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2956076 | Evaluate [imath]\sum_{m=1}^{\infty} \frac{1}{2m+s}\frac{1}{2m+r}[/imath]
[imath]S = \sum_{m=1}^{\infty} \frac{1}{2m+s}\frac{1}{2m+r}[/imath] [imath]r>s>0[/imath] and both [imath]r,s[/imath] integers. write [imath]x=2m+s[/imath] and [imath]y=r-s[/imath]. [imath]\begin{align} &S = \sum_{m=1}^{\infty} \frac{1}{2m+s}\frac{1}{2m+r} = \sum_{x=s+2}^{\infty} \frac{1}{x}\frac{1}{x+y} = \sum_{x=s+2}^{\infty} \left( \frac{1}{xy} - \frac{1}{y(x+y)} \right)=\\ &\frac{1}{y}\sum_{x=s+2}^{\infty} \left( \frac{1}{x} - \frac{1}{x+y} \right) = \frac{1}{r-s}\sum_{x=s+2}^{\infty} \left( \frac{1}{x} - \frac{1}{x+r-s} \right) \end{align}[/imath] From this post Evaluating the sum (telescoping series): [imath]\sum_{k=s+1}^{\infty}\left( \frac{1}{k} - \frac{1}{k+r-s}\right)[/imath]. [imath]\begin{align} S = \frac{1}{r-s}\sum_{x=s+2}^{\infty} \left( \frac{1}{x} - \frac{1}{x+r-s} \right) =\frac{1}{r-s} \left( \frac{1}{s+2} + \frac{1}{s+3} \cdots + \frac{1}{r}\right) \end{align}[/imath] Is the above correct? I asked this another day but I could understand nothing of the answers. Another thing is: I assumed that both [imath]r,s[/imath] are integers, but, If I assumed to be both odd numbers, would something change? I did some tests in Wolfram, and it seems that for both [imath]r,s[/imath] odd integers and [imath]r>s>0[/imath], the sum [imath]S[/imath] will be a rational number. This is a follow up question to: Find the sum of [imath]\sum_{k=0}^{\infty} 1/[(r+2k+2)(s+2k+2)][/imath] The answer in the link use the Digamma function. I wanted to know If the answer in the current post is correct. | 2954661 | Find the sum of [imath]\sum_{k=0}^{\infty} 1/[(r+2k+2)(s+2k+2)][/imath]
[imath]\begin{align} \sum_{k=0}^{\infty} \frac{1}{(r+2k+2)(s+2k+2)} = \sum_{m=1}^{\infty} \frac{1}{(r+2m)(s+2m)} \end{align}[/imath] [imath]r>s>0[/imath]. [imath]r[/imath] is a odd number, but I don't think this gonna be relevant. The fact that [imath]r>s[/imath] is given because it's a telescopic series (I think). I wanted to know where does this series converge to and what happens when [imath]r[/imath] is odd and [imath]s[/imath] is either odd or even, but I don't know how to evaluate this sum. |
2386905 | Show that [imath]\{ f \ast f: f \in \mathcal{E}'(\mathbb{R}^n) \} \subsetneq \mathcal{E}'(\mathbb{R}^n)[/imath]
I wonder if somebody could help me understand the following solution: [imath]\textbf{Problem:}[/imath] Let [imath]F=\{ f \ast f: f \in \mathcal{E}'(\mathbb{R}^n) \} \subset \mathcal{E}'(\mathbb{R}^n)[/imath]. Show that [imath]F \neq \mathcal{E}'(\mathbb{R}^n)[/imath] ([imath]f\ast f[/imath] is the convolution of [imath]f[/imath] with it self and [imath]\mathcal{E'} [/imath] is the space of distributions with compact support in [imath]\mathbb{R}^n[/imath]). [imath]\textbf{Solution:}[/imath] Assume the contrary. Then [imath]\delta_1 -\delta_0 \in F[/imath] so there is an [imath]f \in \mathcal{E'}[/imath] such that [imath]f \ast f= \delta_1 -\delta_0[/imath] Since [imath]f \in \mathcal{E'}[/imath] such the Fourier transform of [imath]f[/imath], [imath]\hat{f}[/imath], is complex analytic. Then [imath] (\hat{f}(\xi))^2= e^{-i\xi}-1 \in \mathcal{O}({\xi}) \implies \hat{f}(0)=0[/imath]. I fail to see why this is a contradiction (also I don't get the last implication). Does anybody understand this? | 2335033 | Schwartz Distribution Theory: (Non-)Surjectivity of [imath]f \mapsto f*f[/imath] in [imath] \mathcal E '[/imath]
For a compactly supported distribution [imath]f[/imath], I want to show the non-surjectivity of the map [imath]H:f \mapsto f*f \in \mathcal{E}'[/imath]. My attempt: I have chosen to show that this is not the case by showing that [imath]\text{image } (H)[/imath] does not contain [imath] \delta_0'[/imath]. For instance I restrict myself to [imath]d=1[/imath]. Suppose that [imath]H[/imath] is surjective. Applying the Fourier transform on the convolution, I get [imath](\hat{f}(z)) ^2 = i z[/imath]. The theorem of Paley-Wiener-Schwartz guarantees that [imath](\hat{f}(z)) [/imath] should be an entire function. However, as it is well-known, no such entire function exists: If this were true, then [imath]e^{i \pi /4}\hat{f}(z)[/imath] would be the square root function, which cannot be well defined on [imath]\mathbb C[/imath], and even less so entire. Q1: Is there another way of achieving the result (i.e. with a choice significantly different than [imath]\delta_0'[/imath], and with an approach different than the one using the Fourier transform)? Q2: Can I generalize this without much effort in the case of [imath]d \ge 1[/imath] (probably using the tensor product)? Edit: The source is an exam at Lund University [problem 3]. |
2957287 | Find a probability mass function of a random variable
[imath]\color{red}{Attempt} [/imath] We start with [imath]k=1[/imath], [imath]P(X=1)[/imath] is the probability that one letter have been put in the correct envelope. Our sample space size is [imath]n[/imath] and since there is only one way that one letter must have been put into the correct envelope and the rest [imath]n-1[/imath] incorrectly and so we see that [imath]P(X=1)= \dfrac{(n-1)!}{n}[/imath], now for [imath]P(X=2)[/imath] it becomes more complicated, so far I know that [imath]{n \choose 2}[/imath] is the size of the sample space and now we want to count the number of ways in which 2 letters must have been put in the correct envelope. First, of all, the [imath]n-2[/imath] letters that have been put incorrectly we have to count them and we have [imath](n-2)![/imath] and then the 2 letters that are put correctly this is done in one way thus [imath] P(X=2) = \frac{(n-2)!}{{n \choose 2} }[/imath] so, in general, we have [imath] P(X=k) = \frac{(n-k)!}{n \choose k } [/imath] is this correct? | 2465803 | What's the probability that a given permutation has exactly $k$ fixed points.
Given a random permutation [imath]\sigma \in S_n[/imath] from [imath]$[n]\to[n]$[/imath] in a uniform probability space, what is the probability that [imath]\sigma [/imath] has exactly [imath]$k$[/imath] fixed points for a given [imath]$k$[/imath] between [imath]1[/imath] and [imath]n[/imath]? In other words: what is the probability that [imath]\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i [/imath] for [imath]$\ i\in \{1,...,k\}$[/imath] and for every [imath]$y \notin \{x_1 , ... , x_k\}$[/imath] we get [imath]\sigma(y) \neq y[/imath]. I saw that [imath]\lim_{n \to \infty } prob(A_0) = e^{-1}[/imath] using Inclusion–exclusion principle and i belive that for a given k : [imath]\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}[/imath] but I am not sure how to show it. *[imath]A_k[/imath] stands for the event "k". |
2957192 | Example of a principal prime ideal containing a proper prime non-zero ideal.
It is true that if [imath]A[/imath] is UFD and [imath]I=\left\langle p \right\rangle[/imath] is a principal prime ideal, it does not exist a proper prime non-zero ideal contained in [imath]I[/imath], but this property does not hold for general rings (commutative rings with 1). I would appreciate some examples, that is, a ring [imath]A[/imath] and prime ideals [imath]\mathfrak{p}, \left\langle p \right\rangle[/imath] such that [imath]\begin{equation} 0 \subsetneq \mathfrak{p} \subsetneq \left\langle p \right\rangle \end{equation}[/imath] | 180907 | The height of a principal prime ideal
A formal consequence of Krull's principal ideal theorem is the following: If [imath]A[/imath] is a Noetherian ring, and [imath]I[/imath] is an ideal generated by [imath]r[/imath] elements, then any prime ideal which is minimal among those that contain [imath]I[/imath] has height at most [imath]r[/imath]. This statement implies that for Noetherian rings, principal prime ideals have height at most [imath]1[/imath]. My question is if this is true for any ring, i.e., is a principal prime ideal of a ring always of height at most [imath]1[/imath]? The above question is clearly true if the following statement is true: If every maximal ideal of a ring is finitely generated, then the ring is Noetherian. (Note that this statement is true if we replace maximal ideals by prime ideals) However, I am not sure if this latter assertion is true, although I do not have a counterexample for it. |
2957101 | Convergence of [imath]E(|X|^k)[/imath] and [imath]\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i)[/imath]
I am asked to prove the following question regarding convergence of [imath]k^\text{th}[/imath] moment of a random variable: [imath]E(|X|^k) < \infty[/imath] if and only if [imath]\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i) < \infty[/imath] for any integer [imath]k > 1[/imath]. I have managed to prove the direction that given [imath]\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i) < \infty[/imath], we have [imath]E(|X|^k) < \infty[/imath]. But I am stuck at the other direction. The proof I have come up with now is as follow: Construct [imath]X_i^k(\omega) = (i-1)^kI_{A_i}[/imath] where [imath]A_i = \{i-1 \leq |X| < i\}[/imath]. Then [imath]|X|^k \geq \sum_{i=1}^{\infty}X_i^k[/imath], and we have [imath]E(|X|^k) \geq E(\sum_{i=1}^{\infty}X_i^k)[/imath]. Since [imath]\sum_{i=1}^{\infty}X_i^k[/imath] is a discrete random varaible taking value [imath](i-1)^k[/imath] with probability [imath]P(i-1 \leq |X| < i)[/imath]. We have [imath] E(|X|^k) \geq \sum_{i=1}^{\infty}(i-1)^kP(i-1 \leq |X| < i). [/imath] I know the next step is to somehow show that the RHS is greater than [imath]\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i)[/imath], but I am stuck at this point and I don't know how to proceed. Can anyone give some hints? Thanks very much! | 2954219 | characterization of the existence of higher order expectation.
Consider a non-neg r.v [imath]X[/imath] and fix a positive integer [imath]n[/imath], I am trying to show that a necessary and sufficient condition for the [imath]n[/imath]th order expectation to exist is that [imath]\sum_{i\in\mathbb{N}}i^{n-1}P(X\ge i)[/imath] converges. Attempt: We have a well established relation, [imath]\sum_i P(X\ge i)\le EX < 1+\sum_iP(X \ge i)[/imath], I am convinced this is used but i am unsure how to to incorporate [imath]i^{n-1}[/imath] . If we consider a partition of the real line given by [imath]A_i=\{i-1\le X < i\}[/imath] and then let choose [imath]X_i=(i-1)I_{A_i}[/imath] then the series is a discrete r.v. with [imath]E(\sum_i X_i)=\sum iP(X\ge i)[/imath] which is similar to the form we need it in. But im unsure where to go from here. |
2954222 | Proof by Induction of [imath]\frac{d^n}{dx^n} (\ln(x))[/imath] = [imath]\frac{(-1)^{n-1}(n-1)!}{x^n}[/imath] for [imath]n\geq1[/imath].
