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2758014 | Find [imath]n[/imath] such that [imath] \binom{2n+1}{n+2} = \binom{2n+1}{5} [/imath]
find the [imath]n[/imath] : ?[imath] \binom{2n+1}{n+2} = \binom{2n+1}{5} [/imath] My try : [imath]\dfrac{(2n+1)!}{(n+2)!\times (n-1)!}=\dfrac{(2n+1)!}{5!\times (2n-4)!}[/imath] [imath]5!\times (2n-4)!=(n+2)!\times (n-1)![/imath] now what ? | 2756253 | Solve the equation [imath]\binom{2n+1}{n+2} = \binom{2n+1}{5} [/imath]
Clearly, [imath]n = 3,6[/imath] are solutions of the following equation. [imath] \binom{2n+1}{n+2} = \binom{2n+1}{5} [/imath] Can we find other values for [imath]n[/imath]? |
2757685 | Re: Find all primes [imath]p[/imath] such that [imath]\phi(x)=x^{13}[/imath] is a homomorphism [imath]\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z[/imath].
Note: This is NOT a duplicate of this question. The Question: Find all primes [imath]p[/imath] such that [imath]\phi(x)=x^{13}[/imath] is a homomorphism [imath]\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z[/imath] My Thoughts: OK, so now I understand that I am supposed to find all primes [imath]p[/imath] such that [imath](x+y)^{13} \equiv x^{13}+y^{13} \pmod p \quad \forall \; x,y \in \Bbb Z/p\Bbb Z[/imath] I expanded out the binomial to get \begin{align} \ & \iff \sum_{k=0}^{13}C_k^{13}x^ky^{13-k} \equiv x^{13}+y^{13}\pmod p \quad \forall \; x,y \in \Bbb Z/p\Bbb Z \\ \ & \iff 13x^{12}y+78x^{11}y^2+\cdots+78x^2y^{11}+13xy^{12} \equiv 0 \pmod p \quad \forall \; x,y \in \Bbb Z/p\Bbb Z \end{align} Clearly, [imath]p=13[/imath] is a solution. But [imath]p=2[/imath] also works, just by inspection. How can I prove that there is no other solution? | 2757655 | Find all primes [imath]p[/imath] such that [imath]\phi(x)=x^{13}[/imath] is a homomorphism [imath]\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z[/imath].
The Question: Find all primes [imath]p[/imath] such that [imath]\phi(x)=x^{13}[/imath] is a homomorphism [imath]\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z[/imath] My Thoughts: So from what I understand \begin{align} \ & \phi:\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z \quad\text{is a homomorphism} \\ \ \iff & \phi(xy) \equiv \phi(x)\phi(y) \pmod p \quad \forall \; x,y\\ \ \iff & (xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y \end{align} But isn't [imath](xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y[/imath] always true, regardless of [imath]p[/imath]? Or is there something I am not understanding? |
2247366 | Solution of Cauchy problem
[imath]\begin{cases} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} = 0, &x\in\mathbb{R},\ t>0\\ u(x,0) = u_0(x), &x\in\mathbb{R}. \end{cases}[/imath] Which choices of the following functions for [imath]u_0[/imath] yield a [imath]C^1[/imath] Solution [imath]u(x,t)[/imath] for all [imath]x\in\mathbb{R}[/imath] and [imath]t>0[/imath] [imath]1)[/imath] [imath]u_0 (x) = \frac{1}{1+ x^2}[/imath] [imath]2)[/imath] [imath]u_0(x) = x[/imath] [imath]3)[/imath] [imath]u_0(x) = 1+x^2[/imath] [imath]4)[/imath] [imath]u_0(x) = 1+2x[/imath] My answer key says the answer is [imath]2[/imath] and [imath]4[/imath]. My question is why we should reject option [imath]1[/imath] and [imath]3[/imath]? [imath]\frac{dt}{1} =\frac{dx}{u}= \frac{du}{0}[/imath] gives me [imath]u=c_1[/imath] and [imath]ut-x=c_2[/imath] where [imath]c_1[/imath] and [imath]c_2[/imath] are constants. Now by condition given in Option [imath]1[/imath] I get a relation between constants I.e [imath]c_1= \frac{1}{1+(c_2)^2}.[/imath] Then I have the solution [imath]u = \frac{1}{1+ (ut-x)^2}.[/imath] Then why we reject option [imath]1[/imath]? | 1797945 | Cauchy Problem for inviscid Burgers' equation
Consider the Cauchy Problem of finding [imath]u(x,t)[/imath] such that [imath]\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0,x\in\mathbb{R},t>0[/imath] [imath]u(x,0) = u_0(x), x\in\mathbb{R}[/imath] Which choices of the following functions for [imath]u_{0}[/imath] yield a [imath]C^{1}[/imath] solution [imath]u(x,t)[/imath] for all [imath]x\in\mathbb{R},t>0[/imath] [imath]u_{0}(x)=\frac{1}{1+x^{2}}[/imath] [imath]u_{0}(x)=x[/imath] [imath]u_{0}(x)=1+x^{2}[/imath] [imath]u_{0}(x)=1+2x[/imath]. If I use the existence and uniqueness theorem for Cauchy problem i get the corresponding determinant is non zero so all are true according to me. But in answer key only option 2nd and 4th is given. Please help me to solve the problem. Thanks a lot. |
2757826 | What is [imath]\lim_{n\to\infty}n!^{1/n}[/imath]?
I am struggling to find [imath]\lim_{n\rightarrow\infty}n!^{1/n}[/imath] being trying this for ages, I know that: [imath]\lim_{n\rightarrow\infty} n^{{1}/{n}} = 1[/imath] which might be useful for using pinching theorem. but I can't think of a way to 'trap' the function underneath. | 1051147 | Evaluate [imath]\displaystyle \lim_{n\to \infty}\sqrt[n]{n!}[/imath]
I am trying to evaluate the [imath]\lim(\sqrt[n]{n!})[/imath] using 2 theorems (2 proofs) Theorem 1: Let [imath]\{c_n\}[/imath] be any sequence in [imath]\mathbb{R}^+[/imath]. Then, [imath]\displaystyle \underline{\lim}\frac{c_{n+1}}{c_n}\leq \underline{\lim}\sqrt[n]{c_n}[/imath] and [imath]\displaystyle \overline{\lim}\sqrt[n]{c_n}\leq \overline{\lim}\frac{c_{n+1}}{c_n}[/imath]. so with 1. I have [imath]\frac{(n+1)!}{n!}[/imath] =[imath]n+1[/imath] which is [imath]\overline{\lim}=\infty[/imath] and 2 with [imath]\sqrt[n]{n!}\geq\sqrt[n]{(n/2)^{n/2}}[/imath]=[imath]\sqrt{\frac{n}{2}}[/imath] which is [imath]\overline{\lim}=\infty[/imath] is it valid? P.S I was not using theorem 1 right |
2758439 | Finding the limit of decreasing function
Given [imath]x_1=\arctan2>x_2>x_3\dots\\ \sin(x_n-x_{n-1})=\frac1{2^{n+1}}\sin x_n \sin{x_{n-1}} [/imath] Find [imath]\cot x_n[/imath] and prove [imath]\lim_{x\to\infty}x_n=\frac\pi4[/imath] I found that [imath]\cot x_n=\cot x_{n-1}-(1/{2^{n+1}})[/imath] Im not sure how to proceed with the limit. If I put [imath]x_n=x_{n-1}[/imath] in above eq it evaluate to [imath]0=0[/imath] | 2755844 | Find [imath]cot\space x_n[/imath] if [imath]sin(x_{n+1}-x_n)+2^{-(n+1)}sin\space x_nsin \space x_{n+1}=0[/imath] for [imath]n\ge1[/imath]
Also Given Suppose [imath]x_1=tan^{-1}2>x_2>x_3>......[/imath] are positive real numbers I cannot understand as to how I should approach but I have a hunch which is a bit hand-wavy way to prove it but tried to show that [imath]lim_{n \to \infty}x_n =\dfrac{\pi}{4}[/imath] (For the proof see my edits in @hypernova's answer) But can anybody prove it and then use it to find [imath]cot \space x_n[/imath] please? |
2758458 | Integration of Lambert W function logarithm.
The infinite power tower or infinite tetration [imath]{\displaystyle z^{z^{z^{\cdot ^{\cdot ^{\cdot }}}}}\!}[/imath] is generally expressed in terms of the Lambert W function in the following form: [imath]{\displaystyle {\frac {W(-\ln(z))}{-\ln(z)}}}[/imath] Considering the integral [imath]\displaystyle \int -\frac{W(-\ln (z))}{\ln (z)} \, dz,[/imath] Does this integral have a closed form solution? Is there a series expansion for this integral? | 2758284 | What is the antiderivative of x^x^x...?
What is [imath]\int{x^{x^{x^{x^{\dots }}}}dx}[/imath]? I managed to get it down to a complicated series after using [imath]u[/imath]-substitution, but I want to see other methods of approach, or if my answer was right. First I let [imath]u=x^{x^{x^{x^{\dots }}}}[/imath], and [imath]du[/imath] would be, in terms of [imath]u[/imath], [imath]du= \frac{u}{\frac{1-ln(u)}{u}e^{ln(u)/u}}[/imath] This makes the integral of [imath]\int{x^{x^{x^{x^{\dots }}}}dx}=\int \frac{1 - ln(u)}{u}{e^{ln(u)/u} }du[/imath] Now making [imath]w = ln(u), dw[/imath] becomes [imath]1/u[/imath] and the integral becomes [imath]\int e^{w/e^{w}}(1-w)dw[/imath] From there, I distributed and made [imath]2[/imath] integrals of series, using the Taylor polynomial for [imath]e^x[/imath]. From there, I integrated and back substituted. I was only trying to find the indefinite integral. I'm still a junior in high school, so let me know if the answer is beyond my scope of understanding. |
2758449 | Example of a convergent pointwise sequence of function and not uniformly convergent
We say that a sequence [imath]f_n(x)[/imath] is pointwise convergent to [imath]f(x)[/imath] on the set [imath]E[/imath] when for every [imath]x\in E[/imath] and [imath]\epsilon >0[/imath] there exists an integer [imath]N[/imath] such that the distance between [imath]f_n(x)[/imath] and [imath]f(x)[/imath] is smaller than [imath]\epsilon[/imath]. We say that a sequence [imath]f_n[/imath] is uniformly convergent to [imath]f[/imath] on the set [imath]E[/imath] if for all [imath]\epsilon >0[/imath] there exists an integer [imath]N[/imath] such that the distance between [imath]f_n[/imath] and [imath]f[/imath] is smaller than [imath]\epsilon[/imath]. The main difference between the aforementioned definitions is that [imath]N[/imath] dependant on [imath]x[/imath] in the context of pointwise convergence, that is for each [imath]x[/imath] in [imath]E[/imath] there exists a specific [imath]N[/imath] associated to [imath]x[/imath] while in the case of uniform convergence there exists a universal [imath]N[/imath] for which the [imath]f_n(x)[/imath] converges to [imath]f(x)[/imath] uniformly with respect to [imath]x[/imath]. Can you please give me an example of sequence of function that converges pointwise and not unifomly to [imath]f(x)[/imath]. Thanks in advance | 992022 | Sequence which converges pointwise but not uniformly?
it might be simple but I don't find a sequence [imath]f_n: [0,1] \rightarrow \mathbb{R}, n \in \mathbb{N}[/imath] that converges pointwise but not uniformly. First I thought it could be [imath]f_n(x) = \frac{x}{n}[/imath] but it is not right, is it? Thanks for help! |
2757519 | Let [imath]n=(6t+1)(12t+1)(18t+1)[/imath]. Show that [imath]a^{n-1} \equiv 1\pmod n[/imath] if [imath](a,n)=1[/imath]
The Question: Let [imath]n=(6t+1)(12t+1)(18t+1)[/imath], where [imath]t\in \Bbb Z[/imath] and [imath]6t+1[/imath], [imath]12t+1[/imath], [imath]18t+1[/imath] are prime, and [imath](a,n)=1[/imath]. Show that [imath]a^{n-1} \equiv 1\pmod n[/imath]. My Thoughts: This blatantly looks like Fermat's Little Theorem, but I can't seem to find a way to adapt the proof of it for this question. In particular, the proof of FLT fails in the last step in this case when you divide by [imath](n-1)![/imath]: [imath]a^{n-1}(n-1)! \equiv (n-1)! \pmod n \implies a^{p-1} \equiv 1 \pmod n[/imath] is a false statement because [imath](n-1)![/imath] is in fact a multiple of [imath]n[/imath]. Any hints? | 546156 | Suppose k is an integer such that [imath]6k + 1[/imath], [imath]12k + 1[/imath], [imath]18k + 1[/imath] are primes. Show that [imath]n = (6k + 1)(12k + 1)(18k + 1)[/imath] is a Carmichael number.
Suppose k is an integer such that [imath]6k + 1[/imath], [imath]12k + 1[/imath], [imath]18k + 1[/imath] are primes. Show that [imath]n = (6k + 1)(12k + 1)(18k + 1)[/imath] is a Carmichael number. This is a question on a past exam that I find difficulty to answer. Can anyone help? |
2758742 | On a closed form for [imath]\int_{-\infty}^\infty\frac{dx}{\left(1+x^2\right)^p}[/imath]
Consider the following function of a real variable [imath]p[/imath] , defined for [imath]p>\frac{1}{2}[/imath]: [imath]I(p) = \int_{-\infty}^\infty\frac{dx}{\left(1+x^2\right)^p}[/imath] Playing around in Wolfram Alpha, I have conjectured that we have the following closed form: [imath]I(p) = \sqrt{\pi}\frac{\Gamma\left(p-\frac{1}{2}\right)}{\Gamma\left(p\right)}[/imath] This agrees with a number of known results, such as: [imath]I(1) = \pi[/imath] [imath]I(2) = \frac{\pi}{2}[/imath] [imath]I(3) = \frac{3\pi}{8}[/imath] It also agrees with [imath]I\left(\frac{1}{2}\right)[/imath] being divergent, which follows from the fact that the antiderivative of [imath]\left(1+x²\right)^{-\frac{1}{2}}[/imath] is [imath]\text{arsinh }(x)[/imath] and the fact that [imath]\lim\limits_{x\rightarrow\pm\infty}{\sinh(x)} = \pm\infty[/imath]. Furthermore, this closed form agrees with any non-integer [imath]p\geq\frac{1}{2}[/imath] that I have tried to evaluate, such as [imath]I\left(\frac{3}{5}\right) = \sqrt{\pi}\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{3}{5}\right)}[/imath]. However, I have not been able to prove this result. I'm thinking that perhaps the residue theorem might work, but I have no idea how to treat poles of a non-integer order in the denominator. Any ideas? | 1324331 | Closed form for [imath]\int_0^\infty\frac{1}{(1+x^2)^s}\,dx[/imath] when [imath]s\in (0.5,\infty)\setminus\mathbb{N}[/imath]
I know that the improper integral [imath] \int_0^\infty\frac{1}{(1+x^2)^s}\,dx [/imath] is convergent for [imath]s>0.5[/imath] and divergent otherwise. Furthermore, it has a closed form for [imath]s \in \mathbb{N}[/imath] (this can be obtained using the Residue Theorem). My question is to know whether it has a closed form for any non-integer [imath]s >0.5[/imath]. |
2758885 | Why can't the indefinite integral [imath]\int\frac{\sin(x)}{x}\mathrm dx[/imath] be found?
I came across a list of functions in my calculus textbook whose indefinite integral cannot be found. It was written that the integral [imath]\int \frac{\sin(x)}{x} dx[/imath] cannot be evaluated without any explanation as to why. I did some research over the internet and found out that the definite integral [imath]\int_{0}^{\infty} \frac{\sin(x)}{x} dx[/imath] can be evaluated using Laplace transformation and is equal to [imath]\pi /2[/imath]. But I still couldn't find answer to my original question. I read somewhere that the integral cannot be expressed using 'elementary functions'. A little help is appreciated, I'm in Calc 1 going advanced than my course but I am sorry if my post shows lack of research. Thank you! | 2887819 | Why does [imath]\frac{\sin x}{x}[/imath] have no anti-derivative
I've been messing around with indefinite integrals. Watching some youtube videos and I found the Sinc function and that it has no finite Anti-derivative. Desmos being my favourite program ever I decided to graph to the equations [imath]y=\frac{\sin(x)}{x}[/imath] and [imath]y=\int_0^x\frac{\sin(a)}{a}da[/imath] (only way I could do indefinite integrals in Desmos) If you do this you might notice [imath]y=\frac{\sin(x)}{x}[/imath] looks very similar to [imath]y =\sinh(x)[/imath] and [imath]y=\int_0^x\frac{\sin(a)}{a}da[/imath] looks very similar to [imath]y=\arctan(x)[/imath] I wondered how we know that [imath]\frac{\sin(x)}{x}[/imath] has no finite anti-derivative because these simple functions give seemingly accurate approximations Best approximation I could get for it in the short time I had was: [imath]\int \sinh(x)-\frac{d}{dx}\bigg[\frac{\sin(x)}{x}\bigg]-\frac{\pi}{2}[/imath] or [imath]2\arctan(e^x)+\frac{\sin(x)}{x^2}-\frac{\cos(x)}{x}-\frac{\pi}{2}[/imath] (In all the examples I'm assume [imath]C=0[/imath] because it's not important and I'm assuming any points with [imath]\frac{0}{0}[/imath] evaluate to their limits as [imath]x[/imath] tends to [imath]0[/imath]) |
2758965 | Show [imath]\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}\,\forall x\in(-1,1)[/imath].
Show [imath]\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}\,\forall x\in(-1,1)[/imath]. Which value does [imath]\sum_{n=1}^{\infty}\frac{(-1)^n}{n}[/imath] take? Now because I skipped forward in my (personal) textbook I know that I could tackle this using knowledge of the Maclaurin/Taylor series. However, it was not covered in the lecture (yet) and my mind is fixated on using Maclaurin/Taylor (which I'm not allowed to use)! Can anybody show me an alternate approach that I will probably feel very stupid for not seeing? | 1075259 | Computing the limit of an alternating series,
I am looking at the series [imath] \sum_{n=1}^\infty\frac{(-1)^n}{n}.[/imath] This series converges (conditionally) by the alternating series test. How can I compute its limit, which is equal to -log(2)? a) I considered [imath]I_n = \int_{0}^1[/imath] [imath]\frac{x^n}{1+x} dx[/imath] -- and showed that this goes to 0, as n goes to infinity (use dominated convergence theorem). b) I computed [[imath]I_k[/imath] + [imath]I_{k-1}[/imath]] (for k [imath]\ge[/imath] 1), and it equals [imath]\frac{1}{k}[/imath]. I should be using my results from a) and b) to compute the above limit, which is part c) of the question that I am working on. So, I have [imath] \sum_{n=1}^\infty\frac{(-1)^n}{n} = \sum_{n=1}^\infty(-1)^n[I_n+I_{n-1}].[/imath] Now I'm stuck. How should I proceed from here? I checked, by computing explicitly, that [imath]-I_0[/imath] = -log(2). But I don't know how to arrive at [imath]-I_0[/imath] from the summation that I'm currently at. Thanks in advance, |
2759308 | Three numbers are chosen from the first [imath]n[/imath] natural numbers. What is the probability that they can be sides of a triangle?
I wonder if anybody can help me solve the following problem: Three numbers are chosen from the first [imath]n[/imath] natural numbers. What is the probability that they can be sides of a triangle? NOTE: The sides can be equal. | 94965 | Probability that a triangle can be formed from a permutation of three edges of random length
What is the probability that a particular set of integer edge lengths selected from an interval [imath][1, N][/imath], can form a triangle? How might this extend to the case where one selects real number edge lengths from the unit interval? I'd also be very interested in any comments on what the probability distribution would look like for the distances between the triangles vertices. |
2759975 | Prove that [imath]T[/imath] is invertible if [imath](T\circ T)+T −2I_V = 0[/imath]
I'm having difficulty proving the following: Let [imath]T:V \to V[/imath] be a linear transformation such that [imath](T\circ T)+T −2I_V = 0[/imath]. Prove that [imath]T[/imath] is invertible. I think that T might be equal to the identity transformation itself ([imath]I[/imath]) but I'm not sure how to get to it from the datum. | 476746 | How do i prove invertibility of this linear transformation
Let [imath]T[/imath] be a linear transformation from a vector space [imath]V[/imath] over reals into [imath]V[/imath] such that [imath]T-T^2=I[/imath]. Show that [imath]T[/imath] is invertible |
2760034 | Find all [imath]n[/imath] such that i. [imath]\phi(n)=\frac n2[/imath], ii. [imath]\phi(n)=4[/imath].
