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3016424 | Proof that [imath]0=1[/imath]?
I recently saw the following "proof" online, and couldn't pinpoint where the mistake was made: From a well known property, [imath]1+2+3+\cdots = -\frac{1}{12}.[/imath] Multiplying both sides by [imath]-1,[/imath] we get [imath]-1-2-3-\cdots = \frac{1}{12}.[/imath] We can thus rearrange these equations as follows: [imath]\begin{align*} 1+2+3+4+\cdots= \, -\frac{1}{12} \\ -1-2-3-\cdots= \; \: \, \, \frac{1}{12} \\ -1-2-3-\cdots= \; \: \, \, \frac{1}{12} \\ 1+2+\cdots = -\frac{1}{12} \end{align*}[/imath] Adding, the RHS clearly sums to [imath]0[/imath], while the LHS yields [imath]1[/imath], seemingly yielding that [imath]0=1[/imath]. Where did this proof go wrong? | 1030399 | Interpretation of Ramanujan summation of infinite divergent series
I am not mathematician by any means so this question might be rather stupid. I came across this Wikipedia article on Ramanujan's summation and found this bewildering formula, [imath]1 + 2 + 3 + \dots = - \frac1{12}[/imath] The article also says that "Ramanujan summation of a divergent series is not a sum in the traditional sense". I am wondering then what this summation actually implies? Is there any interpretation of this confounding result in the physical world that someone not so math-savvy can understand? |
1504169 | Prove that [imath]\sqrt[3]5 - \sqrt[4]3[/imath] is Irrational
I've gone many directions and they all fail. The sum of two irrationals doesn't need to be irrational. I found a proof saying: if irrational [imath]x,y[/imath] have a rational sum [imath]x+y[/imath], then [imath]x-y[/imath] is irrational, or vice versa. However, in this case [imath]x+y[/imath] and [imath]x-y[/imath] are irrational. I must have misinterpreted the proof maybe. Is the sum and difference of two irrationals always irrational? I recognize [imath]n\in\Bbb Q\implies \exists a,b\in\Bbb Z~(b\neq 0) : n=\dfrac ab[/imath] Also the product of a rational and irrational number are irrational so [imath]a = b(\sqrt[3]5 - \sqrt[4]3)[/imath] where [imath]a,b\in \mathbb{Z}[/imath]. I've looked on this basis as well, but so far fruitless. Thank you, Julian | 1595280 | [imath]\sqrt[31]{12} +\sqrt[12]{31}[/imath] is irrational
Prove that [imath]\sqrt[31]{12} +\sqrt[12]{31}[/imath] is irrational. I would assume that [imath]\sqrt[31]{12} +\sqrt[12]{31}[/imath] is rational and try to find a contradiction. However, I don't know where to start. Can someone give me a tip on how to approach this problem? |
2886138 | Name of this recursively defined sequence of prime numbers?
I was wondering if the following sequence has a name: Let [imath]P[/imath] be the set of prime numbers, and [imath]p: \mathbb{N} \rightarrow P[/imath] be the (increasing) enumerating function of the set. Let's define [imath]\alpha_1 = 2, \alpha_{n+1} = p(\alpha_n)[/imath], i.e.,the [imath]\alpha_n-[/imath]prime number. Is there a name for this sequence? The first terms are [imath]2,3,5,11,...[/imath] I know they code one-branch-trees in the Cappello (non standard) codification of rooted trees. (in fact that's why I'm interested in them) | 2404614 | Prime numbers the rank of which is also a prime.
[imath]127[/imath] has an interesting property: It is the [imath]31[/imath]st prime number and its rank ([imath]31[/imath]) is also a prime. [imath]31[/imath] is the [imath]11[/imath]th prime so its rank is also a prime. [imath]11[/imath] is also a prime number with a rank ([imath]5[/imath]) that is also a prime. [imath]5[/imath] is the 3rd prime number and and so its rank ([imath]3[/imath]) is also a prime. And finally [imath]3[/imath] is the [imath]2[/imath]nd prime so its rank is also a prime... Is there a name for primes whose rank (index in the prime series) is also a prime? That is: If [imath]a_i[/imath] is the [imath]i[/imath]th prime number, then [imath]i[/imath] is a also a prime. How about numbers like [imath]127[/imath] where going down ranks of ranks always produce prime numbers (down to rank [imath]2[/imath] of course)? The first [imath]11[/imath] :-) primes in this (infinite) series would be: [imath]3, 5, 11, 31, 127, 709, 5381, 52711, 648391, 9737333, 174440041, \dotsc[/imath] Many thanks! |
3017011 | Elementary Number Theory - Divisibility
If [imath]a \mid r[/imath] and [imath]b \mid r[/imath] show that [imath]\operatorname{lcm}(a,b) \mid r[/imath] by using only elementary properties of numbers. | 1851179 | Prove that If [imath]m'[/imath] is a common multiple of [imath]s[/imath] and [imath]t[/imath], then [imath]m | m'[/imath]. Here [imath]m[/imath] is the LCM of [imath]s[/imath] and [imath]t[/imath].
Prove that If [imath]m'[/imath] is a common multiple of [imath]s[/imath] and [imath]t[/imath], then [imath]m | m'[/imath]. Here [imath]m[/imath] is the LCM of [imath]s[/imath] and [imath]t[/imath]. Although the statement is intuitively clear to me I don't know how to prove. |
3017369 | Can [imath]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4[/imath] have positive integer solutions.
Can [imath]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4[/imath] have positive integer solutions. It's just a problem from some forum. I translate it into this form. Tried some triples but fail. Also I can't prove it doesn't exist integer solutions. | 2655281 | Find all positive integral soludions of [imath]\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4[/imath]
Given an equation [imath]\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4[/imath], how to solve this problem in positive integers? I've tried to assume [imath]a\le b\le c[/imath] and that [imath]b=a+k_1, c=a+k_2[/imath]. So the equation become [imath]\frac{a}{2a+k_1+k_2} + \frac{a+k_1}{2a+k_2} + \frac{a+k_2}{2a+k_1} = 4[/imath] or equivalently, [imath]\frac{1}{2+\frac{k_1}{a}+\frac{k_2}{a}} + \frac{1+\frac{k_1}{a}}{2+\frac{k_2}{a}} + \frac{1+\frac{k_2}{a}}{2+\frac{k_1}{a}} = 4[/imath] Now let [imath]x= \frac{k_1}{a}, y= \frac{k_2}{a}[/imath], it is sufficient to find all positive rational solutions of [imath]\frac{1}{2+x+y} + \frac{1+x}{2+y} + \frac{1+y}{2+x} = 4[/imath]. |
1200832 | Prove that [imath]\sqrt{3}+ \sqrt{5}+ \sqrt{7}[/imath] is irrational
How can I prove that [imath]\sqrt{3}+ \sqrt{5}+ \sqrt{7}[/imath] is irrational? I know that [imath]\sqrt{3}, \sqrt{5}[/imath] and [imath]\sqrt{7}[/imath] are all irrational and that [imath]\sqrt{3}+\sqrt{5}[/imath], [imath]\sqrt{3}+\sqrt{7}[/imath], [imath]\sqrt{5}+\sqrt{7}[/imath] are all irrational, too. But how can I prove that [imath]\sqrt{3}+ \sqrt{5}+ \sqrt{7}[/imath] is irrational? | 1384514 | [imath]p,q,r[/imath] primes, [imath]\sqrt{p}+\sqrt{q}+\sqrt{r}[/imath] is irrational.
I want to prove that for [imath]p,q,r[/imath] different primes, [imath]\sqrt{p}+\sqrt{q}+\sqrt{r}[/imath] is irrational. Is the following proof correct? If [imath]\sqrt{p}+\sqrt{q}+\sqrt{r}[/imath] is rational, then [imath](\sqrt{p}+\sqrt{q}+\sqrt{r})^2[/imath] is rational, thus [imath]p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}[/imath] is rational, therefore [imath]\sqrt{pq}+\sqrt{pr}+\sqrt{qr}[/imath] is rational. If [imath]\sqrt{pq}+\sqrt{pr}+\sqrt{qr}[/imath] is rational, then [imath](\sqrt{pq}+\sqrt{pr}+\sqrt{qr})^2[/imath] is rational, therefore [imath]pq+qr+pr+\sqrt{p^2qr}+\sqrt{pq^2r}+\sqrt{pqr^2}[/imath] is rational, therefore [imath]p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}[/imath] is rational. Now suppose [imath]p<q<r[/imath]. If [imath]p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}[/imath] and [imath]\sqrt{pq}+\sqrt{pr}+\sqrt{qr}[/imath] are rational, then [imath]p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}-p(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})[/imath] is rational, therefore [imath](q-p)\sqrt{pr}+(r-p)\sqrt{pq}[/imath] is rational. If [imath](q-p)\sqrt{pr}+(r-p)\sqrt{pq}[/imath] is rational, then [imath]((q-p)\sqrt{pr}+(r-p)\sqrt{pq})^2[/imath] is rational, thus [imath](q-p)^2pr+2(q-p)(r-p)\sqrt{p^2qr}+(r-p)^2pq[/imath] is rational, thus [imath]\sqrt{qr}[/imath] is rational. But [imath]q,r[/imath] are distinct primes, thus [imath]qr[/imath] can't be a square. Thus [imath]\sqrt{qr}[/imath] is irrational. Contradiction. Also, is there an easier proof? |
3017632 | If [imath]f[/imath] is continuous and [imath]\int_a^b[f(x)]^2=0[/imath] prove that [imath]f(x)=0[/imath] for [imath]x\in[a,b][/imath]
Intuitively, I know [imath][f(x)]^2\ge0[/imath] so the integral would be the area above the x-axis and it would result in [imath][f(x)]^2=0[/imath] and then [imath]f(x)=0[/imath]. How do I prove this with proper terms. This is a reimann integral. | 1091465 | Is it true that [imath]\int_{a}^b h^2=0\implies h=0[/imath]?
I am working on a few examples where my author said this is true if [imath]h[/imath] is continuous. Can someone explain why? Just because the integral is [imath]0[/imath] doesn't mean the integrand is zero as well. [imath]\int_{a}^b h^2=0\implies h=0[/imath]? |
3017840 | Find all holomorphic functions that satisfy a condition
Find all holomorphic funtion [imath]f:B(0,1)\mapsto B(1,4)[/imath] s.t. [imath]f(0)=3[/imath] and [imath]f(1/2)=1[/imath] [imath]B(a,r)[/imath] is the open ball with centre a and radius r. I think that maybe Schwarz lemma will help, but dont know how. Thanks! | 857508 | Holomorphic function [imath]f:B(0,1) \to B(1,4)[/imath] with [imath]f(0)=3[/imath] and [imath]f(\frac{1}{2})=1[/imath]
Problem Find all holomorphic functions [imath]f:B(0,1) \to B(1,4)[/imath] that verify [imath]f(0)=3[/imath] and [imath]f(\frac{1}{2})=1[/imath]. I have no idea how to attack this problem,since it comes after an exercise related to Schwarz lemma, I thought that maybe the idea is to somehow apply the lemma in this problem as well, but if that is the case, I don't know how could I use it. I would appreciate suggestions. |
2881772 | Show if [imath]Q ∩[0,1][/imath] is compact?
Show if [imath]Q ∩[0,1][/imath] is compact. It is bounded. It contains it limit point [imath]1[/imath], therefore it is closed = > compact. What is wrong with this conclusion? | 1807405 | Is [imath][0,1] \cap \Bbb Q[/imath] a compact subset of [imath]\Bbb Q[/imath]?
Is [imath][0,1] \cap \Bbb Q[/imath] a compact subset of [imath]\Bbb Q[/imath]? I get the feeling it isn't compact, but I can't figure out a way to prove this. I understand that, in [imath]\Bbb Q[/imath], any open set [imath]U = \Bbb Q \cap O_i[/imath] such that [imath]O_i \subseteq R[/imath]. But I can't figure out a way to construct a finite set that doesn't contain the intersection set. |
309486 | Integral form of [imath]2\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}[/imath]
Being inspired by this post, I've wondered if the infinite series below may be expressed as an intregral. I'm very curious about that. [imath]2\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}[/imath] I'd appreciate your feedback here. How should I ponder over this case as a simple student? Where should I start from and what techniques I need to think of? Thanks! | 308343 | Find [imath]\sum\limits_{k\, \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}[/imath]
How to find [imath]\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}[/imath] Here we find [imath]\displaystyle\sum_{k=1}^{\infty} \frac{2(k^2-1)}{k^4+k^2+1}=1[/imath] and we know that [imath]\displaystyle\sum_{k \,\text{odd}} + \sum_{k \,\text{even}}=1.[/imath] Can we use this information to find sum? Or, maybe we can find it on other way? We can say that our sum is equal to [imath]\color{red}{2}\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}=\sum_{k=1}^{\infty} \left(\frac{1-4k}{4k^2-2k+1}+\frac{4k-3}{4k^2-6k+3}\right),[/imath] but I don't think we can see something from that. EDIT: Actually, Wolfram find that [imath]2\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}=\pi \text{sech}\left(\frac{\sqrt{3} \pi}{2}\right).[/imath] It's pretty nice closed form. |
3018481 | Radius of convergence of [imath]\sum_0^{\infty} n!x^{n^2}[/imath]
What can be said about the radius of convergence of the poower series [imath]\sum_0^{\infty} n!x^{n^2}[/imath] I know that [imath]\limsup_{n\to\infty}(n!)^{\frac1{n}}\to\infty[/imath]. Is that of any use here? Should I use ratio test or root test? Any hints? Thanks beforehand. | 2187375 | Convergence radius of a power series 3
Find all x for which [imath]\sum_{n=1}^n x^{n^2} n![/imath] is convergent So I tried using the ratio test and got [imath]\lim_{n\to \infty} x^{2n+1}(n+1)[/imath] but I don't know how to proceed from that. I also tried using the root test which gave me [imath]\lim_{n\to \infty} x^{n}(n!)^{1/n}[/imath] which wasn't really helpful either... |
3018161 | Find the set of interstion of [imath] \ (\mathbb{Z}_p \setminus \mathbb{Z}) \cap \mathbb{Q}[/imath] shwoing its elements
This question is from [imath]\text{p-adic numbers}.[/imath] My questions are- [imath](1)[/imath] Is the set [imath] \ (\mathbb{Z}_p \setminus \mathbb{Z}) \cap \mathbb{Q}[/imath] non-empty? If non-empty, then what are the elements or the intersection set? [imath](2)[/imath] Is the set [imath] \ (\mathbb{Z}_p \setminus \mathbb{Z}) \cap \mathbb{Q}_p [/imath] non-empty? If non-empty what are the elements or the intersection set? I can not conclude the answer. Please someone help me with details answer or at least hints. | 3017028 | Is the set [imath] \ (\mathbb{Z}_p \setminus \mathbb{Z}) \cap \mathbb{Q}[/imath] non-empty?
[imath]\text{p-adic numbers}:[/imath] My questions are- [imath](1)[/imath] Is the set [imath] \ (\mathbb{Z}_p \setminus \mathbb{Z}) \cap \mathbb{Q}[/imath] non-empty? [imath](2)[/imath] Is the set [imath] \ (\mathbb{Z}_p \setminus \mathbb{Z}) \cap \mathbb{Q}_p [/imath] non-empty? [imath](3)[/imath] If non-empty , then what are the intersection sets ? I can not conclude the answer. Please someone help me with details answer or at least hints. |
3018294 | Prove that MN-NM is Singular.
Let M and N be Square matrix, Such that [imath]M^2+N^2=MN[/imath]. Then Prove that [imath]MN-NM[/imath] is Singular. So basically I have to prove: det[imath](MN-NM)=0[/imath], I tried to prove this by multiplying the given condition by the inverse of Matrices [imath]M[/imath] and [imath]N[/imath], but could not come to the answer. Could anyone please give a hint? | 1768707 | [imath]A^2+B^2=AB[/imath] and [imath]BA-AB[/imath] is non-singular
The question is: Are there square matrices [imath]A,B[/imath] over [imath]\mathbb{C}[/imath] s.t. [imath]A^2+B^2=AB[/imath] and [imath]BA-AB[/imath] is non-singular? From [imath]A^2+B^2=AB[/imath] one could obtain [imath]A^3+B^3=0[/imath]. Can we get something from this? Edit: [imath]A^2+B^2=AB\implies\\ A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\ A^3+(A^2+B^2)B=A^2B \implies \\ A^3+A^2B+B^3=A^2B \implies \\ A^3+B^3=0 [/imath] |
3018983 | How to re-write an equation
So I have this equation [imath]4y_{k+1} = 2y_k[/imath]. If i divide both sides by 4, I get [imath]y_{k+1}=\frac12y_k[/imath] However, the equation can be re-written as so that [imath]y_{1}=\frac12y_0[/imath] and [imath]y_{2}=\frac12 y_1=\frac1{2^2}y_0[/imath] and [imath]y_{3}=\frac12 y_2=\frac1{2^3}y_0[/imath] [imath]\cdots[/imath] This is a difference equation, where the initial condition first is y=0 and then y=2. Can somebody please explain how , from [imath]y_{k+1}=\frac12y_k[/imath] i can get to the following equations? | 3018899 | How to solve a difference equation?
So I have the difference equation [imath]4y_{k+1} = 2y_k[/imath]. I need to solve this equation, first with the initial condition [imath]y_0 = 0[/imath] and then with the initial condition [imath]y_0 = 2[/imath]. The professor wrote on the answer sheet that applying iteratively the recurrence relation it is easy to check that: [imath]y_k= (1/2)^k y_0 [/imath]. Since I missed the lecture where he explained this, can somebody explain how to come about with this question? I imagine that once I get the equation, then the initial condition [imath]y_0 = 0[/imath] gives us the particular solution [imath]y_k = 0[/imath]. Thus the other initial condition [imath]y_0 = 2[/imath] is verified by the solution [imath]y_k = 2(1/2)^k[/imath]. Any help on how to get to the equation [imath]y_k= (1/2)^k y_0[/imath] will be appreciated. |
3019486 | a conjecture on the binary operation of multiplication
Give a conjecture describing the values of [imath]n[/imath] for which all of the nonzero elements of [imath]Z_n = {0, 1, 2, . . . , n − 1}[/imath] have multiplicative inverses. I am guessing the point of not having [imath]0[/imath] included is so nonprime numbers of [imath]n[/imath] will now have inverses, but I want to make sure this is the case. | 76671 | Prove that: set [imath]\{1, 2, 3, ..., n - 1\}[/imath] is group under multiplication modulo [imath]n[/imath]?
Prove that: The set [imath]\{1, 2, 3, ..., n - 1\}[/imath] is a group under multiplication modulo [imath]n[/imath] if and only if [imath]n[/imath] is a prime number without using Euler's phi function. |
2149857 | Problem regarding radius of convergence of [imath]\sum n!x^{n^2}[/imath].
[imath]\sum_{n=0}^\infty n!x^{n^2}[/imath] Can anyone tell me what will be the radius of convergence for this power series. According to me it is [imath]0[/imath]. If you can please show me how to go about it if I am wrong. | 2149160 | how to calculate radius of convergence
The question is to calculate the radius of convergence of [imath]\sum\limits_{n=0}^\infty n!x^{n^2}[/imath]. I dont know how to calculate R.O.C for such type of questions involving [imath]n^2[/imath] as power of x. How do we do this? Any help appreciated. |
3017498 | Proving [imath][0,1]\cap\mathbb{Q}[/imath] is not compact in [imath]\mathbb{Q}[/imath]
Is [imath][0,1]\cap\mathbb{Q}[/imath] a compact subset of [imath]\mathbb{Q}[/imath]. No it is not. Since [imath]\mathbb{Q}[/imath] is dense in [imath]\mathbb{R}[/imath] for any open set in [0,1] there is a rational that belongs to that set. Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite, since it be composed of infinite number of rationals. Question: Is this proof right? Thanks in advance! | 2200414 | An open covering of [imath]\mathbb{Q} \cap [0,1][/imath] that does not contain any finite subcovering
Consider the topological subspace [imath]\mathbb{Q} \cap [0,1][/imath] endowed with the usual topology of [imath][0,1][/imath], since [imath][0,1][/imath] is Hausdorff and that [imath]\mathbb{Q} \cap [0,1][/imath] is not closed, we conclude that [imath]\mathbb{Q} \cap [0,1][/imath] is not compact, i.e., there exist an open covering [imath]\{O_i\}_{i \in I}[/imath] of [imath]\mathbb{Q} \cap [0,1][/imath] such that for any finite index set [imath]J \subseteq I[/imath], [imath]\displaystyle \bigcup_{j \in J} O_j \subset \mathbb{Q} \cap [0,1][/imath], where the inclusion is strict. My question is if there is an explicite example of such open covering? |
3019753 | Proving that the adjoint of a general linear operator times said operator is a Hermitian on a complex vector space
With L being a linear operator on a complex vector space [imath]V[/imath] show that [imath]L^{_\dagger}L[/imath] is Hermitian with non negative eigenvalues I don't really have much of an idea how to do this exercise in such a general example. The proof of non negative eigenvalues is especially eluding me. Any ideas? | 3018591 | Eigenvalues of [imath]L^\dagger L[/imath] are positive
Let [imath]L[/imath] be a linear operator on a complex vector space [imath]V[/imath], how to show that eigenvalues of [imath]L^\dagger L[/imath] are positive? I guess this might be a statement taught in a functional analysis class or an advanced linear algebra class? |
3019781 | If [imath]p,q[/imath] are primes, then there are only two groups of size [imath]pq[/imath] up to an isomorphism?