Conjecture for [imath]\ln(x)[/imath] and [imath]\frac{d^ny}{dx^n}[/imath] showed how to find the conjecture for [imath]ln(x)[/imath] after testing multiple values. For this question that I have, I have to prove using induction that [imath]\frac{d^ny}{dx^n} (\ln(x)) = \frac{(-1)^{n-1}(n-1)!}{x^n}[/imath] for [imath]n\geq1[/imath]. I proved the base case, where [imath]n=1[/imath]. Then I tried to start the induction step, assuming the claim holds for some [imath]n\geq1[/imath]. I realized that I need to show that [imath]\frac{d^{n+1}y}{dx^{n+1}} (\ln(x)) = \frac{(-1)^{n}(n)!}{x^{n+1}},[/imath] but trying to continue from here has been tricky. | 2954161 | Conjecture for [imath]\ln(x)[/imath] and [imath]\frac{d^ny}{dx^n}[/imath]
I'm not really sure where to start, I found the first, second, third and fourth derivatives of [imath]\ln(x)[/imath] to be [imath]\frac{1}{x}[/imath],-[imath]\frac{1}{x^2}[/imath], [imath]\frac{2}{x^3}[/imath], and -[imath]\frac{6}{x^4}[/imath], respectively. Letting [imath]n[/imath] be a natural number, I have to formulate a conjecture for a formula for [imath]\frac{d^ny}{dx^n}[/imath]. Afterward, I have to use mathematical induction to prove the conjecture. |
2958695 | What does Hilbert's second axiom of connection mean?
Here are the first 2 axioms of connection from Hilbert's Book: I, 1. Two distinct points [imath]A[/imath] and [imath]B[/imath] always completely determine a straight line [imath]a[/imath]. We write [imath]AB = a[/imath] or [imath]BA = a[/imath]. Instead of “determine,” we may also employ other forms of expression; for example, we may say [imath]A[/imath] “lies upon” [imath]a[/imath] , [imath]A[/imath] “is a point of” [imath]a[/imath] , [imath]a[/imath] “goes through” [imath]A[/imath] “and through” [imath]B[/imath] , [imath]a[/imath] “joins” [imath]A[/imath] “and” or “with” [imath]B[/imath] , etc. If [imath]A[/imath] lies upon a and at the same time upon another straight line [imath]b[/imath] , we make use also of the expression: “The straight lines” [imath]a[/imath] “and” [imath]b[/imath] “have the point [imath]A[/imath] in common,” etc. I, 2 . Any two distinct points of a straight line completely determine that line; that is, if [imath]AB = a[/imath] and [imath]AC = a[/imath], where [imath]B \not= C[/imath], then is also [imath]BC = a[/imath]. My interpretation of the first axiom is, For any 2 distinct points, there exists a straight line which is 'completely determined' by them. I admit that I am not sure what 'determines' means, and its difference from 'completely determines', but that doesn't matter, because those notions are exactly what the connection axioms are trying to define. The first axiom then states that the line which is determined by the points [imath]A[/imath] and [imath]B[/imath], we call it [imath]AB[/imath], and since [imath]AB[/imath] is by definition a line we assign it lower-case [imath]a[/imath]. so [imath]AB =a[/imath]. remark 1) The first axiom doesn't state that [imath]A,B[/imath] determine exactly one line, but rather states they determines at least one line. I am still confused why it's not 'Two distinct points A and B always completely determine exactly one straight line a', but I think this is the purpose of the second axiom. Let's now analyze the second axiom: I, 2 . Any two distinct points of a straight line completely determine that line; that is, if [imath]AB = a[/imath] and [imath]AC = a[/imath], where [imath]B \not= C[/imath], then is also BC = a . The hypothesis of the axiom is that there are two distinct points [imath]A,C[/imath] of a straight line [imath]a[/imath]. The definition of a point being 'of a line' is defined in the paragraph after the first axiom. By definition, this means that for [imath]A[/imath], there is necessarily another point [imath]B[/imath] such that [imath]AB[/imath] completely determines [imath]a[/imath] or [imath]AB=a[/imath]. This is also true for [imath]C[/imath], and we call that point [imath]D[/imath]. Now the axiom states that [imath]A,C[/imath] completely determines [imath]a[/imath], so that [imath]AB=CD=a=AC[/imath]. The second axiom is just last equality. But the second sentence(of the second axiom) states another thing. Also I want to remark that [imath]AC[/imath] is not necessarily [imath]a[/imath]. We know that [imath]A,C[/imath] completely determine [imath]a[/imath] from the second axiom, but we still aren't sure that [imath]A,C[/imath] determine exactly one line, so that my last equation above is not necessarily true, right? To put it bluntly, the first sentence of the second axiom says: [imath](1) \quad AB=a \land CD=a \to AC=a[/imath] But the next sentence says [imath](2) \quad AB=a \land BC=a \to AC=a[/imath] But these are not the same. I looked it up on Wikipedia, and the second axiom turns out to be the uniqueness, but clearly statement in the book says a different thing from that of the Wikipedia, right? My question is, which one is the real second axiom, and if they are all the consequences of each other, can you show how? Also is there anything fundamentally wrong with my thought process? | 2778914 | A confusion about the second connection axiom of Euclidean Geometry
In the book of Foundations of Geometry by Hilbert, at page 2, it is stated that I, 1. Two distinct points A and B always completely determine a straight line a. We write AB = a or BA = a. I, 2. Any two distinct points of a straight line completely determine that line; that is, if AB = a and AC = a, where [imath]B \not = C[/imath], then is also BC = a. However, what I understand from "Any two distinct points of a straight line completely determine that line" is that given a line [imath]a[/imath], and points [imath]A,B,C[/imath] on that line, we have [imath]AB = AC=BC = a[/imath], but the logical formulation states, i.e the part after "that is", that given any 3 point [imath]A,B,C[/imath] s.t [imath]AB = AC = a[/imath] implies [imath]BC = a[/imath] (of course provided that [imath]B \not = C[/imath] in both cases), so is there any explanation how should we understand Any two distinct points of a straight line completely determine that line as if AB = a and AC = a, where [imath]B \not = C[/imath], then is also BC = a. |
2955964 | Exponent of the group [imath]GL(n,\mathbb F_2)[/imath]
Let [imath]A\in GL_n(\mathbb F_2)[/imath] be an element of order greater than or equal to [imath]2^n-1[/imath] . Then is it true that order of [imath]A[/imath] is [imath]2^n-1[/imath] ? I know that [imath]|GL_n(\mathbb F_2)|=(2^n-1)(2^n-2^2)...(2^n-2^{n-1})[/imath]. But I'm unable to proceed further. Please help. | 292620 | Exponent of [imath]GL(n,q)[/imath].
Another exponent problem. [imath]GL(n,q)[/imath] is the group of invertible [imath]n\times n[/imath] matrices over the finite field [imath]GF(q)[/imath], where [imath]q[/imath] is a prime power. I am trying to figure out the exponent of this group. Ideas: We'd be doin real good if we could find the set of element orders [imath]\{o(A):A\in GL(n,q)\}[/imath], so I am looking for things to help me figure this out. The order of [imath]GL(n,q)[/imath] is [imath]|GL(n,q)|=\prod_{k=0}^{n-1}\left(q^n-q^k\right)[/imath] The set of prime factors of this number is a little unpredictable so I'm not sure whether I can use this. If [imath]q[/imath] is prime, the maximum order of an element in [imath]GL(n,q)[/imath] is at most [imath]p^n-1[/imath] (and this bound is strict I think). Also I guess the maximal power of [imath]p[/imath] that occurs as an order of this group is [imath]p^{\lceil \log_pn\rceil}[/imath]. (source) I am aware of this, which I believe implies that [imath]\text{Exp}(GL(2,p^n))=\text{lcm}\{p(p^{2n}-1),p-1\}[/imath]. I am trying to think of how to extend it to the general case but I'm not sure. And I'm stuck; can't think of anything else that might help. |
2958952 | Collection of Nonempty Subsets of [imath]\mathbb{Z^{+}}[/imath] a Class or Set?
In ZFC, is the collection [imath]\mathcal{N}[/imath] of nonempty subsets of [imath]\mathbb{Z^{+}}[/imath] a class or set? Please correct me if I am wrong, but so far as I understand the notion of class, a class is a collection of objects which satisfy a certain property [imath]P[/imath]. In this example, let [imath]P[/imath] be the property that [imath]A[/imath] is a nonempty subset of [imath]\mathbb{Z^{+}}[/imath]. Take [imath]N=\{S|PS \}[/imath]. Then is [imath]N[/imath] not just our collection [imath]\mathcal{N}[/imath], and is [imath]N[/imath] not a class? | 2284709 | ZFC with class terms: is any set also a class?
In ZFC with class terms a class is merely a unary predicate [imath]\varphi[/imath] of the language which we then write as [imath]\{X\mid\varphi(X)\}[/imath] which only suggests that we have built a collection out of these objects. We write [imath]Y\in\{X\mid \varphi(X)\}[/imath] for [imath]\varphi(Y)[/imath]. Further we say that such a class (term) [imath]\mathcal C[/imath] is a set if [imath]\exists X\forall Y[Y\in X\leftrightarrow Y\in\mathcal C].[/imath] In NBG and similar set theories (which build on classes instead of sets) every set is also a class. But is this true for this ZFC approach? For me it seems that a class is just a formula. And there are sets that I cannot fully describe using such a formula, e.g. a set given to me by the axiom of choice. Are sets and classes in ZFC (with class terms) just different concept, each one not a "sub-concept" of the other? After Asaf's answer: I do not feel very well when writing [imath]\{X\mid X\in A\}[/imath] for arbitrary sets [imath]A[/imath] because for me [imath]A[/imath] is not a symbol of the language. However, when my set [imath]A[/imath] is definable by a predicate [imath]\varphi[/imath] with [imath]\mathrm{ZFC}\vdash \exists A \varphi(A) \quad\text{ and }\quad \mathrm{ZFC}\vdash\varphi(A)\wedge \varphi(B)\to A=B,[/imath] then I could write [imath]\{X\mid \forall A[\varphi(A)\to X\in A]\}[/imath], and this would feel okay. But my problem are the sets for which there is no such [imath]\varphi[/imath]. I do not know how I can include such an abbreviating notion (like [imath]\{X\mid X\in A\}[/imath]) into a formal proof without feeling not quite sure what I am actually doing. |
2958959 | If [imath]x>0, y>0,x+y=\frac{\pi}{3}[/imath] then maximum value of [imath]\tan x\tan y[/imath]
If [imath]x>0, y>0[/imath] and [imath]x+y=\frac{\pi}{3}[/imath], then find the maximum value of [imath]\tan x\tan y[/imath] My Attempt [imath]x>0, y>0, x+y=\frac{\pi}{3}\implies x, y[/imath] in [imath]1^\text{st}[/imath] quadrant. [imath]\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0[/imath] [imath] \tan x+\tan y\geq2\sqrt{\tan x.\tan y}\implies \tan^2x+\tan^2y+2\tan x\tan y\geq 4\tan x\tan y\\ 2\tan x\tan y\leq\tan^2x+\tan^2y\implies \color{red}{?} [/imath] or [imath] 1-\tan x\tan y=\frac{\tan x+\tan y}{\tan(x+y)}\implies\tan x\tan y=1-\frac{\tan x+\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}\\ \tan x\tan y=1-\frac{\frac{\tan x+\sqrt{3}\tan^2x+\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}=1-\frac{1+\tan^2x}{1+\sqrt{3}\tan x}\leq\color{red}{?}\\ =\frac{1+\sqrt{3}\tan x-1-\tan^2x}{1+\sqrt{3}\tan x}=\frac{\tan x(\sqrt{3}-\tan x)}{1+\sqrt{3}\tan x}=\color{red}{?} [/imath] Note: I prefer not to do differentiation | 1207197 | If [imath]A,B>0[/imath] and [imath]A+B = \frac{\pi}{3},[/imath] Then find Maximum value of [imath]\tan A\cdot \tan B[/imath].
