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2405696 | Dimension of the space of all [imath]m\times n[/imath] matrices over [imath]F[/imath]
Prove that the space of all [imath]m*n[/imath] matrices over the field [imath]F[/imath] has dimension [imath]mn[/imath], by exhibiting a basis for this space. I know that it is too easy to prove but cannot remember any basis for this space. please help me. | 344382 | dim[imath](V)[/imath] = [imath]n[/imath], dim[imath](W)[/imath] = [imath]m[/imath] [imath]\implies[/imath] dim([imath]L(V,W)[/imath]) = [imath]nm[/imath]
I am reading Hoffman & Kunze's chapter on linear transformations, with a view towards understanding dual spaces. (I primarily want to read Calculus on Manifolds; in the first chapter of that book, Spivak marks one exercise concerned with dual spaces as "soon-to-be-important". I never studied dual spaces in my linear algebra course, and I want to be sure I follow his development.) H&K state the theorem that the space of linear transformations [imath]L(V, W)[/imath] from [imath]n[/imath]-dimensional [imath]V[/imath] to [imath]m[/imath] dimensional [imath]W[/imath] has dimension [imath]nm[/imath]. I am having a hard time understanding their proof; they basically just juggle [imath]\Sigma[/imath]-notation. I somewhat follow the steps, which loosely are as follows: you let [imath]\mathcal{B} = \{\alpha_1, ... \alpha_n\}[/imath] be a basis for [imath]V[/imath], and [imath]\mathcal{B'} = \{\beta_1, ... \beta_m\}[/imath] a basis for [imath]W[/imath]. For [imath]1 \leq p \leq m[/imath], [imath]1 \leq q \leq n[/imath], define [imath]E^{(p,q)} : V \to W : c_1\alpha_1 + ... + c_q\alpha_q ... +c_n\alpha_n \to c_q\beta_p;[/imath] that is, let [imath]E^{(p,q)}(\alpha_i) = \left\{ \begin{array}{} 0, & i \neq q \\ \beta_p, & i = q \end{array} \right.[/imath] and put [imath]E^{(p,q)}(\gamma) = \sum_{i=1}^n c_iE^{(p,q)}(\alpha_i).[/imath] Fix some arbitrary l.t. [imath]T: V \to W[/imath] such that [imath]T(\alpha_i) = \sum_{p=1}^m A_{(p,i)} \beta_p = A_{(1,i)}\beta_1 + A_{(2,i)}\beta_2 + ... +A_{(m,i)}\beta_m[/imath] for some weights [imath]A_{(1,i)}, A_{(2,i)}, ... A_{(m,i)}, 1 \leq i \leq n[/imath]. Now consider [imath] T'(\gamma) := \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}E^{(p,q)}(\gamma).[/imath] On the one hand, \begin{align} T'(\gamma) = \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}E^{(p,q)}(\gamma) &= \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}E^{(p,q)}(c_1\alpha_1 + ... + c_q\alpha_q ... +c_n\alpha_n)\\ & = \sum_{p=1}^m\sum_{q=1}^nA_{(p,q)}c_q \beta_p\\ & = \sum_{p=1}^m (\sum_{q=1}^nA_{(p,q)}c_q) \beta_p \end{align} On the other hand, [imath] T(\gamma) = \sum_{i=1}^n c_iT(\alpha_i) = \sum_{i=1}^n c_i\sum_{p=1}^m A_{(p,i)}\beta_p = \sum_{p=1}^m(\sum_{i=1}^nc_iA_{(p,i)})\beta_p[/imath] so [imath]T' = T[/imath]. Then you show that all the [imath]E^{whatever}[/imath] are linearly independent, and you are "done". The problem with this is that I have no idea what's going on here. When I focus, I can make sense of the [imath]\Sigma[/imath]-notation, but otherwise I don't follow the proof. So, I have to ask: What is the intuition for why the theorem is true? Can anybody make the proof given more intuitive? (For example, what are the [imath]E^{whatever}[/imath] really doing? How did H&K know that we'd have [imath]T' = T[/imath] (how did they think of [imath]T'[/imath])? |
2399594 | An inverse question inspired by Cauchy–Schwarz inequality
I have the following question, whose inverse question can be done by the well-known Cauchy–Schwarz inequality. But I do not know how to solve this question: Suppose that [imath]\{a_n\}_{n=1}^\infty[/imath] is a sequence of real numbers such that \begin{equation} \sum_{n=1}^\infty a_n b_n \quad \text{is a convergent series whenever}\quad\sum_{n=1}^\infty b_n^2<\infty. \end{equation} Show that [imath]\sum_{n=1}^\infty a_n^2<\infty[/imath]. | 1131859 | If [imath](a_n)[/imath] is such that [imath]\sum_{n=1}^\infty a_nb_n[/imath] converges for every [imath]b\in\ell_2[/imath], then [imath]a\in\ell_2[/imath]
Please help me with this question. I've been thinking about it for almost two days. Let [imath]a_n[/imath] a real series that have the following property: for every series [imath]b_n[/imath] in [imath]l_2[/imath]: [imath]\sum_{n=1}^\infty a_nb_n[/imath] converges. prove that [imath]a_n[/imath] in [imath]l_2[/imath]. |
2407233 | Does the Epsilon-Delta Definition of a Limit Work with Closed Intervals?
I was thinking about the epsilon-delta definition of a limit and began to wonder if it would work with closed intervals instead of open intervals. That is, for some function [imath]f[/imath], would the folowing definition work? [imath]\forall \epsilon>0\,\exists\delta>0\,(0<\lvert x-c\rvert≤\delta \, \implies \lvert f(x)-L\rvert≤\epsilon) [/imath] I've been trying to think of a counterexample, but can't seem to find one. edit: If it does work, why choose open intervals rather than closed? | 264148 | Why does the definition of limits of a function have strict inequality?
Definition (As written in Michael Spivak's Calculus) The function [imath]f[/imath] approaches a limit [imath]l[/imath] near [imath]a[/imath] means: for every [imath]\epsilon >0[/imath] there is some [imath]\delta > 0[/imath] such that, for all [imath]x[/imath], if [imath]0<|x-a|<\delta[/imath], then [imath]|f(x)-l|<\epsilon[/imath]. my question is: why can't it be: [imath]0<|x-a|\leq \delta,|f(x)-l|\leq \epsilon[/imath] After looking at limits of functions for a long time just to grasp it's meaning and using the definition quite a lot solving homework I realized I keep writing the same inequality without really understanding why. The only explanation given in Spivak's book for this part of the definition goes over it without explaining the inequality. I tried looking for an explanation myself but wasn't really able to find anything wrong with it. Is it also possible to write the definition like that or is there a problem with that? (first non-homework related question :p) |
2407379 | Proof writing for polynomials
Let the polynomial [imath]f(x)=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n[/imath] have integral coefficients. If there exists four distinct integers [imath]a,b,c[/imath] and [imath]d[/imath] such that [imath]f(a)=f(b)=f(c)=f(d)=5[/imath] show that there is no integer [imath]k[/imath] such that [imath]f(k)=8[/imath] I have tried to prove it but I somehow feel that my proof is incorrect please point out the errors and suggest some other way to do the question. This how I proceeded [imath]\frac{x^{n+1}-1}{x-1}=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n [/imath] [imath]\frac{a^{n+1}-1}{a-1}=5[/imath] hence [imath]\frac{a^{n+1}-1}{5}=a-1[/imath] Since [imath]a[/imath] is an integer [imath]a-1[/imath] will also be an integer hence [imath]5|a^{n+1}-1[/imath] Using fermat's little theorem we know that [imath]a^4-1\equiv_5 0[/imath] Since [imath]f(a)=5[/imath] we get [imath]n+1=4[/imath] Let k be an integer such that [imath]f(k)=8[/imath] [imath]\frac{a^4-1}{8}=a-1[/imath] Since [imath]a-1[/imath] is an integer [imath]8|a^4-1[/imath] We will consider two cases when [imath]a[/imath] is an even integer when [imath] a[/imath] is an odd integer If [imath]a[/imath] is even then [imath]a^4-1[/imath] will be odd and it not be divisible by [imath]8[/imath] If [imath]a[/imath] is odd then it will be relatively prime to [imath]8[/imath] hence [imath]8[/imath] will be Carmichael number [imath]a^7-1\equiv0(mod8)[/imath] which proves that [imath]a^4-1[/imath] is not divisible by 8. [imath]\therefore[/imath] there is no integer k such that [imath]f(k)=8[/imath] | 885532 | For [imath]5[/imath] distinct integers [imath]a_i[/imath], [imath]1\le i\le5[/imath], [imath]f(a_i)=2[/imath]. Find an integer b (if it exists) such that f(b) = [imath]9[/imath].
Here's an interesting question I came across.The person who gave it to me told me that it should not take more than [imath]3[/imath] minutes to solve this question. But I could not find any definite solution :( Maybe you guys could give me some help.Any hints would be appreciated.Here's the question: Let [imath]f(x)[/imath] be a polynomial with integer coefficients. Suppose for [imath]5[/imath] distinct integers [imath]a_i[/imath], [imath]1\le i\le5[/imath], [imath]f(a_i)=2[/imath]. Find an integer b (if it exists) such that f(b) = [imath]9[/imath]. |
2407816 | If [imath]P(x)[/imath] is a polynomial over the Real Field such that for an integer [imath]k[/imath], [imath]P(k)[/imath] is an integer, then show that [imath]n!P(x)[/imath] is an integer polynomial
Statement: If [imath]\mathcal{P(x)} \in \mathbb{R}[x] \ni~ \forall~ k\in\mathbb{Z}, \mathcal{P(k)}\in \mathbb{Z}[/imath], then show that [imath]n!\mathcal{P(x)}[/imath] is an integer polynomial. I have the following question regarding the question: 1) The question doesn't seem complete for some reason. Isn't it that [imath]\exists[/imath] particular [imath]p[/imath] such that for any [imath]n\geq p, ~n!\mathcal{P(x)} \in \mathbb{Z}[/imath]? I tried proving the statement. But it was quite unsuccessful. I tried to construct another polynomial [imath]\mathcal{Q(x)}[/imath] with rational coefficients so as to show that both are equal. But that wasn't going the way I wished. And I wasn't getting the [imath]n![/imath]. Please check Point [imath]1[/imath] above as well. | 1867542 | To show that [imath]n!f(x) \in \mathbb{Z}[x][/imath]
If [imath]f(x)[/imath] is a polynomial such that if [imath]y \in\mathbb{Z}[/imath], then [imath]f(y)\in\mathbb{Z}[/imath], show that there exist [imath]n[/imath] such that [imath]n!f(x)\in\mathbb{Z}[x][/imath] |
2406679 | Find all primes that can be represented by [imath]n^3+1[/imath], [imath]n\in \Bbb N[/imath]
Find all primes that can be represented by [imath]n^3+1[/imath], [imath]n\in \Bbb N[/imath]. This is a problem from a math olympiad contest, no answer provided. For sure 2 is one of them (when [imath]n=1[/imath]), are there others? | 1130488 | Prove [imath]\forall n\geq 2,n\in\mathbb{Z}[/imath], [imath](n+1)\mid(n^3+1)[/imath]
Question: Prove [imath]\forall n\geq 2,n\in\mathbb{Z}[/imath], [imath](n+1)\mid(n^3+1)[/imath] I know that it is possible to solve by factoring [imath]n^3+1[/imath] and showing that [imath]n+1[/imath] is a multiple, but I would like to show this via induction. Attempt: Using induction: Let [imath]n=2:3\mid 9[/imath], so true for [imath]n=2[/imath] (1) Assume true for [imath]n=k:(k+1)\mid(k^3+1)[/imath], so [imath](k+1)r=k^3+1[/imath] for some [imath]r\in\mathbb{Z}[/imath] (2) Let [imath]n=k+1:(k+2)\mid(k+1)^3+1[/imath], so [imath](k+2)s=(k+1)^3+1[/imath] for some [imath]s\in\mathbb{Z}[/imath] From (2) we have [imath](k+1)s+s=(k+1)^3+1[/imath], subbing (1) into (2), I get [imath]k^3+1+s=(k+1)^3+1[/imath] Therefore [imath]k^3+s=(k+1)^3[/imath]. Setting [imath]s=3k^2+3k+1[/imath], this holds. Where am I going wrong, is the proof valid, any help would be greatly appreciated. |
2408139 | Question in separable polynomials
Let me ask something about separable polynomials. I found the following definitions: A polynomial over [imath]F[/imath] is called separable if it has no multiple roots (i.e. all its roots are distinct). (Dummit & Foote) A polynomial over [imath]F[/imath] is called separable if every irreducible factor has not multiple roots (the irreducible factors may not be distinct. (Rotman) My question is: Are these two definitions equivalent? Let me give an example which looks strange: Lets take the polynomial [imath]f(x)= (x^2-2)^n \in \Bbb{Q}[x][/imath], with [imath]n\in \Bbb{N}^{\geq 2}[/imath]. Then, according to def. 1 [imath]f(x)=(x-\sqrt2)^n(x+\sqrt2)^n [/imath] so the roots are multiple, so it is not separable over [imath]\Bbb{Q}[/imath]. But, according to def. 2 [imath]f(x)= \underbrace{(x-\sqrt2)(x+\sqrt2)\cdots (x-\sqrt2)(x+\sqrt2)\,}_\text{$n$ times} [/imath] so no one of the irreducible factors has root with multiplicity [imath]=1[/imath]. So, it is separable over [imath]\Bbb{Q}[/imath]. How can this happen? Do I misunderstand something? | 1872751 | Separable polynomial definition (Confused)
I understand there is a new and old definition of separable polynomial (https://en.wikipedia.org/wiki/Separable_polynomial). For the following definition: a) A polynomial over [imath]F[/imath] is said to be separable if it has no multiple roots. (given in Dummit & Foote) b) a polynomial P(X) over a given field K is separable if its roots are distinct in an algebraic closure of K (Wikipedia) c) A polynomial f in F[X] is a separable polynomial if and only if every irreducible factor of f in F[X] has distinct roots (https://en.wikipedia.org/wiki/Separable_extension#cite_note-6) Question) Which is the correct definition? I can see that (b) implies (c), for instance, but not sure if they are equivalent. Thanks for help. |
2408590 | Quaternions question
Got a question about Quaternions. I can kind of get that for [imath]q=a+b{\bf i}+c{\bf j}+d{\bf k}[/imath], [imath]\bf i,j,k[/imath] are square roots of -1, what I don't get is why [imath]{\bf ijk} = -1[/imath] ? Any enlightenments? :) | 2400940 | How do I prove that [imath]ij = k?[/imath]
If I can prove that [imath]ij = k[/imath], given that [imath]i^2 = j^2 = k^2 = -1[/imath], then it will be easy to prove the other quaternion formulas. However, I'm having a lot of trouble getting past this step. I started by setting up the following equation, such that [imath](a, b, c, d) \in \mathbb{R}[/imath] [imath]ij = a + bi + cj + dk[/imath] Obviously, I need to prove that [imath]a = b = c = 0 \land d= 1[/imath]. I've done all sorts of wacky things with this equation and I haven't gotten any closer to proving that [imath]ij = k[/imath]. I would greatly appreciate it if somebody could point me in the right direction. Thanks in advance! |
2408651 | A question on localization
Let [imath]D[/imath] be an integral domain , let [imath]S[/imath] be a multiplicatively closed subset of [imath]D[/imath] not containing zero , then w.l.o.g. by natural embeddings , we can think of [imath]D[/imath] as a subring of [imath]S^{-1} D[/imath] which is further a subring of the fraction field of [imath]D[/imath] . Now if for every [imath]q \in S^{-1} D , \exists [/imath] monic [imath]P_q(X) \in D[X] [/imath] such that [imath]P_q(q)=0[/imath] , then is it true that every element of [imath]S[/imath] is a unit of [imath]D[/imath] ? | 2292351 | Ring of fractions of [imath]R[/imath] is integral over [imath]R[/imath]
Let [imath]R[/imath] be a commutative ring and let [imath]S\subseteq R[/imath] be a multiplicative closed subset containing 1. I need to show that if [imath]S^{-1}R[/imath] is integral over [imath]R[/imath], then [imath]S^{-1}R=R[/imath]. I don't see how to get started. |
2409072 | There are exactly two abelian groups of order [imath]p^2[/imath]
Let [imath]p[/imath] be a prime number. up to isomorphism, there is exactly two abelian groups of order [imath]p^2[/imath]. It is easy to see those two are [imath]\mathbb{Z}_{p^2}[/imath] and [imath]\mathbb{Z}_{p}\times\mathbb{Z}_{p}[/imath]. It is easy to see that if their exist another one then it should not be cyclic. But how to prove exactly? | 1746584 | Help understanding Fundamental Theorem of Finite Abelian Groups
The book I'm reading defines the Fundamental Theorem of Finite Abelian Groups in the usual way that a finite abelian group is isomorphic to a direct product of cyclic groups [imath]\mathbb{Z}_{{p_1}^{e_1}} \times \cdots \times \mathbb{Z}_{{p_r}^{e_r}}[/imath], but then it also states that the group is isomorphic to [imath]\mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_t}[/imath] where for each [imath]i = 1,...,t-1[/imath] we have [imath]m_i | m_{i+1}[/imath]. How can this be if the groups orders are supposed to be relatively prime. If anyone can give an example of how this works I would greatly appreciate it. |
2408965 | Can a weak form of the union axiom be proved in ZF set theory?
I read somewhere that the statement "[imath]x \cup y[/imath] exists for every [imath]x[/imath] and [imath]y[/imath]" can be proven in [imath]ZFC[/imath] set theory without using the full strength of the union axiom, which says that arbitrary unions exist, not just binary unions. Can that statement also be proven from [imath]ZF[/imath] set theory? Is this an open problem? | 498256 | Can we prove the existence of [imath]A\cup B[/imath] without the union axiom?
If [imath]A[/imath] and [imath]B[/imath] are sets, then [imath]A\cup B[/imath] is defined as follows: [imath]A\cup B:=\{x: x\in A \, \, \,\text{ or } \, \, \, x\in B\}.[/imath] In ZFC, [imath]A\cup B[/imath] exists because [imath]\bigcup\{A,B\}[/imath] exists by union axiom, and it is known that axiom of union is independent of rest of ZFC. If there is set such that [imath]A\subset C[/imath] and [imath]B\subset C[/imath], then we can define [imath]A\cup B[/imath] using the [imath]C[/imath] and seperation axiom. But can I prove the existence of [imath]C[/imath] in ZF-Union? Thanks for any help. |
2409125 | Find the value of [imath]\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}[/imath].
I am stuck with the following problem : Find the value of [imath]\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}[/imath] My try : Let [imath]p=\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}\implies \log p=\lim_{x \to 0}{\frac {1}{x^2}}\log\left(\frac {\sin x}{x}\right)[/imath]...After applying l'hospitals rule few times I get [imath]\log p= \lim_{x \to 0}\frac{1-\sec^2x}{4x \tan x+2x^2\sec^2x}[/imath].. we can again apply l'hospitals rule ,but the calculations get bigger and bigger..Is there any other easier way around or I am missing something? | 632527 | Finding the limit [imath]\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}[/imath]
Find the following limit: [imath]\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}[/imath] Well I tried to do the [imath]\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)[/imath] then apply LHR but I seem to get to endless dervivations... There's got to be a more simple approach. |
2410077 | Sum of an arithmetic infinite series
I have to find the sum of the series : [imath]1+ \frac{2}{6} + \frac{2 \times 5}{6 \times 12} + \frac{2 \times 5 \times 8}{6 \times 12 \times 18} + \cdots[/imath].. I find out the [imath]nth[/imath] term to be [imath]u_n = \frac{2 \times 5 \times 8\times \cdots (3n+2)}{6 \times 12 \times 18 \cdots 6n}[/imath]. After finding the [imath]nth[/imath] term I was unable to proceed. Any insight will be happily accepted.Also looking for any tips or advices for solving this kind of problems. Thank you. | 1600414 | Finding the sum of the infinite series whose general term is not easy to visualize: [imath]\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots[/imath]
I am to find out the sum of infinite series:- [imath]\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...............[/imath] I can not figure out the general term of this series. It is looking like a power series as follows:- [imath]\frac{1}{6}+\frac{5}{6^2\cdot2!}+\frac{5\cdot8}{6^3\cdot3!}+\frac{5\cdot8\cdot11}{6^4\cdot4!}+.....[/imath] So how to solve it and is there any easy way to find out the general term of such type of series? |
2410114 | Show that the identity permutation cannot be expressed as the product of an odd number of transpositions.