Let [imath]p,q[/imath] be primes. Show there are exactly two groups of order [imath]pq[/imath], up to an isomorphism [imath]\mathbb{Z}_p\times\mathbb{Z}_q\simeq\mathbb{Z}_{pq}[/imath] I am not too sure what this question is asking or if it even makes sense. I am rather new to group theory, and would appreciate it if someone could guide me through this problem, or direct me to the actual question incase the above makes no sense. I did some research and it has to do with something regarding the Sylow Theorem's. The above "[imath]\times[/imath]" indicates a direct product. I am however, not familiar with semi-direct products. I apologize regarding the ambiguity of the question and the lack of work. Any help would be much appreciated. Thanks in advance! | 1889482 | Classify all groups with order $pq$, $p$ is not equal to $q$
I am reading the following material: http://www.math.ucla.edu/~iacoley/hw/algqual.pdf . At page 5, there is a proof classifying all groups with order [imath]pq[/imath](they are all primes and [imath]$p$[/imath] is not equal to [imath]$q$[/imath]). I am lost in the last two sentence, which claims the semidirect product are all isomorphic. Could someone help me to understand it? Thanks! |
2098034 | Prove [imath]\frac1{(2a+b+c)^2}+\frac1{(a+2b+c)^2}+\frac1{(a+b+2c)^2}\le\frac3{16}[/imath] when [imath]a+b+c=\frac1a+\frac1b+\frac1c[/imath]
Prove: if [imath]a+b+c=\frac1a+\frac1b+\frac1c,\quad0<a,b,c,\in\mathbb R[/imath] then [imath]\frac1{(2a+b+c)^2}+\frac1{(a+2b+c)^2}+\frac1{(a+b+2c)^2}\le\frac3{16}[/imath] All I tried was [imath]2a+b+c=a+\frac1a+\frac1b+\frac1c[/imath] [imath]LHS=a^2b^2c^2\left(\frac1{((a^2+1)bc+ca+ab)^2}+\frac1{((b^2+1)ca+ab+bc)^2}+\frac1{((c^2+1)ab+bc+ca)^2}\right)[/imath] that didn't help at all. Thanks. | 591134 | Let a,b,c be positive real number, proof.
Let a,b,c be positive real number, such that [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c[/imath]. Prove that : [imath]\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2} \leq \frac{3}{16}[/imath] Can anyone help me how to deal with it? |
1339785 | [imath]GL_n(\mathcal{F})[/imath] is a finite group if and only if [imath]\mathcal{F}[/imath] is finite
Show that [imath]GL_n(\mathcal{F})[/imath] is a finite group if and only if [imath]\mathcal{F}[/imath] has a finite number of elements. These are my thoughts. The order of the group [imath]GL_n(\mathcal{F})[/imath] is [imath]|GL_n(\mathcal{F})| = n^2[/imath] because the matrices in the set are [imath]n.n[/imath] matrices and which would mean each matrix has [imath]n^2[/imath] entries. I thought of somehow arguing [imath]\mathcal{F}[/imath] has [imath]n^2[/imath] elements which would show it's finite. Perhaps I need to show [imath]\mathcal{F}[/imath] is a finite. Could you clarify if I'm on the right track or not. Could you help get started on th right track on what the main idea of this proof? I'm stuck and don't if my assumptions are right and how to use them. | 619832 | Show that [imath]GL_n(\mathbb{F})[/imath] is a finite group iff [imath]\mathbb{F}[/imath] has a finite number of elements
I would like to know if my proof below is correct. Problem Show that [imath]GL_n(\mathbb{F})[/imath] is a finite group iff [imath]\mathbb{F}[/imath] has a finite number of elements. Solution If [imath]\mathbb{F}[/imath] is a finite field and say [imath]\vert \mathbb{F} \vert = m[/imath], then the set of all [imath]n \times n[/imath] matrices, which is a superset of [imath]GL_n(\mathbb{F})[/imath], where the elements comes from [imath]\mathbb{F}[/imath] has cardinality [imath]m^{n^2}[/imath]. Hence, [imath]\vert GL_n(\mathbb{F}) \vert < m^{n^2}[/imath] and thereby [imath]GL_n(\mathbb{F})[/imath] is a finite group. If [imath]GL_n(\mathbb{F})[/imath] is a finite group, note that for every non-zero element [imath]a \in \mathbb{F} -\{0\}[/imath], the matrix [imath]aI_{n \times n}\in GL_n(\mathbb{F})[/imath], where [imath]I_{n \times n}[/imath] is the identity matrix with the multiplicative identity along the diagonal. This means [imath]\vert \mathbb{F} \vert \leq GL_n(\mathbb{F}) + 1[/imath] and thereby [imath]\mathbb{F}[/imath] is a finite field. Thanks |
3020671 | How many [imath]\alpha \in S_n[/imath] are there with [imath]\alpha^2 = 1[/imath]?
How many [imath]\alpha \in S_n[/imath] are there with [imath]\alpha^2 = 1[/imath]? I want to use just linear algebra, I have done it using combinatorics but in class that does not work, can I get a little help? What I know is that (i j) = (j i) and (i j)(k I) = (k l)(i j). | 583506 | How many [imath]\alpha \in S_n[/imath] are such that [imath]\alpha^2 = 1[/imath]?
This is not for homework, but I am not great at counting arguments and would like some feedback. The question asks Let [imath]$n \in \mathbb{N}$[/imath]. How many [imath]\alpha \in S_n[/imath] are there such that [imath]\alpha^2 = 1[/imath]? I know that, if [imath]\alpha^2 = 1[/imath], then either [imath]\alpha = 1[/imath] or [imath]\alpha[/imath] is the product of disjoint transpositions. If [imath]\alpha = (i, j)[/imath] is a single transposition, then there are [imath]\frac{1}{2^1 \cdot 1!} \binom{n}{2}[/imath] such [imath]\alpha[/imath] (the [imath]2^1[/imath] and [imath]1![/imath] are put in the denominator to help in noticing the pattern later). If [imath]\alpha = (i, j)(k, l)[/imath] is the product of [imath]2[/imath] disjoint transpositions, then there are [imath]\frac{1}{2^2 \cdot 2!} \binom{n}{2} \binom{n-2}{2}[/imath] such [imath]\alpha[/imath], where the [imath]2^2[/imath] appears in the denominator to account for the cyclic permutations of each transposition, and the [imath]2![/imath] appears to account for the permutation of the transpositions themselves. If [imath]\alpha[/imath] is the product of [imath]3[/imath] disjoint transpositions, then there are [imath]\frac{1}{2^3 \cdot 3!} \binom{n}{2} \binom{n-2}{2} \binom{n-4}{2}[/imath] such [imath]\alpha[/imath]. Extrapolating from this, I find that the total number of [imath]\alpha \in S_n[/imath] such that [imath]\alpha^2 = 1[/imath] is [imath] 1 + \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor} \frac{1}{2^i \cdot i!} \prod_{k=0}^{i-1} \binom{n-2k}{2}. [/imath] Does this look OK? It looks like a rather ugly answer to me, so I have my doubts. Any input would be welcomed. |
3020728 | Prove that the product of the roots of [imath]x^{\log_{2016}x}*\sqrt{2016}=x^{2016}[/imath] is a Natural Number
Prove that the product of the roots of [imath]x^{\log_{2016}x}*\sqrt{2016}=x^{2016}[/imath] is a Natural Number. This is my solution: by putting [imath]log_{2016}[/imath] on both sides we get: [imath]\log^2{_{2016}}x-2016\log_{2016}x=-\log_{2016}\sqrt{2016}[/imath] then by putting [imath]\log_{2016}x[/imath] in front of bracket on left side, then removing [imath]\log_{2016}[/imath] from both sides and squaring the equation, and [imath]t=\log_{2016}x[/imath] [imath]t^2-4032t+2016^2-2016=0[/imath] solving for [imath]t_{1/2}[/imath] we get: [imath]x_1 = 2016^{12\left(168+\sqrt{14}\right)}[/imath] [imath]x_2 = 2016^{12\left(168-\sqrt{14}\right)}[/imath] [imath]x_1 * x_2 = 2016^{4032}[/imath] Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it? | 3020078 | The product of the two roots of [imath]\sqrt{2014}x^{\log_{2014} x}=x^{2014}[/imath] is an integer. Find its units digit
The product of the two roots of [imath]\sqrt{2014}x^{\log_{2014} x}=x^{2014}[/imath] is an integer. Find its units digit. I'm quite unable to solve the problem given. I have no idea how to deal with that [imath]\sqrt{2014}[/imath] term nor the logarithms in the exponent. Is it even necessary to solve for each individual root or is there some other way to progress through the problem? I'm really just a novice so I'd appreciate if anyone could show me the process or processes, if multiple. Thank you very much. |
3013087 | Martingale in probability and Statistics
Let [imath](X_n)[/imath] be a supermartingale such that [imath]EX_n[/imath] is constant. show that [imath]X_n[/imath] is a martingale. My problem here in this question is how can I use the constant expectation concept given to prove the statement. | 177471 | Supermartingale with constant Expectation is a martingale
In my lecture notes they use the fact, that every supermartingale [imath](M_t)[/imath] for which the map [imath]t\mapsto E[M_t][/imath] is constant is already a martingale. Unfortunately I can't prove it. Some help would be appreciated. By definition of a supermartingale we have: [imath]E[M_0]\ge E[M_s]\ge E[M_t][/imath] for [imath]t\ge s\ge 0[/imath]. I also know that [imath]M_s\ge E[M_t|\mathcal{F}_s][/imath]. If I would take expectation I would get an equality by assumption. However I do not see how this helps here to prove [imath]M_s= E[M_t|\mathcal{F}_s][/imath] |
1035092 | Periodic function such that [imath]f(x)+f'(x)\geq 0[/imath] is non-negative.
How does one prove that a real valued periodic, continuously differentiable function [imath]f(x)[/imath] such that [imath]f(x)+f'(x)\geq 0[/imath] must be non negative? I have no counterexample, but do not know how to prove this with the Mean Value theorem. | 3020024 | [imath]f[/imath] is T periodic and [imath]f(x) + f'(x) \ge 0 \Rightarrow f(x) \ge 0[/imath]
Let [imath]f: \Bbb R \to \Bbb R[/imath] be a function such that [imath]f'(x)[/imath] exists and is continuous over [imath]\Bbb R[/imath]. Moreover, let there be a [imath]T > 0[/imath] such that [imath]f(x + T) = f(x)[/imath] for all [imath]x \in \Bbb R[/imath] and let [imath]f(x) + f'(x)\ge 0[/imath] for all [imath]x \in \Bbb R[/imath]. Show that [imath]f(x) \ge 0[/imath] for all [imath]x \in \Bbb R[/imath]. My attempt: [imath]f(x) \ge 0 \iff f(x) \ge f'(x) - f'(x) \iff f(x) + f'(x) \ge f'(x)[/imath]. Thus, it is enoguh to show that [imath]0 \ge f'(x)[/imath]. [imath]\iff 0 \ge \lim_{h\to0}\frac{f(x + h) - f(x)}{h}[/imath] I do not know how to proceed from here. I know that [imath]f'[/imath] also has a periodicity of T but I do not know how to use that here. Am I on the right track? How can I use the periodicity of [imath]f[/imath] to solve the problem? |
3020653 | Find the line integral of [imath]\int\frac{e^{iz}}{z}\, dz[/imath]
This is a homework problem, I need to calculate [imath]\int\frac{e^{iz}}{z}\, dz[/imath] over some curves, but I can find a useful parametrization of [imath]z[/imath] that make this calculations not so difficult. For example I have to find the integral over the arc in the superior semi-plane of [imath]re^{i\theta}[/imath] from [imath]r[/imath] to [imath]-r[/imath] I would like to use a polar parametrization but I end with [imath]\int_0^{\pi}ie^{re^{i\theta}}\, d\theta[/imath], and then I don't know how to proceed, also a hint of a parametrization for lines would be useful. | 443418 | Calculating [imath]\int\frac{e^{iz}}{z}\, dz[/imath] on the semi-circle given by [imath]re^{i\theta}[/imath] where [imath]\theta:\,0\to\pi[/imath]
As a part of an exercise I need to calculate [imath] \lim_{r\to0}\int_{\sigma_{r}}\frac{e^{iz}}{z}\, dz [/imath] Where [imath] \sigma_{r}:\,[0,\pi]\to\mathbb{C} [/imath] [imath] \sigma_{r}(t)=re^{it} [/imath] I know how to calculate the integral on the full circle ([imath]|z|=r[/imath]) using Cauchy integral formula: [imath]2\pi ie^{iz}|_{z=0}=2\pi i[/imath] I have checked if [imath] (\overline{\frac{e^{iz}}{z})}=\frac{e^{iz}}{z} [/imath] so that the integral I want to calculate is [imath]\frac{1}{2}\cdot2\pi i=\pi i[/imath] but I got that that the equality above does not hold. Can someone please suggest a way of calculating this integral ? |
3020560 | What will be the pdf of [imath]X+Y[/imath] if [imath]X[/imath] and [imath]Y[/imath] are iid from Cauchy?
Suppose [imath]X[/imath] and [imath]Y[/imath] follow Cauchy distribution independent of each other. What will be the pdf of [imath]X+Y[/imath]? What I got by using convolution theorem is that the density [imath]g[/imath] of [imath]X+Y[/imath] is [imath]:[/imath] [imath]g(x) = \int_{-\infty}^{\infty} f(y) f(x-y)\ \mathrm {dy}[/imath] where [imath]f[/imath] is the density of the Cauchy distribution given by [imath]f(x)=\frac {1} {\pi ({1+ x^2})},x \in \Bbb R[/imath]. Then the whole integration becomes [imath]\frac {1} {\pi^2} \int_{-\infty}^{\infty} \frac {\mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.[/imath] Now how do I solve this integral? Please help me in this regard. Thank you very much. | 1264531 | Proving the sum of two independent Cauchy Random Variables is Cauchy
Is there any method to show that the sum of two independent Cauchy random variables is Cauchy? I know that it can be derived using Characteristic Functions, but the point is, I have not yet learnt Characteristic Functions. I do not know anything about Complex Analysis, Residue Theorem, etc. I would want to prove the statement only using Real Calculus. Feel free to use Double Integrals if you please. On searching, I found this. However, I was wondering if I could get some help directly on the convolution formula: [imath]$$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx=\int_{-\infty}^\infty\frac{1}{\pi^2}.\frac{1}{1+x^2}.\frac{1}{1+(z-x)^2}dx\tag{1}$$[/imath] Here I have supposed that [imath]X,Y[/imath] are Independent Standard Cauchy. But I think the general formula can be derived easily after some substitutions. I need some help on how to proceed from [imath](1)[/imath]. EDIT: Just as what the hint in the hyperlink said, I got the answer using that hint. However, I am not quite sure that the hint is algebraically correct. Maybe there has been some typing mistake in the book. |
2284695 | Is [imath]I=(x,y)[/imath] seen as an [imath]\mathbb C[x,y]-[/imath]module is free
Is [imath]I=(x,y)[/imath] seen as an [imath]\mathbb C[x,y]-[/imath]module is free ? I would say that [imath]\{x,y\}[/imath] is a basis but since [imath]x[/imath] and [imath]y[/imath] are in [imath]\mathbb C[x,y][/imath], may be it's not true. | 95431 | The ideal [imath](x,y)[/imath] is not a free [imath]K[x,y][/imath]-module
Given a field [imath]K[/imath] we have the polynomial ring [imath]K[x,y][/imath] in [imath]2[/imath] variables, which is also a left module (over itself). How can we prove that the ideal [imath](x,y)[/imath] is not a free module? |
3015423 | number theory lcm
prove if [imath]a, b ,c[/imath] are positive integers, then [imath]lcm(a,lcm(b,c))=lcm(a,b,c)=lcm(lcm(a,b),c)[/imath] [imath]lcm[/imath] is least common multiple My thought is to show that they have common divisors but not sure how to go about it. | 254704 | LCM is associative: [imath]\ \text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))[/imath]
Definition: The least common multiple of [imath]a.b\in\Bbb{Z}[/imath] is the smallest [imath]k\in\Bbb{N}[/imath] that is divisible by both a and b. Proposition: [imath]\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))\tag{2}[/imath] Proof: We want to show [imath]\text{lcm}(\text{lcm}(a,b),c)\mid\text{lcm}(a,\text{lcm}(b,c))[/imath] and [imath]\text{lcm}(a,\text{lcm}(b,c))\mid \text{lcm}(\text{lcm}(a,b),c)[/imath] such that [imath]\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))[/imath]. Let [imath]k=\text{lcm}(\text{lcm}(a,b),c)[/imath], then [imath]c\mid k[/imath] and [imath]\text{lcm}(a,b)\mid k[/imath], but if we let [imath]l=\text{lcm}(a,b)[/imath], then [imath]a\mid l\,\,\text{and}\,\, b\mid l[/imath]. Therefore [imath]c\mid k\,\,,a\mid k\,\,\text{and}\,\, b\mid k[/imath], but then [imath]\text{lcm}(b,c)\mid k[/imath]. Thus [imath]\text{lcm}(a,\text{lcm}(b,c))\mid k[/imath] Question: Does this makes sense, ifso can I do the same for [imath]\text{lcm}(\text{lcm}(a,b),c)\mid \text{lcm}(a,\text{lcm}(b,c))[/imath]? |
3021494 | A finite field [imath]F[/imath] such that [imath]F[\sqrt{2}][/imath] and [imath]F[\sqrt{3}][/imath] are not isomorphic (as fields)
Assuming [imath]F[/imath] is a finite field such that [imath]F[\sqrt{2}][/imath] and [imath]F[\sqrt{3}][/imath] are both fields, I am trying to prove that they must both be isomorphic. Or is there a counterexample? Is there a counterexample where [imath]F[/imath] has prime order? I have looked hard for one and cannot find any. For the avoidance of doubt [imath]F[\sqrt{2}] := F[X]/(X^2-2)[/imath], so assume that [imath]X^2 -2[/imath] and [imath]X^2-3[/imath] are irreducible over [imath]F[/imath]. | 169372 | When is [imath]\mathbb{F}_p[x]/(x^2-2)\simeq\mathbb{F}_p[x]/(x^2-3)[/imath] for small primes?
I've been considering the rings [imath]R_1=\mathbb{F}_p[x]/(x^2-2)[/imath] and [imath]R_2=\mathbb{F}_p[x]/(x^2-3)[/imath], where [imath]\mathbb{F}_p=\mathbb{Z}/(p)[/imath]. I'm trying to figure out if they're isomorphic (as rings I suppose) or not for primes [imath]p=2,5,11[/imath]. I don't think they are for [imath]p=11[/imath], since [imath]x^2-2[/imath] is irreducible over [imath]\mathbb{F}_{11}[/imath], so [imath]R_1[/imath] is a field. But [imath]x^2-3[/imath] has [imath]x=5,6[/imath] as solution in [imath]\mathbb{F}_{11}[/imath], so [imath]R_2[/imath] is not even a domain. For [imath]p=5[/imath], both polynomials are irreducible, so both rings are fields with [imath]25[/imath] elements. I know from my previous studies that any two finite fields of the same order are isomorphic, but I'm curious if there is a simpler way to show the isomorphism in this case, without resorting to that theorem. For [imath]p=2[/imath], neither ring is even a domain as [imath]x=0[/imath] is a solution of [imath]x^2-2[/imath] and [imath]x=1[/imath] is a solution for [imath]x^2-3[/imath], but I'm not sure how to proceed after that. Thank you for any help. |
3021290 | Why are the fundamental and anti-fundamental representation in [imath]SL(2,\mathbb{C})[/imath] not equivalent?
I am currently learning symmetries/group theory and I learnt that the fundamental representation and the anti-fundamental representation of [imath]SL(2,\mathbb{C})[/imath] are not equivalent. This means that no similarity transformation can map one of them to the other. My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent. Can anyone please explain the explanation of my professor, or perhaps give another explanation? | 3012312 | Why are the fundamental and anti-fundamental representation in [imath]\text{SL}(2,\mathbb{C})[/imath] not equivalent?
I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of [imath]\text{SL}(2,\mathbb{C})[/imath], [imath]2 \times 2[/imath] matrix with determinant of [imath]1[/imath], are not equivalent. This means that no similarity transformation can map one of them to the other. My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent. Can anyone please explain the explanation of my professor, or perhaps give another explanation? |
3021300 | How to check if a matrix is diagonalizable?