If [imath]A,B>0[/imath] and [imath]\displaystyle A+B = \frac{\pi}{3},[/imath] Then find Maximum value of [imath]\tan A\cdot \tan B[/imath]. [imath]\bf{My\; Try::}[/imath] Given [imath]\displaystyle A+ B = \frac{\pi}{3}[/imath] and [imath]A,B>0[/imath]. So we can say [imath]\displaystyle 0< A,B<\frac{\pi}{3}[/imath]. Now taking [imath]\tan [/imath] on both side, we get [imath]\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).[/imath] So [imath]\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}[/imath]. Now Let [imath]\displaystyle \tan A\cdot \tan B=y\;,[/imath] Then [imath]\displaystyle \tan B = \frac{y}{\tan A}.[/imath] So [imath]\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A[/imath] So equation [imath]\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0[/imath] Now for real values of [imath]y\;,[/imath] Given equation has real roots. So its [imath]\bf{Discrimnant>0}[/imath] So [imath]\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0[/imath] So we get [imath]3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0[/imath] So we get [imath]\displaystyle y\leq \frac{1}{3}\cup y\geq 3[/imath], But above we get [imath]\displaystyle 0<A,B<\frac{\pi}{3}[/imath] So We Get [imath]\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.[/imath] My Question is can we solve above question using [imath]\bf{A.M\geq G.M}[/imath] Inequality or Power Mean equality. Thanks |
2959053 | A function [imath]f\colon E\to F[/imath] between finite sets is injective iff it is surjective iff it is bijective
i cant solve this exercise, if you can help or give me a lead. I would like to also say its not marked. Let [imath]f[/imath] an application from [imath]E[/imath] to [imath]F[/imath], with [imath]\#E=\#F=n[/imath]. Show that these propositions are equivalent: [imath]f[/imath] is injective [imath]f[/imath] is surjective [imath]f[/imath] is bijective Going from 2 to 3 is fairly simple, and from 3 to 1 is automatic but I can't go from 1 to 2. Thank you, cordially, JF | 2401604 | Suppose [imath]X[/imath] is a finite set and [imath]f : X \to X[/imath] is a function. Then [imath]f[/imath] is injective if and only if [imath]f[/imath] is surjective.
Suppose [imath]X[/imath] is a set and [imath]f : X \to X[/imath] is a function. If [imath]X[/imath] is a finite set, prove that [imath]f[/imath] is injective if and only if [imath]f[/imath] is surjective. Show that when [imath]X[/imath] is an infinite set, the statement is not true in general. For that, let [imath]X = \mathbb{R}[/imath] and find a function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] which is injective but not surjective, and a function [imath]g: \mathbb{R} \to \mathbb{R}[/imath] which is surjective but not injective My attempt at the first part Suppose that the function is surjective but not injective. Let [imath]a,b \in X[/imath] such that [imath]f(a)=f(b)[/imath] but [imath]a \not =b[/imath]. Now since this is a finite set mapping to itself then there are not enough elements in [imath]X[/imath] to map to [imath]X[/imath] since two elements were used to map to one element, thus the function can't be surjective. Therefore if the function is surjective it is also injective. Now suppose that the function is injective but not surjective. Let [imath]a \in X[/imath] such that [imath]f(b)\not =a[/imath] for all [imath]b \in X[/imath]. Since the function was assusmed to be injective each [imath]a \in X[/imath] must map to an element of [imath]X[/imath], thus must have two elemetns in [imath]X[/imath] mapping to the same element in [imath]X[/imath]. Hence the function can't be injective, therefore if the function is injective it must be surjective. Would an exponetial function work for injective but not surjective since can't map to negative numbers. And an example of a function that is surjective but not injective could be a log function since none of the negative numbers get mapped to anything? Also I'm having trouble seeing why it has to be finite for the first part to work. I understand it has something to do with the pigeonhole principle but its not that clear to me. |
2959427 | Show that an operation of elements in a finite group contains each elements exactly once
Finite group [imath]G[/imath] contains elements [imath]g_1,g_2,...,g_n[/imath]. Show [imath]g_1g,g_2g,...,g_ng[/imath] contains each element of G exactly once. I know that by definition [imath]g_ig[/imath] is an element of group [imath]G[/imath]. How can I prove by contradiction that [imath]g_ig[/imath] does not equal [imath]g_jg[/imath]? | 2959400 | Show that an operation of elements in a finite group contains each element exactly once
Finite group [imath]G[/imath] contains elements [imath]g_1,g_2,...,g_n[/imath]. Show that [imath]g_1g,g_2g,...,g_ng[/imath] consists of every element from [imath]G[/imath], where [imath]g\in G[/imath]. I know that the operation of a finite group gives back each element of the group once, but how can I prove this? I know I have to show that some [imath]g_ig[/imath] is in the group and some [imath]g_ig[/imath] does not equal some [imath]g_jg[/imath] |
2958864 | Prove that the interval [imath][0,1)[/imath] and [imath][0,2) \cup [3,4)[/imath] have the same cardinality
By definition, I know I have to show there exists a [imath]1[/imath]-[imath]1[/imath] onto map [imath]f:A\to B[/imath] I am pretty stuck on how to go to the process of proving this. I understand the basic definitions of 1-1 and onto. I recall basic definitions of [imath]1[/imath]-[imath]1[/imath] and onto from high school, I know a function is [imath]1[/imath]-[imath]1[/imath] if it passes the horizontal and vertical line test and a function is onto if for all [imath]Y[/imath] there exists an [imath]x[/imath] in [imath]X[/imath] such that [imath]f(x) = y[/imath], but I am really not sure how to go about proving this at all. Could someone maybe point me in the direction of an example to help me understand each step in the process? | 2397547 | Prove that the interval [imath] \ [0,2)[/imath] and [imath] \ [5,6) \cup [7,8)[/imath] have the same cardinality
Question: Prove that the interval [imath][0,2)[/imath] and [imath][5,6) \cup [7,8)[/imath] have the same cardinality by constructing a bijection between the two sets. You can add a graph to support your argument. I tried graphing a function with domain [imath][0,2)[/imath] and target space [imath][5,6) \cup [7,8)[/imath] but I can't come up with an explicit formula. Will the function be linear? |
2959719 | Prove [imath]x^4 +x+1[/imath] is irreducible in [imath]\mathbb{Z}/2\mathbb{Z}[x][/imath]
[imath] x^4 +x+1[/imath] in [imath]\mathbb{Z}/2\mathbb{Z}[x][/imath] is an irreducible polynomial. So far we have only treated quadratic and cubic polynomials, which are irreducible if they do not have any zeros. However, now I want to show that [imath]x^4 + x+1[/imath] in [imath]\mathbb{Z}/2\mathbb{Z}[x][/imath] is irreducible, I cannot go about checking if it has any zeros, this does not guarantee irreducibility. Is there any clever approach or do I need to determine all the polynomials of lower degree that are irreducible and show that upon division there is always a remainder? [imath] \{x, x+1 ,x^2+x+1, x^3 +x+1, x^3 +x^2+1\} [/imath] are the polynomials I immediately thought of. | 2442115 | Show the polynomial is irreducible in [imath]\textbf Z_2[/imath]
Show that [imath]p(\gamma) = \gamma^{4} + \gamma + 1[/imath] is irreducible over [imath]\textbf Z_2[/imath]. Okay, I would just like to see if i'm on the right track for this problem. [imath]\gamma^{4}+\gamma+1=(\gamma +b)(c\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)[/imath] So then [imath]a = c = 1[/imath]. So [imath]\gamma^{4}+\gamma+1=(\gamma +b)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)[/imath] Let [imath]b = 0[/imath] [imath]\gamma^{4} + \gamma+1=(\gamma)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)[/imath] Let [imath]\gamma = 0[/imath] Then we see [imath]1 = 0[/imath] which is a contradiction. Let [imath]b = 1[/imath] [imath]\gamma^{4}+\gamma+1=(\gamma+1)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)[/imath] Take [imath]\gamma = 1[/imath] And again [imath]1=0[/imath] So our polynomial is irreducible over [imath]\textbf Z_2[/imath] Is there another way I should be computing this? Maybe with Long division of polynomials? I'm quite lost.. |
2959737 | Why isn't [imath]\gcd(x^2+3x+2,x^2+x)=(x+1)[/imath]? [unit normalization of gcds]
Excuse me for the confusing title. I was asked to find [imath]gcd(x^2+3x+2,x^2+x)[/imath] What i did is i factorized both polynomials [imath]x^2+x=(x+1)x[/imath] [imath]x^2+3x+2=(x+1)(x+2)[/imath] So i expected the gcd to be [imath]x+1[/imath] But using the euclidean algorithm i found out the gcd to be [imath]2x+2[/imath]. Why is factorizing wrong? Is it because [imath]K[X][/imath] is not factorial ? Would the euclidean algorithm also work if the polynomials are in [imath]\Bbb Z[X][/imath] ??? | 2939195 | Why integer GCDs are positive? [unit normalization of GCDs]
The definition in my text reads, An integer [imath]d[/imath] is said to be the greatest common divisor of two non-zero integers [imath]a[/imath] and [imath]b[/imath] iff, [imath]d|a[/imath] and [imath]d|b[/imath] and if [imath]k[/imath] is any other common divisor of [imath]a[/imath] and [imath]b[/imath] then [imath]k|d[/imath] Now here's the thing, if [imath]d|a[/imath] and [imath]d|b[/imath] then surely [imath]-d|a[/imath] and [imath]-d|b[/imath] as well, also [imath]k|-d[/imath] What I take from this? GCD is not unique! That is if [imath]\mathrm{gcd}(12,8)= 4[/imath] then by the definition, [imath]\mathrm{gcd}(12,8) = -4[/imath] as well. Yet I never ever seen a negative gcd. Someone please explain. Maybe, [imath]4>-4[/imath], and we want the "greatest common factor" so...? But that still doesn't justifies the definition. |
2959364 | If [imath](b,c) = 1[/imath], then [imath](a,bc) = (a,b)(a,c)[/imath]
I do not know how to solve the following: Show that If [imath](b,c) = 1[/imath], then [imath](a,bc) = (a,b)(a,c)[/imath] Thanks. | 1885955 | Show that if [imath](b,c)=1[/imath], then [imath](a,bc)=(a,b)(a,c)[/imath].