Show that the identity permutation cannot be expressed as the product of an odd number of transpositions. For example consider the example of [imath]S_3,[/imath] [imath]\sigma_{id} = (123)(321) = (1 2) (13)(32)(31)[/imath], which means even number of permutations. In general let [imath]\sigma_{id}= \sigma_1.\sigma_1\cdots \sigma_k [/imath], I need to show if I rewrite each [imath]\sigma_i[/imath] as transpositions then odd number of transpositions will be there. Thank You. | 1147308 | Prove that if the identity is written as the product of [imath]r[/imath] transpositions, then [imath]r[/imath] is an even number
Theorem. If the identity is written as the product of [imath]r[/imath] transpositions, [imath]id=τ_1τ_2\dots τ_r[/imath], then [imath]r[/imath] is an even number. Proof. We will employ induction on [imath]r[/imath]. A transposition cannot be the identity; hence, [imath]r > 1[/imath]. If [imath]r = 2[/imath], then we are done. Suppose that [imath]r > 2[/imath]. In this case the product of the last two transpositions, [imath]τ_{r−1}τ_r[/imath], must be one of the following cases: [imath](ab)(ab) = id\\ (bc)(ab) = (ac)(bc)\\ (cd)(ab) = (ab)(cd)\\ (ac)(ab) = (ab)(bc),[/imath] where [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath] are distinct. The first equation simply says that a transposition is its own inverse. If this case occurs, delete [imath]τ_{r−1} τ_r[/imath] from the product to obtain [imath]id=τ_1τ_2···τ_{r−3}τ_{r−2}[/imath]. By induction [imath]r − 2[/imath] is even; hence, [imath]r[/imath] must be even. In each of the other three cases, we can replace [imath]τ_{r−1} τ_r[/imath] with the right-hand side of the corresponding equation to obtain a new product of [imath]r[/imath] transpositions for the identity. In this new product the last occurrence of a will be in the next-to-the-last transposition. We can continue this process with [imath]τ_{r−2} τ_{r−1}[/imath] to obtain either a product of [imath]r − 2[/imath] transpositions or a new product of [imath]r[/imath] transpositions where the last occurrence of a is in [imath]τ_{r−2}[/imath]. If the identity is the product of [imath]r − 2[/imath] transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with [imath]τ_{r−3}τ_{r−2}[/imath]. At some point either we will have two adjacent, identical transpositions canceling each other out or a will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix a in this instance. Therefore, the identity permutation must be the product of [imath]r − 2[/imath] transpositions and, again by our induction hypothesis, we are done. Question: That is the proof written in the book, but unfortunately after more than [imath]10[/imath] times reading I don't understand the last two paragraph (i.e., from "In each of the other three cases"). Is there another simpler proof or could someone please clarify what is going on? Thank you. |
2410448 | Understanding the question from Shiryaev: [imath]B_m[/imath] be the event in which each of the events [imath]A_1,\dots,A_n[/imath] occurs exactly [imath]m[/imath] times
§2, exercise 5, p.17: Let [imath]A_1,\dots,A_n[/imath] be events, and define [imath]S_0,S_1,\dots,S_n[/imath] as follows: [imath]S_0=1[/imath] [imath]\begin{alignedat}{1}\begin{split}S_{r}=\sum_{J_{r}}P\left(A_{k_{1}}\cap\cdots\cap > A_{k_{r}}\right),\end{split} & 1\leq r\leq n\end{alignedat}[/imath] where the sum is over the unordered subsets [imath]J_{r}=\left[k_{1},\dots,k_{r}\right][/imath] of [imath]\left\{ 1,\dots,n\right\}[/imath] . Let [imath]B_m[/imath] be the event in which each of the events [imath]A_1,\dots,A_n[/imath] occurs exactly [imath]m[/imath] times. Show that [imath]P\left(B_{m}\right)=\sum_{r=m}^{n}\left(-1\right)^{r-m}C_{r}^{m}S_{r}[/imath] I don't understand what [imath]B_m[/imath] is!? (my understandings lead to wrong result, tested with particular values of [imath]m[/imath], and [imath]n[/imath]) The original Russian texts: Пусть [imath]B_{m}[/imath] - событие, состоящее в том, что одновременно произойдет в точности [imath]m[/imath] событий из [imath]A_{1},\dots,A_{n}[/imath]. What is the Russian texts actual meaning? (I want to know if there was problem with translation or it's my fault) I found an answer from a German forum, which said that: (google translator): This is indeed a bit incomprehensible, because here m * n events are combined into one but, I don't know how to interpret the question like that. P/S: I found an answer, my question could be marked as duplicated: The Probability [imath]P_{[m]}[/imath] that exactly [imath]m[/imath] among the [imath]N[/imath] events [imath]A_1,\dots,A_N[/imath] occur simultaneously Indeed, I found that the combined result of the online translators (Google, Yandex, Bing, PROMT) is closed to the exercise posted in the link above: machine translation: Let [imath]B_{m}[/imath] be the event, which consists simultaneous occurrence of exactly [imath]m[/imath] events of [imath]A_{1},\dots,A_{n}[/imath] | 716775 | The Probability [imath]P_{[m]}[/imath] that exactly [imath]m[/imath] among the [imath]N[/imath] events [imath]A_1,\dots,A_N[/imath] occur simultaneously
For [imath]1\le i_1,i_2,...,i_k\le N[/imath] denote [imath]p_{i_1,...,i_k}=\Pr(A_{i_1}\cap A_{i_2}\cap\dots\cap A_{i_k}),[/imath] [imath]S_k=\sum\limits_{1\le i_1\le\dots\le i_k\le N}p_{i_1,\dots,i_k}.[/imath] Show that the probability [imath]P_{[m]}[/imath] that exactly [imath]m[/imath] among the [imath]N[/imath] events [imath]A_1,\dots,A_N[/imath] occur simultaneously is given by [imath]P_{[m]}=S_m-\binom{m+1}{m}S_{m+1}+\binom{m+2}{m}S_{m+2}-\dots (-1)^{N-m}\binom{N}{m}S_{N}.[/imath] We've already proven the inclusion exclusion formula: [imath]\Pr\left(\bigcup\limits_{i=1}^NA_i\right)=S_1-S_2+S_3-\dots (-1)^{N-1}S_N,[/imath] but the problem is the factors [imath]\binom{m+1}{m}...[/imath] before [imath]S_i[/imath]'s, I cannot really show that. |
2410495 | Proof that [imath]x^y=y^x[/imath] is only true for the ordered pairs [imath](2,4)[/imath] and [imath](4,2)[/imath]
I wish to prove that the binary relation [imath]R_1=\{(x,y):x\in\mathbb{Z}^+,y\in\mathbb{Z}^+,x\neq y[/imath] and [imath]x^y=y^x\}[/imath] is a finite set comprising just the two elements (2,4) and (4,2). One can easily verify that [imath]2^4=16=4^2[/imath], implying that indeed [imath](2,4)\in R_1[/imath] and [imath](4,2)\in R_1[/imath], but proving that the set contains no other elements seems to be a little trickier. I would suppose the proof should begin: Assume [imath]x^y=y^x[/imath], and one would arrive at the conclusion that either [imath]x[/imath] must be [imath]2[/imath] and [imath]y[/imath] must be [imath]4[/imath] or vice versa, but I am not entirely certain how one should proceed from such an assumption. Your input is highly appreciated. The problem was found in the book "Rings, Fields and Groups: An Introduction to Abstract Algebra" by R.B.J.T. Allenby (Second edition pp. 58). | 2264576 | Is there only one pair of unequal rational numbers with [imath]n^m = m^n[/imath]?
While talking with a friend about the number [imath]16[/imath], I began thinking about what two positive rational numbers, [imath]n[/imath] and [imath]m[/imath], satisfy the condition that [imath]n^m[/imath] = [imath]m^n[/imath] where [imath]n[/imath] and [imath]m[/imath] are not equal. [imath]2[/imath] and [imath]4[/imath], as discussed with my friend, are the first pair of integers to do this, and seem to be the only pair to do this after brute forcing combinations in a python script. It seems like there should be more possible pairs, so are [imath]2[/imath] and [imath]4[/imath] the only positive rational numbers to satisfy these conditions? |
2409490 | Limit involving inner products
Let [imath]A[/imath] be an [imath]n \times n[/imath] real symmetric matrix with eigenvalues [imath]\lambda_1,\ldots,\lambda_n.[/imath] Find the set of [imath]v \in \mathbb{R}^n[/imath] such that [imath]\lim_{k\to\infty}\left<A^{2k}v,v\right>^{1/k}[/imath] exists and find the set of possible limits. Since [imath]A[/imath] is symmetric there exists an orthonormal basis [imath]v_1,\ldots,v_n[/imath] of eigenvectors of [imath]A[/imath]. Let [imath]\lambda_j[/imath] be the eigenvector corresponding to [imath]v_j[/imath]. If [imath]v \in \mathbb{R}^n[/imath], then [imath]v = c_1v_1 + \cdots c_nv_n, c_j \in \mathbb{R}.[/imath] Since [imath]<v_i, v_j> = 0[/imath] for [imath]i \neq j[/imath] and [imath]= 1[/imath] if [imath]i = j[/imath], and [imath]A^{2k}v_j = \lambda_j^{2k}v_j[/imath], by the linearity of the inner product the limit becomes [imath]\lim_{k\to\infty} \left(c_1^2\lambda_1^{2k} + \cdots c_n^2\lambda_n^{2k}\right)^{1/k}.[/imath] This is where I'm stuck. I don't know how to evaluate the limit. I tried taking logs and using L'Hopital's rule, but that only gave another limit I couldn't compute. Since [imath]c_1^2\lambda_1^{2k} + \cdots + c_n^2\lambda_n^{2k} \leq (c_1^{2/k}\lambda_1^2 + \cdots + c_n^{2/k}\lambda_n^2)^k,[/imath] I know that if the limit exists it's bounded above by [imath]\lambda_1^2 + \cdots \lambda_n^2[/imath]. Any help proceeding would be great. | 195060 | Limit of Matrices
Given that [imath]A[/imath] is symmetric [imath]nxn[/imath] matrix. Show that [imath]\lim_{k \rightarrow \infty} (x^tA^{2k}x)^{1/k}[/imath] exists for all [imath]x \in R^n[/imath] and possible limit values are the eigenvalues of A. Since A is symmetric you can find an orthonormal basis. So A is similar to the diagonal matrix B. So you get [imath]\lim_{k \rightarrow \infty} (\sum_{i=0}^n x_i^2 \lambda_i^{2k})^{1/k}[/imath]. So I can show that this last thing is bounded above and below, but it isn't monotone in k, so how should I show that it's convergent and find what it could possibly converge to? |
2411669 | Convergence of [imath]\sum_{n=1}^\infty \frac{\sin(2^n)}{n}[/imath]
Does the series [imath] \sum_{n=1}^\infty \frac{\sin(2^n)}{n} [/imath] converge? Based on Dirichlet's test, series like [imath] \sum_{n=1}^\infty \frac{\sin(a[n])}{n} [/imath] converges when [imath]a[n][/imath] is an arithmetic sequence. Different from this case, where [imath]2^n[/imath] is a geometric sequence. Not clear if Dirichet's test is still applicable. | 1976417 | A community project: prove (or disprove) that [imath]\sum_{n\geq 1}\frac{\sin(2^n)}{n}[/imath] is convergent
As the title says, I would like to launch a community project for proving that the series [imath]\sum_{n\geq 1}\frac{\sin(2^n)}{n}[/imath] is convergent. An extensive list of considerations follows. The first fact is that the inequality [imath] \sum_{n=1}^{N}\sin(2^n)\ll N^{1-\varepsilon}\qquad\text{or}\qquad\sum_{n=1}^{N}e^{2^n i}\ll \frac{N}{\log(N)^{1+\varepsilon}} \tag{1}[/imath] for some [imath]\varepsilon>0[/imath] is enough to prove the claim by Abel summation. In the same spirit, it is quite common to employ Weyl's inequality / Van der Corput's trick to prove the convergence of [imath]\sum_{n\geq 1}\frac{\sin(n^p)}{n}[/imath]. In our case, however, we do not have an additive base of [imath]\mathbb{N}[/imath] made by perfect powers associated with some exponent, hence an additive base of finite order, so the estimation of the exponential sums appearing in the right side of [imath](1)[/imath] is more difficult. Assuming that for an infinite number of primes [imath]p[/imath] the element [imath]2[/imath] is a generator of [imath]\mathbb{Z}/(p\mathbb{Z})^*[/imath] (Legendre's conjecture), we may probably regard [imath][1,N][/imath] as a subset of [imath]\mathbb{Z}/(p\mathbb{Z})[/imath] (for a huge [imath]p[/imath]) and prove there is enough cancellation to grant [imath](1)[/imath]. However, Legendre's conjecture seems quite out-of-reach at the moment. Integral extimations techniques, that turned out to be pretty effective in other contexts, are almost ineffective here, since [imath]\sin(2^x)[/imath] oscillates too fast, so that there is no reason for expecting that [imath] \sum_{n=M+1}^{M+N}\sin(2^n)\approx \int_{M}^{N+M}\sin(2^x)\,dx, [/imath] so, even if [imath]\lim_{N\to +\infty}\int_{1}^{N}\sin(2^x)\,dx[/imath] is convergent by Dirichlet's test, there is little use of that. However, the series [imath]g(\xi)=\sum_{n\geq 1}\frac{\sin(2^n\xi)}{n}[/imath] is convergent for almost every [imath]\xi[/imath], since [imath]g\in L^2\left(-\pi,\pi\right)[/imath]. Additionally, the statement there is enough cancellation to ensure [imath](1)[/imath] appears to be equivalent to (or, at least, a consequence of) the statement the digits [imath]0[/imath] and [imath]1[/imath] in the binary representation of [imath]\pi[/imath] are equidistributed. It is hard to believe that is not the case, and the Bailey-Borwein-Plouffe formula [imath]\pi=\sum_{k\geq 0}\frac{1}{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right)\tag{2}[/imath] may provide a way to prove it. |
2411302 | Finding a [imath]\text C^\infty[/imath] function which is zero just on a given closed set
If [imath]\text E[/imath] is an arbitrary closed set in [imath]\mathbf R^n[/imath], show that there is an [imath]\text f\in \text C^\infty(\mathbf R^n)[/imath] such that [imath]\text f(x)=0~[/imath]for every [imath]x \in \text E[/imath] and [imath]\text f(x)>0[/imath] for every other [imath]x\in \mathbf R^n[/imath]. My approach is as follows : Since [imath]\text E[/imath] is given to be closed hence [imath]\text E^c=\text V[/imath] will be an open set in [imath]\mathbf R^n[/imath] and hence [imath]\text V[/imath] can be written as countable union of open sets [imath]\{\text V_i\}_{i=1}^{\infty}[/imath] and hence one can have a partition of unity [imath]\{\psi_i\}_{i=1}^{\infty}\subset \text C^\infty (\text V)[/imath] such that [imath]\text {supp}(\psi_i)\subset \text V_i[/imath]. Then if i take [imath]f=\Sigma_{i=1}^{\infty} \psi_i[/imath] then my this [imath]\text f[/imath] is in [imath]\text C^\infty (\text V)[/imath] and [imath]f(x)>0[/imath] in [imath]\text E^c[/imath] and [imath]f(x)=0[/imath] in [imath]\text E[/imath]. Now i am just left with to show that [imath]f\in \text C^\infty (\mathbf R^n)[/imath] or how can i generalize my [imath]f[/imath] to get the desired result. Any type of help will be appreciated. Thanks in advance. | 791248 | Every closed subset [imath]E\subseteq \mathbb{R}^n[/imath] is the zero point set of a smooth function
In Walter Rudin's Principles of mathematical analysis Exercise 5.21, it is proved that for any closed subset [imath]E\subseteq \mathbb{R}[/imath], there exists a smooth function [imath]f[/imath] on [imath]\mathbb{R}[/imath] such that [imath]E=\{x\in \mathbb{R}\mid f(x)=0\}[/imath]. Since [imath]E^c[/imath] is open in [imath]\mathbb{R}[/imath], [imath]E^c[/imath] is a countable disjoint union of open intervals [imath](a_i,b_i)[/imath]. On an interval [imath](a,b)[/imath], if we let [imath]f(x)=\exp(\frac{1}{(x-a)(x-b)})[/imath], [imath]x\in(a,b)[/imath]; [imath]f(x)=0[/imath], [imath]x\notin(a,b)[/imath], ([imath]f(x)=\exp(\frac{1}{(x-a)})[/imath] or [imath]f(x)=\exp(\frac{1}{(x-b)})[/imath] on [imath](a,\infty)[/imath] and [imath](-\infty,b)[/imath] resp.), then [imath]f[/imath] is smooth. Hence we can get a smooth function on [imath]\mathbb{R}[/imath] such that [imath]E[/imath] is the zero point sets of [imath]f[/imath]. (1). How about a closed subset [imath]E[/imath] of [imath]\mathbb{R}^n[/imath]? is it true that for any closed subset [imath]E\subseteq \mathbb{R}^n[/imath], there exists a smooth function [imath]f[/imath] on [imath]\mathbb{R}^n[/imath] such that [imath]E=\{x\in \mathbb{R}^n\mid f(x)=0\}[/imath]? An open set in [imath]\mathbb{R}^n[/imath] cannot be written as a disjoint union of countable open balls hence the proof above is not valid. The partition of unity only claims that for an open set [imath]U[/imath], there exists open [imath]V[/imath] in [imath]U[/imath] such that [imath]\bar V\subseteq U[/imath] and a smooth function [imath]f[/imath] on [imath]\mathbb{R}^n[/imath] such that [imath]supp f\subseteq U[/imath], [imath]f|_V=1[/imath]. (2). The taylor series of [imath]f(x)=\exp(\frac{1}{(x-a)(x-b)})[/imath] does not converge around points [imath]a,b[/imath] hence [imath]f[/imath] is not analytic at [imath]a,b[/imath]. Is it true that for any closed subset [imath]E\subseteq \mathbb{R}[/imath], there exists an analytic function [imath]f[/imath] on [imath]\mathbb{R}[/imath] such that [imath]E=\{x\in \mathbb{R}\mid f(x)=0\}[/imath]? |
2411945 | a geometric problem about area
This is a problem which baffled me for years. Many possibly know the solution. See the figure below. [imath]ABCD [/imath] is some arbitrary quadrilateral. Each side is divided evenly into three parts. It is asserted that the quadrilateral shaded has an area 1/9 of that of the [imath]ABCD[/imath]. My first idea was using affine or projective geometry. But the affine transform cannot transform [imath]ABCD[/imath] into a square. The projective transform can, but it does not preserve the ratio (no longer evenly divided). So, could you provide the solution for me? | 465986 | Trisect a quadrilateral into a [imath]9[/imath]-grid; the middle has [imath]1/9[/imath] the area
Trisect sides of a quadrilateral and connect the points to have nine quadrilaterals, as can be seen in the figure. Prove that the middle quadrilateral area is one ninth of the whole area. |
2412099 | Prove that [imath]4((p-1)!+1)\equiv -p\mod{(p(p+2)})[/imath]
I have found this interesting theorem but with no proof: "[imath]p[/imath] and [imath]p+2[/imath] are twin primes iff [imath]4((p-1)!+1)\equiv -p\mod{(p(p+2)})[/imath]" I have tried to prove it with classical N.T. theorems but I just get nowhere. Any advise? Thanks in advance. | 145160 | [imath](m, m+2)[/imath] is twin prime, iff [imath]4((m-1)! + 1) \equiv -m \pmod {m(m+2)}[/imath]
The Wiki page on Twin Primes says The pair [imath](m, m+2)[/imath] is twin prime, iff [imath]4((m-1)! + 1) \equiv -m \pmod {m(m+2)}[/imath]. This is obviously connected to Wilson's Theorem. Can anybody provide a proof for that, along the lines of thought given here, Wilson's Theorem/Proofs/Prime modulus – another proof? |
2411659 | Non-Injective differentiable map between manifolds
Let [imath]M[/imath] and [imath]N[/imath] be two smooth manifolds with dimensions m and n, respectively, with [imath]m>n[/imath]. Let [imath]f:M\to N[/imath] a smooth map ([imath]C^\infty[/imath]). Show that this map can't be injective. My try: I was wondering how could I reduce this problem to the tangent space, using the differentiability condition, to work with vector spaces, where I know how to proof this result. But I was unable to do it, so any help will be useful. | 2411339 | Differentiable injective function betweem manifolds
Let [imath]f:M\rightarrow N[/imath] an injective differentiable functions between the manifolds M and N. Prove that [imath]\dim M\leq \dim N[/imath]. Can anyone give me a hint to solve this theorem? I tried to relate the injectivity of [imath]f[/imath] with the injectivity of its differential but that is not gonna work. Thanks! |
2412657 | Prove [imath](U \otimes V) \otimes W \cong U \otimes (V \otimes W)[/imath]
Prove [imath](U \otimes V) \otimes W \cong U \otimes (V \otimes W)[/imath]. [imath]U, V, W[/imath] are finite dimensional vector spaces over a field. How can I prove that using the universal property of the tensor product? Is there an isomorphism [imath]\phi: (U \otimes V) \otimes W \to U \otimes (V \otimes W)[/imath] that satisfies [imath]\phi((u \otimes v) \otimes w) = u \otimes (v \otimes w)[/imath]? Thank you in advance. Edit: Is this a valid proof? Let [imath](e_1, ..., e_n)[/imath] be a basis of [imath]U[/imath], [imath](f_1, ..., f_m)[/imath] be a basis of [imath]V[/imath], [imath](g_1, ..., g_r)[/imath] be a basis of [imath]W[/imath]. Then [imath](((e_i \otimes f_j), w_k))_{1 \leq i \leq n, 1 \leq j \leq m, 1\leq k \leq r}[/imath] is a basis of [imath](U \otimes V) \times W[/imath]. Define [imath]\phi: (U \otimes V) \times W \to U \otimes (V \otimes W)[/imath] by [imath]\phi((e_i \otimes f_j), w_k) := e_i \otimes (f_j \otimes g_k)[/imath]. [imath]\phi[/imath] is bilinear. Claim: [imath](U \otimes (V \otimes W), \phi)[/imath] is a tensor product of [imath]U \otimes V[/imath] and [imath]W[/imath]. Then the statement follows. To show this by the universal property let [imath]\phi': (U \otimes V) \times W \to P[/imath] be bilinear. Then there is exactly one morphism [imath]\gamma: U \otimes (V \otimes W) \to P[/imath] given by [imath]\gamma(e_i \otimes (f_j \otimes g_k)) = \phi'(e_i \otimes f_j, g_k)[/imath], since [imath]((e_i \otimes (f_j \otimes g_k)))_{1 \leq i \leq n, 1 \leq j \leq m, 1\leq k \leq r}[/imath] is a basis of [imath]U \otimes (V \otimes W) \to P[/imath]. | 437706 | Proving that Tensor Product is Associative
I want to show that [imath]X\otimes(Y\otimes Z)[/imath] is isomorphic to [imath](X\otimes Y)\otimes Z[/imath]. Intuitively I think I should just choose bases [imath]\{e_{i}\}_{i\in I}, \{f_{j}\}_{j\in J}[/imath], and [imath]\{g_{k}\}_{k\in K}[/imath] for [imath]X,Y,Z[/imath] and map [imath]e_{i}\otimes(f_{j}\otimes g_{k})\mapsto (e_{i}\otimes f_{j})\otimes g_{k}[/imath] This defines a bijective correspondence between basis elements and so should induce a vector space isomorphism. Is there a way to do this using the universal property of the tensor product? [imath]\bf{\text{Context:}}[/imath] I am working through Introduction to Tensor Products of Banach Spaces by Raymond Ryan, and am working on exercises at the end of the first chapter. I'll try to summarize what I have available. We defined the space [imath]X\otimes Y[/imath] to be linear functionals on the space [imath]B(X\times Y)[/imath] of bilinear maps on [imath]X\times Y[/imath]. That is, [imath]x\otimes y:A\mapsto A(x,y)[/imath] for [imath]A\in B(X\times Y)[/imath]. We have stated and proved [imath]\bf{\text{Universal Property of Tensor Products}}[/imath]: Let [imath]X,Y,Z[/imath] be vector spaces. For every bilinear [imath]A:X\times Y\to Z[/imath] there is a unique linear map [imath]\hat{A}:X\otimes Y\to Z[/imath] such that [imath]\hat{A}(x\otimes y) = A(x,y)[/imath]. Next we proved that the Tensor product is unique up to isomorphism (in the sense of having this property). We did not define any higher tensor product structure such as [imath]\otimes_{i\in I}X_{i}[/imath]. |
2412965 | What curve does this represent as t varies over [−1, 1]?