For [imath]n\ge1[/imath], let [imath]\phi(n)[/imath] be the number of positive integers [imath]\le n[/imath], which are relatively prime to [imath]n[/imath] i. Find all [imath]n[/imath] such that [imath]\phi(n)=\frac n2[/imath] ii. Find all [imath]n[/imath] such that [imath]\phi(n)=4[/imath]. I know the formula for suker totient function but that is not telling me anything. Please help. | 295732 | For which Natural [imath]n\ge2: \phi(n)=n/2[/imath]
For which Natural [imath]n\ge2[/imath] does this occur with?: [imath]\phi(n)=n/2[/imath] |
2760064 | Limit of [imath]\sum_{k=2}^{\infty} \frac{1}{k^2-1}[/imath]
My calculator suggests that the limit of this series is 0.75. [imath]\sum_{k=2}^{\infty} \frac{1}{k^2-1}[/imath] How can this be proved. I was thinking about a telescop sum but that didn't help me I appreciate any help you could provide | 993890 | How to find the sum of the series [imath]\sum_{k=2}^\infty \frac{1}{k^2-1}[/imath]?
I have this problem : [imath]S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}[/imath] My solution [imath]S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...][/imath] I think that the sum should be [imath]\frac{1}{2}[/imath] since the limit of : [imath]-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}[/imath] But that wrong. Any ideas? |
2760101 | How to prove that if [imath]f^3 = f[/imath], [imath]f[/imath] is diagonalizable?
Prove that Let [imath]V[/imath] be a finite dim. vector space over a field of characteristic zero, and [imath]f: V \to V[/imath] be a linear map.Then if [imath]f^3 = f,[/imath] then [imath]f[/imath] is diagonalizable. Since [imath]f[/imath] is zero of the polynomial [imath]p(x) = x^3 - x[/imath] the minimal polynomial [imath]m_f[/imath] should divide [imath]p[/imath], hence [imath]m_f[/imath] can be the following polynomials only: [imath]m(x) = (x-1) \\ = x \\ = x+ 1 \\ =x^2 - x \\ = x^2 + x \\ = x^2 - 1 \\ =x(x-1)(x+1) [/imath] But, I couldn't show that if [imath]m[/imath] is either of the followings [imath] x^2 + x [/imath] & [imath] x^2 - 1 [/imath] & [imath] x(x-1)(x+1) [/imath], it should be diagonalizable. So for these 3 cases how can we prove that [imath]f[/imath] should be diagonalizable ? | 2511311 | Proving that if [imath]A[/imath] is a [imath]8\times 8[/imath] matrix over [imath]\mathbb{R}[/imath] and [imath]A^3=A[/imath], then [imath]A[/imath] is diagonalizable.
If [imath]A[/imath] is a [imath]8\times 8[/imath] matrix over [imath]\mathbb{R}[/imath] and [imath]A^3=A[/imath] then prove that [imath]A[/imath] is diagonalizable. I have got that the minimal polynomial of [imath]A[/imath] may be [imath]x^3-x[/imath] or [imath]x[/imath] or [imath]x(x+1)[/imath] or [imath]x(x-1)[/imath] in the 2nd case it is not possible in the 3rd and 4th cases the matrix will be the identity matrix or identity matrix multiplied by a scalar! But I cannot reach any further. |
2761094 | Show that if [imath]X[/imath] is homotopically equivalent to [imath]Y[/imath], then [imath]\widetilde{X}[/imath] is homotopically equivalent to [imath]\widetilde{Y}[/imath]
Suppose that [imath]X,Y[/imath] are arc-connected and locally arc-connected spaces and that [imath]p:\widetilde{X}\to X[/imath] and [imath]q:\widetilde{Y}\to Y[/imath] are universal covering of [imath]X[/imath] and [imath]Y[/imath] respectively. Show that if [imath]X[/imath] is homotopically equivalent to [imath]Y[/imath], then [imath]\widetilde{X}[/imath] is homotopically equivalent to [imath]\widetilde{Y}[/imath] I have to find a homotopic equivalence from [imath]\widetilde{X}[/imath] to [imath]\widetilde{Y}[/imath], I know that there is a homotopic equivalence [imath]f:X\to Y[/imath] and therefore [imath]f_*:\pi_1(X)\to \pi_1(Y)[/imath] is an isomorphism, besides [imath]\widetilde{X}[/imath] and [imath]\widetilde{Y}[/imath] are the universal covers then [imath]\pi_1(\widetilde{X})=\{1\}[/imath] and [imath]\pi_1(\widetilde{Y})=\{1\}[/imath] and I have the next diagram but I do not know what else to do. Could someone help me please?, could I take [imath]g=q^{-1}\circ f\circ p[/imath] ? Thank you very much. | 2069936 | Universal covering spaces of homotopy equivalent spaces are homotopy equivalent
I'm working on an exercise from Hatcher's Algebraic Topology (exercise 1.3.8): Let [imath]\tilde X[/imath] and [imath]\tilde Y[/imath] be simply-connected covering spaces of the path-connected, locally path-connected spaces [imath]X[/imath] and [imath]Y[/imath]. Show that if [imath]X \simeq Y[/imath] then [imath]\tilde X \simeq \tilde Y[/imath]. The exercise recommends using the following result from earlier in the textbook: A map [imath]f: X \to Y[/imath] is a homotopy equivalence if there exist maps [imath]g, h: Y \to X[/imath] such that [imath]fg \simeq \mathbb{1}[/imath] and [imath]hf \simeq \mathbb{1}[/imath]. This problem has been solved here and also here, but both proofs include a step that seems to be a bit of a jump. It's possible I'm missing a very simple argument, but this is the one piece I'm having trouble with. Here's what I have so far: Let [imath]p : \tilde X \to X[/imath] and [imath]q: \tilde Y \to Y[/imath] be the respective covering maps, and let [imath]f : X \to Y[/imath] and [imath]g : Y \to X[/imath] be homotopy inverses (so [imath]gf \simeq \mathbb{1}_X[/imath] and [imath]fg \simeq \mathbb{1}_Y[/imath]). Consider the map [imath]fp : \tilde X \to Y[/imath], and the induced homomorphism [imath](fp)_*: \pi_1(\tilde X) \to \pi_1(Y)[/imath]. Since [imath]\tilde X[/imath] and [imath]\tilde Y[/imath] are simply-connected, [imath](fp)_*(\pi_1(\tilde X)) = 0 = q_*(\pi_1(\tilde Y))[/imath], so by the lifting criterion (prop. 1.33 in Hatcher), we have a lift [imath]\tilde f : \tilde X \to \tilde Y[/imath] of [imath]fp : \tilde X \to Y[/imath]. We similarly have a lift [imath]\tilde g : \tilde Y \to \tilde X[/imath] of [imath]gq: \tilde Y \to X[/imath]. This gives us that [imath]p\tilde g \tilde f = gq\tilde f = gfp \simeq p[/imath]. Call this homotopy [imath]h_t : \tilde X \to X[/imath], so that [imath]h_0 = p\tilde g \tilde f[/imath] and [imath]h_1 = p[/imath]. By the homotopy lifting property, since [imath]\tilde g \tilde f : \tilde X \to \tilde X[/imath] lifts [imath]p \tilde g \tilde f[/imath], there is a homotopy [imath]\tilde h_t[/imath] between [imath]\tilde g \tilde f[/imath] and a lift [imath]\tilde p[/imath] of the covering map [imath]p : \tilde X \to X[/imath]. In particular, [imath]p\tilde p = p[/imath]. Here's my problem: In the first of the links I shared, they claim that [imath]\tilde p[/imath] is a deck transformation of [imath]\tilde X[/imath] (ie a homeomorphism with the property that [imath]p \tilde p = p[/imath]). In the second of the links, they claim something stronger: that by uniqueness of lifts, in fact [imath]\tilde p = \mathbb{1}_{\tilde X}[/imath]. The trouble with the first of these answers is that it's not clear to me that [imath]\tilde p : \tilde X \to \tilde X[/imath] is a homeomorphism. The trouble with the second is that uniqueness of lifts only applies if we can prove that [imath]\tilde p[/imath] fixes a point of [imath]\tilde X[/imath], which is also not clear. (Plus [imath]\tilde p = \mathbb{1}_\tilde X[/imath] seems like too much to ask for because then one could similarly show that [imath]\tilde q = \mathbb{1}_{\tilde Y}[/imath], which gives that [imath]\tilde g \tilde f \simeq \mathbb{1}_\tilde{X}[/imath] and [imath]\tilde f \tilde g \simeq \mathbb{1}_\tilde{Y}[/imath], and we are done without using the recommended hint.) The proof can easily be completed after this is taken care of, but I'm having trouble justifying this. Is there a good reason [imath]\tilde p : \tilde X \to \tilde X[/imath] is a homeomorphism, or better, the identity map of [imath]\tilde X[/imath]? Thanks! |
2762250 | [imath]{\left( \int_{0}^{1} f(x)\right)}^2 \le 2\int_{0}^{1} x{(f(x))}^2[/imath]
Let f be a non-decreasing, integrable function defined on [0, 1]. Show that [imath]{\left( \int_{0}^{1} f(x)\right)}^2 \le 2\int_{0}^{1} x{(f(x))}^2[/imath] I tried using by parts , but it is of no help, [imath]{\left( \int_{0}^{1} f(x)\right)}^2 \le\int_{0}^{1} {(f(x))}^2[/imath] Any hints? | 182785 | Show that [imath]\left(\int_0^1 f(x)\,dx\right)^2\leq2\int_0^1x\, f(x)^2\,dx [/imath]
Let [imath]f[/imath] be a nondecreasing, integrable, function defined on [imath][0, 1][/imath]. Show that [imath]\left(\int_0^1 f(x)\,dx\right)^2\leq2\int_0^1x \,f(x)^2\,dx [/imath] |
2757147 | Open subsets of the connected sum [imath]M_1\# M_2[/imath]
I'm trying to solve a problem in John Lee's ITM (Problem 4-19), but seems that i need helps now. Here's the problem : Let [imath]M_1 \# M_2[/imath] be a connected sum of [imath]n[/imath]-manifolds [imath]M_1[/imath] and [imath]M_2[/imath]. Show that there are open subsets [imath]U_1,U_2 \subseteq M_1 \# M_2[/imath] and points [imath]p_i \in M_i[/imath] such that [imath]U_i \approx M_i \smallsetminus \{p_i\}[/imath], [imath]U_1 \cap U_2 \approx \mathbb{R}^n \smallsetminus \{0\}[/imath], and [imath]U_1\cup U_2 = M_1 \# M_2[/imath]. The definition of the connected sum of [imath]M_1\# M_2[/imath] is the adjunction space of [imath]M_1' \cup_f M_2'[/imath] under a homeomorphism [imath]f : \partial M_1' \to \partial M_2'[/imath], where [imath]M_i' = M_i \smallsetminus B_i[/imath] and [imath]B_i \subseteq M_i[/imath] is a regular coordinate ball of [imath]M_i[/imath]. Regular coordinate ball [imath]B_i \subseteq M_i[/imath] is a coordinate ball with homeomorphism [imath]\varphi : B' \to \varphi(B')=B_{r'}(0)[/imath], where [imath]B' \supseteq B[/imath], such that [imath]\varphi(B) = B_r(0)[/imath] and [imath]\varphi(\bar{B}) = \bar{B}_r(0)[/imath] for [imath]r'>r>0[/imath]. If we denote the embeddings as [imath]e_i : M_i' \to M_1 \# M_2[/imath], and the larger open subsets contain the coordinate balls [imath]B_i[/imath] as [imath]B_i'[/imath], i'm guessing that the desired open subsets are [imath]U_1 = e_1(M_1')\cup e_2(B_2' \smallsetminus B_2)[/imath] and [imath]U_2 = e_1(B_1'\smallsetminus B_1) \cup e_2(M_2')[/imath]. But i having trouble to show that [imath]U_i \approx M_i\smallsetminus \{p_i\}[/imath]. Actually i managed to show that (with [imath]p_i[/imath] is chosen as the "center" of coordinate ball [imath]B_i[/imath]), but my solution is quite long. So i'm not sure it is correct (i'll add my solution here if needed). If these choices of [imath]U_1,U_2[/imath] is correct i really appreciate if somebody tell me some hint to show that [imath]U_i \approx M_i \smallsetminus \{p_i\}[/imath]. If it is not then i would like to know what the correct choices look like. Thank you. [imath]\textbf{Edit : }[/imath]I think i already got what i want here. It has a nice answer. | 2622775 | Nice neighborhoods of each "piece" in a manifold connected sum
This is problem 4.19 in Lee's Topological Manifolds. I have been working on this problem for a couple days now, and just a need a hint in the right direction. If [imath]M[/imath] and [imath]N[/imath] are two [imath]n[/imath]-manifolds, and [imath]B_1\subset M[/imath] and [imath]B_2\subset N[/imath] are two open, regular coordinate balls (definition below), the connected sum [imath]M\#N[/imath] is the quotient of the disjoint union [imath](M-B_1)\displaystyle\sqcup (N-B_2)[/imath] by the relation that identifies points on the spherical boundaries of each component via some homeomorphism [imath]h[/imath]. Now I want to show that there are two open sets [imath]U,V\subset M\#N[/imath], such that: [imath]U\cong M-\{p\}[/imath] and [imath]V\cong N-\{q\}[/imath], for some points [imath]p\in M[/imath] and [imath]q\in N[/imath] [imath]U\cap V\cong S^{n-1}\times\mathbb{R}[/imath] [imath]U\cup V=M\#N[/imath] Here's my intuition: take a larger coordinate ball [imath]D[/imath] in [imath]N[/imath], containing [imath]B_2[/imath], which works because [imath]B_2[/imath] is regular. [imath](M-B_1)\sqcup (D-B_2)[/imath] is a saturated open set, so its image in [imath]M\#N[/imath] is open. This is what I want to be [imath]U[/imath]. Now I've managed to define a homeomorphism [imath]g[/imath] from [imath]D-B_2[/imath] to [imath]\overline{\mathbb{B}}_1(0)-\{0\}[/imath] (the punctured closed unit ball). If I can then map that punctured unit ball to [imath]\overline{B}_1-\{p\}[/imath], I can try and show the map [imath]f:\ (M-B_1)\sqcup(D-B_2)\rightarrow M-\{p\}[/imath] [imath] f(x)=\begin{cases} x& x\in M-B_1\\ g(x)& x\in D-B_2\\ \end{cases} [/imath] is a quotient map, and then use the uniqueness of quotients to show [imath]U[/imath] is homeomorphic to [imath]M-\{p\}[/imath]. However, this is too hard for me to do, because I need to somehow incorporate [imath]h[/imath] in the mix, so that the identifications of [imath]f[/imath] match those of [imath]h[/imath]. Can someone provide a hint here? I feel like this is a lot of work for one problem, and I understand intuitively what to do here, but getting all the details right is proving to be too much. I'm also open to hints about part 2. as well (3. is easy). Definition: A coordinate ball [imath]B\subset M[/imath] is called regular if it has a neighborhood [imath]B'\supset B[/imath], such that there is a homeomorphism [imath]k: B'\rightarrow \mathbb{B}_s(0)[/imath] (this is the open ball of radius [imath]s[/imath] around [imath]0[/imath]). Under this homeomorphism, [imath]B[/imath] goes to [imath]\mathbb{B}_r(0)[/imath] for some [imath]0<r<s[/imath], and [imath]\overline{B}[/imath] goes to [imath]\overline{\mathbb{B}}_r(0)[/imath]. |
2762252 | entire function that misses [imath][0,1][/imath] is constant
Let [imath]f:\mathbb{C}\to\mathbb{C}[/imath] be an entire function. Suppose that [imath]\forall z\in\mathbb{C}:f(z)\not\in[0,1][/imath]. Prove that [imath]f[/imath] is a constant function. I've heard about Picard's little theorem, but we didn't cover it and I am interested if there is another proof for this special case. | 914469 | If [imath]f(\mathbb{C})\subset \mathbb{C}-[0,1][/imath] then [imath]f[/imath] is constant
If [imath]f:\mathbb{C}\longrightarrow\mathbb{C}[/imath] is an entire function such that [imath]f(z)\neq w[/imath] for all [imath]z\in \mathbb{C}[/imath] and for all [imath]w\in [0,1]\subset \mathbb{R}[/imath], how to prove that [imath]f[/imath] is constant (without using Picard's little theorem). Any hint would be appreciated. |
2762717 | About the integral [imath]\int_{0}^{1}\frac{\arctan(x)\,dx}{\sqrt{x(1-x^2)}}[/imath] and related hypergeometric functions
I will be honest about it: the following integral is involved in my current investigations about the interplay between Fourier-Legendre series expansions, hypergeometric functions of the [imath]\phantom{}_3 F_2[/imath] and [imath]\phantom{}_4 F_3[/imath] kind and polylogarithms. Significative contributions will be rewarded through a proper mention of the author. The value of [imath]\phantom{}_3 F_2\left(\tfrac{1}{2},\tfrac{3}{4},1;\tfrac{5}{4},\tfrac{3}{2};-1\right)[/imath] is related to the integral [imath] \mathcal{J}_1 = \int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx, [/imath] which on its turn (by Feynman's trick) is related to [imath] \mathcal{J}_2 = \int_{0}^{1}\frac{1-\phantom{}_2 F_1\left(-\frac{1}{4},1;\frac{1}{4};-x^2\right)}{x^2}\,dx.[/imath] I would be glad to know how to convert [imath]\mathcal{J}_1[/imath] (or [imath]\mathcal{J}_2[/imath]) into a combination of standard mathematical constants and polylogarithms. It most certainly can be done since [imath]\arctan(x)=\text{Im}\log(1+ix)[/imath], but my version of Mathematica is not collaborating with me as I wished. I will be away from keyboard for a few days (time to work for the Italian most celebrated mathematical contest), I hope to find some nice insight after my return. | 2448270 | An elementary proof of [imath]\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx = \frac{1}{32}\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^2[/imath]
When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of [imath]\sum_{n\geq 0}\binom{2n}{n}^2\frac{1}{16^n(4n+1)}=\frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4 [/imath] is: [imath]\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx = \tfrac{1}{32}\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^2\tag{A}[/imath] which might be regarded as a sort of Ahmed's integral under steroids. I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of [imath]K(x)[/imath] of the form [imath]\int_{0}^{1}K(x)\,x^{m\pm 1/4}\,dx[/imath], which are associated with peculiar hypergeometric functions. |
2761686 | Bounded sequence of functions uniformly converges to bounded function
I'm working on the following question: True or False (and explain why or give counterexample). Suppose [imath](f_n)[/imath] uniformly converge to [imath]f[/imath]. If each [imath]f_n[/imath] is bounded then [imath]f[/imath] is bounded. I think this is true, but I'm having trouble showing it. In particular, suppose the [imath]M_n[/imath] is a bound for each [imath]f_n[/imath] then: [imath]|f_n-f| < \epsilon \\ -f_n - \epsilon < f < \epsilon + f_n \\ -M_n - \epsilon < f < \epsilon + M_n[/imath] There's nothing controlling the growth of the [imath]M_n[/imath]'s though. If [imath]f_n[/imath] are uniformly converging, I think the bounds should be close too, but I'm having trouble arguing this. Thoughts? This question is similar, but both answers skip the key detail that I'm hung up on. | 1226604 | If [imath]\{f_n\}[/imath] converges uniformly and each [imath]f_n[/imath] is bounded, show that [imath]\exists M>0[/imath] s.t. [imath]|f_n(x)| \leq M[/imath]
Problem Statement: Let [imath]\{f_n\}[/imath] converge uniformly on a set [imath]E[/imath]. Suppose that each [imath]f_n[/imath] is bounded. Prove that there exists an [imath]M > 0[/imath] s.t. [imath]|f_n(x)| \leq M[/imath] for all [imath]x \in E[/imath] and all [imath]n = 1,2, ...[/imath]. Proof: If each [imath]f_n[/imath] is bounded by some [imath]M_n > 0[/imath] then can't we just take [imath]M = \text{max}\{M_n\}[/imath]? I don't see what's so difficult here. Am I not using the fact that [imath]\{f_n\}[/imath] converges uniformly? |
2762896 | Prove that the only integer solution of the equation [imath]a^2 + b^2 + c^2 = a^2 b^2[/imath] is [imath]a = b = c = 0[/imath].