Problem: Show that if [imath](b,c)=1[/imath], then for any integer [imath]a[/imath] we have [imath](a,bc)=(a,b)(a,c)[/imath]. [Hint: Prove that each member of the alleged equation divides the other.] This is the way I proved it (without the hint): [imath](a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)[/imath] But how should I prove it using that hint or at least a similar proof. |
2959957 | How to prove CP^1 homeomorphic two-dimensional sphere
I'm a complete newbie to this section: book are quite complicated in terms. Also, there is a problem with formal building homeomorphism. So, I need to prove that [imath]CP^1 \simeq S^2 [/imath]. How to do it formally? By the way, as I understand, that [imath]RP^1 \simeq S^1 [/imath] as identification of diametrically opposite points. Maybe my problem use similar solution. | 323116 | [imath]\mathbb{C}\mathbb{P}^1[/imath] is homeomorphic to [imath]S^2[/imath]
I just read on wikipedia that the the complex projective line is homeomorphic to the riemann sphere. How do I prove this? But, before that I have an extremely silly doubt that has been eating me. In the complex projective line, antipodal points are identified, but in the Riemann sphere they are clearly distinct. Then how can the two be homeomorphic. I found out the following map from the complex projective line to [imath]S^2[/imath]. But how do I prove this is a homeomorphism? Finding the inverse of this functions seems complicated, and then how do I prove that the inverse map of a open subset is a open subset, and vice versa? [imath]f:\mathbb{C}\mathbb{P}^1 \to S^2 = [z:w] \to \frac{(2Re(w\bar{z}), 2Im(w\bar{z}),|w|^2-|z|^2)}{(|w|^2+|z|^2)} [/imath] |
2957163 | Rings with decomposition are sums of ideals generated by idempotents
I'm working on a question out of T.Y. Lam's book that has me thrown. Let [imath]B_1 \dots, B_n[/imath] be left ideals (resp. ideals) in a ring [imath]R[/imath]. Show that [imath]R=B_1 \oplus \dots \oplus B_n[/imath] iff there exists idempotents (resp. central idempotent) [imath]e_1, \dots ,e_n[/imath] with sum 1 such that [imath]e_ie_j=0[/imath] whenever [imath]i \neq j[/imath], and [imath]B_i=R e_i[/imath] for all [imath]i[/imath]. In the case where the [imath]B_i[/imath]'s are ideals, if [imath]R=B_1 \oplus \dots \oplus B_n[/imath], then each [imath]B_i[/imath] is a ring with identity [imath]e_i[/imath], and we have an isomorphism between [imath]R[/imath] and the direct product of rings [imath]B_1 \times \dots \times B_n[/imath]. Show that any isomorphism of [imath]R[/imath] with a finite direct product of rings arises in this was. Attempt Supposing we have such idempotents there is an obvious map [imath]R \to \oplus Re_i[/imath] since [imath]r=r \cdot 1=r \sum e_i[/imath]. Hence, [imath]r \mapsto r \sum e_i[/imath]. This map is easily shown to be bijective, however, demonstrating that this is a homomorphism seems to be problematic. Consider [imath]r_1r_2 \mapsto r_1r_2(\sum e_i) \text{ but } r_1(\sum e_i)r_2(\sum e_i) \neq r_1r_2 \sum e_i???[/imath] so perhaps this isn't the correct map? Conversely, supposing we have such a decomposition then clearly [imath]1=\sum b_i[/imath] so let [imath]e_i = b_i[/imath]. I'm guessing the fact that 1 is trivially an idempotent will show that the [imath]e_i[/imath] are but I'm stuck there. We have [imath]1=1^2=(\sum e_i)^2[/imath] and similar to the above the map [imath]r=r*1=r\sum e_i[/imath] jumps out at me. | 2437788 | If [imath]A\simeq\prod A_i[/imath] then [imath]\sum e_i=1[/imath]
Given [imath]A[/imath] commutative ring with unity. If we have rings [imath]A_1,\dots, A_n[/imath]such that [imath]A\simeq A_1\times \dots \times A_n[/imath] there are [imath]e_1,\dots,e_n[/imath] idempotent elements of [imath]A[/imath] such that [imath]e_1+\dots+e_n = 1[/imath] and [imath]e_i e_j =0[/imath] if [imath]i\neq j[/imath]. I'm not sure how general the proof should be written... The isomorphic relation suggest that if [imath]f[/imath] defines the isomorphism then there is a [imath](a_1,\dots, a_n)\in \prod A_i[/imath] such that [imath]f(a_1,\dots,a_n)=1\in A[/imath]. Could it work thinking of the idemmpotent terms as [imath]e_i=(0,\dots,0, a_i,0, \dots,0)?[/imath] This should satisfy [imath]e_i e_j =0[/imath] if [imath]i\neq j[/imath]. But I have two problems with it (1) I don't see why [imath]e_i[/imath] should idempotent as no assumptions in regard to the existence of idempotent elements in [imath]A_i[/imath] has been made. (2) The sum [imath]\sum e_i=1[/imath] has no reason to be true. |
2958156 | Probability of [imath]P1[/imath] winning championship
Two players [imath]P_1[/imath] and [imath]P_2[/imath] are playing the final of a chess championship,which consists of a series of matches.Probability of [imath]P_1[/imath] winning a match is [imath]\frac{2}{3}[/imath] and that of [imath]P_2[/imath] is [imath]\frac{1}{3}[/imath].The winner will be the one who is ahead by two games as compared to the other player and wins atleast [imath]6[/imath] games.Now if the player [imath]P_2[/imath] wins first 4 matches,prove that the probability of [imath]P_1[/imath] winning the championship is [imath]\frac{1088}{3645}[/imath] I have couple of doubts in this question after following the solution given by "Aretino" . Find the probability of [imath]P_1[/imath] winning the championship Its assumed that After a Parity occurs, Probability of winning [imath]P1[/imath] is [imath]p_0[/imath]. But there are different parity situations right? For example we have [imath]P_2 P_2 P_2P_2 P_2 P_1P_1P_1P_1P_1[/imath] also a parity situation.why cant we consider this case? am i missing something here? | 2174613 | Probability to win.
Two players [imath]A[/imath] and [imath]B[/imath] are playing the final of chess championship which contains a series of matches. The probability that [imath]A[/imath] wins is [imath]{2\over 3}[/imath]. The probability that [imath]B[/imath] wins is [imath]{1\over 3}[/imath]. The winner will be the one who is ahead by 2 games as compared to the other and wins at least 6 games. If the player [imath]B[/imath] wins the first 4 matches, find the probability that [imath]A[/imath] wins the championship. My attempt: We should find the following probabilities: [imath]A[/imath] wins 6 matches given [imath]B[/imath] has won the first 4. [imath]A[/imath] wins 7 matches given [imath]B[/imath] has won the first 4 and wins one out of the next 8. But I am not getting any patterns. |
2961574 | Proving [imath]A \cong C[/imath] if [imath]A \cong B[/imath] and [imath]B\cong C[/imath].
Question: Let [imath]A[/imath], [imath]B[/imath], [imath]C[/imath] be groups. Assume that [imath]A \cong B[/imath] and [imath]B\cong C[/imath]. Prove that [imath]A \cong C[/imath]. Proof: Let [imath]\phi: A \to B[/imath] and [imath]\sigma: B \to C[/imath] Suppose [imath]\sigma \circ \phi: A \to C[/imath] [imath]i). One-to-one:[/imath] Then [imath]\sigma \circ \phi[/imath] is one-to-one. Let [imath]a, b \in A[/imath] Suppose [imath](\sigma \circ \phi)(a) = (\sigma \circ \phi)(b)[/imath] [imath]\Rightarrow[/imath] [imath]\sigma (\phi(a)) = \sigma (\phi(b))[/imath] (Since [imath]\sigma[/imath] is one-to-one.) [imath]\Rightarrow[/imath] [imath](\phi(a)) = (\phi(b))[/imath] (Since [imath]\phi[/imath] is one-to-one.) [imath]\Rightarrow[/imath] [imath]a = b[/imath] [imath]ii). Onto:[/imath] Then [imath]\sigma \circ \phi[/imath] is onto. Let [imath]c \in C[/imath]. Because [imath]\sigma[/imath] is onto there exists [imath]b \in B[/imath] such that [imath]\sigma(b) = c[/imath]. Also since [imath]\phi[/imath] is onto there exists a [imath]a \in A[/imath] such that [imath]\phi(a) = b[/imath]. Then [imath]\sigma(\phi(a)) = \sigma(b) = c[/imath] Then [imath]\sigma \circ \phi[/imath] is onto. [imath]iii).Opreation - Perserving:[/imath] Let [imath]x_1, x_2 \in A[/imath]. Then [imath](\sigma \circ \phi) = \sigma(\phi(x_1, x_2)) = \sigma(\phi(x_1)\phi(x_2))[/imath] Because [imath]\phi(x_1) \in B[/imath] and [imath]\phi(x_2) \in B[/imath], [imath]\sigma(\phi(x_1)\sigma(\phi(x_2)) = (\sigma \circ \phi)(x_1)(\sigma \circ \phi)(x_2)[/imath] Then [imath]\sigma \circ \phi[/imath] is operation-preserving. [imath]\therefore A \cong C[/imath] I know it isn't the most elegant proof, but I was wondering if this is totally correct or if I missed something or did something incorrectly. Any help appreciated! | 1493279 | Show that [imath]\sigma\circ\phi : G_1 \to G_3[/imath] is an isomorphism
Let [imath]G_1, G_2[/imath] and [imath]G_3[/imath] be groups. Let [imath]\phi: G_1 \to G_2[/imath] and [imath]\sigma: G_2 \to G_3[/imath] be isomorphisms of groups. Show that [imath]\sigma\circ\phi: G_1 \to G_3[/imath] is an isomorphism. I understand to prove the composition is a homomorphism (operation preserving) and a bijection. I need help with notation and how to show the operation preserving, onto and one-to-one. Operation preserving: [imath]\sigma\circ\phi(ab)=\sigma(\phi(ab))=\sigma(\phi(a)\phi(b))=(\sigma\circ\phi(a))(\sigma\circ\phi(b))[/imath] Am I on the right track? |
1827820 | Proof that [imath]\overline{P(z)} = P(\overline{z})[/imath] for polynomial [imath]P[/imath] with real coefficients
Let [imath] a_0, a_1, a_2, a_3, \ldots , a_n \quad (n \ge 1)[/imath] denote real numbers, and let [imath]z[/imath] be any complex number. With the aid of [imath] \overline {z_1 +z_2+ \ldots +z_n} = \overline z_1 +\overline z_2+ \ldots + \overline z_n [/imath] and [imath] \overline {z_1 z_2 \ldots z_n} = \overline z_1 \overline z_2 \ldots \overline z_n, [/imath] Show that [imath] \overline {a_0 + a_1z+ a_2z^2+ \ldots + a_nz^n} = a_0 + a_1\overline z + a_2\overline z^2+ \ldots + a_n \overline z^n.[/imath] I was also told to use [imath]z[/imath] is a real number if and only if [imath]\overline z = z[/imath] and is [imath]z[/imath] is either real or pure imaginary if and only if [imath] \overline z ^2 = z^2[/imath] Any hints as to where to go? Does this proof take mathematical induction? I was thinking I could factor out [imath]a_0, a_1, etc)[/imath] and then use the fact that [imath]z[/imath] is real if and only if [imath]\overline z = z[/imath] Then say by the other statement above about the sum of the conjugates of complex numbers the two must be equivalent. I feel like this is right but it is not a very elegant proof. | 1221867 | Show if complex function [imath]f(z)[/imath] contains all real coefficients then [imath]f(\bar z)[/imath] = [imath]\bar {f(z)}[/imath]
Note: Suppose [imath]f(z)[/imath] is a polynomial or a proper/strictly proper rational function. Can someone demonstrate why if all coefficient of [imath]f(z)[/imath] are real then relation [imath]f(\bar z)[/imath] = [imath]\bar {f(z)}[/imath] hold Is there a theorem for this result? |
2961786 | Inequality regarding sum of rational powers of positive real numbers
Let [imath]a,b \ge 0[/imath] and [imath]\frac{m}{n} \le 1[/imath] then [imath]a^{\frac{m}{n}} + b^{\frac{m}{n}} \ge (a+b)^{\frac{m}{n}}[/imath]. How to prove this inequality. If [imath]\frac{m}{n} = \frac{1}{2}[/imath], I know how to prove this. The function [imath]f(x) = x^\frac{m}{n}, x \ge 0[/imath] is increasing for [imath]\frac{m}{n} \le 1[/imath]. How can I use the above fact? Can you help me out. This question : For [imath]a, b \geq 0[/imath], [imath]0 < x < 1[/imath], show [imath](a+b)^x \leq a^x + b^x[/imath] has an answer for [imath]a+b = 1[/imath]. | 1993528 | For [imath]a, b \geq 0[/imath], [imath]0 < x < 1[/imath], show [imath](a+b)^x \leq a^x + b^x[/imath]
Let [imath]a, b[/imath] be positive real numbers. Let [imath]0 < x < 1[/imath]. Show that [imath](a+b)^x < a^x + b^x[/imath]. The function [imath]f(x) = (a+b)^x -a^x - b^x[/imath] has [imath]f(0) = -1[/imath], [imath]f(1) = 0[/imath] and it seems to be an increasing function. But I am unable to show that it is increasing function. |
2962222 | Ideals and Field
Let [imath]A \neq0[/imath] be a ring. Then the following are equivalent i) A is a field ii) the only ideals in A are [imath]0[/imath] and [imath](1)[/imath] iii) every homomorphism of A into a non-zero ring B is injective. I have shown [imath]\ \ \ i)\rightarrow ii) \ \ [/imath] and [imath]\ \ \ iii)\rightarrow i) \ \ [/imath] But couldnt do [imath]\ \ \ ii)\rightarrow iii) \ \ [/imath] My thoughts: I think it is enough to show that for a homomorphism [imath]\phi: A\rightarrow B,\ \ \ [/imath] [imath]Ker \ \phi=(0)[/imath] but I dont know how can I show [imath]Ker \ \phi\neq(1)[/imath] which automatically gives us [imath]Ker \ \phi=(0)\ [/imath] as [imath]\ Ker \ \phi[/imath] can only be [imath](1)[/imath] or [imath](0)[/imath] | 820314 | All homomorphisms from a simple ring to a non-zero ring are injective
Let [imath]R[/imath] be a simple ring and [imath]T[/imath] be a non-zero ring. Let [imath]f\colon R \rightarrow T[/imath] be a ring homomorphism. Show [imath]f[/imath] is injective. Proof: [imath]\ker f \lhd R[/imath], so [imath]\ker f=R[/imath] or [imath]\ker f=\{0\}[/imath]. If [imath]\ker f=R[/imath] then [imath]R/R \cong \{0\}[/imath], but [imath]T \neq \{0\}[/imath], which is a contradiction. I can't see why does the last part means contradiction. Thanks for any assistance! |
2961815 | Why [imath] D_n[/imath] is not simple ?