Let [imath]x(t) = \frac{1-t^2}{1 + t^2}[/imath] and [imath]y(t) = \frac{2t}{1+t^2}[/imath] What curve does this represent as [imath]t[/imath] varies over [imath][−1, 1][/imath]? My attempt: I suspect it represent parabola. I know that by trigonometry formula [imath]\sin2A = \frac{2\tan A}{1+\tan^2 A} =\frac{2t}{1+t^2}[/imath]. But here I used the graph of [imath]\sin2A[/imath], it look parabola, and [imath]\cos2A = \frac{1-\tan^2A}{1+\tan^2A} =\frac{1-t^2}{1+t^2}[/imath], similarly it also look like parabola. If anybody help me ,,i would be very thankful,, | 255615 | Parametric Equation of a Circle Using a Line
Consider the unit circle [imath] x^2+y^2=1. [/imath] How can I parametrize it using the line [imath]y=m(x+1)[/imath], where [imath]m[/imath] is its slope? |
1000886 | I am having trouble with this integral from the 2012 MIT Integration Bee: [imath]\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}[/imath]
[imath]\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}} [/imath] Could someone explain to me how to integrate this integral. Thank you and cheers. | 1892802 | Stuck in integration: [imath]\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}[/imath]
[imath]\displaystyle\int {\frac{dx}{( 1+\sqrt {x})\sqrt{(x-{x}^2)}}}[/imath] [imath]=\displaystyle\int\frac{(1-\sqrt x)}{(1+x)\sqrt{x-x^2}}\,dx[/imath] [imath]=\displaystyle\int\frac{(1-\sqrt x+x-x)}{(1+x)\sqrt{x-x^2}}\,dx[/imath] [imath]=\displaystyle\int\frac{\,dx}{\sqrt{x-x^2}}-\displaystyle\int\frac{\sqrt x(1+\sqrt x)}{\sqrt x(1+x)\sqrt{1-x}}\,dx[/imath] I have tried the above integration a lot . But not able to solve further . Please give me some hint how to proceed or some other method |
2406796 | Composition of absolutely continuous function on [imath]\mathbb{R}[/imath]
Is it true that the composition of two absolutely continuous functions on the real line is absolutely continuous? I feel like this should be a resounding no, however, I'm unsure of any quick counterexamples to this claim. Can anyone think of one? | 1887750 | Composition of Absolutely Continuous Functions
I'm aware through a plethora of counterexamples that for two functions [imath]f[/imath], [imath]g[/imath] that are absolutely continuous, their composition is not necessarily so. However, before seeing these examples, I had written out what I thought was a proof confirming that it was true, and I'm not sure where my error is. My proof is as follows: Assume that [imath]f[/imath] and [imath]g[/imath] are absolutely continuous. Let [imath]\epsilon_f > 0[/imath]. Since [imath]f[/imath] is AC, there exists a [imath]\delta_f[/imath] such that for any finite disjoint collection [imath]\{(a_k,b_k)\}[/imath] [imath] \sum_{k=1}^n [b_k - a_k] < \delta_f \Longrightarrow \sum_{k=1}^n |f(b_k) - f(a_k)| < \epsilon_f[/imath] Since [imath]g[/imath] is absolutely continuous as well, let [imath]\epsilon_g = \delta_f[/imath]. Then there exists [imath]\delta_g[/imath] such that [imath] \sum_{k=1}^n [b_k - a_k] < \delta_g \Longrightarrow \sum_{k=1}^n |g(b_k) - g(a_k)| < \epsilon_g = \delta_f \Longrightarrow \sum_{k=1}^n |f(g(b_k)) - f(g(a_k)| < \epsilon_f [/imath] Thus, the composition is AC. I realize that the only difference between this is that in the composition, for the domain of [imath]f[/imath], I'm assuming that the sum [imath]|g(b_k) - g(a_k)|[/imath] in absolute value is less than [imath]\delta_f[/imath]. However, I'm not quite sure why that absolute value should matter in the case of continuity. It's left out of the original definition because it's assumed that [imath]b_k > a_k[/imath]. Other than that, though, I don't see any errors in logic. Thoughts? |
2254808 | How to evaluate this logarithmic integral?
[imath] \displaystyle \int_{0}^{1}\dfrac{\ln(x)\ln(1+x)\ln(1+x+x^{2})}{(1-x)(1+x^{2})}\,dx[/imath] I posted this on I&S but there was no response , so posting it here. Wolfram gave [imath]-0.223434[/imath], can it be evaluated in terms of special functions functions? | 2242141 | An integral of a rational function of logarithm and nonlinear arguments
This problem was posted in I&S [imath] \int_{0}^{1} \dfrac{\log x \log (1+x) \log (1+x+x^{2})}{(1-x)(1+x^{2})}\,\mathrm{d}x \approx -0.223434.[/imath] I am not sure if there exists a closed form but it seems worth trying. I am completely clueless on how to start with this beast. It is worth saying that[imath]1-x^3= (1-x)(1+x+x^2).[/imath] That seems to go no where. I think the integral can be represented as the derivative of the integral representation of the Hypergeometric function but I am not comfortable with that. Any ideas ? |
624670 | Find max: [imath]\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}[/imath]
For [imath]a,b,c>0[/imath] and [imath]abc=1[/imath]. Find max: [imath]\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}[/imath] | 1371054 | Olympiad inequality [imath]\frac{a}{2a + b} + \frac{b}{2b + c} + \frac{c}{2c + a} \leq 1[/imath].
I want to prove that for [imath]a, b, c > 0[/imath] we have [imath]\frac{a}{2a + b} + \frac{b}{2b + c} + \frac{c}{2c + a} \leq 1[/imath]. My approach: I know that each of the individual terms is lesser than [imath]\frac{1}{2}[/imath] because of it's form. I am familiar with the Cauchy-schwarz inequality and the AM-GM-HM inequality. I tried using AM-GM but could not get anywhere because of the way the inequality is structured. Similarly I tried using Cauchy-schwarz as well. I just need some intuition/hints on how to actually reduce this to a feasible form to solve and not the actual answer because it does not provide me with the necessary intuition which I need as I am a beginner when it comes inequalities. If I can't use these 2 inequalities, is this problem best tackled with some other inequality? |
2413202 | A good substitution to prove [imath]\frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}=\frac{f''(c)}{g''(c)}[/imath].
Let [imath]f'(x)[/imath] and [imath]g'(x)[/imath] satisfy the hypothesis of mean value theorem, then prove that [imath]\frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}=\frac{f''(c)}{g''(c)}[/imath] where [imath]g''(c)\neq 0[/imath]. I have tried various things like Cauchy MVT with [imath]f'(x)[/imath] and [imath]g'(x)[/imath], substituting [imath]h(x)=\dfrac{1}{g'(x)}[/imath] ,[imath]h(x)=f'(x)+\dfrac{a}{g'(x)}[/imath] where [imath]a\in\mathbb{R}[/imath], but it didn't lead anywhere. Any help will be appreciated. Please note that I am allowed to use only Cauchy's MVT, LMVT and Rolle's theorem. | 2406383 | Prove that [imath]a, \frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}=\frac{f"(c)}{g"(c)}[/imath]
Prove that if [imath]a<c<b[/imath],and [imath]f'(x)[/imath] is differentiable on [imath] (a,b)[/imath],then [imath]\frac{f(b)-f(a)-(b-a)f'(a)}{g(b)-g(a)-(b-a)g'(a)}=\frac{f''(c)}{g''(c)}[/imath] I tried to construct an auxiliary function [imath]h(x)=(f(b)-f(a)-(b-a)f'(a))g'(x)-f'(x)(g(b)-g(a)-(b-a)g'(a))[/imath] I tried to use Rolle's theorem but I am not getting [imath]h(a)=h(b)[/imath] |
2409790 | Normal Rectangular Hyperbola-Extension
I have already posted this question , although I have solved it , but during discussion , i was asked whether such type of hyperbola exist. Now my question is normal to the hyperbola [imath]xy=c^2[/imath] is drawn from point P,Q and R which meets on point S which lies on [imath]xy=c^2[/imath] .Does this type of Hyperbola exist. Please refer to the following Question Normal(Rectangular Hyperbola) | 2409466 | Three normals of a hyperbola passing through the same point on the curve
The normals at three points [imath]P[/imath], [imath]Q[/imath], [imath]R[/imath] on a rectangular hyperbola [imath]xy = c^2[/imath] intersect at a point on the curve. Prove that the centre of the hyperbola is the centroid of the triangle [imath]PQR[/imath]. I copied the the question word by word. Let assume that the points are (c[imath] t_1[/imath],[imath]\frac{c}{ t_1}[/imath]) , (c[imath] t_2[/imath],[imath]\frac{c}{ t_2}[/imath]) & (c[imath] t_3[/imath],[imath]\frac{c}{t_3}[/imath]), even if I draw the normal, at all the three point how do I intersect three lines at one point. |
2413199 | How to determine if a [imath]\lim\limits_{n \rightarrow \infty}{(1+{ix\over n})^n}[/imath] would be complex
Question Recently, I have been looking at complex limits, The most famous being [imath]e^{ix}[/imath]=[imath]\lim\limits_{n \rightarrow \infty}{(1+{ix\over n})^n}[/imath]. An example would be that when [imath]x = \pi[/imath] we know that the answer will be -1. But how do you determine this due to the fact that you can always [imath]+1[/imath] which will determine the outcome. I am fully aware that you are able to do this via the [imath]i\cdot \sin(a \ln b) +\cos(a\ln b)[/imath] however, how can you prove this via a limit, because if you test it on a calculator, most of the time you'll end up with some imaginary part. Specifically I have been looking at the representation of [imath]\sin x={ie^{-ix}\over 2}-{ie^{ix}\over 2}[/imath]. Everyone would be safe to assume that [imath]\sin x[/imath] is always real, but when you apply a limit then how can you determine if it is only real or imaginary and real? | 1557074 | Do the polynomials [imath](1+z/n)^n[/imath] converge compactly to [imath]e^z[/imath] on [imath]\mathbb{C}[/imath]?
The question is Do the polynomials [imath]p_n(x)=(1+z/n)^n[/imath] converge compactly (or uniformly on compact subsets) to [imath]e^z[/imath] on [imath]\mathbb{C}[/imath]? I thought about expanding [imath]p_n(z)=\sum_{k=0}^n a_k^{(n)}z^k[/imath] where [imath]a_k^{(n)}=\binom{n}{k}\frac{1}{n^k}=\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)[/imath] and trying to show that [imath]\frac{1}{k!}-a_k^{(n)}[/imath] decreases sufficiently fast on any closed ball. That is, I tried to show [imath]\lim_{n\rightarrow\infty}\max_{z\in\overline{B_0(A)}}\left|\sum_{k=0}^n\frac{z^k}{k!}-p_n(z)\right|=0[/imath] for any fixed [imath]A>0[/imath], but I had difficulty with this approach. Any help is appreciated. |
1973885 | The memoryless property of exponential distribution
Let [imath]Z[/imath] be an exponential random variable and [imath]R[/imath] an independent nonnegative random variable. Show that [imath] P(Z-R > u | Z > R ) = P(Z > u) [/imath] From here, I can use Bayes Theorem to split the Left Hand Side, and I know that I have to integrate them. However, I'm not sure how to exactly to proceed from here. I know that [imath]u \geq 0[/imath]. Any help would be greatly appreciated. Thanks! | 1749546 | memoryless property of exponential distributions with random variables
It is true that [imath]P(X>t+s|X>t)=P(X>s)[/imath] for certain values [imath]t[/imath] and [imath]s[/imath]. However, how can I show that this still holds if: [imath]T[/imath] is a continuous random variable. That is [imath]P(X>T+s|X>T)=P(X>s)[/imath] Both [imath]T[/imath] and [imath]S[/imath] are continuous RVs: [imath]P(X>T+S|X>T)=P(X>S)[/imath] All the random variables are independent. |
2414545 | chebyshev expansion
Express First kind Chebyshev polynomial in terms of monomials I have a function as (1) [imath]f=\sum_i^n a_iT_i[/imath] , where [imath]T_i[/imath] are first kind Chebyshev polynomial of order "i". I want to express [imath]f[/imath] in terms of monomials (powers of x). Hence, I need to write [imath]T_i[/imath] in terms of monomials.This can be done using following recursive formula: [imath]T_0(x)=1; T_1(x)=x; T_{n+1}(x)=2xT_n(x)−T_{n−1}(x)[/imath] for [imath]x \in [-1, 1][/imath]. So function [imath]f[/imath] in terms of monomials and the original function in (1) should match over [imath]x\in [-1 1][/imath] However as "i" increases ([imath]i\ge45[/imath]), the two functions do not match. For example, following Matlab plots function [imath]f=T_{50}[/imath] both in monomials space and Chebyshev space. clc;clear;close all; % f=T_n; n=50; x=[-pi:0.1:0]; plot(cos(x),cos((n-1)x),'k-') syms x; Tr(1,1)=1+0*x; Tr(2,1)=x; for i=2:n-1; Tr(i+1,1)= expand(2*x*Tr(i,1)-Tr(i-1,1));end; hold on;for x=-1:0.01:1; plot(x,eval(Tr(end)),'--bo'); end; | 2413957 | First kind Chebyshev polynomial to Monomials
Express First kind Chebyshev polynomial in terms of monomials First kind Chebyshev polynomial of order n ([imath]T_n[/imath]) is defined in terms of cosine function as follow: 1) [imath]T_n(\cos x)=\cos n x[/imath] Instead of working with cosine functions, I want to express Chebyshev polynomial in terms of monomials (powers of x) using the following recursion formula: [imath]T_0(x)=1; T_1(x)=x; T_{n+1}(x)=2xT_n(x)−T_{n−1}(x)[/imath] for [imath]x \in [-1, 1][/imath]. For example, first kind Chebyshev polynomial of order 2 can be expressed as: Usig formula 1) [imath]T_2(cosx)=cos(2x)[/imath], Usig recursive formula) [imath]T_2(x)=2xT_1(x)-T_0(x)=2x^2-1[/imath] If we plot [imath]cos2x[/imath] vesrus [imath]cosx, cosx\in [-1,1][/imath] and also [imath]2x^2-1[/imath] versus [imath]x, x\in [-1, 1][/imath], the two plot should match. However as "n" increases ([imath]n\ge45[/imath]), the [imath]T_n[/imath] obtained as a polynomial by the recursive formula doesn't match the Chebyshev polynomial of formula (1). Why is that????? The recursion formula is not valid for higher orders???? here is my Matlab code that compares two formulas: clc;clear;close all; syms x; n=45; Tr(1,1)=1+0*x; Tr(2,1)=x; for i=2:n; Tr(i+1,1)= expand(2*x*Tr(i,1)-Tr(i-1,1));end; hold on;for x=-1:0.01:1; plot(x,eval(Tr(n)),'--bo'); end; t=[0:0.01:2*pi]; plot(cos(t),cos((n-1)*t),'k*-') |
2414490 | proof of [imath]\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)[/imath] without solving [imath]\int\frac{1}{x^2+1}dx[/imath]
So I know [imath]\int\frac{1}{x^2+1}dx=\arctan(x)+c[/imath] Now when i tried to integral [imath]\frac{1}{x^2+1}[/imath] in a different way i got this: [imath]\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies \begin{cases} A+B=0 \\[2ex] A-B=\frac{1}{i}=-i \end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}[/imath] So in the end i get:[imath]\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx[/imath] After integrating this i get [imath]\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)[/imath] after comparing this to [imath]\arctan(x)[/imath] at [imath]x=0[/imath] i find [imath]C=-i\pi[/imath] [imath]\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)[/imath] now my question is how can i prove that without using the integral | 1479167 | Proving [imath]\rm \frac i 2 \ln\frac {x+i}{x-i}=\arctan x[/imath] .
Proving [imath]\rm \frac i 2 \ln\frac {x+i}{x-i} =\arctan x[/imath] I'd like to prove this identity without taking the derivatives and integrating, what are some cool ways to prove this? |
2414945 | Proof of a Ramanujan Statement
I recently came across the approximation [imath]\ln n! \approx n\ln n - n + \frac{1}{6}\ln(n(1+4n(1+2n)))+\frac{1}{2}\ln \pi[/imath] Supposedly this was written in Ramanujan's last notebook, but I can't seem to find it anywhere else, nevermind a proof for it. Does anyone know if it was ever proven, and if so where such a proof is? | 152342 | Ramanujan's approximation to factorial
I saw this approximation for the factorial given by Ramanujan as [imath]\log(n!) \approx n \log n - n + \frac{\log(n(1+4n(1+2n)))}{6} + \frac{\log(\pi)}{2}[/imath] in wikipedia, which claims the approximation is superior to Stirling's approximation. I tried to locate the reference but unfortunately I could not. I would appreciate if someone can throw light on how this asymptotic is obtained and the order of the error. |
2415117 | Why is the set [imath]\{ (z_1,z_2,z_3) \in \mathbb C^3:z_1^2 +z_2^2+z_3^2=1 \}[/imath] not compact?
Why is the set [imath]\{ (z_1,z_2,z_3) \in \mathbb C^3:z_1^2 +z_2^2+z_3^2=1 \}[/imath] not compact in the Euclidean topology? If we consider [imath]\mathbb C^3[/imath] as [imath]\mathbb R^6[/imath] and [imath]z_n=x_n+iy_n \text{(n=1,2,3)}[/imath] then I think the set will become closed and it is already bounded . Although why it is not compact? Please someone help.. Thank you.. | 1196523 | Is the set [imath]S=\left\{\left(z_1,z_2\right)\in \mathbb C\times \mathbb C:z_1^2+z_2^2=1\right\}.[/imath] compact?
Consider the set [imath]S=\left\{\left(z_1,z_2\right)\in \mathbb C\times \mathbb C:z_1^2+z_2^2=1\right\}.[/imath] Is this set compact in [imath]\mathbb C^2[/imath] ? As [imath]\mathbb C^2[/imath] is a finite dimensional space so a subset of it is compact if and only if it is closed and bounded. But [imath]z_1^2+z_2^2=1[/imath] gives [imath](x_1^2+x_2^2)-(y_1^2+y_2^2)=1[/imath] and [imath]x_1y_1+x_2y_2=0[/imath] , where [imath]z_1=x_1+iy_1[/imath] and [imath]z_2=x_2+iy_2[/imath]. But from here how I can show that the set is closed and bounded ? |
2415295 | [imath]n^{3pq} - n \equiv 0[/imath] (mod [imath]3pq[/imath]), where p and q are prime and [imath]n[/imath] is any integer Find the minimum possible value of [imath]p+q[/imath]
This problem was asked in a local Olympiad. I do not know how to proceed because I did not think Fermats little theorem is applicable to composite numbers. Any help would be appreciated. | 2401393 | Let [imath]p[/imath], [imath]q[/imath] be prime numbers such that [imath]n^{3pq} – n[/imath] is a multiple of [imath]3pq[/imath] for all positive integers [imath]n[/imath]. Find the least possible value of [imath]p + q[/imath].