So I have this matrix [imath] \begin{pmatrix} -2 & 0 & 0 \\ 3 & 1 & -6 \\ 0 & 0 & -2 \\ \end{pmatrix} [/imath] Finding Nul(A+2Id) gives me [imath] \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & -6 \\ 0 & 0 & 0 \\ \end{pmatrix} [/imath] On the answers sheet it says to solve the homogeneous system associated with A + 2Id which gives us Nul(A+2Id)=Span [imath] \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 1 & 0 \\ \end{pmatrix} [/imath] I am completely lost on how to find the Span Nul(A+2Id)=Span [imath] \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 1 & 0 \\ \end{pmatrix} [/imath] Can somebody please help me with the steps in order to get to this solution ? Moreover, on the answers sheet is states Since the dimension of the eigenspace E(λ1) is equal to 2 we observe that the ma- trix A is diagonalizable. How do i know from this solution that the eigenspace E(λ1) is equal to 2? Thanks! | 3021268 | How to check if a matrix is diagonizable?
So i have this [imath]3\times 3[/imath] matrix [imath] \begin{pmatrix} -2 & 0 & 0 \\ 3 & 1 & -6 \\ 0 & 0 & -2 \\ \end{pmatrix} [/imath] I want to check if the matrix is diagonizable. First thing i do is find the roots : namely [imath]\lambda_1 = −2[/imath] of multiplicity [imath]2,[/imath] and [imath]\lambda_2 = 1[/imath] of multiplicity [imath]1.[/imath] In order to ensure that the matrix is diagonalizable we need to find a basis of [imath]\mathbb{R}^3[/imath] whose elements are eigenvectors for [imath]A.[/imath] In particular, since we have the eigenvalue [imath]\lambda_1=-2[/imath] having multiplicity [imath]2[/imath] we need to check that the dimension of the eigenspace [imath]E(\lambda_1)[/imath] is equal to [imath]2.[/imath] We compute the eigenspaces of [imath]A.[/imath] [imath]\DeclareMathOperator\Nul{Nul}[/imath]On the answer sheet is says that we need to find [imath]\Nul(A+2Id)[/imath]... but why [imath]2Id?[/imath] Isn't is supposed to be [imath]-2,[/imath] so [imath]\Nul(-2Id-A)[/imath]?? I do not understand this [imath]\Nul(A+2Id).[/imath] I always thought that we should compute [imath]\Nul(-2Id-A),[/imath] so what do i need to do ? [imath]\Nul(-2Id-A)[/imath] gives me a different result from [imath]\Nul(A+2Id).[/imath] Does that mean that since [imath]-2[/imath] has a multiplicity of [imath]2,[/imath] we need to work with [imath]2[/imath] and not [imath]-2[/imath]? Somebody please help with a clear and simple explanation ! Thanks! |
3021541 | Show [imath]f(x) = x^2 \sin (x^{-3/2}), x\in (0,1] [/imath], [imath]f(0) = 0[/imath] is of bounded variation without using improper integral
Show [imath]f(x) = x^2 \sin (x^{-3/2}), x\in (0,1] [/imath], [imath]f(0) = 0[/imath] is of bounded variation. Try [imath]\forall P = \{0 = x_0 < \cdots, < x_n = 1\}[/imath], [imath]\exists c_k \in (x_{k-1}, x_k)[/imath] s.t. [imath]f(x_k) - f(x_{k-1}) = f'(c_k) (x_k - x_{k-1})$$(\because MVT)[/imath]. Thus, [imath] V(f, P) = \sum |f(x_k) - f(x_{k-1})| = \sum |f'(c_k)| (x_k - x_{k-1}) [/imath] I think the next step should allow me to prove the RHS is less than some value. However, observing the following, [imath] f'(x) = \begin{cases} 2x \sin(x^{-3/2}) - \frac{3}{2} x^{-1/2} \cos(x^{-3/2}) & (x \neq 0 ) \\ 0 & (x=0) \end{cases} [/imath] I have [imath]|f'(x)| \le 2x + \frac{3}{2}x^{-1/2}[/imath], but RHS is not bounded on [imath][0,1][/imath]. Any help, including one that excludes the use of improper integral, will be appreciated. | 1996653 | When is [imath]F(x)=x^a\sin(x^{-b})[/imath] with [imath]F(0)=0[/imath] of bounded variation on [imath][0,1][/imath]?
I'm trying to show that [imath]F(x)=x^a\sin\left(x^{-b}\right)[/imath] for [imath]0<x \leq 1[/imath] and [imath]F(0)=0[/imath] has bounded variation only if [imath]a>b[/imath]. I know I have to show there exist an [imath]M< \infty[/imath] such that for any partition [imath]0=t_0<t_1<...<t_n=1[/imath] we have [imath]\sum_{j=1}^N |F(t_j)-F(t_{j-1})|<M \iff |F(t_1)| + \sum_{j=2}^N |F(t_j)-F(t_{j-1})|<M .[/imath] I'm stuck here. |
3021765 | Can I use Eisenstein's criterion to show [imath]x^{4}-2x^{3}+2x^{2}+x+4[/imath] is reducible over [imath]\mathbb{Q}[/imath]?
Can I use Eisenstein's criterion to show [imath]x^{4}-2x^{3}+2x^{2}+x+4[/imath] is reducible over [imath]\mathbb{Q}[/imath]? Can I say that there does not exist a prime that divides both [imath]2[/imath] and [imath]1[/imath]? Or is there another way to show it is reducible? Do I just use rational root test? | 1535196 | Factoring/Reducing a polynomial [imath]x^4 -2x^3 + 2x^2 + x + 4[/imath]
The problem asks to determine whether or not [imath]x^4 -2x^3 + 2x^2 + x + 4[/imath] is reducible in [imath]\mathbb{Q}[x][/imath]. I tried using the fact that if it is reducible (solution manual said it is) then it is reducible in [imath]\mathbb{Z}[x][/imath]. That didn't work for me. Eisenstein's criteria isn't applicable. |
3022358 | How to show [imath]f(x)[/imath] is [imath]0[/imath] in following problem?
Let [imath]f[/imath] be a continuous function defined on [imath][a, b][/imath]. Assume that there exist constants [imath]α[/imath] and [imath]β[/imath] with [imath](α ≠ β)[/imath] such that [imath]\alpha\int_a^x f(u)du + β\int_x^bf(u)du = 0 [/imath] for all [imath]x[/imath] belonging to [imath][a,b][/imath]. Show that [imath]f(x) = 0 [/imath] for all [imath]x[/imath] belonging to [imath][a,b][/imath]. My attempt: if we take [imath]x = a[/imath], then we get [imath]\int_a^bf(x)dx = 0[/imath]. However this does not imply [imath]f(x)[/imath] is [imath]0[/imath]. | 1462024 | How to show [imath]\alpha\int_a^cf(x)dx+\beta \int_c^b f(x)dx =0[/imath] means [imath]f(x)=0[/imath] everywhere [imath]?[/imath]
Question : Let [imath]f\ :\ [a,b]\rightarrow \mathbb R[/imath] be continuous . And there exist constants [imath]\alpha[/imath] [imath]\beta[/imath] such that [imath]\forall c\in [a,b][/imath] [imath]\alpha \int_a^c f(x)dx +\beta \int_c^b f(x)dx =0 [/imath] Prove that [imath]f=0[/imath] allover [imath][a,b][/imath]. Attempt :(not much useful ) Since the equation holds for all [imath]c\in [a,b][/imath] , if take [imath]c=a[/imath] then we have [imath]\beta \int_a^b f(x)dx =0 [/imath] And taking [imath]c=b[/imath] we get [imath]\alpha \int_a^b f(x)dx=0 [/imath] See if both [imath]\alpha=0[/imath] and [imath]\beta=0[/imath] hold then there is nothing to do . If only one of them is [imath]0[/imath] and the other is non-zero then from the last two equations we get [imath]\int_a^b f(x)dx =0 [/imath] and this does not ensure [imath]f[/imath]'s being identically [imath]0[/imath] on [imath][a,b][/imath] (for example : [imath]f(x)=sin x[/imath] and [imath]a=0[/imath],[imath]b=2\pi [/imath] ) . So our only possibility is that both [imath]\alpha \neq 0[/imath] and [imath]\beta \neq 0[/imath] must hold . Now I cannot proceed any further from this point . Please give me some hints as to what to do next . Thaanks. |
3022382 | Positive definite matrix with an interesting estimate
Let [imath]A=(a_{ij}) \in \mathbb{R}^{n \times n}[/imath] be a symmetric matrix. For all [imath]i=1, \dots ,n[/imath] we have [imath]a_{ii} > \sum_{i \ne j} \vert{a_{ij}}\vert[/imath]. I now have to show that [imath]A[/imath] is positive definite. I tried to look at the general case [imath]x^\top Ax[/imath], using the estimate above. Unfortunately without any luck... In the end I got something like [imath]x^\top Ax > \sum_{m=1}^n (x_m ^2 + x_m \min\{x_k : k \ne m\}) \sum_{m \ne k}a_{mk}[/imath]. I am unfortunately not allowed to use Gershgorins Circle Theorem as we did not discuss it up to this date :) Any ideas to prove the statement? ~Cedric :) | 87528 | A practical way to check if a matrix is positive-definite
Let [imath]A[/imath] be a symmetric [imath]n\times n[/imath] matrix. I found a method on the web to check if [imath]A[/imath] is positive definite: [imath]A[/imath] is positive-definite if all the diagonal entries are positive, and each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row/column. I couldn't find a proof for this statement. I also couldn't find a reference in my linear algebra books. I've a few questions. How do we prove the above statement? Is the following slightly weaker statement true? A symmetric matrix [imath]A[/imath] is positive-definite if all the diagonal entries are positive, each diagonal entry is greater than or equal to the sum of the absolute values of all other entries in the corresponding row/column, and there exists one diagonal entry which is strictly greater than the sum of the absolute values of all other entries in the corresponding row/column. |
3022924 | Calculating determinant of symmetric [imath]n\times n[/imath] matrix by induction
In a calculation I need the determinant of [imath]\mathbf{I}_n+(a_ia_j)_{1\leq i,j\leq n}[/imath] for given real numbers [imath]a_1,\ldots,a_n[/imath]. I conjecture that the determinant equals [imath]1+\sum_{k=1}^na_k^2[/imath], which seems reasonable after making calculations for [imath]n[/imath] up to [imath]5[/imath]. However, I did not manage to proof this formula. I tried proving the formula with induction, but any way I tried working out the induction step using linearity of the first row/column did not work. I'm just looking for some help towards a proof, so an inductive proof is not necessary. Any help is much appreciated. | 1101657 | Find the determinant of the following matrix
Find the determinant of the following matrix: [imath]A = \begin{bmatrix} 1+x_1^2 &x_1x_2 & ... & x_1x_n \\ x_2x_1&1+x_2^2 &... & x_2x_n\\ ...& ... & ... &... \\ x_nx_1& x_nx_2 &... & 1+x_n^2 \end{bmatrix}[/imath] I computed for the case [imath]n=2[/imath], and [imath]n=3[/imath] and guessed that [imath]\det(A)[/imath] should be [imath] 1+\sum_{i=1}^n x_i^2 [/imath] but not sure how to proceed for any [imath]n[/imath]. |
3023124 | Real line [imath]\mathbb R^1[/imath] with the countable complement topology is not compact
The real line with the countable complement topology is not compact. I create an open cover of [imath]\mathbb R[/imath] by: Let [imath]\mathbb Q[/imath] denote the rationals. Let an open cover be defined be the set of [imath]\mathbb Q-\{q\} [/imath], where [imath]q \in \mathbb Q[/imath]. Then this set is countable. I am unsure how to proceed from here. And is this a good example of an open cover that doesn't have a finite subcover? Which will lead to my conclusion of not compact? | 904669 | Non-compactness of [imath]\mathbb{R}[/imath] with the cocountable topology
Is [imath](\mathbb{R},\tau_{co})[/imath] compact where [imath]\tau_{co}[/imath] is the cocountable topology on [imath]\mathbb{R}[/imath]? I have the answer of my teacher but I'd like to see another one so I can understand better how people find the answer intuitively. He says: [imath]\mathbb{N}-\{n\}[/imath] is infinite countable for all [imath]n\in\mathbb{N}[/imath]. Then [imath]V_n = \mathbb{R}-(\mathbb{N}-\{n\})=(\mathbb{R}-\mathbb{N})\cup\{n\}\in\tau_{co}[/imath] for all [imath]n\in\mathbb{N}[/imath]. Then [imath]U=\{ V_n : n\in\mathbb{N}\}[/imath] is a open covering of [imath](\mathbb{R},\tau_{co})[/imath]. [imath]U[/imath] does not contain a countable subcover since [imath]V_n\cap\mathbb{N}=\{n\}[/imath] for each [imath]n\in\mathbb{N}[/imath] and [imath]\mathbb{N}[/imath] is infinite. How would you solve that problem? |
3018492 | Automata theory - How many Nerode equivalent classes for language [imath]L_k[/imath]
Given a constant [imath]k[/imath] we will define the language: [imath]L_k =\{x\#y \mid x \in \{0,1\}^k, y \in \{0,1\}^* \text{ and } x \not=y\}[/imath] How many Nerode equivalent class does [imath]L_k[/imath] have? I need to show the most tight bounds I can find. Can anyone give me a clue? Thanks [Note for those unfamiliar with the term: Nerode equivalence is defined as [imath]s_1Rs_2[/imath] if and only if [imath]\forall t\in \Sigma^* s_1t\in L \iff s_2t\in L[/imath].] | 3021245 | How many Nerode equivalence classes does the language [imath]x\neq y[/imath] have?
I have a language [imath]L_k[/imath] over the alphabet [imath]\Sigma=\{0,1,\#\}[/imath] defined as follows: [imath]\begin{equation} L_k=\{x\#y|x\in\{0,1\}^k,y\in \{0,1\}^*\wedge x\neq y\} \end{equation}[/imath] I would like to find all the Nerode equivalence classes for this language. Since this definition is somewhat hard to google I'm including it here. The Nerode equivalence [imath]R_L[/imath] on a language [imath]L[/imath] over an alphabet [imath]\Sigma[/imath] is defined as follows. For [imath]s_1,s_2\in \Sigma^*[/imath] [imath]s_1R_L s_2[/imath] if and only if [imath]\forall t\in\Sigma^* s_1t\in L\iff s_2t\in L[/imath]. If we have [imath]s_1\in\Sigma^h[/imath] where [imath]0<h\leq k[/imath] such an [imath]s_1[/imath] must be in it's own class. We can show this by contradiction in parts. Suppose [imath]s_2\in\Sigma^*[/imath] and [imath]s_2\neq s_1[/imath]. Case 1) [imath]|s_2|\neq |s_1|[/imath] Choose [imath]t=0^{h-k}\#[/imath] then [imath]s_1t=s_10^{h-k}\# \in L_k[/imath] but [imath]s_2t=s_20^{h-k}\#r\not\in L_k[/imath] since [imath]s_2t[/imath] has a [imath]\#[/imath] somewhere other than at the [imath]k+1[/imath]'st position. Case 2) [imath]|s_2|=|s_1|=h[/imath] Define [imath]t=0^{h-k}\#s_10^{h-k}[/imath]. Then [imath]s_1t=s_10^{h-k}\#s_10^{h-k}\not \in L_k[/imath] since the sides around the [imath]\#[/imath] are equal, but [imath]s_2t=s_20^{h-k}\#s_10^{h-k} \in L_k[/imath] since [imath]s_1\neq s_2[/imath] and so the padding stays unequal. Now how do I proceed for [imath]s_1[/imath] longer than [imath]k[/imath]? Any hints or help appreciated. |
3024338 | Show that [imath]\lim_{n\to\infty}((n+1)^a - n^a) = 0[/imath] for [imath]0 < a < 1[/imath]
Show that: [imath] \lim_{n\to\infty}((n+1)^a - n^a) = 0 [/imath] where: [imath] \begin{cases} 0 < a < 1 \\ n \in \mathbb N \end{cases} [/imath] I've started with: [imath] \begin{cases} a = \frac{1}{1+r}\\ r\in \mathbb R \\ r > 0 \end{cases} [/imath] So the expression may be rewritten as: [imath] x_n = \left(n+1\right)^{\frac{1}{1+r}} - n^{\frac{1}{1+r}} = \sqrt[{1+r}]{n+1} - \sqrt[{1+r}]{n} [/imath] Looks like I need to rationalize the expression to proceed, but I'm not sure how to do it. What is the proper way of doing this? I guess I could use series expansion here, but that requires using derivatives which i want to keep away from for this problem. | 3004042 | limit of [imath]\lim\limits_{x \to \infty}((x+1)^a - x^a)[/imath]
Hello everybody, i couldn't figure out how to demonstrate this equation without L'Hopital i wonder if its possible. [imath]a \in (0,1)[/imath] [imath]\lim\limits_{x \to \infty}((x+1)^a - x^a) = 0[/imath] |
3024584 | Is operation [imath]x*y = \frac{xy}{x+y+1}[/imath] associative?
In Charles Pinter's A Book of Abstract Algebra, Chapter 2: Operations, Section B. Properties of Operations, Question 7 part ii, an operation * is defined on the set of real numbers: [imath]x * y = \frac{xy}{x+y+1} [/imath] I need to find if it is associative, so: [imath] \begin{align*} x*(y*z) &= x* \frac{yz}{y+z+1} \\\\ &= \frac{\frac{xyz}{y+z+1}}{x+\frac{yz}{y+z+1}+1} \\\\ &= \frac{\frac{xyz}{y+z+1}}{\frac{xy+xz+x+yz+y+z+1}{y+z+1}} \\\\ &= \frac{xyz}{xy+xz+yz+x+y+z+1} \\\\ \end{align*} [/imath] Then the same answer for [imath](x*y)*z[/imath]. This is not the same as the answer given at the end: Answer at back What have I overlooked? | 651396 | Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2?
I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative: [imath] x * y = \frac{xy}{x+y+1}[/imath] The book solution: [imath] (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}[/imath] [imath] x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) = \frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}[/imath] However the [imath](x+y+1)[/imath] and [imath](y+z+1)[/imath] terms appear to actually cancel out: [imath] (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} [/imath] [imath] x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) =\frac{x\left(\frac{yz}{y+z+1}\right)}{x+\frac{yz}{y+z+1} + 1} [/imath] So the denominators cancel out, the two groupings are equivalent, the operator is associative. This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something. Sorry for bothering people with such a basic question. |
3023775 | Show that [imath]\langle x,y \rangle = 0[/imath].
Suppose that [imath]\left (X, \langle \cdot,\cdot \rangle \right)[/imath] be a complex inner product space. Let [imath]x,y \in X[/imath] be such that [imath]\|\alpha x + \beta y \|^2 = \|\alpha x\|^2 + \|\beta y\|^2[/imath] for all pairs [imath]\alpha, \beta \in \Bbb C[/imath] [imath]([/imath] where [imath]\|\cdot\|[/imath] is the norm induced by the inner product on [imath]X[/imath] [imath]).[/imath] Then [imath]\langle x,y \rangle = 0[/imath] i.e. [imath]x \perp y.[/imath] How do I show that? Please help me in this regard. Thank you very much. | 2193386 | Proving orthogonality of vectors in a complex vector space
I'm taking a course in Real Analysis and have come across the following question which I initially thought looked quite simple: Let [imath]X[/imath] be an inner product space over [imath]\mathbb{K}[/imath] ([imath]\mathbb{K} = \mathbb{C}[/imath] or [imath]\mathbb{R}[/imath]) Show that x and y are orthogonal if and only if [imath]||\lambda x + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2[/imath] for all [imath]\mu, \lambda \in \mathbb{K}[/imath]. I have proven the statement in one direction (starting with orthogonality and ending up with [imath]||\lambda x + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2[/imath]). However, I seem to be getting stuck trying to go in the other direction. The following is what I have so far: [imath]||\lambda x + \mu y||^2 = ||\lambda x||^2 + 2\bar{\lambda}\mu Re<x,y> + ||\mu y||^2[/imath] Then we must have 2[imath]\bar{\lambda}\mu Re<x,y> = 0[/imath] for the desired equality to hold, and so [imath]Re<x,y> = 0[/imath]. However, for orthogonality I want [imath]<x,y> = 0[/imath] and can't see how I can make this next step. Any help would be greatly appreciated. |
2525446 | Hypergeometric distribution convergences binomial distribution
I don't understand the following statement: Let [imath]X_N[/imath] , N [imath]=[/imath] 1,2, ... random variables with hypergeometric distribution; Parameter ([imath]N[/imath],[P[imath]N[/imath]],n). Show that: [imath]\lim_{N\to \infty} P( X_N = k ) = \binom{n}{k} p^k (1-p)^{n-k}[/imath] Does this means that the hypergeometric distribution approaches binomial distribution for N [imath]\rightarrow[/imath] [imath]\infty[/imath] ? Can somebody give me only the beginning of this proof? | 330553 | Proof that the hypergeometric distribution with large [imath]N[/imath] approaches the binomial distribution.
I have this problem on a textbook that doesn't have a solution. It is: Let [imath]f(x)=\frac{\binom{r}{x} \binom{N-r}{n-x}}{\binom{N}{n}}\;,[/imath] and keep [imath]p=\dfrac{r}{N}[/imath] fixed. Prove that [imath]\lim_{N \rightarrow \infty} f(x)=\binom{n}{x} p^x (1-p)^{n-x}\;.[/imath] Although I can find lots of examples using the binomial to approximate the hypergeometric for very large values of [imath]N[/imath], I couldn't find a full proof of this online. |
2977749 | Is it true that [imath]n^r(X_n-X)\xrightarrow{d}Z[/imath] implies [imath]X_n\xrightarrow{d}X[/imath]?