Any tips are welcome! I have tried various methods, but to no avail. I could use a pointer as to the right direction. Thank you. | 1445553 | Discrete mathematics - Find all integer solutions of the equation [imath]a^2+ b^2 + c^2=a^2 b^2[/imath].
Find all integer solutions of the equation [imath]a^2+ b^2 + c^2=a^2 b^2[/imath]. This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.) |
2763327 | [imath]\sum \exp(-n^{ 1/3})[/imath] series convergence
[imath]\sum_n \exp(-n^{ 1/3})[/imath] Should be convergent by ratio test. However it gives me that [imath]a_{n+1}/a_{n}=\exp(n ^{1/3}-(n+1)^{1/3}) \rightarrow 1,n \rightarrow +\infty [/imath] so the limit ratio test fails. Thank you for suggestions. | 2709418 | Prove that [imath]\sum_{n=1}^{\infty}\exp\left(-n^{\varepsilon}\right)[/imath] converges.
I have the problem to prove, that the following series converges: \begin{equation} \sum_{n=1}^{\infty}\exp\left(-n^{\varepsilon}\right), \end{equation} where [imath]\varepsilon > 0[/imath]. I tried everything. I tried the ratio and root test, but I don't come to a solution. Can someone help me? Thanks! |
1069418 | Find the integral solutions to [imath] x^2+y^2+z^2=x^2y^2[/imath]
I am unfamiliar with this type of problem. How does one solve this and under what category of math does this fall under. Find the integral solutions for [imath]x^2+y^2+z^2=x^2y^2[/imath] | 1809884 | Show that the equation [imath]x^2+y^2+z^2=x^2y^2[/imath] has no integer solution,except [imath]x=y=z=0[/imath]
Show that the equation [imath]x^2+y^2+z^2=x^2y^2[/imath] has no integer solution,except [imath]x=y=z=0.[/imath] Let one of the [imath]x,y,z[/imath] be even number.Let [imath]x=2p[/imath] [imath]x^2+y^2+z^2=x^2y^2[/imath] This gives [imath]y^2+z^2[/imath] is also even,which means either [imath]y,z[/imath] are both even or [imath]y,z[/imath] are both odd. So either [imath]x=2p,y=2q,z=2r[/imath] is the solution or [imath]x=2p,y=2q+1,z=2r+1[/imath] is the solution.I put [imath]x=2p,y=2q+1,z=2r+1[/imath] in the equation [imath]x^2+y^2+z^2=x^2y^2[/imath] and see that it does not satisfy both sides,so it is not the solution. When I put [imath]x=2p,y=2q,z=2r[/imath],i get [imath]p^2+q^2+r^2=4p^2q^2[/imath] I am confused here how to prove that [imath]p^2+q^2+r^2=4p^2q^2[/imath] does not hold true. Also i cannot solve when [imath]x[/imath] is odd which is the second case. Is my method correct?Please help me complete the solution. |
2762667 | Prove that the irreducible polynomial for [imath]\sqrt2+\sqrt3[/imath] over [imath]\mathbb{Q}[/imath] is reducible modulo [imath]p[/imath] for every prime [imath]p[/imath]
I am attempting to solve part (d) of the following problem from Artin's algebra book. I have already solved parts (a) through (c). (a) Let [imath]p[/imath] be an odd prime. Prove that exactly half of the elements of [imath]\mathbb{F}_p^\times[/imath] are squares and that if [imath]α[/imath] and [imath]β[/imath] are nonsquares, then [imath]αβ[/imath] is a square. (b) Prove the same assertion for any finite field of odd order. (c) Prove that in a finite field of even order, every element is a square. (d) Prove that the irreducible polynomial for [imath]γ=\sqrt2+\sqrt3[/imath] over [imath]\mathbb{Q}[/imath] is redubile modulo [imath]p[/imath] for every prime [imath]p[/imath]. Now, I have shown that [imath]g(x)=x^4-10x^2+1\in \mathbb{Q}[x][/imath] is the irreducible polynomial for [imath]γ[/imath]. Let [imath]p[/imath] be a prime. Let [imath]f(x)=x^4-10x^2+1 \in \mathbb{F}_p[x][/imath]. I want to show [imath]f[/imath] is reducible. I see that either 2, 4, or 6 is equal to [imath]s^2[/imath] for some [imath]s\in\mathbb{F}_p[/imath]. I think it will be useful to proceed by cases here. It someone could show me a complete, elementary, simple proof for the [imath]s^2=2[/imath] case, I should be able to finish the rest on my own. | 1875878 | Proving that [imath]x^4 - 10x^2 + 1[/imath] is not irreducible over [imath]\mathbb{Z}_p[/imath] for any prime [imath]p[/imath].
So I have seen the similar question and answers on here for [imath]x^4 +1[/imath], but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with. I managed to prove it for the primes such that [imath]p \equiv 1, 7 \pmod 8[/imath], by noting that [imath]2[/imath] is a square modulo those primes and thus [imath]x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)[/imath] for those [imath]\mathbb{Z}_p[/imath]... however, trying to get a similar result for [imath]3 \pmod 8[/imath] and [imath]5 \pmod 8[/imath] has been stumping me for a long time, I am having a hard time making [imath]q^2 = -1[/imath] and [imath]q^2 = -2[/imath] give me something factorable... I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it? |
1227601 | Geometric interpretation of hyperbolic functions and the hyperbolic angle/argument
I've been reading up on hyperbolic functions and was wondering if there was a geometric definition for the hyperbolic angle and hyperbolic function. In particular I was reading this: Alternative definition of hyperbolic cosine without relying on exponential function and this: Proofs of Hyperbolic Functions My question is this - when we consider trigonometric functions, we treat them as points on a unit circle where (conventionally) the angle is formed between the x-axis and a ray drawn to some other point. I believe that for hyperbolic functions, we take the ray dividing the hyperbola in half, and another ray, and the angle between them is the hyperbolic angle. For the hyperbola [imath]x^2 - y^2 = 1[/imath], somehow, the area of the sector is half the angle. Is there a proof for this? Also, for the hyperbola [imath]y = 1/x[/imath], the area of the sector is equal to the angle. Is there a proof here? Why are the two cases different? | 757091 | Alternative definition of hyperbolic cosine without relying on exponential function
Ordinary trigonometric functions are defined independently of exponential function, and then shown to be related to it by Euler's formula. Can one define hyperbolic cosine so that the formula [imath]\cosh{x}=\dfrac{e^x+e^{-x}}{2}[/imath] becomes something to be proven? |
2763281 | Product of all primitive nth roots of unity is 1?
Is the product of all primitive nth roots of unity equal to 1? Equivalently, is [imath]\Phi_n(0) = 1[/imath] for [imath]n>2[/imath]? More exactly, I'm trying to prove that [imath]x^{\phi(n)} \Phi_n(x^{-1}) = \Phi_n(x).[/imath] Since [imath]\Phi_n(x) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left(x - e^{\frac{2k\pi i}{n}} \right),[/imath] and since the inverses of all primitive nth roots is still the set of primitive nth roots, we have [imath]\Phi_n(x) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left(x - e^{-\frac{2k\pi i}{n}} \right)[/imath] and so [imath]\Phi_n(x^{-1}) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left(x^{-1} - e^{-\frac{2k\pi i}{n}} \right) = \Phi_n(x) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left( \dfrac{e^{\frac{2k\pi i}{n}} - x }{xe^{\frac{2k\pi i}{n}}}\right). [/imath] I have basically finished, because [imath]\phi(n)[/imath] is even so I can switch around the terms in the parantheses, but I still need the products of primitive nth roots of unity to be equal to 1. | 2757761 | Does there exist a formula for product of the primitive [imath] n [/imath]th roots of unity.
I know there is a formula for the sum of the primitive [imath] n [/imath]th roots of unity which is the Mobius function of [imath] n [/imath]. See: The Möbius function is the sum of the primitive [imath]n[/imath]th roots of unity. I am curious about the existence of a formula for the product of the primitive [imath] n [/imath]th roots of unity. |
2763504 | Convergence of [imath]a_{n+2} = \sqrt{a_n} + \sqrt{a_{n+1}}[/imath]
Let [imath]a_1[/imath] and [imath]a_2[/imath] be positive numbers and suppose that the sequence {[imath]a_n[/imath]} is defined recursively by [imath]a_{n+2} = √a_n + √a_{n+1}[/imath]. Show that this sequence is convergent. So, I have been able to show the convergence taking three different cases namely, Case 1: Both [imath]a_1[/imath] , [imath]a_2[/imath] <4, then I proved that the sequence will be monotonically increasing as well as bounded and so converging Case 2: Both [imath]a_1[/imath], [imath]a_2[/imath] >4, in this case the sequence is monotonically decreasing and bounded and hence convergent. Case 3: One of [imath]a_1[/imath] and [imath]a_2[/imath] is <4 & other >4, lets say, [imath]a_1[/imath]<4<[imath]a_2[/imath] in this case sequence will alternatively increase and decrease i.e. [imath]a_{2n-1}[/imath] will be increasing and [imath]a_{2n}[/imath] will be decreasing and [imath]a_{2n}-a_{2n-1}[/imath] will converge to Zero. My question is that is there any general method through which we don't have to take all these cases and can prove the convergence of series in more generality? | 574549 | How to prove this limit exists: [imath]a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}[/imath]
Question: Consider a sequence [imath]\{a_{n}\}[/imath] such that [imath]a_{1},a_{2}>0[/imath], and for all [imath]n \in \mathbb{N}[/imath] we have: [imath]a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}[/imath] Prove that : [imath]\displaystyle\lim_{n\to\infty}a_{n}[/imath] exists and find this limit. My work: If this limit exists, let [imath]\displaystyle\lim_{n\to\infty}a_{n}=x>0[/imath]; then we have [imath]x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4[/imath] But I can't see how to prove the limit of [imath]\{a_{n}\}[/imath] exists. Thank you for you help! |
2763723 | the number of group homomorphism
How many group homomorphism are there from [imath] \mathbb{Z}_{20} [/imath] onto [imath] \mathbb{Z}_{8} [/imath]? How many are there to [imath] \mathbb{Z}_{8} [/imath] ? In first question is onto but second question is to what is the difference between two question? | 1902661 | How many homomorphisms are there from [imath]\Bbb Z_{20}[/imath] onto [imath]\Bbb Z_{8}[/imath]? How many are there to [imath]\Bbb Z_{8}[/imath]?
How many homomorphisms are there from [imath]\Bbb Z_{20}[/imath] onto [imath]\Bbb Z_{8}[/imath]? How many are there to [imath]\Bbb Z_{8}[/imath]? I can see that there are no onto homomorphisms because [imath]|\Bbb Z_{20}|/ |\Bbb Z_{8}|[/imath] is not an integer, but for "to" I can't figure how to solve this. I can see that the trivial homomorphism and the natural one [imath]\phi(x) = x\mod8[/imath] work, but I can't see how I would find others. I can also see that [imath]|\phi(\Bbb Z_{20})| = 1, 2, [/imath] or [imath]4[/imath] because it divides [imath]\Bbb Z_{20}[/imath] and [imath]\Bbb Z_{8}[/imath], but from here I'm stuck too. |
2765520 | Show that the matrix [imath]\big(\frac{1}{a_i + a_j}\big)_{n\times n}[/imath] is positive semidefinite
Given positive real numbers [imath]a_1,\dots ,a_n[/imath], how can I prove that the symmetric matrix composed of the entries [imath]\frac{1}{a_i + a_j}[/imath] is positive semi-definite *this is not a hw question but a self study question from the matrix theory book by xinghzi zhan. | 1627332 | Postive-semidefiniteness of matrix with entries [imath]1/(a_i+a_j)[/imath]
Let [imath]a_1, \ldots, a_n[/imath] be a set of positive numbers. Define a matrix [imath]M_{ij} = \frac{1}{a_i+a_j}[/imath]. I'm trying to prove that [imath]M[/imath] is positive-semidefinite. The hint says to use the fact that [imath]\int_{0}^{\infty} e^{-sx}\; dx = \frac{1}{s}[/imath] if [imath]s > 0[/imath]. However I don't know how this hint is useful. I've tried choosing an arbitrary vector [imath]x[/imath] and substituting [imath]x^{\intercal}Mx = \sum_{i}\sum_{j} \frac{x_ix_j}{a_i+a_j}[/imath] into [imath]s[/imath] and using properties of exponents to simplify the equation into something that is clearly positive, but without any luck. The denominator [imath]\frac{1}{a_i+a_j}[/imath] is simply too difficult to work with. At this point I think I'm just missing some trick that I don't know. Any help would be appreciated. |
2765592 | How can [imath]e^{i\pi}+1[/imath] be zero?
So, most of us are familliar with Euler’s equation stating that [imath]e^{i\pi}+1=0[/imath]. But I was wondering: how can an irrational number to the power of another irrational number equal a whole integer? And if that works, then how can [imath]e^{i\pi}+1=0[/imath] if [imath]i[/imath] isn’t even real? | 326150 | How would you explain why [imath]e^{i\pi}+1=0[/imath] to a middle school student?
Hi I was asked by a friends child who is in middle school why [imath]e^{i\pi}+1=0[/imath]. Now I couldn't think of a way to explain it so he would understand. Albert Einstein once said “If you can't explain it simply, you don't understand it well enough” so how would I explain it. |
2763621 | Prove the following equality involving [imath]f'(0)[/imath]
I was doing the following problem: [imath]f[/imath] is an analytic function defined in a neighborhood of [imath]0[/imath], and [imath]f'(0)\neq 0[/imath]. Show that [imath]\frac{1}{f'(0)}=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{f(z)-f(0)}dz,[/imath] provided that [imath]r>0[/imath] is sufficiently small. I was trying to use the Cauchy integral formula for the inverse of [imath]f[/imath], but it doesn't seem to give the desired integral. Any help would be appreciated. | 2704787 | Prove that [imath]\frac{1}{f'(a)} = \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz[/imath]
Let [imath]f[/imath] be analytic on an open set [imath]U[/imath], let [imath]a \in U[/imath], and [imath]f'(a)\neq 0[/imath]. Show that \begin{align*} \frac{1}{f'(a)} = \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz \end{align*} where [imath]C[/imath] is a small circle centered at a. I thought that this is very similar with Cauchy's integral formula. But, I can't apply the Cauchy's integral formula. How to apply that? Any help is appreciated.. Thank you! |
2765706 | Show that [imath]\operatorname{rank}(A)=\operatorname{rank}(B)[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be two [imath]n\times n [/imath] real matrices, satisfying [imath]A^2=A,\,\, B^2=B[/imath]. Suppose [imath]A+B-I[/imath] is invertible. Show that [imath]\operatorname{rank}(A)=\operatorname{rank}(B)[/imath]. Since [imath] A^2=A,\,\, B^2=B[/imath] we see that [imath]A[/imath], [imath]B[/imath] are either singular matrices or matrices with determinant [imath]1[/imath]. Any ideas on how to proceed from here? | 178016 | Show that [imath]\operatorname{rank}(A) = \operatorname{rank}(B)[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be [imath]n\times n[/imath] real matrices such that [imath]A^2=A[/imath] and [imath]B^2=B[/imath]. Suppose that [imath]I-(A+B)[/imath] is invertible . Show that [imath]\operatorname{Rank}(A)=\operatorname{Rank}(B)[/imath]. I proceed in this way: Note that [imath]A(I-(A+B))=A-A^2-AB=A-A-AB=-AB[/imath] and similarly [imath]B(I-(A+B))=-AB[/imath]. So [imath]A(I-(A+B))=B(I-(A+B))[/imath]. Since [imath]I-(A+B)[/imath] is invertible we get [imath]A=B \Rightarrow \operatorname{Rank}(A)=\operatorname{Rank}(B)[/imath]. |
2766036 | Convergence of Sequence of arithmetic mean
If [imath]a_{n+2}=\frac {a_{n+1}+a_{n}}{2}[/imath] [imath]\forall n[/imath]>0 ,I have to show that [imath]a_{n}\to \frac {a_{1}+2a_{2}}{3}[/imath]. I don't know this problem is easy or difficult as intially I was posting here my query about question but at the time of writting problem I got this solution .If any mistake please tell me .. | 1566479 | Find the limit of the recursive sequence
Definition of the sequence : [imath]a_1=a;\\a_2=b;\\[/imath] and [imath]\ \ \ a_{n+2}={{a_n+a_{n+1}}\over2}[/imath] for [imath]n\geq 1[/imath]. Find the limit of this sequence in terms of [imath]a[/imath] and [imath]b[/imath]. Now in this case , taking [imath]\lim_{n\rightarrow \infty}[/imath] on both sides does not help at all . So what I did was try to find the [imath]n[/imath]-th term in therms of [imath]a[/imath] and [imath]b[/imath] and then take [imath]n[/imath] to [imath]\infty.[/imath] So , using the given recursion formula I wrote down like [imath]7[/imath]-[imath]8[/imath] terms, grouped them in various ways but still could not get the pattern of the [imath]n[/imath]-th term. Terribly sorry for the lack of work/context in this post of mine but I tried for like an hour to figure things out but can't really put those scribbles down here. Please give me some hints as to how to find the answer . Thank you. |
2766523 | [imath]|\{1≤x≤p^2:p^2│x^{p-1}-1\}|=p-1[/imath]
Let [imath]p[/imath] be a prime number. Let [imath]S_p=\{1≤x≤p^2:p^2│x^{p-1}-1\}[/imath]. Prove that [imath]|S_p|=p-1[/imath]. I managed to prove that [imath]|S_p|\geq p-1[/imath]. I took [imath]g[/imath], a primitive root modulo [imath]p^2[/imath], and then proved that [imath]\langle g^p\rangle\subseteq S_p[/imath] ... Now, in order to finish, it is needed to show that [imath]|S_p|\leq p-1[/imath], for this I wanted to say something about number roots of the polynomial [imath]x^{p-1}-1[/imath] over [imath]\mathbb{F}_p[/imath], but it doesn't help since we are dealing with numbers in [imath]\mathbb{Z}/p^2\mathbb{Z}[/imath]. | 1724859 | Show that the congruence [imath]x^{p-1} \equiv 1 \pmod {p^2}[/imath] has precisely [imath]p-1[/imath] solutions modulo [imath]p^2[/imath]
Let [imath]p[/imath] be a prime number. Show that the congruence [imath]x^{p-1} \equiv 1 \pmod {p^2}[/imath] has precisely [imath]p-1[/imath] solutions modulo [imath]p^2[/imath] I know the statment above is true for modulo [imath]p[/imath], but how is this imply to modulo [imath]p^2[/imath] |
1538405 | A question about the number of points of the preimage of a regular value.
I would like to have some explanation about a fact which is written in the first chapter of Milnor's book "topology, from the differentiable viewpoint". The fact is the following: Suppose [imath]M,N[/imath] be two differentiable manifolds of the same dimension. Let [imath]f: M \rightarrow N[/imath] be a differentiable map with [imath]M[/imath] compact and let [imath]y \in N[/imath] be a regular value. We define #[imath]f^{-1}(y)[/imath] to be the number of points in [imath]f^{-1}(y)[/imath], thus [imath]f^{-1}(y)[/imath] does not contain any critical point. And now the part which I do not understand: the book says that an observation that can be made about #[imath]f^{-1}(y)[/imath] is that is locally constant as a function of [imath]y[/imath] (where [imath]y[/imath] ranges only through regular values). I don't understand how to figure out this fact, I tough considering #[imath]f^{-1}: f^{-1}(y) \rightarrow \mathbb{N}[/imath], but still remains unclear to me why it should be a locally constant function. Could someone please can explain to me this fact maybe also (if I am not asking too much) giving a simple example. | 936272 | A doubt from Milnor's "Topology from a Differentiable Viewpoint".
