]f(3)=8[/imath]. Prove that there exists [imath]\xi \in (-3, 3)[/imath] such that [imath]f'''(\xi) \ge 1[/imath]. |
2748197 | Show that any four polynomials in [imath]P_2[/imath] are linearly dependent
[I attempted this question but just wanted to check if I'm on the right track. I started the question by : Let { ([imath]a_o[/imath]+$a_1[imath]x^2$), ($a_o$-$a_1[/imath]x^2$), ($a_1[imath]x^2$-$a_0$), ($a_1[/imath]x^2$-[imath]a_2[/imath]x) ∈ [imath]P_2[/imath] Followed by putting in vector form and into a matrix, reducing it to row echelon form and comparing the rank to the number of columns to determine it is lenearly dependent. Am I right to do it this way? | 473853 | n+1 vectors in [imath]\mathbb{R}^n[/imath] cannot be linearly independent
I was looking for a short snazzy proof on the following statement: n+1 vectors in [imath]\mathbb{R}^n[/imath] cannot be linearly independent A student of mine asked this today morning and I couldn't come up with a proof solely from the definition of linear independence. From a higher level perspective, I explained that if I put the vectors in a matrix then if the only null space entry is the zero vector, then the vectors are independent but since we have one extra column than row and that the row and column rank are equal, there is no way we can have [imath]n+1[/imath] as the rank of the matrix and hence from Rank-Nullity theorem, the dimension of the Nullspace is at least one which implies that there is a combination of the vectors where not all the scalar multiples in the definition are 0 but yet we get a zero as the linear combination. The student hasn't completely learnt the fundamental subspaces yet so I am not sure he grasped what I was saying. Is there a cleaner proof? EDIT: I am stunned how many beautiful answers I got with so much diversity. |
2748592 | Finding a derivative with [imath]\log[/imath]
[imath]f(x)=\log{\frac{x}{1+\sqrt{5-x^2}}}[/imath] I have to find the derivative of [imath]f(x)[/imath]. Please tell me if my steps are correct. [imath]\frac{1+\sqrt{5-x^2}}{x}×\frac{1}{\ln 10}×\frac{1+\sqrt{5-x^2}-x×\frac{1}{2\sqrt{5-x^2}}}{(1+\sqrt{5-x^2})^2}×\frac{1}{2\sqrt{5-x^2}}×(-2x)[/imath] | 2748629 | Stuck calculating the derivative of [imath]f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}[/imath].
I have to calculate the derivative of this: [imath]f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}[/imath] But I'm stuck. This is the point where I have arrived: [imath]f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}[/imath] How can I simplify? I didn't include all the passages. |
2749151 | Deduce the number of divisors, [imath]d(n)[/imath] is multiplicative and obtain a formula of [imath]d(n)[/imath] in terms of prime decomposition
I have been given a question which begins with me deducing [imath]d(n)[/imath] is multiplicative, I know [imath]d(n)= \sum_{d|n} 1,[/imath] and obtain a formula of [imath]d(n)[/imath] in terms of prime decomposition (once again I know is [imath]p_1^{a_1}p_2^{a_2}\cdot\cdot p_k^{a_k}[/imath]). I am aware that in order for a function to be multiplicative [imath]f(nm)=f(n)f(m)[/imath] where [imath](m,n)=1[/imath] and f is non zero. The final part is deduce all positive integers [imath]n[/imath] for when; [imath]\bullet[/imath] [imath]d(n)[/imath] is odd [imath]\bullet[/imath] [imath]d(n)= p_0,[/imath] where [imath]p_0[/imath] is a fixed prime. I have deduced [imath]d(n)[/imath] is multiplicative by letting [imath]n=p_1^{a_1}p_2^{a_2}\cdot\cdot p_k^{a_k}[/imath], where [imath]d(p_1^{a_1}p_2^{a_2}\cdot\cdot p_k^{a_k})= (a_1+1)\cdot \cdot(a_k+1)=d(p_1^{a_1})\cdot d(p_2^{a_2})\cdot \cdot \cdot d(p_k^{a_k})[/imath] as we know the divisors of [imath]n[/imath] are [imath]d=p_1^{b_1}p_2^{b_2}\cdot\cdot p_k^{b_k} [/imath] where eack [imath]b_k[/imath] gives a distinct divisor in the form of [imath](a_k+1)[/imath]. It is the next bit I do not understand. | 992614 | Calculating [imath]\tau(n)[/imath], the number of positive divisors of [imath]n \in \mathbb N[/imath]
So I have these parts to this problem: If [imath]n \in \mathbb N[/imath], then let [imath]\tau(n)[/imath] be the number of positive divisors of [imath]n[/imath]. For example, [imath]\tau(6) = 4[/imath]. (a) If [imath]n= p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}[/imath] where [imath]p_i[/imath] are distinct primes and [imath]a_i\in \mathbb N[/imath], then prove that the number of positive divisors of [imath]\tau(n)=(a_1+1)(a_2+1)\ldots(a_k+1)[/imath]. (b) Find the number of positive divisors of [imath]1,\!633,\!500[/imath]. (c) Prove or disprove that [imath]\tau(nm) = \tau(n)\tau(m)[/imath] for all [imath]m[/imath], [imath]n\in \mathbb N[/imath]. I'm not sure how to prove part a. For part b, I know I can just do a series of divisions to find that answer, so I got [imath]2^2\times 5^3\times 3^3\times11^2[/imath] for the divisors, so I think [imath]\tau(1,\!633,\!500)[/imath] would be 4 in this case. For part c, I think I can do a counterexample of 4 and 6, where [imath]\tau(4)=3[/imath], and [imath]\tau(6)=4[/imath], but [imath]\tau(24)[/imath] is not equal to 12, which is [imath]\tau(4)\tau(6)[/imath], if I did my math right. So, can anyone help me on part a and tell me if I am doing b and c wrong? |
2749352 | Irreducible Polynomial [imath]g = X^4 + X + 1[/imath] in [imath]\mathbb{F}_2[/imath]
I have the Irreducible Polynomial [imath]g = X^4 + X + 1[/imath] over [imath]\mathbb{F}_2 E[/imath], the extension of [imath]\mathbb{F}_2 = \{0,1\}[/imath] with root [imath]\alpha[/imath] of [imath]g[/imath]. My previous problems have asked me to find the number of elements in [imath]E[/imath] and see if they may all be represented in the for [imath]a^n[/imath] with [imath]n \in N[/imath] (I believe the answer to these are 16 and no respectively) I am struggling however to find all the roots of [imath]g[/imath] in [imath]E[/imath] expressed in the form [imath]\nu + \mu\alpha + \lambda \alpha^2 + \gamma \alpha^3[/imath] as my final problem has posed. Any help would be kindly appreciated. | 1591756 | How to find the roots of [imath] x^4 + x + 1[/imath] in the extension of [imath]\Bbb F_2[/imath]?