Recently a exam called PRMO 2017 was conducted. Question 28 went as follows, Let [imath]p[/imath], [imath]q[/imath] be prime numbers such that [imath]n^{3pq} – n[/imath] is a multiple of [imath]3pq[/imath] for all positive integers [imath]n[/imath]. Find the least possible value of [imath]p + q[/imath]. This question was considered to be quite tough. Many people are saying it was too tough to be put in a exam which is open for 14 year olds. How can it be solved? I have not studied number theory, so I really couldn't attempt this. Thanks. |
2409577 | Prove [imath] ∀a, b, c ∈ Z,[/imath] if [imath]ab + ac ≡ 3[/imath] (mod 6) then b [imath]\not\equiv[/imath] c (mod 6).
Prove [imath]∀a, b, c ∈ Z,[/imath] if [imath]ab + ac ≡ 3[/imath] (mod 6) then b [imath]\not\equiv[/imath] c (mod 6). New to proof work and my problem is I have some general feelings about this problem but don't know how to express it since I'm not familiar enough with modular arithmetic theory. I've moved forward using the contrapositve to work with. Prove [imath]∀a, b, c ∈ Z,[/imath] if b [imath]\equiv[/imath] c (mod 6). Then [imath]ab + ac \not\equiv 3[/imath] (mod 6) I know congruent integers b and c imply [imath]\frac{b-c}{6}[/imath] being perfectly divisible and I know multiplying a same constant [imath]a[/imath] to both numbers does not change their congruence before summing them. Basically, I'm stuck here. I think that [imath]ab + ac \not\equiv 3[/imath] (mod 6) is true because I think congruent integers [imath]b[/imath] and [imath]c[/imath] multiplied by a same constant [imath]a[/imath] can only take an answer of 0, 1 and 2 in the equation [imath]ab + ac[/imath](mod 6). Obviously, me thinking things is not very mathematical! Plus I have a strong feeling I'm just flat out wrong somewhere, any ideas or solutions towards proving this statement would be very much stuided and appreciated. :) | 2415315 | Prove [imath]\forall a,b,c \in \mathbb Z[/imath], if [imath]ab+ac\equiv 3\pmod 6[/imath] then [imath]b \not\equiv c \pmod 6[/imath] by contraposition
Prove by contraposition: [imath](\forall a,b,c\in\mathbb Z[/imath]) if [imath]ab+ac\equiv 3 \pmod 6[/imath] then [imath]b \not\equiv c \pmod 6.[/imath] So, by contraposition，it is [imath]\forall a,b,c\in\mathbb Z,[/imath] if [imath]b \equiv c \pmod 6[/imath], then [imath]ab+ac\not\equiv 3 \pmod 6.[/imath] No idea for next step |
2415343 | Let [imath]A[/imath], [imath]B[/imath] be normal subgroups and [imath]A\cap B=\{e\}[/imath]. Prove that [imath]ab=ba[/imath] for all [imath]a\in A[/imath], [imath]b\in B[/imath]
Let [imath]A[/imath], [imath]B[/imath] be normal subgroups and [imath]A\cap B=\{e\}[/imath]. Prove that [imath]ab=ba[/imath] for all [imath]a\in A[/imath], [imath]b\in B[/imath]. As soon as [imath]A[/imath] and [imath]B[/imath] are normal subgroups, one can get that [imath]aB=Ba[/imath] and [imath]bA=Ab[/imath] for all [imath]a\in A[/imath], [imath]b\in B[/imath]. Hence, there are such [imath]a'\in A[/imath] and [imath]b'\in B[/imath] for [imath]a\in A[/imath] and [imath]b\in B[/imath] such that [imath]ab=b'a[/imath] and [imath]ba=a'b[/imath], so we need to proof that [imath]a=a'[/imath] and [imath]b=b'[/imath]. How do I use second condition that subgroups intersection is [imath]\{e\}[/imath]? | 2393299 | Commutator of two normal subgroups with trivial intersections
So I tried looking everywhere but I'm really lost with this proof. I need to to show that: Let [imath]H[/imath], [imath]K[/imath] be normal subgroups of a group [imath]G[/imath], satisfying [imath]H\cap K = \{e\}[/imath]. Let [imath]h\in H[/imath], [imath]k \in K[/imath]. (a) Prove that the commutator [imath]w = hkh^{-1}k^{−1}[/imath] satisfies [imath]w \in H[/imath] and [imath]w \in K[/imath]. Then conclude that [imath]w = e[/imath] (b) Prove that [imath]hk = kh[/imath], for all [imath]h \in H[/imath], [imath]k \in K[/imath]. |
2415415 | Prove that [imath] \lim\limits_{x\to \infty}xf(x)=0 [/imath]
I just need a hint not a whole solution please. Problem: Let [imath]f ∈ L_1(0, ∞)[/imath] be monotone. Prove that [imath] \lim\limits_{x\to \infty}xf(x)=0 [/imath] | 55649 | Does [imath]f[/imath] monotone and [imath]f\in L_{1}([a,\infty))[/imath] imply [imath]\lim_{t\to\infty} t f(t)=0[/imath]?
I want to show that if [imath]f[/imath] is non-increasing and [imath]f\in L_{1}([a,\infty),m)[/imath] where [imath]m[/imath] is Lebesgue measure then [imath]\lim_{t\to\infty} t f(t)=0[/imath]. So far I've been able to show that [imath]f\geq 0[/imath] and that [imath]\lim_{t\to\infty} f(t)=0[/imath]. Since monotone functions are differentiable a.e. I thought about using integration by parts but couldn't get anywhere with that. Any hints or suggestions would be greatly appreciated. |
2415413 | What happens at the boundary of the disk of convergence?
I got stuck in solving a question of power series in complex analysis in Conway's exercise. Which is as follows [imath]:[/imath] Find the radius of convergence of the power series [imath]:[/imath] [imath]\sum_{n=1}^{\infty} \frac {(-1)^n} {n} z^{n(n+1)}.[/imath] What happens at the boundary of the disk of convergence? I found that the radius of convergence to be [imath]1[/imath] and hence the disk of convergence is [imath]|z|=1[/imath]. Now I found at [imath]z=1,-1[/imath] the series is convergent by Leibnitz test. Now how can I check the point of convergence (if any) in the disk of convergence other than [imath]1[/imath] and [imath]-1[/imath]. I tried with Dirchilet's test. But I couldn't actually figure out how is [imath]\sum_{k=1}^{n} z^{k(k+1)}[/imath] look like? Is it bounded for arbitrary [imath]n[/imath] for all [imath]z[/imath] on [imath]\partial B(0;1)[/imath] except at [imath]1[/imath] and [imath]-1[/imath]? Please give me some suggestions. Thank you in advance. | 2415727 | What can be said about the convergence on [imath]|z|=1[/imath]?
I already found the radius of convergence of the power series [imath]:[/imath] [imath]\sum_{n=1}^{\infty} \frac {(-1)^n} {n} z^{n(n+1)}.[/imath] What happens at the boundary of disk of convergence? I found that the radius of convergence to be [imath]1[/imath] and hence the disk of convergence is [imath]|z|<1[/imath]. Now I found at [imath]z=1[/imath] the series converges. Now how can I proceed? Please help me. Thank you in advance. |
2416165 | Prove that[imath] \frac{(n^2)!}{(n!)^n} \in[/imath] Z
Using permutation or otherwise, prove that[imath] \frac{(n^2)!}{(n!)^n}[/imath]an integer, where n is a positive integer. Putting n=1,2,3.... I can prove that above quantity is an integer but this is not correct as it needs to be proved in a methodological way. | 467574 | Using permutation or otherwise, prove that [imath]\frac{(n^2)!}{(n!)^n}[/imath] is an integer,where [imath]n[/imath] is a positive integer.
Using permutation or otherwise, prove that [imath]\displaystyle \frac{(n^2)!}{(n!)^n}[/imath] is an integer,where [imath]n[/imath] is a positive integer. I have no idea how to prove this..!!I am not able to even start this Can u give some hints or the solution.!cheers.!! |
2414292 | Numerical range of the sum of matrices
Let [imath]A \in \mathbb{R}^{n \times n}[/imath] be nonsymmetric positive definite, if [imath]A[/imath] can be decomposed as [imath]A = A_1 \oplus A_2[/imath], where [imath]A_1 \in \mathbb{R}^{p \times p}[/imath] and [imath]A_2 \in \mathbb{R}^{q \times q}[/imath], [imath]p+q=n[/imath], it is known that \begin{align} W(A) = Co(W(A_1) \cup W(A_2)) \end{align} where [imath]W(A) = \left\{\frac{(Av,v)}{(v,v)}, 0 \ne v \in \mathbb{C}^n \right\}[/imath] is the numerical range anc [imath]Co()[/imath] is the convex hull. Is there a similar property for when [imath]A = R_1^T A_1 R_1 + R_2^T A_2 R_2[/imath] but [imath]p + q > n[/imath]? [imath]R_1:\mathbb{R}^n \rightarrow \mathbb{R}^p[/imath] and [imath]R_2:\mathbb{R}^n \rightarrow \mathbb{R}^q[/imath] are orthogonal projections such that their ranges have a non null intersection and the union of their ranges is [imath]\mathbb{R}^n[/imath], i.e. [imath]R_1 R_2 \ne \varnothing[/imath], [imath]R_2 R_1 \ne \varnothing[/imath] and [imath]R_1^TR_1 + R_2^T R_2[/imath] has full rank. I know of [imath]W(A) \subseteq W(R_1^T A_1 R_1) + W(R_2^T A_2 R_2)[/imath] where the sum is element-wise. But I know that if [imath]A_1[/imath] has no kernel in [imath]\mathbb{R}^p[/imath] and [imath]A_2[/imath] has no kernel in [imath]\mathbb{R}^q[/imath], then [imath]A[/imath] has no kernel in [imath]\mathbb{R}^n[/imath], and this doesn't show up in the formula since [imath]R_1^T A_1 R_1[/imath] and [imath]R_2^T A_2 R_2[/imath] obviously have a kernel in [imath]\mathbb{R}^n[/imath]. | 2416236 | Overlapping positive definite block matrices
Let [imath]A \in \mathbb{R}^{p \times p}[/imath] and [imath]B \in \mathbb{R}^{q \times q}[/imath] be nonsymmetric positive definite matrices in the sense that [imath]v^T A v > 0, \forall v \in \mathbb{R}^p[/imath] and [imath]v^T B v > 0, \forall v \in \mathbb{R}^q[/imath]. I know that [imath]A \oplus B[/imath], i.e. a matrix with a block structure \begin{align} \pmatrix{A & 0 \\ 0 & B} \end{align} is positive definite and I also know that if [imath]p=q[/imath], then [imath]A+B[/imath] is positive definite as well. Can I say that as well if the blocks overlap partially? When? For instance a matrix as follows: \begin{align} \pmatrix{a_{11} & a_{12} & a_{13} & a_{14} & 0& 0\\ a_{21} & a_{22} & a_{23} & a_{24}& 0& 0\\ a_{31} & a_{32} & a_{33} + b_{11} & a_{34} + b_{12} &b_{13}&b_{14}\\ a_{41} & a_{42} & a_{43} + b_{21} & a_{44} + b_{22} &b_{23}&b_{24}\\ 0 & 0 & b_{31} & b_{32} &b_{33}&b_{34} \\ 0 & 0 & b_{41} & b_{42} &b_{43}&b_{44}} \end{align} |
2416281 | Eigenvalues of the adjoint representation of a matrix with distinct eigenvalues
This is question 1.6 from Humphrey's "Introduction to Lie Algebras and Representation Theory": Suppose [imath]x \in \mathfrak{gl}(n,\mathrm{F})[/imath] has [imath]n[/imath] distinct eigenvalues [imath]a_1,\ldots,a_n[/imath]. The eigenvalues of the adjoint representation [imath]\mathrm{ad}(x)[/imath] are then precisely the [imath]n^2[/imath] scalars [imath]a_i - a_j[/imath] for [imath]1\leq i,j \leq n[/imath], which of course need not to be distinct. How do I go on proving this? Any help will be much appreciated | 464450 | Eigenvalues of [imath]\operatorname{ad}x[/imath]
Let [imath]x\in \operatorname{gl}(n,F)[/imath] have [imath]n[/imath] distinct eigenvalues [imath]a_1,\ldots,a_n[/imath] in [imath]F[/imath]. Prove that the eigenvalues of [imath]\text{ad }x[/imath] are precisely the [imath]n^2[/imath] scalars [imath]a_i-a_j[/imath] ([imath]1\leq i,j\leq n[/imath]), which of course need not be distinct. So we can represent [imath]x[/imath] by an [imath]n\times n[/imath] matrix [imath]X[/imath]. We have [imath]Xv_1=a_1v_1,\ldots, Xv_n=a_nv_n[/imath] for eigenvectors [imath]v_1,\ldots,v_n[/imath]. Now, [imath]\operatorname{ad}x[/imath] takes [imath]y\in \operatorname{gl}(n,F)[/imath] to [imath]xy-yx[/imath]. I need to show that some [imath]y[/imath] is taken to a scalar multiple of itself, where that scalar is [imath]a_i-a_j[/imath]. What could be that [imath]y[/imath]? |
2409287 | Finding the largest square factor of an integer without complete factorization
When faced with the problem of reducing a fraction to lowest terms, it appears that currently conventional textbooks say one should do something like this: [imath] \frac{408}{684} = \frac{2\times2\times2\times3\times17}{2\times2\times3\times3\times 19} = \frac{17}{19}, [/imath] i.e. factor both the numerator and the denominator and then cancel. (But maybe I'm mistaken as to what the most up-to-date textbooks say.) This becomes somewhat more onerous with things like [imath]\dfrac{49494}{21583}.[/imath] And considerably more onerous in some other cases whose details I leave for now to the reader's imagination. [imath] \frac{49494}{21583} = \frac{2\times3\times73\times113}{113\times191} = \frac{438}{191}. [/imath] Finding the prime factors in a case like [imath]113\times191[/imath] But Euclid's algorithm, introduced around 300 BC and thus the oldest algorithm still in standard use today, tells us that \begin{align} & \gcd(49494,21583) = \gcd(6328,21583) = \gcd(6328,2599) \\[10pt] = {} & \gcd(1130,2599) = \gcd(1130,339) = \gcd(113,339) = \gcd(113,0) = 113, \end{align} so that [imath] \frac{49494}{21583} = \frac{113\times438}{113\times191}. [/imath] No need to search for prime factors. Divisiblity of the numerator by [imath]73[/imath] is irrelevant to the problem and isn't hinted at by Euclid's algorithm. So my question involves another problem for which complete factorization seems to be the standard recommended method: [imath] \sqrt{4536} = \sqrt{2\times2\times2\times3\times3\times3\times3\times7} = 2\times3\times3\times\sqrt{2\times7} = 18\sqrt{14}. [/imath] What about things like [imath]\displaystyle\sqrt{159501017} = \sqrt{13\times13\times83\times83\times137}\text{ ?}[/imath] An efficient algorithm for finding the greatest square factor would tell us that it is [imath]1164241 = (1079)^2[/imath] without bringing in the irrelevant divisibility of that number by [imath]13[/imath] and [imath]83.[/imath] Is there such an algorithm? | 871766 | How to efficiently find the largest perfect square dividing a given large integer?
Given a number [imath]n[/imath]. I need to find the largest [imath]q[/imath] such that [imath]q^2[/imath] divides [imath]n[/imath]. I need the fastest method to find [imath]q[/imath]. [imath]q[/imath] can be any number prime or composite. At present I am factorizing the number [imath]n[/imath] to find the highest number [imath]q[/imath]. I need a better approach which does not involve to factorize the number [imath]n[/imath] completely or by some other approach which is less than [imath]O(\sqrt n)[/imath]. Constraint: [imath]n\leq 10^{18}[/imath] some test cases: if [imath]n=180[/imath] then [imath]q=6[/imath] if [imath]n=17[/imath] then [imath]q=1[/imath] if [imath]n=10000[/imath] then [imath]q=100[/imath] |
2416720 | In a ring, is a [imath](ab)^{-1}=b^{-1}a^{-1}[/imath]?
I vaguely remember seeing this theorem somewhere but am having trouble finding a proof. It is difficult to search for a proof because it is difficult in general for symbolic identities and terms like "inverse", "multiplication", "ring", and "distributive" are quite general in ring theory. I have gotten this far: [imath]1=\left(ab\right)^{-1}\left(ab\right)\quad1=bb^{-1}a^{-1}a[/imath] [imath]\left(ab\right)^{-1}\left(ab\right)=bb^{-1}a^{-1}a[/imath] [imath]\left(ab\right)^{-1}\left(ab\right)a^{-1}=bb^{-1}a^{-1}aa^{-1}[/imath] [imath]\left(ab\right)^{-1}\left(ab\right)a^{-1}=bb^{-1}a^{-1}[/imath] [imath]b^{-1}\left(ab\right)^{-1}\left(ab\right)a^{-1}=b^{-1}bb^{-1}a^{-1}[/imath] [imath]b^{-1}\left(ab\right)^{-1}\left(ab\right)a^{-1}=b^{-1}a^{-1}[/imath] Without commutativity, I cannot simply rearrange the [imath]\mathrm{LHS}[/imath] to an expression where I can eliminate [imath](ab)[/imath]. I may be taking the wrong approach, so any help (including starting over from the beginning) is welcome. | 52969 | Check my proof that [imath](ab)^{-1} = b^{-1} a^{-1}[/imath]
The following question is problem Pinter's Abstract Algebra. And to put things in context: [imath]G[/imath] is a group and [imath]a, b[/imath] are elements of [imath]G[/imath]. I want to show [imath](ab)^{-1}[/imath] = [imath]b^{-1}a^{-1}[/imath]. I originally thought of proving the fact in the following manner: \begin{align*} (ab)^{-1}(ab) &= e \newline (ab)^{-1}(ab)(b^{-1}) &= (e)(b^{-1}) \newline (ab)^{-1}(a)(bb^{-1}) &= (b^{-1}) \newline (ab)^{-1}(a)(e) &= (b^{-1}) \newline (ab)^{-1}(a) &= (b^{-1}) \newline (ab)^{-1}(a)(a^{-1}) &= (b^{-1})(a^{-1}) \newline (ab)^{-1}(e) &= (b^{-1})(a^{-1}) \newline (ab)^{-1} &= (b^{-1})(a^{-1}) \newline \end{align*} I know this may seem extremely inefficient to most, and I know there is a shorter way. But would this be considered a legitimate proof? Thanks in advance! |
2416739 | All the permutations that commute with (1 2)(3 4 5) in [imath]S_5[/imath]
Determine all the permutations [imath]\sigma \in S_5[/imath] that commute with the permutation [imath]\alpha = (1 \space 2)(3 \space 4 \space 5).[/imath] | 922995 | Find eight elements in [imath]S_6[/imath] that commute with [imath](12)(34)(56)[/imath]
This is a homework problem and I'm having trouble just getting a basic understanding. I understand that [imath]S_6[/imath] is a symmetric group of degree [imath]6[/imath]. I'm not sure how to start looking for elements. |
714678 | Solve the congruence system [imath]\ x\equiv m_i-1 \pmod{m_i}\,[/imath] for [imath]\,i = 1,\ldots, k[/imath]
Find natural number [imath]x[/imath] so that [imath]x\equiv 9\pmod{10},\quad x\equiv8\pmod9,\quad ...,\quad x\equiv 1\pmod2[/imath] | 3010391 | modular arithmetic problem puzzle
If you have: [imath]N\equiv 2\mod 3 ,\: N\equiv 3 \mod 4 , \: N\equiv 4 \mod 5 [/imath] What is N? |
2416718 | Prove: [imath]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\implies \frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}=\frac{1}{(x+y+z)^5}[/imath]
Prove that for all real numbers [imath]x,y,z[/imath]: [imath]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\implies \frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}=\frac{1}{(x+y+z)^5}.[/imath] | 1426119 | If [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}[/imath] then [imath]\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.[/imath]
Suppose that [imath]\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}[/imath] . Then , prove that [imath]\displaystyle\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.[/imath] Attempt : From the given relation , [imath]\displaystyle \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}=-2[/imath].........(1) Now I want to calculate , [imath]\displaystyle \frac{a^5}{b^5}+\frac{b^5}{a^5}+\frac{b^5}{c^5}+\frac{c^5}{b^5}+\frac{a^5}{c^5}+\frac{c^5}{a^5}[/imath]. I tried by expanding pair of terms and putting the value of (1), but it can't help... |
2417482 | Does a ring isomorphism preserves the structure of a field?