This is a doubt from Milnor's "Topology from a Differentiable Viewpoint". For a smooth [imath]f:M\to N[/imath], with [imath]M[/imath] compact, and a regular value [imath]y\in N[/imath], we define [imath]n(f^{-1}(y))[/imath] to be the number of points in [imath]f^{-1}(y)[/imath]. It is easy to see that [imath]n(f^{-1}(y))[/imath] is finite. It can also be seen that [imath]n(f^{-1}(y))[/imath] is locally constant; i.e. there is a neighbourhood of [imath]y[/imath] such that for every point [imath]y'[/imath] in that neighbourhood, [imath]n(f^{-1}(y'))=n(f^{-1}(y))[/imath]. The proof for the statement in bold is the following: Let [imath]x_1,x_2,\dots,x_n[/imath] be the points of [imath]f^{-1}(y)[/imath], and choose disjoint neighbourhoods [imath]U_1,U_2,\dots,U_n[/imath] of these points, which are mapped diffeomorphically into neighbourhoods [imath]V_1,V_2,\dots,V_n[/imath] of [imath]y[/imath]. The required neighbourhood is [imath](V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)[/imath]. I don't understand how [imath](V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)[/imath] is open. I know that [imath](V_1\cap V_2\cap\dots V_n)[/imath] is open. However, how can one prove that [imath]f(M-U_1-U_2-\dots U_n)[/imath] is closed? I can see this working if [imath]f[/imath] is surjective, but not otherwise. |
2766475 | p-value, intuition about type-I error=[imath]\alpha[/imath]
In hypothesis testing consider this situation: Define [imath]\alpha=P\{\text{type I error}\}=P\{\text{ Rejecting } { H_0}\text{ when } H_0\text{ is true}\}[/imath]. Also define [imath]p-[/imath]value like this What is the probability of observing a data which is similar to the one on hand, or more extreme, if [imath]H_0[/imath] happens to be true. This probability will be called the [imath]p-[/imath]value. How do I get some inutition about this: A test with a small [imath]p-[/imath]value indicates that the null hypothesis is less plausible than the alternative hypothesis and in this case [imath]H_0[/imath] is rejected. | 2748749 | Why small p-value rejects [imath]H_0[/imath]
I can't understand why people seek small [imath]p[/imath]-value. I mean, [imath]p[/imath]-value is the smallest level at which [imath]H_0[/imath] can be rejected. Making level small we minimize the probability of type [imath]1[/imath] error, which as I think means that we are getting more evidence against [imath]H_0[/imath], since now the probability of accidentally rejecting [imath]H_0[/imath] when it is true decreases [imath]=>[/imath] probability of accepting [imath]H_0[/imath] when it is true is becoming bigger, because [imath]P_{H_0}(test = 1)=1-P_{H_0}(test = 0)[/imath] When [imath]p[/imath]-value is small we reject [imath]H_0[/imath], but it seems to me that when [imath]p[/imath]-value is small it must be vice versa, because the small [imath]p[/imath]-value corresponds to the higher probability of accepting [imath]H_0[/imath] when it is true. I saw the answers that you suggested... but none of them helps. I know the 'interpretation' of p-value, I can imagine it as an area under curve too, but I can't understand the relation to what I wrote above. |
2765789 | Are complex numbers subject to different rules of math?
From what I know, the rule to distribute exponents is like: [imath](a b)^x = a^x b^x[/imath] Thus, if [imath]a = \sqrt 2[/imath] and [imath]b = \sqrt 3[/imath], [imath]ab = \sqrt 6[/imath]. However, the imaginary unit [imath]i = \sqrt{-1}[/imath] has a different behavior, because if I take [imath]a = -2[/imath], [imath]b = -3[/imath] and [imath]x = 1/2[/imath]: [imath]{[(-2) (-3)]}^{1/2} = (-2)^{1/2} (-3)^{1/2} = \sqrt{-2} \sqrt{-3} = i\sqrt 2 i \sqrt 3 = - \sqrt 6[/imath] Although, before I learned complex numbers, I thought, [imath]{[(-2) (-3)]}^{1/2} = [6]^{1/2} = \sqrt 6[/imath] What am I making wrong here and which is the right answer? | 2714618 | Why [imath]e^{i(π/3)} \ne -1[/imath]?
I understand why [imath]e^{i\pi} = -1[/imath]and as a result [imath]e^{i2\pi}=\left(e^{i\pi}\right)^2=1.[/imath] These results can be confirmed using Euler's formula But why does [imath]e^{i\pi/3}\neq -1[/imath] as we can write it [imath](e^{i\pi})^{1/3}[/imath] and as such [imath](-1)^{1/3}[/imath] the cubic root of [imath]-1[/imath] is [imath]-1[/imath] itself. Yet using Euler's formula we get [imath]e^{i\pi/3}=\frac{1}{2}+i\frac{\sqrt 3}{2}[/imath]. |
2767559 | Closure of Connected Set is Connected
I wanted To show that if E is connected subset then [imath]\bar E[/imath] is also Connected By Using Definition. There is one answer regarding this question but that does not use Definition So this question is not repeated .My attempt: On contrary [imath]\bar E[/imath] is disconnected so [imath]\bar E[/imath]=[imath]A \cup B[/imath] Where A and B are disjoint open set .I had only info about E is connected that means E can not be written like above . From Hint:E[imath]\subset[/imath] [imath]\bar E[/imath]=[imath]A \cup B[/imath] So E=[imath](A\cap E) \cup (B\cap E)[/imath] [imath](A\cap E)[/imath] and [imath](B\cap E)[/imath] are both open and disjoint Because In case x[imath]\in(A\cap E) \cap (B\cap E)[/imath] Then x[imath]\in A \cap B[/imath] Which contradicts with Given .Is this is enough to show ? | 513660 | If [imath]E[/imath] is a connected space, then so is [imath]\overline{E}[/imath].
I have to prove that if [imath]E[/imath] is a connected space, then so is [imath]\overline{E}[/imath] (the closure of [imath]E[/imath]) a connected space. I tried to prove the contrapositive. So suppose that [imath]\overline{E}[/imath] is not connected in a metric space [imath]X[/imath]. This implies there are [imath]A\subset X[/imath] and [imath]B\subset X[/imath] such that [imath]A\cap B=\emptyset[/imath], [imath]A\cap \overline{E}\neq \emptyset[/imath], [imath]B\cap \overline{E}\neq \emptyset[/imath] and [imath]\overline{E}\subset A\cup B[/imath]. I will try to prove that this [imath]A[/imath] and [imath]B[/imath] will also work for [imath]E[/imath]. Suppose that [imath]A\cap E=\emptyset[/imath] and [imath]B\cap E=\emptyset[/imath], this implies that [imath]A\cup B[/imath] is a subset of the limit points of [imath]E[/imath], because [imath]A\cap \overline{E}\neq \emptyset[/imath] and [imath]B\cap \overline{E}\neq \emptyset[/imath]. This contradicts the fact that [imath]\overline{E}\subset A\cup B[/imath], so we have to conclude that [imath]A\cap E\neq \emptyset\neq B\cap E[/imath]. Since [imath]E\subset\overline{E}\subset A\cup B[/imath], we have that [imath]E[/imath] is also not connected. My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? Thanks in advance! Edit1: The definition we use for a connected space [imath]E[/imath] in a metric space [imath]X[/imath] is that if [imath]E[/imath] is connected, then there are no open [imath]A,B\subset X[/imath] such that [imath]A\cap B=\emptyset[/imath], [imath]A\cap E\neq \emptyset\neq B\cap E[/imath] and [imath]E\subset A\cup B[/imath]. Edit2: [imath]A[/imath] and [imath]B[/imath] must be open. |
2767417 | Invertible matrices with integer entries has to be permutation matrices
Let [imath]A[/imath] and [imath]B[/imath] be [imath]n\times n[/imath] matrices with integer entries. Show that if [imath]B=A^{-1}[/imath] then [imath]A[/imath] and [imath]B[/imath] are permutation matrices (matrix obtained by permuting rows of the identity matrix). If the entries are non negative integers then it is quite easy to show, but if the entries also involve negative integers then I couldn't show it. | 2139766 | Matrices that can only be permutation matrices
How to prove that if [imath]A[/imath] and [imath]B[/imath] are two [imath]n×n[/imath] matrices with non negative integer entries such that [imath]AB=I[/imath].Then how to show [imath]A[/imath] and [imath]B[/imath] are permutation matrices.Permutation matrices are matrices with the columns just a permutation of the identity matrix. How do i proceed.Can somebody hint. |
2767828 | [imath]f'(0)[/imath] when [imath]f(x)=x^2\sin(1/x)[/imath]
[imath]f'(0)[/imath] when [imath]f(x)=x^2\sin(1/x)[/imath], [imath]f(0)=0[/imath] This is presented as an MCQ and answer is given as "[imath]f[/imath] is differentiable at every [imath] x[/imath] but [imath]f'[/imath] is discontinous at [imath]0[/imath] " I calculated LHD and RHD at [imath]0[/imath] as [imath]0[/imath] and [imath]f'(x)=2x\sin(1/x)-\cos(1/x)[/imath] which is not defined at 0 thus the answer should be "[imath]f[/imath] is not differentiable at [imath]0[/imath]" Is this a misprint or am I missing something? | 2212011 | Continuity of differentiable function [imath]x^2\sin(1/x)[/imath]
Assume this function has a value [imath]0[/imath] at [imath]x=0[/imath]. If the right hand derivative of a derivable function is equal to the right hand limit of the derivative function (same for the left), why aren't functions like the one above continuously differentiable? For a function to be continuous, limits should be finite and equal to value at that particular point. For this function derivative using first principle yields [imath]0[/imath]. The RHD, LHD is [imath]0[/imath], I think. So why isn't the derivative continuous? For finding the RHD and LHD, I have seen answers resorting to using known derivatives. My question is: why don't we use the first principle itself? Why do we get different answers then? |
2767996 | Proving [imath]\text{Gal}(K/F) \cong \text{Gal}(K/E_1) \times \text{Gal}(K/E_2)[/imath]
Let [imath]K/F[/imath] be a Galois extension. Also, let [imath]E_1[/imath] and [imath]E_2[/imath] be two Galois extensions of [imath]F[/imath] in [imath]K[/imath] with [imath]E_1E_2 = K[/imath] and [imath]E_1 \cap E_2 = F[/imath]. Prove that [imath]G = \text{Gal}(K/F) \cong \text{Gal}(K/E_1) \times \text{Gal}(K/E_2)[/imath]. I think this is just an application of results of the Fundamental theorem of Galois theory (and maybe the Chinese Remainder Theorem?), but I'm uncertain on how to prove it. Since we know [imath]E_1/F[/imath] and [imath]E_2/F[/imath] are Galois extensions, we have [imath]\text{Gal}(K/E_1)[/imath] and [imath]\text{Gal}(K/E_1)[/imath] are normal subgroups of [imath]G[/imath]. We get [imath]\text{Gal}(E_1/F) \cong G/\text{Gal}(K/E_1)[/imath] and similarly [imath]\text{Gal}(E_2/F) \cong G/\text{Gal}(K/E_2)[/imath]. I don't know what to do from here, or if these statements if even help. Thoughts? | 35429 | Galois group of the product of two fields being the product of Galois groups?
I'm just wondering if the following theorem is reasonable and whether the proof is makes sense or not? Also I have an application of it which I was trying to do earlier but I made a lot of mistakes so I would be very grateful if anyone would give some comments on this in case I have some very strong misunderstandings about Galois theory. Theorem Let [imath]E_1[/imath] and [imath]E_2[/imath] be Galois field extensions of [imath]F[/imath] with trivial intersection ([imath]E_1 \cap E_2 = F[/imath]), then [imath]E_1 E_2[/imath] has Galois group [imath]\text{Gal}(E_1/F) \times \text{Gal}(E_2/F)[/imath]. Proof [imath]E_1[/imath] and [imath]E_2[/imath] are splitting fields (over [imath]F[/imath]) of some polynomials [imath]f_1[/imath] and [imath]f_2[/imath] so [imath]E_1 E_2[/imath] is the splitting field of [imath]f_1 f_2[/imath] over [imath]F[/imath] and therefore it is Galois and we are justified using the "Gal" notation. Let [imath]\sigma_1 \in \text{Gal}(E_1/F)[/imath] and [imath]\sigma_2 \in \text{Gal}(E_2/F)[/imath] then by the fundamental theorem of Galois theory we can lift these both into elements of [imath]\text{Gal}(E_1 E_2/F)[/imath]. The claim is that [imath]\sigma_1 \sigma_2 = \sigma_2 \sigma_1[/imath], to see this consider an element [imath]\alpha \in E_1[/imath], we have both [imath]\sigma_2 \alpha = \alpha[/imath] and [imath]\sigma_2 \sigma_1 \alpha = \sigma_1 \alpha[/imath] because it is invariant with respect to [imath]\sigma_2[/imath], but we can combine these equalities to get [imath]\sigma_1 \sigma_2 \alpha = \sigma_2 \sigma_1 \alpha[/imath]. Both groups [imath]\text{Gal}(E_i/F)[/imath] ([imath]i = 1,2[/imath]) are included in [imath]\text{Gal}(E_1 E_2/F)[/imath] and the degrees say there are no more elements, furthermore since they commute we can conclude [imath]\text{Gal}(E_1 E_2/F) = \text{Gal}(E_1/F) \times \text{Gal}(E_2/F)[/imath]. So hopefully that theorem is, if not correct, fixable.. and can be used to prove that [imath]\mathbb Q (\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})[/imath] has Galois group [imath]C_2^m[/imath] ([imath]m \le n[/imath]) over [imath]\mathbb Q[/imath]? |
1196461 | Why is de Rham cohomology invariant under deformation retraction, or, more generally, homotopy equivalence?
For example, I know that [imath]H^k(S^n \times I^m) = H^k(S^n)[/imath] (where [imath]I[/imath] is an open interval) because there is a deformation retraction from [imath]S^n \times I^m[/imath] to [imath]S^n[/imath]. Why are the cohomology groups invariant under deformation retractions? | 129819 | Proof that two spaces that are homotopic have the same de Rham cohomology
I know this is true, but how do I prove it? Specifically, I'm trying to calculate the de Rham cohomology of the 3-sphere by using the Mayer-Vietoris sequence and covering [imath]S^3[/imath] with two hemispherical sets [imath]U, V[/imath], such that [imath]U[/imath] intersects [imath]V[/imath] in a "2-band", an equatorial band homotopic to a 2-sphere. |
2769277 | Let [imath]p[/imath] and [imath]q[/imath] be distinct primes. [imath]\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}[/imath]
Let [imath]p[/imath] and [imath]q[/imath] be distinct primes. Then the number of positive integer solutions of the equation [imath]\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}[/imath] are: I have no idea how to start. Any hints would be helpful. | 1343056 | Solve [imath]\frac1x+\frac1y=\frac1{pq}[/imath]
For [imath]x,y\in\mathbb{N}[/imath] how many ordered pairs [imath]\left(x,y\right)[/imath] satisfy [imath]\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{pq}[/imath] where [imath]p,q[/imath] are distinct primes? |
2768590 | If f is uniformly differentialable on [imath](a,b)[/imath] then [imath]f'[/imath] is continuous on [imath](a,b)[/imath]?
If f is uniformly differentialable on [imath](a,b)[/imath] then [imath]f'[/imath] is continuous on [imath](a,b)[/imath]? I just knew the definition A function [imath]f:(a,b)\to R[/imath] is said to be uniformly differentialable iff f is differntiable on [imath](a,b)[/imath], and for each [imath]\epsilon>0[/imath] there is [imath]\delta>0[/imath] such that [imath]0<|x-y|<\delta[/imath] and for any [imath]x,y \in (a,b)[/imath] imply that [imath]\left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|<\epsilon[/imath] Is that this statement is true : f is uniformly differentialable on [imath](a,b)[/imath] then [imath]f'[/imath] is continuous on [imath](a,b)[/imath]? | 2231535 | show that if [imath]f[/imath] is uniformly differentiable then prove that [imath]f'[/imath] is continuous
How would I go about showing that if [imath]f[/imath] is uniformly differentiable then [imath]f'[/imath] is continuous. my attempt: A differentiable function [imath]f:[a,b]\to \Bbb R[/imath] is said to be uniformly differentiable on [imath][a,b][/imath] if [imath]\forall \epsilon>0 \exists \delta>0:\forall x,y\in[a,b][/imath] we have [imath]0<|x-y|<\delta\implies |\frac{f(x)-f(y)}{x-y}-f'(y)|<\epsilon[/imath] I don't know where to go from here, any help would be highly appreciated. |
2769368 | Modified alternating harmonic series
I am having trouble characterizing the types of alternating with-differing-period harmonic series which converge. For example, [imath]\sum_{n\in \mathbb{N}} (-1)^n\frac{1}{n}[/imath] converges, but what about [imath]1+\frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} \dots[/imath] ([imath]2[/imath] positive terms and [imath]1[/imath] negative term). Can we characterize how far the gaps can be between negative terms? | 463991 | Does the series [imath]1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots[/imath] converge?
Does the following variant of the harmonic series converge? If it diverges (which I think it does), can I know if it diverges to [imath]\infty[/imath] or has no limit? Note that the series is not alternating in the classical sense of the word. [imath]1 + \frac{1}{2} -\frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots[/imath] The generic term of the series would have to be something like, [imath]a_n = \left\{\begin{array}{ll} -\frac{1}{n}, & \text{if } 3 \mid n \\ \;\, \,\, \frac{1}{n}, & \text{otherwise}\end{array}\right.[/imath] I'm not sure if it's very helpful. The terms divisible by 3 are negative, and the others are positive. Is there a way to decide and prove whether an alternating series of this sort (e.g. with a period other than 2) converges? Or one where terms are positive or negative according to some other rule? Almost all convergence tests I've come across are generally limited to either simple alternating series or where all the terms are positive. |
2769340 | Test for convergence of the series [imath]\sum\limits_{n=3}^\infty\frac{1}{(\ln \ln n)^{\ln n}}[/imath]
Test for convergence of the series [imath]\sum\limits_{n=3}^\infty\frac{1}{(\ln \ln n)^{\ln n}}[/imath] I test the series by this steps [imath](\ln \ln n)^{\ln n} = (e^{\ln \ln \ln n})^{\ln n}= (e^{ \ln n})^{\ln \ln \ln n }=n^{\ln \ln \ln n}[/imath]. Then [imath](\ln{ {\ln \ln{n}}}) > 2[/imath]. Thus, [imath]\frac{1}{(\ln \ln n)^{\ln n}} < \frac{1}{n^2}[/imath] and thus the series converges by the comparison test. Is this work..!? are there some wrongs ..!? | 1034789 | Nature of the series [imath]\sum\limits_{n=3}^{\infty} \frac{1}{(\log\log n)^{\log n}}[/imath]
Does [imath]\sum\limits_{n=3}^{\infty} \dfrac{1}{(\log\log n)^{\log n}}[/imath] converge or diverge ? I tried some tests , but nothing conclusive is coming . Pleas help |
2769757 | How many even integers between [imath]100[/imath] and [imath]1000[/imath] have distinct digits?
How many even integers between [imath]100[/imath] and [imath]1000[/imath] have distinct digits? I tried: [imath]9 \cdot 9 \cdot 5=405[/imath] [imath]9[/imath] [[imath]9,8,7,6,5,4,3,2,1[/imath] ([imath]9[/imath])] [imath]9[/imath] [[imath]9, 8, 7, 6, 5, 4, 3, 2, 1, 0[/imath] ([imath]10[/imath])] - the first one picked ([imath]1[/imath]) [imath]5[/imath] ([imath]0,2,4,6,8[/imath]) But the answer of exercise is [imath]328[/imath]. I don't know why. | 1435569 | How many positive integers less than 1000 have distinct digits and are even?