Let [imath]h[/imath] the irreducible polynomial over [imath]\Bbb F_2 = \{0, 1\}[/imath] [imath]h = x^4 +x +1[/imath] Let [imath]E=\Bbb F_2(\alpha)[/imath], i.e. the field [imath]\Bbb F_2[/imath] extended with [imath]\alpha[/imath], where [imath]\alpha[/imath] is a root of [imath]h[/imath]. How do we determine the numbers of elements in the extension [imath]E[/imath]? Can every non-zero element in the extension be expressed in the form [imath](\alpha)^n[/imath] for some [imath]n\in\Bbb N[/imath] ? How do we find all the roots of [imath]h[/imath] over the extension [imath]E[/imath]? |
2746994 | Finding the State Space Model of a linearized Equation of satellite
I have been trying for days to find how to convert the linearized Equation of a satellite given by the equation attached in the image to STATE SPACE equation. Please I have been struggling for days. I am a newbie in control system. Please do help out. [imath] \ddot\varepsilon+D\,\dot\varepsilon+K\,\varepsilon=B\,u \tag{1} [/imath] Where [imath] \varepsilon= \begin{pmatrix} r \\ \theta-\omega\,t \end{pmatrix} = \begin{pmatrix} \text{radial-deviation-from-R} \\ \text{angular-position} \end{pmatrix} [/imath] [imath] u= \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} \text{radial-thrust} \\ \text{tangential-thrust} \end{pmatrix} [/imath] [imath] D= \begin{pmatrix} 2\omega\over\ R & 0 \\ 0 & 0 \end{pmatrix} [/imath] [imath] K= \begin{pmatrix} -2\omega^2 & 0 \\ 0 & 0 \end{pmatrix} [/imath] [imath] B= \begin{pmatrix} 1\over m & 0 \\ 0 & 1\over m \end{pmatrix} [/imath] Want to convert above equation [imath](1)[/imath] to state space equation of the form [imath] \dot x = A\,x+B\,u [/imath] Also if [imath]y=\dot\theta[/imath] is observed at the output what will be the transfer function? Satellite Equation | 70993 | How to express a 2nd order ODE as 1st order ODE's?
Express the 2nd order ODE [imath]\begin{align}\mathrm d_t^2 u:=\frac{\mathrm d^2 u}{\mathrm dt^2}&=\sin(u)+\cos(\omega t)\qquad \omega \in \mathbb Z /\{0\} \\u(0)&=a\\\mathrm d_t u(0)&=b\end{align}[/imath] as a system of 1st order ODEs and verify there exists a global solution by invoking the global existence and uniqueness theorems. I'm not sure how to express second order ODEs as first order ODEs, any tips? |
2739395 | Irreducible factors of characteristic polynomial and minimal polynomial are same?
I've seen in my linear algebra textbook that one can prove that the irreducible factors of a characteristic polynomial and minimal polynomial are the same using Primary Decomposition Theorem, but I have no idea how this happens. So the fact that irreducible factors of a minimal polynomial are irreducible factors of a characteristic polynomial is trivial due to Cayley-Hamilton Theorem, however the converse does not seem so easy. I think I found a solution using field extensions and eigen values, but I want to try to prove this using Primary Decompostion Theorem. I've found Showing that minimal polynomial has the same irreducible factors as characteristic polynomial this question, it says, Let the minimal polynomial of a linear transfomation [imath]T:V \to V[/imath] over field F be [imath]m = {f_1}^{m_1} ... {f_n}^{m_n}[/imath]. Now restrict [imath]T[/imath] to [imath]ker({f_i}^{m_i})[/imath], and use the Cayley-Hamilton Theorem, and the fact that [imath]f_i[/imath] is irreducible to prove the statement. How ever I don't see how this happens, if I let [imath]W_i = ker({f_i}^{m_i})[/imath], I can show that the minimal polynomial of [imath]T|_{W_i}[/imath] is [imath]{f_i}^{m_i}[/imath] but using Cayley-Hamilton only gives that the characteristic polynomial of [imath]T|_{W_i}[/imath]is a multiple of [imath]{f_i}^{m_i}[/imath], and I don't think this really helps. I'm thinking that I have to use some unique properties of a characteristic polynomial other than the fact that it's an annihilator, since not every annihilators have the same irreuducible factors as minimal polynomials. Can you help me with this? +) Okay so to be exact, there's a Lemma on my Textbook: Let [imath]V[/imath] be a vector space over field [imath]F[/imath], and suppose [imath]T[/imath] is a linear transformation from vector space [imath]V[/imath] to [imath]V[/imath], and let [imath]f_1(t), ...f_k(t)[/imath] be mutually relatively prime monic polynomials over field [imath]F[/imath]. if [imath]f_1(t)f_2(t)...f_k(t)[/imath] is an annihilator of [imath]T[/imath], [imath]V[/imath] can be decomposed into direct sums of [imath]kerf_i(T)[/imath]. and there's a exercise that says to prove the following using the Lemma above: Let [imath]V[/imath] be a vector space over field [imath]F[/imath], and suppose [imath]T[/imath] is a linear transformation from [imath]V[/imath] to [imath]V[/imath], then the 'set of monic irreducible divisor over field F' is the same for the characteristic polynomial of [imath]T[/imath] and the minimal polynomial of [imath]T[/imath]. I have no knowledge about modules, and I know characteristic polynomial as [imath]det(tI-T)[/imath] and minimal polynomial as the minimal annihilator of T. This isn't my homework or anything, I just got stuck while studying with this textbook on myself :(( Any help? | 825848 | Showing that minimal polynomial has the same irreducible factors as characteristic polynomial
I'm trying to show that the minimal polynomial of a linear transformation [imath]T:V \to V[/imath] over some field [imath]k[/imath] has the same irreducible factors as the characteristic polynomial of [imath]T[/imath]. So if [imath]m = {f_1}^{m_1} ... {f_n}^{m_n}[/imath] then [imath]\chi = {f_1}^{d_1} ... {f_n}^{d_n}[/imath] with [imath]f_i[/imath] irreducible and [imath]m_i \le d_i[/imath]. Now I've managed to prove this using the primary decomposition theorem and then restricting [imath]T[/imath] to [imath]ker({f_i}^{m_i})[/imath] and then using the fact that the minimal polynomial must divide the characteristic polynomial (Cayley-Hamilton) and then the irreducibility of [imath]f_i[/imath] gives us the result. However I would like to be able to prove this directly using facts about polynomials/fields without relying on the primary decomposition theorem for vector spaces. Is this fact about polynomials true in general? We know that [imath]m[/imath] divides [imath]\chi[/imath] and so certainly [imath]\chi = {f_1}^{m_1} ... {f_n}^{m_n} \times g[/imath] but then how do we show that [imath]g[/imath] must have only [imath]f_i[/imath] as it's factors? I'm guessing I need to use the fact that they share the same roots. And I'm also guessing that it depends on [imath]k[/imath], i.e. if [imath]k[/imath] is algebraically closed then it is easy because the polynomials split completely into linear factors. Help is much appreciated, Thanks |
2749547 | Proof Question- Calculus
I've been trying to prove the following: Lets have a function continuous and derivable for any x that belongs to the real numbers; and for any [imath]x[/imath] and [imath]h: f(x+h) -f(x) = hf'(x)[/imath]. So, I have to prove that [imath]f(x)= ax+b[/imath], where a and b are constants. I tried to use the Mean Value Theorem because it has its similarities but I kept getting nowhere. It is obvious that the function satisfies the equation but that ain't a proof. Thanks. | 1943224 | How would I prove that this function is affine?