The problem is the following one: Let [imath]f: A \rightarrow B[/imath] a ring isomorphism. If A is a field, is B a field? I think It's false, but I can't think of a counterexample, and if it's true I have no idea of how to prove it. | 1748642 | Isomorphisms preserving integral domains and fields
Let [imath]R \simeq S[/imath] be isomorphic commutative rings with unity. Prove the following: a). If [imath]R[/imath] is an integral domain then [imath]S[/imath] is an integral domain. For this, I said that If I let [imath]f: R \to S[/imath] be an isomorphism and [imath](R, + , \circ), (S, \oplus, \ast)[/imath], then let [imath]x, y\in R[/imath], such that [imath]x \circ y = 0_R[/imath], because [imath]f[/imath] is an isomorphism, [imath]f(x \circ y) = f(x) \ast f(y) = f(0_R)[/imath]. If I let [imath]R[/imath] be an integral domain then, as per the definition [imath]R[/imath] is commutative, has unity, and no zero divisors. So [imath]f(x \circ y) = f(x) \ast f(y) = f(0_R)[/imath] where we have either [imath]f(x \circ y) = f(0_R) \ast f(y) = f(0_R)[/imath] or [imath]f(x \circ y) = f(x) \ast f(0_R) = f(0_R)[/imath]. Since [imath]f[/imath] is an isomomorphism, let [imath]s = f(x), t = f(y)| s,t \in S[/imath]. Then [imath]s \ast t = f(0_R)[/imath]. By definition, [imath]S[/imath] also has unity so [imath]0_S[/imath] exists. If we map [imath]0_R \to 0_S[/imath]. Then either [imath]s = 0_S[/imath] or [imath]t = 0_S[/imath]. So then, since [imath]S[/imath] has unity, is commutative, and has a no zero divisors as we showed above, it is also an Integral domain. I'm not sure I have this correctly. b). If [imath]R[/imath] is a field then [imath]S[/imath] is a field. I don't really understand how to do this one. Since [imath]S[/imath] is an integral domain, it has a [imath]0_S[/imath]. So I need to show that [imath]S[/imath] distributes [imath]\ast[/imath] over [imath]\oplus[/imath]. Any help with these is appreciated |
2416964 | Is the union of a chain of topologies a topology?
Let's consider the following inclusion chain of topologies on space [imath]X[/imath]: [imath]\tau_1\subset\tau_2\subset\cdots\subset\tau_n\subset\cdots[/imath]. Let [imath]\tau=\bigcup_{n=1}^\infty \tau_n[/imath]. Is [imath]\tau[/imath] a topology? Obviously , the intersection of any two sets from [imath]\tau[/imath] belongs to [imath]\tau[/imath]. However, it is not clear whether [imath]\bigcup_{n=1}^\infty A_n\in\tau[/imath] where [imath]A_n\in\tau_n[/imath]. I think, in general [imath]\tau[/imath] is not a topology but cannot find a counterexample. | 468343 | Union of ascending chain of Topologies
Let [imath]\lbrace \tau_n:n\in \mathbb{N}\rbrace[/imath] be an ascending chain of topologies on a nonempty set [imath]X[/imath]. Then is [imath]\bigcup\limits^{\infty}_{n=0}\tau_n[/imath] a topology on [imath]X[/imath]? I have a strong notion that it may not be a topology. But I am not getting the proper counter example. Please help. I know this must be a topology if the collection is finite. |
2400974 | Is [imath]1/\infty=0{}[/imath]?
If [imath]1\div0=\infty[/imath], then [imath]1\div(1\div0)=0[/imath]. Does this mean [imath]1\div\infty=0[/imath]? Or is it always close to zero? Please let me know what is correct: either it's zero or always close to zero. | 1013287 | [imath]1/\infty[/imath] is zero or infinitesimal?
Since [imath]\infty>0[/imath], so [imath]1/\infty >0[/imath], thus I think [imath]1/\infty[/imath] should be infinitesimal, but the calculus book says [imath]\lim_{x \to \infty} \frac{1}{x}= 0[/imath] So is [imath]1/\infty[/imath] zero or infinitesimal ? P.S. I mean are [imath]1/\infty[/imath] and [imath]\lim_{x \to \infty} 1/x[/imath] the same thing here? |
2418054 | [imath]a\equiv b\pmod n[/imath] iff [imath]a \bmod m = b \bmod m[/imath]
I'm sort of stuck on this proof I know that [imath]a\equiv b\pmod n[/imath] means [imath]a - b = mk[/imath] for some integer [imath]k[/imath] and by the Division algorithm [imath]a = mq_1 + r_1[/imath] [imath]b = mq_2 + r_2[/imath] [imath]a - b = m(q_1 - q_2) + (r_1 - r_2)[/imath] Once I get to the line above I get stuck how do I show that [imath]r_1 - r_2[/imath] are actually equal and the same remainder? I've looked at other proofs that just say the remainder for [imath]a[/imath] and [imath]b[/imath] equal to [imath]r[/imath]. I feel like i'm missing something extremely obvious. | 2395053 | Why are naturals < modulus equal if congruent? [uniqueness of remainder]
Let [imath]a[/imath], [imath]b[/imath], and [imath]n[/imath] be integers with [imath]0 \le a \lt n[/imath] and [imath]0 \le b \lt n[/imath]. Suppose [imath]a \equiv b \pmod n[/imath]. Then [imath]a-b=0[/imath] and [imath]a=b[/imath]. Why does the congruence of two non negative integers imply that they must be equal? |
2418142 | Show identity for [imath]e^x[/imath]
I want to show that, using the definition [imath]e^x := \lim_{n \to \infty} (1+\frac{x}{n})^n[/imath], the following identity holds: [imath] e^{x+y}=e^xe^y[/imath] This is what I tried: \begin{align} e^xe^y &=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n \\ &= \lim_{n \to \infty}\left(\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)\right)^n \\ &=\lim_{n \to \infty}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n \end{align} However, this seems to be more than [imath]e^{x+y}=\lim_{n \to \infty}(1+\frac{x+y}{n})^n[/imath]. How can I resolve this seeming contradiction to the upper identity? | 1451221 | Show that [imath]e^{x+y}=e^xe^y[/imath] using [imath]e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n[/imath].
I was looking for a proof of [imath]e^{x+y}=e^xe^y[/imath] using the fact that [imath]e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.[/imath] So I have that [imath]\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}[/imath] But I can't get [imath]\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.[/imath] Any idea ? |
2417602 | LP Maximising a negative objective function
How can I turn this into canonical form and then use the two-phase simplex method to solve it? [imath]max\quad -2x_1-x_2-x_3 \\ \text{subject to: }-x_1-x_3 \le -1 \\ -x_1+x_2 \le -2 \\ -x_2+x_3 \le -1 \\ x_1,x_2,x_3 \ge 0 [/imath] When I put it into the tableau with only slack variables there are no negative coefficients in the bottom room to find the column to pivot around | 2415062 | Linear programming - Maximizing negative objective function
How can I turn this into canonical form and then use the two-phase simplex method to solve it? Would I need to add slack variable AND surplus variables? [imath]\max\quad -2x_1-x_2-x_3 \\ \text{subject to: }-x_1-x_3 \le -1 \\ -x_1+x_2 \le -2 \\ -x_2+x_3 \le -1 \\ x_1,x_2,x_3 \ge 0 [/imath] When I put it into the tableau with only slack variables there are no negative coefficients in the bottom room to find the column to pivot around. |
2418056 | Show that if [imath]M[/imath] is closed ,then [imath](M^\perp)^\perp=M[/imath]
let [imath]M[/imath] be a subspace of a normed space [imath]X[/imath] and [imath]N[/imath] a subspace of its dual [imath]X^{*}[/imath]. we define [imath]M^\perp=\{f\in X^{*}:f(x)=0,\forall x\in M\}[/imath] and [imath]N^\perp=\{x\in X:f(x)=0,\forall f\in N\}[/imath] Show that if [imath]M[/imath] is cloesd ,then [imath](M^\perp)^\perp=M[/imath] i think [imath]M\subset(M^\perp)^\perp[/imath] is easy. but how to do other way. any hint | 1135883 | Showing [imath](M^\perp)^\perp=\overline{M}[/imath]
I have a question about a step in proving [imath](M^\perp)^\perp=\overline{M}[/imath] where [imath]M[/imath] is a linear subspace of a normed vector space [imath]E[/imath]. And [imath]M^\perp=\{f\in E^*|\langle f,x\rangle =0\}[/imath] This is the proof from Brezis book: Proof: Assuming the other direction, show [imath]\subset[/imath] direction. Suppose there exists [imath]x_0\in (M^\perp)^{\perp}[/imath] such that [imath]x_0\notin \overline{M}[/imath]. Then, Hahn-Banach says there exists a hyperplane that strictly separates [imath]\{x_0\}[/imath] and [imath]\overline{M}[/imath]. Thus there are some [imath]f\in E^*[/imath] and [imath]\alpha\in\mathbb{R}[/imath] such that [imath] \langle f,x\rangle <\alpha<\langle f, x_0\rangle, \forall x\in M[/imath] Since [imath]M[/imath] is a linear space it follows that [imath]\langle f,x\rangle =0[/imath](WHY??), for all [imath]x[/imath] and [imath]\langle f, x_0\rangle >0[/imath]. Therefore [imath]f\in M^\perp[/imath] and consequently [imath]\langle f, x_0\rangle =0[/imath], a contradiction. Can someone tell me why the place (WHY??) is true? I couldn't find out the line follows from the fact that [imath]M[/imath] is a linear space. Thanks |
2418912 | Prove that a power series[imath]\sum_{n=0}^\infty a_nz^n[/imath] which is convergent for any [imath] z \in \Bbb N[/imath], converges for any [imath]z \in \Bbb C[/imath].
Proof: The power series [imath]\sum_{n=0}^\infty a_nz^n[/imath] which converges for any [imath] z \in \Bbb N[/imath], then the series converges absolutely for ay [imath]z_0[/imath] s.t. [imath]|z_0|<|z|[/imath]. Let [imath]z_\alpha \in \Bbb C[/imath], [imath]\forall[/imath] [imath]z_\alpha[/imath], [imath]\exists z \in \Bbb N[/imath] s.t. [imath]|z_\alpha|<|z|[/imath]. Since the series converges for any [imath]z \in \Bbb N[/imath], using the proposition above, the series converges absolutely for any [imath]z_\alpha \in \Bbb C[/imath]. Question. I am not totally sure about my proof, is correct or not? If not, how can i improved it? | 2416505 | Prove that a power series [imath]\sum_{n=0}^{\infty}{{a_n}{z^n}}[/imath] which converges for any [imath]z \in \mathbb{N}[/imath], converges for any [imath]z \in \mathbb{C}[/imath].
PROBLEM Prove that a power series [imath]\sum_{n=0}^{\infty}{{a_n}{z^n}}[/imath] which converges for any [imath]z \in \mathbb{N}[/imath], converges for any [imath]z \in \mathbb{C}[/imath]. MY ATTEMPT Since the power series [imath]\sum_{n=0}^{\infty}{{a_n}{z^n}}[/imath] converges for any [imath]z \in \mathbb{N}[/imath], then [imath]\sum_{n=0}^{\infty}{{a_n}{z^n}}[/imath] can be considered (?) to be a polynomial in [imath]\mathbb{C}[/imath]. Hence, [imath]\sum_{n=0}^{\infty}{{a_n}{z^n}}[/imath] is entire, and therefore continuous in [imath]\mathbb{C}[/imath] (since differentiability implies continuity). It follows that [imath]\sum_{n=0}^{\infty}{{a_n}{z^n}}[/imath] converges for any [imath]z \in \mathbb{C}[/imath]. QUESTION I am aware that, while I may have (some) ideas for a (general) proof, admittedly there are several grey areas (i.e., gaps) that need to be filled. Would you be kind enough as to fill in these gaps, by providing useful hints? |
2419016 | Every maximal ideal is a prime ideal
The question formuled in the exam was exactly: ''Every maximal ideal is a prime ideal'' Maximal and prime ideals are defined for a commutative ring R, but the proof I have for maximal ideal [imath]\Rightarrow[/imath] prime ideal needs that R is a commutative and unitary ring, because it uses that [imath]R/I[/imath] is field iff [imath]I[/imath] is maximal, and that only happens if R is both commutative and unitary. Then my question is: Is there a proof for maximal ideal [imath]\Rightarrow[/imath] prime ideal for R a commutative ring (not neccesarily unitary)? If not, there must be a counterexample with a maximal ideal which isn't prime in a R commutative (not unitary) ring, right? I know there are more posts like this, but the difference is that those posts suppose R as a commutative and unitary ring (that's the difference with this one, just to be clear). | 169188 | A maximal ideal is always a prime ideal?
A maximal ideal is always a prime ideal, and the quotient ring is always a field. In general, not all prime ideals are maximal. 1 In [imath]2\mathbb{Z}[/imath], [imath]4 \mathbb{Z} [/imath] is a maximal ideal. Nevertheless it is not prime because [imath]2 \cdot 2 \in 4\mathbb{Z}[/imath] but [imath]2 \notin 4\mathbb{Z}[/imath]. What is that is misunderstand? |
2419202 | Smallest subring of rationals containing [imath] p/q[/imath], where [imath] p[/imath] and [imath]q [/imath] are relatively prime
I am trying to find the smallest subring of rationals containing [imath]p/q[/imath], where p,q are relatively prime. I am constructing the subring from [imath]p/q [/imath] and I think rationals of form [imath]pZ/q^k[/imath] where k is whole number is the answer. Am I wrong? Is there any general solution for [imath]p/q[/imath]? Edit: I am looking for a general solution for any [imath]p/q[/imath]. There is no constraint that 1 should be in the subring. | 2388877 | Find all rings [imath]R[/imath] satisfying [imath]\mathbb Z \subset R \subset \mathbb Z[\frac {1}{2}][/imath]
Find all rings [imath]R[/imath] satisfying [imath]\mathbb Z \subset R \subset \mathbb Z[\frac {1}{2}][/imath]. Solution: Let [imath]f \in R[/imath]. Then [imath]f= \frac {m}{2^n}[/imath] for some integers [imath]m,n[/imath]. Since [imath]2^{n-1} \in R[/imath], hence [imath]\frac {m}{2} \in R[/imath]. Hence one gets that [imath] \frac {1}{2} \in R[/imath], so [imath]R=\mathbb Z[\frac {1}{2}][/imath], so there is no ring [imath]R[/imath] satisfying given conditions. What are the other possible solutions of this problem? P.S: You can give a solution using any theorem from commutative algebra also. |
2419216 | Determinant of a specific [imath]n\times n[/imath] matrix
Let [imath]M_n= \begin{bmatrix} 1 & 2 & 3 & ... & n\\ 2 & 2 & 3 & ... & n\\ 3 & 3 & 3 & ... & n\\ \vdots & \vdots & \vdots & \ddots & n\\ n & n & n & n & n \end{bmatrix}[/imath] What's [imath]\det(M_n)[/imath] ? Seems like it is one of these exercises virtually impossible to do without the one trick, which is why I ask this question. Computing the first few determinants yields: [imath]\det(M_1)=1[/imath] [imath]\det(M_2)=-2[/imath] [imath]\det(M_3)=3[/imath] so if I were to make a wild guess, I would say [imath]\det(M_n)=n(-1)^{n+1}[/imath] The induction process didn't work too well for me because I can't find the determinant once a row and a column have been added to pad the matrix to dimension [imath]n+1[/imath]. I tried to use the well-known formula for the determinant by cutting the inside in four parts ([imath]M_n[/imath] in the upper-left, [imath](n+1)[/imath] in the lower-right, and the [imath](n+1)[/imath] row and column, but at the end I must compute the determinant of a sum of matrices which didn't help... I don't know the name of this specific matrix, so maybe this is a repost, but if this is the case, I can't find the original one I'd be grateful if someone could provide some hints. | 2013663 | If [imath]a_{ij}=\max(i,j)[/imath], calculate the determinant of [imath]A[/imath]
If [imath]A[/imath] is an [imath]n \times n[/imath] real matrix and [imath]a_{ij}=\max(i,j)[/imath] for [imath]i,j = 1,2,\dots,n[/imath], calculate the determinant of [imath]A[/imath]. So, we know that [imath]A=\begin{pmatrix} 1 & 2 & 3 & \dots & n\\ 2 & 2 & 3 & \dots & n\\ 3 & 3 & 3 & \dots & n\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ n& n & n & \dots & n \end{pmatrix}[/imath] but what do I do after? |
2419820 | Why are there [imath]p^n[/imath] elements in the quotient field [imath]\mathbb{F}_p [x]/ (f)[/imath]
Let [imath]f[/imath] be a irreducible polynomial in [imath]\mathbb{F}_p[x][/imath] of degree [imath]n[/imath]. Why are there [imath]p^n[/imath] elements in the quotient field [imath]\mathbb{F}_p [x]/ (f)[/imath] I am having some difficulty convincing myself this is true. Any help or insight is deeply appreciated | 82420 | Polynomials, finite fields and cardinality/dimension considerations
Can someone please give me a hint how to prove that the quotient ring [imath]\mathbb{F}_p[x]/{\langle f\rangle}[/imath], where [imath]f[/imath] is a irreducible polynomial of degree [imath]k[/imath] and [imath]p[/imath] is a prime (and [imath]\langle f\rangle[/imath] the ideal generated by [imath]f[/imath]), has cardinality [imath]p^k[/imath] and how to determine the dimension of [imath]\mathbb{F}_p[x]/{\langle f\rangle}[/imath] as a vector space ? |
2419998 | Taking the square root of both sides of a congruence.
Quite probably a silly question, but if I have: [imath]a^{2} \equiv 1 \pmod n[/imath], is this the same as [imath]a \equiv 1 \pmod n[/imath]? | 104961 | Number of solutions of [imath]x^2=1[/imath] in [imath]\mathbb{Z}/n\mathbb{Z}[/imath]
Next is what I have worked out to the moment. [imath]1[/imath] and [imath]-1[/imath] are roots for all [imath]n[/imath]. [imath]x \in \mathbb{Z}/n\mathbb{Z},\ [/imath] [imath]x^2\equiv1 \Leftrightarrow (x-1)(x+1)\equiv0 \Leftrightarrow \exists k \in \mathbb{Z}/n\mathbb{Z}: k(k+2)\equiv0 [/imath]. But how can it be applied to find other roots? |
2420217 | Why [imath]\int \frac{1}{x}[/imath] is [imath]\ln|x| + C[/imath] not [imath]\ln(x) + C[/imath]?
I am reading about indefinite integrals and I have a question regarding the use of the modulus in integrating [imath]\frac{1}{x}[/imath]. The book I am reading gives me the following explanation - The book later gives no further explanation, which has left me confused. I have found no reasonable explanation online, so I would like to ask why this is the case in integration. What does the book cite as "difficulties that may arise"? | 1034109 | Why does my professor say that writing [imath]\int \frac 1x \mathrm{d}x = \ln|x| + C[/imath] is wrong?
My professor says that writing this is convenient [imath]\int \frac 1x \mathrm{d}x = \ln|x| + C\tag{1}[/imath] but wrong, since it should be written as: [imath]\int \frac 1x \mathrm{d}x = \begin{cases}\ln x + C &x > 0\quad(\star)\\[0.2em] \ln(-x) + C &x < 0\end{cases}[/imath] I was wondering why is that the case. I thought that the two were equivalent, as one can see by the definition of absolute value. In [imath](\star)[/imath] the equality sign is dropped because the logarithm is not defined in [imath]0[/imath], but that would be the case with [imath](1)[/imath] as well. |
2419875 | Subgroup of integers
I know that the only possible subgroups of [imath]\mathbb Z[/imath] are of the form [imath]m\mathbb Z[/imath]. But how can I prove that these are the only possible subgroups? | 1301499 | Proof that [imath]n\Bbb Z \leq \Bbb Z[/imath] and are the only subgroups of [imath]\Bbb Z[/imath]
My challenge is Prove that if [imath]n = 0,1,2,\ldots[/imath] and [imath]n\Bbb Z = \lbrace nk: k \in \Bbb Z \rbrace[/imath], show that [imath]n\Bbb Z[/imath] is a subgroup of [imath]\Bbb Z[/imath] and are the only subgroups. I handled the first point by saying that i) Identity is 0, and [imath]0\in n\Bbb Z[/imath] and [imath]0 \in \Bbb Z[/imath] ii) For any [imath]a,b \in n\Bbb Z[/imath] [imath]a+b = na_2 + nb_2 = n(a_2+b_2) \in n \Bbb Z[/imath] iii) For any [imath]na \in n\Bbb Z[/imath], its inverse [imath]n(-a) \in n\Bbb Z[/imath]. For the second point, I said that all of the subgroups should be infinite, because for any finite group the largest element [imath]a+a \notin H[/imath], where [imath]H[/imath] is the potential subgroup, which is a contradiction. So [imath]H[/imath] must be of infinite order. For any [imath]a \in H, -a[/imath] should also be in [imath]H[/imath]. The difference between any [imath]a,b \in H[/imath] should also be in [imath]H[/imath], i.e. [imath](a-b) \in H[/imath]. This leaves only the whole group or any group of the form [imath]n\Bbb Z[/imath]. Is my proof complete? |
108071 | Units of Gaussian integers
How can we show that [imath]\pm 1, \pm i[/imath] are the only units in the ring of Gaussian integers, [imath]\mathbb Z[i][/imath]? Thank you. | 2660709 | Units in ring theory
How to determine [imath]U( Z[i])[/imath] I tried this [imath](a+bi)(c+di) = 1,[/imath] where [imath]a,b,c,d[/imath] are integers. and compared real no. With real |
2420499 | Solving [imath]\int\limits_0^\infty\frac{\ln x}{\left(x^2+4\right)^2}\text{d}x[/imath] using residues.