Why [imath]D_n[/imath] is not simple ? i was asking this question to my professor . My professor said me that [imath]D_n[/imath] is not simple because subgroup genereated by [imath]< \frac{R_{360}}{n}>[/imath] is normal in [imath]D_n[/imath]. Actually i didn't understand the meaning Any hints/solution will be appreciated thanks u | 2205865 | Some Subgroup of Dihedral Group is Normal
I ran into this question when I was studying for my abstract algebra midterm. Show that the subgroup [imath]H[/imath] of rotations is normal in the dihedral group [imath]D_n[/imath]. Find the quotient group [imath]D_n/H[/imath]. I'm not quite sure where to begin. I know that for a Dihedral group of [imath]n\geq 3[/imath], then [imath]r^n=1[/imath] where [imath]r[/imath] is a rotation, and [imath]s^2=1[/imath] where [imath]s[/imath] is a reflection, and [imath]srs=r^{-1}[/imath]. I was not sure how to prove something is a normal subgroup from here. Any advice, thanks! |
2961854 | Show that [imath]\text{trace}(A)\leq \sqrt{n}\sqrt{\text{trace}(AA^t)}[/imath] where [imath]A[/imath] is in [imath]M_{n}(\mathbb{R}).[/imath]
Show that [imath]\text{trace}(A)\leq \sqrt{n}\sqrt{\text{trace}(AA^t)}[/imath] where [imath]A[/imath] is in [imath]M_{n}(\mathbb{R}).[/imath] We know that [imath]\text{trace}(A) =\sum_{i}\lambda_i[/imath] where [imath]\lambda_i[/imath] are eigenvalues of [imath]A.[/imath] Then I was wondering if we could show that the eigenvalues of [imath]AA^t[/imath] are [imath]\lambda_i^2[/imath] then we could apply the AM-GM inequality to complete the proof. So we know that [imath]A[/imath] and [imath]A^t[/imath] have the same eigenvalues, but can we use this conclude that [imath]AA^t[/imath] have [imath]\lambda_i^2[/imath] as eigenvalues? | 1226356 | Show the Trace[imath](B)^{2} \leq[/imath] nTrace[imath](B^{T}B)[/imath]
The following definition is needed for my actual question: For [imath]A, B \in \mathcal{M}_{n \times n}[/imath] define [imath] \langle A, B \rangle = \text{Trace}(B^{T}A) = \sum_{j=1}^{n}\sum_{i=1}^{n}b_{ij} \, a_{ij} [/imath] And here is the actual question: For [imath]B \in \mathcal{M}_{n \times n}[/imath] show that [imath] \text{Trace}(B)^{2} \:\leq\: n \,\text{Trace}(B^{T}B) [/imath] I know I have to use the Cauchy-Schwarz Inequality, but I wasn't exactly sure how to. Here is my attempt: Since I only have one matrix I decided to use the identity matrix [imath] \begin{align} \text{Trace}(B)^{2} \:&\leq\: n \,\text{Trace}(B^{T}B) \\ \text{Trace}(I^{T}B)^{2} \:&\leq\: n \, \langle B, B \rangle \\ \langle I_{n}, B \rangle^{2} \:&\leq\: n \, \Vert B \Vert^{2} \\ \langle I_{n}, B \rangle \:&\leq\: \sqrt{n} \, \Vert B \Vert \end{align} [/imath] I feel like I either did it incorrect, or that I'm close but am missing something. |
2961702 | Induction proof: [imath]\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}[/imath] [imath]<2[/imath]
Prove by induction the following. [imath]\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.[/imath] Caveat: The [imath]<[/imath] will be hard to work with directly. Instead, the equation above can be written in the form, [imath]\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}=2-\text{something}.[/imath] First, find the "something" and then use that form of the equation to prove the assertion. I can't seem to figure out the form that this equation can be written as. Also, once I find the form how would I do the proof. I understand it involves using a Basic Step and an Induction Step | 2393411 | Prove Finite Sum
To prove (1) [imath]\sum_{i=1}^n \dfrac {i}{2^i} <2[/imath] Someone shows that (2) [imath]\sum_{i=1}^n \dfrac {i}{2^i} = 2-\dfrac{n+2}{2^n}[/imath] so we can prove (1). Can anyone guide me how from (1) they come up with (2) please ? On what base they find (2) ? |
2962490 | Is there a Hamiltonian path? (i.e. can the general associative law be solved with Graph Theory?)
Consider the different bracketings of the summation [imath]1+2+3+4+5\,.[/imath] I've listed them all below: [imath] 1+(2+(3+(4+5)))\,,\quad (1+((2+3)+4))+5\\ 1+(2+((3+4)+5))\,,\quad (1+(2+(3+4)))+5\\ 1+((2+(3+4))+5)\,,\quad ((1+2)+(3+4))+5\\ 1+(((2+3)+4)+5)\,,\quad (1+2)+((3+4)+5)\\ 1+((2+3)+(4+5))\,,\quad (1+2)+(3+(4+5))\\ (1+(2+3))+(4+5)\,,\quad ((1+2)+3)+(4+5)\\ ((1+(2+3))+4)+5\,,\quad (((1+2)+3)+4)+5 [/imath] As we go down the first column, go up to the top of the second, and down the second column, we see that each of these bracketings are connected by a single use of the associative law. These different bracketings also exhaust the possibilities for [imath]1+2+3+4+5[/imath]. A similar pattern can be found for [imath]1+2+3[/imath] trivially and for [imath]1+2+3+4[/imath] almost trivially. These inspire the following graph-theoretic question: Consider the graph whose nodes consist of the various "valid bracketings" of the summation underneath [imath]1+2+\cdots+n.[/imath] (In general, the number of nodes for this graph will be the [imath]n-1[/imath]st Catalan number). Two nodes will be connected by an edge if we may travel between the two via one application of the associative law. Is there a Hamiltonian path in this graph? The graph for [imath]1+2+3+4+5[/imath] is a 14-node graph where every node has a degree of three. | 1408646 | Hamiltonian cycles in associahedron graphs
Let two distinct fully parenthesized products of [imath]n[/imath] symbols be called adjacent provided one of them may be obtained from the other by a single application of the associative law. Such graphs may be viewed as "associahedrons", though my question doesn't involve the graph theory approach so much, and in my case I assume for fixed [imath]n[/imath] that the graph contains as vertices all the parenthesized products. For example, for [imath]n=4[/imath] there are five vertices, each adjacent to [imath]2[/imath] others, and so essentially two possible Hamiltonian cycles. I found that in the case of [imath]n=5[/imath] it was possible to find a Hamiltonian cycle going through all [imath]14[/imath] vertices. (Here I think there are more than one such cycle...) My question is what happens for larger [imath]n.[/imath] I tried [imath]n=6[/imath] and got close, proceeding by using cycles from previous cases and "sewing them up" into a complete cycle. However I got lost in the details. So I'm curious about the [imath]n=6[/imath] case, and in general for higher [imath]n[/imath] is there a known cutoff beyond which it is known no Hamiltonian cycle exists. Associahedron reference at Wiki is here Here's the list I found for the [imath]n=5[/imath] case: [imath] (((12)3)4)5 \\ ((1(23))4)5 \\ (1((23)4))5 \\ (1(2(34)))5 \\ ((12)(34))5 \\ (12)((34)5) \\ (12)(3(45)) \\ 1(2(3(45))) \\ 1(2((34)5)) \\ 1((2(34))5) \\ 1(((23)4)5) \\ 1((23)(45)) \\ (1(23))(45) \\ ((12)3)(45) [/imath] Each is adjacent to the one below in the sense of one use of associative law, including the final one back to the first, to complete the cycle. |
2962456 | Help with finding [imath]\int_0^\infty \frac{\arctan(x)\log(x^2+1)}{x(x^2+1)}[/imath]
I am trying to evaluate [imath]\int_0^\infty \frac{\arctan(x)\log(x^2+1)}{x(x^2+1)}[/imath] I have tried differentiating under the integral sign, with different parameters, as well as contour integration, but I have not succeeded with neither of them. Wolfram Alpha gives a numerical value [imath]\approx 0.754694[/imath], but I suspect that there is a more exact answer. I would be very glad to get some hints on how to solve it. | 2128300 | A closed form for a triple integral with sines and cosines
[imath]\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz[/imath] I saw this integral [imath]I[/imath] posted on a page on Facebook . The author claims that there is a closed form for it. My Attempt This can be rewritten as [imath]3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz[/imath] Now consider [imath]F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz[/imath] Taking the derivative [imath]F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz[/imath] By symmetry we have [imath]F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2[/imath] Using W|A I got [imath]F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)[/imath] By integeration we have [imath]F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da[/imath] Let [imath]x = 2/a[/imath] [imath]\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx[/imath] Question I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ? |
2962942 | Simplify this expression: [imath] \left((p\to q) \land (q \to r)\right)\to (p \to r)[/imath]
how to prove that expression is a tautology by using law of logic?? Thank you in instant | 2954837 | How to prove that [imath][(p \to q) \land (q \to r)] \to (p \to r)[/imath] is a tautology without using the truth table?
I am looking for a way to prove that the statement, [imath][(p \to q) \land (q \to r)] \to (p \to r)[/imath], is a tautology without the help of the truth table. By using only Laws and Theorems like De Morgan's Law, Domination Law, etc. Also, I can't use the rules of inference. Please help, thank you. |
2962797 | Let [imath]p>0[/imath] be a constant. Prove that if [imath]E{|X|}^p<\infty[/imath], then [imath]x^pP(|X| > x)\to 0(x\to \infty)[/imath].
My idea is [imath]x^pP(|X|>x)=x^pEI(|X|>x)\le x^p E\frac{|X|^p}{x^p}I(|X|>x)\le E|X|^p<\infty[/imath].Then I have no idea. Is there anyone can help me? | 2711613 | Find limit of a random variable: [imath]\lim_\limits{s\to\infty} s^n P(|Y| > s)[/imath].