Let [imath]f[/imath] be a differentiable function such that for every [imath]x[/imath] and [imath]h[/imath] it holds that [imath]f(x+h)-f(x)=hf'(x)[/imath]. Prove that [imath]f(x)=kx+n[/imath] where [imath]k[/imath] and [imath]n[/imath] are constants. I get it why this is true, and I tried to prove it somehow, but I can't seem to prove it rigorously with analysis. Any ideas? |
2750409 | Field extensions isomorphic to subfields of each other
Suppose [imath]K_1,K_2[/imath] are field extensions (not necessarily finite) of a field [imath]F[/imath] and that [imath]K_1[/imath] is [imath]F[/imath]-isomorphic to a subfield of [imath]K_2[/imath] (i.e. there is a field homomorphism [imath]K_1 \to K_2[/imath] which injects [imath]K_1[/imath] into [imath]K_2[/imath] and is the identity on [imath]F[/imath]) and [imath]K_2[/imath] is [imath]F[/imath]-isomorphic to a subfield of [imath]K_1[/imath]. Is this enough to say that [imath]K_1[/imath] and [imath]K_2[/imath] are isomorphic as fields? | 1280184 | Field that is a subfield of own of its subfields
Let [imath]K[/imath] and [imath]L[/imath] be fields. We have homomorphisms [imath]f: K \to L[/imath] and [imath]g: L \to K[/imath]. Are [imath]K[/imath] and [imath]L[/imath] necessarily isomorphic? |
2750250 | Why is it if [imath]\epsilon > 0[/imath] is arbitrary implies that [imath]d(p,p') = 0[/imath] if [imath]d(p,p') < \epsilon[/imath]?
I know this is such a basic question and its embarrassing to ask but I just don't understand why Rudin concludes such a thing. In theorem 3.2 (a) he proves that if two sequences converge to different limits then the limits are actually the same as follow: [imath] \epsilon > 0 [/imath] [imath] n \geq N \implies d(p_n,p) < \frac{\epsilon}{2}[/imath] [imath] n \geq N' \implies d(p_n,p') < \frac{\epsilon}{2}[/imath] let [imath]n = \max(N,N')[/imath]: [imath] d(p,p') \leq d(p,p_n) + d(p_n,p') < \epsilon [/imath] everything so far made sense. The last step of the proof is what I don't understand: since [imath]\epsilon[/imath] arbitrary, we conclude that [imath]d(p,p') = 0[/imath] thats the part that I don't understand. The proof started assuming that [imath] \epsilon > 0[/imath] so of course the proof didn't conclude by plugging in zero. In chapter 2 there was such a big emphasis that the points in the neighborhoods to the limit points are different from the limit point. See this proof I'd conclude there is always a positive difference between p and p' but its never exactly zero (just like in chapter 2 and limit points). Of course I assume I am wrong and there some subtle but important point that I don't understand and wanted to clarify it. Can someone clarify why my assertion that the difference is always positive (of course by assumption) is wrong? If anything I'd conclude that [imath]p[/imath] and [imath]p'[/imath] are limits of each other but not that they are the same as Rudin concludes or that their distance is zero. What did I miss? note: I do know limit points and limits aren't the same, was just saying that to provide context for my confusion. | 2452084 | Theorem 3.2 Rudin -- How is 0 concluded?
I'm having trouble understanding why the arbitrariness of [imath]\epsilon[/imath] allows us to conclude that [imath]d(p,p')<0[/imath]. It seems we could likely conclude a value such as [imath]\frac {\epsilon}{100}[/imath] couldn't we?? The other idea that would normally work is the limit (as [imath]n[/imath] approaches [imath]\infty[/imath], [imath]p[/imath] approaches [imath]p'[/imath]) but that would mean we are further in another sequence which would have same problem Thanks in advance [imath]\ [/imath] [imath]\ [/imath] [imath]\ [/imath] Definition of Convergence A sequence [imath]\{ p_n \}[/imath] in a metric space [imath]X[/imath] is said to converge if there us a point [imath]p \in X[/imath] with the following property: For every [imath] \epsilon>0[/imath] there is an integer [imath]N[/imath] such that [imath]n \ge N[/imath] implies distance function [imath]d(p_n, p) < \epsilon.[/imath] |
2750890 | [imath]T[/imath] is compact iff [imath]\langle Te_i,e_j\rangle\to 0[/imath] as [imath]i,j\to\infty[/imath]
Let [imath]H[/imath] be a separable Hilbert space and [imath]T:H\to H[/imath] be a bounded operator. Suppose [imath](e_i)[/imath] is a countable orthonormal basis in [imath]H[/imath]. Show that [imath]T[/imath] is compact if and only if [imath]\langle Te_i,e_j\rangle\to 0[/imath] as [imath]i,j\to\infty[/imath]. Note: I could prove that if [imath]T[/imath] is compact then [imath]\langle Te_i,e_j\rangle\to 0[/imath]. I am stuck at the converse part. | 402324 | Criteria of compactness of an operator
Suppose [imath]K[/imath] is a linear operator in a separable Hilbert space [imath]H[/imath] such that for any Hilbert basis [imath]\{e_i\}[/imath] of [imath]H[/imath] we have [imath]\lim_{i,j \to \infty} (Ke_i,e_j) = 0[/imath]. Is it true that [imath]K[/imath] is compact? Thanks in advance for any help. |
2751150 | Prove that the sum of two continuous functions is continuous using epsilon delta.
