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2498895 | Prove that [imath]A \times B = B \times A [/imath] if and only if [imath]A = B \lor A = \emptyset \lor B = \emptyset[/imath]
Prove that [imath]A \times B = B \times A [/imath] if and only if [imath]A = B \lor A = \emptyset \lor B = \emptyset [/imath] So, I need to prove that [imath]A\times B = B \times A \iff A = B \lor A = \emptyset \lor B = \emptyset[/imath] It is easy to show the right-to-left implication. The problem begins when I want to show the left-to-right one. I tried using the axiom of extentionality, but that did not work. What kind of trick should I use to prove that the relation from left to right holds as well? | 1751807 | If [imath]A[/imath] and [imath]B[/imath] are nonempty sets, prove that [imath]A \times B = B \times A[/imath] if and only if [imath]A = B[/imath]
If [imath]A[/imath] and [imath]B[/imath] are nonempty sets, prove that [imath]A \times B = B \times A[/imath] if and only if [imath]A = B[/imath]. Proving the first direction of this is easy. That is, if [imath]A = B[/imath] then it is obvious that [imath]A^2 = A^2[/imath]. I am wondering how to prove the other direction now. Maybe proving it by contradiction will be easier than directly. |
2499236 | Analysis: Show that there exists an [imath]x[/imath] so that [imath]f'(x)=0[/imath]. (Derivative at point is equal to[imath] 0[/imath])
Question: Look at already answered question. Same. | 1128690 | Prove there exists [imath]a\in \Bbb{R}[/imath] such that [imath]f'(a)=0[/imath].
Let [imath]f[/imath] be differentiable on [imath]\Bbb{R}[/imath] and let [imath]\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0[/imath]. Prove there exists [imath]a\in \Bbb{R}[/imath] such that [imath]f'(a)=0[/imath]. Attempt: If [imath]f[/imath] is constant we are done. Otherwise, [imath]f[/imath] is not constant. Suppose there is no such [imath]a[/imath]. Then, in particular, [imath]f[/imath] has no extremum [imath]\implies[/imath] [imath]f[/imath] has no point where is goes from increasing to decreasing or from decreasing to increasing. [imath]f[/imath] is continuous since it is differentiable. Hence, since [imath]\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0[/imath], [imath]f[/imath] must be constant. A contradiction. Therefore either [imath]f[/imath] is constant and [imath]f'(x)=0[/imath] everywhere, or [imath]f[/imath] has an extremum at [imath]x=a[/imath] and [imath]f'(a)=0[/imath]. I feel like my attempt is not formal enough, any correction? I would appreciate your help. |
1949143 | An interesting exercise about converging positive series, involving [imath]\sum_{n\geq 1}a_n^{\frac{n-1}{n}}[/imath]
Yesterday I stumbled across an interesting exercise (Indam test 2014, Exercise B3): (Ex) Given a positive sequence [imath]\{a_n\}_{n\geq 1}[/imath] such that [imath]\sum_{n\geq 1}a_n[/imath] is convergent, prove that [imath] \sum_{n\geq 1}a_n^{\frac{n-1}{n}}[/imath] is convergent, too. My proof exploits an idea from Carleman's inequality. We have: [imath] a_n^{\frac{n-1}{n}}=\text{GM}\left(\frac{1}{n},2a_n,\frac{3}{2}a_n,\ldots,\frac{n}{n-1}a_n\right) [/imath] and by the AM-GM inequality [imath] a_n^{\frac{n-1}{n}}\leq \frac{1}{n}\left(\frac{1}{n}+a_n\sum_{k=1}^{n-1}\frac{k+1}{k}\right)\leq \frac{1}{n^2}+\left(1+\frac{\log n}{n}\right)a_n [/imath] hence [imath] \sum_{n\geq 1}a_n^{\frac{n-1}{n}}\color{red}{\leq} \frac{\pi^2}{6}+\left(1+\frac{1}{e}\right)\sum_{n\geq 1}a_n.[/imath] Now my actual Question: Is there a simpler proof of (Ex), maybe through Holder's inequality, maybe exploiting the approximations [imath] \sum_{m<n\leq 2m}a_n^{\frac{2m-1}{2m}}\approx \sum_{m<n\leq 2m}a_n^{\frac{n-1}{n}}\approx \sum_{m<n\leq 2m}a_n^{\frac{m-1}{m}}[/imath] "blocking" the exponents over small summation sub-ranges? | 2891237 | If [imath]\sum a_n[/imath] converges then [imath]\sum a_n^{1-1/n}[/imath] converges.
Let [imath]a_n[/imath] be a sequence of positive reals such that [imath]\sum_1^\infty a_n[/imath] is finite then show that [imath]\sum_1^\infty a_n^{1-\frac{1}{n}}[/imath] is also finite. N.B Is this statement true? I tried to use the limit comparison test but it's inconclusive. The converse is obviously true which makes it an "iff" statement. No well-known test seems to be working. Any help is appreciated. |
2500095 | Proving [imath]\mathbb{Z}[\sqrt{6}][/imath] is a PID
I'm having trouble proving that [imath]\mathbb{Z}[\sqrt{6}][/imath] is a PID. I thought that the way to go would be to prove that it's a Euclidean Domain, but I got stuck there as well. Any suggestions on how to proceed? I'm looking for a way to prove that it's a PID without having to prove that it's a Euclidean Domain | 124484 | Show [imath]\mathbb{Z}[\sqrt{6}][/imath] is a Euclidean domain
I'm attempting to modify the proof the [imath]\mathbb{Z}[\sqrt{2}][/imath] is a Euclidean domain to prove a similar result for [imath]\mathbb{Z}[\sqrt{6}][/imath]. The idea is to prove that [imath]\mathbb{Q}[\sqrt{6}][/imath] is Euclidean which will then give the result for [imath]\mathbb{Z}[\sqrt{6}][/imath]. According to Dummit and Foote's Abstract Algebra this should have norm [imath]d(a+b\sqrt{6})=|a^2-6b^2|[/imath]. The result for [imath]\mathbb{Z}[\sqrt{2}][/imath] relies on the fact that for any element [imath]x[/imath] in [imath]\mathbb{Q}[\sqrt{2}][/imath] there is an element [imath]x'[/imath] in [imath]\mathbb{Z}[\sqrt{2}][/imath] with [imath]d(x-x')<1[/imath], which is then used to show that for [imath]x=qy+r[/imath] we have [imath] d(r)=d(x-qy)=d\left(y\left(\frac{x}{y}-q\right)\right)=d(y)d\left(\frac{x}{y}-q\right)<d(y) [/imath] But in this case, I'm having a hard time showing that we can find an element with [imath]d(x-x')<1[/imath] for [imath]x\in\mathbb{Q}[\sqrt{6}], x'\in\mathbb{Z}[\sqrt{6}][/imath]. Consider [imath]x=a+b\sqrt{6}\in\mathbb{Q}[\sqrt{6}][/imath]. The furthest this element can be away from any [imath]x'=a'+b'\sqrt{6}\in\mathbb{Z}[\sqrt{6}][/imath] is when [imath]|a-a'|=\frac{1}{2}=|b-b'| \Rightarrow[/imath] [imath] d(x-x')=\left|\left(\frac{1}{2}\right)^2-6\left(\frac{1}{2}\right)^2\right|=\frac{5}{4}>1 [/imath] What can I do to get around this issue? |
2499869 | Path-connected components of the configuration space [imath]F_n(R^d)[/imath]
Let [imath]X[/imath] be a topological space, for [imath]n \geq 1[/imath] we define [imath]F_n(X) = \lbrace{ (x_1,...,x_n) \in X^n | x_i \neq x_j for i \neq j \rbrace} [/imath] the configuration space of n points of X. My question is about the path-connected components of [imath]F_n(\mathbb{R}^d)[/imath]. I know that : 1- for [imath]n=1[/imath], [imath]F_1(X) = X[/imath] and hence [imath]\pi_0(F_1(\mathbb{R}^d)) = \pi_0(\mathbb{R}^d) = \lbrace \mathbb{R}^d \rbrace \: \forall d \geq 1 [/imath] 2- [imath]\forall n \geq 1[/imath], [imath]F_n(\mathbb{R}) = \bigcup \lbrace x_i < x_j \: or \: x_i > x_j \: | \: \forall \: 0 \leq i,j \geq n \rbrace [/imath] Hence [imath]\pi_0(F_n(\mathbb{R})) = \lbrace x_i < x_j \: or \: x_i > x_j \: | \: \forall \: 0 \leq i,j \geq n \rbrace [/imath] I am unable to see how the path-connected componnents of [imath]F_n(\mathbb{R}^d[/imath]) look like for all [imath]n \geq 1[/imath] and for all [imath]d \geq 1[/imath]... | 1197590 | Is the following set (path) connected?
This is a homework question. [imath]d,n\ge 2[/imath]. Let [imath]L=\{(x_1,...,x_n)\in (\mathbb{R}^d)^n: x_i\in \mathbb{R}^d, x_i\ne x_j \forall i\ne j\}[/imath]. I tend to think it is not path connected because if you assume [imath]d=1[/imath], this is the case. But I cannot prove for [imath]d\ge 2[/imath]. Any help will be appreciated! :) |
2501058 | Why is [imath]\sqrt x \times \sqrt y = \sqrt {xy}[/imath]
Sorry I do not know latex. Why is [imath]\sqrt x \times \sqrt y = \sqrt {xy}[/imath]? It also applies for division, but why not addition and subtraction? i.e., why is [imath]\sqrt x + \sqrt y[/imath] not equal to [imath]\sqrt {x+y}[/imath]? Thank you for editing and answering. | 627966 | Proof of identity [imath]\sqrt {xy} = \sqrt x \sqrt y[/imath] for [imath]x,y \in \mathbb R^+[/imath]
Proof of identity [imath]\sqrt {xy} = \sqrt x \sqrt y[/imath] for [imath]x,y \in \mathbb R^+[/imath] I've been looking at the stated identity, which makes sense in [imath]\mathbb R^+[/imath] but fails in [imath]\mathbb R[/imath], since [imath]\sqrt {-1 \cdot -1} \neq \sqrt {-1} \sqrt {-1}[/imath]. How does one prove this identity ? Suppose we have [imath]x,y \in \mathbb R^+[/imath] then [imath]\sqrt{xy}^2 = xy = (\sqrt x \sqrt y)^2[/imath]. Idea 1: The square-root function is bijective (monotonic increasing) and has inverse [imath]X^2[/imath]. This in turn implies [imath]\sqrt {xy} = \sqrt x \sqrt y[/imath] Idea 2: The equation [imath]X^2 = xy[/imath] has at most [imath]2[/imath] solutions in [imath]\mathbb C[/imath], which implies [imath]\sqrt x \sqrt y[/imath] must be equal to [imath]\pm \sqrt {xy}[/imath]. Are these ideas rigorous enough ? Is there some simpler way of proving this ? |
2501178 | Question about greatest common divisor.
Let [imath]a,b\in \mathbb{N}[/imath] and [imath]\gcd(a,b) = 1[/imath] Is it correct that [imath]\gcd(a^2,b^2) = 1[/imath] as well? And if so, how do I prove it? | 512924 | Prove that if [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a^2,b^2)=1[/imath]
So, if [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a^2,b^2)=1[/imath] means [imath]1=ax+by[/imath], and want to show [imath]a^2x+b^2y=1[/imath]. By squaring [imath]1=ax+by[/imath] both sides, I get, [imath]1=(ax)^2+b(2axby+by^2)[/imath]. It doesn't help my proof. Please help me with this proof. |
2501208 | Prove that [imath](1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4[/imath]
If a,b,c are positive prove that [imath](1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4[/imath] My approach I tried (1+a)(1+b)(1+c) =1+[a+b+c+ab+bc+ac+abc] Find the AM of Square Bracket, them AM [imath]\ge[/imath] GM But if I add 1 then AM>GM. Please help me with my approach | 1876539 | Prove that if [imath] a,b,c > 0 [/imath], then [imath] [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) [/imath].
Problem. Prove that if [imath] a,b,c > 0 [/imath], then [imath] [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) [/imath]. I don’t know how to solve this problem... What I can think of is to just simplify this inequality: [imath] \left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} > a^{4} b^{4} c^{4}. [/imath] How can I proceed with solving this problem? Note: This is a question of sequence and series, specifically AM-GM-HM inequality... |
2501417 | Let [imath]p\geq 5[/imath] be a prime. Show that [imath]\sum_{k=0}^{\frac {p-1}{2}} {\binom p k}3^k \equiv 2^p-1 \pmod {p^2}[/imath]
Let [imath]p\geq 5[/imath] be a prime. Show that [imath]\sum_{k=0}^{\frac {p-1}{2}} {\binom p k}3^k \equiv 2^p-1 \pmod {p^2}[/imath] I couldn't prove that [imath]p^2[/imath] divides it. My solution that [imath]\text {mod } p[/imath] works: Since each term except [imath]{\binom p 0} 3^0[/imath] is divisible by [imath]p[/imath], we are left to show that [imath]1+pk \equiv 2^p-1 \pmod p[/imath]. By Fermat's Little Theorem, [imath]p\mid 2^p-2[/imath], so the rest follows clearly. | 2428619 | [imath]\sum\limits_{k=0}^{\frac{p-1}2}3^k\binom{p}{k}\equiv 2^p-1\pmod{p^2}[/imath]
Let [imath]p[/imath] be prime and [imath]p\ge5[/imath]. My friend askes me the following [imath]\sum\limits_{k=0}^{\frac{p-1}2}3^k\binom{p}{k}\equiv 2^p-1\pmod{p^2}[/imath]. Here [imath]\binom{p}{k}=\frac{p!}{k!(p-k)!}[/imath] for any [imath]k=0,\ldots,p[/imath]. We define [imath]0!=1[/imath]. I think this is true, but I have no idea to attempt it. |
2502245 | Integrable functions vanish at infinity a.e.?
Suppose that [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] is integrable with repsect to Lebesgue measure. Prove that for almost every [imath]x\in[0,1][/imath] the sequence [imath]\{f(x+n)\}_{n=1}^\infty[/imath] tends to [imath]0[/imath] as [imath]n\rightarrow\infty[/imath]. Let [imath]g_n(x)=f(x+n)[/imath] for [imath]x\in\mathbb{R}[/imath]. Then above claim is equivalent to show that [imath]g_n\rightarrow0[/imath] pointwise on [imath][0,1][/imath] almost every. I'm puzzled by this problem. Could someone give useful hints or solution? | 2045509 | For [imath]f :\mathbb{R} \rightarrow \mathbb{R}\displaystyle[/imath] we have [imath]{\lim_{n \to +\infty} f(x+n) =0}[/imath] almost everywhere
I want to show that for [imath]f: \mathbb{R} \rightarrow\mathbb{R}[/imath] Lebesgue integrable [imath]\displaystyle{\lim_{n \to +\infty} f(x+n) =0}[/imath] almost everywhere. It is clear to me that this statement is true and I have tried to prove it by contradiction, by assuming that either [imath]\displaystyle{\lim_{n \to +\infty} f(x+n) =\infty}[/imath] or[imath]\displaystyle{\lim_{n \to +\infty} f(x+n) =K}[/imath] with [imath]K\in \mathbb{R}[/imath] . In the first case we can follow that [imath]|f|=\infty[/imath] so that [imath]\int_\mathbb{R}f=\infty[/imath] and we have a contradiction. In the other case we know that there must be an [imath]a\in \mathbb{R}[/imath] so that [imath]\forall x>a[/imath] we have [imath]f_+(x)>K/2[/imath] and by that: [imath]\int_{{\mathbb{R}}+}|f|>\int_{\mathbb{R}_+}f_+=\int_{[0,a]}f_++ \int_{(a,\infty)}f_+>\int_{[0,a]}f_++\int_{(a,\infty)}K/2=\infty[/imath] Would this be enough to show the statement? Can I even make the assumption that the limit of [imath]f(x+n)[/imath] "exists"? Any hints would be great. |
2497089 | Uniform distribution limit using characteristic functions
I've been working on the following problem: Let [imath](X_j)_{j\geq 1}[/imath] be independent random variables having the uniform [imath](-j,j)[/imath] distribution. a) Prove that [imath]\lim_{n\to \infty} \frac{S_n}{n^{3/2}}=Z [/imath] in distribution, where [imath]Z\sim N(0,\frac{1}{9})[/imath] Where [imath]S_n=\sum_{i=1}^n X_i[/imath] [imath]\textbf{Attempt at solution}[/imath] 1) The characteristic function of [imath]X_j[/imath] is [imath]\phi_{X_j}(t)=\frac{sin(tj)}{tj}[/imath] 2) Then, the characteristic function of [imath]\frac{S_n(t)}{n^{3/2}}[/imath] is [imath]\phi_{\frac{S_n}{n^{3/2}}}(t)=\prod_{i=1}^n\frac{sin(\frac{t}{n^{3/2}}i)}{\frac{t}{n^{3/2}}i}[/imath]. 3) If one can show that the limit as [imath]n\to \infty[/imath] of the above expression is [imath]e^{-u^2/18}[/imath] then that should finish the proof. However, I don't see an easy way to evaluate that limit. | 245414 | How can I show [imath]\lim_{n\to\infty}\prod_{j}^n\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=e^{-u^2/18}[/imath]?
This is from and exercise of probability theory: Let [imath](X_j)_{j\geq 1}[/imath] be independent and let [imath]X_j[/imath] have the uniform distribution on [imath](-j,j)[/imath]. Show that [imath] \lim_{n\to\infty}\frac{S_n}{n^{3/2}}=Z\sim N(0,\frac{1}{9}) [/imath] in distribution. In terms of characteristic functions, it suffices to show that [imath] \lim_{n\to\infty}\prod_{j=1}^n\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=e^{-u^2/18}. [/imath] Take [imath]\log[/imath] on both side one gets: [imath] \lim_{n\to\infty}\sum_{j=1}^n\log\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=-\frac{u^2}{18}. [/imath] How can this limit be proved? |
1934477 | Prove [imath]\lim_{x\to a}\sqrt{x}=\sqrt{a}[/imath] by definition Stewart solution error
Let [imath]a>0[/imath]. I want to show [imath]\lim_{x\to a}\sqrt{x}=\sqrt{a}[/imath] Purpose of this question was to investigate Stewart's online solution, which turns out to be incorrect. Their solution I don't understand the solution on their website. Are they saying that [imath]1/x < x [/imath]? Seems that way to me by the way they choose delta. My solution : [imath]|x-a|< a \implies \frac{1}{\sqrt{x} + \sqrt{a}} < \frac{1}{\sqrt{a}}[/imath] and choose delta accordingly. Is that correct? Thanks for the help. | 703587 | [imath]\epsilon - \delta[/imath] proof that [imath]\lim_{x \to a} \sqrt x = \sqrt a[/imath]
I am trying to prove, [imath]\lim_{x \to a} \sqrt x = \sqrt a[/imath] As per the definition of limit for every [imath]\epsilon > 0[/imath] there is some [imath]\delta > 0[/imath] such that [imath] 0 < |x-a| < \delta[/imath] implies [imath]|f(x) - \sqrt a| < \epsilon[/imath] [imath]|\sqrt x - \sqrt a| = \frac {|x -a|}{\sqrt x + \sqrt a} [/imath] since [imath] 0 < |x-a| < \delta[/imath] [imath]\frac {|x -a|}{\sqrt x + \sqrt a} < \delta[/imath] Can I stop my proof at this point since I found [imath]\epsilon[/imath] ( which is [imath]\delta[/imath] in this case) |
2502476 | How to solve [imath]\int ^{\infty}_{-\infty}\dfrac {\sin x}{x+i}dx[/imath]
should we use residue and construct a contour? I can't imagine how to make it. | 554810 | Calculating [imath]\int_{- \infty}^{\infty} \frac{\sin x dx}{x+i} [/imath]
I'm having trouble calculating the integral [imath]\int_{- \infty}^{\infty} \frac{\sin x}{x+i}dx [/imath] using residue calculus. I've previously encountered expressions of the form [imath]\int_{- \infty}^{\infty} f(x) \sin x dx [/imath] where you would consider [imath]f(z)e^{iz}[/imath] on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with [imath]e^{iz}[/imath] in my complex function because [imath]\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}...[/imath] How to remedy this? I'm not sure if substituting [imath]\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})[/imath] and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work |
2502766 | Prove: If u, v, w are integers such that [imath]u^2 + v^2 = w^2[/imath] , then [imath]u[/imath] and [imath]v[/imath] can’t be both odd.
I am trying to solve this problem, But I'm unable to get a single hint can you help me to solve this problem. If [imath]u[/imath], [imath]v[/imath], [imath]w[/imath] are integers such that [imath]u^2 + v^2 = w^2[/imath] , then [imath]u[/imath] and [imath]v[/imath] can’t be both odd. | 1676265 | If [imath]a^2+b^2=c^2[/imath] then [imath]a[/imath] or [imath]b[/imath] is even.