[imath]Y[/imath] is a random variable such that [imath]{E[|Y|^n]}<\infty[/imath] for [imath]n>0[/imath]. Find [imath]\lim_\limits{s\to\infty} s^n P(|Y| > s)[/imath]. The answer key gives the answer [imath]0[/imath] without any explanation. I would assume this involves the use of Cauchy-Schwarz inequality and the fact that [imath]E[Y^p]=\int p \, t^{p-1} \, S_x \,\mathrm{d}t[/imath], [imath]S_x[/imath] being the survivor function. Any help would be appreciated thanks. |
2963080 | Show that this set is a sigma algebra
Let [imath](X, \mathcal A)[/imath] be a measurable set, [imath]Y[/imath] a subset of [imath]X[/imath] Show that [imath]\mathcal A_Y$$= { A \cap Y : A \in \mathcal A } [/imath] is a [imath]\sigma[/imath]-algebra on [imath]Y[/imath] | 598591 | Proving a set is a Sigma Algebra
Let [imath]D[/imath] be a [imath]\sigma[/imath]-algebra of subsets of a set [imath]X[/imath], and let [imath]Y[/imath] be an arbitrary subset of [imath]X[/imath]. Let [imath]B= \{ A\cap Y | A \in D \}[/imath]. Show that [imath]B[/imath] is a [imath]\sigma[/imath]-algebra of subsets of [imath]Y[/imath]. If we let [imath]A=X[/imath], then clearly [imath]Y \in B[/imath]. Suppose [imath]G \in B[/imath]. Show [imath]G^c \in B[/imath]. So [imath]G = A\cap Y[/imath] for some [imath]A \in D[/imath]. So [imath]G^c = A^c \cup Y^c[/imath] I'm stuck here! Not sure how to show [imath]G^c \in B[/imath] Let [imath]G1,G2,G3,.... \in B[/imath]. Not sure how to show that the infinite union of the [imath]G_i[/imath]'s is an element of [imath]B[/imath]. |
2963119 | Sum of series [imath]\sum_{n=0}^\infty \frac{1}{n!(n+2)}[/imath]
I'm sure that this sum we should calculate with telescoping, but what algebraic conversions I must do? [imath]\sum_{n=0}^\infty \frac{1}{n!(n+2)}[/imath] | 581603 | Evaluate [imath]\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots[/imath]
Question is to Evaluate : [imath]\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots[/imath] what all i could do is : [imath]\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}[/imath] I have [imath]\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}[/imath] Now, I am not able to see what [imath]\sum_{n=1}^{\infty} \frac{n}{(n+2)!}[/imath] would be. I would be thankful if some one can help me to clear this. Thank you. |
2963233 | Prove that [imath]\sqrt{a}+ \sqrt{b} + \sqrt{c} + \sqrt{d} \le 10 [/imath]
This problem is from Antonio Caminha's book "Números reais" (real numbers), a Brazillian book for contest math preparation. Problem: If [imath]a[/imath],[imath]b[/imath],[imath]c[/imath] and [imath]d[/imath] are non-negative real numbers such that [imath]a \le 1[/imath], [imath]a + b \le 5[/imath], [imath]a + b + c\le 14[/imath] and [imath]a + b + c +d\le 30[/imath], prove that [imath]\sqrt{a}+ \sqrt{b} + \sqrt{c} + \sqrt{d} \le 10 [/imath] In the problem, it suggests to use the Abel's inequality. But I couldn't find out the solution, I just noticed that if we take [imath]a=1[/imath], [imath]b=4[/imath], [imath]c=9[/imath] and [imath]d=16[/imath], we have [imath]\sqrt{a}+ \sqrt{b} + \sqrt{c} + \sqrt{d} = 10 [/imath]. The problem is already solved here: Inequality question. However, there the problem is solved using advanced math (like polyhedral domain, stationary points, etc). Apparantly, this problem can be solved just with elementary math (things at the level of Abel's inequality, AM–GM inequality, etc), since the book is at high school level or even lower. | 871861 | Inequality question
[imath]a,b,c,d\ge 0[/imath] [imath]a\le 1[/imath] [imath]a+b\le 5[/imath] [imath]a+b+c\le 14[/imath] [imath]a+b+c+d\le 30[/imath] Prove that [imath]\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10[/imath]. We can subtract inequalities to get the answer, but that is wrong... I can't think of any another method... Any hints or suggestions will be helpful. |
2962626 | Exponential Matrix and commutator
I need help with a question. Let [imath]A[/imath] and [imath]B[/imath] be real or complex-valued matrices. Define [imath][A,B] = BA-AB[/imath]. Prove that if [imath][A,[A,B]]=[B,[A,B]]=0[/imath] then, for every [imath]t \in \mathbb{R}[/imath]: [imath]e^{tB}e^{tA}=e^{t(A+B)}e^{\frac{t^2}{2}[A,B]}[/imath] Hint: Verify that [imath]\Phi(t)=e^{-t(A+B)}e^{tA}e^{tB}[/imath] is a solution of [imath]X'=t[A,B]X[/imath]. I know how to prove that if [imath]\Phi[/imath] is a solution of the ODE then the identity holds. But I think I am miscalculating the derivative of [imath]\Phi[/imath]. I can't see how that [imath]t[/imath] would appear in the expression of the derivative. | 2370932 | Proof of a matrix exponential identity
Let [imath]A,B \in M_n(\mathbb{R})[/imath], and suppose [imath][A,[A,B]] = [B,[A,B]] = 0[/imath], where [imath][A,B] = BA-AB[/imath]. We want to show [imath] e^{At}e^{Bt}=e^{(A+B)t}e^{[A,B]t^2/2}, \quad \forall t\in \mathbb{R}. [/imath] I'm given a hint to show that [imath]x(t) =e^{-(A+B)t}e^{Bt}e^{At}x_0[/imath] is a solution to the ODE [imath]x'= t[A,B]x[/imath] for each [imath]x_0\in \mathbb{R}[/imath]. I haven't been able to prove the hint, let alone see why being a solution helps prove the identity. Taking the derivative of [imath]x(t)[/imath] didn't lead to much, and moreover there's no [imath]t[/imath] outside of the exponent, and substituting it into the RHS of the ODE also didn't help. The RHS of the identity to be shown solves [imath]x' = (A+B+t[A,B])x[/imath], which maybe gives some intuition about the ODE in the hint, but I'm not really sure where to go. |
2952894 | Show that, for square matrices [imath]A[/imath] and [imath]B[/imath], [imath]A+B=AB[/imath] implies [imath]AB=BA[/imath].
Let [imath]A[/imath] and [imath]B[/imath] be two [imath]n[/imath]-by-[imath]n[/imath] real matrices such that [imath]A+B = AB[/imath]. How do I prove that [imath]AB= BA[/imath]? I have tried using the trace function on [imath]A+B-AB[/imath]. But I could not get any Ideas. Kindly provide me with hints. | 818832 | Matrices A+B=AB implies A commutes with B
[imath]A[/imath] and [imath]B[/imath] are [imath]n\times n[/imath] matrices and [imath]A+B=AB[/imath]. I have an interesting proof that this implies [imath]A[/imath] commutes with [imath]B[/imath], but the proof only works when [imath]||B|| \lt 1[/imath]. I'm looking for a way to salvage the proof so that it works for all [imath]B[/imath]. [imath]A=AB-B[/imath] [imath]A=(A-I)B[/imath] by repeatedly substituting [imath]A[/imath] with [imath](A-I)B[/imath] on the RHS, [imath]A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i[/imath] Since the space of [imath]n\times n[/imath] matrices is Banach (hence complete) under the standard operator norm, the limit [imath]\sum_{i=1}^{\infty}B^i[/imath] is well defined given [imath]||B|| \lt 1[/imath]. Hence, taking the limit as [imath]N \rightarrow \infty[/imath], [imath]A = -\sum_{i=1}^{\infty}B^i[/imath] Now the LHS commutes with B, hence so does the RHS. Is there some trick to extend the method to work with any [imath]B[/imath] where [imath]||B|| \geq 1[/imath]? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged. |
2964232 | How to disprove the statement that n points divide a circle in [imath]2^{n-1}[/imath] regions?
The statement is: n points on a circle divide it in [imath]2^{n-1}[/imath] regions. I have used geometry to figure out that indeed, this statement is false, but I was wondering how to dis-prove this statement using simple and logical mathematics. What I did for this, is if [imath]n[/imath] points, then there will be [imath]^nC_2[/imath] lines. Intersections in the lines can be [imath]0[/imath] or [imath]n-1[/imath] or [imath]n[/imath] only. Am I going in the right direction and how should I go on with this to find something about the regions the lines divide the circle into. Edit: I don't want to dis-prove this with a counter example, I am thinking of doing this in the way I have started, with logic and simple mathematics. | 961537 | Number or regions formed when [imath]n[/imath] points on a circle are joined
The maximum number [imath]R_{n}[/imath] of regions formed when [imath]n[/imath] points on a circle are joined in pairs is [imath]\frac{1}{24}\left(n^{4}-6n^{3}+23n^{2}-18n+24\right)[/imath]. This is a fact that I have read in several essays on the dangers of jumping to conclusions in mathematics. In your opinion, what's the quickest (and/or nicest) way to prove this formula? Professor Paul Zeitz explains somewhere an especially convincing way to tackle this pearl following an idea which a certain high schooler in one of his talks (lectures?) came up with; yet, I don't remember where it was that I actually read this... Let me thank you in advance for your insightful replies. |
2964315 | [imath]\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx[/imath]
[imath]\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx[/imath] I tried putting [imath]x=\sin \theta[/imath] and changing the limits from [imath]0[/imath] to [imath]\frac{\pi}{2}[/imath]and i got [imath]\int_0^\frac{\pi}{2}\sin^2\theta\ln \sin \theta d\theta[/imath] i cannot further solve it | 1559380 | Seemingly Simple Integral [imath]\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx[/imath].
Evaluate [imath]\int_0^1 f(x) dx[/imath] where [imath]f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}[/imath] I started off with the substitution [imath]x=\sin y[/imath], which resulted in the integrand reducing to [imath]\sin^2y\cdot \ln (\sin y) dy[/imath] Then I used the property of definite integrals that [imath]\int_a^b f(x) dx = \int_a^b f(a+b-x) dx[/imath] Then too it wasn't getting simplified. I tried [imath]e^z=\sin x[/imath], but this gave no headway because after a while I reached a complete full-stop. How should I go about this? |
2964519 | Chinese remainder theorem with quadratics
I'm struggling to understand how the Chinese remainder theorem works on quadratics (though I know how it works in the simpler cases). For example in my number theory book there's an example with [imath]f(x) = x^{2} + x + 7[/imath] and we need to find all the solutions to [imath]f(x) \equiv 0[/imath] (mod [imath]189[/imath]) given that the roots mod [imath]27[/imath] are [imath]4, 13, 22[/imath] and the roots mod [imath]7[/imath] are [imath]0[/imath] and [imath]6[/imath]. There's a solution in the book to the example but I can't understand it very well. It says: "...we find that [imath]x \equiv a_{1}[/imath] (mod [imath]27[/imath]) and [imath]x \equiv a_{2}[/imath] mod([imath]7[/imath]) if and only if [imath]x \equiv 28 \cdot a_{1} - 27 \cdot a_{2}[/imath] (mod [imath]189[/imath])." And then the solution (there are six) follows easily by plugging in the solutions provided in the mod [imath]27[/imath] and mod [imath]7[/imath] cases listed above. But my main struggle is understanding how this equation [imath]x \equiv 28 \cdot a_{1} - 27 \cdot a_{2}[/imath] (mod [imath]189[/imath]) was derived, and how to get this equation in general for these problems (I think it involves Chinese remainder theorem and Euclidean algorithm but I can't figure it out). | 2052001 | Quadratic congruence in [imath] \mathbb{Z}/187\mathbb{Z} [/imath]
Given [imath]x^2 - \overline{51}x - \overline{43} = \overline{0}[/imath]. Solve it in [imath]\mathbb{Z}/187\mathbb{Z}[/imath]. First of all, does the [imath]\overline{x}[/imath] mean that [imath]\overline x = \{x + z \ | z\in I\}[/imath]? I am getting really confused, because 187 is composite. I think I should somehow decompose this equation into two equations and solve them in [imath]\mathbb{Z}/11\mathbb{Z}[/imath] and [imath]\mathbb{Z}/17\mathbb{Z}[/imath], but I'm not sure. And what is the pattern to solve things like that? |
2964573 | Solve: if [imath]\tan (a+b-c)/\tan (a-b+c)=\tan(c)/\tan(b) [/imath], prove that [imath]\sin 2a + \sin 2b + \sin 2c = 0[/imath]
My try : I can proceed from here but as you all know that it will take more than 1 page , but I want a short approach for it. | 2783259 | If [imath]\frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}[/imath], then [imath]\sin(b-c)=0[/imath] or [imath]\sin 2a + \sin2b + \sin 2c=0[/imath]
If [imath]\frac{\tan (\alpha + \beta - \gamma )}{\tan (\alpha - \beta + \gamma )}=\frac{\tan \gamma}{\tan \beta}[/imath] then prove [imath]\sin (\beta - \gamma)=0[/imath] or [imath]\sin 2\alpha + \sin2\beta + \sin 2\gamma =0[/imath] here my try goes - To prove [imath]\sin (\beta - \gamma)=0[/imath], sin x = 0, only when x = 0 or 360 degree, but for the sake of simplicity let's take x = 0, then we've to prove that [imath]\beta - \gamma =0[/imath] or [imath]\beta = \gamma[/imath]. Now, If [imath]\beta = \alpha[/imath], then first equation will be [imath]\frac{\tan \alpha}{\tan \alpha}= \frac{\tan\gamma}{\tan\beta}[/imath] or [imath]\frac{\tan\gamma}{\tan\beta}=1[/imath] and [imath]\frac{\tan x}{\tan y}=1[/imath], only if [imath]\tan x = \tan y[/imath] So, also in this case [imath]\tan \gamma[/imath] must be equal to [imath]\tan \beta[/imath] so, as to prove that [imath]\sin (\beta - \gamma)=0[/imath]. So, how to prove that [imath]\beta = \gamma[/imath]? and how to solve the second part of the question here? |
2964411 | Solution to equation [imath]2^x [/imath] = [imath]x^8[/imath]
I was able to solve this equation using graphical methods, but cannot figure out a mathematical solution to the equation. What approach should be taken to solve it? | 1937149 | Intuition for why equations of the form [imath]k^x=x^c[/imath] are not solvable trivially?