[imath]\lim _{x\rightarrow a} f(x)+g(x) = f(a)+g(a)[/imath] Let [imath]ε>0[/imath] be given Since [imath]f[/imath] and [imath]g[/imath] are continuous, [imath]|f(x)-f(a)|< ε[/imath] when [imath]0<|x-a|< \delta_f [/imath] and [imath]|g(x)-g(a)|< ε[/imath] [imath]\ [/imath] when [imath]0<|x-a|< _g[/imath] Let [imath]_h[/imath] be defined as [imath]\min(_g ,_f)[/imath] and [imath]h(x)[/imath] be defined as [imath]f(x)+g(x)[/imath] [imath]\ |f(x)-f(a)|+|g(x)-g(a)|< 2ε[/imath] [imath]\ |f(x)-f(a) + g(x)-g(a)|< 2ε[/imath] [imath]\ |h(x)-h(a)|< 2ε[/imath] We can replace [imath]_f[/imath] and [imath]_g[/imath] by [imath]_h[/imath] to get [imath]|f(x)-f(a)| < ε[/imath] when [imath]0<|x-a|<_h[/imath] and [imath]\ |g(x)-g(a)|< ε[/imath] when [imath]0<|x-a|<_h[/imath] How do I prove that [imath]\ |h(x)-h(a)|< ε[/imath] instead of [imath]\ |h(x)-h(a)|< 2ε[/imath]? | 2750160 | Prove that the sum of two continuous functions is continuous
[imath]\lim _{x\rightarrow a} f(x)+g(x) = f(a)+g(a)[/imath] Let [imath]ε>0[/imath] be given Since [imath]f[/imath] and [imath]g[/imath] are continuous, [imath]|f(x)-f(a)|< ε[/imath] when [imath]0<|x-a|< \delta_f [/imath] and [imath]|g(x)-g(a)|< ε[/imath] [imath]\ [/imath] when [imath]0<|x-a|< _g[/imath] Let [imath]_h[/imath] be defined as [imath]\min(_g ,_f)[/imath] and [imath]h(x)[/imath] be defined as [imath]f(x)+g(x)[/imath] [imath]\ |f(x)-f(a)|+|g(x)-g(a)|< 2ε[/imath] (1) [imath]\ |f(x)-f(a) + g(x)-g(a)|< 2ε[/imath] [imath]\ |h(x)-h(a)|< 2ε[/imath] We can replace [imath]_f[/imath] and [imath]_g[/imath] by [imath]_h[/imath] to get [imath]|f(x)-f(a)| < ε[/imath] when [imath]0<|x-a|<_h[/imath] and [imath]\ |g(x)-g(a)|< ε[/imath] when [imath]0<|x-a|<_h[/imath] Since (1) is true when [imath]0<|x-a|<_h[/imath], we can find a [imath]_h[/imath] for every value of [imath]2ε[/imath] for every 2ε>0 there is a δ>0 such that |x−a|<δ⟹∣h(x)−h(a)∣<2ε and for every 2ε there exists an ε so for every ε there exists a δ. Is the proof logically correct and is it worded correctly? |
2751924 | Gaussian probability law
for [imath]X[/imath] and [imath]Y[/imath], two independent random variables following the densities [imath]f_X (x) = xe^{-\frac{x^2}{ 2}} 1_{]0,+\infty[} (x)[/imath] and [imath]f_Y (y) = \frac{1}{\pi \sqrt{1-y^2}} 1_{]-1,1[} (y)[/imath] , I found that [imath](X,Y)[/imath] follows [imath] \frac{1}{\pi } e^{-\frac{x^2+y^2}{ 2}}1_{\mathbb{R}}(x)1_{\mathbb{R}_+^*}(y)[/imath]. Is it the standard normal law, though we have a factor [imath]\frac{1}{\pi} [/imath] instead of [imath]\frac{1}{\sqrt{\pi}}[/imath] in the density? Thank you! | 2735818 | Density of random variables
Let [imath]X[/imath] and [imath]Y[/imath] be independent random variables with respective densities: [imath]f_X (x) = xe^{-\frac{x^2}{ 2}} 1_{]0,+\infty[} (x)[/imath] and [imath] f_Y (y) = \frac{1}{\pi \sqrt{1-y^2}} 1_{]-1,1[} (y)[/imath] Let [imath]U:=XY[/imath] and [imath]V:=X\sqrt{1-Y^2}[/imath] I try to check whether U and V are independent and which law they follow. What I did: After some calculations, I found the density of [imath](U,V)[/imath]: [imath] f_{(U,V)}(u,v)=\frac{1}{\pi } e^{-\frac{u^2}{ 2}}1_{\mathbb{R}}e^{-\frac{v^2}{ 2}}1_{\mathbb{R}_+^* }[/imath] Which mean that [imath](U,V)[/imath] follows a [imath] \mathcal{N}(0,1)[/imath] law (because [imath]u^2+v^2=x^2[/imath]) and that U and V are independent because [imath]f_{(U,V)}[/imath] can be factored as the products of densities of [imath]U[/imath] and [imath]V[/imath] My question: There is a factor [imath]\frac{1}{\pi}[/imath] and I don't know to which density it will belong: to [imath]U[/imath] density or to [imath]V[/imath] density? Many thanks! |
1410095 | Notation Question [imath]n[/imath] [imath] < <[/imath] [imath]m[/imath]
What does the following notation mean: [imath]n< < m[/imath] , where [imath]n[/imath] and [imath]m[/imath] are numbers? | 1069343 | Notation question: [imath]\ll[/imath]
I was perusing http://mathworld.wolfram.com/HighlyCompositeNumber.html and saw the following at the end: Nicholas proved that there exists a constant [imath]c_2>0[/imath] such that [imath]Q(x) \ll (\ln x)^{c_2}[/imath]. What does the [imath]\ll[/imath] mean in this context? (fyi, [imath]Q(x)[/imath] is the number of highly composite numbers less than or equal to [imath]x[/imath]) |
759818 | Lie bracket of vector fields on [imath]R^2[/imath]
Compute the Lie bracket[imath]\Big[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\Big][/imath] on [imath]R^2[/imath] Can you help me please? | 359356 | Lie bracket of vector fields on [imath]\Bbb R^{n}[/imath]
Please show how to solve? I am stack with lie bracket. Thank you. |
2751633 | Sum to [imath]n[/imath] terms of the given series
Find the sum to [imath]n[/imath] terms of the given series: [imath]0.3+0.33+0.333+0.3333+\cdots[/imath] My Attempt: Let [imath]S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}[/imath] [imath]=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots[/imath] [imath]=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right][/imath] How do I continue from here? | 765660 | Expression generating [imath]\left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right)[/imath]
I'm looking for a closed-form expression (in terms of [imath]n[/imath]), that will give the sequence [imath] (s_n) = \left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right). [/imath] Can anyone think of one? I made a related post to this question several minutes ago but I realized I was interpreting the sequence wrong. |
2752131 | A general formula for the second degree derivative?