I am having trouble proving this directly. I know that it is easy to prove by contradiction by assuming both [imath]a[/imath] and [imath]b[/imath] to be odd, but how should I start to try to prove this? This is a homework problem and I am just looking for help getting started, not the full answer. |
2499739 | How to compute: [imath] \lim\limits_{a\to \infty}\frac1a\int_1^a a^{\frac1x} dx[/imath]
Help me to compute the following limit [imath] \lim_{a\to \infty}\frac1a\int_1^a a^{\frac1x} dx[/imath] I failed to apply the mean value theorem. And this the feature i had in mind when I saw this first Any Hint? | 1424234 | Computing [imath]\lim_{A\to\infty} \frac{1}{A} \int\limits_1^A \! A^{\frac{1}{x}} \, \mathrm{d}x.[/imath]
On this year's IMC there was this problem: Compute [imath] \lim_{A\to\infty} \frac{1}{A} \int\limits_1^A \! A^{\frac{1}{x}} \, \mathrm{d}x. [/imath] In addition to the two official solutions, I am curious as to if there exist other solutions. I heard that someone wrote that this is actually a probability density function (just for this one line, that person got 6 out of 10 points), so it would be great to see an answer involving this claim. Also, I thought of turning this integral (using substitutions) to something I can evaluate using the Gamma function. Could someone hint a substitution which could lead to that? EDIT: The official solutions can be found here, Problem 7. |
2503281 | No Nontrivial Homomorphism from [imath]\Bbb{Q}[/imath] to [imath]S_3[/imath]?
I am working through an example in Hungerford. In this example, he asserts that it's not difficult to show that the only homomorphism from [imath]\Bbb{Q}[/imath], regarded as an additive group, to [imath]S_3[/imath] is the trivial homomorphism. I am having trouble seeing this. Treat [imath]\Bbb{Q}[/imath] as an additive group, and let [imath]G[/imath] be some group (written multiplicatively). If [imath]f : \Bbb{Q} \to G[/imath] a homomorphism, I believe that [imath]f(p/q) = f(1)^{-pq}[/imath], so that the homomorphism is determined by how it maps [imath]1[/imath] (at least is what I proved on my own; it may be wrong). In our case [imath]G = S_3[/imath]. If [imath]f(1)= e[/imath], then [imath]f[/imath] is the trivial homomorphism. If [imath]f(1) \neq e[/imath], then... Not sure what to do at this point. I could use a hint. | 1807381 | Homomorphism from [imath](\Bbb Q,+)[/imath] to a finite group
Prove that if [imath]f[/imath] is a homomorphism from [imath](\Bbb Q,+)[/imath] to a finite group [imath]G[/imath] then [imath]f(q)=e_G[/imath] for all [imath]q[/imath] in [imath]\Bbb Q[/imath]. I attempted the following: Firsty I reason that [imath]f(1)[/imath] generates the entire image, because [imath]f(p/q)=p/qf(1)[/imath]. But since [imath]f[\Bbb Q][/imath] is an infinite subgroup of [imath]G[/imath] and [imath]G[/imath] itself is finite we are forced to let [imath]f(1)=e_G[/imath]. Is this ok? |
2503306 | Suppose [imath]g{^n}=e[/imath]. Show the order of [imath]g[/imath] divides [imath]n[/imath]
Suppose [imath]g{^n}[/imath]=e. Show the order of [imath]g[/imath] divides [imath]n[/imath]. Would I use Eulers Theorem???; [imath]a{^{\phi p}}[/imath] [imath]\equiv1 \pmod p[/imath] [imath]a{^{p-1}}\equiv1 \pmod p[/imath] [imath]a{^p}\equiv a\pmod p[/imath] So then I would have [imath]g{^n}\equiv g\pmod n[/imath] then I think you use the [imath]\gcd[/imath], which states [imath]\gcd(a,b) = 1[/imath] or [imath]a=nq+r[/imath] and [imath]b=nq+r[/imath] which is [imath]a\equiv b\pmod n[/imath]?? | 1778084 | Show that [imath]g^k=e[/imath] if and only if the order of [imath]g[/imath] is finite and divides [imath]k[/imath].
Let [imath]G[/imath] be a multiplicatively written group and let [imath]g\in G[/imath]. Show that [imath]g^k=e[/imath] if and only if the order of [imath]g[/imath] is finite and divides [imath]k[/imath]. Solution Suppose [imath]\operatorname{ord}(g)[/imath] divides [imath]k[/imath] and let [imath]\operatorname{ord}(g)=n.[/imath] So [imath]k=tn[/imath] for some [imath]t\in\mathbb{Z}.[/imath] Thus [imath]g^k=g^{tn}=(g^{n)^t}=e.[/imath] Now suppose [imath]g^k=e.[/imath] Where do I go from here? Edit: Suppose [imath]g^k=e[/imath] and suppose [imath]\operatorname{ord}(g)=n.[/imath] Now [imath]k=nq+r[/imath] for some [imath]0\leq r<n.[/imath] This now gives [imath]e=g^k=g^{nq+r}=g^{nq}g^r=g^r.[/imath] However, we have a contradiction as [imath]n[/imath] is the smallest integer for which [imath]g^r=e.[/imath] Thus [imath]r=0[/imath] and so [imath]n[/imath] divides [imath]k[/imath]. Is this correct? |
2498961 | Checking riemann curvature tensor [imath]C^\infty(M)[/imath] linear in [imath]Z[/imath] with [imath]R(X,Y)Z=\nabla_X\nabla_YZ-\nabla Y\nabla_XZ-\nabla_{[X,Y]}Z[/imath]
Let [imath]M[/imath] be a smooth manifold. Given affine connection [imath]\nabla:\Gamma(TM)\times\Gamma(TM)\to\Gamma(TM)[/imath], I want to check that curvature tensor field [imath]R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z[/imath] is [imath]Z[/imath] [imath]C^\infty(M)[/imath] linear. So I want to check [imath]R(X,Y)fZ=fR(X,Y)Z[/imath]. I can surely check this in local coordinates. Q: Is there any trick I can use to simplify my checking? | 722516 | Prove the Curvature Tensor is a Tensor
For an affine connection [imath]\nabla[/imath], prove the curvature R [imath]R(X,Y,Z,\alpha)=\alpha(\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z -\nabla_{[X,Y]}Z)[/imath] with [imath]X,Y,Z[/imath] vector fields and [imath]\alpha[/imath] a co-vector, is a tensor. So I realise the aim is probably to show that each of the three terms in the expression for [imath]R[/imath] are tensors themselves as then the result will obviously follow. However, I'm not too sure how to show each of these terms are tensors. Do I need to expand out each covariant derivate to get a bunch of connection coefficients and then see some cancellations? Still though, I'm not too sure what I'll want to see after all of this so that I can say "…therefore [imath]R[/imath] is a tensor". Also, very much related to the question, I'm a bit confused with the difference between [imath]\nabla_Y Z[/imath] [imath]\space \space [/imath] and [imath]\space \space [/imath] [imath]\nabla_{\mu}\omega[/imath]. The second one I know as the covariant derivative of a 1-form, but what is meant by the covariant derivate of a vector [imath]\textit{field}[/imath]? As in, what's the difference between the [imath]\nabla_Y[/imath] and [imath]\nabla_{\mu}[/imath]? Edit: In answering this last question, I have realised that [imath]\nabla_YZ=Y^{\mu}\nabla_{\mu}Z^{\nu}[/imath] and [imath]\nabla_{\mu}\omega=\nabla_{\mu}\omega_{\nu}[/imath]. |
977782 | nature of characteristic polynomial
Let [imath]A\in M_3(\mathbb R)[/imath] which is not a diagonal matrix.Let [imath]p[/imath] be a polynomial in one variable with real coefficients & of degree 3 such that [imath]p(A)=0[/imath]. Can we conclude that [imath]p=cp_A [/imath] where [imath]c\in \mathbb R [/imath] and [imath]p_A[/imath] is acharacteristic polynomial of [imath]A[/imath]?If not what kind of roots(real/complex) p should have in order to make such a conclusion. | 615690 | Diagonalisation and characteristic polynomial
Let [imath]A\in M_3(\Bbb R)[/imath] which is not a diagonal matrix. Let [imath]P[/imath] be a polynomial (in one variable), with real coefficients and of degree 3 such that [imath]P(A) = 0[/imath]. Pick out the true statements: a. [imath]P = cP_A[/imath] where [imath]c \in\Bbb R[/imath] and [imath]P_A[/imath] is the characteristic polynomial of [imath]A[/imath]; b. if [imath]P[/imath] has a complex root (i.e. a root with non-zero imaginary part), then [imath]P = cP_A[/imath], with [imath]c[/imath] and [imath]P_A[/imath] as above; c. if [imath]P[/imath] has a complex root, then [imath]A[/imath] is diagonalizable over [imath]\Bbb C[/imath]. I guess b and c are true although I am unable to disprove a. Because A could satisfy [imath](X-A)[/imath] but this is not a polynomial. Now if [imath]P[/imath] has a complex root, then its conjugate too is its root as the polynomial has real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since [imath]A[/imath] satisfies the polynomial the only difference between the characteristic polynomial and [imath]P[/imath] has to be a constant. |
2503614 | Dimension of Vector Space and Subspace Proof.
Let V be a vector space, and W a subspace of V. Prove that if [imath]dim(V)=n[/imath], then [imath]dim(W)\le dim(V)[/imath]. I know this is supposed to be a proof by contradiction, but I'm not sure on how to tackle it. | 2494291 | Prove that if W is a subspace of a finite dimensional vector space V, then dim(W) ≤ dim(V).
My attempt at a solution: Proof Given a vector space [imath]V[/imath], let [imath]S={v_1,...,v_n}[/imath] be a basis for [imath]V[/imath]. Consequently [imath]\dim(V)=n[/imath]. Now given [imath]W[/imath], a subspace of [imath]V[/imath], it will be shown that [imath]\dim(W)\le \dim V[/imath] First, consider the case where [imath]W=[/imath]{0}. Then [imath]dim(W)=0 \le\ n[/imath], Subsequently, consider the case for [imath]W=V[/imath]. Then it follows that [imath]dim(W)=dim(V)=n[/imath]. Finally, consider the case where [imath]W \subset V[/imath] and without loss of generality assume [imath]v_n \notin W[/imath]. Let [imath]T={v_1,...,v_k}[/imath] where [imath]k\lt n[/imath] be a basis for the subspace W. Now, since S is a basis for V, for every [imath]v_i \in S[/imath], [imath]c_1v_1+...+c_nv_n=0[/imath] has only the trivial solution, thus none of the vectors in S can be expressed as linear combinations of the other vectors in S. It follows that the set T is not a basis for V as [imath]v_n[/imath] cannot be generated from T. The [imath]dim(W)[/imath] is given by the number of vectors in the basis. Therefore, [imath]dim(W)=k \lt n[/imath] Therefore, it has been shown that if [imath]W \subseteq V[/imath] then [imath]dim(W) \le dim(V)[/imath] |
2502825 | Euclidean algorithm and GL2(Z)
How can I use the Euclidean algorithm to prove the matrices [imath] \begin{pmatrix} 1&1\\ 0&1\\ \end{pmatrix} \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} [/imath] are generating GL(2,Z)? | 976315 | Generators of [imath]\text{GL}_{2}(\mathbb{Z})[/imath] group, good reference book?
Does anyone know, where I can find a reference (preferably a book) which says that the general linear group [imath]\text{GL}_{2}(\mathbb{Z})[/imath] is generated by the set [imath]\left\{\begin{bmatrix} 1&0\\0&-1\end{bmatrix}, \begin{bmatrix} 0&1\\1&0\end{bmatrix}, \begin{bmatrix} 1&1\\0&1\end{bmatrix}\right\}[/imath] Thank you in advance. |
1118792 | Shortcut for determining equivalence relations?
Is there a short cut to determine the number of equivalence relations on the set [imath]\{1,2,3,4\}[/imath]? I mean I could do that manually but for a larger set it becomes annoying. Is there a general way to partition it? | 21393 | Number of equivalence relations on a set
If a set has [imath]n[/imath] elements then what are maximum number of equivalence classes and equivalence relations possible on it? |
2503770 | Proof of [imath]\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}r^n = n![/imath]
I want to provide proof for the following equation: [imath]\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}r^n = n![/imath] I think it resembles the typical way of representing the Principle of Exclusion & Inclusion: [imath]\sum_{\emptyset \ne J \in [n]}(-1)^{\mid J\mid-1}\mid\cap_{j\in J}Z_j\mid[/imath] However, still vague to let [imath]Z_j[/imath] to be which set to connect two different representations. any advice or pinch of hints? | 1862571 | Combinatorial proof of [imath]\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n![/imath], using inclusion-exclusion
If [imath]l[/imath] and [imath]n[/imath] are any positive integers, is there a proof of the identity [imath]\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!\;[/imath] which uses the Inclusion-Exclusion Principle? (If necessary, restrict to the case where [imath]l\ge n[/imath].) This question is closely related to Expressing a factorial as difference of powers: [imath]\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n![/imath]? and also Proof of the summation [imath]n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k[/imath]? |
2503607 | If [imath]G_1,\cdots,G_n,\cdots[/imath] are metrizable topological groups, then [imath]G= \prod_{n\in \mathbb N} G_n[/imath] is also a topological group.
Let [imath]G_1,\cdots, G_n, \cdots[/imath] metrizable topological groups. Prove that [imath]G= \prod_{n\in \mathbb N} G_n[/imath] is also a topological group. My book does not define the operations that we should have in [imath]G[/imath]. So I will state them (as they naturally should be) and give an attempt to show that they are continuous. Define [imath]\mu: G\times G \rightarrow G[/imath] given by [imath]\{(a_i)_{i\in \mathbb N}, (b_i)_{i\in \mathbb N}\} \mapsto (a_i\cdot b_i)_{i\in \mathbb N}[/imath] and [imath]\phi: G\rightarrow G[/imath] given by [imath](a_i)_{i\in \mathbb N} \mapsto (a_i^{-1})_{i\in \mathbb N}[/imath]. I will only prove that [imath]\mu[/imath] is continuous (to the other function the argument is similar). It is then sufficient to show that for every [imath]j\in \mathbb N[/imath] the function [imath]\mu_i = p_i \circ \mu[/imath] is continuous. Hence, given [imath]j[/imath] we see that [imath]\mu_j((a_i)_{i\in \mathbb N}, (b_i)_{i\in \mathbb N}) = a_j\cdot b_j = m(a_j,b_j)[/imath], where [imath]m: G_i \times G_j\rightarrow G_j[/imath]. Since [imath]G_j[/imath] is a topological group, we have (by definition) that [imath]m[/imath] is continuous, which shows that [imath]\mu_j[/imath] is also continuous. Hence comes the continuity of [imath]\mu[/imath]. Is this proof fine? Am i missing something? | 2309319 | Is the topological product [imath]\Pi \{G_\alpha: \alpha < \kappa\} [/imath] a topological group?
Let [imath]\mathcal G= \{G_\alpha: \alpha < \kappa\}[/imath] be a family of topological groups. My question is this: Is the topological product [imath]\Pi \{G_\alpha: \alpha < \kappa\} [/imath] a topological group? Thanks ahead. |
2504106 | Find the value of p for which the function f(x)=[imath]\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^{5}-3x+ln5[/imath] decreases [imath]\forall[/imath]x [imath]\in\mathbb{R}[/imath]
Question Find the value of p for which the function f(x)=[imath]\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^{5}-3x+ln5[/imath] decreases [imath]\forall[/imath]x [imath]\in\mathbb{R}[/imath] MY Approach[imath]\Longrightarrow[/imath]f'[imath]\left(x\right)[/imath]= $\left(\frac{\sqrt{p+4}}{1-p}-1\right)5x^{4}-3[imath]\leq$0$\Longrightarrow[/imath]\left(\frac{\sqrt{p+4}}{1-p}-1\right)\leq[imath]\frac{3}{5x^{4}}[/imath]\Longrightarrow[imath]\left(\frac{\sqrt{p+4}}{1-p}-1\right)<0[/imath]\Longrightarrow$x $\in[imath]\left(-\infty,\frac{3-\sqrt{21}}{2}\right)[/imath]\bigcup$ [imath]\left(\frac{3+\sqrt{21}}{2},\infty\right)[/imath] But Book Says Answer is $\left[-4,\frac{3-\sqrt{21}}{2}\right][imath]\bigcup[/imath]\left(1,\infty\right)$ | 897969 | When is [imath]f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5[/imath] decreasing [imath]\forall\; x[/imath]?
When is [imath]f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5[/imath] decreasing [imath]\forall\; x[/imath]? Diffrentiating: [imath]f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3[/imath] If [imath]f[/imath] is decreasing, [imath]f'<0[/imath]: [imath]5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3<0\implies x^4<\frac3{5\left(\frac{\sqrt{p+4}}{1-p}-1\right)}[/imath] Since if [imath]\left(\frac{\sqrt{p+4}}{1-p}-1\right)\to0^+\implies x^4<+\infty[/imath], which holds forallx, So:[imath]p=\frac{3\pm\sqrt{21}}2[/imath]which doesn't ply with textbook-answer, since it mentions an domain for p. |
1446675 | Show that elements are not conjugate to their inverse in groups of odd order.
Show that if G is a group of odd order, then no [imath]x\in G[/imath] other than the identity is conjugate to its inverse. We can't have elements of order 2, since by Lagrange theorem this would mean we would have the subgroup generated by that element would have to divide the order of the group, but G is odd so this can't happen. Hence we can't have [imath]x = x^{-1}[/imath]. Now suppose there exists element [imath]x\in G[/imath] such that [imath]x^{-1}=gxg^{-1}[/imath] for some [imath]g\in G[/imath]. Suppose that [imath]y\in\operatorname{Orb}_G(x)[/imath], so we have [imath]y=qxq^{-1}[/imath]. Then [imath]y^{-1}=qx^{-1}q^{-1}=qgxg^{-1}q^{-1}[/imath], which implies that [imath]y^{-1}\in\operatorname{Orb}_G(x)[/imath]. So for each [imath]y\in\operatorname{Orb}_G(x)[/imath] we will also have [imath]y^{-1}\in \operatorname{Orb}_G(x)[/imath], so [imath]\operatorname{Orb}_G(x)[/imath] must be even, but by orbit-stabilizer theorem we have [imath]|\operatorname{Orb}_G(x)|=[G:C_G(x)][/imath], where [imath]C_G(i)[/imath] is the centralizer of group [imath]G[/imath] for element [imath]x[/imath], but by Lagrange this must divide order of group [imath]G[/imath], which is odd, so this can't happen. Therefore our original assumption is wrong and we get our result. | 197828 | An element of a group [imath]G[/imath] is not conjugate to its inverse if [imath]\lvert G\rvert[/imath] is odd
Prove that if [imath]G[/imath] is a finite group of odd order, then no [imath]x\in[/imath][imath]G[/imath] , other than [imath]x=1[/imath], is conjugate to its inverse. This question is from Advanced Modern Algebra (exer 2.79) by Joseph J. Rotman. The hint states that if [imath]x[/imath] and [imath]x^{-1}[/imath] are conjugate, how many elements are in [imath]x^{G}[/imath]? What I know so far: [imath]\left\lvert x^{G}\right\rvert[/imath] is odd (greater than 1, otherwise it's in the center) and is a divisor of |[imath]G[/imath]| |Z([imath]G[/imath])| has a common factor (other than 1) with size of the orbit [imath]\left\lvert x^{G}\right\rvert[/imath] so that the center is not just the identity. This is from the class equation. The centralizer of [imath]x[/imath] has odd size and [imath]\lvert C_{G}(x) \rvert \cdot \lvert x^{G}\rvert=\lvert G\rvert[/imath] I don't see the implication of [imath]x[/imath] and its inverse being conjugate has other than their orbits having the same size and laying in the same conjugacy class. Is the info I have useful? Any help would be appreciated. |
2504379 | Show that [imath]C_0([0,\infty))[/imath] is complete.
Let [imath]C_b([0,\infty))[/imath] be the space of bounded continuous functions on [imath][0,\infty)[/imath]. On [imath]C_b([0,\infty))[/imath] we define a norm by [imath]\|f\|_\infty = \sup\{|f(x)| : x\in [0,\infty)\}[/imath]. Furthermore, define \begin{equation} C_0([0,\infty)) = \{f \in C_b([0,\infty)):\lim\limits_{n \to \infty}f(x) = 0\} \end{equation} Exercise: Show that [imath]C_0([0,\infty)), \|\cdot\|_\infty)[/imath] is complete. Hint: you can use that [imath](C_b([0,\infty)),\|\cdot\|_\infty)[/imath] is a complete space. I know that: If [imath](C_b([0,\infty)),\|\cdot\|_\infty)[/imath] is a complete space, [imath]C_0([0,\infty)), \|\cdot\|_\infty)[/imath] is complete if [imath]C_0([0,\infty))[/imath] is closed in [imath]C_b([0,\infty))[/imath]. [imath]C_0([0,\infty))[/imath] is closed [imath]\Leftrightarrow[/imath] [imath]B_\epsilon(x) \cap C_0([0,\infty)) \neq \emptyset[/imath] for any [imath]\epsilon > 0[/imath], then [imath]x\in C_0([0,\infty))[/imath]. What I think I should do: Pick [imath]x\in C_b([0,\infty))[/imath] such that [imath]B_\epsilon(x)\cap C_0([0,\infty)) \neq \emptyset[/imath] for every [imath]\epsilon >0[/imath]. Then; show that [imath]x\in C_0([0,\infty))[/imath] as well. Question: How do I show that [imath](C_0([0,\infty)), \|\cdot\|_\infty)[/imath] is complete by using my approach and showing that [imath]x\in C_0([0,\infty))[/imath] as well? | 2491344 | Showing that [imath](C_0([0,\infty)), \|\cdot\|_\infty)[/imath] is complete.