I see that without this last requirement [imath]n^r(X_n-X)[/imath] would blow up in probability, how to prove this formally? By contradiction maybe? Like assuming it doesn't converge in probability and hence finding a subsequence such that [imath]\mathbb{P}(|X_n-X|>\varepsilon)>\Delta[/imath] for all [imath]k[/imath], and then showing that this implies [imath]n^r(X_n-X)[/imath] to go to infinity in probability? | 2974774 | Convergence in distribution with given rate implies convergence in probability
Question: If [imath]n^r(X_n-X)[/imath] converges in distribution for some [imath]r>0[/imath], do we have that [imath]X_n[/imath] converges to [imath]X[/imath] in probability? As [imath]P(n^r|X_n-X|>\varepsilon)\to P(|Z|>\varepsilon)[/imath] (where [imath]Z[/imath] is the random variable to which [imath]n^r(X_n-X)[/imath] converges), can we say something about [imath]P(|X_n-X|>\varepsilon)[/imath]? |
2726515 | Smooth approximation of non-negative Sobolev function
I need a help to solve this question: Let [imath]u\in H^{1}_{0}(\Omega)[/imath] with [imath]u\geq 0[/imath]. Can I find a sequence of smooth non-negative functions converging to [imath]u[/imath] in [imath]H^{1}_{0}(\Omega)[/imath]? Thank you in advance. | 405255 | About smooth approximation in a Sobolev space
I want to prove the following fact : Consider [imath]\Omega \subset R^n[/imath] a bounded and open set. Let [imath]v \in H^{1}_{0}(\Omega)[/imath] a nonnegative function. Then exists a sequence [imath]v_m[/imath] in [imath]C^{\infty}_{0}(\Omega)[/imath] of nongative functions converging to [imath]v[/imath] in [imath]H^{1}( \Omega)[/imath]. I know how to prove this fact (I think this fact can help): If [imath]u_m[/imath] is a sequence converging to [imath]u[/imath] in [imath]H^1(\Omega)[/imath] , then [imath]{u^{+}_m} \rightarrow u^+[/imath]. Someone can give me a hint ? Thanks in advance |
1089478 | Is the question in the Munkres's topology book wrong?
At the end of cheapter [imath]8.1[/imath], [imath]4)[/imath] Given spaces [imath]X[/imath] and [imath]Y[/imath], let [imath][X,Y][/imath] denote the set of homotopy classes of maps of [imath]X[/imath] into [imath]Y[/imath]. [imath]b)[/imath] Show that if [imath]Y[/imath] is path connected, the set [imath][I,Y][/imath] has a single element.([imath]I=[0,1][/imath]) Claim[imath]1:[/imath] If [imath]Y[/imath] contains a hole then the loop containing the hole can not be homotopic to the ones which is not containing. So is it wrong ? Claim[imath]2:[/imath] Beside that, when I first try to prove the claim, Let [imath]f:I\to Y[/imath] be a continious map. Set [imath]F:I\times I\to Y[/imath] [imath](x,t)\mapsto f(x(1-t)) [/imath] Then [imath]f[/imath] is homotopic to a constant map. As [imath]Y[/imath] is path connected all constant map is homotopic to each other. Thus we are done. But I am not sure about niether of them. | 2919829 | The set of homotopy classes
So I came across this exercise: Show that if [imath]Y[/imath] is path-connected then the set of homotopy classes [imath][I,Y][/imath] of maps of [imath]I=[0,1][/imath] into Y has a single point. But I am confused here. If [imath]Y=S^1[/imath] then not all loops are homotopic! I mean, isn't true that [imath]\pi_1(Y)\subset [I,Y][/imath]? Or maybe I don't understand the set [imath][I,Y][/imath] well! |
3025492 | How to show that [imath](0,1)[/imath] and [imath][0,1)[/imath] are not homeomorphic?
How to show that [imath](0,1)[/imath] and [imath][0,1)[/imath] are not homeomorphic? My attempt: On contrary suppose it is then there exist [imath]f:[0,1)\to (0,1)[/imath] which is continuous But inverse image of open set is open So contradiction Is my argument is correct? | 2074259 | How can I show that a bijection between [imath][0,1)[/imath] and [imath](0,1)[/imath] cannot be continuous?
Suppose [imath]f:[0,1)\to (0,1)[/imath] is bijective. Prove that [imath]f[/imath] is not continuous. I know that [imath](0,1)[/imath] and [imath][0,1)[/imath] are not homeomorphic spaces because of the connected property. Can we then conclude that [imath]f[/imath] is not continuous because of that? |
3025356 | Prove that [imath](X,\Vert {\cdot}\Vert_2)[/imath] is not complete where [imath]\Vert f\Vert_2=\left(\int_{-2}^{2}|f(t)|^2 dt\right)^{1/2}[/imath]
Prove that [imath](X,\Vert {\cdot}\Vert_2)[/imath] is not complete where [imath]X=C[-2,2][/imath] and [imath]\begin{align}\Vert f\Vert_2=\left(\int_{-2}^{2}|f(t)|^2 dt\right)^{1/2}.\end{align}[/imath] MY TRIAL It suffices to produce a Cauchy sequence in [imath]C[-2,2][/imath] that does not converge to a point in [imath]C[-2,2][/imath]. Consider the function defined by [imath]\begin{align} f_n(t)=\begin{cases}0,& \text{if}\;-2\leq t\leq 0,\\nt,& \text{if}\;0\leq t\leq \frac{1}{n},\\1 &\text{if}\;\frac{1}{n}\leq t\leq 2.\end{cases} \end{align}[/imath] Clearly, the above function is continuous by Pasting Lemma. Next, we show that the function is Cauchy. Let [imath]m,n\in \Bbb{N}[/imath] such that [imath]m\geq n.[/imath] Then, we show that [imath]\begin{align} \Vert f_n-f_m\Vert_2=\left(\int_{-2}^{2}|f_m(t)-f_n(t)|^2 dt\right)^{1/2} \to 0,\; \text{as}\;n\to \infty.\end{align}[/imath] Now, [imath]\begin{align} \Vert f_n-f_m\Vert_2\leq\left(\int_{-2}^{2}|f_m(t)||f_m(t)-f_n(t)| +|f_n(t)||f_m(t)-f_n(t)| dt\right)^{1/2} \end{align}[/imath] We know that [imath]\begin{align} \int_{-2}^{2}|f(t)| dt\leq\sqrt{4}\left(\int_{-2}^{2}|f(t)|^2 dt\right)^{1/2}\end{align}[/imath] So, [imath]\begin{align} \Vert f_n-f_m\Vert_2\leq\left(\int_{-2}^{2}|f_m(t)||f_m(t)-f_n(t)|dt +\int_{-2}^{2}|f_n(t)||f_m(t)-f_n(t)| dt\right)^{1/2} \end{align}[/imath] I'm stuck here and don't know how to proceed. Any help please? | 2691510 | Show that [imath](C([a,b]),\|{\cdot}\|_2)[/imath] is not a complete normed vector space
Problem: Show that [imath](C([a,b]),\|{\cdot}\|_2)[/imath] is not a complete normed vector space. I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general [imath][a,b][/imath]. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case? Attempt: In order to show that [imath](C([a,b]),\|{\cdot}\|_2)[/imath] is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in [imath]C([a,b])[/imath]. In particular, it will suffice to show that our chosen Cauchy sequence converges under [imath]\|\cdot\|_2[/imath] to a discontinuous function. Consider the following sequence of functions, [imath](f_n)_{n=1}^\infty\subset C([a,b])[/imath], given by, [imath]f_n(x)=\begin{cases} 0,\,\text{when}\,\,x\in[a,\frac{b-a}2-\frac{2^{-n}}{b-a}] \\[2ex] 1,\,\text{when}\,\,x\in[\frac{b-a}2+\frac{2^{-n}}{b-a},b] \\[2ex] \frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},\,\text{when}\,\,x\in[\frac{b-a}2-\frac{2^{-n}}{b-a},\frac{b-a}2+\frac{2^{-n}}{b-a}] \end{cases} [/imath] The definition of [imath]f_n(x)[/imath] when [imath]x\in[\frac{b-a}2-\frac{2^{-n}}{b-a},\frac{b-a}2+\frac{2^{-n}}{b-a}][/imath] is the linear interpolant over that particular subinterval of [imath][a,b][/imath]. I claim that this is a Cauchy sequence. We consider: [imath]\|f_n-f_m\|_2^2=\int_a^b|f_n(x)-f_m(x)|^2dx[/imath] [imath]=\int_a^{\frac{b-a}2-\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+\int_{\frac{b-a}2+\frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx[/imath] [imath]=0+\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0[/imath] The [imath]0[/imath]'s for the first and last integrals because we have [imath]f_n(x)-f_m(x)=0-0[/imath] and [imath]f_n(x)-f_m(x)=1-1[/imath] on each respectively. As to the middle integral, here is what I've gotten so far: [imath]\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx[/imath] Based in the example I am following, supposing that [imath]m\ge n[/imath], I think that I should be able to argue that the whole thing is bound above by [imath]\frac{2^{-n}}{b-a}[/imath] which would mean that [imath]\|f_n-f_m\|_2^2\le\frac{2^{-n}}{b-a}\to0[/imath] as [imath]n\to\infty[/imath]. I don't quite see how I'm meant to bound it above by [imath]\frac{2^{-n}}{b-a}[/imath], as I keep tripping up over applying that [imath]m\ge n[/imath], leading me to question whether the approach I have employed here is a fruitful one after all. |
3025337 | What is the inverse of X modulo [imath]1 + X + X^2 + X^3 + X^4[/imath]?
What is the inverse of X modulo [imath]1 + X + X^2 + X^3 + X^4[/imath]? Is there any open softwares to calculate such things easily? | 2959859 | Let [imath]\theta[/imath] be a root of [imath]p(x)=x^3+9x+6[/imath], find the inverse of [imath]1+\theta[/imath] in [imath]\mathbb{Q(\theta)}[/imath]
Let [imath]\theta[/imath] be a root of [imath]p(x)=x^3+9x+6[/imath], find the inverse of [imath]1+\theta[/imath] in [imath]\mathbb{Q(\theta)}[/imath]. So problems like this really annoy me but I did crappy on the last homework after making a lot of arithmetic mistakes so I want to run everything by you guys. This one isn't actually for homework it's just a suggested practice problem but okay lets go!!! So I'm not sure if the method I used to do this was standard or not, I don't remember my professor showing me, but what I did first was used the canonical euclidean algorithm in [imath]\mathbb{Q[x]}[/imath] and wrote: [imath]x^3+9x+6[/imath] [imath]=(1+x)(x^2-x+10-\frac{4}{x+1})[/imath] [imath]=(1+x)(x^2-x+10)-4[/imath] the reduced modulo the minimal polynomial of [imath]\theta[/imath] i.e. [imath]x^3+9x+6[/imath] and so [imath]4=(1+x)(x^2-x+10)[/imath] [imath]\rightarrow[/imath] [imath]1= \frac{1}{4}(1+x)(x^2-x+10)[/imath] and sooo ah [imath](\theta^2-\theta+10)\frac{1}{4}[/imath] is what I believe is the inverse of [imath](1+\theta)[/imath] in the quotient field... Like i said I hate these problems but ehh is this correct? On a lighter note it's finally getting cool enough for me to go on proper runs here and so that's pretty bomb! |
3024999 | ¿How to solve this functional equation: [imath]xf(y)+yf(x)=(x+y)f(x)f(y)[/imath]? Need some help!
So I've been reading about functional equations and how to solve them. I found a pretty interesting problem (for me) but I think I need some help, some hint. I've never worked with this kind of problems. Well, the problem is this: Find all functions [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] such that for all [imath]x, y \in \mathbb{R}[/imath] the following equation holds: [imath]xf(y)+yf(x)=(x+y)f(x)f(y)[/imath] I've tried with some substitutions like [imath]x=a[/imath] and [imath]y=-a[/imath] for all [imath]a \neq 0[/imath] it follows that [imath]f(a)=f(-a)[/imath], so [imath]f[/imath] is an even function. After that, i realized that setting [imath]x=y=a[/imath] for [imath]a \neq 0[/imath] it follows that [imath]\Biggl(f(a)-\frac{1}{2} \Biggr)^{2}=\frac{1}{4}[/imath] and then [imath]f(x)=1[/imath] or [imath]f(x)=0[/imath] for all [imath]x \neq 0[/imath]. I think I'm almost done but I still don't know that happens with [imath]f(0)[/imath]. I would appreciate any hint to solve this problem, ¿Could you recommend any book for this kind of exercises? , ¡Thanks in advance! | 819565 | Find all [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]\forall x,y\in\mathbb R[/imath] the given equality holds: [imath]xf(y)+yf(x)=(x+y)f(x)f(y)[/imath].
Find all [imath]f:\mathbb R\to \mathbb R[/imath] such that [imath]\forall x,y\in\mathbb R[/imath] the given equality holds: [imath]xf(y)+yf(x)=(x+y)f(x)f(y)[/imath] My try: whenever [imath]y=0[/imath], we have [imath]x\cdot f(0)\cdot(1-f(x))=0[/imath] So the solutions can be either generated by: [imath]f(x)=\begin{cases}c & c\in\mathbb R & x=0\\ 1 & x\neq 0\end{cases}[/imath] where [imath]c[/imath] generate an infinite amount of solutions, or it can be any function satisfying [imath]f(0)=0[/imath]. However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown). Note that this generates the trivial solutions [imath]f(x)=0[/imath] and [imath]f(x)=1[/imath] as well. Is my reasoning correct? If so, how can I check whether all the generated functions satisfy the initial equation? Off-topic, but do [imath]f^2(x)[/imath] and [imath]f(x)^2[/imath] denote the same concept? My teacher uses the former while my book uses the latter (for squaring [imath]f(x)[/imath]). Which one is correct, or are both of them correct? |
3025663 | Prove [imath]\sum\limits_{r=0}^n{n\choose r}^2={2n\choose n}[/imath]
The question is to prove:[imath]\sum\limits_{r=0}^n{n\choose r}^2={2n\choose n}[/imath] using the binomial expansion of [imath](1+x)^n[/imath] and [imath](1+x^{-1})^n[/imath]. So firstly: [imath](1+x)^n=\sum_{i=0}^n\frac{n!.x^n}{i!.(n-i)!}[/imath] [imath](1+x^{-1})^n=\sum_{i=0}^n\frac{n!.x^{-n}}{i!.(n-i)!}[/imath] and we can work out that: [imath]\sum_{r=0}^n\left[\begin{pmatrix}n\\r\end{pmatrix}\right]^2=\sum_{r=0}^n\left(\frac{n!}{r!(n-r)!}\right)^2[/imath] but I am unsure of where to go from here | 793256 | Non-inductive, not combinatorial proof of [imath]\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n[/imath]
I've seen the identity [imath]\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n[/imath] used here recently. I checked for proofs here http://www.proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer... So my question is : do you know how to prove directly through computation that [imath]\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n[/imath] ? |
3026164 | How do you turn any quadratic into squared form
For example, [imath]x^2 + 5x + 7[/imath] is [imath](x + 2.5)^2 + 0.75[/imath] but how would you figure that out? It's useful for proving any quadratic is greater than 0 but it's not always easy to find so. Thanks! edit: Sorry I'm dumb I didn't see the + 0.75, this is just the vertex form. | 1307025 | I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?)
I was taught that you could only complete the square of a quadratic if the coefficient on the [imath]x^2[/imath] term is 1. However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the [imath]x[/imath] term by twice the square root of the coefficient of the [imath]x^2[/imath] term. So, if you have [imath]ax^2 + bx + c[/imath], your perfect square would be [imath](\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2[/imath] If [imath]a[/imath] is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there. For example: In the equation [imath]5x^2 + 6x + 5 = 0[/imath], we could do: [imath]25x^2 + 30x + 25 = 0[/imath] [imath](5x+3)^2 = -16[/imath] [imath]5x+3 = \pm4i[/imath] [imath]x = \pm \frac{4i}{5} - \frac{3}{5}[/imath] My questions are: -Is this correct? -Is there ever an advantage to using the quadratic formula? -Are there quadratics that are unsolvable this way? |
3026258 | Understanding the concept of conditional probability
We have [imath]X_1,..,[/imath] indepdentnt random variables with common distribution [imath]F(x)[/imath] and [imath]N[/imath] geometric rv independent of the X_i's . Let [imath]M = \max ( X_1,...,X_N)[/imath]. Im trying to understand the following: Im having trouble understanding the first and third equality. This is how I view it for the third equality [imath] P(M \leq x, N=n \mid N > 1 ) = \frac{P(M \leq x, N=n, N>1 )}{P(N > 1) } = \frac{P(M\leq x \mid N=n, N>1)P(N=n, N>1)}{P(N>1)} = \frac{P(M\leq x \mid N=n, N>1) P(N=n \mid N>1)P(N>1)}{P(N>1)}= P(M\leq x \mid N=n, N>1) P(N=n \mid N>1) [/imath] Is this the correct reasoning? Also, the first equality follows by definition? | 3025962 | Trying to derive a result on conditional probability
We have [imath]X_1,...[/imath] indepdent random variables with common distribution [imath]F[/imath] and [imath]N[/imath] is a geometric random variable independent of [imath]X_i[/imath]'s. Let [imath]M = \max(X_1,...X_N)[/imath]. Im trying to convince myself that [imath] P( M \leq x \mid N > 1 ) = F(x) P(M \leq x ) [/imath] I understand that since [imath]X_1[/imath] is independent of all [imath]X_i's[/imath] and from [imath]N[/imath], when we can pull it out of the probability and we have [imath] P(X_1 \leq x) P( \max(X_2,....,X_N ) \leq x \mid N > 1 ) [/imath] And then we have [imath] F(x) P( \max( X_1,...,X_{N-1} \leq x \mid N > 1 ) [/imath] How does [imath] P( \max( X_1,...,X_{N-1} \leq x \mid N > 1 ) = P( M \leq x ) [/imath] ?? |
3026644 | Why does the equation [imath]x^4 -17 = 2y^2[/imath] have no rational solution?
I am interested in understanding why [imath]x^4 - 17 = 2 y^2[/imath] has no rational solutions. My first approach was to show that every rational solution leads to an integral solution but that doesn't seem to be the case. | 688325 | Show that the curve [imath]2Y^2 = X^4 - 17[/imath] has no points in [imath]\mathbb{Q}[/imath]
There is a hint to show that if there were points in [imath]\mathbb{Q}[/imath], then there would exist [imath]r,s, t \in \mathbb{Z}[/imath] with [imath]\gcd(r, t) = 1[/imath] such that [imath]2s^2 = t^4 - 17r^4[/imath] , and then show that any prime dividing [imath]s[/imath] is a quadratic residue modulo [imath]17[/imath]. I'm stuck on the first bit - if you suppose [imath]\left(\frac{a}{b}, \frac{c}{d}\right)[/imath] is a point on the curve in [imath]\mathbb{Q}[/imath] (so [imath]a, b, c, d \in \mathbb{Z}[/imath]) then you get to [imath]2a^2d^4 = b^2c^4 - 17b^2d^4[/imath], but I'm not sure where to go from here since you don't know that [imath]b[/imath] is a square? |
3026829 | Prove equivalence of particular areas in generic quadrilateral
I am a university student taking a course in Mathematics Teaching, and one assignment is the following problem, to be solved using only Euclidean geometry as is presented in Euclid's Elements. Unfortunately I have no idea where to start. While I am required to present a very formalized solution, as if it were a proposition in Euclid's book, I am confident in my ability to produce it if I had just an idea for a solution to this problem. Here is the problem: Consider a quadrilateral [imath]ABCD[/imath], and [imath]M_1 ... M_4[/imath] the midpoints of its sides [imath]AB, ..., DA[/imath] respectively. Draw the segments [imath]AM_3, BM_4, ...[/imath] as in the picture. Show that the triangles [imath]t_1, ...., t_4[/imath] have total area equal to that of the quadrilateral [imath]q_1[/imath] Thank you. | 1401311 | Quadrilateral's area problem
I have some troubles with this problem : Let [imath]ABCD[/imath] be a convex quadrilateral. [imath]M[/imath], [imath]N[/imath], [imath]P[/imath] and [imath]Q[/imath] are the midpoints of the sides [imath]AB[/imath], [imath]BC[/imath], [imath]CD[/imath] and [imath]AD[/imath]. [imath]AN[/imath], [imath]BP[/imath], [imath]MD[/imath] and [imath]CQ[/imath] are interescting in [imath]X[/imath], [imath]Y[/imath], [imath]Z[/imath] and [imath]T[/imath] like in the figure below. Prove that [imath][XYZT] = [AMX] + [BYN] + [CZP] + [DTQ][/imath]. It is noted with [imath][ABC][/imath] the area of [imath]ΔABC[/imath]. Since [imath]M[/imath], [imath]N[/imath], [imath]P[/imath] and [imath]Q[/imath] are midpoints, the first thing that came in my mind was the median property : the median divides a triangle in two echivalent triangles (with the same area). I would appreciate some suggestions. Thanks! |
3026517 | A problem about three sequences.