Is the method I wrote for the second degree derivative formula is correct? Can you verify? [imath]\begin{align} f''(x) &=\lim_{\Delta x \to 0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x} \tag{1}\\[4pt] &= \lim_{\Delta x \to 0} \left[\frac{\lim_{\Delta x\to 0}\frac{f(x+2 \Delta x)-f(x+\Delta x)}{\Delta x}-\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x} \right] \tag{2}\\[4pt] &= \lim_{\Delta x \to 0} \left[\frac{\frac{f(x+2 \Delta x)-f(x+\Delta x)}{\Delta x}-\frac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x} \right] \tag{3}\\[4pt] &=\lim_{\Delta x \to 0}\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{{\Delta x}^2} \tag{4}\\[4pt] \Longrightarrow \quad f''(x) &=\lim_{\Delta x \to 0}\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{{\Delta x}^2} \tag{5}\\[4pt] \Longrightarrow \quad f''(x-\Delta x) &=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-2f(x)+f(x-\Delta x)}{{\Delta x}^2} \tag{6}\\[4pt] \Longrightarrow \quad f''(x) &=\lim_{\Delta x \to 0}\frac{f(x-\Delta x)+f(x+\Delta x)-2f(x)}{{\Delta x}^2} \tag{7} \end{align}[/imath] REMARK Q: My solution is different (especially the last steps) than the one shown here. Therefore, it can not be evaluated under the category "Duplicate". | 2133984 | If [imath]f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}[/imath], show that [imath]\lim_{s \to \infty} f(s)/s = 1[/imath].
If [imath]f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}[/imath], show that [imath]\lim_{s \to \infty} f(s)/s = 1[/imath]. This function comes up in the parameterization of the solutions to [imath]x^y = y^x[/imath]. See for example, here: Are there real solutions to [imath]x^y = y^x = 3[/imath] where [imath]y \neq x[/imath]? If we write [imath]y = rx[/imath], we get [imath]x = r^{1/(r-1)}[/imath] and [imath]y=r^{r/(r-1)}[/imath]. Then [imath]x^y =(r^{1/(r-1)})^{r^{r/(r-1)}} =r^{r^{r/(r-1)}/(r-1)} [/imath]. Finally, if we write [imath]r = 1+s[/imath], this is [imath]f(s) = (1+s)^{(1+s)^{1+1/s}/s} [/imath]. Wolfy says that, around [imath]s=0[/imath], [imath]f(s) =e^e + \dfrac{e^{1 + e} s^2}{24} - \dfrac{e^{1 + e} s^3}{24} + \dfrac{e^{1 + e} (219 + 5 e) s^4}{5760} + O(s^5) [/imath] and, around [imath]s = \infty[/imath], [imath]f(s) =s + (\log^2(1/s) - \log(1/s) + 1) + \dfrac{\log^4(1/s) - 3 \log^3(1/s) + 5 \log^2(1/s) - 6 \log(1/s) + 2}{2 s} + O((1/s)^2), [/imath] calling this a generalized Puiseux series. My question is: How to show that that expansion at [imath]s = \infty[/imath] is correct. I would be satisfied with a proof that, as the title says, [imath]\lim_{s \to \infty} \dfrac{f(s)}{s} = 1 [/imath]. It might help to note that [imath]\dfrac{y}{x} = r[/imath] and [imath]\dfrac{r}{s} = \dfrac{1+s}{s} = 1+\dfrac1{s} [/imath]. |
2752822 | [imath]\int_{0}^{+\infty} \frac{dx}{f(x)}[/imath] converges if and only if [imath]\int_{0}^{+\infty} \frac{x}{F(x)}dx[/imath] converges
Suppose [imath]f[/imath] is positive and increasing on [imath][0, +\infty)[/imath]. Let [imath]F(x) = \displaystyle\int_{0}^{x}f(t)dt[/imath]. How do we show that [imath]\displaystyle \int_{0}^{+\infty} \frac{dx}{f(x)}[/imath] converges if and only if [imath]\displaystyle\int_{0}^{+\infty} \frac{x}{F(x)}dx[/imath] converges? I was thinking of using the comparison test, but I can't really find a way to connect the two integrals. | 2748249 | Convergence of Integrals - A Problem
Suppose [imath]f[/imath] is positive and increasing on [imath][0, +\infty)[/imath]. Suppose that $F(x)=[imath]\int_0^x f(t)dt$. Prove that [/imath]\int_0^{+\infty} \frac {1}{f(x)} dx[imath] converges if and only if [/imath]\int_0^{+\infty} \frac {x}{F(x)} dx$[imath] converges.[/imath] My thoughts, and attempt at solution - Since f[imath] is increasing, [/imath]F(x)=\int_0^xf(t)\,dt\le\int_0^xf(x)\,dt=xf(x).[imath] We can rearrange that to [/imath]\frac1{f(x)}\le\frac{x}{F(x)}.[imath] Hence if the integral of [/imath]\frac{x}{F(x)}[imath] converges, then so does the integral of [/imath]\frac1{f(x)}.$ However, the above working seems incomplete as it bounds the integral from the upper side only. In fact, it shows that if the second integral converges, then the first one may converge, but not vice versa. How do I place a lower bound on the integral, and go about the rest of the solution? Haven't been able to proceed from here, please help. |
2753286 | Using contour integration to evaluate this integral
I am trying to show that [imath]\int_0^{\infty} \frac {\sin{\phi_1 x}} x \frac {\sin{\phi_2 x}} x \cdots \frac {\sin{\phi_n x}} x \cos{a_1 x} \cdots \cos{a_m x} \frac {\sin ax} x \, \mathrm d x = \frac {\pi} 2 \phi_1 \phi_2 \cdots \phi_n,[/imath] if [imath]\phi_1,\phi_2, \ldots, \phi_n, a_1,a_2, \ldots, a_m[/imath] are real, [imath]a[/imath] is positive and [imath]a>|\phi_1|+|\phi_2|+\cdots + |\phi_n| + |a_1| + \cdots + |a_m|.[/imath] As with [imath]\int_0^\infty \frac {\sin x} x\,\mathrm d x[/imath] and [imath]\int_0^\infty \left(\frac{\sin x} x\right)^2\,\mathrm d x[/imath] I have tried integrating over a semi-circular contour with an indentation at the origin and then using Jordan's lemma, which should still work as the integrand is an even function. But when I expand the sines and cosines out in terms of exponentials I cannot decide whether the semi-circle should be in the upper-half or the lower-half plane since I cannot know what the signs of the constants in the exponents are. Hints on how to go about this would thus be appreciated. | 2637235 | Evaluate [imath]\int_0^\infty\frac{\sin(\varphi_1x)}x\frac{\sin\varphi_2x}x\cdots\frac{\sin\varphi_nx}x\frac{\sin(ax)}x\cos(a_1x)\cdots\cos(a_mx) \, dx[/imath]
How to evaluate [imath] \int_0^\infty \frac{\sin(\varphi_1x)}{x}\frac{\sin\varphi_2x}{x} \cdots \frac{\sin\varphi_nx}{x} \frac{\sin(ax)}{x}\cos(a_1x) \cdots \cos(a_mx) \, dx \text{ ?} [/imath] For small [imath]n[/imath] and [imath]m[/imath] it's simple (setting [imath]\sin(kx)=\dfrac{e^{ikx}-e^{-ikx}}{2i}[/imath] and using Jordan's lemma, but for arbitrary [imath]n[/imath] the calculation is too tedious. Surely there must be some nice trick here? Thanks |
2753847 | What am I doing wrong while finding the Laurent series of [imath]\frac{1}{\sin(z)}[/imath]?