I need to proof that [imath](C_0([0,\infty)), \ \|\cdot\|_\infty)[/imath] is complete. I think that I did it correctly (or atleast am thinking in the right direction), but I still struggle with writing it down correctly. I used that [imath](C_b([0,\infty)), \|\cdot\|_\infty)[/imath] is complete, so therefore as [imath]C_0[0,\infty) \subset C_b[0,\infty) [/imath], I wanted to use that if [imath]A[/imath] is a closed subset of a complete space [imath](M,d)[/imath], that [imath](A,d)[/imath] is complete. Thus, it came down to showing that [imath]C_0[0,\infty)[/imath] is closed. Let [imath]f_n(k)[/imath] be a sequence of elements out of [imath]C_0[0,\infty)[/imath] that converges to an element [imath]f[/imath] out of [imath]C_b[0,\infty)[/imath]. I need to proof that [imath]f(k) \in C_0[0,\infty)[/imath]. Per the definition of convergence, we know that [imath]\|f_n(k) - f(k) \|_\infty \rightarrow 0[/imath] whenever [imath]n \rightarrow \infty[/imath]. We use this to find [imath] |f_n(k)-f(k)| \leq\|f_n(k) - f(k)\|_\infty \rightarrow 0[/imath]. Then we consider [imath]|f(k)| = |f(k) - f_n(k) +f_n(k)| \leq |f(k)-f_n(k)| + |f_n(k)|[/imath]. If we take limits on both sides, we get: [imath]\lim_{k\to \infty} |f(k)| \leq \lim_{k\to \infty} |f(k)-f_n(k)| + \lim_{k\to \infty} |f_n(k)| = 0 + 0 = 0[/imath] (the first term follows from what we have previously stated and the second term follows from the fact that [imath]f_n(k)[/imath] is an element of [imath]C_0[/imath]. Thus, [imath]f \in C_0[/imath] and therefore [imath]C_0[/imath] is closed and therefore it is complete. My questions are the following; What exactly does it mean to be a sequence of functions? I used [imath]f_n(k)[/imath] and [imath]f(k)[/imath] in my proof because I saw that it was used in another exercise, but I'm not sure what it means to write that [imath]k[/imath] down. Is the structure of my proof sound? Is there a better way to proof such statements? Thanks for reading, K. Kamal |
2500737 | About an inequality in inner product space with the best approximation property
I tried to prove an inequality in an inner product space [imath]X[/imath] which is as follows, Let [imath]M[/imath] be a linear set in an inner product space [imath]X[/imath], and $x[imath]∈[/imath]X$ such that [imath]ρ[/imath] = $inf[imath]_y[/imath]_∈[imath]_M$ $\|[/imath]x[imath]-[/imath]y[imath]\|$. Prove that for any $y[/imath]_1$, $y[imath]_2$ $∈$ $M$, $\|[/imath]y[imath]_1[/imath]-[imath]y[/imath]_2[imath]\|$ $\leq$ $([/imath]\|$x$-[imath]y[/imath]_1[imath]\|[/imath]^2[imath]-[/imath]ρ[imath]^2[/imath])[imath]^1[/imath]^/[imath]^2$ $+[/imath]([imath]\|$x$-[/imath]y[imath]_2[/imath]\|[imath]^2[/imath]-[imath]ρ[/imath]^2[imath])[/imath]^1[imath]^/[/imath]^2$. I tried to use the property of inner product and best approximation but somewhere lost in the middle and couldn't get the desired result. How should I proceed? | 2499111 | Inner Product Space Inequality
Let [imath]X[/imath] be an inner product space, let [imath]L\subset X[/imath] be a vector subspace, let [imath]x\in X[/imath], let [imath]d=\inf_{y\in L}\left\|x-y\right\|[/imath], and let [imath]y_1,y_2\in L[/imath]. Could someone please provide a hint of how to show that [imath]\left\|y_1-y_2\right\|\leq\sqrt{\left\|y_1-x\right\|^2-d^2}+\sqrt{\left\|y_2-x\right\|^2-d^2}\tag*{?}[/imath] I have tried the parallelogram law and orthogonal decomposition to no avail. The case where [imath]x\in\overline L[/imath] is straightforward. |
2504277 | Prove [imath] \frac{1}{x+1} \le \ln(1+\frac{1}{x}) \le \frac{1}{x}[/imath] with usage of the definition of a logarithmic function
Use the defintion of a logarithmic function to prove [imath] \frac{1}{x+1} \le \ln(1+\frac{1}{x}) \le \frac{1}{x}[/imath] for any positive integer x. ( Using defintion : [imath]e^y=\ln^{-1}y[/imath] ) Attempt: Using the definition, [imath]e^y=\ln^{-1}y[/imath] , I tried to prove the right hand side of the equality first by [imath] 1 +\frac{1}{x} \gt \frac{1}{x}[/imath], [imath]\ln(1 +\frac{1}{x})\lt \frac{1}{x} \Rightarrow e^{\ln(\frac{x+1}{x})} \le e^{\frac{1}{x}}[/imath] but I am really stuck and lost on how I should do it so that one of the sides there is ln and the other side is the value and also for the equality to be [imath]\le[/imath] as [imath] 1 +\frac{1}{x} \gt \frac{1}{x}[/imath] is strictly less than and is not equal. | 229599 | Prove [imath]\frac 1{x+1}<\ln\left(\frac{x+1}x\right)<0.5\left(\frac 1x+\frac 1{x+1}\right)[/imath]
I looking for help with proving the following inequality. Any relevant logarithmic identities would be great. Tried differentiating and taking limits and I'm lost as to how to approach this. [imath]\frac 1{x+1}<\ln\left(\frac{x+1}x\right)<0.5\left(\frac 1x+\frac 1{x+1}\right)[/imath] |
2504449 | Show that [imath]\lim_{n\to\infty} a^{\frac1n}=1[/imath]
Any ideas about how to show this neatly? I have seen a way using inequality between geometric and arithmetic mean. Is there a simpler way? | 1122350 | Proving [imath]\lim_{n\to\infty} a^{\frac{1}{n}}=1[/imath] by definition of limit
given a sequence [imath]a_n=a^{\frac{1}{n}}[/imath] for [imath]n\in\mathbb{N}^*[/imath], [imath]a\in\mathbb{R},a>1[/imath] then proof that [imath]\lim\limits_{n\to+\infty}a_n=1[/imath] by definition. proof: given [imath]a_n=a^{\frac{1}{n}}[/imath] for [imath]a\in\mathbb{R},a>1[/imath]. for [imath]n\in\mathbb{N}^*,n+1>n\Rightarrow \frac{1}{n+1}<\frac{1}{n}[/imath] and because [imath]a>1[/imath] we gets [imath]a^{\frac{1}{n+1}}<a^{\frac{1}{n}}[/imath], since [imath]\frac{1}{n+1}>0[/imath], then [imath]1=a^{0}<a^{\frac{1}{n+1}}<a^{\frac{1}{n}}\Rightarrow 1<a_{n+1}<a_{n}[/imath]. then we proof that [imath]\lim\limits_{n\to+\infty}a_n=1[/imath] we need to show that [imath]\forall\epsilon>0,\exists N,\forall n>N,|a_n-1|<\epsilon[/imath] then for [imath]\epsilon>0[/imath], choose [imath]N=\frac{1}{\log_a(\epsilon+1)}[/imath], since [imath]a>1[/imath] we gets that [imath]\forall n>N,1<a^{\frac{1}{n}}<a^\frac{1}{N}\Rightarrow0<a^{\frac{1}{n}}-1<a^{\frac{1}{N}}-1\Rightarrow |a^{\frac{1}{n}}-1|<|a^{\frac{1}{N}}-1|=|a^{\log_a(\epsilon+1)}-1|=|\epsilon+1-1|=\epsilon\Rightarrow |a^{\frac{1}{n}}-1|<\epsilon[/imath] wich implies that [imath]\lim\limits_{n\to+\infty}a_n=1\square[/imath] my proof to the limit is correct? |
242798 | Given 3 distinct primes {[imath]p,q,r[/imath]}, then [imath]|G|=pqr \implies G[/imath] not simple
Here's a question I've been asked; Given distinct primes [imath]p,q,r[/imath], show that any group [imath]G[/imath] of order [imath]pqr[/imath] is not simple. So far, my idea has been to individually check each possible proper subgroup, since by Lagrange there are only [imath]\dbinom{3}{2} + \dbinom{3}{1} + 1[/imath] of them, and show that at least one of them other than the trivial group is normal. Surely there is a smarter way, since generalizing this approach wouldn't work so well if I wanted to show similar results for groups who's order is a product of [imath]n[/imath] distinct primes. Is there some other approach? Thanks for any insights | 542377 | [imath]|G| = pqr[/imath] with [imath]p[/imath], [imath]q[/imath] and [imath]r[/imath] distinct primes. Show G is not simple.
[imath]|G| = pqr[/imath] with [imath]p[/imath], [imath]q[/imath] and [imath]r[/imath] distinct primes. Show G is not simple. I know this might have been asked and answered before. I just wanted someone to tell me if my argument is OK: Let [imath]|G| = pqr[/imath], and assume [imath]p < q < r[/imath]. We have at least one Sylow-[imath]q[/imath] subgroup (call it [imath]Q[/imath]) and at least another Sylow-[imath]r[/imath] subgroup (call it [imath]R[/imath]). Consider the subgroup [imath]K = [/imath] [imath]<Q, R>[/imath]. Now [imath]K[/imath] is not the whole group because there is no element of order [imath]p[/imath] in there. Further, [imath]|K|[/imath] must equal [imath]qr[/imath] because nothing smaller is possible by Lagrange's theorem. Therefore [G : K] = p, which is the smallest prime dividing G. Hence by a standard theorem (for instance, see corollary 4.5, p. 44, Isaacs' Algebra), K must be normal. Hence [imath]G[/imath] is not simple. Thanks! WH |
2504374 | If [imath]A[/imath] is a nowhere dense subset of a topological space [imath]X[/imath], then [imath]X\setminus A[/imath] is dense in [imath]X[/imath].
If [imath]A[/imath] is a nowhere dense subset of a topological space [imath]X[/imath], then [imath]X-A[/imath] is dense in [imath]X[/imath]. [imath](\overline{A})^\circ = \emptyset[/imath] ([imath]\because[/imath] [imath]A[/imath] is nowhere dense). We want to prove, [imath]\overline{X-A}=X[/imath] How can I prove for arbitrary topological space? Please help me. If you have short alternate solutions, please share with me. | 829752 | How to show the that a set [imath]A[/imath] nowhere dense is equivalent to the complement of [imath]A[/imath] containing a dense open set?
I was reading a textbook and saw an alternative formulation of nowhere dense. I am not sure how to prove this alternate formulation below: The Normal Nowhere Dense Statement: Let [imath]X[/imath] be a metric space. A subset [imath]A ⊆ X[/imath] is called nowhere dense in [imath]X[/imath] if the interior of the closure of [imath]A[/imath] is empty, i.e. [imath](\overline{A})^{\circ} = ∅[/imath]. Otherwise put, [imath]A[/imath] is nowhere dense iff it is contained in a closed set with empty interior. Alternate Formulation: "Passing to complements, we can say equivalently that [imath]A[/imath] is nowhere dense iff its complement contains a dense open set." Does anyone know how I can prove this? It seems rather painfully straightforward but I am not sure how to show it exactly. Thank you! |
2503430 | Find [imath]f(x)[/imath] such that [imath]f(\frac{3}{x})+f(\frac{x}{2})=\frac{9}{x^2}+\frac{x^2}{4}-2[/imath]
How to find [imath]f(x)[/imath] from below relation ? [imath]f\left(\frac{3}{x}\right)+f\left(\frac{x}{2}\right)=\frac{9}{x^2}+\frac{x^2}{4}-2[/imath] I have two question about this .I find [imath]f(x)=x^2-1[/imath] that work fine ,but [imath]\\(1)[/imath] how can be sure that there is no other solutions ? [imath]\\(2)[/imath] How can find [imath]f(x)[/imath] in general ? | 2502655 | let [imath]f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2[/imath] then find [imath]f(x)[/imath]
let [imath]f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2[/imath] then find [imath]f(x)[/imath] My Try : [imath]f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1[/imath] So we have : [imath]f(x)=x^2-1[/imath] it is right ?Is there another answer? |
2504720 | Rigorous proof of [imath]n\sum_{d|n}\frac{1}{d}=\sum_{d|n}d[/imath] for [imath]n\in\mathbb N[/imath].
Let [imath]n\in\mathbb N[/imath] and concider the sum [imath]\sum_{d|n}\frac{n}{d}.[/imath] I understand that dividing [imath]n[/imath] by any of it's divisors is also a divisor of [imath]n[/imath] so intuitively it would make sense that [imath]\{\frac n d\ :\ d|n \}=\{d\ :\ d|n\}[/imath], from which the desired equality in the title is acquired. but I cannot "see" how this can be proven rigorously. Any help or insight would be appreciated! Edit: In many sources, including the question this one is marked as a possible duplicate to, it is not explained why the identity holds, it's just used. If it actually is explained, it's not clear to me, at least. I am interested in an actual explanation, that's why my question differs from the one marked as duplicate. | 157419 | Show that [imath]\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}[/imath] for every positive integer [imath]n[/imath].
Show that [imath]\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}[/imath] for every positive integer [imath]n[/imath]. where [imath]\sigma (n)[/imath] is the sum of all the divisors of [imath]n[/imath] and [imath]\sum\nolimits_{d|n} f(d)[/imath] is the summation of [imath]f[/imath] at each [imath]d[/imath] where [imath]d[/imath] is the divisor of [imath]n[/imath]. I have written [imath]n=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}.......p_k^{\alpha_k}[/imath] then:- [imath]\begin {align*} \sum\nolimits_{d|n} \frac{1}{d}&=\frac{d_2.d_3......d_m+d_1.d_3......d_m+........+d_1.d_2.d_3......d_{m-1}}{d_1.d_2.d_3......d_m} \\&=\frac{d_2.d_3......d_m+d_1.d_3......d_m+........+d_1.d_2.d_3......d_{m-1}}{p_1^{1+2+...+\alpha_1}p_2^{1+2+...+\alpha_2}p_3^{1+2+....+\alpha_3}.......p_k^{1+2+....+\alpha_k}} \\ \end{align*}[/imath] where [imath]d_i[/imath] is some divisor among the [imath]m[/imath] divisors. Then I cannot comprehend the numerator so that to get the desired result. Also suggest some other approches to this question. |
2504260 | Problems on Legendre Symbols
I ran in to this problem today, and I seem to have trouble. Suppose [imath]p[/imath] is prime, [imath]p = 1\mod4[/imath], and that [imath]a^2+b^2=p[/imath] with [imath]a[/imath] odd and positive. Show that [imath](\frac{a}{p}) = 1[/imath]. How can I show that this holds true? | 1781773 | when p is sum of two square integer, prove (a/p) which is legendre symbol = 1
Let a,b be integers and p be an odd prime. if [imath]p[/imath]=[imath]a^2+b^2[/imath] and a is odd, prove [imath](a/p)[/imath] which is legendre symbol = [imath]1[/imath] what i have done is that : because p and a are odd, b must be even and p is the form of [imath]4k+1[/imath] ([imath]k[/imath] is integer) and after this, how to prove it ? |
2504560 | Prove the union [imath]\bigcup A_n[/imath] has empty interior in [imath]X[/imath]
Let [imath]X[/imath] be a Hausdorff compact space; let [imath]\{A_n\}[/imath] be a countable collection of closed set of [imath]X[/imath]. Show that if each set [imath]A_n[/imath] has empty interior in [imath]X[/imath], then the union [imath]\bigcup A_n[/imath] has empty interior in [imath]X[/imath]. My solution: Suppose that, [imath]U=[/imath] interior of [imath]\bigcup A_n[/imath] is nonempty. Then any open subset of [imath]U[/imath] can not have isolated point, since then all interior of [imath]A_n[/imath] can not be empty. (If [imath]U[/imath] have isolated point, then for some [imath]x\in X[/imath] with [imath]\{x\}\subset U[/imath] is an open set, then it must be a subset of some [imath]A_n[/imath], since [imath]\{x\}[/imath] do not intersects with [imath]X-A_n[/imath]) Let, [imath]x_0[/imath] and [imath]x_1[/imath] (where [imath]x_1\in A_{1}[/imath]) are two points in [imath]U[/imath]. Then there exists two disjoint open sets in [imath]X[/imath], say, [imath]B_1[/imath] and [imath]C_1[/imath]. Take, [imath]B_2=U\cap A^c_{1}\cap C_1[/imath], then [imath]x_1\notin \bar B_2[/imath]. Since [imath]B_2[/imath] is an open set containing [imath]U[/imath], there is a point [imath]x_2\in B_2[/imath] with [imath]x_2\in A_{k_2}[/imath] for some [imath]k_2[/imath], for which, in the same way, we can get an open set [imath]C_2[/imath] such that [imath]x_2\notin \bar B_3[/imath] with [imath]B_3=U\cap A^c_{k_2}\cap C_2[/imath] In this way, we can find sequence [imath]\bar B_2\supset \bar B_3\supset\dots[/imath] Since, [imath]X[/imath] is compact, [imath]\bigcap_{n\geq 2} \bar B_n[/imath] is nonempty. Then there exists a point [imath]x[/imath] which is an element of interior of [imath]A_n[/imath], for some [imath]n[/imath]. A contradiction. Is my solution ok? | 2505370 | Show that the union [imath]\bigcup A_n[/imath] has empty interior in [imath]X[/imath]
Let [imath]X[/imath] be a Hausdorff compact space; let [imath]\{A_n\}[/imath] be a countable collection of closed set of [imath]X[/imath]. Show that if each set [imath]A_n[/imath] has empty interior in [imath]X[/imath], then the union [imath]\bigcup A_n[/imath] has empty interior in [imath]X[/imath]. My solution: Suppose that, [imath]U=[/imath] interior of [imath]\bigcup A_n[/imath] is nonempty. Then any open subset of [imath]U[/imath] can not have isolated point, since then all interior of [imath]A_n[/imath] can not be empty. (If [imath]U[/imath] have isolated point, then for some [imath]x\in X[/imath] with [imath]\{x\}\subset U[/imath] is an open set, then it must be a subset of some [imath]A_n[/imath], since [imath]\{x\}[/imath] do not intersects with [imath]X-A_n[/imath]). Let, [imath]x_1\in A_1[/imath]. Choose a point [imath]x_2\in U[/imath] that do not contain in [imath]A_1[/imath]. Now choose disjoint sets [imath]W_1[/imath] and [imath]W_2[/imath] about [imath]x_1[/imath] and [imath]x_2[/imath]. Now, take [imath]V_1=W_2\cap U\cap A^c_1[/imath]. Then, [imath]x_1[/imath] do not contain in [imath]\bar V_1[/imath]. Again do this: There exist an [imath]k_1[/imath] such that [imath]x_2\in A_{k_1}[/imath]. Choose a point [imath]x_3\in V_1[/imath] that do not contain in [imath]A_{k_1}[/imath]. Now choose disjoint sets [imath]W_3[/imath] and [imath]W_4[/imath] about [imath]x_2[/imath] and [imath]x_3[/imath]. Now, take [imath]V_2=W_4\cap U\cap A^c_{k_1}[/imath]. Then, [imath]x_2[/imath] do not contain in [imath]\bar V_2[/imath]. In this way we can find a collection of closed sets [imath]\{\bar V_n\}[/imath] such that [imath]\bar V_1\supset \bar V_2\supset\bar V_3\supset\dots[/imath] Note that, since [imath]X[/imath] is compact, [imath]\bigcap \bar V_n[/imath] is nonempty. Let, [imath]x\in \bigcap \bar V_n[/imath]. Interior of [imath]A_n[/imath] is empty and [imath]x\notin A_n[/imath] for all [imath]n[/imath] leads to a contradiction. Is this solution ok? Any help appreciated. Sorry that I published same question twice. I marked the other one duplicate. |
2505498 | Having difficulty solving ODE: [imath]y' = \frac{1}{2} a y^2 + b y - 1[/imath] with [imath]y(0)=0[/imath]
I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve: [imath]y' = \frac{1}{2} a y^2 + b y - 1[/imath]with [imath]y(0)=0[/imath] The solution is given as [imath]y(x) = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma},[/imath] with [imath]\Gamma = \sqrt{b^2 + 2a}[/imath] I have tried numerous approaches, among other Bernoulli differential equations and variable separable methods. I keep on getting stuck on solving the integral that I encounter, and if I am able to solve it it looks nowhere near the solution as provided by the professor. If someone could show me the steps I would be very thankfull! I have invested too many hours of my time on this question now and giving up is no more option! | 2504760 | Help solving variable separable ODE: [imath]y' = \frac{1}{2} a y^2 + b y - 1[/imath] with [imath]y(0)=0[/imath]
I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve: [imath]y' = \frac{1}{2} a y^2 + b y - 1[/imath]with [imath]y(0)=0[/imath] The solution is given as [imath]y(x) = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma},[/imath] with [imath]\Gamma = \sqrt{b^2 + 2a}[/imath] I am really getting stuck at this exercise and would love to have someone show me how this solution is derived. One thing I did find out is that this ODE is variable separable. That is, [imath]y' = g(x)h(y) = (1) \cdot (\frac{1}{2} ay^2 + by - 1),[/imath] and therefore the solution would result from solving [imath]\int \frac{1}{\frac{1}{2} ay^2 + by - 1} dy = \int dx + C,[/imath] where C is clearly zero because [imath]y(0) = 0[/imath]. I am now getting stuck at solving the left integral. Could anyone please show me the steps? UPDATE So I came quite far with @LutzL solution, however my answers seems to slightly deviate from the solution given above. These are the steps I performed (continuing from @LutzL's answer): You complete the square [imath]\frac12ay^2+by-1=\frac12a(y+\frac ba)^2-1-\frac{b^2}{2a}[/imath] and use this to inspire the change of coordinates [imath]u=ay+b[/imath] leading to [imath] \int \frac{dy}{\frac12ay^2+by-1}=\int\frac{2\,du}{u^2-2a-b^2} [/imath] and for that your integral tables should give a form using the inverse hyperbolic tangent. Or you perform a partial fraction decomposition for [imath] \frac{2Γ}{u^2-Γ^2}=-\frac{1}{u+Γ}+\frac{1}{u-Γ} [/imath] and find the corresponding logarithmic anti-derivatives, [imath] \ln|u-Γ|-\ln|u+Γ|=Γx+c,\\ \frac{u-Γ}{u+Γ}=Ce^{Γx},\ C=\pm e^c [/imath] which you now can easily solve for [imath]u[/imath] and then [imath]y[/imath]. Given that [imath]y(0) = 0[/imath] we have [imath]u(y(0)) = u(0) = b[/imath] and therefore the final equation becomes [imath]\frac{u(0)-Γ}{u(0)+Γ}= \frac{b-Γ}{b+Γ}=Ce^{Γ\cdot0} = Ce^{Γ \cdot 0} = C[/imath] Now by first isolating [imath]u[/imath] I get [imath]u - \Gamma = u C e^{\Gamma x} + \Gamma C e^{\Gamma x} \Rightarrow \\ u \left( 1 - C e^{ \Gamma x} \right) = \Gamma \left( 1 + C e^{\Gamma x} \right) \Rightarrow \\ u = \frac{\Gamma \left( 1 + C e^{\Gamma x} \right)}{\left( 1 - C e^{ \Gamma x} \right)}[/imath] Now substituting u and C gives [imath]ay + b= \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right)}{\left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)} \Rightarrow \\ y = \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right) - b \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}{a \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}[/imath] Now using the fact that [imath]\Gamma = \sqrt{b^2 + 2a} \Rightarrow a = \frac{(\Gamma + b)(\Gamma - b)}{2}[/imath] we get that [imath]y = \frac{2 \left( \Gamma \left( 1 + \frac{b-\Gamma }{b+\Gamma } e^{\Gamma x} \right) - b \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right) \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\ = \frac{2 \left( (\Gamma - b) + (\Gamma + b) \frac{b-\Gamma }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\ = \frac{2 \left( (\Gamma - b) - (\Gamma + b) \frac{\Gamma - b }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)}[/imath] Now cancelling the terms [imath](b + \Gamma)[/imath] and [imath](b - \Gamma)[/imath] wherever possible and multiplying denominator and nominator by -1 gives [imath]y = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(\Gamma + b) \left( \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} - 1 \right)} [/imath] So clearly, I got the nominator right, but I can not seem to get the denominator to equal [imath](b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma[/imath]. Can someone rescue me and show me what I did wrong? Maybe it helps if I say that x is always positive? |
2499406 | Affine transformation read order
I would like to ask a question about affine matrix transformations, specifically an explanation about why these two ways to interpret them are equivalent. Suppose I have (in 2D for simplicity) an usual translation matrix T and a rotation matrix R as follows: [imath] T=\begin{bmatrix}1 & 0 & tx\\0 & 1 & ty\\ 0 & 0 & 1\end{bmatrix} R=\begin{bmatrix}\cos \theta & -\sin\theta & 0\\\sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{bmatrix} [/imath] Suppose for example: [imath] tx=0,ty=2,θ=45∘ [/imath] Now if I transform a quad applying translation and rotation to every vertex in this order: R * T * Vertex (Column vector), I obtain this: Reading the transforms left-to-right both translation and rotation transforms are done around the origin, considering T first and R after. However this is equivalent to considering the same equation reading it right-to-left R first and T after, transforming the quad relative to itself and not the origin: The result is the same. What is a mathematical explanation about why these two ways of reading the same affine transform composition are equivalent? | 2498235 | Affine transformation read order
I would like to ask a question about affine matrix transformations, specifically a mathematical explanation about why these two ways to interpret them are equivalent. Suppose I have a translation matrix T (for simplicity 2D) and a rotation matrix R as follows: [imath] T= \begin{bmatrix} 1 & 0 & tx \\ 0 & 1 & ty \\ 0 & 0 & 1 \\ \end{bmatrix} R= \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} [/imath] Suppose: [imath] tx = 0 , ty = 2, \theta = 45^{ \circ } [/imath] Now if I transform a quad applying translation and rotation to every vertex in this order: R * T * Vertex (Column vector), I obtain this: Since both translation and rotation transforms are done around the origin, considering T first and R after. However this is equivalent to considering R first and T after modifying the quad relative to itself and not the origin: The result is the same. Why is this equivalent? |
2505299 | How to prove that [imath]S_n = n^{2}[/imath]?