I am not looking for an answer on this. Just need to clarify why my approach is failing - [imath]N_1 + N_2 + N_3[/imath], i.e. single digit, double digit, 3 digit single [imath]= 2, 4, 6, 8[/imath], i.e 4 double = X non-zero [imath]= 8 \cdot 4 = 32[/imath] X zero [imath]= 9 \cdot 1 = 9[/imath] Now the confusing part three digit, breaking into 4 cases X zero zero [imath]= 9[/imath] X nz nz [imath]= 7 \cdot 8 \cdot 4 = 224[/imath] X z nz [imath]= 8 \cdot 1 \cdot 4 = 32[/imath] X nz z [imath]= 8 \cdot 9 \cdot 1 = 72[/imath] Three digit total comes to [imath]= 9 + 224 + 32 + 72 = 337[/imath]. This answer is wrong and it should be [imath]328[/imath]. What am I missing in the logic? Please suggest. |
2769332 | Q: Sequence definition of non-convergence (Not diverging definition)
I have that convergence requires that the following be true [imath] \forall\epsilon>0\exists N: n\ge N:|a_n-a|<\epsilon [/imath] for some sequence [imath]\{a_n\}_{n\in\mathbb{N}}[/imath] that converges to [imath]a[/imath] for [imath]n\to\infty[/imath]. I want to use the reverse of this requirement. I do not know how to express it but I am guessing [imath] \exists\epsilon>0\forall N:n\ge N:|a_n-a|\ge\epsilon [/imath] it is expressed like the above. Can you help me out? | 545669 | Negating the Definition of a Convergent Sequence to Find the Definition of a Divergent Sequence
My task is to write a precise mathematical statement that "the sequence [imath](a_n)[/imath] does not converge to a number [imath]\mathscr l[/imath]" So, I have my definition of a convergent sequence: "[imath]\forall\varepsilon>0[/imath] [imath]\exists N\in\Bbb R[/imath] such that [imath]|x_n -\mathscr l|<\varepsilon[/imath] [imath]\forall n \in \Bbb N[/imath] with [imath]n>N[/imath]" Would the correct negation of this be "[imath]\forall\varepsilon>0[/imath] [imath]\exists N\in\Bbb R[/imath] such that [imath]|x_n -\mathscr l|>\varepsilon[/imath] [imath]\forall n \in \Bbb N[/imath] with [imath]n>N[/imath]"? It doesn't seem that this is the answer as the next part of my task is to prove that a sequence is divergent using my formed proof, but it'd be difficult to do since it's a general proof of divergence and not just a proof that [imath](a_n)[/imath] doesn't converge a specific number [imath]\mathscr l[/imath] Perhaps I should find a prove that [imath](a_n)[/imath] tends to [imath]\pm\infty[/imath]? This is more simple but it does not include monotone sequences such as [imath]x_n:=(-1)^n[/imath]. Can someone assist me with this task? All comments and answers are appreciated. |
2770345 | Given the sum of this series, find the sum of another series.
I've stumbled across this problem that asks, given this series whose sum is that: [imath]\sum_{n=1}^∞ \frac{1}{n^2} = \frac{{\pi^2}}{6}[/imath] Find the sum of this series: [imath]\sum_{n=1}^∞ \frac{1}{(2n-1)^2} [/imath] The second function is just the odd function of the first function, however, I do not know where to proceed from there. | 402451 | Evaluating a summation of inverse squares over odd indices
[imath] \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}[/imath] I want to evaluate this sum when [imath]n[/imath] takes only odd values. |
2769890 | Why is [imath]A\rightarrow \operatorname{im}(f)[/imath] an epimorphism in an abelian category
This is an exercise in Ravil Vakil's Algebraic Geometry notes labeled hard Let [imath]f: A\rightarrow B[/imath] be a morphism in an abelian category. Then [imath]f[/imath] factors through a morphism [imath]f':A\rightarrow \operatorname{im}(f)[/imath] (recall that [imath]\operatorname{im}(f)[/imath] is the (domain of the) kernel of the cokernel of [imath]f[/imath]). Why is [imath]f'[/imath] an epimorphism? That is why does [imath]g\circ f'=0[/imath] imply [imath]g=0[/imath] for any [imath]g: \operatorname{im}(f)\rightarrow C[/imath] ? | 1884741 | [imath]A \longrightarrow \text{Im} f[/imath] is an epimorphism and a cokernel of ker [imath]f \longrightarrow A[/imath] in every abelian category
How can I verify the following statement? The image of a morphism [imath]f : A \longrightarrow B[/imath] is defined as [imath]\text{Im}(f) = \text{Ke}r(\text{coke}r f)[/imath] whenever it exists (e.g., in every abelian category). The morphism [imath]f : A \longrightarrow B[/imath] factors uniquely through [imath]\text{Im} f \longrightarrow B[/imath] whenever [imath]\text{Im} f[/imath] exists, and [imath]A \longrightarrow \text{Im} f[/imath] is an epimorphism and a cokernel of ker [imath]f \longrightarrow A[/imath] in every abelian category. |
2251510 | Is [imath]\mathbb Z[X]/(X^2-1)[/imath] a domain?
I have a doubt on something. I have a theorem that says that if [imath]R[/imath] is a ring, then [imath]R/I[/imath] is a domain iff [imath]I[/imath] is prime. Since [imath]X^2-1[/imath] is note prime, [imath]\mathbb Z[X]/(X^2-1)[/imath] is not a domain. But, [imath]X^2-1=(X-1)(X+1)[/imath], and since [imath]X-1[/imath] and [imath]X+1[/imath] are prime in [imath]\mathbb Z[X][/imath], there are coprime, and thus, by the chinese remainder [imath]\mathbb Z[X]/(X^2-1)\cong \mathbb Z[X]/(X-1)\times \mathbb Z[X]/(X+1),[/imath] since [imath]\mathbb Z[X]/(X-1)\times \mathbb Z[X]/(X+1)[/imath] is a domain, then so is [imath]\mathbb Z[X/(X^2-1)[/imath]. What is the mistakes here ? | 1750425 | Prove or disprove that the ring [imath]\mathbb{Z}[x]/(x^2-1)[/imath] is an integral domain
Prove or disprove that the ring [imath]\mathbb{Z}[x]/(x^2-1)[/imath] is an integral domain. It is easy to see that [imath]\mathbb{Z}[x]/(f)=\{P+(f): \deg(P)<\deg(f)\}[/imath]. If [imath]\deg(P) \geq \deg(f)[/imath], there exists [imath]q,r \in \mathbb{Z}[x][/imath] such that [imath]P=qf+r[/imath] with [imath]r=0[/imath] or [imath]\deg(r)< \deg(f) \implies P+(f)=qf+r+(f)=r+(f)[/imath]. So [imath]\mathbb{Z}[x]/(f) = \{0,1, x, (x-1), (x+1)\}[/imath]. But [imath](x-1)(x+1)= x^2-1 \equiv 0 \mod (x^2-1)[/imath]. Conclusion : Our ring is not an integral domain. Question : Am I right? Otherwise, is anyone could help me on the ambiguity of this problem? |
2770574 | How do I find the sum of the series -1^2-2^2+3^2+4^2-5^2… upto 4n terms?
I tried by giving [imath] S = \sum_{k=0}^{n-1} \left((4k+3)^2+(4k+4)^2-(4k+1)^2-(4k+2)^2\right) [/imath] but I am stuck here. I have no idea what to do next. The answer in my book says 4n(n+1). How can I get it? I tried expanding (4k+1)^2, etc. and got [imath] \sum_{k=0}^{n-1} (8k+20) [/imath], I tried to further expand this by taking it as [imath] 8\sum_{k=0}^{n-1}k + 20n = 8(n-1)(n)/2 + 20n = 4n^2+16n= 4n(n+4) [/imath] which is not the right answer. What have I done wrong? What must I do now? | 474980 | Find the sum of the series [imath]1^2-2^2+3^2-4^2+...-(2n)^2[/imath]
Find the sum of the series [imath]1^2-2^2+3^2-4^2+...-(2n)^2[/imath] I tried rewriting it as [imath]\sum_{r=1}^{2n}-1^{n+1}(r^2)[/imath] but it didn't help. Also, looked at re-arranging as [imath]1^2+3^2+5^2+7^2+...+(2n-1)^2[/imath] and [imath]-2^2-4-6^2-8^2-...-(2n)^2[/imath] Still couldn't get to the given answer of [imath]-n(2n+1)[/imath] |
2770556 | Will [imath]Z[X] / (x^2 +1)[/imath] be a PID?
Will [imath]Z[X] / (x^2 +1)[/imath] be a PID? My Try : It is not a PID. If [imath]R[/imath] = [imath]Z[X] / (x^2 +1)[/imath] were a PID then [imath]2[/imath] will be an irreducible element in [imath]R[/imath] and [imath]R/(2)[/imath] is a field. But [imath]x[/imath] will not have an inverse element in [imath]R/(2)[/imath]. contradiction Have I gone wrong anywhere? Please correct me if I have. Thank You in Advance. | 610126 | Is [imath]\mathbb{Z}[x]/(x^2+1)[/imath] isomorphic to [imath]\mathbb{Z}[i][/imath]?
Is [imath]\mathbb{Z}[x]/(x^2+1)[/imath] isomorphic to [imath]\mathbb{Z}[i][/imath]? My attempt is that try to define a mapping [imath]g[/imath] from [imath]\mathbb{Z}[x][/imath] to [imath]\mathbb{Z}[i][/imath] by [imath]g(f(x))= f(i)[/imath], for [imath]f(x)\in\mathbb{Z}[x][/imath]. If it is possible then [imath]\ker g[/imath] is [imath](x^2+1)[/imath]? Am I on the right track? Please Help. |
2770759 | If [imath]M[/imath] is a subspace of a Hilbert space, then [imath](M^\bot)^\bot[/imath] is the closure of [imath]M[/imath].
I have found a few questions similar to this one, namely If [imath]M[/imath] is a closed subspace of an Hilbert space [imath]H[/imath], then [imath]M^{\perp\perp}=M[/imath], but I am not assuming that [imath]M[/imath] is closed, so I believe this warrants a separate question. I know that I would like to use the orthogonal composition theorem, but I am not sure how to arrive there. Any hints for what proof approach I should take are appreciated. Is it as simple as a double inclusion proof? | 1043940 | Double orthogonal complement is equal to topological closure
So I'm in an advanced Linear Algebra class and we just moved into Hilbert spaces and such, and I'm struggling with this question. Let [imath]A[/imath] be a nonempty subset of a Hilbert space [imath]H[/imath]. Denote by [imath]\operatorname{span}(A)[/imath] the linear subspace of all finite linear combinations of vectors in [imath]A[/imath], and by [imath]\overline{\operatorname{span}(A)}[/imath] the topological closure of [imath]\operatorname{span}(A)[/imath] with respect to [imath]\|\cdot\|[/imath]. Also, let [imath]A^⊥ = \{h ∈ H : \langle h,f \rangle = 0, ∀f ∈ A\}[/imath] and [imath]A^{⊥⊥} = (A^⊥)^⊥[/imath]. Use orthogonal projection to prove that [imath]A^{⊥⊥} =\overline{\operatorname{span}(A)}[/imath]. The thing that trips me up is that we don't know much about [imath]A[/imath], like if I knew a little more, perhaps to show it's closed, then I can do the direct sum decomposition blah blah, but it also confuses me why we're using the complement of [imath]\operatorname{span}(A)[/imath]. Is it possible to show that [imath]\operatorname{span}(A)[/imath] is closed, then go from there? I know it might look a bit like a duplicate, but all the questions I find don't refer to orthogonal projection at all. Any hints would be greatly appreciated! |
2535703 | Prove there is no homomorphism from [imath]Z_{16} \oplus Z_2 [/imath] ont [imath]Z_4 \oplus Z_4[/imath]
Prove there is no homomorphism from [imath]Z_{16} \oplus Z_2 [/imath] onto [imath]Z_4 \oplus Z_4[/imath] I have no idea at all how to attempt such question. I have read some solutions in which we find the cardinality of [imath]kerm\phi[/imath] for some [imath]\phi mapping [/imath] that is onto using the first theorem of homomorphism. After that I don't understand anything Is there a general approach to these questions ? | 2354279 | No homomorphism from [imath]Z_{16}\oplus Z_{2}[/imath] onto [imath]Z_{4}\oplus Z_{4}[/imath].
For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours. Prove that no homomorphism exists from [imath]Z_{16}\oplus Z_2[/imath] onto [imath]Z_4\oplus Z_4[/imath]. Assume a homomorphism [imath]\phi[/imath] exists from [imath]Z_{16}\oplus Z_2[/imath] onto [imath]Z_4\oplus Z_4[/imath]. Then [imath]\phi[/imath] is an isomorphism. Note that the [imath]\left | \ker \phi \right |=2[/imath] The [imath]\ker \phi[/imath] is also a normal subgroup for [imath]Z_{16}\oplus Z_{2}[/imath]. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups. By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements. Hence, [imath]\ker\phi[/imath] = [imath]\left \{ (0,0),(8,0) \right \}[/imath] ,[imath]\left \{ (0,0),(0,8) \right \}[/imath],[imath]\left \{ (0,0),(8,1) \right \}[/imath], possibly. By the First isomorphism theorem: [imath]\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2[/imath] [imath]\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2[/imath] [imath]g\ker \phi \mapsto \phi(g)=\Psi(g\ker \phi)[/imath] Any help is appreciated. Edit: I know that [imath](2,0)[/imath] has order [imath]8[/imath] in [imath]Z_{16}\oplus Z_{2}[/imath] but any elements in [imath]Z_{4}\oplus Z_{4}[/imath] does not have order [imath]8[/imath] which would have solved the question at the outset. But I would like a different route. |
2770952 | Show [imath]f(z)=\int_{c} \frac{g(w)}{w-z}[/imath] [imath]dw[/imath] is analytic on any domain [imath]D[/imath] not containing points of [imath]C[/imath]
I know how to show that [imath]f(z)[/imath] is continuous but I'm stuck at showing that [imath]f(z)[/imath] is analytic. [imath]f(z)[/imath] is continuous: [imath]|f(z)-f(a)|=\left|\int_{c} \frac{g(w)}{w-z}-\frac{g(w)}{w-a} dw\right|=\left|\int_{c} \frac{g(w)(z-a)}{(w-z)(w-a)}dw\right|[/imath] By ML-Formula: [imath]\left|\int_{c} \frac{g(w)(z-a)}{(w-z)(w-a)}dw\right|\leq L(C)M|z-a|[/imath] where [imath]L(c)[/imath] is the length of the curve and M is bound of [imath]\frac{g(w)}{(w-z)(w-a)}[/imath] over the curve [imath]C[/imath] which exists since it is continuous on [imath]C[/imath]. Hence, we can let [imath]\delta = \dfrac{\epsilon}{L(C)(M+1)}[/imath] Any hints or solutions as to show [imath]f(z)[/imath] is analytic? | 836454 | How can i prove this problem about Cauchy integral?
Let [imath]V[/imath] be an open connected subset of [imath]\mathbb{C}[/imath]. Let [imath]f:\rightarrow \mathbb{C}[/imath] be a function continuous on a contour [imath]C[/imath]. Define [imath]F(z)=\int_C \frac{f(\zeta)}{\zeta - z} d\zeta[/imath] on [imath]V[/imath]. My question is: How can i show that [imath]F[/imath] is analytic at every point except [imath]C[/imath]? How can I show that [imath]F'(z)=\int_C \frac{f(\zeta)}{(\zeta - z)^2} d\zeta[/imath] where [imath]z[/imath] not in [imath]C[/imath]? |
2770664 | Show that [imath]Y_1=\frac{X_{(1)}}{X_{(2)}},Y_2=\frac{X_{(2)}}{X_{(3)}},\dots, Y_{n-1}=\frac{X_{(n-1)}}{X_{(n)}}[/imath], and [imath]Y_{(n)}=X_{(n)}[/imath] are independent
I am into order statistics lately, and I have a problem here. Let [imath]X_1,X_2,..,X_n[/imath] be a random sample from [imath]f(x)=1 , 0<x<1[/imath]. Show that [imath]Y_1=\frac{X_{(1)}}{X_{(2)}},Y_2=\frac{X_{(2)}}{X_{(3)}},\dots, Y_{n-1}=\frac{X_{(n-1)}}{X_{(n)}}[/imath], and [imath]Y_n=X_{(n)}[/imath] are independent. I know all the procedures to do the problem.The only thing is that the Jacobian of transformation is coming pretty bad and is quite tedious to calculate. Do you have a simpler method of calculating the Jacobian? | 2770213 | Showing that [imath]\frac{X_{(i)}}{X_{(n)}},i=1,2,...,n-1[/imath] and [imath]X_{(n)}[/imath] are independent for a population with df [imath]F(y)=y^{\theta}[/imath]
Let [imath]X_1,X_2,...,X_n[/imath] be i.i.d with df [imath]F(y)=y^{\theta}, 0<y<1, \theta>0[/imath]. Show that [imath]\frac{X_{(i)}}{X_{(n)}}[/imath], for [imath]i=1,2,...,n-1[/imath] and [imath]X_{(n)}[/imath] are independent. I found the population pdf to be [imath]f(x)=\theta x^{\theta -1} I_{0<x<1}[/imath] Now, [imath]f(x_1,x_2,...x_n)=\theta^n\prod_{i=1}^{n}{x_{i}}^{\theta - 1}[/imath] Now, the joint distribution of the order statistics: [imath]f_{1,2,..,n}(x_1,...x_n)=n! \theta^n\prod_{i=1}^{n}{x_{i}}^{\theta - 1} I_{0<x_1<x_2,...<x_n<1}[/imath] Now , we take the transformation [imath]Y_i=\frac{X_i}{X_n} [/imath] for [imath]i=1,2,..,n-1[/imath] and [imath]Y_n=X_n[/imath] Have I done correctly so far? Also, I found out the Jacobian of transformation [imath]={y_n}^{n-1}[/imath]. I am also finding it difficult to get the ranges of [imath]Y_i[/imath]'s. Help! |
2768281 | Determine if the following series converges or diverges: [imath]\sum _{n=1}^{\infty }\ln\left(\frac{n}{n+1}\right)[/imath]
I'm having trouble understanding if the following series converges or diverges: [imath] \sum _{n=1}^{\infty }\ln\left(\frac{n}{n+1}\right) [/imath] I've noticed that [imath]\lim _{x\to \infty \:}\ln \left(\frac{x}{x+1}\right) = 0[/imath] and therefore I can't deduce that it diverges, but other than that I'm really not sure what is the right way to go here. Any help is appreciated | 1747382 | Does [imath]\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})[/imath] converge/diverges??
Does this series converge or diverge? [imath]\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})[/imath] my thought is that, I can break it down to [imath]\sum_{k=1}^{\infty}\ln(k) - \sum_{k=1}^{\infty}\ln(k+1)[/imath] then maybe using comparison test or something? But I don't know exactly how to prove whether this series converges or diverges. Any help would be appreciated! |
1669836 | How many triplets [imath]\left(m,n,p\right)[/imath] where [imath]p[/imath] is prime satisfy [imath]m^2-3mn+p^2n^2=12p[/imath]?
How many triplets [imath]\left(m,n,p\right)[/imath] where [imath]p[/imath] is prime satisfy [imath]m^2-3mn+p^2n^2=12p[/imath]? I've tried taking both sides of the equation modulo 2, 3, and 4, but it hasn't gotten me very far. | 946121 | Diophantine equation with a prime number
Suppose [imath]m,n[/imath] are integers and [imath]p[/imath] is a prime number. Find [imath](p,n,m)[/imath] if [imath]m^2-3mn+n^2p^2 = 12p[/imath]. I have tried to use the quadratic formula on m,n,p but the three turned to be really messy. Also I tried to factor and take advantage of the fact [imath]p[/imath] is prime but that didn't really help. |
2771240 | Given a field, the field of algebraic elements over it in an algebraically closed field extension is again algebraically closed?
Let [imath]\mathbb F[/imath] be a field and [imath]\mathbb K [/imath] be an extension field of [imath]\mathbb F[/imath] such that [imath]\mathbb K[/imath] is algebraically closed. Let [imath]\mathbb L[/imath] be the field of all elements of [imath]\mathbb K[/imath] which are algebraic over [imath]\mathbb F[/imath]. Then [imath]\mathbb L_{|\mathbb F}[/imath] is an algebraic extension. My question is : Is [imath]\mathbb L[/imath] algebraically closed ? I am trying to prove the existence of algebraic closure, so please don't assume that every field has an algebraic closure. | 27647 | Is the sub-field of algebraic elements of a field extension of [imath]K[/imath] containing roots of polynomials over [imath]K[/imath] algebraically closed?
If I have a field [imath]K[/imath] and an extension [imath]L[/imath] of [imath]K[/imath] such that all (non-constant) polynomials in [imath]K[X][/imath] have a root in [imath]L[/imath], is the set of algebraic elements of [imath]L[/imath] over [imath]K[/imath] (the sub-field of all the elements of [imath]L[/imath] which are roots of a polynomial in [imath]K[X][/imath]) algebraically closed ? Do you have a counterexample ? |
2771517 | Let [imath]f:\mathbb{R}\rightarrow[0,\infty][/imath] be measurable that [imath]\int f(x)dx=0[/imath], then [imath]f(x)=0[/imath] for a.e.