I'm trying to compute the Laurent series of [imath]\frac{1}{\sin(z)}[/imath] at [imath]z_0=0[/imath]. From what I've seen on the internet this is given as [imath]f(z)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+...[/imath] My attempt: [imath]f(z)=\frac{1}{\sin(z)}[/imath] can be rewritten as [imath]f(z)=\frac{1}{z}\frac{z}{\sin(z)}[/imath] We can express [imath]\sin(z)[/imath] as [imath]\sin(z)=z-\frac{z^2}{3!}+\frac{z^4}{5!}+...[/imath] [imath]\therefore \frac{z}{\sin(z)}=(\frac{\sin(z)}{z})^{-1}=(1-\frac{z}{3!}+\frac{z^3}{5!}+...)^{-1}[/imath] To get this to equal what I know it should, then we should have [imath](\frac{\sin(z)}{z})^{-1}=(1+\frac{z}{3!}+\frac{7z^3}{360}+...)[/imath] but how can this be what do I do with my negative sign and how do I obtain the [imath]\frac{7}{360}[/imath] coefficient? | 648923 | Calculate Laurent series for [imath]1/ \sin(z)[/imath]
How can calculate Laurent series for [imath]f(z)=1/ \sin(z) [/imath] ?? I searched for it and found only the final result, is there a simple way to explain it ? |
2754173 | If [imath]x+1[/imath] is a factor of [imath]ax^4 + bx^2 + c,[/imath] what is the value of [imath]a + b + c?[/imath]
The same question was posed here Find the value of a+b+c but I cannot make sense of the answer. My long division must be incorrect as I ended up with [imath]a+b+c[/imath] which is definitely wrong. I added [imath]0[/imath] to represent [imath]x^3[/imath] and [imath]x^1[/imath] terms. I know the remainder should be zero if [imath]x+1[/imath] is a factor. I am not putting two and two together. | 696646 | Find the value of a+b+c
If [imath]x+1[/imath] is a factor of [imath]ax^4 + bx^2 + c[/imath], find the value of [imath]a + b + c[/imath]? I know that it is equal to zero, but I have to know How to do it. |
2753973 | In need of theorem recommendations to prove that f(z) maps [imath]\Bbb C \setminus B[/imath] maps to [imath]\Bbb C \setminus \{0\}[/imath]
Given a bounded set B of the complex plane, I want to prove that [imath]f(z)=e^z[/imath] maps [imath]\Bbb C \setminus B[/imath] ont [imath]\Bbb C \setminus \{0\}[/imath]. I have no idea where to start with this. Does anyone have any recommendations on what theorems would be useful here ? | 2240570 | Prove that [imath]f(z) = e^z[/imath] maps [imath]\mathbb{C} \setminus B [/imath] onto [imath]\mathbb{C} \setminus \{0\}[/imath].
Prove that [imath]f(z) = e^z[/imath] maps [imath]\mathbb{C} \setminus B [/imath] onto [imath]\mathbb{C} \setminus \{0\}[/imath] where [imath]B[/imath] is a bounded subset of [imath]\mathbb{C}[/imath]. I have no idea how to even attempt this question so I would greatly appreciate some hints. |
2754673 | Existence of a normal p-Sylow subgroup when [imath]\lvert\mkern1mu G\mkern1mu \rvert=q^{2}\mkern1mu p[/imath]
If [imath]\lvert\,G\,\rvert=p^{2}\cdot\,q[/imath] where [imath]q,p[/imath] are primes, How can i show [imath]G[/imath] has a normal p-sylow subgroup ? I tried working with the sylow theorem but i cant reach any contradiction when [imath]Np = q[/imath] and [imath]Nq = p,p^2[/imath] Also tried searching the same question but couldn't find in Stack Exchange | 210663 | Showing that a Sylow subgroup of a group of order [imath]p^{2}q[/imath] is normal.