If [imath]\{x\}[/imath] denotes the fractional part of [imath]x[/imath] and for [imath]n\in \mathbb{N}[/imath] we define the set [imath]A_{n}=\left\{k \in \mathbb{N}: \{\frac{n}{k}\} \geq \frac{1}{2} \right\}[/imath]. Then how can I try to calculate the sum: [imath] S_{n}=\sum_{k \in A_{n}} \varphi(k)[/imath] where [imath]\varphi[/imath] is the Euler's totient function. I've tried calculating the sum by hand for integers [imath]1\leq n\leq 10[/imath] and noticed that [imath]S_{n} = n^{2}[/imath] for these, but other than this, I have not found any other patterns. | 168150 | Problem on Euler's Phi function
Let [imath]S(n)[/imath] be [imath]S(n)=\left\{k\;\left|\;\left\{\frac{n}{k}\right\}\right.\geq \frac{1}{2}\right\}[/imath],where [imath]\{x\}[/imath] is the fractional part of [imath]x[/imath] Prove that : \begin{align} \sum_{k\in S(n)} \varphi(k)=n^2 \end{align} maybe the fact \begin{align} \sum_{k \leq 2n}\left\lfloor{\frac{n}{k}+\frac{1}{2}}\right\rfloor \varphi(k)=\frac{3n^2+n}{2}\end{align} and \begin{align} \sum_{k \leq 2n}\left\lfloor{\frac{n}{k}}\right\rfloor \varphi(k)=\frac{n^2+n}{2}\end{align} can help, but I can't prove these two equations Edited:Sorry that my question is to prove the two equation before, because I've already known how to relate the problem with these two equations, and I've also known how to prove the second equation from Identity involving Euler's totient function: [imath]\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}[/imath] but I can't solve the first one with the similar method. |
2505809 | Prove that if [imath]x[/imath] is divisible by [imath]4[/imath], then [imath]x=a^2-b^2[/imath]
If [imath]x\in Z[/imath] and divisible by [imath]4[/imath], then there are existing [imath]a,b \in Z[/imath] , so: [imath]x=a^2-b^2[/imath] I had been trying to figure out which way this should be solved. First I found out that if [imath]x=20[/imath], then [imath]a^2 = 6^2[/imath] and [imath]b^2 = 4^2[/imath] (of course [imath]20=36-16[/imath] ). now I'm not sure if i could place those examples as an answer, or I need to prove it with some "proving tricks". anyway, I thought about that [imath]a^2 - b^2 = (a+b)(a-b)[/imath] and [imath]4\mid x[/imath] is also [imath]x=4k[/imath] for some [imath]k[/imath] integer. | 491914 | If [imath]a[/imath] is divisible by [imath]4[/imath], then there exist int [imath]b[/imath] and [imath]c[/imath] such that [imath]a = b^2 - c^2[/imath]
I want to prove this: Prove that if [imath]a \in \mathbb{Z}[/imath] is divisible by [imath]4[/imath], then there exist [imath]b[/imath] and [imath]c[/imath] where [imath]b,c \in \mathbb{Z}[/imath] such that [imath]a = b^2 - c^2[/imath] I want to prove this directily: [imath]4\mid a [/imath], [imath]a\in \mathbb{Z} \implies \exists b,c \in \mathbb{Z}, \quad[/imath] s.t. [imath]a=b^2-c^2[/imath] Let [imath]k\in \mathbb{Z}[/imath]. Then [imath]4=a\cdot k[/imath] and [imath]k[/imath] can divide [imath]4[/imath]. Hence, [imath]\frac{4}{k} =b^2 - c^2 = (b-c)(b+c) \implies \dfrac{4}{k\cdot(b-c)} = b+c[/imath] Is this proof correct? |
2500094 | Showing [imath]a^n + b^n \ge (a+b)^n[/imath] for [imath]a,b > 0[/imath] and [imath]0 \le n \le 1[/imath]
Initially I had to show that for [imath]0\le n \le 1[/imath] and [imath]a,b > 0[/imath]: [imath]a^n + b^n \ge (a+b)^n[/imath] I made [imath]u(x) = a^x + b^x - (a+b)^x[/imath] [imath]u'(x) = a^x \log a + b^x \log b - (a+b)^x \log (a+b)[/imath] Now I get stuck. Cannot find where to go. please help me. | 1016534 | Prove [imath](|x| + |y|)^p \le |x|^p + |y|^p[/imath] for [imath]x,y \in \mathbb R[/imath] and [imath]p \in (0,1][/imath].
Prove [imath](|x| + |y|)^p \le |x|^p + |y|^p[/imath] for [imath]x,y \in \mathbb R[/imath] and [imath]p \in (0,1][/imath]. A hint is given that for [imath]0 \le \alpha \le \beta[/imath] there exist a number [imath]\xi \in (\beta, \alpha + \beta)[/imath] s.t. [imath](\alpha + \beta)^p - \beta^p = p \xi^{p-1}\alpha \le \alpha^{p}[/imath]. I see that by proving the above hint the result holds for positive numbers. In order to prove the above hint I've considered the Intermediate value theorem, however I've not yet succeeded. |
2502481 | Prove that Space of Lipschitz continous functions is Banach space
Let [imath]X[/imath] be the vector space of Lipschitz continous functions, [imath][0, 1] \rightarrow \mathbb{R}[/imath]. For [imath]x\in X[/imath] set [imath]\Vert x \Vert_{Lip}=\vert x(0)\vert+sup_{s\neq t}\left\vert \frac{x(s)-x(t)}{s-t}\right\vert.[/imath] I need to prove: [imath]\Vert x \Vert_{\infty}\leq\Vert x \Vert_{Lip}[/imath] for [imath]x\in X.[/imath] [imath]\left ( X,\Vert.\Vert_{Lip} \right )[/imath] is a Banach space. I have tried to make an estimation of [imath]\Vert x \Vert_{\infty}[/imath] and compare it with [imath]\Vert x \Vert_{Lip}[/imath] but i could't reach that far. As to the Point 2. i know that you start with a Cauchy sequence and need to prove that it converges in [imath]X.[/imath] Unfortunately i couldnt come further either. I will appreciate any comment or help. Thanks. | 1569788 | the space of lipschitz function is complete with respect to some norm
Let [imath]V[/imath] be the space of real valued lipschitz functions over [imath][a,b][/imath],we define: [imath]M_f=sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|}[/imath] and lipschitz norm: [imath]||f||_{Lip}=|f(a)|+M_f[/imath] prove that [imath]V[/imath] with lipschitz norm is a complete normed vector space. it is easy to see [imath]V[/imath] is a vector space,but what about the completeness? is there any hint? thank you very much. |
2399164 | Which of the following statements are correct? (topology)
consider [imath]S^1 =\{(x,y)\in \mathbb{R}^2 : x^2 + y^2 =1\}[/imath] , [imath]D=\{(x,y)\in \mathbb{R}^2: x^2 + y^2 \leq 1\}[/imath] and [imath]E = \{(x,y)\in \mathbb{R}^2: 2x^2 + 3y^2 \leq 1\}[/imath] Which of the following statements are true? a. If [imath]f : D \to S^1[/imath] is a continuous mapping, then there exists [imath]x \in S^1[/imath] such that [imath]f(x)=x[/imath]. b. If [imath]f: S^1 \to S^1[/imath] is a continuous mapping, then there exists [imath]x \in S^1[/imath] such that [imath]f(x)=x[/imath]. c. If [imath]f : E \to E[/imath] is a continuous mapping, then there exists [imath]x \in E[/imath] such that [imath]f(x) = x[/imath]. i was using the intermediate theorem, that from my point of view , all option a b and c are correct by intermediate theorem. But im doubting about my answer. If anbody help me i would be very thankful to him. | 186117 | another topology multiple choice
Let [imath]S^1 = \{(x, y) \in \Bbb R^2 : x^2 + y^2 = 1 \}.[/imath] Let [imath]D = \{(x, y) \in\Bbb R^2 : x^2 + y^2 \le 1 \}[/imath] and [imath]E = \{(x, y) \in\Bbb R^2 : 2x^2 + 3y^2 \le 1\}[/imath] be also considered as subspaces of [imath]\Bbb R^2.[/imath] Which of the following statements are true? a. If [imath]f : D \to S^1[/imath] is a continuous mapping, then there exists [imath]x \in S^1[/imath] such that [imath]f(x) = x[/imath]. b. If [imath]f : S^1 \to S^1[/imath] is a continuous mapping, then there exists [imath]x \in S^1[/imath] such that [imath]f(x) = x[/imath]. c. If [imath]f : E \to E[/imath] is a continuous mapping, then there exists [imath]x \in E[/imath] such that [imath]f(x) = x[/imath]. |
2505714 | Showing Cantor set is uncountable...
[imath]C[/imath] is cantor set on [imath][0,1][/imath]. [Definition: [imath]A_0[/imath] is the interval [imath][0,1][/imath]. [imath]A_n=A_{n-1}-\bigcup_{k=0}^{\infty}\Big(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\Big),\space\space n=1,2,\dots[/imath] The intersection [imath]C=\bigcap_{n=0}^{\infty} A_n[/imath] is called Cantor set on [imath][0,1][/imath]] (a) Show that [imath]C[/imath] is totally disconnected. (b) Show that [imath]C[/imath] is compact. (c) Show that each set [imath]A_n[/imath] is union of finitely many disjoint closed interval of length [imath]1/3^n[/imath], end show that the end points of this interval lie in [imath]C[/imath]. (d) Show that [imath]C[/imath] has no isolated points. (e) Conclude that [imath]C[/imath] is uncountable. I have no problem of showing the first four ((a), (b), (c) and (d)) part of the proof. But got stuck at (e). I know this theorem: Let [imath]X[/imath] be the nonempty compact Hausdorff space. If [imath]X[/imath] has no isolated points, then [imath]X[/imath] is uncountable. How can I show that [imath]C[/imath] is Hausdorff subspace of [imath][0,1][/imath]? | 1269425 | The cantor set is uncountable
I am reading a proof that the cantor set is uncountable and I don't understand it. Hopefully someone can help me. Let [imath]C[/imath] be the Cantor set and [imath]x\in C[/imath]. Then there exists unique [imath]x_k\in \{0,2\}[/imath] such that [imath]x=\sum_{k\in \mathbb N}\frac{x_k}{3^k}[/imath]. Conversely every [imath]x[/imath] with this representation lies in C. If [imath]C[/imath] would be countable then there would exists a injective map [imath]f:C\rightarrow \mathbb N[/imath]. Let [imath]f^{-1}(i)=\sum_{k\in \mathbb N}\frac{x_{ik}}{3^k}[/imath] or [imath]f^{-1}(i)=\emptyset[/imath]. Set [imath]z_k=2-x_{i_k}[/imath], then z=[imath]\sum_{k\in \mathbb N}\frac{z_k}{3^k}[/imath] Then [imath]z[/imath] is in the cantor set but z isn't in the pre-image of [imath]f[/imath]. Questions: Shouldn't it be [imath]z_k=2-x_k[/imath]? I understand that z lies in C but why it is not in the pre image of f? Thanks in advance |
2506708 | paint [imath]\mathbb{R}^2[/imath] with two colors
Can we paint [imath]\mathbb{R}^2[/imath] with two colors (say, black and white) so that any horizontal line pass through countable black points and any vertical line pass through countable white points? Update. I was amazed that this is a classical problem! Actually what I wanted to do is using set theory to prove that there exist incomparable Turing degrees. | 1872557 | Equivalence between CH and existence of certain sets in [imath]\mathbb{R}^2[/imath]
The problem is to prove the following equivalence: [imath]2^{\aleph_0}=\aleph_1 \iff \exists\ A,B\in\mathbb{R}^2:\\ \textrm{a)}A\cup B=\mathbb{R}^2 \\\textrm{b)}\ \forall\ \textrm{lines } l\ \textrm{parallel to x-axis, } |A\cap l|\leq\aleph_0 \\\textrm{c)}\ \forall\ \textrm{lines } l\ \textrm{parallel to y-axis, } |B\cap l|\leq\aleph_0[/imath] I think I know how to prove "[imath]\impliedby[/imath]": Let's assume the RHS and that [imath]2^{\aleph_0}\geq\aleph_2[/imath]. Now let's take a sequence [imath]\langle x_{\alpha}:\ \alpha<\omega_1\rangle[/imath]. Since [imath]\bigcup_\limits{\alpha<\omega_1}|B\cap x_{\alpha}|\leq \aleph_1\cdot\aleph_0=\aleph_1[/imath] there is such a [imath]y[/imath]-coordinate that neither of [imath](x_{\alpha},y)[/imath] is in [imath]B[/imath]. But hence we obtain [imath]\aleph_1[/imath] point which all lie on the same line parallel to [imath]x[/imath]-axis. Since [imath]A\cup B=\mathbb{R}^2[/imath] they all must be contained in [imath]A[/imath] [imath]-[/imath] a contradiction as [imath]A[/imath] has at most countably many of those points. But what about the other implication? Do I contruct these sets by induction? Over all lines or something else? I haven't come up with anything yet, perhaps you could help me. |
2506773 | [imath]x\in \mathbb{R}\setminus\{0\}[/imath] and [imath]x + \frac{1}{x}[/imath] is an integer. Prove [imath]\forall n\in \mathbb{N}[/imath] [imath]x^n + \frac{1}{x^n}[/imath] is also an integer.
Let [imath]x[/imath] be a non-zero real number such that [imath]x + \frac{1}{x}[/imath] is an integer. Prove that [imath]\forall n\in \mathbb{N}[/imath] the number [imath]x + \frac{1}{x}[/imath] is also an integer. Attempt at solution using induction: base case: [imath]n = 1[/imath] then [imath]x^n + \frac{1}{x^n} = x + \frac{1}{x}[/imath] which is an integer Inductive assumption: Assume that for some [imath]k\in \mathbb{N} : x^k + \frac{1}{x^k}[/imath] We have to show that [imath] x^{k+1} + \frac{1}{x^{k+1}}[/imath] is an integer. [imath] x^{k+1} + \frac{1}{x^{k+1}} = x^k\cdot x + \frac{1}{x^k\cdot x} = \frac{(x^k\cdot x)\cdot (x^k\cdot x)+1}{x^k\cdot x}[/imath] I can't find a way to seperate [imath]x[/imath] and [imath]\frac{1}{x}[/imath] from the term so I can use the inductive assumption. Edit: I don't think that this question should count as a duplicate since the linked question is asking to specifically solve another problem and one of the answers of that question utilize the proof from this question in their answer but that question itself is different and not related to this question since that question can be solved without using this proof as other answers of that question don't include it. | 936479 | Proving that [imath]\frac{\phi^{400}+1}{\phi^{200}}[/imath] is an integer.
How do we prove that [imath]\dfrac{\phi^{400}+1}{\phi^{200}}[/imath] is an integer, where [imath]\phi[/imath] is the golden ratio? This appeared in an answer to a question I asked previously, but I do not see how to prove this.. |
2506813 | Stirling number of the first kind equality
The question I am looking at asks to use a combinatorial argument to show that [imath] \begin{bmatrix} n \\ 2 \\ \end{bmatrix} = \frac{n!}{2} \sum_{m=1}^{n-1} \frac{1}{m(n-m)}[/imath] I tried to use the recurrence relation for Stirling numbers of the first kind but couldn't find a way to make this work, help appreciated thanks. | 2503049 | Proof concerning Stirling numbers of the first kind
How do I show that [imath]{ n\brack 2} = \frac{n!}{2} \sum_{m=1}^{n-1} \frac{1}{m(n-m)}[/imath] using a combinatorial argument? My definition of [imath]{n \brack 2}[/imath] is "the number of elements of [imath]S_n[/imath] which can be decomposed as the product of exactly [imath]2[/imath] disjoint cycles". |
2507014 | Divergence of reciprocal of primes, Euler
On Wikipedia at link currently is: \begin{align} \ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right) = -\sum_p \ln \left( 1-\frac{1}{p}\right) \\ & {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\ & {} = \sum_{p}\frac{1}{p}+ \sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p ^3} + \sum_{p}\frac{1}{4p^4} + \cdots \\ & {} = \left( \sum_p \frac{1}{p} \right) + K \end{align} And then Wikipedia says that [imath]K<1[/imath], without any explanation. How do we know that [imath]\sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p ^3} + \sum_{p}\frac{1}{4p^4} + \cdots[/imath] is equal to a constant [imath]K<1[/imath]? | 2442234 | Prove [imath]\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots[/imath] converges
Prove [imath]\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots[/imath] converges, where the sums are for all primes [imath]p[/imath]. I've found in this link that [imath]\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots[/imath] converges to [imath]K[/imath], where [imath]K<1[/imath]. I know that [imath]\sum_p \frac{1}{p^2}\le \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6},[/imath] because see Basel problem and specific values of Riemann zeta function. Therefore also [imath]\sum_p \frac{1}{p^k}\le \sum_p \frac{1}{p^2}\le\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}[/imath] for all [imath]k\ge 3[/imath], [imath]k\in\mathbb Z.[/imath] |
2506950 | How to calculate [imath]\int_0^\infty \left(\frac{\sin x}{x}\right)^2\,dx[/imath]
This is my exercise in Complex Analysis. [imath] \int_0^\infty \left(\frac{\sin x}{x}\right)^2\,dx = ? [/imath] I tried solving this by [imath](\sin x)^2=1-\cos(2x)[/imath] but I realized that [imath]\int_0^\infty \frac{dx}{x^2}[/imath] does not converge. Can someone help me? Thank you so much. | 646096 | Calculating [imath]\int_0^\infty \frac {\sin^2x}{x^2}dx[/imath] using the Residue Theorem.