How can you evaluate the following integral using residues? [imath]\int\limits_0^\infty\frac{\ln x}{\left(x^2+4\right)^2}\text{d}x[/imath] | 117605 | Evaluating [imath]\int\limits_0^\infty \frac{\log x} {(1+x^2)^2} dx[/imath] with residue theory
I need a little help with this question, please! I have to evaluate the real convergent improper integrals using RESIDUE THEORY (vital that I use this), using the following contour: [imath]\int\limits_0^\infty \frac{\log x} {(1+x^2)^2} dx[/imath] Using this contour: [imath]R>1[/imath] and [imath]r<1[/imath] |
2419992 | Let [imath]d = \gcd(a,b)[/imath]. If [imath]a = da'[/imath] and [imath]b = db'[/imath], show that [imath]\gcd(a',b')=1[/imath].
Let [imath]d = \gcd(a,b)[/imath]. If [imath]a = da'[/imath] and [imath]b = db'[/imath], show that [imath]\gcd(a',b')=1[/imath]. So far I concluded that [imath]d[/imath] divides both [imath]a[/imath] and [imath]b[/imath], and their remainders are zero. I don't know what to do next, someone please help me. Thank you. | 645330 | GCD of the already GCD
Say [imath]a[/imath] and [imath]b[/imath] are integers. [imath]\gcd(a,b)[/imath] is then [imath]d[/imath]. Now if [imath]a[/imath] equals [imath]dm[/imath] for some integer [imath]m[/imath] and b equals [imath]dn[/imath] for some integer n, how come the gcd of this m and n is always 1? |
2420778 | Given [imath]x,y\in\mathbb{Q}[/imath] such that [imath]x, \exists r\in\mathbb{Q} : x[/imath]
Given two postive rational numbers [imath]x[/imath] and [imath]y[/imath] such that [imath]x<y[/imath], how do you prove that there exists a rational number [imath]r[/imath] such that [imath]x<r^2<y[/imath]. I am aware that since [imath]\mathbb{Q}[/imath] is dense in [imath]\mathbb{R}[/imath], and [imath]x[/imath] & [imath]y[/imath] are positive, there exists a rational number [imath]r[/imath] such that [imath]\sqrt{x}<r<\sqrt{y}[/imath]. But I would like a proof which gives such a rational number, while just staying within the context of the rational number system. | 236997 | How to prove the existence of [imath]b[/imath] in [imath]Q[/imath] such that [imath]a in Q?[/imath]
I would like to prove the existence of [imath]b \in \mathbb Q[/imath] such that [imath]a<b^2<c[/imath] for any given [imath]a,c \in \mathbb Q[/imath] with [imath]a,c>0[/imath] I want to use the statement above to prove a statement in a link I thought that '[imath]b[/imath]' must be exist. But, in my opinion, [imath]\sqrt{}[/imath] can't not be used because [imath]\sqrt{a}[/imath] or [imath]\sqrt{c}[/imath] may not exist in [imath]\mathbb Q[/imath] for some [imath]a[/imath] and [imath]c[/imath]. I couldn't find a clue to prove the statement before the real number is constructed from [imath]\mathbb Q[/imath]. Would you help me to prove that? Thanks all for replying and pointing out errors. |
2420662 | About groups [imath]G_1,G_2,H[/imath] such that [imath]G_1\times H \cong G_2\times H[/imath]
Are there any groups [imath]G_1,G_2,H[/imath] such that [imath]G_1\times H \cong G_2\times H[/imath], but [imath]G_1[/imath] is not isomorphic to [imath]G_2[/imath]? | 2165999 | Cancellation Law for External direct product
Prove or disprove the following assertion. Let [imath]G, H,[/imath] and [imath]K[/imath] be groups. If [imath]G\times K \cong H \times K[/imath], then [imath]G\cong H[/imath]. |
2420431 | Dot product formula proof
If two vectors [imath]\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}[/imath] and [imath]\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}[/imath], have angle [imath]\theta[/imath] between them then the dot product (scalar product) of [imath]\vec{A}[/imath] and [imath]\vec{B}[/imath] is [imath]\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta[/imath] [imath]\vec{A}\cdot\vec{B} =\left(A_x^2 +A_y^2+A_z^2\right)^{1/2}\left(B_x^2 +B_y^2 +B_z^2\right)^{1/2} \cos \theta [/imath] [imath]A_x B_x+ A_yB_y+A_zB_z=\left(A_x^2 +A_y^2+A_z^2\right)^{1/2}\left(B_x^2 +B_y^2 +B_z^2\right)^{1/2} \cos \theta [/imath] Can we mathematically derive the last equation from right to left or left to right? Is there any proof stating that the equation to be practically and mathematically true? | 2380217 | Why are the two dot product definitions equal?
I have an intuitive understanding of why [imath]a\dot{}b=|a||b|\cos{\theta}[/imath] geometrically. The projection of one vector onto another makes sense to me when explaining the origin of this geometric definition. What I don't understand is why [imath]a\dot{}b=a_xb_x + a_yb_y = |a||b|\cos{\theta}[/imath]. How does the algebraic version of the dot product connect to the geometric version? Can you derive the algebraic definition from the geometric? I read the answers to this question, but the proofs seem to depend on the actual algebraic definition to arrive at it. My main question is, why are the two definitions really equal? |
2368577 | how to solve the inequality [imath]\lvert w \sigma^{'}{(wa+b)}\rvert \gt 1[/imath]?
when i read the book of neural networks and deep learning at chapter 5 (the URL is,the related content of chapter five )i have encounter a problem to try to resolve the problem . when we training the deep neural network ,the vanishing gradient problem is often encountered. so we need to adjust the w(weight) parameter to fit the problem,but what's the scope of w.intuitive ,the larger w is ,the problem fit better. but the w is related with the [imath]\sigma'(z) = \sigma'(wa +b)[/imath],when w is larger,the value of [imath]\sigma'(z) = \sigma'(wa +b)[/imath] is smaller( the activity function is sigmod [imath]\sigma(x)=\frac{1}{1+e^{-x}}[/imath] and [imath]\sigma'(x) =\sigma(x)(1-\sigma(x))[/imath] ),and the product [imath]|w \sigma'(wa+b)|[/imath] is changing to bigger or smaller, we do not know so we need to calculate the scope of w,so that make sure with a larger w , the product also become larger ,only with that we can solve the vanishing gradient problem . the question's content is Consider the product[imath] |w \sigma'(wa+b)|[/imath]. Suppose [imath]|wσ′(wa+b)|≥1[/imath]. (1) Argue that this can only ever occur if [imath]|w|≥4[/imath]. (2) Supposing that [imath]|w|≥4[/imath], consider the set of input activations a for which [imath]|wσ′(wa+b)|≥1[/imath]. Show that the set of a satisfying that constraint can range over an interval no greater in width than \begin{eqnarray} \frac{2}{|w|} \ln\left( \frac{|w|(1+\sqrt{1-4/|w|})}{2}-1\right). \tag{123}\end{eqnarray} (3) Show numerically that the above expression bounding the width of the range is greatest at [imath]|w|≈6.9[/imath], where it takes a value [imath]≈0.45[/imath]. And so even given that everything lines up just perfectly, we still have a fairly narrow range of input activations which can avoid the vanishing gradient problem. Although it has related the deep learning ,but I know this is only a mathematical problems, the only need to do is solving equations and the inequation . due to my level of math, i can't solve this problem ,can somebody help me to solve the problem . thanks very much. and i have another problem can the inequation and the equation in this problem solved by MATLAB? | 1851727 | Find the interval over which this function is greater than 1
Consider the product [imath]\left|w\sigma'(wa+b)\right|[/imath], where [imath]\sigma(x) = \frac{1}{1+e^{-x}}[/imath]. Suppose that [imath]|w| \geq 4[/imath], and [imath]|w\sigma'(wa+b)|\geq 1[/imath]. Show that the set of [imath]a[/imath] satisfying that constraint can range over an interval no greater in width than [imath]\frac{2}{|w|}\ln\Bigg( \frac{|w|(1+\sqrt{1−4/|w|}}{2}−1\Bigg).[/imath] Not quite sure where to start, I tried to work with the original equation: [imath]e^{wa+b}/\left(e^{wa+b}+1\right)^2 \geq 1/|w| \implies wa+b - 2\ln\left(e^{wa+b}+1\right) \geq -\ln(w).[/imath] However, this doesn't seem to be going anywhere, and I'm just not sure how to go about finding the width of an [imath]\textit{interval}[/imath]. Appreciate any suggestions/solutions, thanks! |
2419814 | Integration of [imath]\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}[/imath] with the help of signal properties.
Find the integral [imath]\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}[/imath] where [imath]B[/imath] is a constant. This integration i tried by normal method that gives the result [imath]\frac{\pi B}{2\sqrt2}[/imath] but it goes so lengthy like this is there any signal properties that i use to solve it in simple steps? | 1511717 | How to compute [imath]\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.[/imath]
How to compute [imath]\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.[/imath] I'm interested in more ways of computing this integral. There is always the straight forward method to decompose into simple elements [imath]\dfrac{1}{(1 + t^{4})}[/imath] it works but it is tedious. If someone has a faster and clever method, I'm interested :) Update : My quesion is different than this question because all the solution that state there is talk about the the straight forward method to decompose into simple elements [imath]\dfrac{1}{(1 + t^{4})}[/imath] as i said before i'm not intersted in that way |
2421073 | How to prove [imath] a + b \sqrt[3]{2} + c\sqrt[3]{4} =0[/imath] iff [imath]a=b=c=0 [/imath]
From the extension of my previous question For [imath]a,b,c\in \mathbb{Q}[/imath] \begin{align} a + b \sqrt[3]{2} + c\sqrt[3]{4} =0 \quad \Leftrightarrow \quad a=b=c=0 \end{align} again also one direction is easy, but how about other direction? | 120489 | Given that [imath]a+b\sqrt[3]{2} +c\sqrt[3]{4} =0[/imath], where [imath]a,b,c[/imath] are integers. Show [imath]a=b=c=0[/imath]
Given that [imath]\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}[/imath], where [imath]a,b,c[/imath] are integers. Show [imath]a=b=c=0[/imath] Do I use modular arithmetic? |
2421056 | Show the set of all functions [imath]f : A \rightarrow \{0,1\}[/imath] is equinumerous with the power set of [imath]A.[/imath]
I am having some difficultly establishing a bijection from the set of functions [imath] f: A\rightarrow \{0,1\},[/imath] to the power set of [imath]A.[/imath] For the case where [imath]A[/imath] is finite, I found that [imath]2^{|A|}[/imath] is the total amount of functions from [imath]A[/imath], and that [imath]|P(A)| = 2^{|A|}[/imath] as well. For the cases where [imath]A[/imath] is denumerable or uncountable I am unsure how to proceed. I know by Cantor's theorem that [imath]|A| < |P(A)|[/imath]. | 486830 | Prove that there is a bijection between the set of all subsets of [imath]X[/imath], [imath]P(X)[/imath], and the set of functions from [imath]X[/imath] to [imath]\{0,1\}[/imath].
Given any set [imath]X[/imath], let [imath]P(X)[/imath] be the set of all subsets of [imath]X[/imath], and let [imath]\{0,1\}^X[/imath] be the set of all functions [imath]X \rightarrow \{0,1\}[/imath]. Construct a bijection (and its inverse) between P(X) and [imath]\{0,1\}^X[/imath]. Let us define [imath]f: P(X) \rightarrow \{0,1\}^X[/imath] by [imath]f(\alpha) = g_{\alpha}[/imath] where [imath]\alpha \in P(X)[/imath] and [imath]g_{\alpha} : X \rightarrow \{0,1\}[/imath] is defined by [imath]g_{\alpha}(x) =\begin{cases} 1 & x \in \alpha \\ 0 & x \in X-\alpha \end{cases}[/imath] We need to check if this function is well-defined. Let [imath]\alpha, \beta \in P(X)[/imath]. If [imath]\alpha=\beta[/imath], then [imath]\alpha[/imath] and [imath]\beta[/imath] correspond to the same set, and so correspond to the same function since both [imath]g_{\alpha}[/imath] and [imath]g_{\beta}[/imath] send that set to 1 and the ([imath]X-[/imath]that set) to 0. If we have a function [imath]g_{\alpha} \in \{0,1\}[/imath], then [imath]\alpha[/imath] is a set in [imath]X[/imath] that is sent to 1. But since [imath]P(X)[/imath] is the set of all sets of [imath]X[/imath], we must have [imath]\alpha \in P(X)[/imath]. So the function is surjective. Finally, if [imath]g_{\alpha}=g_{\beta}[/imath] then both [imath]g_{\alpha}[/imath] and [imath]g_{\beta}[/imath] send the set [imath]\alpha[/imath] or [imath]\beta[/imath] to 1. So [imath]\alpha[/imath] and [imath]\beta[/imath] must be same set. In other words, the function is injective; and hence bijective. The inverse function is [imath]f^{-1}: \{0,1\}^X \rightarrow P(X)[/imath] where [imath]f^{-1}(g_{\alpha}) = \alpha[/imath]. Another way to prove this is to first construct the inverse function, show that it is well defined, and then show that [imath]ff^{-1}[/imath] and [imath]f^{-1}f[/imath] are the identity functions, right? So if we let [imath]g_{\alpha}=g_{\beta}[/imath], then [imath]\alpha=\beta[/imath], because otherwise [imath]g_{\alpha} (\beta)=0[/imath] and [imath]g_{\beta} (\beta)=1[/imath]; a contradiction. Now we have [imath]f \circ f^{-1} (g_{\alpha}) = f(\alpha) = g_{\alpha}[/imath] and [imath]f^{-1} \circ f(\alpha) = f^{-1}(g_{\alpha}) = \alpha[/imath] , and so [imath]f[/imath] and [imath]f^{-1}[/imath] must be bijective. |
2421293 | Is [imath]{\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=4[/imath] correct?
My teacher asked me this question. But I think this is wrong. Anyone found this, before me? I do not know. Anyway, Is my solution correct? [imath]{x^{x^{x^{x^x}}}}^{...}=2[/imath] [imath]x^2=2[/imath] [imath]x=\sqrt 2[/imath] [imath]{\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=2[/imath] Now, Let, [imath]{x^{x^{x^{x^x}}}}^{...}=4[/imath] [imath]x^4=4[/imath] [imath]x=\sqrt2[/imath] [imath]{\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=4[/imath] Contradiction! | 2444941 | Two solutions to one number.
I met a question which said : Find the value of [imath]\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{.^{.^{.^{.^{.}}}}}}}}[/imath] Now to start I declared [imath]y=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{.^{.^{.^{.^{.}}}}}}}}[/imath] Now this implies that [imath]y=\sqrt2^y[/imath] Now solving this equation we get [imath]y=2,4[/imath] but then how can a single number have two values. So where am I going wrong? Thank you :) |
2420908 | Number of paths in a complete graph
I'm aware of the previous question which has been answered : Number of unique paths in a complete graph with n verticies Today , I found another formula for the number of paths between two vertices in a complete graph [imath] \space \ \lfloor e \times (p-2) !\rfloor [/imath] such that [imath]p[/imath] is number of vertices in complete graph . I don't know how [imath]e[/imath] and floor function have appeared in that . In the other words , I'm looking for the proof . | 2406920 | Total number paths between two nodes in a complete graph
In a complete graph total number of paths between two nodes is equal to: [imath]\lfloor(P-2)!e\rfloor[/imath] This formula doesn't make sense for me at all, specially I don't know how [imath]{e}[/imath] plays a role in this formula. could anyone prove that simply with enough explanation? |
2421083 | math induction help?
I've been stuck on this question for a good while now, and I think I need some pointers. [imath](1+x)^n = nC_0 + nC_1x + nC_2x^2 ... + nC_{n-1}x^{n-1} + nCnx^n[/imath] where n greater than or equal to 1 I'm at the point where I replace n with k+1 and solve, but no matter what I do I cannot get the sides of the equation to look the same. | 2205908 | Solving binomial theorem via induction
I'm trying to prove binomial theorem by induction, but I'm a little stuck. I would look at online resources as this problem has been done many times, but the version I am trying to prove the binomial theorem in a different form. [imath](1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k[/imath] I'm mostly confused as to how I can make the left side be equivalent to a summation, any help is appreciated. Try to hint me along! |
2421691 | prove that a graph with degree [imath]\frac{n-1}{2}[/imath] has a diameter of at most 2
I try to solve the following: let [imath]G[/imath] be a simple graph with minimum degree of [imath]\frac{(n-1)}{2}[/imath] with [imath]n[/imath] vertices. prove that [imath]G[/imath] is connected and that the diameter of [imath]G[/imath] is at most [imath]2[/imath] I Know how to prove that the graph is connected - assume that we take some connected component K, for each [imath]v\in k[/imath] we have that [imath]deg(v)=\frac{n-1}{2}[/imath] so that means that we have at least [imath]\frac{n-1}{2}+1=\frac{n+1}{2}[/imath] vertices which implies that if we have more then one connected component we have more than [imath]n[/imath] vertices - a contradiction But how to prove the diameter part? my way of thinking was to compute how many vertices can I reach after "two moves" trail but I got some not conclusive result: [imath]\frac{n-1}{2} + \frac{n-1}{2}\cdot ( \frac{n-1}{2} -1)[/imath] [imath]\Downarrow[/imath] [imath]=\frac{(n-1)^2}{4}[/imath] which is not bigger than [imath]n-1[/imath] for all [imath]n[/imath] I know how to work around this but this seems "lame" I am sure there is some other elegant way. | 1980006 | Determine the diameter of a graph of order [imath]n \geq 3[/imath] where [imath]\delta (G) \geq \frac{n-1}{2}[/imath].
Let [imath]G[/imath] be a graph of order [imath]n \geq 3[/imath] such that [imath]\delta (G) \geq \frac{n-1}{2}.[/imath] Determine the diameter of [imath]G[/imath]. So [imath]\delta (G)[/imath] is the minimum degree of a vertex in the graph [imath]G[/imath]. How can I approach such a question? I know that the diameter of a graph is the "largest shortest-path" between any pair of vertices in a graph, or in other words, the maximum distance between all possible pairs of vertices in [imath]G[/imath]. Also, do I have to make use of the maximum degree of [imath]G[/imath] to solve this problem? |
2422049 | Subgroup of a given index
The following problem is from the book "Finite Group Theory" by Martin Isaacs. (3.B.12) Let [imath]G[/imath] be solvable and assume that [imath]\Phi(G) = 1[/imath]. Let [imath]M[/imath] be a maximal subgroup of [imath]G[/imath] and suppose that [imath]H \leq M[/imath]. Show that [imath]G[/imath] has a subgroup with index equal to [imath][M:H][/imath] I've been trying to solve this problem for quite some time, but all the time I'm coming up empty. As the lecture is about complements, especially the Schur-Zassenhaus Theorem I tried to prove that [imath]M[/imath] has a normal complemented [imath]L[/imath] in [imath]G[/imath], as then [imath][G:HL] = [M:H][/imath] and [imath]HL[/imath] is the wanted subgroup. Unfortunatelly I can't seem to prove this one. I took a look at two cases. In the first assume that there is a minimal normal subgroup [imath]L[/imath] of [imath]G[/imath], s.t. [imath]L \not \leq M[/imath]. Then by the maximality of [imath]M[/imath] we get [imath]G = ML[/imath]. Also we have [imath]L \cap M \unlhd M[/imath] and [imath]L \cap M \unlhd L[/imath] (as [imath]L[/imath] is elementary abelian, as [imath]G[/imath] is solvable), which gives us that [imath]L \cap M \unlhd G[/imath]. As [imath]L \cap M \not = L[/imath] from the minimality of [imath]L[/imath] we get [imath]L \cap M = 1[/imath], so [imath]L[/imath] is a complement of [imath]M[/imath]. But I can't solve the case when all minimal normal subgroups of [imath]G[/imath] are in [imath]M[/imath]. I tried using induction to solve this question, but the problem arrises when [imath]H[/imath] doesn't contain any normal non-trivial subgroup of [imath]G[/imath], as then I can't reduce the problem to a factor group from smaller order. Anyway I beleive the problem has to do something with a minimal normal group, as if for some [imath]T \leq G, [G:T] = [M:H][/imath], then [imath]|T| = p^k \cdot |H|[/imath]. But I can't prove the existence of a minimal normal group of order [imath]p^k[/imath], nor I can prove that it intersects trivially with [imath]H[/imath]. I notice that I'm not using the condition [imath]\Phi(G) = 1[/imath], but I can't see it how. The only way I managed to achieve something by using it is to prove that every minimal normal subgroup of [imath]G[/imath] has a complement in [imath]G[/imath], but unfortunatelly this doesn't prove that every maximal subgroup is complemented. | 33274 | A problem from Isaacs' FGT
This is problem 3B.12. from the afore mentioned book. It asks to show that, if [imath]G[/imath] is a solvable Frattini-free group (that is, [imath]Φ(G)=1[/imath]) and [imath]H[/imath] is a subgroup of some maximal subgroup [imath]M[/imath] of [imath]G[/imath], then [imath]G[/imath] contains a subgroup of index [imath]|M:H|[/imath]. It is easy to see that, if [imath]N \unlhd G[/imath] and [imath]M\cap N=1[/imath], then [imath]G=MN[/imath], [imath]M\cong G/N [/imath] and [imath]H\cong K/N [/imath] for some [imath] K \leqslant G [/imath], where [imath]K[/imath] contains [imath]N[/imath]. Then [imath]|G:K|=|G/N:K/N|=|M:H| [/imath] and we are done. However, the hypothesis will not necessarily hold. But, if [imath]N[/imath] is minimal among normal subgroups of [imath]G[/imath] (in the sense of containment), then [imath]N[/imath] is elementary abelian, hence nilpotent, which implies that [imath]M \cap N \lneq N_G( M \cap N ) \cap N [/imath]. Thus, [imath]M \lneq N_G( M \cap N )[/imath] and by the maximality of [imath]M[/imath] we deduce that [imath]M \cap N \unlhd G[/imath]. So, for this particular [imath]M[/imath], if it is not true that there exists some normal subgroup of [imath]G[/imath] with which it intersects trivially, then [imath]M \cap N = N[/imath] by the minimality of [imath]N[/imath]. Hence [imath]M[/imath] contains every minimal subgroup of [imath]G[/imath]. How should I proceed from here? |
2421778 | Sum of areas of triangles
Let [imath]ABC[/imath] an equilateral triangle and [imath]O[/imath] a point in interior of [imath]ABC[/imath]. Consider [imath]M, N, P[/imath] the projections on the sides [imath]AB, BC, AC[/imath]. Then the sum of areas of triangles [imath] AOM, BON, COP[/imath] is the same like the sum of areas of triangles [imath]BOM, CON, AOP[/imath]. How to start? I try to find an elementary solution. | 651119 | Inside an equilateral triangle [imath]ABC[/imath],an arbitrary point [imath]P[/imath] is taken from which the perpendiculars [imath]PD,PE[/imath] and [imath]PF[/imath] are dropped onto the sides...