Suppose that I have a group [imath]G[/imath] of order [imath]p^{2}q[/imath] for two distinct primes [imath]p[/imath] and [imath]q[/imath]. I need to first show that one of its Sylow subgroups is normal. I start by letting [imath]H[/imath] be a Sylow [imath]p[/imath]-subgroup and [imath]K[/imath] be a Sylow [imath]q[/imath]-subgroup. If [imath]K[/imath] is not normal, then letting [imath]r[/imath] denote the number of Sylow [imath]q[/imath]-subgroups, I have that [imath]r|p^{2}[/imath] and [imath]r\equiv 1[/imath](mod [imath]q[/imath]). I also have by the second Sylow theorem that [imath]r\neq 1[/imath]. So [imath]r = p[/imath] or [imath]p^{2}[/imath]. If [imath]r = p^{2}[/imath], I can write out the Sylow [imath]q[/imath]-subgroups, and count their combined elements to show that [imath]H[/imath] is uniquely determined and then again imply Sylow number [imath]2[/imath] to obtain that [imath]H[/imath] is normal. But if [imath]r = p[/imath], then I cannot see how to proceed. Can anyone give any advice? Thank you. ( I am happy to provide more detail for the [imath]r = p^{2}[/imath] case if it is desired or appropriate. ) |
2755437 | Proving that an ellipsoid is convex
I'm trying to prove that an ellipsoid defined by [imath]E:=\{(x,y,z) \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1 \}[/imath] with [imath]a,b,c \gt 0[/imath] is convex, but I'm having trouble with it. I tried with the definition of convexity, but I reach a point of [imath] \lambda \left( \frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} + \frac{x_3^2}{c^2} \right) + 2 \lambda ( 1 - \lambda) \left( \frac{x_1y_1}{a^2} + \frac{x_2y_2}{b^2} + \frac{x_3y_3}{c^2} \right) + (1 - \lambda)^2 \left( \frac{y_1^2}{a^2} + \frac{y_2^2}{b^2} + \frac{y_3^2}{c^2} \right) \le 1 [/imath] and I don't know how to proceed from there (or even if it's right). I think that there might be another (easier) way to do this, but i can't come up with it, so if someone could help me with this proof I'll be really grateful [imath]:)[/imath] | 2429995 | Prove that the ellipsoid [imath]E = \{ (x,y,z) \in R^3 \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1 \}[/imath] is convex
I know what an ellipsoid is, but I do not know how to approach how to prove that one is convex. Any advice would be greatly appreciated. |
2754793 | How to check a Lie algebra homomorphism
Let [imath]L[/imath] be a Lie algebra and [imath]A[/imath] be its subalgebra. Let consider [imath]h : A \to Der(L)[/imath] defined by [imath]h(a)(l)=[l,a][/imath] then how to show that [imath]h[/imath] is Lie algebra homomorphism? According to the definition of Lie algebras homomorphisms [imath]h[/imath] must satisfies [imath] h([a_1,a_2])=[h(a_1),h(a_2)][/imath]. Using Jacobi identity, we need to prove that [imath][[l_1,l_2]a][/imath] equals to [imath][[l_1,a],[l_2,a]][/imath] ? I have problem with calculation of Jacobi identity! Edit: There are useful answers to this question: Checking a Lie homomorphism has been described carefully in Adjoint map is Lie homomorphism and Adjoint map is a Lie homomorphism. | 1339289 | Adjoint map is Lie homomorphism
The Jacobi identity of a Lie algebra says that [imath]ad: \mathfrak g \to End(\mathfrak g)[/imath] is a derivation. I am a bit emberassed but what is the easieast way to see that for every [imath]X \in \mathfrak g[/imath], [imath]ad_X: \mathfrak g \to \mathfrak g, Y \mapsto [X,Y][/imath] is a Lie homomorphism, i.e., [imath][ad_X Y, ad_X Z] = ad_X([Y,Z])[/imath] ? Edit: The identity in the above line is wrong. It is [imath]ad_X[/imath] a derivation (endomorphism satisfying the product rule), not a Lie homomorphism (respecting the Lie bracket). I was a bit confused here, sorry. |
2755091 | Prove that [imath]7^n - 1[/imath] is divisible by 6
I know that I have to prove this with the induction formula. If proved the first condition i.e. [imath]n=6[/imath] which is divisible by [imath]6[/imath]. But I got stuck on how to proceed with the second condition i.e [imath]k[/imath]. | 1131164 | Divisibility Proof with Induction - Stuck on Induction Step
I'm working on a problem that's given me the run around for about a weekend. The statement: For all [imath]m[/imath] greater than or equal to [imath]2[/imath] and for all [imath]n[/imath] greater than or equal to [imath]0[/imath], [imath]m - 1[/imath] divides [imath]m^n - 1[/imath]. Using induction over [imath]n[/imath], the base step comes easy since [imath]m^n - 1[/imath] is [imath]0[/imath] when [imath]n = 0[/imath]. My induction hypothesis is letting [imath]k \geq 0[/imath] and assuming that [imath]m - 1[/imath] divides [imath]m^k - 1[/imath]. In order to show that [imath]m - 1[/imath] divides [imath]m^{k+1} - 1[/imath], I obviously need to use the induction hypothesis. However, no matter where I try to use the fact that [imath]m^k - 1 = (m-1)a[/imath] for some [imath]a[/imath] in the integers, the expression [imath]m^{k+1} - 1[/imath] always becomes more difficult to get to [imath]m^{k+1} - 1[/imath] being equal to [imath](m-1)b[/imath] for some [imath]b[/imath] in the integers. In other words, I can't figure out any actually helpful way to apply the induction hypothesis with the goal of proving the next step. Any help would be appreciated! |
2755690 | Are the eigenvectors of [imath]A^2[/imath] the same as the eigenvectors for [imath]A[/imath]? Are the lambdas just squared?
Given matrix [imath]A[/imath], we square it. We found the eigenvalues and eigenvectors for [imath]A[/imath]. If we square [imath]A[/imath] (or it to a power of any number [imath]n[/imath]), do the eigenvalues become squares (or to the power of any number [imath]n[/imath]), while the eigenvectors remain the same? | 767835 | Proving Eigenvalue squared is Eigenvalue of [imath]A^2[/imath]
The question is: Prove that if [imath]\lambda[/imath] is an eigenvalue of a matrix A with corresponding eigenvector x, then [imath]\lambda^2[/imath] is an eigenvalue of [imath]A^2[/imath] with corresponding eigenvector x. I assume I need to start with the equation [imath]Ax=\lambda x[/imath] and end up with [imath]A^2 x=\lambda^2 x[/imath] but between those I am kind of lost. I have manipulated the equations several different ways and just can't seem to end up where I need to be. Help would be greatly appreciated as I believe this will be on a test tomorrow. |
2756714 | Find conjugacy classes of [imath]S_3[/imath]
I am trying to understand the best way to find the conjugacy classes of [imath]S_3[/imath]. What I am doing to find these is first consider what elements would be in the center of the group [imath]Z(S_3)[/imath]. Then, after that, I try finding the conjugacy classes for each element of [imath]S_3[/imath] that is not in the group. Namely, I try finding the conjugacy class of 2 cycles, so for [imath](12)[/imath] and after that I try finding the conjugacy class for 3 cycles like [imath](123)[/imath]. Doing all the work of computing the conjugacies, I found taht the conjugacy class of [imath](12)[/imath] is [imath]\{(12),(13),(23)\}[/imath] and the conjugacy class of [imath](123)[/imath] is [imath]\{(123),(132)\}[/imath]. Now my question is, is there a more efficient way to find these conjugacy classes, without computing all the conjugacies? For instance for the group [imath]A4[/imath], I think this approach is not efficient and definitely harder to apply. Thanks! | 102170 | Is there a systematic way of finding the conjugacy class and/or centralizer of an element?
Is there a systematic way of finding the conjugacy class and centralizer of an element? Could the task be simplified if we are working with "special groups" such as [imath]S_n[/imath] or [imath]A_n[/imath]? Are there any intuitive approaches? Thanks. |
2756422 | What's the meaning of [imath]\pi_0[/imath] in Algebraic Topology?
In my general topology class, we introduced some basic algebraic topology. My professor mentioned the term [imath]\pi_0[/imath] together with path components of [imath]X[/imath], but I forgot what he said. Does anyone know what is it? And he also mentioned [imath]\pi_{\leq 1}[/imath] and [imath]\pi_2[/imath]. What are these things? We use Munkres' Topology. In section 52, he mentions "There are indeed groups [imath]\pi_n (X,x_0)[/imath] for all [imath]n \in \Bbb Z_+[/imath], but we shall not study them in this book. They are part of the general subject called homotopy theory." | 1596822 | Zeroth homotopy group: what exactly is it?