I am trying to compute the following integral using the Residue Theorem but am quite stuck: [imath]\int_0^\infty \frac{\sin^2x}{x^2}dx[/imath] I have tried applying Jordan's lemma, having written [imath]\sin(x)[/imath] as [imath]\dfrac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}[/imath], to not much avail. I have also tried using a rectangular integration path but it didn't get me far. I'd be grateful for an insightful advice. |
2475094 | Divisibility within a block of natural numbers
I have solved the simpler case of this question (with just [imath]a[/imath] and [imath]b[/imath] and a [imath]b[/imath] length block), but no answer on this site has ever properly answered the generalisation: Let [imath]a,b,c \in \Bbb{N}.\space [/imath]Suppose [imath]a<b<c[/imath]. Does every block of [imath]c[/imath] consecutive naturals contain three distinct naturals whose product is divisible by [imath]abc[/imath]? I would be very grateful for any help. | 2405567 | Divisibility in blocks of natural numbers
For [imath]a<b[/imath], prove that every block of [imath]b[/imath] consecutive naturals, there are two distinct naturals whose product is divisible by [imath]ab[/imath]. [I believe I can do this: see below] Now, for [imath]a<b<c[/imath], is it true that in every block of [imath]c[/imath] consecutive naturals there are three distinct naturals whose product is divisible by [imath]abc[/imath]? Informal attempt for part 1: In [imath]b[/imath] consecutive integers there must be a multiple of [imath]a[/imath] and a multiple of [imath]b[/imath], since [imath]a<b[/imath]. If they are the same, I've shown that you can select the next multiple of [imath]a[/imath] in the block and thus we have a product divisible by [imath]ab[/imath]. EDIT: there isn't necessarily another multiple of [imath]a[/imath] but I can now show a multiple of [imath]gcd(a,b)[/imath]. Please answer both parts of the question, if you would like to, and certainly criticise my informal attempt where necessary. I am more interested in the second part however: my friend who gave this to me was ambiguous about whether it's actually true, so possibly I actually need to prove it's false. I have noticed this question has been asked before, but one of the questions was abandoned because the questioner was rude, and for the other, no answer was actually reached, so I was hoping another attempt might be made. Thank you very much for your consideration. |
2507615 | Show that infinitely many members of [imath]1,11,111,1111,\dots[/imath] are divisible by [imath]2^{2017}+1[/imath].
Show that infinitely many members of [imath]1,11,111,1111,\dots[/imath] are divisible by [imath]2^{2017}+1[/imath]. I know that the sequence above is equal to [imath]a_n = \frac{10^n-1}{9}[/imath] for [imath]n = 1,2,3,\dots[/imath]. I also feel that I should use the pigeonhole principle but do not know how to attack the problem. | 2507238 | Prove that there are infinitely many integers of the form [imath]111\ldots111[/imath] that are multiple of [imath]n[/imath].
Let [imath]n[/imath] be a positive odd integer which is not a multiple of [imath]5[/imath]. Prove that there are infinitely many integers of the form [imath]111\ldots111[/imath] that are multiple of [imath]n[/imath]. I learned about Euler's theorem and [imath]\phi[/imath] function, but I don't know how to solve this problem. |
2507636 | How to solve this ( and general )logarithmic Equation
[imath] 4^{x+1} - 6^x - 2*9^{x+1} = 0[/imath] I recently stumbled across this logarithmic Equation and really I have no clue how to solve this. Also , please provide a tactical way to approach such questions | 1356236 | What is the solution to the equation [imath]9^x - 6^x - 2\cdot 4^x = 0 [/imath]?
I want to solve: [imath]9^x - 6^x - 2\cdot 4^x = 0 [/imath] I was able to get to the equation below by substituting [imath]a[/imath] for [imath]3^x[/imath] and [imath]b[/imath] for [imath]2^x[/imath]: [imath] a^2 - ab - 2b^2 = 0 [/imath] And then I tried \begin{align}x &= \log_3{a} \\ x &= \log_2{b} \\ \log_3{a} &=\log_2{b}\end{align} But I don't know what to do after this point. Any help is appreciated. |
2499066 | Probability a bit in a bit string is 1 after swapping
Stuck on a homework question, so I could use all the help I could get. Let [imath]x = x(1), \dots , x(n)[/imath] be a bit string containing exactly [imath]m[/imath] occurrences of 1. Consider the following operation on [imath]x[/imath]: we choose a random pair of indices [imath](i,j),[/imath] and we swap [imath]x(i)[/imath] and [imath]x(j)[/imath] so that [imath]x'(i) = x(j),[/imath] [imath]x'(j) = x(i),[/imath] while [imath]x'(k) = x(k)[/imath] if [imath]k \neq i[/imath] and [imath]k \neq j.[/imath] (If [imath]i = j,[/imath] therefore, then we change nothing.) Let [imath]X_1 = x,[/imath] and let [imath]X_2, \dots, X_N[/imath] be obtained by such a sequence of operations (always swapping a new random pair) that so [imath]X_{r+1} = X_r[/imath]. The number of 1s remains [imath]m[/imath] in each iteration. Show for each [imath]i[/imath], we have [imath]P(X_N (i) = 1) \rightarrow \frac{m}{n}[/imath] as [imath]N \rightarrow \infty[/imath]. We're given this hint: Consider the last time [imath]i[/imath] was swapped. I've gathered that the probability that [imath]i[/imath] is swapped on any given iteration is [imath]1-(1- \frac{1}{n})^2[/imath]. I've also figured out that the probability [imath]i[/imath] is 1 after a swap is [imath]\frac{m}{n}[/imath], as there are m choices for i to change to (including itself) after being selected for a swap, but I'm not sure how to apply this. | 2498972 | Bit swapping probability question
Let [imath]x = x(1), \dots , x(n)[/imath] be a bit string containing exactly [imath]m[/imath] occurrences of 1. Consider the following operation on [imath]x[/imath]: we choose a random pair of indices [imath](i,j),[/imath] and we swap [imath]x(i)[/imath] and [imath]x(j)[/imath] so that [imath]x'(i) = x(j),[/imath] [imath]x'(j) = x(i),[/imath] while [imath]x'(k) = x(k)[/imath] if [imath]k \neq i[/imath] and [imath]k \neq j.[/imath] (If [imath]i = j,[/imath] therefore, then we change nothing.) Let [imath]X_1 = x,[/imath] and let [imath]X_2, \dots, X_N[/imath] be obtained by such a sequence of operations (always swapping a new random pair) that so [imath]X_{r+1} = X_r[/imath]. The number of 1s remains [imath]m[/imath] in each iteration. Show for each [imath]i[/imath], we have [imath]P(X_N (i) = 1) \rightarrow \frac{m}{n}[/imath] as [imath]N \rightarrow \infty[/imath]. What's confusing me is that it seems as though the probability on any iteration of [imath]X_j[/imath] that [imath]X_j(i) = 1[/imath] is always [imath]\frac{m}{n}[/imath], which would make the problem trivial. |
2507629 | Proving that this function has the same value for all integers [imath]\geq4[/imath].
My teacher gave us this question: A function [imath]f[/imath] has the property [imath]f(x+y) = f(xy) [/imath] [imath]\forall x, y\geq4; x,y\epsilon Z[/imath] [imath]f(8)=9[/imath] Find [imath]f(9)[/imath]. I know the solution to this question. [imath]9=f(8) = f(4+4) = f(16)=f(8+8)=f(64)=f(16\cdot4)=f(20)=f(4\cdot5)=f(9)=9[/imath] But I tried this for other values and everytime it came [imath]9[/imath]. I think this will be true for every integer [imath]\geq 4[/imath]. Intuition tells me that I can reach to any such integer from any such other integer by performing those operations. Is the function [imath]9[/imath] for every integer [imath]\geq 4[/imath]? If so, what would be its rigorous proof? I think it may need some number theory. Edit: I realized that if my assumption is true, it is only possible for [imath]n\geq8[/imath]. | 1668442 | Prove that [imath]f(x)=8[/imath] for all natural numbers [imath]x\ge{8}[/imath]
A function [imath]f[/imath] is such that [imath]f(a+b)=f(ab)[/imath] for all natural numbers [imath]a,b\ge{4}[/imath] and [imath]f(8)=8[/imath]. Prove that [imath]f(x)=8[/imath] for all natural numbers [imath]x\ge{8}[/imath] |
2507928 | Why does [imath]A_6[/imath] 6-cycle from [imath]S_6[/imath] which also has even number of 2-cycles and hence is even permutation
For [imath]S_6[/imath],the possible orders are 1,2,3,4,5,6. While for [imath]A_6[/imath], the possible orders are 1,2,3,4,5 but not 6. Why does [imath]A_6[/imath] neglects the 6-cycle in [imath]S_n[/imath] which also has 0 number of 2-cycles and hence an even permutation . I was going to upload the image of possible orders of [imath]S_6[/imath]. but error pops up that I can't upload it . | 319979 | How to write permutations as product of disjoint cycles and transpositions
[imath]\sigma=\begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 4&2&9&10&6&5&11&7&8&1&3 \end{pmatrix}[/imath] (1) I am asked to write this permutation in [imath]S_{11}[/imath] as a product of disjoint cycles and also as a product of transpositions. (2) Also find the order of the element. Is this permutation even or odd? i think these are the disjoint cycles [imath]E_{1}=(1,4,10)[/imath], [imath]\;\operatorname{order}E_1 =3[/imath] [imath]E_{2}= (3,9,8,7,11),\;[/imath] [imath]\operatorname{order}E_{2} =5[/imath] [imath]E_{3}=(5,6),\;[/imath] [imath]\operatorname{order}E_{3} =2[/imath] [imath]S_{11}[/imath]= [imath]E_{1} \cdot E_{2}\cdot E_{3}[/imath] [imath]\operatorname{order}E_{3}[/imath] is even so the order of the permutation is even. Why are they asking this? and what is the significance of it being even or odd? transpositions i have read this a couple times but the one example in my textbook is rather unclear, i am not sure what this means? i think the transposition for [imath]E_{1}[/imath] is [imath](1,4)(1,10)[/imath] but im not sure what this means. |
2507387 | Existence of a cofree functor (right adjoint)
Given is the forgetful functor: [imath]U: Grp \rightarrow Set[/imath]. I've already seen that we can construct a left adjoint [imath]F: Set \rightarrow Grp[/imath] to this functor by introducing the free functor, which associates to any set [imath]X \in Set[/imath] the free group [imath]F(X)[/imath]. Now I want to check if there is a right adjoint to our forgetful functor, let's say it is the functor [imath]R: Set \rightarrow Grp[/imath]. If we had a right adjoint, we'd have a bijection between hom-sets, namely [imath]Hom_{Grp}(R(X), G) \simeq Hom_{Set}(X, U(G))[/imath] for any set [imath]X \in Set[/imath] and [imath]G \in Grp[/imath]. I guess there is no such a right adjoint to the forgetful functor [imath]U[/imath] but I don't know how to show it. Can anybody help me, please? Thank you very much! | 294330 | Existence of a functor [imath]\mathsf{Sets} \to \mathsf{Groups}[/imath] that admits a left adjoint
The forgetful functor [imath]\mathsf{Groups} \to \mathsf{Sets}[/imath] admits a left adjoint, namely, "forming the free group" functor. I was wondering if this has a left adjoint. This seems unlikely, but I don't have a proof. Is there a functor [imath]\mathsf{Sets} \to \mathsf{Groups}[/imath] that admits a left adjoint at all? |
2507312 | Proving product rule identity using index notation
[imath]\nabla\cdot(\mathbf{u}\times\mathbf{v}) = (\nabla\times\mathbf{u})\cdot\mathbf{v} - (\nabla\times\mathbf{v})\cdot\mathbf{u}[/imath] Hi, the above is a vector equation, where u and v are vectors. I am trying to prove this identity using index notation. I am able to get the first term of the right-hand side, but I don't see where the second term with the minus in front comes from. Any help? Thanks! EDIT: I don't think this is a duplicate of that thread since the question in that thread has a scalar f and a vector v, whereas my problem has two vectors u and v, and no scalars. | 2117661 | Verify the following relationship: [imath]\nabla \cdot (a \times b) = b \cdot \nabla \times a - a \cdot \nabla \times b [/imath]
Verify the vector identity: [imath]\nabla \cdot (a \times b) = b \cdot \nabla \times a - a \cdot \nabla \times b [/imath] Given that: [imath]a = (R_a, S_a, T_a)[/imath], [imath]b = (R_b, S_b, T_b)[/imath] and [imath]\nabla = (\frac{\partial} {\partial x},\frac{\partial} {\partial y},\frac{\partial} {\partial z} )[/imath]. Where [imath]R_i, S_i, T_i[/imath] have continuous partial derivatives. I attempted it on paper but cannot get it to work. Can anyone show me how this is done? My working: LHS: [imath]\nabla \cdot (a \times b) = \frac{\partial}{\partial x}(S_a T_b - S_b T_a) + \frac{\partial}{\partial y}(T_a R_b - R_a T_b) + \frac{\partial}{\partial z}(R_a S_b - R_b S_a) [/imath] RHS: [imath]\nabla \times a = (\frac{\partial}{\partial y} T_a - \frac{\partial}{\partial z} S_a)\hat i - (\frac{\partial}{\partial x} T_a - \frac{\partial}{\partial z} R_a)\hat j + (\frac{\partial}{\partial x} S_a - \frac{\partial}{\partial y} R_a)\hat k[/imath] [imath]b \cdot (\nabla \times a) = R_b(\frac{\partial}{\partial y} T_a - \frac{\partial}{\partial z} S_a) - S_b(\frac{\partial}{\partial x} T_a - \frac{\partial}{\partial z} R_a) + T_b(\frac{\partial}{\partial x} S_a - \frac{\partial}{\partial y} R_a)[/imath] Now I computed the other part of the RHS as well however I can't see how I would be able to maninipulate it into the LHS. Anyone have any idea? |
2508338 | Show that these 5 lines are concurrent using vectors
Pentagon [imath]ABCDE[/imath] is inscribed in a circle. For any edge of [imath]ABCDE[/imath], we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent. I'm supposed to use vectors for this problem since I just finished a unit on vectors. I have no idea where to start. Can I get some starter tips? | 2437084 | Vector Applications in Euclidean Geometry
Pentagon [imath]ABCDE[/imath] is inscribed in a circle. For any edge of [imath]ABCDE[/imath], we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent. We just finished a unit on vectors in my Pre-Calculus class but I have no idea on how to do this final problem. Any help would be greatly appreciated! |
2508411 | Evaluate: [imath]\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}[/imath]
Evaluate: [imath]\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}[/imath] My Attempt: [imath]=\lim_{y\to x} \dfrac {y\sec y - x\sec x}{y-x}[/imath] [imath]=\lim_{y\to x} \dfrac {y\cos x-x\cos y}{\cos x\cos y \cdot (y-x)}[/imath] | 2413300 | Evaluate [imath]\lim_{y\to 0} \dfrac {(x+y)\sec (x+y) - x\sec x}{y}[/imath]
Evaluate [imath]\lim_{y\to 0} \dfrac {(x+y)\sec (x+y) - x\sec x}{y}[/imath]. My Attempt: [imath]=\lim_{y\to 0} \dfrac {(x+y) \sec (x+y) - x\sec x}{y}[/imath] [imath]=\lim_{y\to 0} \dfrac {x\sec (x+y) + y\sec (x+y) - x\sec x}{y}[/imath] [imath]=\lim_{y\to 0} \dfrac {x(\sec (x+y) - \sec x) + y\sec (x+y)}{y}[/imath] |
1787648 | Subspace of topological space has lower dimension
The dimension of a topological space [imath]X[/imath] is defined to be the length of the maximal chain of closed irreducible subsets [imath]\varnothing \neq X_0 \subsetneq X_1 \subsetneq ... \subsetneq X_n \subset X[/imath]. Let [imath]A[/imath] be a subset of topological space [imath]X[/imath], how can I show that dim[imath]A[/imath] [imath]\leq[/imath] dim[imath]X[/imath]? I can show that given a chain of closed irreducible subsets of [imath]A[/imath], [imath]A \cap X_0 \subsetneq ... \subsetneq A \cap X_n[/imath], where [imath]X_0,...,X_n[/imath] are closed subsets of X, we can get a chain of closed subsets [imath]X_0 \cap... \cap X_n \subsetneq ... \subsetneq X_{n-1} \cap X_n \subsetneq X_n[/imath] of [imath]X[/imath] of the same length, but I am having trouble showing the subsets are irreducible. Any help is appreciated! | 497192 | The dimension of a subspace is less than dimension of the whole space.
Let [imath]X[/imath] be topological space and [imath]Y\subset X[/imath]. The goal is to show that [imath]\dim Y\leq \dim X[/imath]. Here we use this definition of dimension. Let [imath]Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_{n}[/imath] be a chain of strictly increasing closed irreducible subsets of [imath]X[/imath] then [imath]\dim X[/imath] is the supremum over all such chains. This is what I've got so far. I'm just stuck on one last bit. A hint would be nice. Let [imath]Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_n[/imath] be a chain of irreducible closed subsets of [imath]Y[/imath]. Since each [imath]Y_i[/imath] is irreducible, we just need to close them in [imath]X[/imath]. There exists [imath]F_i[/imath] closed in [imath]X[/imath] such that [imath]Y_i=F_i\cap Y[/imath]. Then if we closed each set in [imath]X[/imath], [imath]\overline{F_0\cap Y}\subset\overline{F_1\cap Y}\subset \cdots \subset \overline{F_n\cap Y}[/imath] is a chain of irreducible closed subsets of [imath]X[/imath]. I just need to show that they are distinct. I really don't know where to proceed. A hint would be appreciated. Thanks. |
2507433 | Does [imath]x^2 - y^2 = 12345678[/imath] (x,y are integers) have any solutions?