Let [imath]\{a_k\}_{k\ge 0}, \{b_k\}_{k\ge 0}, \{\xi_k\}_{k\ge 0}[/imath] are non-negative sequences, for all [imath]k\ge 0[/imath], [imath]a_{k+1}^2 \le (a_k + b_k)^2 - \xi_k^2[/imath]. (1) Prove that [imath]\sum_{i=1}^k \xi_i^2 \le \left( a_1 + \sum_{i=0}^k b_i \right)^2[/imath]. (2) If [imath]\{b_k\}_{k\ge 0}[/imath] satisfy that [imath]\sum_{k=0}^{\infty} b_k^2 < +\infty[/imath], then [imath]\lim\limits_{k\to \infty} \frac{1}{k}\sum_{i=1}^k \xi_i^2 = 0[/imath]. Any hints are appreciated, thanks for your help. | 2901014 | Prove the inequality and limitation
Suppose [imath]\{a_k\}[/imath] ,[imath]\{b_k\}[/imath] and [imath]\{\xi_k\}[/imath] are non negative, and for all [imath]k\ge 0[/imath] , we have [imath]a_{k+1}^2\le(a_k+b_k)^2-\xi_k^2[/imath] Prove 1.[imath]\sum_{i=1}^{k}\xi_i^2\le(a_1+\sum_{i=0}^kb_i)^2[/imath] 2. If [imath]\{b_k\}[/imath] also satisfy [imath]\sum_{k=0}^{\infty}b_k^2\lt+\infty [/imath], then [imath]\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^k\xi_i^2=0[/imath] After doing this transformation:[imath]\xi_k^2\le(a_k+b_k)^2-a_{k+1}^2[/imath] [imath]\Rightarrow[/imath] [imath]\sum_{i=1}^{k}\xi_k^2\le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+\cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})[/imath] then I can't move forward any more. |
3026967 | How to change the system in 4x4 system of first-order equations?
Consider the coupled spring-mass system with a frictionless table, two masses [imath]m_1[/imath] and [imath]m_2[/imath], and three springs with spring constants [imath]k_1, k_2[/imath], and [imath]k_3[/imath] respectively. The equation of motion for the system are given by: [imath]y_1''=-\frac{(k_1+k_2)}{m_1}y_1+\frac{k_2}{m_1}y_2[/imath] [imath]y_2''=\frac{k_2}{m_2}y_1-\frac{(k_1+k_2)}{m_2}y_2[/imath] Assume that the masses are [imath]m_1 = 2[/imath], [imath]m_2 = 9/4[/imath], and the spring constants are [imath]k_1=1,k_2=3,k_3=15/4[/imath]. a)use 4x4 system of first order equations to model this system of two second order equations. (hint: [imath]x_1=y_1,x_2=y_2,x_3=y_1',x_4=y_2'[/imath]) | 3026925 | How to change the system in four first order equations?
Consider the coupled spring-mass system with a frictionless table, two masses [imath]m_1[/imath] and [imath]m_2[/imath], and three springs with spring constants [imath]k_1, k_2[/imath], and [imath]k_3[/imath] respectively. The equation of motion for the system are given by: [imath]m_1\frac{d^2x_1}{dt^2}=-(k_1+k_2)x_1+k_2x_2[/imath] [imath]m_2\frac{d^2x_2}{dt^2}=k_2x_1-(k_2+k_3)x_2[/imath] Assume that the masses are [imath]m_1 = 2[/imath], [imath]m_2 = \frac{9}{4}[/imath], and the spring constants are [imath]k_1=1,k_2=3,k_3=\frac{15}{4}[/imath]. I substitute all value in the system above, and able to get [imath]\frac{d^2x_1}{dt^2}=-2x_1+\frac{3}{2}x_2[/imath] [imath]\frac{d^2x_2}{dt^2}=\frac{4}{3}x_1-3x_2[/imath] Then how to convert the system about into four first order equations? |
3027031 | Prove/Disprove [imath]\sum a_n[/imath] and [imath]\sum b_n[/imath] converge together iff [imath]\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}/b_n } = 1[/imath]
Prove/Disprove that if [imath]\sum a_n[/imath] and [imath]\sum b_n[/imath] are some series and [imath]\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}/b_n } = 1[/imath] then the series converge together or not converge together. This doesn't seem to be correct to me so maybe there is some counter example. i know this is true if both series are strictly nonnegative (from the first series comparison test) - how does it change if both are negative (or alternate?) | 594661 | True/False: [imath]\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1[/imath] implies [imath]\sum {{a_n},\sum {{b_n}} } [/imath] converge or diverge together.
[imath]\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1[/imath] Prove the statement implies [imath]\sum {{a_n},\sum {{b_n}} } [/imath] converge or diverge together. My guess the statement is true. if [imath]\sum{{a_n}}[/imath] diverges, then [imath]\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0[/imath] So, [imath]\eqalign{ & \mathop {\lim }\limits_{n \to \infty } {a_n} = L \ne 0 \cr & {{\mathop {\lim }\limits_{n \to \infty } {a_n}} \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow {L \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {b_n} \ne 0 \cr} [/imath] therefore, [imath]\sum {b_n}[/imath] also diverges. What I was not managed to do is proving that the two series converges together. Or maybe the statement is not always true? |
3026831 | Why is Euler's reasoning correct in his proof that [imath]\sum_n n^{-2}=\frac{\pi^2}{6}[/imath]?
I was reading Euler's proof that [imath]\sum_n n^{-2}=\frac{\pi^2}{6}[/imath] and I don't agree with his reasoning. My issue is outlined below. First, Euler observed that the function [imath]\sin x[/imath] has roots at [imath]x = 0, ±π, ±2π, ±3π, \ldots[/imath]. Next, he observed that the infinite product [imath]x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{(2\pi)^2})(1-\frac{x^2}{(3\pi)^2})\ldots[/imath] also has roots at [imath]x = 0, ±π, ±2π, ±3π, \ldots[/imath]. Euler then believed that these two functions are equivalent. I have a problem with the above statement. Obviously, having the same roots does not imply that the two functions are equivalent. For instance, [imath]f(x)=0[/imath] also has roots at [imath]x = 0, ±π, ±2π, ±3π, \ldots[/imath]. Euler then goes on to use the infinite product and Maclaurin expansion of [imath]\sin x[/imath] to compare coefficients of [imath]x^3[/imath]. But his belief that those two functions are equivalent stains all subsequent parts of the proof. Am I missing something obvious here? Or was Euler not rigourous enough in his proof? | 341065 | A Question On Euler's Proof Of the Basel Problem
I've studied the proof that Euler gave for the famous Basel Problem, and it would seem that while it is technically correct, he does not justify all of his steps properly. Namely, he assumes that [imath]\frac{\sin(x)}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots[/imath] simply because they have the same roots, which is really not a strong enough condition. How do you really show that the equality holds? He then notices that if you use the above equality, and consider it against the Taylor expansion for [imath]\frac{\sin(x)}{x}[/imath], then you can equate the coefficients of the two infinite expansions at each order, and the result of the Basel problem follows. But how do you know that if you have two different expansions for a function, then their coefficients at each order must be equal? I would really appreciate if someone could show me how to make these two intuitive, yet informal, steps rigorous. |
3027082 | Are elements in [imath]A_n[/imath] the squares of elements in [imath]S_n[/imath]?
Tt is true that elements in [imath]A_4[/imath] are squares of elements in [imath]S_4[/imath]. Is it true for [imath]A_n[/imath]? | 1372267 | Characterisation of the squares of the symmetric group
I found out that for [imath]n\le 4[/imath] we have [imath]S_n^2=A_n[/imath] with [imath]G^2[/imath] defined by [imath]G^2:=\{g^2 \mid g\in G\}[/imath] for any group [imath]G[/imath]. Surely we have [imath]S_n^2\subseteq A_n[/imath] for all [imath]n\in\mathbb N[/imath]. Is there a characterisation of the squares of the symmetric group? When is [imath]S_n^2[/imath] a group and what does it look like in general? Edit: So far we found out, that [imath]S_n^2[/imath] generates [imath]A_n[/imath] because each 3-cycle is a square (since we have [imath]g=g^{-2}[/imath]) and [imath]A_n[/imath] is generated by the 3-cycles. So we simplified the original question to the question for which [imath]n[/imath] we have [imath]S_n^2=A_n[/imath] and how to characterise the elements which a missing in case we don't have the equality. |
3026930 | [imath][\mathsf{I},\mathsf{C}][/imath] and [imath][\mathsf{J},\mathsf{C}][/imath] are equivalent if [imath]\mathsf{I}[/imath] and [imath]\mathsf{J}[/imath] are.
Let [imath]\mathsf{I}[/imath] and [imath]\mathsf{J}[/imath] be equivalent categories. Let [imath]\mathsf{C}[/imath] be another category. I need to prove that the categories of functors [imath][\mathsf{I},\mathsf{C}][/imath] and [imath][\mathsf{J},\mathsf{C}][/imath] are equivalent. Let functors [imath]E\colon\mathsf{I}\to\mathsf{J}[/imath], [imath]E'\colon\mathsf{J}\to\mathsf{I}[/imath] together with natural isomorphisms [imath]\eta\colon 1_{\mathsf{I}} \Rightarrow E'E, \epsilon\colon EE' \Rightarrow 1_{\mathsf{J}}[/imath] define an equivalence of categories. Define a functor [imath]y\colon[\mathsf{I},\mathsf{C}]\to[\mathsf{J},\mathsf{C}][/imath] which maps each functor [imath]F\colon\mathsf{I}\to\mathsf{C}[/imath] to [imath]FE'[/imath] and each natural transformation [imath]\alpha\colon F\Rightarrow G[/imath] to the natural transformation [imath]\beta\colon FE'\Rightarrow GE'[/imath] such that for any [imath]j \in \mathsf{J}[/imath] we have [imath]\beta_j = \alpha_{E'(j)}[/imath]. I have proven that this functor is faithful and essentially surjective on objects, but I'm not sure how to prove that it is full. | 627676 | Equivalence of Categories and of Their Functor Categories
Suppose [imath]A, B, C[/imath] are categories. If [imath]A[/imath] and [imath]B[/imath] are equivalent, is it the case that [imath]C^A[/imath] and [imath]C^B[/imath] are equivalent? Also, is it the case that [imath]A^C[/imath] and [imath]B^C[/imath] are equivalent. I first conjectured that [imath]C^A[/imath] and [imath]C^B[/imath] are equivalent, but then as I tried to prove the conjecture, I started to realize that it might not be the case. So now I'm trying to find a counterexample. So far I have not been able to find one. Maybe I'm stuck with "too complicated" ideas. I strongly believe that there are very simple counterexamples. Please help. (For more detail, I got to the point where I have a cube that I want to be commutative, but I can only show that 4 of its sides are commutative squares. That's when I realize the other two sides may not necessarily commute.) And of course, there is the second part to the problem that's not just the dual problem. I would like to know about the special case in which [imath]A[/imath] is a preorder and [imath]B[/imath] is a partial order which is the skeleton of [imath]A[/imath]. This may have nothing to do with the validity of the statement in question at all, but it is the setting that originated this question. |
3027337 | Sum of "n" independent random variables.
Let [imath]X_1,X_2,...,X_n \sim U_{[0,1]}[/imath]. [imath]X_1,..,X_n[/imath] are independent. Let us define : [imath]Y=X_1+X_2+...+X_n[/imath] How to find the PDF of [imath]Y[/imath]? | 619552 | Density of the sum of [imath]n[/imath] uniform(0,1) distributed random variables
I am working on the following problem: Let [imath]X_1, X_2, \ldots, X_n, \ldots[/imath] be iid. random variables, each of Uniform[imath](0,1)[/imath] distribution. Denote by [imath]f_n(x)[/imath] the density of the random variable [imath]S_n := \sum_{k = 1}^n X_k[/imath]. Then \begin{align*} f_n(x) = \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{n}{k} (x-k)^{n-1}, \end{align*} where [imath][x][/imath] denotes the floor function. I tried this, but I think something went wrong: Prove by induction. For [imath]n = 1[/imath] \begin{align*} f_1(x) = \frac{1}{(1-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k} (x-k)^{1-1} &= \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k} =\begin{cases} 1, & 0 \le x < 1 \\ 0, & \text{otherwise} \end{cases} \end{align*} Furthermore \begin{align*} f_{n+1}(x) &= \mathbb P(S_{n+1} = x) = \mathbb P(S_n + X_{n+1} = x) \\ &= \sum_{m = 0}^{\infty} \mathbb P(S_n + X_{n+1} = x \mid X_{n+1} = m) \mathbb P(X_{n+1} = m) \\ &= \sum_{m = 0}^{\infty} \mathbb P(S_n + m = x) \cdot 1_{[0,1]}(m) = \mathbb P(S_n = x) + \mathbb P(S_n = x-1) \\ &= \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \frac{1}{(n-1)!} \sum_{k = 0}^{[x-1]} (-1)^k\binom{n}{k}(x-1-k)^{n-1} \\ &= \frac{1}{(n-1)!}\left(\sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 0}^{[x]-1} (-1)^k\binom{n}{k}(x-1-k)^{n-1}\right) \\ &= \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 1}^{[x]} (-1)^{k-1}\binom{n}{k-1}(x-1-(k-1))^{n-1}\right) \\ &= \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} - (-1)^{k}\binom{n}{k-1}(x-k)^{n-1}\right). \end{align*} How can I solve this? Edit: \begin{align*} f_{n+1}(x) &= (f_n * f_{X_{n+1}})(x) = \int_{-\infty}^\infty f_{n}(y) f_{X_{n+1}}(x-y) \, dy \\ &= \int_{-\infty}^\infty \frac{1}{(n-1)!} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(0,1)}(x-y) \, dy \\ &= \frac{1}{(n-1)!} \int_{-\infty}^{\infty} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(x-1, x)}(y)\, dy \\ &= \frac{1}{(n-1)!} \int_{x-1}^{x} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \, dy \end{align*} I want to swap the sum and the integral, but the sum depends on [imath]y[/imath]. |
3027339 | Proving inequality relation between two complex numbers and a positive real parameter.
Question: Prove that: [imath]|z_1+z_2|^2 \le (1+c)|z_1|^2+(1+{1\over c})|z_2|^2[/imath] where [imath]z_1,z_2[/imath] are complex numbers and [imath]c[/imath] is a positive real parameter. Solution: We can write [imath]|z|^2=z\bar z[/imath] Same way [imath](z_1+z_2)(\bar z_1+\bar z_2) \le (1+c)(z_1)(\bar z_1)+(1+{1\over c})(z_2)(\bar z_2)[/imath] which will yield [imath]z_2\bar z_1+z_1\bar z_2 \le cz_1\bar z_1+{1\over c}z_2\bar z_2[/imath] How to solve after this? Is this question correct? Am I doing correct? Post script: Offtopic This maybe irrelevant with the question. If you want to answer it you may or you may not. As I'm quite new to the topic. Can anyone of you suggest a good book for a beginner as there is nothing left in the book I'm reading. It's just introduction with few problems. This is the last one. How to pronounce [imath]\bar z[/imath]? Some pronounce z bar and some strictly prohibit saying the same. They rather say complex conjugate of z. Thanks for answering any of these questions or providing any hints. | 1598307 | Show that [imath]|z_1 + z_2|^2 < (1+C)|z_1|^2 + \left(1 + \frac{1}{C}\right) |z_2|^2[/imath]
Let [imath]z_1[/imath] and [imath]z_2[/imath] be two complex numbers. Show that there exists [imath]C > 0[/imath] with [imath] |z_1 + z_2|^2 < (1+C)|z_1|^2 + \left(1 + \frac{1}{C}\right) |z_2|^2. [/imath] I tried to simplify the L.H.S and R.H.S, SNF I was finally left to compare between a real number and a complex number I really couldn't think of anything else. Please help. |
3027827 | Show that these curve families are orthogonal: [imath]f(xy) = C[/imath] and [imath]y^2 - x^2 = D[/imath]
Let the function [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] be differentiable and its derivative is never zero. Show that these curves are orthogonal: [imath]f(xy) = C[/imath] [imath] y^2 -x^2 = D[/imath] [imath]C[/imath] and [imath]D[/imath] are real numbers. Edit: I checked the duplicate flag and I do not fully agree with it, for this question is somewhat different. Furthermore, after I posted (as my first post) this question I solved it and was writing an answer for it. It took some time because of the formatting. But then the post closed which made me lose my progress. In conclusion, to the curators who are reading this, do not dismiss questions so easily. Anyway, where is my answer: Let [imath] g(x,y) = xy [/imath] [imath] f\circ g = f(g(x,y)) = f(xy)[/imath] Differentiating with respect to [imath]x[/imath] the first two equations we get: [imath] (f(xy))' = 0 [/imath] [imath] 2yy' -2x = 0 [/imath] Applying the chain rule, and simplifying we get: [imath] (y+xy')f'(xy) = 0 [/imath] [imath] y' = \frac{x}{y} [/imath] Simplifying further the first equation (we can divide by [imath]f'(xy)[/imath] because its never zero) we get: [imath] y' = -\frac{y}{x} = m_1 [/imath] [imath] y' = \frac{x}{y} = m_2[/imath] Since [imath] m_1 = -\frac{1}{m_2} [/imath], we can assume that the tangent lines of the first family of curves is perpendicular to the tangent lines of the second family of curves. Making these curves orthogonal. Edit 2: The duplicate is more alike than I thought, the curator is most likely correct. | 3018519 | Show that the families of hyberbolae [imath]x^2-y^2=a[/imath], [imath]xy=b[/imath] create an orthogonal grid: what am I doing wrong?
Show that the families of hyberbolae [imath]x^2-y^2=a[/imath], [imath]xy=b[/imath] create an orthogonal grid. My idea is to find the intersection points of the hyperbolae, and then take a look at the [imath]y_1' \cdot y_2'[/imath] at those points. It should be [imath]-1[/imath]. From [imath]\begin{cases} |y|=\sqrt{x^2-a} \\ y=\frac{b}{x} \end{cases}[/imath] we get [imath]\sqrt{x^2-a}=\pm \frac{b}{x}[/imath]. My derivatives are [imath]y_1'=\frac{1}{\sqrt{x^2-a}}[/imath], [imath]y_2'= \frac{b}{x^2}[/imath]. [imath]y_1'\cdot y_2' = \pm \frac{x}{b} \cdot \frac{b}{x^2} = \pm \frac{1}{x} \neq -1[/imath]. I will be grateful for a hint or a full solution, but what most concerns me is where did my reasoning go astray. Thank you. |
3027678 | Find a matrix B such that [imath]B^5 = A[/imath]
I am being asked to find a matrix [imath]B[/imath] where [imath]B^5 = A[/imath] [imath]A = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}[/imath] In the first part of the question I was asked to find the eigenvalues & eigenvectors for the matrix which I found successfully. If someone could help me finish this then that would be great. | 1834826 | Find a matrix [imath]B[/imath] such that [imath]B^3 = A[/imath]
[imath]A=\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix}[/imath] Find a matrix [imath]B[/imath] such that [imath]B^3[/imath] = A My attempt: I found [imath]\lambda_1= 1+{\sqrt 2}[/imath] and [imath]\lambda_2= 1-{\sqrt 2}[/imath] I also found their corresponding eigenvectors [imath]\vec v_1 =\begin{pmatrix} \frac{-\sqrt 2}{2} \\ 1 \end{pmatrix}[/imath] and [imath]\vec v_2 = \begin{pmatrix} \frac{\sqrt 2}{2} \\ 1 \end{pmatrix}[/imath] I know the Power function of a matrix formula [imath]A=PDP^{-1}[/imath] Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ? |
3027600 | [imath]\mathbb{F}_{p^d}\subseteq\mathbb{F}_{p^n}[/imath] if and only if [imath]d[/imath] divides [imath]n[/imath]
I am trying to solve the following exercise of Dummit and Foote Book(page # 551). Let [imath]a>1[/imath] be an integer. Prove for any positive integers [imath]n,d[/imath] that [imath]d[/imath] divides [imath]n[/imath] if and only if [imath]a^d-1[/imath] divides [imath]a^n-1[/imath]. Conclude in particular that [imath]\mathbb{F}_{p^d}\subseteq\mathbb{F}_{p^n}[/imath] if and only if [imath]d[/imath] divides [imath]n[/imath]. I did the first part and I know that for all [imath]\alpha\in \mathbb{F}_{p^d}[/imath], [imath]\alpha^{p^d}=\alpha[/imath]. How can I apply the first part for the second? Any help is greatly appreciated. Thank you. | 2934644 | Proving that [imath]GF(p^n)[/imath] contains a unique subfield isomorphic to [imath]GF(p^m)[/imath] if and only if [imath]m[/imath] is a divisor of [imath]n[/imath].