It recently occurred to me that I did not know how to solve equations of the form [imath]k^x=x^c[/imath] for any two constants [imath]k[/imath] and [imath]c[/imath]. After much pain in algebraically manipulating the equation (using logarithms) I confirmed the x is trapped in an exponent and cannot be extracted without some external knowledge. Finding the roots of the derivative creates a similar problem. I then did some research and discovered Lambert's W function, which I haven't the mathematical maturity to understand deeply. (Solve [imath]2^x=x^2[/imath]). I'm looking for some rigorous intuition towards why without numerical methods or Lambert's W function the equation is not solvable for x, or more specifically, a proof as to why this class of function is unsolvable using only logs and rationals? The best answer I have for myself is that because x is both a power and an exponent, either extracting x with a log or removing the exponent with a rational will keep the other side of the equation unsimplified. What is the mathematical language with which to express this idea? As an aside, I'm also wondering what area of mathematics considers issues such as these (this one being on the simpler side). |
2963734 | Showing that two random variables are independent
I have the following problem: Given two independent standard normal random variables, call them [imath]X[/imath] and [imath]Y[/imath], how can I show that [imath]Z = X^2 + Y^2[/imath] and [imath]W=\frac{X}{Y}[/imath] are also independent? I know that because [imath]X[/imath] and [imath]Y[/imath] are standard normal I can write their distributions as [imath]f_X(x) = \frac{1}{2\pi}e^\frac{-(x)^2}{2}[/imath] [imath]f_Y(y) = \frac{1}{2\pi}e^\frac{-(y)^2}{2}[/imath] Then because they are independent I can write their joint probability distribution as [imath]f_{X,Y}(x,y) = \frac{1}{2\pi}e^\frac{-(x^+y^2)}{2}[/imath] Now I know that to show [imath]Z[/imath] and [imath]W[/imath] are independent, I need to show that their joint probability distribution is equal to the product of the marginal distribution functions. But I'm not sure how to find [imath]f_{Z,W}(z,w)[/imath], [imath]f_Z(z)[/imath] and [imath]f_W(w)[/imath]. Any help is much appreciated. | 2959094 | If [imath]X[/imath] and [imath]Y[/imath] are independent [imath]N(0,\sigma^2)[/imath], then [imath]X^2+Y^2[/imath] and [imath]X/Y[/imath] are independent?
If [imath]X[/imath] and [imath]Y[/imath] are independent, then [imath]X^2+Y^2[/imath] and [imath]X/Y[/imath] are independent? I was solving the problem for the case that [imath]X[/imath] and [imath]Y[/imath] are independent [imath]N(0,\sigma^2)[/imath]. So i found that [imath]X^2+Y^2[/imath] is negative exponential, [imath]X/Y[/imath] is Cauchy distribution. How can i prove that [imath]X^2+Y^2[/imath] and [imath]X/Y[/imath] are independent? Also, what it will be if the case for general random variable [imath]X[/imath], [imath]Y[/imath] with [imath]X^2+Y^2[/imath] and [imath]X/Y[/imath] are defined? |
2964172 | find the answer in terms of [imath]a[/imath] and [imath]b[/imath] only ([imath]a, b[/imath] are roots of [imath]\ x^4 + x^3 - 1 = 0[/imath]
If [imath]a[/imath] and [imath]b[/imath] are the two solutions of [imath]\ x^4 + x^3 - 1 = 0[/imath] , what is the solution of [imath]\ x^6 + x^4 + x^3 - x^2 - 1 = 0[/imath] ? Well I am not able to eliminate or convert [imath]\ x^6[/imath]. Please help. | 301823 | If [imath]a[/imath] and [imath]b[/imath] are two roots of [imath]x^4 + x^3 - 1 = 0[/imath] prove that [imath]ab[/imath] is a root of [imath]x^6 + x^4 + x^3 - x^2 - 1[/imath].
If [imath]a[/imath] and [imath]b[/imath] are two roots of [imath]x^4 + x^3 - 1 = 0[/imath] prove that [imath]ab[/imath] is a root of [imath]x^6 + x^4 + x^3 - x^2 - 1[/imath]. Students and I are unsure how to go about this problem. Also will this be a problem I can solve and prove in front of a class in 20 minutes? |
2964966 | Prove V is a convex set
The exercise is as below Let [imath]V = \{x \in \mathbb{R}^2: 2x_1^2+3x_2^2\leq4\}[/imath] Prove: V is convex I started the proof like this: Let [imath]x,y \in V[/imath] be given. Let [imath]\lambda \in [0,1][/imath]. We want to show that [imath]\lambda x + (1-\lambda) y \in V[/imath]. Now, I know that I have to rewrite [imath]2[\lambda x_1 + (1-\lambda)y_1]^2+3[\lambda x_2 + (1-\lambda)y_2]^2[/imath] into something like [imath]\lambda(2x_1^2+3x_2^2) + (1-\lambda)(2y_1^2+3y_2^2) \leq \lambda*4+(1-\lambda)*4 = 4[/imath] But I just can't get my head around it, can someone please make a step-by-step derivation? I am really hopeless at this point. | 2546650 | Proving that the set enclosed by an ellipse is convex
I'm trying to prove that the following set ([imath]B[/imath]) defined by an ellipse is convex, but I am getting stuck. [imath]B = \left\{(x_1,x_2):3(x_1-3)^2+2(x_2-3)^2 \leq 1 \right\} [/imath] My sad attempt: [imath] \textrm{Let }x,y \: \in \: B \:, \: \lambda \: \in \: [0,1][/imath] [imath] \lambda x+(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\: \lambda x_2 +(1-\lambda)y_2 )[/imath] [imath]3(\lambda x_1 + (1-\lambda)y_1 -3)^2+2(\lambda x_2+(1-\lambda)y_2-3)^2=3(\lambda ^2x_1^2 +(1-\lambda)^2y_1^2+9+2\lambda x_1(1-\lambda)y_1-6\lambda x_1 -6(1-\lambda)y_1)+2(\lambda ^2x_2^2 +(1-\lambda)^2y_2^2+9+2\lambda x_2(1-\lambda)y_2-6\lambda x_2 -6(1-\lambda)y_2)[/imath] and this is where I have lost all confidence in proving it. I was able to factor some of the terms into the form [imath](\lambda x_1 -3)^2[/imath], [imath]((1-\lambda) y_1 -3)^2[/imath], [imath](\lambda x_2 -3)^2[/imath], and [imath]((1-\lambda) x_2 -3)^2[/imath], but I don't think that helps the situation any. |
2965856 | Is Euler's formula if-and-only-if?
It's a famous fact that for all connected, planar graphs, Euler observed (but failed, remarkably, to prove) that [imath]V-E+F=2[/imath] However, all of the proofs I have seen (including those I have created myself) are one-way, i.e. they show that if a graph satisfies the above conditions then the polyhedral formula holds (typically because the proof involves gradually deconstructing one's graph in a manner that leaves [imath]V-E+F[/imath] invariant). Does the converse in general hold? That is, if [imath]V-E+F =2[/imath] for some graph, is that enough to conclude that it's planar and connected? | 713232 | Question about Eulers formula [imath]v - e + f = 2[/imath]
Generally the theorem by Euler is stated: If [imath]G[/imath] is connected and planar then [imath]v - e + f = 2[/imath] (where [imath]v[/imath] is the number of vertices, [imath]e[/imath] is the number of edges and [imath]f[/imath] is the number of faces of the graph [imath]G[/imath]). My question is: Is this theorem an equivalence? I.e. is it true that if the equation [imath]v - e + f = 2[/imath] holds for a graph [imath]G[/imath] then [imath]G[/imath] is connected and planar. Really thankful for any help with this. I couldn't find a counter-example myself, (i.e. a graph for which [imath]v - e + f = 2[/imath] holds but is either not connected or not planar) but I guess the theorem would be stated as an equivalence if it were an equivalence... |
2965188 | If [imath]\gcd (a,b) = 1,[/imath] can any integer be written as a linear combination of [imath]a,b?[/imath]
I am thinking about this in the context of the two water jugs problem. I know that a jug of capacity [imath]n[/imath] can be filled if [imath]\gcd (a,b) \mid n.[/imath] Does this have the corollary that any integer can be written as a linear combination of [imath]a[/imath] and [imath]b[/imath] if [imath]\gcd (a,b) = 1?[/imath] | 1257448 | Any integer is an integral linear combination of two coprimes (Bezout identity for gcd)
Let [imath](p,q)=1[/imath] and let [imath]n,x,y \in \Bbb Z[/imath]. Then for [imath]\forall p, q. \exists x,y[/imath] such that [imath] n = xp + yq [/imath] It that valid and provable? Or does it come down to Goldbach's cojecture? |
2965603 | how to prove that given f(x) and g(x) is continuous then [imath]f*g(x)[/imath] is also continuous
Question how to prove that given f(x) and g(x) is continuous then [imath]f*g(x)[/imath] is also continuous All I have is that because we already know that f(x) and g(x) is continuous then [imath]|x-a|<\delta \implies |f(x)-f(a)| [/imath] and [imath]|g(x)-g(a)|<\epsilon[/imath] And I also know that [imath]\lim{_{x\to a}}{f(x)}g(x)=\lim{_{x\to a}}{g(x)}*\lim{_{x\to a}}{f(x)}=f(a)*g(a)[/imath] but I don't know what to do with those information | 1593789 | Prove that the product of two continuous functions is continuous
I was trying to prove an important theorem concerning continuous functions, namely that the product of two continuous functions is continuous. I am nearly at the end of the proof but I do not understand the last step my teacher made... I'm sure some of you is able to help me out :) Thanks in advance! Theorem Let [imath]f, g\colon \Bbb R \to\Bbb R [/imath] be continuous. Then [imath]F\colon \Bbb R\to\Bbb R[/imath] defined by [imath]F(x) = f(x)g(x)[/imath] is continuous. Proof Let [imath]f, g\colon \Bbb R \to\Bbb R [/imath] be given such that [imath]f[/imath] and [imath]g[/imath] are continuous. Let [imath]F\colon \Bbb R\to\Bbb R[/imath] be defined by [imath]F(x) = f(x)g(x)[/imath]. We want to show: [imath]F[/imath] is continuous, that is, for all [imath]a \in\Bbb R[/imath], for every [imath]\epsilon > 0[/imath], there exists some [imath]\delta > 0[/imath] such that for all [imath]x \in \Bbb R[/imath] with [imath]|x - a| < \delta[/imath], [imath]|F(x)-F(a)| < \epsilon[/imath]. Let [imath]a \in\Bbb R[/imath] be given. Let [imath]\epsilon > 0[/imath] be given. Let [imath]\delta_f > 0[/imath] be such that for all [imath]x \in\Bbb R[/imath] with [imath]|x-a| < \delta_f[/imath], [imath]|f(x)-f(a)| < \frac\epsilon{2(|g(a)| + \epsilon)}[/imath]. Such a [imath]\delta_f[/imath] exists since [imath]f[/imath] is continuous. Let [imath]\delta_g > 0[/imath] be such that for all [imath]x \in\Bbb R[/imath] with [imath]|x-a| < \delta_g[/imath], [imath]|g(x)-g(a)| < \frac\epsilon{ 2|f(a)| + 1}[/imath]. Such a [imath]\delta_g[/imath] exists since [imath]g[/imath] is continuous. Let [imath]\delta_3 > 0[/imath] be such that for all [imath]x\in\Bbb R[/imath] with [imath]|x-a| < \delta_3[/imath], [imath]|g(x)-g(a)| < \epsilon[/imath], which implies [imath]|g(x)| < |g(a)| + \epsilon[/imath]. Such a [imath]\delta_3[/imath] exists since [imath]g[/imath] is continuous. Choose [imath]\delta = \min\{\delta_f, \delta_g, \delta_3\}[/imath]. Then for all [imath]x \in\Bbb R[/imath] with [imath]|x-a| < \delta[/imath]: [imath]\begin{align}|F(x) - F(a)| &= |f(x)g(x) - f(a)g(a)| \\ &= |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)| \\ &\le |f(x)g(x) - f(a)g(x)| + |f(a)g(x) - f(a)g(a)|\\ &= |(g(x)[f(x) - f(a)]| + |f(a)[g(x) - g(a)]|\\&= |g(x)| * |f(x)-f(a)| + |f(a)| * |g(x) - g(a)| \\ &< [|g(x)| + \epsilon] \cdot \frac \epsilon {2(|g(a)| + \epsilon)} + |f(a)| \cdot |g(x) - g(a)| \\ &= \frac\epsilon{2 + |f(a)| \cdot |g(x) - g(a)|}\end{align}[/imath] Problem Now we need [imath]|f(a)| \cdot |g(x) - g(a)|[/imath] to be equal to [imath]\frac \epsilon 2[/imath] so that we have [imath]\frac\epsilon2+\frac\epsilon 2= \epsilon[/imath] (which completes the proof). The problem is I do not see why/how [imath]|f(a)| \cdot |g(x) - g(a)|[/imath] is equal to [imath]\frac\epsilon2[/imath]. Or is there maybe an error somewhere? |
2966585 | Euler totient function - [imath]|(Z/nZ)^*|=\phi(n)[/imath]
Prove that [imath]|(Z/nZ)^*|=\phi(n)[/imath] Can someone please direct me to that proof? I couldn’t find it on the internet. Thanks. | 1876938 | Show that number of elements in [imath](Z/nZ)^*[/imath] is [imath]\phi (n)[/imath]
Show that number of elements in [imath](Z/nZ)^*[/imath] is [imath]\phi (n)[/imath]. [imath](Z/nZ)^*[/imath] is set having those classes whose inverse exists. If we prove that [imath](Z/nZ)[/imath] has elements whose inverse exists are those who are relatively prime to n. Then my question will be proved So i write let [imath]gcd(a,n)=1[/imath] [imath]1= as+nt[/imath] [imath]as=1mod(n)[/imath] as ~ 1 so class of a . class of s = 1 i.e inverse of class of a exists Kindly suggest is this correct or some alternative option Thanks |
2966631 | Is it true that every cyclic group of order [imath]n[/imath] contains exactly the same number of subgroups as the number of divisors of [imath]n[/imath]?