This question appeared in KVPY online exam held on Nov 5. This equation is the equation of a hyperbola so any solution must be of the form [imath]a\sec(k),a\tan(k)[/imath]. So no solutions. | 2191177 | Prove that the equation [imath]x^2-y^2 = 2002[/imath] has no integer solution
Prove that the equation [imath]x^2 − y^2 = 2002[/imath] has no integer solution. If I had the equation [imath]x^3 - 7y = 3[/imath] I would easily conclude that for it to have any integer solution then [imath]x^2\equiv3\pmod 7[/imath] must be true. Applying the same logic here, I know that [imath]x² = y^2 +2002[/imath] so I can conclude that for it to have any integer solution then [imath]x^2\equiv2002\pmod y[/imath] must be true. Is this conclusion correct? From that congruence I can prove that the equation hasn't any integer solution without much trouble but I am not sure if that congruence is valid. |
2505732 | Gaussian and Mean Curvatures for a Ruled Surface
We are asked to prove the following theorem found in page 88 of Differential Geometry: Curves, Surfaces, Manifolds by Wolfgang Kühnel. Using standard parameters, calculate the Gaussian curvature and the mean curvature of a ruled surface as follows: [imath]K = -\dfrac {{\lambda}^2} {{{\lambda}^2 +v^2}^2}[/imath] and [imath]H = -\dfrac {1} {2({\lambda}^2 +v^2)^{3/2}} (Jv^2 + \lambda' v + \lambda (\lambda J + F))[/imath] In standard parameters, a ruled surface is [imath]f(u,v) = c(u) + v X(u)[/imath] and [imath]||X|| = ||X'|| = 1[/imath] and [imath]\langle c', X' \rangle = 0[/imath]. Thus, using standard parameters, a ruled surface is, up to Euclidean motions, uniquely determined by the following quantities: [imath]F = \langle c', X\rangle[/imath] [imath]\lambda = \langle c' \times X, X' \rangle = \det (c', X, X')[/imath] [imath]J = \langle X'', X \times X' \rangle = \det (X, X', X'')[/imath] Also, the first fundamental form is given as follows: [imath]I = \begin {pmatrix} \langle c',c' \rangle + v^2 & \langle c', X \rangle \\ \langle c', X \rangle & 1 \end {pmatrix} = \begin {pmatrix} F^2 + {\lambda}^2 + v^2 & F \\ F & 1 \end {pmatrix}[/imath] with [imath]\det (I) = \lambda^2 + v^2[/imath]. So far, I have that [imath]f_u (u,v) = c' + vX'[/imath] and [imath]f_v (u,v) = X[/imath] But I don't know how to proceed from there to get the first fundamental form, the normal vector, and the second fundamental form. | 2507529 | Gaussian and Mean Curvatures for a Ruled Surface
We are asked to prove the following theorem found in page 88 of Differential Geometry: Curves, Surfaces, Manifolds by Wolfgang Kühnel. Using standard parameters, calculate the Gaussian curvature and the mean curvature of a ruled surface as follows: [imath]K = -\dfrac {{\lambda}^2} {{{\lambda}^2 +v^2}^2}[/imath] and [imath]H = -\dfrac {1} {2({\lambda}^2 +v^2)^{3/2}} (Jv^2 + \lambda' v + \lambda (\lambda J + F))[/imath] In standard parameters, a ruled surface is [imath]f(u,v) = c(u) + v X(u)[/imath] and [imath]||X|| = ||X'|| = 1[/imath] and [imath]\langle c', X' \rangle = 0[/imath]. Thus, using standard parameters, a ruled surface is, up to Euclidean motions, uniquely determined by the following quantities: [imath]F = \langle c', X\rangle[/imath] [imath]\lambda = \langle c' \times X, X' \rangle = \det (c', X, X')[/imath] [imath]J = \langle X'', X \times X' \rangle = \det (X, X', X'')[/imath] Also, the first fundamental form is given as follows: [imath]I = \begin {pmatrix} \langle c',c' \rangle + v^2 & \langle c', X \rangle \\ \langle c', X \rangle & 1 \end {pmatrix} = \begin {pmatrix} F^2 + {\lambda}^2 + v^2 & F \\ F & 1 \end {pmatrix}[/imath] with [imath]\det (I) = \lambda^2 + v^2[/imath]. So far, I have that [imath]f_u (u,v) = c' + vX'[/imath] and [imath]f_v (u,v) = X[/imath] Also, [imath]f_{vv} (u,v) =0[/imath] I know the formula for the first fundamental form, the normal vector, and the second fundamental form. However, I don't know how to obtain the dot products and the cross products without having to isolate the components. |
2509054 | Prove that [imath]|G:A|=|G:B||B:A|[/imath] for [imath]A\subseteq B\subseteq G[/imath]
Given [imath]G[/imath] a group and [imath]A,B[/imath] subgroups of [imath]G[/imath] such that [imath]A\subseteq B\subseteq G[/imath], I want to prove [imath]|G:A|=|G:B||B:A|[/imath] where these are the indexes of the respective groups. I have been presented with a proof as follows but get stuck near to the end; Let [imath]Bx_i[/imath] and [imath]Ab_j[/imath] be the set of distinct right cosets in [imath]G[/imath], [imath]B[/imath] respectively for some [imath]i\in I, j\in J[/imath]. Then [imath]|G:B|=|I|[/imath] and [imath]|B:A|=|J|[/imath]. With this set up, the proof claims that [imath]Ab_jx_i[/imath] are all the distinct right cosets of [imath]A[/imath] in [imath]G[/imath]. I understand how the proof has shown that these are the distinct right cosets of [imath]A[/imath], but do not understand the deduction that these are distinct. This is the given argument; Suppose [imath]Ab_jx_i=Ab_kx_l[/imath]. As [imath]Ab_j,Ab_k \subseteq B[/imath], this implies [imath]Bx_i =Bx_l[/imath] and hence [imath]i=l[/imath]. A similar argument is used to show [imath]Ab_j=Ab_k[/imath] implies [imath]j=k[/imath]. Surely we can have two cosets being equal to each other with distinct elements representing each coset and so why would [imath]i=l[/imath] here? eg. if [imath]G=(\mathbb Q^*, \times)[/imath] and [imath]H=\{-1,1\}[/imath] then the right cosets [imath]H(1)=H(-1)=H[/imath]. | 2336415 | How to prove the cosets are distinct?
In Fraleigh, an exercise asks to prove the following theorem: Suppose [imath]H[/imath] and [imath]K[/imath] are subgroups of a group [imath]G[/imath] such that [imath]K \le H \le G[/imath], and suppose [imath](H:K)[/imath] and [imath](G:H)[/imath] are both finite. Then [imath](G:K)[/imath] is finite and [imath](G:K)=(G:H)(H:K)[/imath]. If we let [imath]\{a_iH\mid i=1,\dots,r\}[/imath] be the collection of distinct left cosets of [imath]H[/imath] in [imath]G[/imath] and [imath]\{b_jK\mid j=1,\dots,s\}[/imath] be the collection of distinct left cosets of [imath]K[/imath] in [imath]H[/imath], then the set of left cosets of [imath]K[/imath] in [imath]G[/imath] is [imath]\{(a_ib_j)K\mid i=1,\dots,r; j=1,\dots,s\}[/imath]. My question is how do we show that these are distinct? If we suppose [imath]a_ib_jK=a_pb_qK[/imath], then [imath]a_ib_jk_1=a_pb_qk_2[/imath] for some [imath]k_1,k_2[/imath] in [imath]K[/imath]. Then since [imath]b_jk_1,b_qk_2 \in H[/imath], it follows that [imath]a_i[/imath] and [imath]a_p[/imath] are in the same coset. Now, where do I go from here? Just because they are in the same coset does not mean that [imath]a_i=a_p[/imath], right? |
1401793 | How [imath]\mathbb{R}[/imath] is union of countable collection of disjoint component intervals?
Mathematical analysis by Tom Apostol Definition. let [imath]S[/imath] be an open subset of [imath]\Bbb R[/imath]. An open interval [imath]I[/imath] (which may be finite or infinite) is called component interval of [imath]S[/imath] if [imath]I \subset S[/imath] and if there is no open interval [imath]J \ne I[/imath] such that [imath]I \subseteq J[/imath] and [imath]J \subseteq S[/imath]. Now, I can't understand what it says. If I take [imath]S =(1,2)[/imath] , then is [imath](1,2)[/imath] its only component interval? Then how can [imath]\Bbb R[/imath] be a union of such open intervals? Examples will be very helpful. | 189793 | What's the component interval?
In Apostol, page [imath]51[/imath], he defines what he calls the component interval. I can't find any reference to it on the web. I have some problems with the definition: Let [imath]S \subseteq \mathbb{R}[/imath]. An open interval [imath]I[/imath] of [imath]S[/imath] is a component interval if there does not exist an open interval [imath]J[/imath] of [imath]S[/imath] such that [imath]I \subset J[/imath]. Intuitively, I get that [imath]I[/imath] is the largest possible open interval that is contained in [imath]S[/imath]. I think that the set of all rationals between the end points of [imath]S[/imath], [imath]\{\alpha \in (A,B)\ |\ \alpha \in \mathbb{Q}\}[/imath], is a component interval. Is that true? If [imath]D[/imath] is dense in [imath]S[/imath], is [imath]D[/imath] in general a component interval of [imath]S[/imath]? |
1531799 | Evaluate improper integral [imath]\int_0^\infty \frac{x\sin x}{x^2+1}dx[/imath]
How to prove that [imath]\int_0^\infty \frac{x\sin x}{x^2+1}dx=\frac{\pi}{2e}[/imath] I've tried several basic approaches like substitution and IBP but can't move forward. | 491439 | How would one evaluate [imath]\int\frac{x\sin(x)}{x^2+1}[/imath] over the real line?
Another exam problem I'm looking at is to evaluate the following integral. [imath] \int_{-\infty}^{\infty} \frac{x\hspace{-0.04 in}\cdot\hspace{-0.04 in}\sin(x)}{x^{\hspace{.02 in}2}+1} dx [/imath] This is a complex analysis exam, so the solution probably involves contours. [imath]\:[/imath] Since the integrand is an even function, one could potentially simplify by changing one endpoint to [imath]0[/imath]. However, I have no idea how to make a contour work since the absolute value of the integrand grows exponentially away from the real axis. What does one need to do to evaluate that integral? |
2509340 | Continuity of limit on a compact set implies uniform convergence
Suppose [imath]f_n[/imath] is a sequence of increasing functions defined on [imath][a,b] \to \mathbb{R}[/imath]. Say that [imath]f_n \to f[/imath] pointwise, such that [imath]f[/imath] is continuous. Does this imply that the convergence is also uniform? | 834126 | Sequence of monotone functions converging to a continuous limit, is the convergence uniform?
I'm reading some extreme value theory and in particular regular variation in Resnick's 1987 book Extreme Values, Regular Variation, and Point Processes, and several times he has claimed uniform convergence of a sequence of functions because "monotone functions are converging pointwise to a continuous limit". I am finding this reasoning a little dubious. EDIT: Thanks to some comments below I realized I was confused, I really want each [imath]f_n[/imath] to be a monotone function on [imath][a,b][/imath], so it isn't quite like Dini's theorem. Formally I am wondering: Let [imath]f_n:[a,b]\rightarrow\mathbb{R}[/imath] be a sequence of non-decreasing functions converging pointwise to the continuous function [imath]f[/imath]. Is the convergence uniform? My thoughts: I have been thinking that since the set of discontinuities of each function [imath]f_n[/imath] is at most countable, the set of discontinuities of the entire sequence is at most countable...something something? Any thoughts would be greatly appreciated! |
2509533 | Proving that the intersection of Sylow p-subgroups is normal?
Let [imath]\{K_1, K_2, ..., K_n\}[/imath] be the set of all Sylow [imath]p[/imath]-subgroups in a finite group [imath]G[/imath]. Prove that [imath] H = \cap_{i=1}^{n} K_i [/imath] is a normal subgroup in G. Proof attempt: We will proceed by using induction on [imath]n[/imath]. Suppose [imath]n = 1[/imath]. If there is only one Sylow [imath]p[/imath]-subgroup of G, that group must be normal. So [imath]H = K_1[/imath] is normal. Assume that the result is true for [imath]n \geq 1[/imath] Sylow [imath]p[/imath]-subgroups. Then we want to show it must be true for [imath]n+1[/imath] Sylow [imath]p[/imath]-subgroups. Case 1: There are [imath]n+1[/imath] unique prime factors of [imath]|G|[/imath]. Then each [imath]K_i[/imath] (with [imath]i = 1, ..., n+1[/imath]) has a unique prime (or prime-power) order. By uniqueness of prime factorization, then: [imath] H = \cap_{i=1}^{n+1} K_i = \{e\} [/imath] Clearly, this is a normal subgroup. Case 2: We do not have uniqueness of prime factors. This is where I get stuck. My idea is that this should fall out from the fact that all Sylow p-subgroups are conjugate with each other, but I'm not exactly sure how to proceed. Edit: I see the related question, but I'm hoping to receive feedback on my attempt at the solution, and whether this is a possible alternative way to prove the result. | 1041648 | Intersection of all [imath]p[/imath]-Sylow subgroups is normal
Let [imath]G[/imath] be a finite group, [imath]p[/imath] a prime number that divides [imath]|G|[/imath] and [imath]O_p(G)=\bigcap_{P \in Syl_p(G)}P[/imath]. Prove that 1) [imath]O_p(G) \lhd G[/imath] 2) [imath]O_p(G)[/imath] is maximal among the normal [imath]p[/imath]-subgroups of [imath]G[/imath]. 3) [imath]O_p(G/O_p(G))=1[/imath] I think I could do 2) using the following proposition which I will just enunciate: Proposition Let [imath]P[/imath] be a [imath]p[/imath]-Sylow and [imath]H \subset G[/imath] a [imath]p[/imath]-subgroup. Then there is [imath]x \in G[/imath] such that [imath]xHx^{-1} \subset P[/imath]. Solution to 2): Suppose [imath]O_p(G) \subset H[/imath] for some [imath]H \lhd G[/imath] p-subgroup. By the proposition we have that for each [imath]P[/imath] p-Sylow group, there is [imath]x: H=xHx^{-1} \subset P[/imath]. But then [imath]H \in \bigcap_{P \in Syl_p(G)} P=O_p(G)[/imath], so [imath]O_p(G)=H[/imath], it follows [imath]O_p(G)[/imath] is maximal. I am pretty lost in 1) and 3), I would appreciate suggestions or hints rather than complete answers. Thanks in advance. |
2509528 | Suppose [imath]f'[/imath] is continuous and positive on [imath][a, b][/imath]. Prove [imath]\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1} (y) dy = b f(b) - a f(a)[/imath]
Suppose [imath]f'[/imath] is continuous and positive on [imath][a, b][/imath]. Prove [imath]\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1} (y) dy = b f(b) - a f(a)[/imath] I'm attempting to apply a change of variables but unsure where to start. | 986800 | Prove that:[imath]\int_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) dx[/imath]
I just wanted to ask, if my proof is correct. I haven't seen the equation before, but I think it's quite useful. Let [imath]f[/imath] be an bijective differentiable function. Then the inverse function [imath]f^{-1}[/imath] exists and the following equation holds: [imath]\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx[/imath] Proof. [imath]f[/imath] is an bijective differentiable function and [imath]F[/imath] is an antiderivative of [imath]f[/imath]. First we need to find an antiderivative of [imath]f^{-1}[/imath]. [imath]\int f^{-1}(x) dx[/imath] with substitution [imath]x = f(y)[/imath] yields: [imath]\int y \cdot f'(y) dy = y \cdot f(y) - \int f(y) dy = y \cdot f(y) - F(y)[/imath] resubstitution yields: [imath]\int f^{-1}(x) dx = x \cdot f^{-1}(x) - F(f^{-1}(x))[/imath] hence [imath]\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = \left[x \cdot f^{-1}(x) - F(f^{-1}(x)) \right ]_{f(a)}^{f(b)}[/imath] [imath]=b \cdot f(b) - F(b) - (a \cdot f(a) - F(a)) = F(a) - F(b) + b \cdot f(b) - a \cdot f(a)[/imath] [imath]= \int\limits_{b}^{a} f(x) dx + b \cdot f(b) - a \cdot f(a) = -\int\limits_{a}^{b} f(x) dx + b \cdot f(b) - a \cdot f(a)[/imath] All in all: [imath]\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = -\int\limits_{a}^{b} f(x) dx + b \cdot f(b) - a \cdot f(a)[/imath] which is equal to [imath]\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx[/imath] q.e.d. |
2509540 | Prove that [imath] AB = BA [/imath] if and only if $ \det\begin{bmatrix} b & a - c \\ e & d - f \\ \end{bmatrix} = 0 $
Question: Let [imath]\mathbb{F}[/imath] a commutative field and [imath] a,b,c,d,e,f\in\mathbb{F} [/imath] be scalars and suppose that [imath] A [/imath] and [imath] B [/imath] are the following matrices: [imath] A = \begin{bmatrix} a & b \\ 0 & c \\ \end{bmatrix} [/imath] and [imath] B = \begin{bmatrix} d & e \\ 0 & f \\ \end{bmatrix} [/imath] Prove that [imath] AB = BA [/imath] if and only if [imath]$ \det\begin{pmatrix} \begin{bmatrix} b & a - c \\ e & d - f \\ \end{bmatrix} \end{pmatrix} = 0 $[/imath] My steps: [imath]$$ \det\begin{pmatrix} \begin{bmatrix} b & a - c \\ e & d - f \\ \end{bmatrix} \end{pmatrix} = 0 $$[/imath] [imath] b(d - f) - e(a - c) = 0 [/imath] [imath] bd - bf - ea + ec = 0 [/imath] [imath] bd - bf = ea - ec [/imath] At this point I considered that one of the following things must be true for the solution to hold: [imath] b = e = 0 [/imath] which would mean that [imath] A [/imath] and [imath] B [/imath] are diagonal matrices meaning [imath] AB = BA [/imath] Another condition for the solution to hold would be [imath] b = e, a = d, f = c [/imath] which would mean that [imath] A = B [/imath] hence [imath] AB = BA [/imath] must be true. Would this be the right approach to the problem or is there a point that I am missing? | 1261560 | Show that matrices [imath]A[/imath] and [imath]B[/imath] commute if and only if [imath]\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0.[/imath]
Show that the matrices [imath]A=\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \text{and B = $\begin{bmatrix} d & e \\ 0 & f \end{bmatrix}$}[/imath] commute if and only if [imath]\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0.[/imath] My approach: For [imath]AB=BA[/imath], we must have [imath]AB=\begin{bmatrix} ad & ae+bf \\\ 0 & cf\end{bmatrix}[/imath] and [imath]BA=\begin{bmatrix} ad & bd+ce \\\ 0 & cf\end{bmatrix}[/imath] where [imath]ae+bf=bd+ce[/imath]. We need to show that if [imath]ae+bf=bd+ce[/imath] then [imath]\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0.[/imath] Since [imath]\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0[/imath], we see that: [imath]b(d-f)-e(a-c)=0 \iff bd-bf-ae+ce=0 \iff bd+ce=ae+bf.[/imath] Thus, if [imath]AB=BA[/imath] then [imath]\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0.[/imath] (Showing the converse) If [imath]\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0[/imath] then we have [imath]bd+ce=ae+bf[/imath] and [imath]AB=BA[/imath] if and only if [imath]ae+bf=bd+ce[/imath], which we have. Is this a proper approach am I missing something here? |
2488760 | is the group cyclic, if Normal Subgroup and Quotient group are cyclic.
Prove or disprove that if [imath]H[/imath] is a normal subgroup of a group [imath]G[/imath] such that [imath]H[/imath] and [imath]\frac{G}{H}[/imath] are cyclic, then [imath]G[/imath] is cyclic. I am not sure if the above question has any mistake. I know examples where [imath]\frac{G}{H}[/imath] is cyclic, but [imath]G[/imath] is not cyclic. It is [imath]G = S_3 = \{ I, (12),(23),(31), (123), (132) \}[/imath] and [imath]H = \{I,(123),(132)\}[/imath] H is a normal subgroup of G. [imath]\frac{G}{H}[/imath] is cyclic since order of G/H is 2 and every group of prime order is cyclic. But G is not cyclic. I am doing graduation course. Please give me an example to disprove or proof for above problem. | 2509034 | Prove or disprove: If [imath]H[/imath] and [imath]G/H[/imath] are cyclic, then [imath]G[/imath] is cyclic.
I tried to prove it by proving that if [imath]\frac GH[/imath] is cyclic, there is [imath]gH[/imath] such that [imath](gH)^n=e_{(G/H)}=e_{(G)}*H[/imath]. Since [imath](gH)^n=g^nH[/imath], [imath]g^n=e_G[/imath]. So [imath]G[/imath] is cyclic. I don't think it's correct since I didn't use the condition that [imath]H[/imath] is cyclic but I couldn't figure what's wrong. Would be great if someone can point out what I'm doing wrong and maybe point me in the right direction. Thanks in advance. |
2509692 | [imath]\forall x \in H[/imath] [imath](x,Ax) = (x,Bx) \implies A=B[/imath]
Let [imath]H[/imath] be a complex Hilbert space, let [imath]A,B[/imath] be bounded linear operators on [imath]H[/imath]. I want to show [imath]\forall x \in H[/imath] [imath](x,Ax) = (x,Bx) \implies A=B[/imath]. I can show that [imath]\forall x,y \in H[/imath] [imath](y,Ax) = (y,Bx) \implies A=B[/imath], pretty easily. Is there a way to show that [imath]\forall x \in H[/imath] [imath](x,Ax) = (x,Bx)[/imath] [imath]\implies A=B[/imath] implies [imath]\forall x,y \in H[/imath] [imath](y,Ax) = (y,Bx) \implies A=B[/imath] to finish, or another method? | 1010676 | Proving that if [imath]\langle Ax,x\rangle =0[/imath] for every [imath]x[/imath], then [imath]A[/imath] is the zero operator
I feel kind of dumb but I needed this little claim as a part of a proof I'm writing, and I figured out that I'd better just ask, since I could not find the correct algebraic manipulation needed in order to prove it. So I want to prove this: Suppose [imath]H[/imath] is an Hilbert space, and [imath]A: H\to H[/imath] is a linear operator (if that matters, which I believe it does not, we can assume that [imath]A[/imath] is a projection). Suppose that for every [imath]x\in H[/imath], [imath](Ax,x)=0[/imath]. Then [imath]A=0[/imath]. What I tried to do is a little bit of Cauchy-Schwarz and some algebra but nothing worked out. Thanks! |
2509654 | Find a limit of a sum
Can anyone help me with the problem: [imath]\lim_{n \rightarrow \infty } \sum_{k=0}^{2n} \frac{k}{k + n^{2} } [/imath] In the hints it's adviced to make a Riemann sum from it and calculate the integral but i don't know how to do it | 2504491 | [imath] \lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)[/imath]
Evaluate: [imath] L=\lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)[/imath] My approach: Each term can be written as [imath] \frac k{n^2+k}=\frac {n^2+k-n^2}{n^2+k}=1-\frac {n^2}{n^2+k}[/imath] [imath] \therefore \lim_{n \to \infty}\frac k{n^2+k}=\lim_{n \to \infty}\left(1-\frac {n^2}{n^2+k}\right)=0[/imath] hence, [imath] L=0[/imath] Problem: The correct answer is 1/2, please indicate the flaw in my approach or post a new solution. Thank You |
2509620 | [imath]\lim \inf \frac{x_{n+1}}{x_n} \leq \lim \inf \sqrt[n]{x_n}[/imath] for strictly positive real seq [imath]\{x_n\}[/imath]
Let [imath]\{x_n\}[/imath] be a sequence of strictly positive real numbers. I want to show that [imath]\lim \inf \frac{x_{n+1}}{x_n} \leq \lim \inf \sqrt[n]{x_n} .[/imath] I've tried and failed to assume the contrary to derive a contradiction. any hints? | 1618217 | Prove that [imath]\lim \inf \dfrac{x_{n+1}}{x_n}\leq \lim \inf x_n^{1\over n} \leq \lim \sup x_n^{1\over n} \leq \lim \sup \dfrac{x_{n+1}}{x_n}. [/imath]
If [imath](x_n)[/imath] is any sequence of positive real numbers, then [imath]\lim \inf \dfrac{x_{n+1}}{x_n}\leq \lim \inf x_n^{1\over n} \leq \lim \sup x_n^{1\over n} \leq \lim \sup \dfrac{x_{n+1}}{x_n}. [/imath] I don't understand how am I suppose to use the hint nor what to do afterwards. Hint: If [imath]\lim \sup \dfrac{x_{n+1}}{x_n}-\infty[/imath], right hand inequality is obvious. So suppose [imath]\lim \sup \dfrac{x_{n+1}}{x_n}=M[/imath]. Then [imath]\dfrac{x_{n+1}}{x_n}< M+\epsilon[/imath] [imath]\forall n\geq N[/imath], i.e., [imath]x_{n+k}\leq (M+\epsilon)^kx_n[/imath] for all [imath]k\geq 0[/imath]. |
2507702 | Proving [imath]4^n=\sum_{k=0}^n2^k\binom{2n-k}{n}[/imath]
[imath]4^n = \sum\limits_{k=0}^{n}2^k\cdot{{2n - k} \choose n}[/imath] I tried formal power series, but failed. | 1782432 | How to prove that [imath]\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n[/imath].