I am looking to prove that [imath]GF(p^n)[/imath] contains a unique subfield isomorphic to [imath]GF(p^m)[/imath] if and only if [imath]m[/imath] is a divisor of [imath]n[/imath]. In the Wikipedia article: https://en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness They say that this statement has been proven by E. H. Moore in 1893. How to go about proving the theorem? I know first I need to generate a subset, prove that it's a subfield, and then assume they are two and show it's unique. But how to generate it in the first place? Consider the set of polynomials of the form [imath]x^{p^m}-x=0[/imath]? Is there a way to show that it's a subfield without checking all the axioms? |
3028338 | If [imath]A^2=I[/imath] and [imath]\lambda \not =-1[/imath], then [imath]A=I[/imath]
Given an [imath]n\times n[/imath] matrix [imath]A[/imath] with [imath]A^2=I[/imath], assume that [imath]-1[/imath] is not an eigenvalue of [imath]A[/imath]. Prove [imath]A=I[/imath]. Proof attempt: Since [imath]A^2=I,[/imath] we have [imath]A^{-1}=A[/imath]. Using that fact that [imath]\det(A)=\prod_{i=1}^n\lambda_i[/imath], we see that each of the eigenvalues is nonzero. For any eigenvector [imath]v[/imath] of [imath]A[/imath], we see [imath]Av=\lambda v[/imath], so [imath]\frac{1}{\lambda} v = A^{-1}v=Av[/imath]. This implies [imath]\lambda^2=1[/imath]. By assumption, we conclude that each eigenvalue is [imath]1[/imath]. Hence, [imath]\det A=1[/imath]. (1) How do we know that [imath]A[/imath] actually has any eigenvectors? E.g., the map which rotates a vector in the plane by [imath]\pi/2[/imath] clockwise is invertible but has no eigenvectors. (Perhaps there should be some given information about the underlying field.) (2) How do I use [imath]\lambda=1[/imath] to conclude [imath]A=I[/imath]? I tried using [imath]A=A^{-1} = \frac{C^T}{\det A}=C^T[/imath], where [imath]C[/imath] is the matrix of cofactors of [imath]A[/imath]. From this, I conclude that each entry of [imath]A[/imath] equals its corresponding co-factor -- i.e., [imath]A_{i,j} = \det(\overline{A_{i,j}})[/imath], where [imath]\overline{A_{i,j}}[/imath] is the sub-matrix of [imath]A[/imath] obtained by deleting row [imath]i[/imath] and column [imath]j[/imath]. Using the co-factor expansion of the determinant along row [imath]i[/imath], we also conclude [imath]1=\det A=\sum_{j=1}^n A_{i,j}(-1)^{i+j}\det(\overline{A_{i,j}})=\sum_{j=1}^n(A_{i,j})^2(-1)^{i+j}.[/imath] | 3024796 | How can I prove that a linear map [imath]T:V\to V[/imath] such that [imath]T^2 = I[/imath] is equal to the Identity map?
Suppose [imath]T \in \mathcal{L}(V)[/imath], [imath]T^2=I[/imath], and [imath]-1[/imath] is not an eigenvalue of [imath]T[/imath]. Prove that [imath]T=I[/imath]. I'm not sure where to start, though I know I somehow have to show that [imath]T \circ T[/imath] yields the identity matrix, but there isn't much information given. |
714605 | A question on primitive root of unity
Let n be a positive integer and let [imath]\alpha[/imath] , [imath]\beta[/imath] be primitive n-th roots of unity. a) Show that [imath]\frac{1-\alpha}{1-\beta}[/imath] is an algebraic integer. b) If [imath]n\geq 6[/imath] is divisible by at least two primes ,show that [imath]1-\alpha[/imath] is a unit in the ring [imath]\mathbb Z[\alpha][/imath]. For a) I tried with [imath]\alpha =\omega[/imath] and [imath]\beta= \omega^2[/imath] primitive cube rots of unity and got the polynomial [imath]x^2-x+1[/imath]. But for nth case the calculation is going tedious.Please help. b)Here we have only to show that [imath]\alpha [/imath] is a nilpotent element.Then we are done.But how to show [imath]\alpha[/imath] is nilpotent? Thanks in advance! | 216035 | Why [imath](1-\zeta)[/imath] unit where [imath]\zeta[/imath] is a primitive nth and n divisible by two primes
From Chapter VII of Lang's Algebra. The question asks if [imath]n\geq 6[/imath] and [imath]n[/imath] is divisible by at least two primes, show that [imath]1-\zeta[/imath] is a unit in the ring [imath]\mathbb{Z}[\zeta][/imath] I am having a hard time understanding why this is true. This is in the integral dependence chapter, but that has not given me any inspiration. I have also tried using cyclotonic polynomial to no avail Thanks for any direction. |
3028278 | [imath]f,g[/imath] cont. functions on [imath][a,b][/imath] and [imath]g[/imath] is non-negative. then [imath]\exists c \in [a,b][/imath] such that [imath]\int_{a}^{b} f(x)g(x) dx = f(c) \int_{a}^{b} g(x) dx.[/imath]
let [imath]f,g[/imath] be continuous functions on [imath][a,b][/imath] and suppose that [imath]g[/imath] is non-negative. then there exists apoint [imath]c \in [a,b][/imath] such that [imath]\int_{a}^{b} f(x)g(x) dx = f(c) \int_{a}^{b} g(x) dx.[/imath] I have seen a solution similar to my problem in the link below, except that this solution assumes that [imath]g>0,[/imath] but my problem assumes that [imath]g \geq 0[/imath], does my problem contain a mistake? or the result is true also if g is non negative? Existence of [imath]c[/imath] such that [imath]\int_{a}^{b} f(x)g(x)dx = f(c)\int_{a}^{b} g(x)[/imath] | 1068594 | If [imath]f[/imath] is continuous and [imath]g[/imath] is integrable on [imath][a,b][/imath], with [imath]g(x) \ge 0[/imath] for all [imath]x \in [a,b][/imath] ...
Suppose [imath]f : [a,b] \to \mathbb{R}[/imath] is continuous and [imath]g \in \mathcal{R}[a,b][/imath] with [imath]g(x) \ge 0[/imath] for all [imath]x \in [a,b][/imath]. Show that there exists a [imath]c \in [a,b][/imath] such that [imath]\int_a^b f(x)g(x) \, dx = f(c) \int_a^b g(x) \, dx.[/imath] My attempt: My class proved a theorem which states that [imath]f[/imath] attains its minimum and maximum ([imath]m[/imath] and [imath]M[/imath], respectively), since [imath]f[/imath] is continuous on a closed interval like [imath][a,b][/imath]. So we would have [imath]m \le f(x) \le M[/imath]. Multiplying both sides by [imath]g[/imath] and integrating, we get [imath]m \int_a^b g(x) \, dx \le \int_a^b f(x) g(x) \, dx \le M \int_a^b g(x) \, dx.[/imath] I am stuck after this. Where can I go from here? |
3028668 | Find [imath]\lim_{n\to \infty}((n+1)!)^{\frac{1}{n+1}}-((n)!)^{\frac{1}{n}}.[/imath]
Find [imath]\lim_{n\to \infty}((n+1)!)^{\frac{1}{n+1}}-((n)!)^{\frac{1}{n}}.[/imath] We need to deal the limit [imath]\lim_{n\to \infty} \frac{\log(1)+\log(2)+...+\log(n)}{n}[/imath]. We know that [imath]\lim_{n\to \infty} \log(n)=\infty \implies \lim_{n\to \infty} \frac{\log(1)+\log(2)+...+\log(n)}{n}=\infty[/imath](since, By Cauchy's first theorem on limit). Hence we get [imath]\infty-\infty[/imath]. How do I show that there exists finite limit? | 2020153 | Limit of the sequence [imath]a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}[/imath]
Find the limit of the sequence [imath]a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}.[/imath] The limit is supposed to be [imath]e^{-1}[/imath] and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is [imath]1.[/imath] Please tell me what's going wrong here. Proof: We can write [imath]a_n=\sqrt[n]{n!}(b_n-1),[/imath] where [imath]$b_n=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}.$[/imath] Hence [imath]$$a_n=n\cdot\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}=\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}\cdot\ln{b_n^n.}$$[/imath] But [imath]$\lim \limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=e^{-1},$[/imath] so [imath]b_n\to1[/imath] as [imath]n\to \infty.[/imath] On the other hand, [imath]\lim \limits_{n\to \infty}b_n^n=\lim \limits_{n\to \infty}\frac{(n+1)!}{n!}.\frac{1}{\sqrt[n+1]{(n+1)!}}=e.[/imath] Also [imath]\lim \limits_{n\to \infty}\frac{b_n-1}{\ln{b_n}}=\lim \limits_{n\to \infty}\frac{1}{\ln{[1+(b_n-1)]^{1/(b_n-1)}}}=\frac{1}{\ln e}=1[/imath] And the author goes on to say that the limit is [imath]e^{-1}.[/imath] But doesn't it say that the limit is [imath]1?[/imath] I must have overlooked something so obvious. |
3027852 | How to find the integral submanifold?
Suppose [imath]U \subseteq R^3[/imath] is the subset that all three coordinates are positive. Let [imath]D[/imath] be the distribution on [imath]U[/imath] spanned by two vector fields: [imath]X = y\frac{\partial }{\partial z}-z\frac{\partial }{\partial y}[/imath] , [imath]Y=z\frac{\partial }{\partial x}-x\frac{\partial }{\partial z}[/imath]. Then how to find the integral submanifold? I had shown that [imath][X, Y]_p = \frac{y(p)}{z(p)}Y_p+\frac{x(p)}{z(p)}X_p[/imath], hence [imath]D[/imath] is involutive. I also know how to calculate the integral curve of [imath]X[/imath] and [imath]Y[/imath]. But I dont know how to proceed to calculate the integral submanifold specifically. | 3026335 | find the general integral manifolds of a given distribution
Let [imath]D[/imath] be the distribution on [imath]M=\{(x,y,z),x,y,z>0\}[/imath] generated by [imath]X=y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}[/imath] and [imath]Y=z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}[/imath], show [imath]D[/imath] is involutive and find the general integral manifolds of [imath]D[/imath]. I compute the flows of [imath]X[/imath] and [imath]Y[/imath], there are a lot of [imath]\sin[/imath] and [imath]\cos[/imath] in the form of flows of [imath]X[/imath] and [imath]Y[/imath] , I got [imath]\phi_t(x,y,z)=(x,-z\sin t+y\cos t,y\sin t+z\cos t)[/imath] and [imath]\psi_s(x,y,z)=(z\sin s+x\cos s,y,-x\sin s+z\cos s)[/imath], but to find the integral manifolds at last, I can not eliminate [imath]s[/imath] and [imath]t[/imath] to get a general form |
2863930 | Show, using the definition of derivative, that [imath]f:\mathbb{R}^2\rightarrow\mathbb{R}[/imath] defined by [imath]f(x,y)=xy[/imath] is differentiable everywhere.
This question has the following hint: Find a linear function [imath]L(x,y)[/imath] such that [imath]f(x+h,y+k)-f(x,y)-L(h,k)=hk[/imath]. Note that [imath]|hk|\leq l^2[/imath] where [imath]l:=max\{|h|,|k|\}[/imath]. Does this help you establish that a certain limit is zero? To establish that the limit is zero, provide an [imath]\epsilon-\delta[/imath] proof. My initial post questioned the linear function part; hence the first couple comments. I'm now adding my proposed solution that I'd like to have critiqued. Let [imath]h,k\in\mathbb{R}[/imath] such that [imath]|hk|\leq l^2[/imath] where [imath]l:=\mbox{max}\{|h|,|k|\}[/imath], and suppose [imath]L(h,k)=kx+hy[/imath]. Thus, [imath]\frac{|f(x+h,y+k)-f(x,y)-L(h,k)}{||(h,k)||}=\frac{|(x+h)(y+k)-xy-(kx+hy)|}{\sqrt{h^2+k^2}}=\frac{|hk|}{\sqrt{h^2+k^2}}.[/imath] Then, note that [imath]|h|^2\leq h^2+k^2\Rightarrow|h|\leq\sqrt{h^2+k^2}\Rightarrow\frac{|h|}{\sqrt{h^2+k^2}}\leq1\Rightarrow\frac{|hk|}{\sqrt{h^2+k^2}}\leq|k|.[/imath] Hence, [imath]\frac{|f(x+h,y+k)-f(x,y)-L(h,k)}{||(h,k)||}=\frac{|hk|}{\sqrt{h^2+k^2}}\leq|k|.[/imath] So, [imath]\lim_{(h,k)\rightarrow(0,0)}\frac{|f(x+h,y+k)-f(x,y)-L(h,k)}{||(h,k)||}=0.[/imath] Therefore, by the definition of derivative, [imath]f[/imath] is differentiable everywhere. It seems to me that an [imath]\epsilon-\delta[/imath] proof is unnecessary, but maybe I'm implicitly making an assumption that needs proved? | 2338904 | Prove that [imath]f(x,y)=xy[/imath] is differentiable using the definition
Let [imath]f:\mathbb R^2\to \mathbb R[/imath] defined by [imath]f(x,y)=xy[/imath]. Prove [imath]f[/imath] is differentiable. According to the definition I have to show this [imath]\frac{\|f((x_0,y_0)+(h_1,h_2)) - f(x_0,y_0)-T(h_1\space h_2)\|}{\|(h_1,h_2)\|}\to 0[/imath] But I got stuck in the proof. \begin{align} & \frac{\|f((x_0,y_0)+(h_1,h_2))-f(x_0,y_0)-T(h_1\space h_2)\|}{\|(h_1,h_2\|} \\[10pt] = {} & \frac{\|(x_0+h_1)(y_0+h_2)-x_0y_0-(x,y)(h_1\space h_2)\|}{\sqrt{h_1^2+h_2^2}} \\[10pt] \le {} &\frac{\|x_0h_2+y_0h_1+h_1h_2\|}{\sqrt{h_1^2+h_2^2}}+\frac{\|-(x,y)(h_1\space h_2)\|}{\sqrt{h_1^2+h_2^2}} \end{align} and I think I the first term after the inequality goes to zero, but what happen to the second term? How can it goes to zero too? |
2514308 | If [imath]F\subseteq L\subseteq K[/imath] are fields, then show that [imath][K : F]_s = [K: L]_s[L: F]_s[/imath] and [imath][K: F]_i = [K: L]_i[L: F]_i[/imath]
Prove the following product formulas for separability and inseparability degree: If [imath]F\subseteq L\subseteq K[/imath] are fields, then show that [imath][K : F]_s = [K: L]_s[L: F]_s[/imath] and [imath][K: F]_i = [K: L]_i[L: F]_i[/imath] This question has already been posted twice here show that [imath][K:F]_s=[K:L]_s[L:F]_s[/imath] and [imath][K:F]_i=[K:L]_i[L:F]_i[/imath]. , and here [imath][K : F]_s = [K : L]_s [L : F]_s [/imath] and [imath][K : F]_i = [K : L]_i [L : F]_i [/imath], but in one they do not answer and in another I do not understand the answer because they use another definition for this grade. I am following another book and in that book they define it in the following way: Let [imath]K[/imath] be a finite extension of [imath]F[/imath]. If [imath]S[/imath] and [imath]I[/imath] are the separable and purely inseparable closures of [imath]F[/imath] in [imath]K[/imath], respectively, we define the separable degree [imath][K : F]_s[/imath] of [imath]K/ F[/imath] to be [imath][S : F][/imath] and the insepaTable degree [imath][K : F]_i[/imath] to be [imath][K : S][/imath]. With these definitions, we see that [imath][K: F]_s[K : F]_i = [K : F][/imath]. I then have that [imath][K: L]_s[L: F]_s=[S_1:L][S_2:F][/imath] where [imath]S_1[/imath] and [imath]S_2[/imath] are the separable closure of [imath]L[/imath] in [imath]K[/imath] and [imath]F[/imath] in [imath]L[/imath] respectively, how can I relate this to [imath][K:F]_s=[S:F][/imath]? | 1480158 | Show that [imath][K:F]_s = [K:L]_s [L:F]_s[/imath] and [imath][K:F]_i = [K:L]_i [L:F]_i[/imath].
This is a problem from Patrick Morandi's Field and Galois Theory: Chapter I.4, exercise 15. Let [imath]K[/imath] be a finite extension of [imath]F[/imath]. If [imath]S=\{x\in K \mid x \text{ is separable over } F \}[/imath] and [imath]I=\{x\in K \mid x \text{ is purely inseparable over } F \}[/imath] are the separable and purely inseparable closures of [imath]F[/imath] in [imath]K[/imath], respectively, we define the separable degree [imath][K:F]_s = [S:F][/imath] and the purely inseparable degree [imath][K:F]_i = [K:S][/imath]. Now using these definitions, Prove the following product formulas for separability and inseparability degree: If [imath]F \subseteq L \subseteq K[/imath] are fields, then show that [imath][K:F]_s = [K:L]_s [L:F]_s[/imath] and [imath][K:F]_i = [K:L]_i [L:F]_i[/imath]. Proving just one of the equality will be enough (thanks to Tower law and the property that [imath][K:F]_s [K:F]_i = [K:F][/imath]). I started proving: [imath][K:F]_s=[K:L]_s[L:F]_s[/imath]. Suppose [imath][K:L]_s=[S_1:F][/imath] and [imath][L:F]_s=[S_2:F][/imath] where [imath]S_1[/imath] and [imath]S_2[/imath] are separable closures of [imath]K/L[/imath] and [imath]L/F[/imath] respectively. And if we take [imath]\{a_i \mid i=1,\dots,m\}[/imath] and [imath]\{b_i \mid i=1,\dots,n\}[/imath] are the basis of [imath]K/L[/imath] and [imath]L/F[/imath] respectively. Is it true that [imath]\{a_ib_j \mid i=1,\dots,m, j=1,\dots,n\}[/imath] is a basis of [imath]S/F[/imath]? Then for [imath]x \in S[/imath] how do we proceed? |
3029512 | Integrability of composite functions
Let [imath]f[/imath] be a Riemann-integrable function on a closed interval [imath][a,b] \subset \mathbb{R}[/imath]. Let g be a function on [imath]\mathbb{R}[/imath]. What conditions must g satisfy so that [imath]g \circ f[/imath] is also Riemann-integrable ? Thank you! | 1060834 | Composition of two Riemann integrable functions
Suppose [imath]f,g[/imath] are two Riemann Integrable functions .Is it true that [imath]f\circ g[/imath] is also Riemann Integrable? Trying this for a long time but not getting the answer |
3029721 | Prove that [imath]\int_{0}^{\pi/2} \frac{\sin x}{x} dx[/imath] converges
Prove that [imath]\int_{0}^{\pi/2} \frac{\sin x}{x} dx[/imath] converges. Shall I compare it with 1/x? | 688342 | Is [imath] \int_0^1{\frac{\sin x}{x}dx} [/imath] convergent?
I have this integral: [imath] \int_0^1{\frac{\sin x}{x}dx} [/imath] And I should prove that it is convergent. I have understand that if the resulting area is finite, then this integral is convergent, right? So, I do this: [imath] \int_0^1{\frac{\sin x}{x}dx} = \int_0^1{\sin x\cdot \frac{1}{x}dx} =\\= \begin{bmatrix} -\cos x\cdot \ln x\end{bmatrix}_0^1 = (-\cos1\cdot \ln1)-(\cos0\cdot \ln0) = 0-(-\infty) = \infty [/imath] But, I get an infinit value, which I think is wrong. What am I doing wrong? Ok, so apparently I messed some things up. If I do this again and use the integration by parts method I get this: [imath] \int_0^1{\frac{\sin x}{x}dx} = \int_0^1{\sin x\cdot \frac{1}{x}dx} = \sin x \ln x- \int_0^1{(\cos x\ln x) dx} = \sin x \ln x -\begin{bmatrix} \sin\frac{1}{x} \end{bmatrix}_0^1 \Rightarrow \lim_{t \rightarrow 0} {\int_t^1\frac{\sin x}{x} dx} = \lim_{t \rightarrow 0}{(\sin x \ln x-\frac{sin1}{1}-\frac{\sin t}{t})} [/imath] Is that more correct? And how do I proceed? |
3029746 | "Find the general solution of the Euler homogenous equation [imath]\frac{dy}{dx}=\frac{2y-x}{2x-y}[/imath]."