Inside an equilateral triangle [imath]ABC[/imath],an arbitrary point [imath]P[/imath] is taken from which the perpendiculars [imath]PD,PE[/imath] and [imath]PF[/imath] are dropped onto the sides [imath]BC,CA[/imath] and [imath]AB[/imath],respectively.Show that the ratio [imath]\dfrac{PD+PE+PF}{BD+CE+AF}[/imath] does not depend upon the choice of the point [imath]P[/imath] and find its value. I have no idea how to proceed. I could not do anything beyond calculating the area of [imath]PDB,PCE,PFA[/imath] but here I get nothing. Please help! |
2422698 | Find [imath]\displaystyle{\lim_{n \to \infty}} \sqrt[n]{n!}[/imath] without Stirling's approximation.
Okay I've seen many posts as this one, but in every post it uses Stirling(I can't use it because I've never learnt it), or weird equalities with logs (I also never learnt this) or other methods I'm not allowed to use in my exam. I have to prove that [imath]\lim n!^{(1/n)}[/imath] = [imath]\infty[/imath], only by using D'Alembert, [imath]n^{th}[/imath] root test, Stolz, squeeze theorem or by playing with inequalities etc What I thought is that I could prove that [imath]\lim \frac{1}{n!^{(1/n)}} = 0[/imath] (again only using those basic theorems) and that would mean that [imath]\lim n!^{(1/n)} = \infty[/imath] but I'm not sure if that's correct. Please don't mark as duplicate! I need an answer and I haven't found any post with an answer I can use!! | 514388 | Find [imath]\lim_{n \to \infty} \sqrt[n]{n!}[/imath].
I am playing around with the root/ratio test to practice with series. I just showed that [imath]\sum \frac{1}{n!}[/imath] converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered [imath]\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}[/imath]. I have made the claim that [imath]\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.[/imath] I have begun the proof and I think it is on the right track but some verification would be nice. Proof: We can see that for n=1, this obviously holds, [imath]\left(1!\right)^1\geq 1.[/imath] Now suppose that this is the case for some [imath]n=k[/imath], then we have [imath]\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}[/imath]. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks! |
2423091 | For the field [imath]F=\{a+b\sqrt{3}\}[/imath], find [imath]d[/imath] so that [imath]b\cdot d=1[/imath], where [imath]b,d\in F[/imath]
For the field [imath]F=\{a+b\sqrt{3}\}[/imath], find [imath]d[/imath] so that [imath]b\cdot d=1[/imath], where [imath]b,d\in F[/imath] Let [imath]b=a_1+b_1\sqrt{3}[/imath] Let [imath]d=a_2+b_2\sqrt{3}[/imath] We need to find [imath]a_2,b_2[/imath] such that [imath]b\cdot d=1[/imath] Doing the math, we get that [imath]b\cdot d=(a_1a_2+3b_1b_2)+(a_1b_2+a_2b_1)\sqrt{3}[/imath] But what from here? How do I decide that I want [imath](a_1a_2+3b_1b_2)[/imath] to equal [imath]0[/imath], and so I want [imath](a_1b_2+a_2b_1)[/imath] to be [imath]1[/imath]. Or I want the first part of the expression to be [imath]1[/imath], and the next part to be [imath]0[/imath]. How do I decide? | 2420797 | Show that the set [imath]\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}[/imath] is a field
Show that the set [imath]\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}[/imath] is a field I am having a little trouble proving that for each non-zero [imath]b\in\mathbb{Q}(\sqrt{p})[/imath], there is [imath]d\in\mathbb{Q}[/imath] such that [imath]b\cdot d=1[/imath]. My attempt: Let [imath]b=a_1+b_1\sqrt{p}[/imath] Let [imath]d=a_2+b_2\sqrt{p}[/imath] We want [imath]b\cdot d=(a_1+b_1\sqrt{p})(a_2+b_2\sqrt{p})[/imath] [imath]=a_1a_2+a_1b_2\sqrt{p}+a_2b_1\sqrt{p}+b_1b_2p[/imath] [imath]=(a_1a_2+b_1b_2p)+(a_1b_2+a_2b_1)\sqrt{p}[/imath] [imath]=1[/imath] Am I following the right approach? What should be my next step |
2423194 | Differential Equation Question for a separation of variables problem
I'm struggling with the differential equation [imath]yln(x)dx/dy = ((y+1)/x)^2[/imath] Can anyone help me out? | 2423064 | Differential Equation Separation of Variables
How do I separate the variables (as I assume will work for this equation) in the following equation: [imath]yln(x) \frac {dx}{dy} = (\frac {y+1}{x})^{2}[/imath] I am struggling with getting [imath]x[/imath] out of the term [imath] (\frac {y+1}{x})^{2}[/imath]. |
2423229 | If [imath]X[/imath] is uncountable, then the set of bounded injective functions: [imath]f: X \rightarrow \mathbb{R}[/imath] has empty interior.
If [imath]X[/imath] is uncountable, then the set of bounded injective functions: [imath]f: X \rightarrow \mathbb{R}[/imath] has empty interior. Attempt: In order to prove that the interior is empty, we must show that for any [imath]f[/imath] in the set and for any ball centered on [imath]f[/imath]: [imath]B(f;\epsilon)[/imath], one can find a function [imath]g\in B(f;\epsilon)[/imath] such that [imath]g[/imath] is not injective. Take any [imath]f[/imath] in the set and any [imath]\epsilon>0[/imath]. We note that the image of [imath]f[/imath], [imath]f(X)[/imath], is an uncountable subset of [imath]\mathbb{R}[/imath]. Therefore, it contains at leats one of its accumulation points, say [imath]f(x_0[/imath]), where [imath]x_0\in X[/imath]. Hence it follows that [imath]B(f(x_0);\epsilon/2)\cap f(X) \neq \emptyset[/imath]. We take [imath]f(y_0)\in B(f(x_0);\epsilon/2)\cap f(X)[/imath] such that [imath]f(x_0)<f(y_0)[/imath], where [imath]y_0\in X.[/imath] Define the following function: [imath]g:X\rightarrow \mathbb{R}[/imath] by letting: [imath]g(y_0) = f(x_0)[/imath] and [imath]g(x)=f(x)[/imath] elsewhere. Clearly [imath]g[/imath] is not injective and: [imath]\mbox{sup}_{x\in X}|f(x)-g(x)| = |f(y_0)-g(y_0)|=|f(y_0)-f(x_0)|<\epsilon,[/imath] since we chose [imath]f(y_0)\in B(f(x_0);\epsilon/2)[/imath]. This shows that [imath]g\in B(f;\epsilon)[/imath], [imath]g[/imath] being not injective. Is this fine? Any different idea? | 2228613 | Let [imath]X[/imath] be uncountable. Then the set of injective functions [imath]f\colon X \to \mathbb{R}[/imath] has empty interior in [imath]\mathscr{C}(X;\mathbb{R})[/imath]
Here I'm working with the metric: [imath]d(f,g)=\sup_{x \in X}|f(x)-g(x)|.[/imath] Let's take [imath]h[/imath] in the set and let [imath]\epsilon>0[/imath] be given. I need to prove that there exists another bounded, not injective function [imath]j[/imath] such that [imath]d(j,h)<\epsilon[/imath]. I would be done if there where two images of [imath]h[/imath] to a distance less than [imath]2\epsilon[/imath]. Cause in such case, let them be [imath]h(x_0)[/imath], [imath]h(y_0)[/imath] and suppose that [imath]h(x_0)>h(y_0)[/imath]. Then setting [imath]j(x_0)=h(x_0)-\dfrac{h(x_0)-h(y_0)}{2}[/imath] , [imath]j(y_0)=h(y_0) + \dfrac{h(x_0)-h(y_0)}{2}[/imath] and [imath]j(x)=h(x)[/imath] otherwise, we would have the required. Suppose, by way of contradiction, that no two images are that near, i.e., in each open interval of lenght [imath]2\epsilon[/imath], there is at most one image of [imath]h[/imath]. Since [imath]h[/imath] is bounded, its image can be covered by a finite number of intervals of lenght [imath]2\epsilon[/imath], which implies that there are only finite images of [imath]h[/imath]. Finally, like [imath]h[/imath] was injective on an infinite set [imath]X[/imath], this cannot be. I have my doubts because I didn't use that [imath]X[/imath] was uncountable, just infinite. Is this proof ok? |
2423520 | Integration by Substitution: substituting variable with complex functions
I'm learning integration by the method of substitution. While doing sums, I encountered a case in which I got this expression in the denominator: [imath]\sqrt {a^2 + x^2 } [/imath] Now, I wanted to substitute [imath]x[/imath] as [imath]a i \sin \theta[/imath] where [imath]i = \sqrt {-1} [/imath]. My idea was to take [imath]a[/imath] outside the root, and obtain [imath]\cos \theta [/imath] from the root. Although I've reached the answer, I couldn't find any complex function substitution example in my book, which makes me wonder whether my substitution is correct. Can I substitute a variable with a complex function in integration? | 163946 | Are Complex Substitutions Legal in Integration?
This question has been irritating me for awhile so I thought I'd ask here. Are complex substitutions in integration okay? Can the following substitution used to evaluate the Fresnel integrals: [imath]\int_{0}^{\infty} \sin x^2\, dx=\operatorname {Im}\left( \int_0^\infty\cos x^2\, dx + i\int_0^\infty\sin x^2\, dx\right)=\operatorname {Im}\left(\int_0^\infty \exp(ix^2)\, dx\right)[/imath] Letting [imath]ix^2=-z^2 \implies x=\pm\sqrt{iz^2}=\pm \sqrt{i}z \implies z=\pm \sqrt{-i} x \implies dx = \pm\sqrt{i}\, dz[/imath] Thus the integral becomes [imath]\operatorname {Im}\left(\pm \sqrt{i}\int_0^{\pm\sqrt{-i}\infty} \exp(z^2)\, dz\right)[/imath] This step requires some justification, and I am hoping someone can help me justify this step as well: [imath]\pm \sqrt{i}\int_0^{\pm\sqrt{-i}\infty} \exp(z^2)\, dz=\pm\sqrt{i}\int^\infty_0\exp(z^2)\, dz=\pm\sqrt{i}\left(\frac{\sqrt{\pi}}{2}\right)[/imath] Thus [imath]\operatorname {Im}\left(\int_0^\infty \exp(ix^2)\, dx\right)=\operatorname {Im}\left(\pm\frac{\sqrt{i\pi}}{2}\right)=\operatorname {Im}\left(\pm\frac{(1+i)\sqrt{\pi}}{2\sqrt{2}}\right)=\pm\frac{1}{2}\sqrt{\frac{\pi}{2}}[/imath] We find that the correct answer is the positive part (simply prove the integral is positive, perhaps by showing the integral can be written as an alternating sum of integrals). Can someone help justify this substitution? Is this legal? |
2422356 | Prove that this bilinear form is symmetric or skew-symmetric
There's this bilinear form [imath]f[/imath] such that [imath]\forall x, y \in V \ \ f(x, y) = 0 \implies f(y, x) = 0[/imath]. What I need to prove is that this form is symmetric or skew-symmetric. I don't understand how to approach this. I've tried adding something like [imath]f(x - x, y)[/imath] (which equals to [imath]0[/imath]) and such to [imath]f(x, y)[/imath], and then applying the property above to the result, but didn't get anything. I've also tried looking at [imath]f(x, y) - f(y, x)[/imath] and something like [imath]f(x + y - y, y)[/imath], hoping I'd get some revelation if I try to transform these expressions in various ways, but to no avail. | 656213 | Bilinear form with symmetric "perpendicular" relation is either symmetric or skew-symmetric
Let [imath]b[/imath] be a bilinear form on a finite-dimension vector space [imath]V[/imath] (over a field with char [imath]\neq[/imath] 2) such that for each [imath]x,y\in V[/imath] one has [imath]b(x,y)=0\Leftrightarrow b(y,x)=0[/imath]. Prove that [imath]b[/imath] is symmetric or skew-symmetric. The condition is equal to this: for every vector [imath]x[/imath] co-vectors [imath]b(\cdot,x)[/imath] and [imath]b(x,\cdot)[/imath] have the same kernels, so for some non-zero constant [imath]c:b(\cdot,x)=cb(x,\cdot)[/imath]. But I didn't manage to prove that [imath]c[/imath] is equal to [imath]1[/imath] or [imath]-1[/imath] for every [imath]x[/imath]. |
2424084 | Prove by induction that for [imath]n\geq1[/imath], [imath](a_1+a_2+...+a_n)/n \geq (a_1*a_2*...*a_n)^{1/n}[/imath]
Prove by induction that for [imath]n\geq1[/imath], [imath](a_1+a_2+...+a_n)/n \geq (a_1*a_2*...*a_n)^{1/n}.[/imath] I proved the base case, not sure how to do the inductive step. The values of the sequence are greater than or equal to zero. | 2376862 | Proving AM-GM with induction
I am trying to prove AM-GM with the following steps: Prove that AM-GM holds for two variables Prove that if AM-GM holds for [imath]k[/imath] variables then it holds for [imath]2k[/imath] variables Prove that if AM-GM holds for [imath]k[/imath] variables then it holds for [imath]k-1[/imath] variables If I prove all of these, I will have proved AM-GM. I have proved (1) and (2), but I am still struggling with (3). Proof for (1) By the Trivial Inequality, we know that the first statement is true. Therefore \begin{align} (a-b)^2 \ge 0 \implies\\ a^2-2ab+b^2 \ge 0 \implies \\ a^2+2ab+b^2 \ge 4ab \implies\\ a+b \ge 2\sqrt{ab} \implies\\ \frac{a+b}{2} \ge \sqrt{ab}\hspace{3mm}\blacksquare\\ \end{align} Proof for (2) Let our numbers be [imath]a_1, a_2, a_3, \ldots, a_{2k}[/imath]. Denote [imath]x = \frac{\displaystyle\sum_{i=1}^{k}a_i}{k}, y = \frac{\displaystyle\sum_{i=k+1}^{2k}a_i}{k}[/imath] We know that by AM-GM for 2 variables, that [imath]\frac{x+y}{2}\ge \sqrt{xy}[/imath] By our definition of [imath]x, y[/imath] this means that [imath]\frac{x+y}{2} = \frac{\frac{\displaystyle\sum_{i=1}^{2k}a_i}{k}}{2}=\frac{a_1+a_2+a_3+\ldots+a_{2k}}{2k}[/imath] Again by definition, [imath]\sqrt{xy} = \sqrt{\left(\frac{x_1+x_2+\ldots+x_k}{k}\right)\left(\frac{x_{k+1}+x_{k+2}+\ldots+x_{2k}}{k}\right)}[/imath] By AM-GM for [imath]k[/imath] variables, [imath]\sqrt{\left(\frac{x_1+x_2+\ldots+x_k}{k}\right)\left(\frac{x_{k+1}+x_{k+2}+\ldots+x_{2k}}{k}\right)} \ge \sqrt{\sqrt[k]{x_1x_2\ldots x_k}\sqrt[k]{x_{k+1}x_{k+2}\ldots x_{2k}}}[/imath] This is simply equal to [imath]\sqrt[2k]{x_1x_2x_3\ldots x_{2k}}\hspace{3mm}\blacksquare[/imath] However, I have no idea how to proceed on (3). I tried using the definition of AM and GM but did not come up with anything useful. Feel free to point out any errors or flaws on my proofs for (1) and (2). |
2424295 | Do we need the Axiom of Choice to show that [imath](\mathbb{R}, +)[/imath] and [imath](\mathbb{C}, +)[/imath] are group isomorphic?
Under the axiom of choice, [imath]\mathbb{R}[/imath] and [imath]\mathbb{C}[/imath] are both [imath]2^{\aleph_0}[/imath]-dimensional vector spaces over [imath]\mathbb{Q}[/imath]. Because they have the same dimension, they are isomorphic as vector spaces. That vector space isomorphism is also a group isomorphism between them. The proof is clearly non-constructive. Is there a way to explicity construct such an isomorphism, or do we need the axiom of choice? | 932795 | What is needed to make Euclidean spaces isomorphic as groups?
Consider the abelian groups [imath]G_n=(\mathbb R^n,+)[/imath] for [imath]n\geq1[/imath]. Claim: For any [imath]n[/imath] and [imath]m[/imath] the groups [imath]G_n[/imath] and [imath]G_m[/imath] are isomorphic. This claim is true if one assumes the axiom of choice, and I have sketched a proof below. But this claim seems much weaker than the axiom of choice, which brings me to the following questions: Are there any milder extensions of ZF that make the claim true? If the claim can be proven with some other axioms, what does the proof look like? Are there any known axioms that make the claim false? Does the claim have any interesting (set-theoretical) corollaries? This claim was also considered in this earlier post, but from a different point of view. A sketchy proof of the claim: [imath]\newcommand{\Q}{\mathbb Q}[/imath] The groups [imath]G_n[/imath] are naturally vector spaces over [imath]\Q[/imath]. It is easy to check that the claim is equivalent with [imath]G_n[/imath] and [imath]G_m[/imath] being isomorphic as vector spaces over [imath]\Q[/imath]. All vector spaces have a Hamel basis; let the basis of [imath]G_n[/imath] be [imath]B_n[/imath]. Thus [imath]G_n=\Q^{(B_n)}[/imath] (brackets meaning only finitely many nonzero components). But [imath]|\Q^{(B_n)}|=|B_n|[/imath] since the basis is infinite, so [imath]|B_n|=|G_n|=|\mathbb R|[/imath]. There is a bijection between the bases of the vector spaces [imath]G_n[/imath] and [imath]G_m[/imath] since they have the same cardinality. This gives a linear bijection between them. |
2423351 | Proof verification: elementary Number Theory Proof of [imath]am+bn = 1[/imath]
Can anyone find a hole in my proof or tell me how to make it stronger or simpler? The "lemma above" referenced in the fourth paragraph is a small but solid proof showing that [imath]r[/imath] is strictly less than [imath]b[/imath] in the division algorithm if it's written [imath]a=bq+r[/imath]. [imath]\mathbf {Theorem}[/imath] If [imath]m[/imath] and [imath]n[/imath] are positive integers that are relatively prime, then there are integers [imath]a[/imath] and [imath]b[/imath] such that [imath]am+bn=1[/imath]. [imath]\mathbf {Proof}[/imath] Let [imath]D = \{d \in \Bbb Z : am+bn>0\}[/imath]. Then there is a least element [imath]l[/imath] in D. Since [imath]m[/imath] and [imath]n[/imath] are relatively prime, we need only show that [imath]l[/imath] divides [imath]m[/imath] and [imath]n[/imath]. Suppose [imath]l[/imath] does not divide [imath]m[/imath]. Then there are integers [imath]q[/imath] and [imath]r[/imath] such that [imath]m=ql+r[/imath]. Rewritten, we have that [imath]r=ql-m=q(am+bn)-m=qam-m+qbn=m(qa-1)+n(qb)[/imath]. Thus, [imath]r \in D[/imath]. By the lemma above, [imath]r<l[/imath], contradicting that [imath]l[/imath] is the least element of [imath]D[/imath]. We conclude then that [imath]l[/imath] divides [imath]m[/imath]. Switching [imath]m[/imath] for [imath]n[/imath] gives the same result for [imath]n[/imath]. Since [imath]m[/imath] and [imath]n[/imath] are relatively prime, [imath]l=1[/imath] and it follows that there are integers [imath]a[/imath] and [imath]b[/imath] such that [imath]am+bn=1[/imath]. | 2384326 | Proving if [imath]d_0[/imath] is the smallest positive integer in [imath]S[/imath] then [imath]d_0 = \gcd(a,b)[/imath]
I would appreciate hints to this. I've done part (a) but am unconfident. Wondering how I could approach part (b) Question's comment -- The aim of this question is to use the Division Algorithm and the definition of greatest common divisor (gcd) to show that [imath]d_0 = \gcd(a,b)[/imath]. Question: Let [imath]a,b[/imath] be integers, not both zero, let [imath]S[/imath] be the set of integers defined by [imath]S = \{ ax + by | x,y \in \mathbb{Z}\}[/imath] and let [imath]d_0[/imath] be the smallest positive integer in the set [imath]S[/imath]. Prove: a. If [imath]s\in S[/imath], then [imath]d_0[/imath] is a divisor of [imath]s[/imath]. b. [imath]d_0[/imath] is a divisor of both [imath]a[/imath] and [imath]b[/imath]. c. If [imath]d[/imath] is a divisor of both [imath]a[/imath] and [imath]b[/imath], then [imath]d[/imath] is a divisor of [imath]d_0[/imath]. d. [imath]d_0 =\gcd(a,b),[/imath] and hence there exists integers [imath]x,y[/imath] such that [imath]ax + by = \gcd(a,b)[/imath]. Attempt: Note I don't expect full (or any hints to the problem(s) as there are quite a bit, but I would just appreciate hints. a. Let [imath]s \in S[/imath]. Then [imath]s = d_0 q + r[/imath] for some [imath]q,r \in \mathbb{Z}[/imath]. By definition, [imath]d_0 = ax_0 + by_0[/imath] for some [imath]x_0,y_0 \in \mathbb{Z}[/imath]. [imath]s = ax_s + by_s[/imath] for some [imath]x_s,y_s \in \mathbb{Z}[/imath]. Hence, [imath]ax_s + by_s = q(ax_0 + by_0) + r\\ a(x_s - qx_0) + b(y_s - qy_0) = r.[/imath] Define [imath]x_r:= x_s - qx_0 \in \mathbb{Z}[/imath] and [imath]y_r:= y_s - qy_0 \in \mathbb{Z}[/imath]. Then this shows that [imath]r \in S[/imath] which implies [imath]d_0 | s[/imath]. |
2419004 | [imath](\mathbb{R}^{3},\times)[/imath] is a Lie algebra that does not have Lie subalgebras of dimension 2.