So I've been struggling with this sum for some time and I just can't figure it out. I tried proving by induction that if the sum above is a [imath]S_n[/imath] then [imath]S_{n+1} = 4S_n[/imath], but I didn't really succeed so here I am. Thanks in advance. |
2510008 | Does every integer greater than two equal the number of conjugacy classes of some finite non-abelian group?
Let [imath]k > 2[/imath] be an integer. Is there a finite non-abelian group with exactly [imath]k[/imath] conjugacy classes? The only group with [imath]1[/imath] conjugacy class is the trivial group, and the only group with exactly [imath]2[/imath] conjugacy classes is the group of order [imath]2[/imath], so the restriction [imath]k>2[/imath] is necessary. (The symmetric group [imath]S_3[/imath] has [imath]3[/imath] conjugacy classes, and is the only such non-abelian group.) For small values of [imath]k[/imath], the number of non-abelian groups with exactly [imath]k[/imath] conjugacy classes are known. I had originally come to this question for square-free values of [imath]k[/imath], but upon thinking about it, I realised that there was an even more basic form of the question, whose answer I did not know and could not find, as stated above. I thought that it might be possible to construct a group of the form [imath]G = A\rtimes B[/imath], with [imath]A[/imath] abelian of order [imath]k[/imath], and [imath]B\leq\operatorname{Aut}(A)[/imath] so that [imath]G[/imath] has exactly [imath]k[/imath] conjugacy classes, but I was unable to come up with a general argument. It does not always work for cyclic [imath]A\cong C_k[/imath]. | 203268 | Conjugacy Classes in Finite Groups
Given an integer [imath]n\geq 1[/imath], it is known that there are only finitely many finite groups with exactly [imath]n[/imath] conjugacy classes. Question: For [imath]n\geq 3[/imath], does there exists a finite non-abelian group, with exactly [imath]n[/imath] conjugacy classes? |
2510010 | Why does the Monotone Convergence Theorem require increasing sequence.
The monotone convergence theorem in my text requires [imath]f_n[/imath] to be a non-negative, increasing, pointwise convergence sequence. Why does [imath]f_n[/imath] have to be a decreasing sequence? Is there an example of when the sequence is decreasing it the theorem is not true? | 252452 | Counter example for [imath]F_n \downarrow F[/imath] then [imath]\int f_n d\mu \downarrow \int f d\mu[/imath]?
Is there any counter example for for [imath]F_n \downarrow F[/imath] then [imath]\int F_n \, d\mu \downarrow \int F \, d\mu[/imath]? I came up with the one below but [imath]F_n[/imath] does not go down to [imath]0[/imath] monotonically. I need something that goes monotonically. [imath]F_n = \frac1n \cdot 1_{[0,n]}(x)[/imath] we know that [imath]\int F_n \, d\mu = 1 \text{ and not [/imath]\int0 \, d\mu[imath]}[/imath] |
2510774 | Why are the group axioms called "axioms"?
My book says: A group is an ordered pair [imath](G, \cdot)[/imath] where [imath]G[/imath] is a set and [imath]\cdot[/imath] is a binary operation satisfying the following axioms [emphasis mine]: [imath]\vdots[/imath] Why doesn't it say A group is an ordered pair [imath](G, \cdot)[/imath] where [imath]G[/imath] is a set and [imath]\cdot[/imath] is a binary operation satisfying the following propreties: [imath]\vdots[/imath] As far as I know, axioms are statements which we take for granted; I don't understand why the group axioms are something we have to take for granted; we already know that there exist objects (such as the set of permutations of [imath]3[/imath] objects) which satisfy those propreties. Even if we knew of no objects which satisfy those propreties, then perhaps the axiom should be "There exists an object which satisfies those propreties" rather the axioms being the propreties themselves. | 1801970 | Meaning of the word "axiom"
One usually describes an axiom to be a proposition regarded as self-evidently true without proof. Thus, axioms are propositions we assume to be true and we use them in an axiomatic theory as premises to infer conclusions, which are called "theorems" of this theory. For example, we can use the Peano axioms to prove theorems of arithmetic. This is one meaning of the word "axiom". But I recognized that the word "axiom" is also used in quite different contexts. For example, a group is defined to be an algebraic structure consisting of a set [imath]G[/imath], an operation [imath]G\times G\to G: (a, b)\mapsto ab[/imath], an element [imath]1\in G[/imath] and a mapping [imath]G\to G: a\mapsto a^{-1}[/imath] such that the following conditions, the so-called group axioms, are satisfied: [imath]\forall a, b, c\in G.\ (ab)c=a(bc)[/imath], [imath]\forall a\in G.\ 1a=a=a1[/imath] and [imath]\forall a\in G.\ aa^{-1}=1=a^{-1}a[/imath]. Why are these conditions (that an algebraic structure has to satisfy to be called a group) called axioms? What have these conditions to do with the word "axiom" in the sense specified above? I am really asking about this modern use of the word "axiom" in mathematical jargon. It would be very interesting to see how the modern use of the word "axiom" historically developed from the original meaning. Now, let me give more details why it appears to me that the word is being used in two different meanings: As peter.petrov did, one can argue that group theory is about the conclusions one can draw from the group axioms just as arithmetic is about the conclusions one can draw from the Peano axioms. But in my opinion there is a big difference: while arithmetic is really about natural numbers, the successor operation, addition, multiplication and the "less than" relation, group theory is not just about group elements, the group operation, the identity element and the inverse function. Group theory is rather about models of the group axioms. Thus: The axioms of group theory are not the group axioms, the axioms of group theory are the axioms of set theory. Theorems of arithmetic can be formalized as sentences over the signature (a. k. a. language) [imath]\{0, s, +, \cdot\}[/imath], while theorems of group theory cannot always be formalized as sentences over the signature [imath]\{\cdot, 1, ^{-1}\}[/imath]. Let me give an example: A typical theorem of arithmetic is the case [imath]n=4[/imath] of Fermat's last theorem. It can be formalized as follows over the signature [imath]\{0, s, +, \cdot\}[/imath]: [imath]\neg\exists x\exists y\exists z(x\not = 0\land y\not = 0\land z\not = 0\quad\land\quad x\cdot x\cdot x\cdot x + y\cdot y\cdot y\cdot y = z\cdot z\cdot z\cdot z).[/imath] A typical theorem of group theory is Lagrange's theorem which states that for any finite group G, the order of every subgroup H of G divides the order of G. I think that one cannot formalize this theorem as a sentence over the "group theoretic" signature [imath]\{\cdot, 1, ^{-1}\}[/imath]; or can one? |
267518 | Is there a harmonic function in the whole plane that is postive everywhere?
This is one of the past qualifying exam problems that I was working on. I know that, when we let [imath]z=x+iy[/imath], [imath]{|z|}^2=x^2+y^2[/imath] is not harmonic. I do not know where to start to prove that there is no harmonic function that is positive everywhere. Any help or ideas idea will be really appreciated. Thank you in advance. | 2431880 | Let [imath]U: \mathbb C \to \mathbb R[/imath] be a harmonic function such that [imath]U(z)\ge 0 \ \forall z\in \mathbb C[/imath] ; then [imath]U[/imath] is constant.
Let [imath]U: \mathbb C \to \mathbb R[/imath] be a harmonic function such that [imath]U(z)\ge 0 \ \forall z\in \mathbb C[/imath] ; then [imath]U[/imath] is constant. Let [imath]u:\mathbb{C}\rightarrow\mathbb{R}[/imath] be a harmonic function such that [imath]0\leq f(z)[/imath] for all [imath] z \in \mathbb{C}[/imath]. Prove that [imath]u[/imath] is constant. I don't know about simply connected domain. Can we do it without using that? Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the substantial things covered are Lioville's theorem, Maximum modulus theorem. |
1187699 | The evaluation of [imath]\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{1/x^2} = e^{1/3}[/imath]
Wolfram Alpha says: [imath]\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{1/x^2} = e^{1/3}[/imath] What are the steps leading to this result? Only with the Hospital rule, I have as a result infinity. | 1770823 | How to Find [imath] \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}[/imath].
Can someone help me with this limit? I'm working on it for hours and cant figure it out. [imath] \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}[/imath] I started transforming to the form [imath] \lim_{x\to 0} e^{ {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}} }[/imath] and applied the l'Hopital rule (since indeterminated [imath]\frac00[/imath]), getting: [imath] \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)[/imath] From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me? |
1131069 | Does [imath]\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0[/imath]?
[imath]\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0[/imath] My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =): (Sorry for the small font, but you can zoom in to see better with Firefox) [imath] \begin{align} & \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\ & \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\ & e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\ & e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\ & e^{-\infty} = 0\\ \end{align} [/imath] Edit #1: Continuing after the mistake of the [imath]\tan(x)[/imath]: [imath]\begin{align} & e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\ & e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\ & e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\ \end{align} [/imath] | 447312 | Evaluating: [imath]\lim\limits_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}[/imath]
I am trying to evaluate the following but without result. [imath]\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}[/imath] Can you please give me some hints? I have tried to put log to both sides but it hasn't lead me somewhere... Thanks a lot |
2510119 | The union of [imath]x[/imath] and [imath]y[/imath] axis with subspace topology is not locally euclidean
The union of [imath]x [/imath] axis and the [imath]y[/imath] axis it's not manifold . It's a hausdorff second countable, but it's not locally euclidean. I was tryin to prove it by contradiction, but I can't , the issue sopposed to be with the (0,0). | 2467251 | How to prove a cross is not locally homeomorphic to an euclidean space?
Consider the subset [imath]X = \{(x, 0) \mid -1 \lt x \lt 1\} \cup \{(0, y) \mid -1 \lt y \lt 1\} \subset \Bbb R^2[/imath]. It's just two open segments intersecting at [imath](0, 0)[/imath]. How do you show that [imath]X[/imath], equipped with the subspace topology of [imath](\Bbb R^2, \lvert \lvert \cdot \rvert \rvert_2)[/imath], is not locally homeomorphic to an euclidean space because of the point [imath](0, 0)[/imath] ? |
535704 | [imath]f(z)=f(iz)\forall z[/imath], my question is : is there such [imath]f[/imath] exists?
[imath]f[/imath] be an entire function such that [imath]|f(z)|\le c|z|^3\forall |z|\ge 3,f(z)=f(iz)\forall z[/imath], my question is : is there such [imath]f[/imath] exists? [imath]f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+a_5z^5+\dots=a_0+a_1iz-a_2z^2-ia_3z^3+a_4z^4+ia_5z^5\dots[/imath] [imath]|g(z)|=|{f(z)\over z^3}|\le c\forall |z|\ge 3[/imath] so by Liuvilles [imath]f(z)=kz^3\forall |z|\ge 3[/imath] also [imath]kz^3=-ki z^3[/imath] so [imath]k=0[/imath], so [imath]f(z)=0\forall |z|\ge 3[/imath]? so [imath]f\equiv 0[/imath]? am I wrong anywhere? | 403031 | Entire function [imath]f[/imath] such that [imath]|f(z)| ≤ K |z|^3[/imath] for [imath]|z|\ge1[/imath] and [imath]f(z) = f(iz)[/imath] for every [imath]z∈C[/imath]
Let [imath]f(z)[/imath] be an entire function such that for some constant [imath]K[/imath] , [imath]|f(z)| ≤ K[/imath] [imath]|z|^3<[/imath] for [imath]|z|\ge1[/imath] and [imath]f(z) = f(iz)[/imath] [imath]∀z∈C[/imath] , then which of the following are correct ? (A)[imath] f(z) = Kz^3[/imath] [imath]∀ z∈\Bbb{C}[/imath] (B) [imath]f(z)[/imath] is a constant function (C) [imath]f(z)[/imath] is quadratic function (D) No such [imath]f[/imath] exists My attempt : I can deduce that since [imath]f(z) = f(iz)[/imath] and since [imath]i^4=1 \implies f[/imath] must involve only fourth powers of [imath]z[/imath] . Hence [imath]f(z)=a_1z^4+b_1z^8+c_1z^{16}+ \cdots [/imath]-------(1) Also for [imath]|z| \ge 1 [/imath]: let C. Then [imath]|g(z)| = |f(z)|/|z^3| \le k[/imath] => By the Cauchy estimate theorem we can prove that : [imath]f(z)/z^3[/imath] is a constant function [imath]= c [/imath] [imath]\implies f(z)=cz^3[/imath] for [imath]|z| > 1[/imath] ...........(2) From (1) and (2) ; we get there does not exist a function like this ? My textbook answer says it's a constant function ? |
1939713 | Finding [imath]\int_0^{\infty}xe^{-\lambda_1x} \, dx[/imath] with integration by parts
I'm returning to graduate school 20+ years after taking calculus. As part of a homework assignment in my statistics class, I need to find: [imath]\int_0^\infty xe^{-\lambda_1x}\; dx[/imath] Finding this integral is one step in finding the expected value of a given continuous distribution function, whose solution is the answer to the homework question. Now, I've had a gentle nudge by the instructor that I should use integration by parts to solve this, but even that and some looking around online hasn't helped me. I found: [imath]\int f(x) g'(x) \; dx = f(x) g(x) - \int f'(x) g(x) \;dx[/imath] or [imath]\int u \,dv = uv - \int v\, du[/imath] Choosing [imath]u, dv[/imath] and solving for [imath]du, v[/imath], I get: [imath]u=x, du = dx, dv=e^{-\lambda_1x} \, dx, v=\int e^{-\lambda_1x}=-\frac{1}{\lambda_1} e^{-\lambda_1x}[/imath] I am not at all certain that I got [imath]v[/imath] correct, but using that, I get: [imath]\int xe^{-\lambda_1x} \, dx =\frac{-1}{\lambda_1}e^{-\lambda_1x}[/imath] But when I plug that back into the original formula to answer the original question, I get an answer that doesn't make sense. Can you help me find my mistake? | 508529 | Finding [imath]\int_0^{\infty}xe^{-\lambda x} \, dx[/imath]
How to integrate: [imath]\int_0^\infty x \, \lambda e^{-\lambda x} \, dx[/imath] I tried using integration by parts: Let [imath]u = x[/imath] and [imath]dv = \lambda e^{-\lambda x} \, dx[/imath] Then [imath]du = dx[/imath] And [imath]v = - \lambda e ^{-\lambda x}[/imath] Correct so far? Then [imath]\begin{aligned} uv - \int v \, du &= -\lambda x e^{-\lambda x} - \int (- \lambda e^{- \lambda x}) dx \\ &= -\lambda x e^{-\lambda x} - \lambda e^{-\lambda x} \end{aligned}[/imath] The correct answer from lecture notes UPDATE 2 Let [imath]u=\lambda x[/imath] and [imath]dv = e^{-\lambda x} dx[/imath] Then [imath]du = \lambda \, dx[/imath] [imath]dv = e^{-\lambda x} \, dx[/imath] Let [imath]y = -\lambda x[/imath] Then [imath]dy = -\lambda \, dx[/imath] So [imath]dx = -\frac{1}{\lambda} dy[/imath] [imath]\begin{aligned} v &= \int e^u \cdot - \frac{1}{\lambda} du \\ &= - \frac{1}{\lambda} e^{-\lambda x} \end{aligned}[/imath] But here I seem to have an extra [imath]- \frac{1}{\lambda}[/imath] in [imath]v[/imath]? If I continue using integration by parts, I get: [imath]-x e^{-\lambda x} + \color{red}{\frac{1}{\lambda}} \int e^{-\lambda x} \, dx[/imath] |
1161278 | Show that if [imath]1> x>0[/imath], then $x-1 ≥ \ln(x) ≥ 1−\frac{1}{x}$
Show that if [imath]$0 < x < 1$[/imath], then [imath]$$x-1 ≥ \ln(x) ≥ 1−\frac{1}{x}.$$[/imath] I know how to prove it using the MVT and I can prove it for [imath]x> 1[/imath] but I don't understand how to prove it for [imath]x > 0[/imath] . | 324345 | Intuition behind logarithm inequality: [imath]1 - \frac1x \leq \log x \leq x-1[/imath]
One of fundamental inequalities on logarithm is: [imath] 1 - \frac1x \leq \log x \leq x-1 \quad\text{for all $x > 0$},[/imath] which you may prefer write in the form of [imath] \frac{x}{1+x} \leq \log{(1+x)} \leq x \quad\text{for all $x > -1$}.[/imath] The upper bound is very intuitive -- it's easy to derive from Taylor series as follows: [imath] \log(1+x) = \sum_{i=1}^\infty (-1)^{n+1}\frac{x^n}{n} \leq (-1)^{1+1}\frac{x^1}{1} = x.[/imath] My question is: "what is the intuition behind the lower bound?" I know how to prove the lower bound of [imath]\log (1+x)[/imath] (maybe by checking the derivative of the function [imath]f(x) = \frac{x}{1+x}-\log(1+x)[/imath] and showing it's decreasing) but I'm curious how one can obtain this kind of lower bound. My ultimate goal is to come up with a new lower bound on some logarithm-related function, and I'd like to apply the intuition behind the standard logarithm lower-bound to my setting. |
2511620 | If a group [imath]G[/imath] is infinite, prove that it has a non-trivial subgroup
I am working on some elementary group theory problems and one of them states: "Show that an infinite group [imath]G[/imath] has to contain a non-trivial subgroup, i.e. a subgroup [imath]\neq G[/imath], {[imath]e[/imath]}." My process:[imath]G:\{a,b,...,e,a^{-1},b^{-1}...\}[/imath] So a possible subgroup could be [imath]G_{s}: \{a,a^{-1},e\}[/imath] Since [imath]a*a^{-1}=e[/imath] [imath]a*e=a[/imath] [imath]a^{-1}*e=a^{-1}[/imath] are all contained within the subgroup. Is this approach valid? I feel like I'm missing something... EDIT: Totally forgot about [imath]a*a[/imath], could anyone give me any hints on how to approach this problem? | 1608280 | Let [imath]G[/imath] be a non-trivial group with no non-trivial proper subgroup. Prove that [imath]G[/imath] cannot be infinite group.
Let [imath]G[/imath] be a non-trivial group with no non-trivial proper subgroup. Prove that [imath]G[/imath] cannot be infinite group. A hints is given as order of G is not infinite since [imath]a[/imath] and [imath]a^{-1}[/imath] are only generators. I know that infinite cyclic group has only two generators. But don't know how to prove the problem. |
2511760 | Proof of [imath]f(x)=\sin x +x[/imath] is Surjective
Without plotting graph prove that [imath]f: \mathbb{R}\to \mathbb{R}[/imath] [imath]f(x)=\sin x+x[/imath] is an Onto function Let [imath]y_0[/imath] be any number in codomain Then we require atleast one [imath]x_0[/imath] in Domain such that [imath]y_0=\sin x_0 +x_0[/imath] Now since [imath]\sin x_0[/imath] is bounded in [imath]\left[-1 \:\: 1\right][/imath] we have [imath]x_0=y_0-\sin x_0[/imath] hence there is an [imath]x_0[/imath] in Domain. So [imath]f[/imath] is surjective. But i feel something missing in this proof? Any other way to prove? | 1576588 | Prove without using graphing calculators that [imath]f: \mathbb R\to \mathbb R,\,f(x)=x+\sin x[/imath] is both one-to-one, onto (bijective) function.
Prove that the function [imath]f:\mathbb R\to \mathbb R[/imath] defined by [imath]f(x)=x+\sin x[/imath] for [imath]x\in \mathbb R[/imath] is a bijective function. The codomain of the [imath]f(x)=x+\sin x[/imath] is [imath]\mathbb R[/imath] and the range is also [imath]\mathbb R[/imath]. So this function is an onto function. But I am confused in proving this function is one-to-one. I know about its graph and I know that if a function passes the horizontal line test (i.e horizontal lines should not cut the function at more than one point), then it is a one-to-one function. The graph of this function looks like the graph of [imath]y=x[/imath] with sinusoids going along the [imath]y=x[/imath] line. If I use a graphing calculator at hand, then I can tell that it is a one-to-one function and [imath]f(x)=\frac{x}{2}+\sin x[/imath] or [imath]\frac{x}{3}+\sin x[/imath] functions are not, but in the examination I need to prove this function is one-to-one theoritically, without graphing calculators. I tried the method which we generally use to prove a function is one-to-one but no success. Let [imath]f(x_1)=f(x_2)[/imath] and we have to prove that [imath]x_1=x_2[/imath] in order fot the function to be one-to-one. Let [imath]x_1+\sin x_1=x_2+\sin x_2[/imath] But I am stuck here and could not proceed further. |
2511744 | How many ordered quadruples [imath](a,b,c,d)[/imath] of nonnegative integers are there such that [imath]abcd = 288[/imath]?