Let [imath]f:\mathbb{R}\rightarrow[0,\infty][/imath] be measurable and assume that [imath]\int_{\mathbb{R}}f(x)dx=0[/imath]. Prove that [imath]f(x)=0[/imath] for a.e. [imath]x\in\mathbb{R}[/imath]. I know the proof is simple if the function [imath]f[/imath] is defined to be continuous because then, we can show that the sign is preserved. But in this case, I'm not to assume that the given function is continuous. How can I go about proving this? | 1772239 | [imath]\int f=0[/imath] implies [imath]f=0[/imath] almost everywhere
Let [imath]f[/imath] be nonnegative and measurable. Show that [imath]\int f =0[/imath] implies [imath]f=0[/imath] almost everywhere (except possibly on a set of measure zero). My thoughts. Take a set E with [imath]m(E) \neq 0[/imath]. Suppose [imath]f \neq 0[/imath] on E. That means [imath]\int_E f \neq 0[/imath]. Since [imath]f[/imath] is nonnegative this implies [imath]\int f >0[/imath]. That contradicts the given information, so [imath]f[/imath] must be 0 almost everywhere. |
2771566 | Compute a monster integal
We had to compute the integral [imath]\int \frac{3}{x^3-1} dx[/imath] I am just starting to learn about integration, and I have no idea how to begin. Can you please help me solve this? Even a hint would be very helpful. Thank you in advance! P.S. This question is not really a duplicate because the numerator is 3 not 1, so it is harder. If it were 1, I could do it easily. | 1129726 | How do I evaluate the integral [imath]\int \frac{1}{x^3 -1}dx[/imath]
this is my first question here so I hope I did everything right. Still really new to LaTeX as well[imath]\int \frac{1}{x^3 -1}dx [/imath] I first used partial fractions to decompose this integral into two parts[imath]\int \frac{1}{3(x-1)}dx -\frac{1}{3} \int \frac{x+2}{x^2 + x + 1}dx[/imath] I definitely can solve the first part as [imath]\frac{\ln(x-1)}{3}[/imath] but I'm stuck with the second part as I don't know what to do since the denominator is not easily factor-able. Thanks in advance to anybody taking the time to read this. Edit: I do know that you can complete the square to make the integral prettier, but I don't know how to cancel the numerator or seperate it to relate the integral to [imath]\int \frac{1}{x^2 + 1}[/imath] to get some value of arctan(x) +c |
2771853 | Are the two groups isomorphic?
Let [imath]G,H,K[/imath] be three groups. If [imath]G\times H[/imath] is isomorphic to [imath]G\times K[/imath], then is it true for [imath]H\cong K[/imath]? | 835731 | [imath]G \times H \cong G \times K[/imath] , then [imath] K \cong H[/imath]
I already know that if groups [imath]G,H,K[/imath] are finitely generated abelian groups, following is true. If [imath]G\times K[/imath] is isomorphic to [imath]H\times K[/imath], then [imath]G[/imath] is isomorphic to [imath]H[/imath]. I prove this by uniquness of factorization of finitely generated abelian groups. My questions are If group [imath]G[/imath] is finite,[imath]G\times K[/imath] is isomorphic to [imath]H\times K[/imath], then [imath]G[/imath] is isomorphic to [imath]H[/imath]? Can you give me a easiest proof and it can be proved by projection function on external direct product? If [imath]G,H,K[/imath] are groups. If [imath]G\times K[/imath] is isomorphic to [imath]H\times K[/imath], then [imath]G[/imath] is isomorphic to [imath]H[/imath]. This statement is false. What is a counterexample? *[imath]\times[/imath] is external direct product |
2772167 | Tell whether module of root of such polynomial is smaller than 1
Let's consider such polynomial: [imath] W_n(z)=\sum_{k=1}^n kz^{k-1} [/imath] Tell whether this statement is true for any [imath]n \in \mathbb{N}[/imath]: [imath]W_n(z)=0 \Rightarrow |z|<1[/imath] Here is what I have evaluated: It is obvious that [imath]z[/imath] can't be a real number bigger or equal to [imath]0[/imath]. Also, if [imath]n[/imath] is even, then [imath]z[/imath] can't be a real number smaller than [imath]-1[/imath], because if it was, we would have: [imath]W_n(z)=\sum_{k=1}^n kz^{k-1}=\sum_{k=1}^{\frac{n}{2}}\Big((2k+1)z^{2k}+2kz^{2k-1}\Big)=\sum_{k=1}^{\frac{n}{2}}z^{2k-1}\Big(2kz+2k+z\Big)[/imath] Both [imath]z^{2k-1}[/imath] and [imath]2kz+2k+z[/imath] are smaller than [imath]-1[/imath] so each term of the last sum is bigger than [imath]1[/imath] so the final sum is bigger than [imath]0[/imath]. | 286810 | Show that the zeros of [imath]\sum_{k=1}^n kz^{k-1}[/imath] are inside the unit disc
This is a homework problem that I wasn't able to solve, and I feel a little silly about it (the assignment has been submitted). The problem: Let [imath]P(z) = 1 + 2z + 3z^2 + \cdots + nz^{n-1}[/imath]. By considering [imath](1-z)P(z)[/imath], show that zeros of [imath]P(z)[/imath] are contained inside the unit disc. This is from Chapter 1 of Bak's Complex Analysis. A solution should only use algebraic manipulations of complex numbers (in other words, this problem is very easy to solve by employing the Gauss-Lucas theorem, but that's not "on the table" -- complex differentiation hasn't been defined in the scope of what's permissible). What I've done: [imath]\begin{align*} (1-z)P(z) &= 1+z+z^2+\cdots+z^{n-1}-nz^n \\ &= \frac{z^n-1}{z-1}-nz^n. \end{align*}[/imath] Let [imath]z[/imath] be a zero of [imath]P(z)[/imath]; then it is a zero of [imath](1-z)P(z)[/imath]. Consequently, [imath]\begin{align*} nz^n &= \frac{z^n-1}{z-1} \\ nz^n(z-1) &= z^n-1 \\ nz^n(z-1)+1 &= z^n \end{align*} [/imath] Then [imath]|z^n| = |z|^n = |nz^n(z-1)+1|.[/imath] From the triangle inequality, [imath] \begin{align*} |z|^n &\le |nz^n(z-1)| + |1| \\ |z|^n &\le n|z^n||z-1| + |1| \\ n|z|^n(|z-1|) &\le 1 \\ |z|^n &\le \frac{1}{n|z-1|} \end{align*} [/imath] At this point, I am stuck. I feel like I went down the wrong path, or maybe I just made an arithmetical error, but I can't find where. |
2771901 | Convergence of a Polynomial Series
I recently saw this question about the convergence of [imath]\sum_{n=0}^{\infty} \sqrt[3]{n^3+1} - n[/imath] I tried finding the convergence of this series by factoring out an [imath]n^3[/imath] so that the sum became, [imath]\sum_{n=0}^{\infty} n\sqrt[3]{1 + \frac{1}{n^3}}[/imath] but couldn't figure out any way to simplify and solve this. Is there a simpler or better way to solve this? | 1599120 | Testing the convergence of cube root of some function of n
I have to test the convergence of the following series:- [imath]\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n[/imath] My approach is as follows :- [imath]n^3+1>1=\sqrt[3]{n^3+1}>1=\sqrt[3]{n^3+1}-n>1-n[/imath] Now since[imath]\sum 1-n[/imath] diverges, the series under consideration diverges. Is this right or wrong? |
2772341 | How can I prove that the determinant of the matrix is [imath]0[/imath] only when [imath]a=b=c=d=0[/imath]?
How can I prove that the determinant of the matrix is [imath]0[/imath] only when [imath]a=b=c=d=0[/imath]? I tried with the Laplace Method of Expansion but I cannot solve the final equation. \begin{bmatrix} a & b & c & \ d \\ b & -a & d & \ -c \\ c & -d & -a & \ b \\ d & c & -b & \ -a \\ \end{bmatrix} Thanks in advance! | 2416817 | Prove that [imath]\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2.[/imath]
Let [imath]a, b, c, d \in \mathbb K[/imath] where [imath]\mathbb K[/imath] is a field. Prove that [imath]\det \begin{bmatrix} a & -b & -c & -d\\ b & a & -d & c\\ c & d & a & -b\\ d & -c & b & a \end{bmatrix} = (a^2+b^2+c^2+d^2)^2[/imath] I'm looking for a smart way to solve this problem. If we denote [imath]A = \begin{bmatrix} a & -b \\ b & a \\ \end{bmatrix}[/imath] and [imath]B = \begin{bmatrix} -c & -d \\ -d & c \\ \end{bmatrix}[/imath] we have that [imath] \begin{bmatrix} a & -b & -c & -d\\ b & a & -d & c\\ c & d & a & -b\\ d & -c & b & a \end{bmatrix} = \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix} [/imath] So it's sufficient to proof that [imath] \det \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix} = (\det A - \det B)^2. [/imath] Help? |
2771664 | [imath]|G| = pq[/imath] with [imath]p[/imath], [imath]q[/imath] primes not necessarily distinct: case [imath]p=q[/imath]
I'd like to review the reasoning of this question for the case [imath]p=q[/imath]. It says If [imath]|G|=pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are primes that are not necessarily distinct. Prove that [imath]G[/imath] is abelian or [imath]Z(G)=1[/imath]. This question is already answered here for the case [imath]p \ne q[/imath]. The reasoning is: [imath]Z(G)[/imath] is a subgroup of [imath]G[/imath] and then [imath]|Z(G)|[/imath] is [imath]1,p,q[/imath] or [imath]pq[/imath]. If [imath]|Z(G)| = p[/imath] then [imath]G/Z(G)[/imath] has order [imath]q[/imath] and therefore is cyclic, which means [imath]G[/imath] is abelian and that is a contradiction because in that case we would have [imath]Z(G)=G[/imath]. We conclude [imath]G[/imath] is abelian or [imath]Z(G)=1[/imath]. Now, if [imath]p=q[/imath] what part of the previous reasoning fails? I mean, we still could say "since [imath]|G| = pp[/imath] then [imath]|Z(G)|[/imath] is [imath]1,p[/imath] or [imath]pp[/imath]. If [imath]|Z(G)|=p[/imath] then [imath]G/Z(G)[/imath] has prime order and hence is cyclic" and conclude again [imath]G[/imath] is abelian or [imath]Z(G)=1[/imath]. However if [imath]G[/imath] has order [imath]p^2[/imath] ([imath]p[/imath] prime) then [imath]Z(G) \ne 1[/imath], according to a further result. So as much as possible, without involving the class equation I'd like to know what is wrong with that proof for the case [imath]p=q[/imath] (actually this problem is in Dummit&Foote and it is supposed to be solved without class equation). | 1859272 | Showing order of [imath]Z (G) =1[/imath] or [imath]pq[/imath]
Question : If [imath]|G|=pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are primes that are not necessarily distinct. Prove that the order of [imath]Z (G) =1[/imath] or [imath]pq[/imath]. Showing the order is [imath]pq[/imath] is trivial. I unsure how to start with showing the order is [imath]1[/imath]. Hints are appreciated. Thanks in advance. |
2772787 | Example where [imath]f\in L^{2}(\mathbb{R})[/imath], but for all [imath]p\in[1,2)\cup(2,\infty][/imath] [imath]f\notin L^{p}(\mathbb{R})[/imath].
Construct an example of a function [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] such that [imath]f\in L^{2}(\mathbb{R})[/imath], but such that for all [imath]p\in[1,2)\cup(2,\infty][/imath] [imath]f\notin L^{p}(\mathbb{R})[/imath]. I can see how to go about constructing [imath]f\in L^1(\Bbb{R})[/imath] and [imath]f\notin L^p[/imath] for [imath]p>1[/imath], but this one asks for [imath]f\in L^2[/imath] Or is there a general construction for any specific [imath]p[/imath]? | 675491 | How to find a function in [imath] L_{2}(0,\infty)[/imath] but not [imath]L_{q}(0,\infty)[/imath], [imath]q\neq 2 [/imath]
How to find [imath]x\in L_{2}(0,\infty)[/imath] \ [imath]L_{q}(0,\infty)[/imath] , [imath]q\neq 2 [/imath] i tried [imath]\frac{1}{t.lnt}[/imath] with various degrees on [imath]t[/imath] and [imath]ln(t)[/imath] Could you please help me with this question |
2772890 | If [imath]f[/imath] is continuous and bounded on [imath]\Bbb{R}[/imath], does it mean that [imath]f[/imath] is uniformly continuous on [imath]\Bbb{R}[/imath]?
Please, I need some clarification on this. If [imath]f[/imath] is continuous and bounded on [imath]\Bbb{R}[/imath], does it mean that [imath]f[/imath] is uniformly continuous on [imath]\Bbb{R}[/imath]? | 1368406 | Continuous and bounded imply uniform continuity?
I am thinking about this since couple hour. Is a continuous and bounded function [imath]f:\mathbb R\to\mathbb R[/imath] uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let [imath]\epsilon>0[/imath] and [imath]a\in\mathbb R[/imath]. By continuity, there is a [imath]\delta_a>0[/imath] such that [imath]d(f(x),f(y))<d(f(x),f(a))+d(f(y),f(a))<\epsilon[/imath] if [imath]d(x,y)\leq d(x,a)+d(y,a)<\delta_a.[/imath] Then, if [imath]\inf_a\delta_a>0[/imath] I set [imath]\delta=\inf_a\delta_a[/imath] and we are done since [imath]d(f(x),f(y))<\epsilon[/imath] if [imath]d(x,y)<\delta,[/imath] but I didn't success to prove that [imath]\delta\neq 0[/imath]. |
915631 | Show that [imath]\pm 1[/imath] are the only eigenvalues of the linear operator [imath]f[/imath] as the transpose of a matrix
Let [imath]V[/imath] be the vector space of [imath]n\times n[/imath] matrices over [imath]K[/imath] under addition and let the linear operator [imath]f[/imath] be given by [imath]f(A)=A^{T}[/imath], where [imath]A^T[/imath] denotes the transpose of matrix [imath]A[/imath]. Show that [imath]\pm 1[/imath] are the only eigenvalues of [imath]f[/imath]. Suppose the characteristic of [imath]K[/imath] is not [imath]2[/imath]. What are the dimensions of the eigenspaces [imath]E(1)[/imath] and [imath]E(-1)[/imath]? Eigenvalues are found from equation: [imath]f(v)=\lambda v[/imath], therefore we have [imath]v^T=\lambda v[/imath], I don't understand how this shows [imath]\lambda =\pm 1[/imath]. I know that [imath]\dim E(1)+\dim E(2)=n[/imath] would this suffice for the second part? | 792079 | Eigenvalues of the transpose
I have tried this problem and I'm unsuccessful. Let [imath]T: M(n,n)\rightarrow M(n,n)[/imath] be the mapping defined by [imath] T(A) = A^T[/imath]. Show that the only eigenvalues of [imath]T[/imath] are [imath]\pm 1[/imath]. |
2773066 | Find eigenvalues and eigenspaces of a linear map in [imath]\mathbb{C}_{4\times4}[/imath]
Let [imath]f:\mathbb{C}_{4\times4}\longrightarrow\mathbb{C}_{4\times4}[/imath] be the linear map defined by [imath]f(A)=A+3A^T[/imath]. Determine eigenvalues, the corresponding eigenspaces and their dimension. I cannot start this exercise. Some helps? Thank You | 1899628 | Finding the minimal polynomial, the eigenvalues, and the characteristic polynomial of [imath]T(A)=A+A^t[/imath]
Let [imath]M_n[/imath] denote the vector space of all [imath]n\times n[/imath] real valued matrices and for any matrix [imath]A\in M_n[/imath] let [imath]A^t[/imath] denote its transpose. Define the linear map [imath]T:M_n\rightarrow M_n[/imath] as [imath]T(A)=A+A^t.[/imath] What is the minimal polynomial of [imath]T[/imath]? What are the eigenvalues and the corresponding eigenspaces of [imath]T[/imath]? What is the characteristic polynomial of [imath]T[/imath]? I learned about minimal polynomials of operators from [imath]\textit{Linear Algebra Done Right}[/imath] by Axler. In his description he explains how in order to find the minimal polynomial you must find the smallest positive integer [imath]m[/imath] such that [imath]T^m[/imath] is a linear combination of [imath]I, T, T^2, \ldots, T^{m-1}[/imath]. That is, the smallest integer [imath]m[/imath] such that [imath]T^m = a_0I+a_1T+a_2T^2+ \cdots +a_{m-1}T^{m-1}.[/imath] As for the eigenvalues...since we are working with real-valued matrices, then [imath]A[/imath] and [imath]A^t[/imath] have the same eigenvalues, so if [imath]\lambda \in \mathbb{R}[/imath] is an eigenvalue of [imath]A[/imath] (and hence of [imath]A^t[/imath]) wouldn't [imath]2\lambda[/imath] be an eigenvalue of [imath]A+A^t[/imath]? I apologize if I am saying utter nonsense... |
1892410 | The Möbius function is the sum of the primitive [imath]n[/imath]th roots of unity.
Did you know that the Möbius function [imath]\mu[/imath] is the sum of the primitive nth roots of unity? I want to know about meaning of this. This statement is expressed as, [imath]\mu(n) = \sum_{\substack{k=1 \\ (k,n)=1}}^n \exp\left(\frac{2\pi ik}{n}\right).[/imath] | 737892 | Coefficient of [imath]n[/imath]th cyclotomic polynomial equals [imath]-\mu(n)[/imath]
For a polynomial [imath]f=X^n+a_1X^{n-1}+\ldots+a_n \in \mathbb{Q}[X][/imath] we define [imath]\varphi(f):=a_1 \in \mathbb{Q}[/imath]. Now I want to show that for the [imath]n[/imath]th cyclotomic polynomial [imath]\Phi_n[/imath] it holds that [imath]\varphi(\Phi_n)=-\mu(n)[/imath] where [imath]\mu(n)[/imath] is the Möbius function. What I know is that [imath]\displaystyle\Phi_n=\prod_{d|n} (X^{\frac{n}{d}}-1)^{\mu(d)}[/imath]. |
2769766 | Show that [imath]\cos(x) \le e^{-x^2/2}[/imath] for [imath]0 \le x \lt \pi/2[/imath].
Show that [imath]\cos(x) \le e^{-x^2/2}[/imath] for [imath]0 \le x \lt \pi/2[/imath]. This inequality came up in my solution to Show that the sequence [imath]\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}[/imath] converges. This is in the category of "There should be a number of ways to prove this." Here is one way I came up with. I am interested in seeing how many others there are. [imath](\ln(\cos(x))' =-\tan(x) \le -x[/imath] for [imath]0 \le x \lt \pi/2[/imath]. Integrating from [imath]0[/imath] to [imath]x[/imath], since the two sides are [imath]0[/imath] at [imath]x=0[/imath], [imath]\ln(\cos(x)) \le -x^2/2[/imath] so [imath]\cos(x) \le e^{-x^2/2} [/imath]. | 1795803 | How to see [imath]\cos x \leq \exp(-x^2/2)[/imath] on [imath]x \in [0,\pi/2][/imath]?
Can anyone help me with the above inequality? I tried looking at the series expansion and I guess the answer indeed lies there, but I fail to see it. Thanks |
2773438 | Does [imath]|d(x_n.a)-d(a,x)|\leq d(x_n,x)[/imath]?