Could anyone give me a suggestion to start solving this problem? Proof that [imath](\mathbb{R}^{3},\times)[/imath] is a Lie algebra that does not have Lie subalgebras of dimension 2. | 2198401 | 2-dimensional Lie subgroups of [imath]SO(3)[/imath]
It is well know that [imath]SO(3)[/imath] is a 3-dimensional Lie group. My question is [imath]SO(3)[/imath] don't has 2-dimensional Lie subgroups? |
2424670 | Show [imath] \lim_{t \to \infty} -t^{x}e^{-t} = 0 [/imath]
I have this limit (derived from an improper integral in the larger problem) that WolframAlpha says equals zero. But it doesn't have the steps to get there. How indeed do I show this instead of waving my hands that we can apply l'Hospital's rule to infinity and the denominator will eventually win and the limit will approach zero. Check form ([imath]\infty * 0[/imath]): [imath]\lim_{t \to \infty} -t^{x}e^{-t}[/imath] Rearrange into [imath]\frac\infty\infty[/imath]: [imath]= \lim_{t \to \infty} \frac{-t^x}{e^t}[/imath] Apply l'Hospital's Rule: [imath]= \lim_{t \to \infty} \frac{-xt^{x-1}}{e^t}[/imath] I don't see a clear way of showing this goes to zero, but zero is the correct answer. | 2423848 | Show that [imath]\lim_{z \rightarrow \infty} -z^xe^{-z} = 0[/imath]
Let [imath]x>0[/imath] be a constant. Show that [imath]\lim_{z \rightarrow \infty} -z^xe^{-z} = 0[/imath]. Do I apply L'hopital's rule here? It doesn't seem to help, for example, after one round of differentiating the numerator and denominator, I get [imath]\frac{d}{dz} z^x = xz^{x-1}[/imath] and [imath]\frac{d}{dz}e^z = e^z[/imath]. I am unsure on how to proceed. |
2424959 | How to derive the binomial theorem?
I saw the binomial theorem in some 11th grade book in my library. It was proved by induction. But there was no derivation of it. I do not like to memorize it. So I'm hoping someone can show me a derivation for it and give hint for another way to derive it. It would be great if you try to use high school mathematical tools if possible, so I can understand it. [imath] (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}. [/imath] | 851275 | Proving the Binomial theorem
I am trying to work through the proof I understand it until on this website http://www.mcs.sdsmt.edu/ecorwin/cs251/bin_thm/bin_thm.html : Pascal's identity does not apply for [imath]k = 0[/imath] or for [imath]k = n[/imath]. In our sum, this means we need to split out the [imath]j = 0[/imath] and [imath]j = k + 1[/imath] terms before applying Pascal's identity. Why does k=o and k=n correspond to [imath]j=0[/imath] and [imath]j=k+1[/imath]? Now we have the problem of matching the two sums from the left side with the two from the right side. First, notice that the out front on the right side can be thought of as the [imath]j = 0[/imath] term in the first sum. Next, the y [imath]k+1[/imath] out front on the right side can be thought of as the [imath]j = k + 1[/imath] term in the second sum. Why can you think of these terms equaling [imath]j=0[/imath] in the first sum, why not for the second sum? And the same for [imath]j=k+1[/imath]? |
2417780 | On the structure of a particular equation (maybe Sylvester's)
I sincerely hope this question is interesting for you to answer. Assume that [imath]V\in\mathbb{R}^{1\times k}[/imath], [imath]B_i\in\mathbb{R}^{k\times 1}[/imath], [imath]T_i\in\mathbb{R}^{k\times k}[/imath] with [imath]i=1\dots k[/imath], and, finally [imath]L\in\mathbb{R}^{1\times k}[/imath]. I've the following equation to solve for V (all other matrices are assumed to be known): [imath]\begin{equation} V - VB_1VT_1 - VB_2VT_2 - \dots - VB_kVT_k = L\end{equation}[/imath] I would like to know if such an equation can be classified as a particular equation with known solution, for example, a Sylvester equation (I know that the proposed equation is nonlinear though), or any other type, such that the solution can be found in a closed-form expression. Thanks!!!! | 2417991 | Looking for a hint on the class of a particular equation
I've to solve the following nonlinear matrix equation, and I was wondering if there is any bigger class where to fit such an equation so that I can either: Obtain a closed-form solution (ideal) Use appropriate algorithms to solve it numerically. Assume that [imath]V\in\mathbb{R}^{1\times k}[/imath], [imath]B_i\in\mathbb{R}^{k\times 1}[/imath], [imath]T_i\in\mathbb{R}^{k \times k}[/imath] with [imath]i=1,\dots,k[/imath], and finally [imath]L\in\mathbb{R}^{1\times k}[/imath]. The equation reads: [imath]V - \sum_{i=1}^{k} VB_iVT_i = L[/imath] And it has to be solved for [imath]V[/imath] (all the other matrices are assumed to be known). Any help or suggestion will be extremely appreciated! Thanks!!! |
2423766 | Find the value of [imath]\frac {|B-A| }{|1-\overline AB|}[/imath] where [imath]|B| = 1[/imath].
If [imath]A[/imath] and [imath]B[/imath] are two different complex numbers and [imath]|B| = 1[/imath], find the value of [imath]\frac {|B-A| }{|1-\overline AB|}[/imath] where, as usual, [imath]\overline A[/imath] denotes the conjugate of [imath]A[/imath]. If possible please don't tell me the entire answer just tell me from where to begin. | 1857842 | Find modulus of [imath]\frac{|z_1-z_2|}{|1-(z_1)(\overline{z_2})|}[/imath]
If [imath]z_1[/imath] and [imath]z_2[/imath] are two different complex numbers and [imath]\lvert z_1\rvert=1 [/imath] then find [imath] \frac{\lvert z_1-z_2 \rvert}{\lvert 1-z_1\bar{z_2} \rvert} [/imath] |
2424974 | symbol for the set of integers from 1 to N
Is there a special symbol for the set: [imath] \{1, 2, 3, \dots, n\}[/imath], or [imath]\{x| 1\leq x\leq n, n \in \mathbb{Z} \} [/imath]? | 287251 | About Math notation: the set of the first [imath]n[/imath] natural numbers
Is it true that I can write the set of the the first [imath]n[/imath] natural numbers as [imath][ n] [/imath]? For example, [imath][10]= \{1,2,3,4,5,6,7,8,9,10 \}[/imath]. And in which math context this is used? |
2424700 | Product of two irreducible affine varieties is irreducible.
Let [imath]K[/imath] be any algebraically closed ground field. For [imath]X \subset \mathbb{A}^n[/imath] and [imath]Y \subset \mathbb{A}^{m}[/imath] irreducible affine varities, I am trying to show that the product affine variety [imath]X \times Y[/imath] is also irreducible by showing that the coordinate ring [imath]A(X \times Y) = K[x_1,...,x_{n+m}]/I(X \times Y)[/imath] is an integral domain. I am trying to do this by contraposition but am stuck. Here is what I have: Since [imath]X[/imath] and [imath]Y[/imath] are affine varities so [imath]X=V(S)[/imath] and [imath]Y=V(T)[/imath] for some subsets [imath]S \subset K[x_1,...,x_n][/imath] and [imath]T \subset K[x_1,...,x_m][/imath]. Then [imath]X \times Y = V(P)[/imath] where [imath]P \subset K[x_1,...,x_{n+m}][/imath] is the set of all polynomials from [imath]S[/imath] considered as polynomials in [imath]n+m[/imath] variables by setting the last [imath]m[/imath] variables equal to zero, along with the polynomials from [imath]T[/imath] considered as polynomials in [imath]n+m[/imath] variables by setting the first [imath]n[/imath] variables equal to zero. We also have that [imath]V(P)=V((P))[/imath] where [imath](P)[/imath] is the ideal generated by [imath]P[/imath]. Hence [imath]A(X \times Y)=K[x_1,...,x_{n+m}]/I(X \times Y) = K[x_{1},...,x_{n+m}]/\sqrt(P)[/imath] by Hilbert's Nullstellensatz. Assuming [imath]A(X \times Y)[/imath] is not an integral domain there exits [imath]f,g \neq 0[/imath] such that [imath]f,g \in \sqrt(P)[/imath]; moreover we can assume that this implies that [imath]fg \in (P)[/imath] since, if not, then since for some [imath]k[/imath] we have [imath](fg)^k = f^kg^k \in (P) \subset \sqrt(P)[/imath], [imath]f^k[/imath] and [imath]g^k[/imath] are also zero divisors so we just choose [imath]f=f^k[/imath] and [imath]g=g^k[/imath] instead. I'm not sure if what I've done so far points me in the right direction, but I am at a loss for seeing how to show that [imath]X[/imath] or [imath]Y[/imath] must be reducible. Edit: My question was unfairly marked as a duplicate thread. None of the threads cited as alleged duplicates address how to prove that this is irreducible by showing that the coordinate ring of the product is an integral domain. | 152056 | Is fibre product of varieties irreducible (integral)?
Let [imath]k[/imath] be an algebraically closed field and [imath]X,Y[/imath] varieties (i.e. integral, separated schemes of finite type over [imath]k[/imath]). Is the fibre product [imath]X \times_k Y[/imath] necessary irreducible or integral? I have another more or less related question: If [imath]A,B[/imath] are finitely generated [imath]k[/imath]-algebras which are also integral domains, is [imath]A \otimes_k B[/imath] an integral domain? |
2424524 | A kind of GCD of cyclotomic polynomials
We know that in [imath]\mathbb Q[x][/imath] that two cyclotomic polynomials [imath]\Phi_n(x),\Phi_m(x)[/imath] are relatively prime when [imath]m\neq n[/imath]. This means that there are integer polynomials [imath]a(x),b(x)[/imath] and a positive integer [imath]D[/imath] such that: [imath]a(x)\Phi_n(x)+b(x)\Phi_m(x)=D[/imath] The smallest such [imath]D[/imath] can be thought of as some kind of integer [imath]GCD[/imath] of the polynomials. I'm wondering if there is a good general formula for this? Lets write [imath]\langle m,n\rangle[/imath] to be this [imath]D[/imath]. If [imath]m=1[/imath] then [imath]\langle 1,n\rangle=\Phi_n(1)=\begin{cases}p&n=p^k\\1&\text{otherwise}\end{cases}[/imath] If [imath]m,n>1[/imath] and [imath]\gcd(m,n)=1[/imath] then we easily get that [imath]\langle m,n\rangle = 1[/imath], by the same argument that [imath]\gcd(x^n-1,x^m-1)=x^{\gcd(m,n)}-1[/imath], since the proof of that is entirely in [imath]\mathbb Z[x][/imath]. When [imath]m=p,n=2p[/imath] with [imath]p[/imath] prime, you get [imath]\langle m,n\rangle = 2[/imath], since we get [imath]\Phi_{p}(x)(1-x)+\Phi_{2p}(x)(1+x)=2[/imath]. A related question is, given distinct integers [imath]m,n[/imath], what is the smallest [imath]D[/imath] such that for all integers [imath]k[/imath]: [imath]\gcd(\Phi_m(k),\Phi_n(k))\mid D[/imath] In general, this [imath]D\mid\langle m,n\rangle[/imath] but is not necessarily equal. For example, when [imath]m=p,n=2p[/imath] where [imath]p[/imath] is an odd prime, then [imath]\Phi_m(k)[/imath] is never even, so [imath]D=1[/imath] when [imath]\langle m,n\rangle = 2[/imath]. | 257548 | Cyclotomic Polynomials and GCD
Since Cyclotomic polynomials are irreducible over [imath]\mathbb{Q}[/imath], [imath]\phi_n(x)[/imath], [imath]\phi_m(x)[/imath] are coprime as polynomials in [imath]\mathbb{Z}[x][/imath]. Working over [imath]\mathbb{Q}[/imath], [imath](\phi_n(x)[/imath], [imath]\phi_m(x))=(1)[/imath]. This implies that [imath](\phi_n(x), \phi_m(x))=(c)[/imath] for some [imath]c \in \mathbb{N}[/imath] when this ideal is considered in [imath]\mathbb{Z}[X][/imath]. Can this [imath]c[/imath] be evaluated as a function of [imath]n,m[/imath]? What can be said about [imath]f(x) = gcd(\phi_n(x),\phi_m(x))[/imath], when the polynomials are considered as scalars, i.e. evaluated at some [imath]x[/imath]? [imath]\forall x: f(x) | c[/imath] from question 1, but can something stronger be said? More concretely, what is the image of [imath]f[/imath]? EDIT: This paper by Apostol, provided in the comments to Greg's answer, gives a pretty good answer, which Greg guessed. It basically calculates the resultant of 2 cyclotomic polynomials, which gives a number that is divisible by the optimal constant [imath]c=(\phi_n(x), \phi_m(x))[/imath], and it is either 1 (when [imath]\frac{n}{m}[/imath] is not a prime power) or a power of a prime ([imath]p^{\phi(m)}[/imath] when [imath]\frac{n}{m}[/imath] is a power of [imath]p[/imath]). When it is [imath]1[/imath] (the common case), we have a full answer for both questions. When it is a prime power, we only have an upper bound for the multiplicity ([imath]\phi(m)[/imath], when [imath]m|n[/imath]). This elementary paper gives some weaker result but it is simpler. |
2425686 | Proving that a polynomial is bounded on bounded domains
Definition: A real function [imath]f:X\to\mathbb{R}[/imath] is said to be bounded if there exists a [imath]k>0[/imath] such that [imath]|f(x)|\le k[/imath] Definition: A subset [imath]X[/imath] of a metric space [imath](M, d)[/imath] is said to be bounded if [imath]d(x,y)\le c[/imath] for all [imath]x,y\in X[/imath]. Let [imath]p:\mathbb{R}\to \mathbb{R}[/imath] a real polynomial function of degree [imath]n>0[/imath] given by [imath] p(t) = a_0 + a_1t+ \cdots + a_nt^n [/imath] Show that [imath]p[/imath] is not bounded. Moreover, show that for every bounded subset [imath]X\subset\mathbb{R}[/imath], the restriction [imath]s:X\to \mathbb{R}, s(t) = p(t)[/imath] is bounded. | 2423337 | Proving that any polynomial [imath]p:\mathbb{R}\rightarrow \mathbb{R}[/imath] is not bounded, but its restriction to any bounded set is bounded
Let [imath]p(t) = a_0+\dots +a_nt^n, a_n\neq 0,[/imath] a polynomial with degree [imath]n>0[/imath]. Prove that [imath]p:\mathbb{R}\rightarrow \mathbb{R}[/imath] is not bounded, but for any bounded subset [imath]X\subset \mathbb{R}[/imath] the restriction [imath]p\vert X[/imath] is bounded. I want to prove this in very elementary terms. I mean, without using the notion of limit or even continuity. I can only show this with "strong" results. Is it possible with elementary terms? |
372383 | difficulties in proving Rouche's theorem
This problem came from Schaum's complex variable p.156 Suppose [imath]f(x)[/imath] and [imath](g)[/imath] are analytic inside and on a simple closed curve [imath]C[/imath] and suppose [imath]|g(z)|\lt |f(z)|[/imath] on [imath]C[/imath]. Then [imath]f(z)+g(z)[/imath] and [imath]f(z)[/imath] have the same number of zeros inside [imath]C[/imath]. Here is the proof: Let [imath]F(z)=\frac{g(z)}{f(z)}[/imath], or briefly [imath]g=fF[/imath]. then, if [imath]N_1[/imath] and [imath]N_2[/imath] are the number of zeros inside [imath]C[/imath] of [imath]f+g[/imath] and [imath]f[/imath], respectively, we have by Argument Theorem, using the fact that these function have no poles inside C, [imath]N_1=\frac1{2\pi i}∮\frac{f'+g'}{f+g}dz[/imath] [imath]N_2=\frac{1}{2\pi i}∮\frac{f'}fdz[/imath] then[imath]N_1-N_2=\frac 1{2\pi i}∮\frac{f'+f'F+fF'}{f+fF}dz-\frac 1{2\pi i}∮\frac{f'}{f}dz[/imath][imath]=\frac 1{2\pi i}∮\frac{f'(1+F)+fF'}{f(1+F)}dz-\frac 1{2\pi i}∮\frac{f'}{f}dz[/imath][imath]=\frac 1{2\pi i}∮\left\{\frac{f'}{f}+\frac{F'}{1+F}\right\}dz-\frac 1{2\pi i}∮\frac{f'}{f}dz=∮\frac{F'}{1+F}dz[/imath][imath]=\frac 1{2\pi i}∮F'(1-F+F^2-F^3+...)dz=0[/imath] using the given fact that [imath]|F|\lt 1[/imath] on [imath]C[/imath] so that the series is uniformly convergent on [imath]C[/imath] and term by term integration yields the value zero. Thus, [imath]N_1=N_2[/imath] as required. I don't understand the last step that why it yields to zero. I also don't understand the wiki version. | 388278 | Having trouble understand the proof of Rouché's Theorem
I am trying to understand this proof of Rouché's theorem, but I am missing the logic of the last and most crucial step. Here are the assumptions: Suppose that [imath] f [/imath] and [imath] g [/imath] are analytic inside and on a regular closed curve [imath] \gamma [/imath] and that [imath] |f(z)| \gt |g(z)| [/imath] for all [imath] z > \in \gamma [/imath]. Then [imath] \mathcal{Z}(f + g) = \mathcal{Z}(f) \text{ inside > } \gamma [/imath] The proof goes as follows: First note that [imath] f \neq 0[/imath] on [imath] \gamma [/imath] since otherwise [imath] |g| \lt 0 [/imath] which doesn't make any sense. So, we can write the following: [imath] \mathcal{Z}(f + g) = \frac{1}{2\pi i} \int_{\gamma}\frac{(f+g)'}{(f+g)} = \frac{1}{2\pi i} \int_{\gamma} \frac{(f(1 + \frac{g}{f}))'}{f(1 + \frac{g}{f})} = \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} + \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} [/imath] We know [imath] f [/imath] is analytic, so [imath] \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathcal{Z}(f) [/imath] by the argument principle. What I don't know is, why is [imath] \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} = 0[/imath]? |
2425342 | Is [imath] (A∪B)×(C∩D)=(A×C)∪(B×D)[/imath] true for all sets A,B,C and D?
Is [imath](A\cup B)\times(C\cap D)=(A\times C)\cup (B \times D)[/imath] true for all sets [imath]A[/imath],[imath]B[/imath],[imath]C[/imath] and [imath]D[/imath]? I have no idea what to do here. How could I possibly go about trying to prove this? Or disprove it? It's not exactly the same question with the previous one First I tried to prove that both sides are equal if [imath](A\cup B)\times(C\cap D)\subseteq(A×C)∪(B×D)[/imath] and [imath](A×C)∪(B×D)\subseteq(A\cup B)\times(C\cap D)[/imath] . Thank you! | 2424142 | Is [imath] (A∪B)×(C∩D)=(A∪C)×(B∪D)[/imath] true for all sets A,B,C and D?
Is [imath](A \cup B)\times (C ∩ D)=(A∪C)×(B∪D)[/imath] true for all sets A,B,C and D? I have no idea what to do here. How could I possibly go about trying to prove this? Or disprove it? |
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