I know how to do [imath]a+b+c+d=288[/imath], but I don't have any clue how to get started if it is multiplication. | 2506553 | How many ordered quadruples [imath](w, x, y, z)[/imath] of non-negative integers are there such that [imath]wxyz = 288[/imath]? Why?
I found the prime factorization [imath]288 = 2^5 \cdot 3^2[/imath] and then I tried to set up some algebraic equations, but got stuck. How would we proceed so that we get the answer? Especially when we are only looking at non-negative integers. Any help? |
2510439 | Existence of solutions of Diophantine equations
One of the problems in the book, I am quite confused to show this in a general way. How could I show that [imath]x^2+y^2 = 3(u^2+v^2)[/imath] has no nontrivial integral solutions? It seems like these problems have a general way to show there are solutions or no solutions to them, and I can not do this yet as I just started to learn about Diophantine equations. Helping me understand the way to solve these problems would be greatly appreciated. | 987396 | Prove that [imath]x^2 + y^2 = 3(z^2 + m^2)[/imath] has no solutions in integer
Prove that: [imath] x^2 + y^2 = 3(z^2 + m^2) [/imath] has no solutions in integer Except [imath]0 0 0 0[/imath] |
2511886 | How many ways are there to distribute 5 balls of different colours among three persons with additional conditions?
Problem: The total number of ways in which 5 balls of different colour can be distributed among three persons so that each person gets at least one ball? Source: A book on combinatorics. Was asked in an entrance exam. My try: To simplify the problem, we can distribute [imath]3[/imath] of the [imath]5[/imath] balls beforehand so we can get rid of the constraint. Let us select 3 balls and permute them within these three persons, so the number of ways of doing so [imath] N_1 = C(5,3) + P(3,3) = 60[/imath] Now that we have done so, we have two balls left. Let them be [imath]B_1[/imath] and [imath]B_2[/imath] These can be distributed as below: Let both the remaining balls be with one person, so we have [imath](B_1 B_2, 0, 0)[/imath] and there are [imath]3[/imath] ways of doing so. Let the remaining balls be distributed one by one, so we have [imath](B_1, B_2, 0)[/imath] and there are [imath]3! = 6[/imath] ways of doing so. We can take the total number ways ([imath]N[/imath]) by: [imath]N = N_1 \times 3 + N_1 \times 6[/imath] [imath]N = 60 \times 3 + 60 \times 6[/imath] [imath]N = 540[/imath] This approach seems correct to me but still it gets the wrong answer... The right answer is [imath]150[/imath] ways Please guide me and correct me if I'm wrong. All help appreciated! | 2452609 | In how many ways can [imath]5[/imath] balls of different colours be placed in [imath]3[/imath] boxes of different sizes if no box remains empty?
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty. My attempt:- First choose 3 balls to be placed in 3 boxes so that none of them remain empty in [imath]{{5}\choose{3}}\cdot3! = 60[/imath] ways. Now remaining 2 balls can go into any of the 3 boxes in [imath]3\cdot3 = 9[/imath] ways. Total number of ways [imath]= 60\cdot9 = 540[/imath]. Where am I going wrong ? |
2512101 | [imath]|z+w|=|z|+|w|[/imath] iff [imath]z=cw[/imath]
I'm trying to prove that if [imath]z,w[/imath] are non non-zero complex numbers then [imath]|z+w|=|z|+|w|[/imath] iff there exists a positive real [imath]c[/imath] st [imath]z=cw[/imath]. Now I've proved the leftward implication but I'm having difficulties with the rightward one, so I'd appreciate any hint. NOTE: I can't use the exponential or polar representation of complex numbers nor the notion of scalar product but only the basic algebraic properties of complex numbers. | 116319 | If [imath]z,w\in \mathbb{C}^{\times}[/imath], then [imath]|z+w|=|z|+|w|[/imath] iff [imath]w=tz[/imath] for some [imath]t\gt 0[/imath]
This is Exercise EP [imath]8[/imath] from Fernandez and Bernardes's book Introdução às Funções de uma Variável Complexa (in Portuguese). If [imath]z,w\in \mathbb{C}^{\times}[/imath], then [imath]|z+w|=|z|+|w|[/imath] iff [imath]w=tz[/imath] for some [imath]t\gt 0.[/imath] I supposed that [imath]|z+w|=|z|+|w|[/imath] and by squaring it I got [imath]\operatorname{Re}(z\bar{w})=|z||w|[/imath] and I am stuck. I would appreciate a hint in this question. |
2512182 | Any easier way to do [imath]\int^\infty_0 e^{-(x^2+\frac{\alpha^2}{x^2})\beta^2} = \frac{\sqrt{\pi}}{2\beta}e^{-2\alpha\beta^2}[/imath]
I have a textbook proof which i am finding complex - [imath]\int^\infty_0 e^{-(x^2+\frac{\alpha^2}{x^2})\beta^2} = \frac{\sqrt{\pi}}{2\beta}e^{-2\alpha\beta^2}[/imath] Any simpler way to do this? | 1909074 | Prove [imath]\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{e}^{-2\sqrt{ab}}[/imath]
This is Example 8.4.1 from Chapter 8 (The Normal Integral) of Irresistible Integrals by Boros and Moll. The authors outlined a solution method, which I will provide in full here. My question: is there another way to obtain this result? Here is the solution of Boros and Moll with their notation. Let \begin{equation} \mathrm{L}(a,b) := \int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x \label{eq:iic8-1} \tag{1} \end{equation} Making the substitution [imath]t=x\sqrt{a}[/imath] yields \begin{equation} \int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{ab}{t^{2}}) \mathrm{d} t \label{eq:iic8-2} \tag{2} \end{equation} Letting [imath]ab=c[/imath] we call the integral in equation \eqref{eq:iic8-2} [imath]f(c)[/imath], \begin{equation} f(c) = \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \mathrm{d} t \label{eq:iic8-3} \tag{3} \end{equation} so that \begin{equation} \mathrm{L}(a,b) = \frac{f(ab)}{\sqrt{a}} \label{eq:iic8-4} \tag{4} \end{equation} In equation \eqref{eq:iic8-3} we let [imath]y=\sqrt{c}/t[/imath] \begin{equation} f(c) = \sqrt{c} \int\limits_{0}^{\infty} \mathrm{exp}(-y^{2}-\frac{c}{y^{2}}) y^{-2} \mathrm{d} y \label{eq:iic8-5} \tag{5} \end{equation} Combining equations \eqref{eq:iic8-3} and \eqref{eq:iic8-5}, we have \begin{equation} f(c) = \frac{1}{2} \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \left(1+\frac{\sqrt{c}}{t^{2}} \right) \mathrm{d} t \label{eq:iic8-6} \tag{6} \end{equation} Now we let [imath]s = t - \sqrt{c}/t[/imath] \begin{equation} f(c) = \frac{\mathrm{e}^{- 2\sqrt{c}}}{2} \int\limits_{-\infty}^{\infty} \mathrm{exp}(-s^{2}) \mathrm{d} s = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \lim_{z \to \infty} \mathrm{erf}(z) = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \label{eq:iic8-7} \tag{7} \end{equation} Combining equations \eqref{eq:iic8-1}, \eqref{eq:iic8-4}, and \eqref{eq:iic8-7} yields our result. |
2511914 | How many different/unique 4-letter arrangements are there of the letters in the word MISSISSAUGA
So far I've broken it off to:- [imath]M=1[/imath] [imath]I=2[/imath] [imath]S=4[/imath] [imath]A=2[/imath] [imath]U=1[/imath] [imath]G=1[/imath] However, I don't really know what to do next. An answer (with steps please) would be greatly appreciated. This seems to be a really important question :) | 2509722 | How many different/unique [imath]4[/imath]-letter arrangements are there of the letters in the word Mississauga?
How many different/unique [imath]4[/imath]-letter arrangements are there of the letters in the word Mississauga? I'm thinking we need permutations? I already tried finding each letter and still don't get it... any help? So far I've broken it off to:- [imath]M=1[/imath] [imath]I=2[/imath] [imath]S=4[/imath] [imath]A=2[/imath] [imath]U=1[/imath] [imath]G=1[/imath] However, I don't really know what to do next. |
2512036 | Number of solutions to [imath]x^n=e[/imath] in group G is divisible by n, where n divides o(G)
It is straightforward if G is abelian, but appears to be difficult to prove for non-abelian groups. Since, it has been asked in supplementary exercise of Herstein's group theory, I guess some elementary group theoretic proof exists. Can anyone please give me hint. I will be grateful. | 2069016 | Number of solutions to [imath]x^n=e[/imath] in group [imath]G[/imath] is divisible by [imath]n[/imath]
My question is, can we prove that the number of solutions to [imath]x^n=e[/imath] in group [imath]G[/imath] is divisible by [imath]n[/imath], for [imath]n | |G|[/imath] Apparently the answer is it is indeed true, infact for the equation [imath]x^n = a[/imath] for any [imath]a \in G [/imath] also, however I have had many ideas each of which having a small flaw with them. For example I attempted induction on order of [imath]G[/imath] as follows: Assume true for all groups of order [imath] \leq k[/imath]. Consider [imath]|G| = k+1[/imath]. If [imath]G[/imath] is cyclic, we show fairly easily the result holds. (In fact if I recall correctly there are [imath]n[/imath] solutions for each divisor [imath]n[/imath]). If [imath]G[/imath] is not cyclic, we can look at the subgroups of [imath]G[/imath]. It'd be nice to say the number of solutions in each group is divisible by [imath]n[/imath] by the induction hypothesis and get the desired answer by inclusion-exlusion for solutions in all the subgroups, however we could potentially have a subgroup of order less than [imath]n[/imath] and still contain a solution. So if we were able to show it works for all the subgroups of orders [imath] < n[/imath] collectively we would be done. I have no idea whether this can work though. I did see this question The number of solutions of [imath]x^n = e[/imath] in a finite group is a multiple of n, whenever n divides the group order. however I couldn't find much help on it. From the hint given I managed to show [imath]no. solutions = \sum_{ d|n} \lambda_{i} \phi(d)[/imath] for some [imath]\lambda_{i}[/imath]s, however they could be different so I can't just use the well know summing of euler-phi formula. Any solution/ideas would be very much appreciated. |
2512295 | Simple Induction Proof: [imath]\sum_{i=1}^{n} = \frac{i(i+1)}{2}[/imath]
I want to prove by induction that [imath]\sum_{i=1}^{n} = \frac{i(i+1)}{2}[/imath] I'm getting stuck as to whether this can be done. My base case of 1 says no. | 1104696 | Proof By Induction of a Sum
Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following: [imath]\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right) }{ 2 } .[/imath] If [imath]n=1[/imath] Then [imath]1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1[/imath] Now assume [imath]n=k[/imath]. Thus [imath]\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1[/imath] If now [imath]n=k+1[/imath] [imath]\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}[/imath] Is there anything more that I have to do, or is this it? |
2507660 | Adjoint representation, standard basis - [imath]\mathfrak{sl}(2)[/imath] and Killing form
Let [imath]L[/imath] be the Lie algebra [imath]\mathfrak{sl}_{2}[/imath] (char [imath]{F}[/imath] [imath]\neq[/imath] 2). Take as standard basis for [imath]L[/imath] the three matrices: [imath]x=\begin{pmatrix} 0&1\\0&0\end{pmatrix}, y= \begin{pmatrix} 0&0\\1&0\end{pmatrix}[/imath] and [imath]h=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}[/imath]. When using the adjoint representation, the three matrices become: [imath]ad_{x}=\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}, ad_{y}=\begin{pmatrix} 0&0&0\\-1&0&0\\0&2&0\end{pmatrix}[/imath] and [imath]ad_{h}=\begin{pmatrix} 2&0&0\\0&0&0\\0&0&-2\end{pmatrix}[/imath]. Therefore [imath]\kappa[/imath] (the Killing form) has matrix [imath]\begin{pmatrix} 0&0&4\\0&8&0\\4&0&0\end{pmatrix}[/imath]. Can someone explain how the adjoint and further, the Killing form, are calculated here? I see now how to calculate the adjoint and the Killing form. [imath]ad_{x}(x)[/imath] = [imath][xx -xx] = 0[/imath] [imath]ad_{x}(h) = [xh - hx][/imath] = [imath]\begin{pmatrix} 0&1\\0&0\end{pmatrix} \times \begin{pmatrix} 1&0\\0&-1\end{pmatrix}[/imath] - [imath]\begin{pmatrix} 1&0\\0&-1\end{pmatrix}\times\begin{pmatrix} 0&1\\0&0\end{pmatrix}[/imath] = [imath]\begin{pmatrix} 0&-2\\0&0\end{pmatrix} = -2x[/imath] [imath]ad_{x}(y) = [xy - yx][/imath] = [imath]\begin{pmatrix} 0&1\\0&0\end{pmatrix} \times \begin{pmatrix} 0&0\\1&0\end{pmatrix}[/imath] - [imath]\begin{pmatrix} 0&0\\-1&0\end{pmatrix}\times\begin{pmatrix} 0&1\\0&0\end{pmatrix}[/imath] = [imath]\begin{pmatrix} 1&0\\0&-1\end{pmatrix} = h[/imath] Then, these become the three columns of [imath]ad_{x}[/imath]. Do the same for [imath]y[/imath] and [imath]h[/imath]. For the Killing form, [imath]\kappa(x,x) = tr(ad_{x}ad_{x})[/imath] [imath]\kappa(x,h) = tr(ad_{x}ad_{h})[/imath] [imath]\kappa(x,y) = tr(ad_{x}ad_{y})[/imath] [imath]tr(ad_{x}ad_{x}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix})= tr(\begin{pmatrix} 0&&\\&0&\\&&0\end{pmatrix}) = 0[/imath]. [imath]tr(ad_{x}ad_{h}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 2&0&0\\0&0&0\\0&0&-2\end{pmatrix})= tr(\begin{pmatrix} 0&&\\&0&\\&&0\end{pmatrix}) = 0[/imath]. [imath]tr(ad_{x}ad_{y}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 0&0&0\\-1&0&0\\0&2&0\end{pmatrix})= tr(\begin{pmatrix} 2&&\\&2&\\&&0\end{pmatrix}) = 4[/imath]. Thus, these three entries are the entries for the first row. Repeat for [imath]h[/imath] and [imath]y[/imath]. | 112550 | Basis for adjoint representation of [imath]sl(2,F)[/imath]
Consider the lie algebra [imath]sl(2,F)[/imath] with standard basis [imath]x=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}[/imath], [imath]j=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}[/imath], [imath]h=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}[/imath]. I want to find the casimir element of the adjoint representation of [imath]sl(2,F)[/imath]. How can I go about this? Thanks. |
2512067 | Explaining why a composition of a harmonic function and a Mobius transform is harmonic
I'm not sure how to actually show/explain this. I only recall learning harmonic functions being functions that satisfy Laplace's Equation. Let [imath]f(z)=\frac{1+z}{1-z}[/imath] for all [imath]z\in\mathbb{C}\backslash\{1\}[/imath]. a) Show that [imath]f[/imath] maps [imath]\{z\in\mathbb{C}:|z|<1\}[/imath] onto [imath]\{w\in\mathbb{C}:\mathrm{Re}(w)>0\}[/imath]. b) Explain why [imath]H\circ f[/imath] is harmonic in [imath]\{z\in\mathbb{C}:|z|<1\}[/imath] if [imath]H[/imath] is harmonic in [imath]\{w\in\mathbb{C}:\mathrm{Re}(w)>0\}[/imath]. I've done [imath](a)[/imath] by simply inverting [imath]f[/imath] and using the condition that [imath]|z|<1[/imath]. However, I'm not sure how this is used (if it is) in part (b). | 243193 | If [imath]f:D\to D’[/imath] is analytic and [imath]u: D'\to R[/imath] is harmonic then the composition of [imath]u[/imath] and [imath]f[/imath] is harmonic in [imath]D[/imath]
If [imath]f:D\to D’[/imath] is analytic and [imath]u: D'\to R[/imath] is harmonic then the composition of [imath]u[/imath] and [imath]f[/imath] is harmonic in [imath]D[/imath]. How can I show that the above statement is true/false? Can anyone help me? |
2512573 | [imath]X_{n} \to X [/imath] in distribution, [imath]Y_{n} \to Y[/imath] in distribution but [imath]X_{n} + Y_{n}[/imath] does not converge to [imath]X + Y[/imath] in distribution?!?
From my problem set: Give an example for sequences of random variables [imath]X_{n}[/imath], [imath]X[/imath], [imath]Y_{n}[/imath], [imath]Y[/imath] with [imath]X_{n} \to X[/imath] in distribution, [imath]Y_{n} \to Y[/imath] in distribution but [imath]X_{n} + Y_{n}[/imath] does not converge to [imath]X + Y[/imath] in distribution. Many thanks in advance. | 9580 | Linearity of convergence in distribution of random variables
if [imath]X_n[/imath] converges to [imath]X[/imath] and [imath]Y_n[/imath] converges to [imath]Y[/imath] in distribution, what about [imath]X_n + Y_n [/imath] would that converge to [imath]X+Y[/imath] in distribution ? any ideas how i could prove or disprove this |
2512862 | Prove that n is a prime number if and only if [imath](n−2)! \equiv 1 \pmod n[/imath].
Prove that [imath]n[/imath] is a prime number if and only if [imath](n−2)! \equiv 1 \pmod n[/imath]. I proved the part where we suppose that the latter is true and prove that n is prime. However, I'm stuck on how to prove the latter is true assuming n is prime. I tried following the proof for Wilson's Theorem, but I'm not sure how to translate it into a proof for [imath](n-2)![/imath]. Can anyone show me where to start? | 2124719 | Showing a number [imath]n[/imath] is prime if [imath](n-2)! \equiv 1 \pmod n[/imath]
I need to show that if [imath](n-2)! \equiv 1 \pmod n[/imath] then [imath]n[/imath] is prime. I think that if [imath]n[/imath] is composite, then we'll have every factor of [imath]n[/imath] in [imath](n-2)![/imath], and it would yield that [imath](n-2)! \equiv 0 \pmod n[/imath]. However, I didn't use the fact that it specifically congruent to [imath]1 \bmod n[/imath], so I think I'm getting something fundamental wrong. Is my solution correct? Why do we demand congruence to [imath]1 \bmod n[/imath]? |
540271 | [imath]f[/imath] be an entire function such that [imath]f(0)=0, f(1)=2[/imath]
[imath]f[/imath] be an entire function such that [imath]f(0)=0, f(1)=2[/imath] Then (i) there exist a sequence [imath](z_n )[/imath] such that [imath]|z_n | > n[/imath] and [imath]|f (z_n )| > n.[/imath] (ii) there exist a sequence [imath](z_n )[/imath] such that [imath]|z_n | > n[/imath] and [imath]|f (z_n )| < n.[/imath] (iii) there exist a bounded sequence [imath]z_n [/imath] and [imath]|f (z_n )| > n.[/imath] (iv) there exist a sequence [imath](z_n )[/imath] such that [imath]|z_n | \to 0[/imath] and [imath]f (z_n ) \to 2 .[/imath] [imath]f(z)=z(z+1)[/imath] is an entire and satisfis the condition,but this function does not satisfies (iv) by taking [imath]z_n={1\over n}[/imath], (i) is satisfied by taking [imath]z_n=n+1[/imath], but what am I supposed to do here? what about (ii),(iii)? thanks for the help. | 258639 | [imath]f:\mathbb C\rightarrow \mathbb C[/imath] be an arbitrary analytic function satisfying [imath]f(0)=0[/imath] and [imath]f(1)=2.[/imath]
I am stuck with the following problem: (GATE-Question) Let [imath]f:\mathbb C\rightarrow \mathbb C[/imath] be an arbitrary analytic function satisfying [imath]f(0)=0[/imath] and [imath]f(1)=2.[/imath] Then, which of the following items is correct? (a) there exists a sequence [imath]\{z_{n}\}[/imath] such that [imath]|z_{n}|> n[/imath] and [imath]|f(z_{n})|> n[/imath], (b) there exists a sequence [imath]\{z_{n}\}[/imath] such that [imath]|z_{n}|>n[/imath] and [imath]|f(z_{n})|< n[/imath], (c) there exists a bounded sequence [imath]\{z_{n}\}[/imath] such that [imath]|f(z_{n})|> n[/imath], (d) there exists a sequence [imath]\{z_{n}\}[/imath] such that [imath]z_{n} \rightarrow 0[/imath] and [imath]f(z_{n})\rightarrow 2.[/imath] I do not know how to progress with the problem or what property to use. Please help. Thanks in advance for your time. |
2512971 | How to compute [imath]\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\cos(nx)}{2^n}\bigg)^2dx[/imath]
I am trying to compute: [imath]\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\cos(nx)}{2^n}\bigg)^2dx[/imath] I tired to expand the integrand but it is too messy and lengthy.The Thing is I'm not sure how to deal with the square. If anyone has an idea on how to compute I will appreciate. | 613787 | Compute [imath]\int _0 ^{2\pi} (\sum_{n=1}^\infty \frac {\cos(nx)}{2^n})^2 dx[/imath]
Question: Compute [imath]\int _0 ^{2\pi} \left(\sum_{n=1}^\infty \frac {\cos(nx)}{2^n}\right)^2 dx[/imath] Thoughts: Tried interchanging the integral and the sum, but then the integral turned out to be zero... |
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