Let [imath](M,d)[/imath] be a metric space and [imath]x\in M[/imath], [imath]x_n\to x[/imath] So [imath]|d(x_n,a)-d(a,x)|\leq d(x_n,x)[/imath] is a concluse form the triangle inequallty? | 1490868 | Proof of the Reverse Triangle Inequality
Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know: if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction). Proposition: [imath]|d(x,y)-d(y,z)| \leq d(x,z)[/imath]. Proof: We proceed by cases. Case 1: [imath]d(x,y)-d(y,z) \geq 0[/imath]. Assume by contradiction that [imath]d(x,y)-d(y,z) > d(x,z)[/imath]. Hence, [imath]d(x,y)> d(y,z) + d(x,z)[/imath], contradicting the triangle inequality. Case 2: [imath]d(x,y)-d(y,z) < 0[/imath]. Assume by contradiction that [imath] - d(x,y) + d(y,z) > d(x,z)[/imath]. Hence, [imath]d(x,y) < d(y,z) - d(x,z)[/imath]. By applying the triangle inequality, we have that [imath]d(y,x) \geq d(y,z) + d(x,z)[/imath], which is by symmetry equivalent to [imath]d(x,y) \geq d(y,z) + d(x,z)[/imath], obtaining the desired contradiction. [imath]\square[/imath] As always, any feedback is welcome. Thank you for your time. |
2773886 | Can this proof be applied to C1-functions
In show that a straight line has a Lebesgue measure of zero, does the proof given applies to every [imath]C^1[/imath]-function or does it need some changes for it to be true? More specifically, I want to know if the argument that [imath]λ(K)≤∑_i(b_i−a_i)\frac{2ϵ}{2^i}[/imath] works for [imath]f(x)[/imath] as [imath]C^1[/imath]-function? | 218584 | Continuously Differentiable Curves in [imath]\mathbb{R}^{d}[/imath] and their Lebesgue Measure
Show that the image of the curve [imath]\Gamma\in\mathscr{C}^{1}\left([a,b]\to\mathbb{R}^{d}\right)[/imath] has d-dimensional Lebesgue measure zero (of course, [imath]d\geq2[/imath]). This can be proved using the absolute continuity of [imath]\Gamma'[/imath] (since [imath][a,b][/imath] is compact and [imath]\Gamma'[/imath] is assumed continuous, hence in [imath]L^{1}([a,b])[/imath]) together with the fundamental theorem of calculus to obtain an [imath]\epsilon[/imath]-small cover of [imath]\Gamma[/imath] by balls. But I am trying to prove this using more elementary means (i.e. without integration). Intuitively, since [imath]\Gamma[/imath] is smooth, we ought to (for fine enough partitions) be able to cover [imath]\Gamma[/imath] by boxes which arise from its tangent line. And by taking the partition of [imath][a,b][/imath] to be finer and finer, the "tangent box" cover ought to also get smaller and smaller. More rigorously, the vector-version of the mean value theorem can be applied: [imath]|\Gamma(t_{i-1})-\Gamma(t_{i})|\leq(t_{i}-t_{i-1})|\Gamma'(t_{i}^{\star})|\leq M_{i}\Delta t[/imath] where [imath]t_{i}^{\star}\in(t_{i-1},t_{i})[/imath] and [imath]M_{i}=\sup_{t\in[t_{i-1},t_{i}]}|\Gamma'(t)|[/imath] which exists and is finite since [imath]\Gamma'[/imath] is continuous. But to me, it's not quite clear how to rigorously construct a cover by boxes from here. NOTE: In the proof I mentioned (using integration), essentially you sum the left hand side over all partition intervals of uniform length [imath]\delta[/imath] (which depends on the [imath]\Gamma'[/imath]), and "integrate" the right hand side. Actually, to be more specific, for each [imath]\epsilon>0[/imath] there exists a [imath]\delta>0[/imath] such that [imath]||P||<\delta[/imath] implies [imath]\int_{t_{i-1}}^{t_{i}}|\Gamma'(t)|dt<\epsilon[/imath] (e.g. absolute continuity). This allows you to define numbers [imath]\epsilon_{i}=\sup_{t,\bar{t}\in[t_{i-1},t_{i}]}|\Gamma(t)-\Gamma(\bar{t})|\leq\epsilon[/imath] so that [imath]\sum_{i=1}^{\#P}\epsilon_{i}\leq||\Gamma'||_{L^{1}([a,b])}[/imath]. Then you can use these [imath]\epsilon_{i}[/imath] to put balls at each point [imath]\Gamma(t_{i})[/imath] of radius (say) [imath]2\epsilon_{i}[/imath], thus giving you an [imath]\epsilon[/imath]-small cover. Again though, this is harder to establish without integration theory. |
2773629 | Prove that [imath]|||A||| \leq |||I|||[/imath] for every doubly stochastic matrix [imath]A \in M_n[/imath]
Prove that [imath]|||A||| \leq |||I|||[/imath] for every doubly stochastic matrix [imath]A \in M_n[/imath]. Let [imath]|||\cdot|||[/imath] be a unitarily invariant norm on [imath]M_n[/imath]. Do you (experts) agree with these proofs? Please correct me if I am making any mistakes. Thank you so much in advance. Proof#1: Utilizing Birkhoff-Von Neumann Theorem, the doubly stochastic matrix [imath]A[/imath] can be described by a convex combination of permutation matrices, i.e., [imath]A = \sum_i \alpha_i P_i , [/imath] in which [imath]\sum_i \alpha_i = 1[/imath], [imath]\alpha_i \geq 0 \ \forall i[/imath], and [imath]P_i[/imath] are the permutation matrices (with norm [imath]1[/imath]). So, by utilizing the properties of the matrix norm [imath]|||A||| \leq \sum_i |\alpha_i| \ \underbrace{|||P_i|||}_{=|||I|||} = |||I||| \sum_i \alpha_i = |||I|||. [/imath] Proof#2: See herein Why does [imath]\left\| {\left| A \right|} \right\| \le \left\| {\left| I \right|} \right\|[/imath], for every doubly stochastic matrix [imath]A \in M_n[/imath]? | 1591700 | Why does [imath]\left\| {\left| A \right|} \right\| \le \left\| {\left| I \right|} \right\|[/imath], for every doubly stochastic matrix [imath]A \in M_n[/imath]?
Let [imath]\left\| {\left| . \right|} \right\|[/imath] be a unitarily invariant matrix norm on [imath]M_n[/imath]. Why does [imath]\left\| {\left| A \right|} \right\| \le \left\| {\left| I \right|} \right\|[/imath], for every doubly stochastic matrix [imath]A \in M_n[/imath] ? |
2775186 | Group of order 24 is solvable.
I'm trying to show that any group of order 24 is solvable. I know [imath]G[/imath] is solvable if and only if both it's normal subgroup [imath]N[/imath] and [imath]G/N[/imath] is solvable. I have proved previously that a group of order 24 has a normal subgroup of order 4 or 8. But I'm stuck.. how do I proceed here? How do I show that [imath]N[/imath] is solvable and [imath]G/N[/imath] is also solvable? | 353552 | Group of order [imath]8p[/imath] is solvable, for any prime [imath]p[/imath]
Consider the following question: Show that a group [imath]G[/imath] of order [imath]8p[/imath] is solvable, for any prime [imath]p[/imath]. I am kind of stuck, but here are my first attempts: I chose the series of subroups [imath]G>H_8>H_4>H_2>H_1=\{e\}[/imath] where [imath]H_k[/imath] has order [imath]k[/imath]. All of these subgroups exist due to Sylow's Theorems. We have the quotients [imath]H_8/H_4\cong H_4/H_2 \cong H_2/H_1 \cong \mathbb Z/2\mathbb Z[/imath], so [imath]G>H_8\rhd H_4\rhd H_2\rhd H_1=\{e\}[/imath]. Only the factor [imath]G/H_8 \cong \mathbb Z/p\mathbb Z[/imath] is causing me a headache, because although its of prime order (and hence abelian) I don't know whether [imath]H_8[/imath] is normal in [imath]G[/imath] (unless [imath]p=2[/imath]). If [imath]p\ne 2[/imath], then the number [imath]k[/imath] of Sylow [imath]2[/imath]-subgroups is [imath]1[/imath] mod [imath]2[/imath]. Since [imath]k[/imath] divides [imath]|G|[/imath] we might have [imath]k=1[/imath] or [imath]k=p[/imath]. If we had [imath]k=1[/imath] then [imath]H_8[/imath] would be the only Sylow [imath]2[/imath]-subgroup (and hence normal in [imath]G[/imath]). But how could we show this? Or is there even an easier way to approach this problem? [imath]\ [/imath] edit: I am going to try a different approach: The case [imath]p=2[/imath] is clear. When [imath]p=3[/imath] or [imath]p=7[/imath] the group has order 24 or 56 and since the smallest simple non-abelian group has order 60, the group must also be solvable in these exceptional cases (as Mariano pointed out). In any other case we get a Sylow [imath]p[/imath]-subgroup [imath]H_p[/imath] of order [imath]p[/imath], which is is normal in [imath]G[/imath]. The quotient [imath]G/H_p[/imath] has order [imath]8=2^3[/imath]; and prime power order implies solvable. So [imath]G/H_p[/imath] is solvable and [imath]H_p[/imath] is also solvable. And since the composition factors of [imath]G[/imath] are those of [imath]H_p[/imath] together with those of [imath]G/H_p[/imath], we conclude that [imath]G[/imath] is solvable iff [imath]H_p[/imath] and [imath]G/H_p[/imath] are solvable, which is the case. Hence [imath]G[/imath] is solvable if [imath]|G|=8p[/imath]. |
2775547 | Is [imath]x\sin(\frac{1}{x}),\;0 uniformly continuous?[/imath]
I want to prove or disprove that [imath]x\sin(\frac{1}{x}),\;0<x<\infty[/imath] is uniformly continuous. Can anyone please, help me? | 560486 | Uniform continuity of [imath]f(x) = x \sin{\frac{1}{x}}[/imath] for [imath]x \neq 0[/imath] and [imath]f(0) = 0.[/imath]
For the [imath]f(x) = x \sin{\frac{1}{x}}[/imath] for [imath]x \neq 0[/imath] and [imath]f(0) = 0,[/imath] my text book asks the following questions. (b) Why is [imath]f[/imath] uniformly continuous on any bounded subset of [imath]\mathbb{R}[/imath]? (c) Is [imath]f[/imath] uniformly continuous on [imath]\mathbb{R}[/imath]?? The graph for the function is this. For the question (b), if I take subset between [imath][0.2,0.6][/imath] or the subset where the slope is steep, I don't think the function is uniformly continuous because I think for a given [imath]\epsilon>0[/imath], there is no unique [imath]\delta >0[/imath] for the bounded subset. Therefore, it also cannot be uniformly continuous on [imath]\mathbb{R}.[/imath] However, the questions sounds like the function is uniformly continuous and the book says that it is uniformly continuous. The answer on the book says something but I need more explanation. Thanks. |
1136208 | Prove that [imath]n < 2^n[/imath] for all n holds N.
Prove that [imath]n < 2^n[/imath] for all [imath]n ∈ N[/imath]. By induction. I know how simple is this, but could anyone help and give detailed explanation? Edit: Its [imath]2^n[/imath] NOT 2n | 969367 | Prove by Induction : [imath]n < 2^n[/imath]
So I need to prove the inequality : [imath]n < 2^n[/imath] by Induction. What I have done so far is : Step [imath]1[/imath]: Prove that the statement is true for [imath]n=1[/imath] [imath]1<2^1[/imath] (true) Step [imath]2[/imath]: Prove that, if [imath]p(n)[/imath] is true, then [imath]p(n+1)[/imath] is also true. Assume that [imath]p(n): n < 2^n[/imath] is true. then : (Add [imath]1[/imath] on both side to get [imath]n+1[/imath]) : [imath]n+1 < 2^n +1[/imath] (Multiply the original inequality by [imath]2[/imath] to get [imath]2^{n+1}[/imath] ) : [imath]2^{n+1} < 2^n . 2[/imath] I don't know where to go from here. In fact, I am not sure whether what I did is true or not. I appreciate any insight and help. |
1875397 | Prove [imath]2^n \lt\ n![/imath]
Prove by induction: [imath]A(n) = 2^n \lt\ n![/imath] [imath]A(n+1) = 2^{n+1} \lt\ (n+1)![/imath] The expression below is true for [imath]n = 5[/imath]. [imath]2^n \lt\ \frac{n!}{2}[/imath] Then, we have: [imath]2^n \lt\ \frac{n!}{2} \lt\ n! \lt\ (n+1)![/imath] Which implies that: [imath]2^{n+1} \lt\ (n+1)![/imath] for every [imath]n \ge\ 5[/imath] Is this proof correct? Update: the case is true for [imath]n = 4[/imath]. It's true for [imath]n = 5[/imath], too. I started the induction proof with [imath]n = 5[/imath]. Let be more clear: [imath]A(n) = 2^n \lt\ n![/imath] is true for [imath]n = 5[/imath]. For [imath]n= 5[/imath], the expression [imath]2^n \lt\ \frac{n!}{2}[/imath] is true too. From the expression above, we can deduce: [imath]2^n \lt\ \frac{n!}{2} \lt\ n! \lt\ (n+1)![/imath] [imath]2^n * 2 \lt\ n! \lt\ (n+1)![/imath] Hence: [imath]2^{n+1} \lt\ (n+1)![/imath] I would like to know if there is something wrong with this proof. Remember, I assumed the expression is true for [imath]n=5[/imath], not [imath]n=4[/imath] (which is true too). Update 2: I think is neccessary to prove the expression below by induction first: "For [imath]n= 5[/imath], the expression [imath]2^n \lt\ \frac{n!}{2}[/imath] is true too." We don't know if it's true for [imath] n\gt\ 5[/imath] | 1376162 | Prove by induction that [imath]n! > 2^n[/imath]
Suppose that when [imath]n=k[/imath] [imath](k\geq4)[/imath], we have that [imath]k!>2^k[/imath]. Now, we have to prove that [imath](k+1)!\geq2^{k+1}[/imath] when [imath]n=(k+1)[/imath] [imath](k\geq4)[/imath]. [imath](k+1)! = (k+1)k! > (k+1)2^k \text{ (since }k!>2^k)[/imath] That implies [imath](k+1)!>2^k2[/imath], since [imath](k+1)>2[/imath] because of [imath]k[/imath] is greater than or equal to 4. Therefore [imath](k+1)!>2^{k+1}[/imath] Finally we may conclude that [imath]n!>2^n[/imath] for all integers [imath]n\geq4[/imath] Solved, thank you! |
249743 | Proof by induction - Being stuck before being able to prove anything!
I am currently in the process of learning how to prove statements by induction. For the most part, I understand but I am at most clueless when it comes to prove some inequalities. For example, I am having a hard time proving that: [imath]n < 2^n[/imath] is true. (Even if it's clear that it is indeed true!) In the induction step, I am stuck with: [imath]n + 1 < 2^{n + 1}[/imath] I don't know what would be the assumption I should make from this statement. Anyone could point me in the right direction? Thanks! | 2831779 | Prove by induction for [imath]\forall n\in \Bbb N_1 [/imath] that [imath]2^n \ge 1+n [/imath].
Can someone please help me in this exercise? [imath]2^n \ge 1+n [/imath] What I've done so far is prove for [imath]n=1[/imath]. So: [imath]2^1 \ge 1+1 [/imath] [imath]2 \ge 2 [/imath] Which is true Now, I'm supposing that what I have to prove is for [imath]n+1[/imath]: [imath]2^{n+1} \ge 1+ (n+1)[/imath] But I don't know how to continue it. Thank you for the help |
1349601 | Basic mathematical induction regarding inequalities
These are just the examples from my textbook, but I don't think it did not explain well. One of the problem was to prove the inequality [imath]n<2^n[/imath] for all integers [imath]n[/imath]. I understand we assume if [imath]P(n)[/imath] is true, [imath]P(k+1)[/imath] is true. Thus [imath]k+1< 2^{k+1}[/imath] But in the text book, they added one more line that I don't understand: [imath]k+1<2^k+1 \le 2^k+2^k=2\cdot2^k=2^{k+1}[/imath] I understand [imath]k+1<2^{k+1}[/imath] But how do I know [imath]2^k+1[/imath] is larger than [imath]k+1[/imath], and smaller than [imath]2^{k+1}[/imath]? I have one another example just in case, proving [imath]2^n<n![/imath] using induction. [imath]2^{k+1}=2\cdot 2^k<2\cdot k!<(k+1)k!=(k+1)![/imath] I understand we should start with [imath]2^{k+1}<(k+1)![/imath], but Where did [imath]2\cdot k![/imath] come from? Textbook says it came from inductive hypothesis, but I still don't get it. Please advise. | 1225024 | Prove that [imath]2^n\le n![/imath] for all [imath]n \in \mathbb{N},n\ge4[/imath]
The problem i have is: Prove that [imath]2^n\le n![/imath] for all [imath]n \in \mathbb{N},n\ge4[/imath] Ive been trying to use different examples of similar problems like at: http://web.cacs.louisiana.edu/~mgr/261/induction.html First i show the base case [imath]n=4[/imath] is true. Then assuming [imath]2^k\le k![/imath] for some [imath]k \in \mathbb{N},n\ge4[/imath] For [imath]k+1[/imath] we have [imath]2^{k+1}\le (k+1)![/imath] Rewritten as [imath]2\cdot2^k\le k!\cdot(k+1)[/imath] Can you not simply say [imath]2^k\le k![/imath] from the inductive hypothesis, and [imath]2\lt4\le k\lt k+1[/imath] proving the induction step? I am having trouble following some of what seems to me like unnecessary steps like in the example, but feel like what i did above is wrong as im of course just learning how to use induction. |
2775839 | Showing that a function is continuous for [imath]x[/imath]
If [imath]f(x)[/imath] is continuous at the point [imath]x=0[/imath] and for all real numbers [imath]x[/imath] and [imath]y[/imath], the function [imath]f(x+y)= f(x) + f(y)[/imath]. Show that [imath]f[/imath] is continuous for all values of [imath]x[/imath]. Not sure where to begin this problem. Any info on it would be appreciated. | 774902 | Proof of continuity [imath]f(x+y) = f(x) + f(y)[/imath]
I am trying to prove that if I have [imath]f:\mathbb{R} \to \mathbb{R}[/imath] satisfying [imath]\forall x,y\in\mathbb{R},f(x+y) = f(x) + f(y)[/imath]. Which is assumed continuous at [imath]0[/imath], that [imath]f[/imath] is continuous on [imath]\mathbb{R}[/imath] I am fairly sure it is, for that property seems to be a property of polynomials, and we know polynomials are continuous where defined(all reals) Any ideas for rigor? [imath]|x-a| \lt \delta \implies |f(x+a)-f(x)-f(a)| \lt \epsilon[/imath] No idea where to go from here, this is my first time doing [imath]\epsilon-\delta[/imath] stuff. |
616635 | Prove using induction : [imath]n < 3^n[/imath]
[imath] p(n) = n < 3^n = q(n) [/imath] when [imath]n=1[/imath], [imath]p(n)=1< 3=q(n)[/imath] Assume the result is true for [imath]n=m[/imath] [imath]p(m)=m < 3^m[/imath] when [imath]n = m+1[/imath] [imath]p(m+1) = m+1 < 3^m +1<3*3^m = 3^{m+1}=q(m+1)[/imath] is this correct? I tried up to this extent to solve this. And also tried other methods. But no luck. Can some one show me the way to get this done. Thank you. | 596876 | Prove inequality [imath]n<3^n[/imath] using mathematical induction
Prove that [imath]n<3^n[/imath] where [imath]n \in \mathbb N[/imath], when [imath] n=1 [/imath], I have proved it's true. And assumed when [imath]n = p[/imath] , [imath]p<3^p[/imath] is true. Can any body help me in showing that it is true for [imath]n =p+1[/imath] |
2776515 | Cardinality of set of strict total orders on the reals
I'm revising for an exam - two questions: i) What's the cardinality of the set of strict total orders on [imath]R[/imath]? This is a subset of the power set of [imath]R[/imath] (because [imath]|R|=|R.R|[/imath]), so I know its cardinality is less than this - but I can't find a way to show its cardinality is equal to this, i.e. an injection from the power set into the set of strict total orders. ii) What is the cardinality of the set of countable subsets of F, where F is the set of functions from [imath]R[/imath] to [imath]R[/imath]? Cardinality of F is same as cardinality of power set of [imath]R[/imath] (one way given by reasoning above, and the other way by the fact that [imath]|P(R)|=|2^R|=|R^R|[/imath]. The set of countable subsets of this I am not sure how to calculate though. | 711144 | Cardinality of Orderings of [imath]\mathbb{R}[/imath]
For a finite set [imath]S[/imath] there are [imath]\vert S\vert![/imath] orderings of its elements. What is the cardinality of all orderings of [imath]\mathbb{N}[/imath]? What would [imath]\vert \mathbb{N}\vert![/imath] mean? Is it [imath]\prod_{n\in\mathbb{N}} \aleph_0=\aleph_0^{\aleph_0}=\aleph_1[/imath]? Wikipedia says the "equivalence classes of well-orderings of the natural numbers" has cardinality [imath]\aleph_1[/imath]. But I can't put together in my mind the connection between well-orderings of [imath]\mathbb{N}[/imath] and all orderings of [imath]\mathbb{N}[/imath]. (Are there non well-ordered orderings of [imath]\mathbb{N}[/imath]?) This leads to wondering what the cardinality of all orderings of [imath]\mathbb{R}[/imath] is. (Actually my thought process went in the other direction, from [imath]\mathbb{R}[/imath] to [imath]\mathbb{N}[/imath].) I think knowing the answer to the above question will teach how to determine what [imath]\vert \mathbb{R}\vert ![/imath] means, if anything. I will be happy if you assume the continuum hypothesis to be true for the purposes of this question, but any level of completeness is welcome. |
2776604 | Basic Topology about metric space [imath]R[/imath]
Is there a nonempty perfect set in [imath]R[/imath] which contains no rational number? Its the excercise of Principles of Mathematical Analysis by Walter Rudin. I can't find the answer.I try to construct a set as cantor set but I can't prove it has no rational number or maybe it doesn't exist .Can anyone give me a hint ?I thought abt it for a whole day. | 2776544 | Existence of a special set in [imath]\mathbb{R}^2[/imath]
Let [imath]X[/imath] be a non–empty subset of [imath]\mathbb{R}^2[/imath] satisfying the following properties: Each point of [imath]X[/imath] is a limit point of [imath]X[/imath]. ( Under the distance topology in [imath] \mathbb{R}^2[/imath]) No rational point belongs to [imath]X[/imath]. (We call a rational point to be a point whose two coordinates in [imath]\mathbb{R}^2[/imath] are both rational numbers) [imath]X[/imath] is closed. Does such a set [imath]X[/imath] exist? If such a set [imath]X[/imath] exists, then any rational point in [imath]\mathbb{R}^2[/imath] is not a limit point of [imath]X[/imath]. But I don't know how to go on. Any hint would be of great welcome! Edit: Thanks for everybody's help! Consider another question: What if we change [imath]\mathbb{R}^2[/imath] in the problem into [imath]\mathbb{R}[/imath]? |
2777141 | Show that if F is a finite field, then [imath]F^F=P_F[/imath].