"Find the general solution of the Euler homogenous equation [imath]\frac{dy}{dx}=\frac{2y-x}{2x-y}[/imath]." Through my working out, after introducing a new variable v, where y=xv, I end up having to solve the following: [imath]\frac{1}{2}\int\frac{1}{v-1}dv-\frac{3}{2}\int\frac{1}{v+1}dv[/imath] = [imath]\int\frac{1}{x} dx[/imath] This leads me to the result of [imath]\frac{3}{2}ln\frac{(v-1)^3}{(v+1)}=ln(x)+c[/imath] Before I begin to rearrange this, to get in terms of y, could someone tell me if I am going in the correct direction because it looks as though I'm going to end up with an [imath]e^c[/imath] term, and I am unsure whether this is correct or not. Ideally, could someone provide an answer they get from solving this problem - just so I can see if I am going in the correct direction/know if I am correct once I reach an answer? Thank you. I have been attempting this question for a while now | 3029351 | Find the general solution of the Euler homogenous equation
"Find the general solution of the Euler homogenous equation [imath]\frac{dy}{dx}=\frac{2y-x}{2x-y}[/imath]." Through my working out, after introducing a new variable v, where y=xv, I end up having to solve the following: [imath]\frac{1}{2}\int\frac{1}{v-1}dv-\frac{3}{2}\int\frac{1}{v+1}dv[/imath] = [imath]\int\frac{1}{x} dx[/imath] This leads me to the result of [imath]\frac{3}{2}ln\frac{(v-1)^3}{(v+1)}=ln(x)+c[/imath] Before I begin to rearrange this, to get in terms of y, could someone tell me if I am going in the correct direction because it looks as though I'm going to end up with an [imath]e^c[/imath] term, and I am unsure whether this is correct or not. Ideally, could someone provide an answer they get from solving this problem - just so I can see if I am going in the correct direction/know if I am correct once I reach an answer? Thank you. |
3029295 | Evaluating [imath]\lim_{x\to \infty}(\frac{x}{x-1})^x[/imath]
I am going over a solution given to solving the follow limit, [imath]\lim_{x\to \infty}(\frac{x}{x-1})^x[/imath] The solution continues as follows, Consider raising the function to [imath]e^{ln\cdots}[/imath] We can find the limit as follows, [imath]\lim_{x\to \infty} x \ln(\frac{x}{x-1}) = \lim_{x\to \infty} \frac{\ln(\frac{x}{x-1})}{\frac{1}{x}}[/imath] The solution argues this is just [imath]\frac{0}{0}[/imath] and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is [imath]e[/imath]. However, I don't understand how that expression evaluates to [imath]\frac{0}{0}[/imath], in fact it seems to express [imath]\frac{\ln(\frac{\infty}{\infty})}{0}[/imath] I assume the argument is that [imath]\frac{\infty}{\infty}[/imath] equals 1, and [imath]\ln(1) = 0[/imath], so we have [imath]\frac{0}{0}[/imath]. But I thought we cannot evaluate [imath]\frac{\infty}{\infty}[/imath]? | 1553948 | Prove that [imath]\lim_{x \to \infty}\big(\frac{x}{x-1}\big)^x[/imath] is also [imath]e[/imath].
Trying to make sense out of the idea that [imath]100\%[/imath] continuous decay is [imath]\frac{1}{e}[/imath], I thought about this: You can express [imath]1+\frac{1}{x}[/imath] as [imath]\frac{x+1}{x}[/imath], such that [imath]\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x[/imath] And you can express [imath]1-\frac{1}{x}[/imath] as [imath]\frac{x-1}{x}[/imath], such that [imath]\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}[/imath] Now [imath]\big(\frac{x}{x-1}\big)^x[/imath] and [imath]\big(\frac{x+1}{x}\big)^x[/imath] look very similar, and I can imagine (and see on Mathematica) that [imath]\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x[/imath] also approaches [imath]e[/imath]. However I'm clueless about limit proofs, how do you prove that? EDIT: Just had a last second insight. Can you say that [imath]\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)[/imath] [imath]= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1[/imath] [imath]=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e ?[/imath] (not sure about that last limit transition...I mean I'm pretty sure it's true, just not sure how to write it; I would appreciate feedback on that, thank you) |
3028910 | Show that [imath]\int_{\mathbb{R}} |\mathbb{E}(e^{i \xi X})|^2 \, d\xi < \infty[/imath] implies [imath]P(X=x)=0, \forall x\in\mathbb{R}[/imath].
Let [imath]X[/imath] be a real-valued random variable with [imath]\varphi_X\in L^2(\mathbb{R})[/imath], where [imath]\varphi_X[/imath] denotes the characteristic function of [imath]X[/imath]. Prove that [imath]P(X=x)=0, \forall x\in\mathbb{R}[/imath]. Since [imath]\varphi_X\in L^2(\mathbb{R})[/imath], I believe that implies that [imath]X[/imath] must be continuous, then we use that to conclude that [imath]P(X=x)=0[/imath] for all [imath]x\in\mathbb{R}[/imath]. However, I am having a hard time formalizing this idea. Any help would be appreciated. | 2083508 | If characteristic function is [imath]L^2[/imath] does it mean that the distribution is absolutely continuos?
Suppose, we have a characteristic function [imath]\phi(t)[/imath]. Suppose that \begin{align} \int_{-\infty}^\infty |\phi(t)|^2 dt <\infty \end{align} Does this mean the corresponding probability distribution is absolutely continuous? I found that if the following condition holds \begin{align} \lim _{T \to \infty} \frac{1}{2T}\int_{-T}^T |\phi(t)|^2 dt=0 \end{align} then the distribution is contionous. However, I think they mean continuous and I am not sure if this implies existence of density. |
3029710 | Let [imath]T = \frac{X}{S}[/imath] with X a random exponential variable of parameter lambda and S gamma random variable
Let [imath]T = \frac{X}{S}[/imath] with X a random exponential variable of parameter lambda and S gamma random variable of this form: [imath]\Gamma(2,\lambda)[/imath]. I want to find the density of T. Know, I tried to find [imath]F_T(t)[/imath], so that I can derive and obtain the desired result. This is what I did: [imath] F_T(t) = \mathbb{P}[T \leq t] = \mathbb{P}[\frac{X}{S} \leq t][/imath] Then I am not sure if I should write S = X + Y with X and Y independent random variable of exponential distribution [imath]\lambda[/imath] or if I should continue with the definition. I tried to keep going with the definition: [imath] \mathbb{P}[\frac{X}{S} \leq t] = \mathbb{P}[X \leq tS] [/imath] However now I cannot say that this is equal to [imath]F_X(tS)[/imath] because I have S as random variable, so I tried to write it in this way: [imath]\mathbb{P}[X \leq ts \ \vert S = s][/imath] So that I can use Bayes theorem and write: [imath]\frac{\mathbb{P}[X \leq ts \cap S = s]}{\mathbb{P}[S = s]}[/imath] However here I am not sure on how to proceed. Any idea? | 76175 | Distribution of Ratio of Exponential and Gamma random variable
A recent question (Are there any (pairs of) simple distributions that give rise to a power law ratio?) asked about the distribution of the ratio of two random variables, and the answer accepted there was a reference to Wikipedia which (in simplified and restated form) claims that if [imath]X[/imath] is a Gamma random variable with parameters [imath](n, 1)[/imath] and [imath]Y[/imath] is an exponential random variable with parameter [imath]1[/imath], then [imath]Y/X[/imath] is a Pareto random variable with parameters [imath](1, n)[/imath]. Presumably [imath]X[/imath] and [imath]Y[/imath] need to be independent for this result to hold. But, Wikipedia's page on Pareto random variables doesn't seem to include a statement as to what [imath](1, n)[/imath] means, though based on what it does says, a reasonable interpretation is that the Pareto random variable takes on values in [imath](1,\infty)[/imath] and its complementary CDF decays away as [imath]z^{-n}[/imath]. My question is: what is the intuitive explanation for the ratio [imath]Y/X[/imath] to have value [imath]1[/imath] or more? It would seem that all positive values should occur, and indeed the event [imath]\{Y < X\}[/imath] should have large probability since the Gamma random variable has larger mean than the exponential random variable. I did work out the complementary CDF of [imath]Y/X[/imath] and got [imath](1+z)^{-n}[/imath] for [imath]z > 0[/imath] which is not quite what Wikipedia claims. Added Note: Thanks to Sasha and Didier Piau for confirming my calculation that for [imath]z > 0[/imath], [imath]P\{Y/X > z\} = (1+z)^{-n}[/imath] which of course implies that [imath]P\{Y/X + 1 > z\} = P\{Y/X > z-1\} = (1 + z - 1)^{-n} = z^{-n}~ \text{for}~ z > 1[/imath] and thus it is [imath]Y/X + 1[/imath] which is a Pareto random variable, not [imath]Y/X[/imath] as claimed by Wikipedia. This leads to a simple answer to a question posed by S Huntsman: Are there any (pairs) of simple distributions that give rise to a power law ratio? If [imath]W[/imath] and [imath]X[/imath] are the [imath](n+1)[/imath]-th and [imath]n[/imath]-th arrival times in a (homogeneous) Poisson process, then [imath]W/X[/imath] is a Pareto [imath](1,n)[/imath] random variable: [imath]P\{W/X > a\} = a^{-n}[/imath] for [imath]a > 1[/imath]. I suspect that this result is quite well known in the theory of Poisson processes but I don't have a reference for it. |
3031623 | Normal Subgroup Representation Theory
Let [imath]N \subset G[/imath] be a normal subgroup. Show there exists a finite collection of irreducible representations [imath]\phi_i: G \rightarrow GL(V_i)[/imath] of [imath]G[/imath] such that [imath] N = \bigcap{\ker(\phi_i)}[/imath] | 2594993 | Show that every normal subgroup can be written as an intersection of [imath]\ker \chi[/imath]'s for [imath]\chi \in \operatorname{Irr}(G).[/imath]
The question is as follows: This can be the third part of this question Linked to the results: Let [imath]G[/imath] be a finite group and [imath]N[/imath] a normal subgroup of [imath]G[/imath]. c) Show that [imath]N[/imath] is the intersection of the sets of the form [imath]\ker \xi[/imath] that contain [imath]N[/imath] with [imath]\xi \in \operatorname{Irr}(G)[/imath]. Some attempts: For the finite group [imath]G[/imath] and [imath]\rho[/imath] a representation with induced character [imath]\chi_{\rho}[/imath], we have [imath]\ker \rho = \{ g \in G: \chi_{\rho}(g) = \chi_{\rho}(e) \} [/imath] by Lemma 15.17; Isaacs "Algebra: A Graduate Course". Then it makes sense to define the kernel of the character [imath]\chi[/imath], denoted [imath]\ker \chi[/imath] by [imath]\ker \chi = \{ g \in G: \chi(g) = \chi(e) \}[/imath]. In particular, we know that for every character [imath]\chi[/imath] one has that [imath]\ker \chi \unlhd G.[/imath] For the irreducible characters [imath]\chi^{(\alpha)}[/imath], [imath]\alpha \in \hat{G}[/imath], we give the special symbols [imath]N^{(\alpha)}[/imath] for [imath]\ker \chi^{(\alpha)}[/imath]. Now what we are going to show is that knowing [imath]N^{(\alpha)}[/imath] for every [imath]\alpha \in \hat{G}[/imath] enables one to know [imath]\ker \chi[/imath] for every character [imath]\chi[/imath]. Indeed, if we let [imath]\chi[/imath] be a character with representation as a linear combination of the irreducible characters [imath]\chi = \sum_{\alpha \in \hat{G}} m^{(\alpha)}\chi^{(\alpha)}[/imath], then we have [imath]\ker \chi =\bigcap \{ N^{(\alpha)} : m^{(\alpha)} > 0 \}.[/imath] Because if [imath]\chi^{(\alpha)} (g) = d_{\alpha}[/imath] for every [imath]\alpha[/imath] such that [imath]m^{(\alpha)} > 0[/imath] one sees that [imath]\chi (g) = \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \chi^{(\alpha)} (g) = \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \chi^{(\alpha)} (e) = \chi (e) [/imath] and so [imath]g \in \ker \chi[/imath]. Conversely, since one evidently has that [imath]|\chi^{(\alpha)} (g)| \le d_{\alpha}[/imath] for every [imath]\alpha \in \hat{G}[/imath] we see that for [imath]g \in \ker \chi[/imath] one has that \begin{align} |\chi (g)| &= \left| \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \chi^{(\alpha)} (g) \right|\\ &\le \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \left| \chi^{(\alpha)} (g) \right| \\& \le \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} d_{\alpha} \\ &= \chi(1) = \chi(g) \end{align} from where it follows from this that [imath]\chi^{(\alpha)} (g)[/imath] must be real and so if [imath]\chi^{(\alpha)} (g) \le d_{\alpha}[/imath] for any [imath]\alpha \in \hat{G}[/imath] then this would induce a strict inequality for [imath]\chi(g)[/imath] and [imath]\chi(1)[/imath]. It follows that [imath]\chi^{(\alpha)} (g) = d_{\alpha}[/imath] for any [imath]\alpha \in \hat{G}[/imath] such that [imath]m^{(\alpha)} >0[/imath]. Then the conclusion follows. Can someone please let me know if I am wrong and we cannot get the result through what I wrote? Thanks! |
3030983 | Prove that [imath]\sum_{n=1}^{\infty} \big(\frac{1}{4}\big)^n {2n \choose n}[/imath] diverges
I need to prove that the following series [imath] \sum_{n=1}^{\infty} \bigg(\frac{1}{4}\bigg)^n {2n \choose n}[/imath] diverges. I've managed to change it to the following formula: [imath] 4^n=(1+1)^{2n}=\bigg(\sum_{k=0}^n{n\choose k}\bigg)^2 \text{ and } {2n\choose n}=\sum_{k=0}^n{n\choose k}^2 \text{ so }[/imath] [imath]\sum_{n=1}^\infty \bigg(\frac{1}{4}\bigg)^n {2n \choose n} =\sum_{n=1}^\infty \frac{\sum_{k=0}^n{n\choose k}^2}{\left(\sum_{k=0}^n{n\choose k}\right)^2} [/imath] Where do I go from here? I know that the limit of series' sequence goes to [imath]0[/imath] but that isn't helpful really since I need to prove that it diverges. Also all the ratio tests and root test are inconclusive. I already checked that. Think of 4 as variable [imath]a[/imath]. I need to check for which [imath]a[/imath] this series converges. I already know that for [imath]a>4[/imath] it does, but I need to prove that for [imath]a=4[/imath] it diverges. There was a mistake in my calculating of limit in Raabe's test which concluded that this series is indeed divergent. | 2857889 | Show that [imath]\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}[/imath] is divergent
Given the series [imath]S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}[/imath] I'm trying to prove its divergent but with no luck. It doesn't tend to [imath]\infty[/imath] as [imath]n[/imath] grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than [imath]S[/imath] that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints? Thanks! |
3032220 | Finding [imath]\lim_{x\to 1}\frac{\sqrt[m]x - 1}{\sqrt[n]x - 1}[/imath]
Find the limit, without using L'Hospital's rule [imath] \lim_{x\to 1}\frac{\sqrt[m]x - 1}{\sqrt[n]x - 1}. [/imath] | 981301 | Find [imath]\,\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}[/imath]
How do I calculate [imath]\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}[/imath]. Please help me. Thanks! |
3031364 | Integral of the reciprocal of a complex polynomial
For a polynomial P(z) of degree [imath]n\geq2[/imath], show that there exists some [imath]R_{1}>0[/imath] such that for [imath]R>R_{1}[/imath] it holds that: [imath]\int_{C_{R}}\frac{1}{P(z)}dz=0[/imath] where [imath]C_{R}[/imath] is a circle of radius R with the centre [imath]0[/imath]. I am already aware of the fact that [imath]\lim_{R\to \infty}\int_{C_{R}}\frac{1}{P(z)}dz=0[/imath] and how to go about proving it but I'm not sure how to evaluate the first integral written above without using any limit. Would I maybe use a similar approach as was used to show the latter integral? | 1897089 | Proof that [imath]\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} [/imath] over a closed curve is zero.
Prove that [imath]\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} [/imath] over a closed smooth simple curve that contains all of the roots of the polynomial [imath]P(z)[/imath] is either zero if [imath]n \geq 2[/imath] or equals [imath]\frac{1} {a_0} [/imath] if it's degree is [imath]1[/imath]. Proof: [imath]n\geq 2[/imath] [imath]\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}[/imath] So after partial decomposition of the fraction I'll have something like [imath]\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots[/imath] where [imath]A,B \in \mathbb{C}[/imath] (Is this fact true that [imath]A,B[/imath] wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on different circles with their center at each root and argue that the integral is the same as the sum of those circle integrals. Now since [imath]\frac {1}{z-z_n}[/imath] will be analytic inside the circle when I'm integrating on circles that do not contain [imath]z_n[/imath] some of those integrals will be zero. and now since my curves are of the form [imath]\gamma(t)=z_n+e^{it}[/imath] , [imath]t \in [0,2\pi][/imath] the rest of the integrals will look like [imath]\int_0^{2\pi} \frac{Aie^{ti}} {e^{it}}\,{\rm d}t [/imath] and all of them will be [imath]A_i[/imath]. Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind. |
3028665 | For sufficiently large [imath]n[/imath], Which number is bigger, [imath]2^n[/imath] or [imath]n^{1000}[/imath]?
How do I determine which number is bigger as [imath]n[/imath] gets sufficiently large, [imath]2^n[/imath] or [imath]n^ {1000}[/imath]? It seems to me it is a limit problem so I tried to tackle it that way. [imath]\lim_{n\to \infty} \frac{2^n}{n^{1000}}[/imath] My thoughts are that, after some [imath]n[/imath], the numerator terms will be more than the terms in the denominator so we'll have something like [imath]\frac{\overbrace{2\times 2\times\cdots \times 2}^{1000\text{ factors}}}{n\times n \times \cdots \times n} \times 2^{n-1000}[/imath] At this point, I was thinking of using the fact that [imath]2^n[/imath] grows faster slower than [imath]n![/imath] as [imath]n[/imath] gets larger so the limit, in this case, will be greater than [imath]1[/imath], meaning [imath]2^n[/imath] is bigger than [imath]n^{1000}[/imath] for sufficiently large [imath]n[/imath]. This conclusion is really just a surmise based on a non-concrete formulation. Therefore, I'd appreciate any input on how to tackle this problem. | 31934 | Proof of a binomial theorem based inequality?