What are the elements in the zeroth homotopy group? Also, why does [imath]\pi_0(X)=0[/imath] imply that the space is path-connected? Thanks for the help. I find that zeroth homotopy groups are rarely discussed in literature, hence having some trouble understanding it. I do understand that the elements in [imath]\pi_1(X)[/imath] are loops (homotopy classes of loops), trying to see the relation to [imath]\pi_0[/imath]. |
2757398 | What are the numbers present in the following Arithmetic Progression?
Four different integers form an increasing A.P. One of these numbers is equal to the sum of the squares of the other three numbers. Find the numbers My Attempt: I assumed the four numbers as: [imath](a-3d), (a-d), (a+d), (a+3d)[/imath] but now i can't understand which number to square. | 989047 | How to Solve this Arithmetic Progression Question?
Please help- Four different integers form an increasing AP.One of these numbers is equal to the sum of the squares of the other three numbers.Then- find all the four numbers. I assumed the numbers to be [imath]a,a+d,a+2d,a+3d[/imath] and wlog let [imath]a+3d[/imath] this number and according to question- [imath]a^2+(a+d)^2+(a+2d)^2=a+3d[/imath] but I could not solve the above equation. |
2756564 | Fourier Transform of [imath]|f|[/imath]
I take the definition of the Fourier transform of [imath]f \in L^2 (\mathbb{R}^3)[/imath] to be [imath] \widehat{f} (k) = \int\limits_{\mathbb{R}^3} e^{-2 \pi i k \cdot x} f(x) dx [/imath] Question: Can we say anything useful about [imath]\widehat{|f|}[/imath] in terms of [imath]\widehat{f}[/imath]? | 2685784 | Find the Fourier transform of the absolute value of a function, given its Fourier transform
It is possible to find the following Fourier transform? [imath] \widehat{\vert f\vert}(\xi)[/imath], provided [imath] f [/imath] has Fourier transform. I have not found any information about this statement. |
2757785 | [imath]M/m\ge\sqrt{3}[/imath].
Suppose there are six distinct points in the plane. Let [imath]m[/imath] and [imath]M[/imath] be the minimum and maximum distances among the al possible distances between pairs of points. I have to show [imath]M/m\ge\sqrt{3}[/imath]. I have tried placing the points on the plane in an extremal fashion, but couldn't figure out how to derive such inequality. | 1744983 | Maximum and minimum distance of two points
Consider six distinct points in a plane. Let [imath]m[/imath] and [imath]M[/imath] denote respectively the minimum and the maximum distance between any pair of points. Show that [imath]M/m \geqslant \sqrt{3}[/imath]. |
273463 | Could someone help solve this PDE?
Given, [imath]U_t-U_{xx}-2U_x=0[/imath] Using method of separation of variables, find ALL possible solution. My answer : [imath]U(x,t)=X(x)T(t)[/imath] [imath]T'(t)/T(t)[/imath]=[imath][X''(x)+2X'(x)]/X(x)[/imath]=[imath]\lambda[/imath] From here I'm stuck. Could someone show me some working steps for me to proceed further. | 269746 | PDE separation of variables
Hi could someone guide me this problem It says , [imath] u_t - u_{xx}-2 u_x=0 [/imath] Use the method of separation of variables to find all possible solutions. Could someone help me out for this problem. I'm beginner at PDE. I would be much appreciated if you able to show some partial work so that I can understand. Thanks in advance for taking my consideration |
2757895 | is [imath]\frac{0}{0}[/imath] indeterminate or undefined?
I know in calculus the form [imath]\frac{0}{0}[/imath] is indeterminate.but if it is not calculus is it still indeterminate or undefined in the real number field? P.S. I know that [imath]\frac{1}{0}[/imath] is undefined whether it is calculus or normal arithmetic in the real number system. but in the projectively extensive real number system, [imath]\frac{1}{0}=\infty[/imath] but in this system what would be [imath]\frac{0}{0}[/imath] it becomes [imath]0*\infty[/imath] also an indeterminate form. in that system the problem [imath]\frac{1}{0}[/imath] is solved but [imath]\frac{0}{0}[/imath] remains. P.P.S some answers in the question which has been identified as a duplicate of my Q state the [imath]\frac{0}{0}[/imath] is indeterminate. but in Wikipedia page(division by zero) states it is undefined. and that question was asked not to clarify this ambiguity(indeterminate or undefined) and there is no clear answer to my problem. | 2127695 | Can we say that [imath]\frac{0}{0}[/imath] is every number?
Suppose we have an equation [imath]ab=0[/imath]. This equation is true when statements [imath]a=0[/imath] or [imath]b=0[/imath] are true. If [imath]a=0[/imath], then [imath]b=\frac{0}{0}[/imath]. That means [imath]b[/imath] could be any number for [imath]ab=0[/imath] to be true. If the set which groups all the numbers is the complex set, then [imath]b[/imath] will every number within [imath]\mathbb{C}[/imath], so [imath]\forall z\in\mathbb{C}:b=\frac{0}{0}=z[/imath]. Therefore, [imath]\frac{0}{0}[/imath] is every number. I know it really is not defined as number but conceptually it is every number, right? Is this right, or am I missing something? |
2757451 | Prove that [imath]\mathbb{Z}(\zeta)/(1-\zeta)\cong\mathbb{Z}_p[/imath].
Let be [imath]\zeta[/imath] a [imath]p[/imath]-root ot unity, with [imath]p\ge3[/imath] prime. How to prove that [imath]\mathbb{Z}(\zeta)/(1-\zeta)\cong\mathbb{Z}_p[/imath]? | 2099406 | Isomorphism of [imath]\mathbb Z[\xi]/\left<\xi-1\right>[/imath] with [imath]\xi[/imath] root of unity.
I am trying to prove the following isomorphism but I am not sure how to. Let [imath]\xi=e^{2\pi i/p}[/imath] and [imath]p[/imath] a prime. Show [imath]\mathbb Z[\xi]/\left<\xi-1\right> \cong \mathbb F_p .[/imath] I wanted to create a map [imath]\phi : \mathbb Z \rightarrow \mathbb F_p [/imath] with kernel [imath]<\xi-1>[/imath] but so for I have not have good ideas. Any hint? I know that any element in [imath]<\xi-1>[/imath] is of the form [imath] (a_{p-1}-a_0)+(a_0-a_1)\xi+\cdots+(a_{p-2}-a_{p-1})\xi^{p-1} [/imath] but I do not know how to use this information. Thank you! |
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