The hint is: Let F has as elements [imath]a_1,...,a_n[/imath]. Note that if [imath]f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)[/imath], then [imath]f_i(a_j)=0[/imath] for [imath]i\neq j[/imath] and the value [imath]f_i(a_j)[/imath] can be controlled by the choice of [imath]c\in F[/imath]. Use this to show that every function on F is a polynomial function. First, I need to construct a polynomial [imath]p(x) \in F[x][/imath] such that [imath]p_i(a_i)\neg 0[/imath] and [imath]p_i(a_j)= 0[/imath] When [imath]i\neq j[/imath] and explain why such a polynomial exists. Next I think I need to show for all [imath]a\in F[/imath] that f(a)=p(a). Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you! | 2760866 | If F is a finite field, then [imath]F^F=P_F[/imath].
The hint is: Let F has as elements [imath]a_1,...,a_n[/imath]. Note that if [imath]f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)[/imath], then [imath]f_i(a_j)=0[/imath] for [imath]i\neq j[/imath] and the value [imath]f_i(a_j)[/imath] can be controlled by the choice of [imath]c\in F[/imath]. Use this to show that every function on F is a polynomial function. Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you! |
2105184 | Lévy and Lévy Prokhorov metric
I was reading "On choosing and bounding probability metrics" (https://www.math.hmc.edu/~su/papers.dir/metrics.pdf) and encountered the Lévy distance,defined by [imath] d_L(\mu,\nu):=\inf\lbrace{}\varepsilon>0\,:\, \mu((-\infty,x-\varepsilon])\leq\nu((-\infty,x])\leq\mu((-\infty,x+\varepsilon]), \forall x\in\mathbb{R}\rbrace [/imath] as a special case (on [imath]\mathbb{R}[/imath]) of the Lévy-Prokhorov distance, defined by [imath] d_P(\mu,\nu):=\inf\lbrace\varepsilon>0\,:\,\mu(B)\leq\nu(B^{\varepsilon})+\varepsilon,\forall \text{ measurable }B\subset\mathcal{X}\rbrace [/imath] In the transcript it said that [imath]d_P[/imath] metrized weak convergence. It is clear that [imath]d_L\leq{}d_P.[/imath] Additionally it mentioned that [imath]d_L[/imath] metrizes weak convergence too, which I don't quite understand since it is only bounded from above by Lévy-Prokhorov. For it to metrize weak convergence it should have to be equivalent to [imath]d_P[/imath], which I don't believe it is. Or maybe I'm mistaken. Can anyone explain? Thanks a lot. | 246412 | Prokhorov metric vs. total variation norm
Let [imath](S,d)[/imath] be a metric space and let [imath]\mathcal P(S)[/imath] denote the space of Borel probability measures on [imath]S[/imath] endowed with the Prokhorov metric [imath]\pi:\mathcal P(S)\times \mathcal P(S)\to \mathbb R_+[/imath] given by [imath] \pi(P,Q):=\inf\{\varepsilon\geq 0:P(F)\leq Q(F^\varepsilon)+\varepsilon \text{ for all closed } F\subset S\} [/imath] where the [imath]\varepsilon[/imath]-inflation of a set is given by [imath] F^{\varepsilon} = \{x\in S:d(x,F)<\varepsilon\}.[/imath] Another useful metric on [imath]\mathcal P(S)[/imath] is induced by the total variation norm, i.e. [imath] \rho(P,Q):=\sup\limits_{A\in \mathfrak B(S)}|P(A) - Q(A)| [/imath] where [imath]\mathfrak B(S)[/imath] is the Borel [imath]\sigma[/imath]-algebra on [imath](S,d)[/imath]. I wonder if there are any interesting relations between these two metrics, [imath]\pi[/imath] and [imath]\rho[/imath]. In particular, I know that convergence in [imath]\rho[/imath] implies the weak convergence and hence if [imath]S[/imath] is separable than it implies the convergence in [imath]\pi[/imath]. I wonder, however, if under some additional assumptions it is possible to derive some non-trivial bounds on [imath]\rho[/imath] if I know upper bounds on [imath]\pi[/imath]. Or at least, if it is possible to upper-bound [imath]|P(F) - Q(F)|[/imath] for closed [imath]F[/imath] using [imath]\pi[/imath]. |
2777528 | Does [imath]A\simeq_\varphi B[/imath] imply [imath]A/{\sim}\,\, \simeq B/\varphi(\sim)[/imath]?
Let [imath]A,B[/imath] be two homeomorph topological space under homeomorphism [imath]\varphi[/imath] and let [imath]\sim[/imath] be an equivalence relation on [imath]A[/imath]. Then Does [imath]A\simeq_\varphi B[/imath] imply [imath]A/{\sim}\,\, \simeq B/\varphi(\sim)[/imath]? Where [imath]\varphi(\sim)[/imath] is defined as follow [imath]x,y\in A,\quad x\sim y \iff \varphi(x)\,\varphi(\sim)\,\varphi(y).[/imath] | 1833027 | If [imath]X[/imath] is homeomorphic to [imath]Y[/imath] then is [imath]X/{\sim}[/imath] homeomorphic to [imath]Y/{\sim'}[/imath]?/
Let [imath]f:X\to Y[/imath] be a homeomorphism between topological spaces. Suppose we have an equivalence relation [imath]\sim[/imath] defined on [imath]X[/imath]. Define an equivalence [imath]\sim'[/imath] on [imath]Y[/imath] by [imath]y_1\sim'y_2[/imath] iff [imath]x_1\sim x_2[/imath] where [imath]y_i=f(x_i)[/imath] Is [imath]X/{\sim}[/imath] homeomorphic to [imath]Y/{\sim'}[/imath]? I have a well defined continuous map [imath]F:X/{\sim}\to Y/{\sim'}[/imath] given by [imath][x]\mapsto[f(x)][/imath]. This can easily be seen to be bijective. I am having trouble proving that this is open. I know there is some diagram chasing involved but I am confused. We have the following commutative diagram - [imath]\require{AMScd}[/imath] \begin{CD} X @>f>> Y\\ @Vq_XVV\circlearrowright @VVq_YV\\ X/{\sim}@>>F> Y/{\sim'} \end{CD} If [imath]U[/imath] is open in [imath]X/\sim[/imath] then [imath]q_X^{-1}(U)[/imath] is open in [imath]X[/imath] and [imath]f(q_X^{-1}(U))[/imath] is open in [imath]Y[/imath]. To prove that [imath]F(U)[/imath] is open in [imath]Y/\sim'[/imath] we need to show that [imath]q_Y^{-1}(F(U))[/imath] is open in [imath]Y[/imath]. This will be true if [imath]f(q_X^{-1}(U))=q_Y^{-1}(F(U))[/imath]. But I am having trouble proving this is equality. Any help will be appreciated. Thank you. |
2683852 | Constant Presheaf on Irreducible Space already a Sheaf
Let [imath]X[/imath] be a space, [imath]U \subset X[/imath] an open subset and let [imath]\mathbb{Z}_U[/imath] the constant presheaf on [imath]X[/imath] defined via \begin{equation} \Gamma(V, \mathbb{Z}_U) = \begin{cases} \mathbb{Z} & \text{if } V \subset U, \\ 0 & \text{else} \end{cases} \end{equation} Generally, [imath]\mathbb{Z}_U[/imath] is just a presheaf, therefore becomes just a sheaf [imath]i_!\mathbb{Z}_U[/imath] after sheafification. My question is why is [imath]\mathbb{Z}_U[/imath] already a sheaf, if [imath]X[/imath] is irreducible? | 1088853 | Prove that a presheaf is a sheaf
Let [imath]X[/imath] be a variety. Show that if [imath]X[/imath] is irreducible, then the constant abelian presheaf [imath]\mathcal{F}[/imath] with [imath]\mathcal{F}(U)=\mathbb{Z}[/imath] for every nonempty open subset [imath]U\subseteq X[/imath] and [imath]\mathcal{F}(\emptyset)=0[/imath] is a sheaf. Any leads? What does the word "constant" means here? |
2777624 | To prove basic properties of real numbers using field axioms
To prove [imath]0 <1[/imath] 1Now if [imath]a \leq b[/imath] and [imath]0 \leq c[/imath] then [imath]ac \leq bc[/imath] Assume that [imath]1 \leq 0[/imath]. Let [imath]a=1[/imath] and [imath]b=0[/imath] and [imath]0 \leq c[/imath]. So we get [imath]c \leq 0[/imath] which is a contradiction. Is this correct? To prove if [imath]0 < a[/imath] , then [imath]0 < a^{-1}[/imath] so we have [imath]aa^{-1}=1[/imath]. So [imath]a a ^{-1} \geq 0[/imath]. Again using same axiom as above with [imath]c=1[/imath] we get [imath]0 \leq aa^{-1}[/imath] which implies [imath]a^{-1} > 0[/imath] or [imath]a^{-1}=0[/imath]. But since [imath]a \neq 0 [/imath] so [imath]a^{-1} > 0[/imath] Thanks | 231438 | how do I prove that [imath]1 > 0[/imath] in an ordered field?
I've started studying calculus. As part of studies I've encountered a question. How does one prove that [imath]1 > 0[/imath]? I tried proving it by contradiction by saying that [imath]1 < 0[/imath], but I can't seem to contradict this hypothesis. Any help will be welcomed. |
2778312 | Finding the correct family of equations for the given equation
In my calculus book, (not for homework, I was just looking around), I was given the following equality and was told to find all valid functions that fit it. Find all families of functions that satisfy the following equality: [imath]\bigg(\int f(t)dt\bigg)\bigg(\int {1\over f(t)}dt\bigg)=-1[/imath] My two failed attempts to solve this yielded [imath]f(x)=e^x+C[/imath] and [imath]f(x)=e^{-x}+C[/imath] depending on my methods. What families of functions satisfies this relationship? | 1061109 | Find all functions: [imath]\left ( \int \frac{dx}{f(x)} \right )\left ( \int f(x)dx \right )=c[/imath]
Find all functions [imath]f(x)[/imath] so that: [imath] \left ( \int \frac{dx}{f(x)} \right )\left ( \int f(x)dx \right )=c [/imath] where c is a constant. My attempt was to differentiate both sides but that appears to lead me to nowhere. I would assume that the solution involves some sort of differential equations but I am very likely to be wrong. Thanks in advance! |
2777945 | Express [imath]x^4 + y^4 + x^2 + y^2[/imath] as sum of squares of three polynomials in [imath]x,y[/imath]
I don't know any identity that'd help me simplify it. I know of Brahmagupta's identity and tried using but no good. Any hints? Edit: So far I've tried various things, [imath](x^2+1/2)^2 + (y^2 +1/2)^2 -1/2[/imath] This doesn't get me anywhere. I tried solving in x as a quadratic so as to get an insight or something but that didn't help too. I even tried writing general polynomials of degree too and then getting the solution by comparing coefficients, but that wasn't elegant and was too computational. | 248487 | Find three polynomials whose squares sum up to [imath]x^4 + y^4 + x^2 + y^2[/imath]
Prove that [imath]p(x,y) = x^4 + y^4 + x^2 + y^2[/imath] can be written as a sum of squares of three polynomials over [imath]x,y[/imath] for real numbers. |
2778600 | Is there a generator for [imath](\Bbb R, +)[/imath]?
I know that the generator for [imath](\Bbb Z,+)[/imath] is [imath]<1>[/imath], because you can get any number in [imath]\Bbb Z[/imath] as a sum of [imath]1's[/imath] and [imath]-1's[/imath]. But what about [imath](\Bbb R, +)[/imath] how can one find a generator for this group , or is it impossible ? | 354413 | What are the generators of [imath](\mathbb{R},+)[/imath]?
The real numbers as a group under addition are an infinitely-generated group. I'm not sure what those generators are though (or if it even makes sense to ask the question). For example, could we say that the generators are [imath]1, 0.1,0.01,\dots[/imath] as a real number can be defined as a limit [imath]\sum_i^n w_i 10^{-i}[/imath] for some [imath]w[/imath]? |
2769986 | What are the interior and boundary of [imath]\Bbb R^2 - \{(x,sin(1/x))|x>0\}[/imath] in [imath]\Bbb R^2[/imath]?
In [imath]\Bbb R^2[/imath], is the interior of [imath]\Bbb R^2 - \{(x,sin(1/x))|x>0\}[/imath] itself? The curve is continuous on [imath](0,\infty)[/imath] so its graph is closed, and hence the set is open. Moreover, its boundary is [imath]\{(x,sin(1/x))|x>0\}[/imath]. Am I correct? | 1628757 | Interior of the Graph of [imath]\sin(1/x)[/imath]
Let [imath]f(x)=\sin(1/x)[/imath]. If I'm correct then the boundary of [imath]S=\{(x,f(x)):x\in\mathbb{R}\} \subset \mathbb{R}^2[/imath] is [imath]\{(x,f(x)):x \text{ in domain of } f(x)\}\cup\{(0,0)\}[/imath], since for any [imath]\varepsilon > 0[/imath] and an open ball [imath]B_\varepsilon(0)[/imath], [imath]B_\varepsilon(0)\cap S\ne \emptyset \ne B_\varepsilon(0)\cap [S]^c[/imath]. [I have some doubts about this part though]. For the interior of [imath]S[/imath], it is the empty set, since there does not exist an [imath]\varepsilon > 0[/imath] such that [imath]B_\varepsilon((x,f(x)))\subset S[/imath]. Thus, the closure of [imath]S[/imath] is [imath]S\cup\{(0,0)\}[/imath]. Please let me know if you think this is about right. |
2760029 | Every closed subset of a Banach separable space is separable
Could you give me some hint to prove this assertion? I've found a proof when the closed subset doesn't have isolated points: If [imath]X[/imath] is a separable space, then it admits a numerable dense basis. Now consider the internal points of a closed subset [imath]C[/imath], since it is open, it is a numerable union of elements of the numerable basis. Now, let we consider [imath]C[/imath] a closed subset of [imath]X[/imath]; [imath]C[/imath] doesn't have isolated points, so the closure of the interior of [imath]C[/imath] is the closure of [imath]C[/imath], since we can write the interior of [imath]C[/imath] like a numerable union of elements in the numerable basis, there exists a subset of [imath]C[/imath] with cardinality [imath]\leq \aleph_0[/imath] such that its closure is all [imath]C[/imath], so [imath]C[/imath] is separable. Another approach to prove the assertion is make the projection of [imath]N[/imath] (the numerable dense subset of [imath]X[/imath] whose closure is all [imath]X[/imath]), in the closed subset [imath]C[/imath], but I doubt this thing will works in Hilbert spaces only, because in a Banach space I don't know if [imath]\forall m\in N[/imath] [imath]\arg\inf_{c\in C}{d(m,c)}[/imath] is unique. -The problem is that I can't visualize how [imath]\mathbb{R}\setminus\mathbb{Q} [/imath] is a separable space... who is its numerable dense subset whose closure is all [imath]\mathbb{R}\setminus\mathbb{Q} [/imath]?- | 516886 | Prove that a subset of a separable set is itself separable
The problem statement, all variables and given/known data: Show that if [imath]X[/imath] is a subset of [imath]M[/imath] and [imath](M,d)[/imath] is separable, then [imath](X,d)[/imath] is separable. [This may be a little bit trickier than it looks - [imath]E[/imath] may be a countable dense subset of [imath]M[/imath] with [imath]X\cap E = \varnothing[/imath].] Definitions Per our book: A metric space [imath](M,d)[/imath] is separable if there exists a countable dense [imath]E[/imath] contained in [imath]M[/imath]. [imath]E[/imath] contained in [imath]M[/imath] is dense if [imath]\forall m\in M[/imath], [imath]\forall ε>0\in\mathbb R[/imath], [imath]\exists e\in E[/imath] s.t. [imath]d(m,e) < ε[/imath] The attempt at a solution My best attempt was doomed from the start, because I don't quite understand the hint. My thought process went as follows: Since [imath]X[/imath] is a subset of [imath]M[/imath], [imath]\forall x\in X, x\in M[/imath]. Thus, since [imath]E[/imath] is dense in [imath]M[/imath], [imath]\forall x\in X[/imath], and ε > 0, [imath]\exists e\in E[/imath] st [imath]d(x,e)<ε [/imath]. At this point, I was done, because the set of [imath]e[/imath]'s satisfying the above, is a subset of [imath]E[/imath], a countable set. So a subset of a countable set is dense in [imath]X[/imath], and [imath]X[/imath] is separable. This is incorrect, but I cannot see why. Any help clearing up the confusion would be greatly appreciated. Thanks! Edit: I wish I could upvote all of you for your help! I really appreciate the speedy replies and attempts to make this information clear to me. |
2779839 | Proving an exponent property
I know this might be an easy task but I cannot seem to get anywhere with this proof: I am given this: [imath]\exp(x)=[/imath] [imath]\sum_{n=0}^\infty \frac{x^n}{n!}[/imath]. I have to prove that: [imath]\exp(x+y)=\exp(x)\exp(y)[/imath]. | 285227 | Prove [imath]\exp(x+y) = \exp(x) \exp(y)[/imath] for [imath]\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}[/imath]
I am trying to prove [imath]\exp(x+y) = \exp(x) \exp(y)[/imath]. I may use that [imath]\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}[/imath] I further know how to multiply two power series in one point, i.e. if [imath]f(x) = \sum_{n=0}^\infty c_n(x-a)^n[/imath] and [imath]g(x) = \sum_{k=0}^\infty d_n(x-a)^n[/imath] then [imath] f(x)g(x) = \sum_{n=0}^\infty e_n(x-a)^n [/imath] with [imath] e_n = \sum_{m=0}^n c_md_{n-m} [/imath] |
2779140 | continuous mapping of metrizable space
Let [imath]f[/imath] be a continuous mapping of a metrizable space [imath](X,\tau)[/imath] onto a topological space [imath](Y,\tau_1)[/imath] . Is [imath](Y,\tau_1)[/imath] necessarily metrizable ? | 190303 | A condition for the image of a metrizable space to be metrizable
I noticed this problem on a previous exam that I completely missed, and I was wondering if anyone could help me out. Suppose [imath]f: Y \rightarrow X[/imath] is a continuous mapping of a separable metric space [imath]Y[/imath] onto a compact Hausdorff space [imath]X[/imath]. Show that [imath]X[/imath] is metrizable. Furthermore, give an example of a continuous mapping of a separable metric space onto a mon-metrizable Tychonoff space. Any help would be greatly appreciated! |
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