Let [imath]k \in N, x \gt 0[/imath]. Show that there exists some [imath]n_2 \in \mathbb{N}[/imath] so that [imath]\forall n \geq n_2: (1+x)^n \gt n^k[/imath]. Hint: binomial theorem. My thought on this is first to make the substitution [imath](1+x)=b[/imath] which means [imath]b>1[/imath] and [imath]b^n>1[/imath]. This would also be true if [imath]k=0[/imath] and [imath]n=1[/imath] thus [imath]n_2=1[/imath]. Next step I can think of is using archimedian property [imath]a>0, y \in \mathbb{R}, m \in \mathbb{N}, ma>y[/imath]. This will result in [imath]b^n=(1+x)^n=ma > y = n^k[/imath]. My current idea would be to replace [imath]n[/imath] with some other cleverly devised number OR using induction because of the request for all [imath]n[/imath] and maybe then I can use binomial theorem to finally solve this. What's really throwing me off is that [imath]n[/imath] is the exponent on the left and also the base on the right (that is why I thought about replacing [imath]n[/imath]). Any hints on how to get to the next step? Thanks. |
3014429 | All nilpotent [imath]2 \times 2[/imath] matrices satisfy [imath]A^{2}=0[/imath]
I have problems to show that if [imath]A[/imath] is a [imath]2 \times 2[/imath] matrix and if there exists some positive integer such that [imath]A^{n}=0[/imath] then [imath]A^{2}=0[/imath]. I only showed that [imath]A[/imath] is a singular matrix but nothing else. Thanks any help will be appreciated. | 1200829 | All nilpotent [imath]2\times 2[/imath] matrices
I want to find all nilpotent [imath]2\times 2[/imath] matrices. All nilpotent [imath]2 \times 2[/imath] matrices are similar([imath]A=P^{-1}JP[/imath]) to [imath]J = \begin{bmatrix} 0&1\\0&0\end{bmatrix}[/imath] But how do I find all of these matrices? I do think that the only such cases are [imath]J[/imath] and [imath]J^ T[/imath] |
3014232 | Density question on intersection of two spaces
Let [imath]X[/imath] and [imath]Y[/imath] be two Banach Spaces and [imath]Z[/imath] be a dense subset of [imath]X[/imath] and of [imath]Y[/imath], i.e., [imath]\overline{(Z)}^{\|.\|_X} = X[/imath] and [imath]\overline{(Z)}^{\|.\|_Y} = Y[/imath]. Then can we say that [imath]Z[/imath] is dense in [imath](X \cap Y, \|.\|_{X \cap Y})[/imath], where [imath]\|.\|_{X \cap Y}[/imath] is any norm (to be specific either max norm or "[imath]\|.\|_X + \|.\|_Y[/imath]" norm)? | 3011468 | Dense subset of two Banach spaces also dense in the intersection
My question is: Let [imath] V [/imath] be a vector space (over [imath] \mathbb K\in\{\mathbb{R}, \mathbb{C}\} [/imath]), [imath] X,Y\subseteq V [/imath] two subspaces equipped with norms [imath] \|\cdot\|_X, \|\cdot\|_Y [/imath] such that [imath] (X,\|\cdot\|_X) [/imath] and [imath](Y,\|\cdot\|_Y)[/imath] are Banach spaces and [imath] D\subseteq X\cap Y[/imath]. If [imath] D [/imath] is dense in [imath] (X,\|\cdot\|_X) [/imath] and [imath](Y,\|\cdot\|_Y)[/imath], is [imath] D [/imath] also dense in [imath] X\cap Y [/imath] equipped with [imath] \|\cdot\|:=\|\cdot\|_X + \|\cdot\|_Y [/imath]? At first sight, it seemed very clear to me that this should be true. But I even fail to answer the following (possibly) easier question: Let [imath] X [/imath] be a vector space over [imath] \mathbb K [/imath] equipped with two norms [imath] \|\cdot\|_1, \|\cdot\|_2 [/imath] such that [imath] (X,\|\cdot\|_1) [/imath] and [imath](X,\|\cdot\|_1)[/imath] are Banach spaces and [imath] D\subseteq X[/imath]. If [imath] D [/imath] is dense in [imath] (X,\|\cdot\|_1) [/imath] and [imath](X,\|\cdot\|_2)[/imath], is [imath] D [/imath] also dense in [imath] X [/imath] equipped with [imath] \|\cdot\|:=\|\cdot\|_1 + \|\cdot\|_2 [/imath]? The answer is yes, if [imath] \|\cdot\|_1, \|\cdot\|_2 [/imath] are equivalent, so I tried to think about a counterexample using nonequivalent norms on a specific space and I also found a nice paper about nonisomorphic complete norms (https://www.researchgate.net/publication/226200984_Equivalent_complete_norms_and_positivity) but it didn't helped me so far to construct anything useful for my question. Thanks for your help! |
3032628 | Explain why J is a prime Ideal in Z[x]
I am trying to prove that the ideal [imath]J=<x+1>[/imath] is prime in the ring [imath]\mathbb{Z}[x][/imath]. I know that if the generator is prime, then the ring modulo the generator is an integral domain. I can show that this is true in this case after a bit of work. However, I was wondering if there was a quicker way to do this. Any thoughts? Thanks | 1212767 | [imath]\mathbb Z [X] / (X)[/imath] isomorphic to [imath]\mathbb Z[X] / (X+1)[/imath] isomorphic to [imath]\mathbb Z [X] / (X+2015)[/imath]
I have to prove that [imath]\mathbb Z [X] / (X) \cong \mathbb Z[X] / (X+1) \cong \mathbb Z [X] / (X+2015)[/imath]. I think that one answer could be that [imath]\mathbb Z[X]/(X) \cong \mathbb Z(0)[/imath], [imath]\mathbb Z[X]/(X+1)\cong Z(-1)[/imath] and [imath]\mathbb Z [X] /(X+2015)\cong \mathbb Z(-2015)[/imath], as [imath]0, -1, -2015[/imath] are roots of these polynomials. Also, as [imath]0, -1, -2015\in \mathbb Z[/imath], then all of them are isomorphic to [imath]\mathbb Z [/imath], hence isomorphic between them. But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials. |
3032800 | Use the sine rule to prove trig identity
Using the sine rule: [imath] \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}[/imath] prove, for triangle ABC: [imath]\sin\left(\frac{B-C}{2}\right) = \frac{b-c}{a} \cos\left(\frac A2\right)[/imath] Using the sine rule it's easy to translate the RHS into: [imath]\sin\left(\frac{B-C}{2}\right) = \frac{\sin(B)-\sin(C)}{\sin(A)} \cos\left(\frac A2\right)[/imath] Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS. A nudge in the right direction would really be appreciated. Thanks. | 1394674 | Proof of the identity: [imath]c\sin \frac{A-B}{2} \equiv (a-b) \cos \frac{C}{2}[/imath]
Trigs is not my strongest apparently... I need to prove [imath]c\sin \frac{A-B}{2} = (a-b) \cos \frac{C}{2}[/imath] for a general triangle [imath]ABC[/imath]. Here is what I do, or rather, here is how I fail at proving it: [imath]\cos \frac{C}{2} \equiv \sin \frac{A+B}{2}[/imath], so [imath]\displaystyle{\frac{\sin \frac{A-B}{2}}{\sin \frac{A+B}{2}} \equiv \displaystyle{\frac{a-b}{c}}}[/imath]. This implies: [imath]\displaystyle{\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}} \equiv \frac{a-b}{c}}[/imath]. Now, imagine graphing an angle bisector from angle [imath]A[/imath] and then from angle [imath]B[/imath], the point where they intersect, let's call it [imath]K[/imath]. From that point drop a perpendicular on [imath]AB[/imath], let's call that point [imath]L[/imath]. Hence, [imath]\tan \frac{A}{2} = \frac{KL}{AL}[/imath] and [imath]\tan \frac{B}{2} = \frac{KL}{LB}[/imath]. Plugging those in, gives us: [imath]\frac{LB-AL}{c} \equiv \frac{a-b}{c}[/imath] And now I have no clue how to show that [imath]LB-AL = a-b[/imath]. If you could let me know how to show that and/or you know a better way of proving the identity, please share. |
3032289 | If [imath]0< a < 1[/imath]; show that [imath]\lim na^n[/imath] goes to [imath]0[/imath].
If [imath]0< a < 1[/imath]; show that [imath]\lim na^n[/imath] goes to [imath]0[/imath] as [imath]n[/imath] goes to [imath]\infty[/imath] I know that when [imath]0< a < 1[/imath], [imath]\lim a^n[/imath] goes to [imath]0[/imath] and [imath]\lim n[/imath] goes to [imath]\infty[/imath] [imath]|a^n| <\epsilon[/imath] take [imath]\epsilon_0 = \epsilon/|n|[/imath] then [imath]|a^n| < \epsilon_0 = \epsilon/|n|[/imath], then [imath]|n||a^n|< \epsilon[/imath], and since [imath]|na^n|\le|n||a^n|< \epsilon[/imath], then [imath]|na^n|< \epsilon[/imath] Is this kind of argument sufficient for this case? | 1220951 | if [imath]|a|<1[/imath] so [imath]\lim_{n\to \infty}na^n=0[/imath].
Prove that if [imath]|a|<1[/imath] ([imath]a[/imath] is real) so [imath]\lim_{n\to \infty}na^n=0[/imath]. I know that I need to use squeeze theory (because I have [imath]-1<a<1[/imath]) but I dont see how. thanks |
3033244 | Showing that the Distribution of a Random Variable Must be Standard Normal using Characteristic Functions
Question Let [imath]X[/imath] and [imath]Y[/imath] be i.i.d with means [imath]0[/imath] and variances [imath]1[/imath]. Let [imath]\phi(t)[/imath] be their common characteristic function and suppose that [imath]X+Y[/imath] and [imath]X-Y[/imath] are independent. Show that [imath]\phi(2t)=\phi(t)^3\phi(-t)[/imath] and deduce that [imath]X[/imath] and [imath]Y[/imath] are standard normal random variables. The above question is from Grimmett and Stirzaker. My attempt I was able to show the first part but unable to fully justify the second part. For the first part here is a proof. Note that [imath] \begin{align} \phi(2t)=Ee^{it2X}&=E\exp\{it(X+Y+X-Y)\}\\ &=E\exp\{it(X+Y\}E\exp\{it(X-Y\}\\ &=E\exp(itX)E\exp(itY)E\exp(itX)E\exp(-itY)\\ &=\phi(t)^3\phi(-t) \end{align} [/imath] by the independence assumptions. My Problem For the second part, it is easy to show that [imath]e^{-t^2/2}[/imath] (the cf of a standard normal) satisfies the equation [imath]\phi(2t)=\phi(t)^3\phi(-t)[/imath] but I am unable to show that this is the only choice of [imath]\phi[/imath] that satisfies this equation. I tried taking derivatives and setting up a differential equation but it got messy. Any help is appreciated. | 556030 | [imath]X[/imath] and [imath]Y[/imath] i.i.d., [imath]X+Y[/imath] and [imath]X-Y[/imath] independent, [imath]\mathbb{E}(X)=0 [/imath] and [imath]\mathbb{E}(X^2)=1[/imath]. Show [imath]X \sim N(0,1)[/imath]
[imath]X[/imath] and [imath]Y[/imath] are independent and identically distribued (i.i.d.), [imath]X+Y[/imath] and [imath]X-Y[/imath] are independent, [imath]\mathbb{E}(X)=0[/imath] and [imath]\mathbb{E}(X^2)=1[/imath]. Show that [imath]X\sim N(0,1)[/imath]. We should use characteristic functions to prove this. Any ideas? |
3033627 | Given that [imath]A[/imath] and [imath]B[/imath] be [imath]n \times n[/imath] matrices over [imath]\mathbb{C}[/imath] then choose the correct option
Given that [imath]A[/imath] and [imath]B[/imath] be [imath]n \times n[/imath] matrices over [imath]\mathbb{C}[/imath]. Then choose the correct options [imath](1)[/imath] [imath]AB[/imath] and [imath]BA [/imath] always have the same set of eigenvalues [imath](2)[/imath] If [imath]AB[/imath] and [imath] BA[/imath] have same set of eigenvalue then AB = BA [imath](3)[/imath] If [imath]A ^{-1}[/imath] exist then [imath]AB[/imath] and [imath]BA[/imath] are similar [imath](4)[/imath] The rank of [imath]AB[/imath] is always same as the ranks of [imath]BA[/imath] My attempt: I thinks all option [imath]1,2,3,4[/imath] will be correct if take [imath]A=B= I[/imath] Any hints/solution will be appreciated Thank you! | 2535339 | Let [imath]A[/imath] and [imath]B[/imath] be matrices over [imath]\mathbb C[/imath]. Then pick out the correct statements.
Let [imath]A[/imath] and [imath]B[/imath] be matrices over [imath]\mathbb C[/imath]. Then, [imath]AB[/imath] and [imath]BA[/imath] always have the same set of eigenvalues. If [imath]AB[/imath] and [imath]BA[/imath] have the same set of eigenvalues then [imath]AB=BA[/imath]. If [imath]A^{-1}[/imath] exists then [imath]AB[/imath] and [imath]BA[/imath] are similar. The rank of [imath]AB[/imath] is always the same as the rank of [imath]BA[/imath] . Suppose [imath]AB=BA[/imath] Let [imath]x[/imath] be the eigen vector of [imath]A[/imath] corresponding to the eigenvalue [imath]a[/imath]. [imath]ABx=BAx=aBx \implies Bx[/imath] is the eigen vector of [imath]A[/imath]. If the eigen space corresponding to the eigen values of [imath]A[/imath] is one. Then, [imath]Bx=\lambda x \implies x[/imath] is the eigen vector of [imath]B[/imath]. So [imath]AB[/imath] and [imath]BA[/imath] have same set of eigen values. statement is false. Am I correct? I don't know, How to judge the statement. I don't know, How to judge the statement. Statement is false, I could obtain the counter examples. Please check my answers. Please help me. |
3034295 | Alternate definition of symmetric group action
Let [imath]S_3[/imath] act on [imath]\Re^3[/imath] by the permutation matrices P(σ), where P(σ) permutes the basis vectors: [imath]P(σ)(e_i) = e_{σ(i)}[/imath] Show that, with this formula, [imath]\sigma(x_1,x_2,x_3) = P(\sigma)(x_1e_1 + x_2e_2 + x_3e_3) = (x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)} )[/imath] this is an action of [imath]S_3[/imath] on [imath]\Re^3[/imath]. It seems to me that this is not true, since the inverse of a product flips the order. Am I missing something? | 2166434 | [imath]S_3[/imath] permutes the elements of [imath]\mathbb{R}^3[/imath], show that [imath]\sigma (x_1, x_2, x_3) = ( x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})[/imath].
Let the permutation group [imath]S_3[/imath] act on [imath]\mathbb{R}^3[/imath] by permuting the elements. How can we show that this action can be written as [imath]\sigma (x_1, x_2, x_3) = ( x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})[/imath] for [imath]\sigma \in S_3[/imath], [imath](x_1, x_2, x_3) \in \mathbb{R}^3[/imath]. For instance, if [imath]\sigma = (123)[/imath], we have [imath]\sigma (x_1, x_2, x_3) = (123)(x_1, x_2, x_3) = (x_3, x_1, x_2)[/imath]. Just looking at some examples, this seems so obvious, but I'm not sure how to go about showing it. |
3034188 | How to find a matrix with characteristic equation?
I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY. I need to find a matrix with characteristic equation given by λ² − λ − 1 = 0 I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix. I let A be any 2 × 2 matrix, where a, b, c, d ∈ R. A=[imath] \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} [/imath] The characteristic equation of A is: p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if: −(a + d) = −1 and ad − bc = −1. I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting −(a + d) = −1 and ad − bc = −1 and why , for example, they are not setting λ²=-1? What does that mean? Please be as simple and clear as possible! I really appreciate the help ! Thanks guys Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation . | 1658106 | Determining a matrix from its characteristic polynomial
Let [imath]A\in\mathcal{M}_{n}(K)[/imath], where [imath]K[/imath] is a field. Then, we can obtain the characteristic polynomial of [imath]A[/imath] by simply taking [imath]p(\lambda)=\det(A-\lambda I_n)[/imath], which give us something like [imath]p(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}(\text{tr } A)\lambda^{n-1}+\ldots+\det A.[/imath] Now, how can we obtain the matrix [imath]A[/imath] knowing the characteristic polynomial? Cheers |
3034399 | Let R be a domain with 1. Show that if aR = bR, then au = b for some unit u ∈ R.
Let [imath]R[/imath] be a domain with [imath]1[/imath]. Show that if [imath]aR = bR[/imath], then [imath]au = b[/imath] for some unit [imath]u ∈ R[/imath]. Any hint on how to start this proof? | 355994 | Two principal ideals coincide if and only if their generators are associated
Suppose we have a ring [imath]R[/imath] and [imath](a),(b)[/imath] are both ideals of [imath]R[/imath]. Is it always true that [imath](a)=(b)[/imath] if and only if there exists a unit [imath]c[/imath] such that [imath]a=bc[/imath] (i.e., [imath]a[/imath] and [imath]b[/imath] are associate)? I have already verified that the forwards direction is true. But I have no idea on the backward direction. If it is true, can someone provide a proof to me? Backward direction: Suppose that there exists a unit [imath]c \in R[/imath] such that [imath]a=bc[/imath]. This implies that [imath](a)\subset (b)[/imath]. By using the same thing, (i.e., [imath]b[/imath] and [imath]a[/imath] are associate), there exists a unit [imath]d \in R[/imath] such that [imath]b=ad[/imath]. This implies that [imath](b) \subset (a)[/imath]. is this correct? |
304639 | Extending simplicial complex
Let [imath]X[/imath] be a simplex and [imath]Y\subseteq |X|[/imath] a simplicial complex. Can I construct a simplicial complex [imath]X'\supseteq Y[/imath] s.t. [imath]|X'|=|X|[/imath]? Can I do it without introducing new vertices apart from [imath]V:=X^{(0)}\cup Y^{(0)}[/imath]? My first idea was to start with [imath]X':=Y[/imath] and repeatedly adding, for each [imath]\sigma\subseteq V[/imath] s.t. [imath]\sigma[/imath] is a nondegenerate simplex and [imath]X'\cup \sigma[/imath] is a simplicial complex, [imath]X':=X'\cup\sigma[/imath]. However, does it work? | 3007769 | A non-polyhedral pair [imath](X,A)[/imath] with both [imath]X[/imath] and [imath]A[/imath] polyhedral
Does there exist a topological space [imath]X[/imath] and a closed subspace [imath]A[/imath] such that each of [imath]X[/imath] and [imath]A[/imath] is isomorphic to the topological space of some simplicial complex, but such that there does not exist a pair [imath](S,T)[/imath] consisting of a simplicial complex [imath]S[/imath] and a subcomplex [imath]T[/imath] with [imath](X,A) \cong (|S|,|T|)[/imath]? This is (essentially) problem A2 from Chapter 3 of Spanier's book Algebraic topology. At first I suspected that the intended example was the comb space [imath]A[/imath] inside the unit square [imath]X[/imath], but it seems that [imath]A[/imath] is not polyhedral (if it were, then being compact, it would have to be the space of a finite simplicial complex). It doesn't seem to me that there is an example that has already appeared in Spanier's book, so I am somewhat at a loss as to the intended answer. |
3031331 | expected value of Consecutive numbers
We choose randomly a subset [imath]\mathit S[/imath] of size 25 from the set [imath]\{ 1,2,...,100\}[/imath] , what is the expected value of number of Consecutive numbers in [imath]\mathit S[/imath]? Consecutive numbers in [imath]\mathit S[/imath]: is a pair [imath]\{ i,i+1\}[/imath] where i[imath]\in$$\mathit S[/imath] and i+1[imath]\in$$\mathit S[/imath]. | 3027467 | Probability and expectancy problem
if we choose a size [imath]25[/imath] subset of the set: [imath]\{1,2,....100\}[/imath] what is the expectancy of the number of sequential pairs in the subset? expectancy still confuses me, can anybody help? |
3034736 | What is the real characteristic equation?
My book says that you form the character equation as [imath]\begin{vmatrix} \lambda I - A \end{vmatrix}[/imath]. However I see occasionally on stack exchange and other resources that define the character equation as [imath]\begin{vmatrix}A - \lambda I\end{vmatrix}[/imath]. Why does this appear to be such a trivial matter? Here are some examples below where I have found the latter being used. Geometric multiplicity of repeated Eigenvalues https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors (under "Eigenvalues and the characteristic polynomial" | 1893998 | [imath]|A-\lambda I|[/imath] and [imath]|\lambda I-A|[/imath]
I understand that both can be used to find eigenvalues and eigenvectors as [imath]Ax=\lambda x\iff Ax-\lambda x=0 \iff x(A-\lambda I)=0\iff |A-\lambda I|=0[/imath] And in the same way [imath]Ax=\lambda x\iff 0=\lambda x-AX \iff x(\lambda I-Ax)=0\iff |\lambda I-A|=0[/imath] So the different between the two will be by plus/minus sign in the characteristic polynomial? for example [imath]A=\begin{pmatrix} 2 & 0 &1 \\ 0 & 3 &0\\ 1 &0 &2 \end{pmatrix}[/imath] will be [imath]f_A(\lambda)=(\lambda-3)^{2}(\lambda-1)[/imath] for [imath]|\lambda I-A|[/imath] and [imath]f_A(\lambda)=-(\lambda-3)^{2}(\lambda-1)[/imath] for [imath]|A-\lambda I|[/imath] |
3034613 | Proving [imath]\sin\frac\pi7 \cdot \sin\frac{2 \pi}{7}\cdot\sin\frac{3 \pi}{7} = \frac{\sqrt7}{8}[/imath]
Prove [imath]\sin\frac\pi7 \cdot \sin\frac{2 \pi}{7}\cdot\sin\frac{3 \pi}{7} = \frac{\sqrt7}{8}[/imath] I tried using the product to sum identities but those got me nowhere. Any help would be great. | 1784712 | Evaluation of [imath]\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}[/imath]
Evaluation of [imath]\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = [/imath] [imath]\bf{My\; Try::}[/imath] I have solved Using Direct formula:: [imath]\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}[/imath] Now Put [imath]n=7\;,[/imath] We get [imath]\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}\cdot \sin \frac{4\pi}{7}\cdot \sin \frac{5\pi}{7}\cdot \sin \frac{6\pi}{7}=\frac{7}{2^{7-1}}[/imath] So [imath]\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}[/imath] Now my question is how can we solve it without using Direct Formula, Help me Thanks |
3032247 | Find non-identical matrices such that one cannot be converted into another by rearranging rows and then columns (Counting problem)
Given [imath]N \times M[/imath] binary matrices (matrix containing only [imath]0[/imath]'s and [imath]1[/imath]'s), two matrices are called identical if one can be converted into the other by first permuting the [imath]N[/imath] rows and then permuting the [imath]M[/imath] columns of the resulting matrix. Find the number of non-identical matrices. I'm not really sure of how to begin this problem either. It's some recursion (or composition of more than one recursion) and I can't figure that out. May I get some valid hints? | 22159 | How many [imath]n\times m[/imath] binary matrices are there, up to row and column permutations?
I'm interested in the number of binary matrices of a given size that are distinct with regard to row and column permutations. If [imath]\sim[/imath] is the equivalence relation on [imath]n\times m[/imath] binary matrices such that [imath]A \sim B[/imath] iff one can obtain B from applying a permutation matrix to A, I'm interested in the number of [imath]\sim[/imath]-equivalence classes over all [imath]n\times m[/imath] binary matrices. I know there are [imath]2^{nm}[/imath] binary matrices of size [imath]n\times m[/imath], and [imath]n!m![/imath] possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes. |
3032386 | If [imath]X,Y,Z[/imath] are PAIRWISE independent (i.e. only [imath]X,Y[/imath] and [imath]Y,Z[/imath] and [imath]X,Z[/imath] are) , does this imply that [imath]X[/imath] and [imath](Y,Z)[/imath] are independent?
This question came to my mind when I was reading a theorem on Poisson Processes stating that the sum of two independent Poisson processes, [imath]X_t[/imath] and [imath]Y_t[/imath] is a Poisson process. In the proof, one has to show that [imath]X_t + Y_t[/imath] has independent increments, and in order to prove this one must show the question I posed. Is it true? Are there any counterexamples? | 2047616 | Find three Poisson-distributed random variables, pairwise independent but not mutually independent
I am asked to give an example of three Poisson-distributed random variables which are pairwise independent, but are not mutually independent. I thought of the example about the Intersection where cars are coming from one side and they can go to one of three directions left, right or keep straight, with probabilities [imath]p \ [/imath], [imath]q \ [/imath] and [imath]1-p-q[/imath], respectively. However, I find it hard to prove. |
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