The fundamental theorem of cyclic groups says that a for a cyclic group [imath]G[/imath] where [imath]|G|=n[/imath], for every divisor [imath]d[/imath] of [imath]n[/imath], there is a unique subgroup of [imath]G[/imath] with order [imath]d[/imath]. My question is that whether these are the only subgroups of any cyclic groups? In other words, is it true that [imath]G[/imath] contains exactly [imath]m[/imath] subgroups where [imath]m[/imath] is the number of divisors of [imath]|G|[/imath]? | 2907720 | Show that a finite cyclic group of order [imath]n[/imath] has exactly one subgroup of order [imath]d[/imath] dividing [imath]n[/imath] and these are all the subgroups it has
My attempt Note- Addition is under [imath]mod[/imath] [imath]d[/imath] Let [imath]G = (a^0,a^1,a^2.....a^{n-1})[/imath] [imath]n=kd[/imath] Let [imath]H_d = (a^{0k},a^{1k},a^{2k}......a^{{(d-1)}k})[/imath] For [imath]a^{z_1k},a^{z_2k} \in H_d[/imath] [imath]a^{z_1k}a^{z_2k} = a^{{(z_1+z_2)}k} \in H_d[/imath] [imath]H_d[/imath] is closed For [imath]a^{zk} \in H_d[/imath] [imath]a^{zk} (a^{zk})^{-1}=a^0[/imath] [imath](a^{zk})^{-1} = (a^{-1})^{zk}[/imath] Since [imath]a^{-1}[/imath] is [imath]a^{n-1}[/imath] we can rewrite it as [imath](a^{zk})^{-1} = (a^{n-1})^{zk}=a^{z(n-1)k} \in H_d[/imath] Since [imath](a^{zk})^{-1} \in H_d[/imath] and [imath]H_d[/imath] is closed, [imath]H_d[/imath] is a subgroup. I can easily prove the 3rd part by using Lagrange's theorem. I have no idea how to prove that only [imath]1[/imath] subgroup of order [imath]d[/imath] exists. Maybe by proving that the elements of the [imath]2[/imath] groups have to be the same? I would really appreciate if someone could verify my proof so far and give some hints to complete the proof. |
2966450 | Find the partial sum of the series and the limit of it [imath]\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}[/imath]
Find the partial sum of the series and the limit of it: [imath]\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}[/imath] I did [imath]\lim_{n\to\infty}\frac{1}{2^n}\tan\frac{1}{2^n} = 0[/imath] so we can not say that the series is divergent. I tried to use telescopic sum but I do not know what to do with that [imath]\tan[/imath]. Can you give me any hint how to write it? UPDATE So I did what @lab bhattacharjee recommended and I got that: [imath]\frac{1}{2}\tan x = \frac{1}{\tfrac{1}{\tan\frac{x}{2}}-\tan\frac{x}{2}}\iff\tan x = \frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}=\frac{1+\tan\frac{x}{2}+\tan\frac{x}{2} - 1}{(1+\tan\frac{x}{2})(1-\tan\frac{x}{2})} = \frac{1}{1-\tan\frac{x}{2}}-\frac{1}{1+\tan\frac{x}{2}}.[/imath] What should I do next? | 1488921 | Evaluate [imath]\sum\limits_{n=1}^{\infty} \frac{1}{2^n}\tan\frac{x}{2^n}[/imath]
Evaluate [imath]\sum\limits_{n=1}^{\infty} \frac{1}{2^n}\tan\frac{x}{2^n}[/imath]. My attempt: Let [imath]S[/imath] = [imath]\sum\limits_{n=1}^{\infty} \frac{1}{2^n}[/imath] [imath]tan{\frac{x}{2^n}}[/imath] [imath]\int S dx[/imath] = [imath]\sum\limits_{n=1}^{\infty}[/imath] [imath]\int \frac{1}{2^n}[/imath] [imath]tan{\frac{x}{2^n}} dx[/imath] = [imath]\sum\limits_{n=1}^{\infty}[/imath] [imath]\ln \sec {\frac{x}{2^n}}[/imath] = [imath]\ln(\prod_{n=1}^{\infty}\sec {\frac{x}{2^n}})[/imath] not sure how to proceed with this, Due to the limitations of my school syllabus, I would greatly appreciate a solution involving complex numbers. Thanks |
1000530 | If the product of two continuous functions is zero, must one of the functions be zero?
Suppose that I have two continuous functions [imath]f : \left[ a, b \right] \rightarrow \mathbb{R} \quad \text{and} \quad g : \left[ a, b \right] \rightarrow \mathbb{R}[/imath] and they have the following property [imath]f(x) \times g(x) = 0 \space , \forall x \space \in \left[ a, b \right][/imath] Can I say that one of the functions necessarily has to be equal to [imath]0[/imath]? For example, [imath]f(x) = 0 \space , \forall x \space \in \left[ a, b \right][/imath]. UPDATE: Ok, I can see from the counterexamples that the affirmation is not true, but now I cannot see in which cases it is true. If I let the function [imath]g : \left[ a, b \right] \rightarrow \mathbb{R}[/imath] be any continuous function, then in that case must I have [imath]f(x) = 0[/imath] ? | 677402 | product of two non zero continuous function is zero
Can you give me examples of two functions [imath]f[/imath] and [imath]g[/imath] such that both are non-zero continuous function but their product is zero. |
2966889 | Prove that [imath]G[/imath] is isomorphic to [imath]\text{Inn}(G)[/imath] if [imath]G[/imath] has trivial center [imath]Z(G)[/imath].
I know the isomorphism is [imath]F: G \rightarrow \text{Inn}(G)[/imath] defined as [imath]F(a) = \phi_a[/imath] where [imath]\phi_a(x) = axa^{-1}[/imath] but I'm not sure how to use the assumption (that [imath]G[/imath] has trivial center) to prove injectivity and surjectivity. I got as far as showing it was a homomorphism but just having trouble with the last part. This is what I have for the proof of a bijection: Suppose [imath]G[/imath] is a group such that [imath]Z(G) = \{e\}[/imath] and let [imath]F : G \rightarrow \text{Inn}(G)[/imath] be defined by [imath]F(a) = \phi_a[/imath]. I claim that [imath]F[/imath] is injective by proving that if [imath]F(a) = F(b)[/imath], then [imath]a=b[/imath] (i.e. that [imath]F(a)[/imath] and [imath]F(b)[/imath] are the same function). Suppose for some [imath]a,b,x \in G[/imath] with [imath]x \neq e[/imath] that [imath]F(a)(x) = F(b)(x) \implies \phi_a(x) = \phi_b(x) \implies axa^{-1} = bxb^{-1} \implies ax = bxb^{-1} a \implies x = a^{-1} bxb^{-1} a \implies a^{-1} b = e \implies b = a[/imath]. Do I need to make any other assumptions on my choice of [imath]x[/imath] (should I even assume [imath]x \neq e[/imath]?) or [imath]a[/imath] and [imath]b[/imath]? We have not discussed quotient groups or kernel and I would like to prove it by showing that [imath]F[/imath] is a bijective homomorphism. | 2962740 | Detail in bijection between Inn(G) and G/Z(G)?
I've seen other posts (Inner automorphisms form a normal subgroup of [imath]\operatorname{Aut}(G)[/imath]) about the topic of [imath]\operatorname{Inn}(G) \simeq G/Z(G)[/imath], but what I want to ask is a detail. When we ensure that [imath]\operatorname{Inn}(G) \simeq G/Z(G)[/imath], we want to say that there exists some isomorphism [imath]F[/imath] such that: [imath]F: G/Z(G) \rightarrow \operatorname{Inn}(G), \quad F(gz) = \tau_g[/imath] where [imath]g \in G, z \in Z(G)[/imath] and [imath]\tau_g(h) = ghg^{-1}[/imath] for all [imath]h \in G[/imath] But is this [imath]F[/imath] really an isomorphism? I'm not pretty sure due to there is not ONE [imath]\tau_g[/imath] for ONE [imath]gz[/imath], what we have is a total of [imath]|Z(G)| = \text{order of }Z(G)[/imath] elements of the form [imath]gz[/imath] for just ONE [imath]\tau_g[/imath] due to [imath]\tau_g = \tau_{gz}[/imath] if [imath]z \in Z(G)[/imath] |
2966312 | Prove inequality for complex numbers
Can anyone help me or at least give me some hints how to prove following inequality : [imath] \lvert z-a\rvert \le \lvert \lvert a\rvert-\lvert z\rvert\rvert + \lvert z\rvert*\lvert arg (z/a)\rvert [/imath] for any [imath] z, a \in \Bbb C, a \neq 0[/imath] | 500541 | How to prove this inequality? [imath] | z-1 | \le | | z | -1 | + | z| \cdot | \arg z | [/imath]
If [imath]z[/imath] is any non-zero complex number, how to prove the following inequality? [imath] | z-1 | \le | | z | -1 | + | z| \cdot | \arg z | [/imath] Hints please! |
2967701 | How solve this lim?
I need to know the limit of this sequence: [imath]\lim (a^n +b^n)^{1/n}[/imath] for [imath]a,b>0[/imath] I suppose that this limits will be in function of a and b, but how I can solve it? | 1490722 | Simple Calculus Proof
I'm having some difficulty with this proof. Could you help me? Let [imath]a,b>0[/imath] and [imath]c_n = (a^n + b^n)^{1/n}[/imath]. Show that [imath]\lim_{n\to \infty}c_n = \max \{a,b\}.[/imath] Thank you. |
2967747 | Solve the limit: [imath] \lim\limits_{n\to\infty}\prod\limits_{1\leq k\leq n}\left( 1+\frac{k}{n} \right)^{\frac{1}{k}} .[/imath]
Solve the following limit: [imath] \lim\limits_{n\to\infty}\prod\limits_{1\leq k\leq n}\left( 1+\frac{k}{n} \right)^{\frac{1}{k}} .[/imath] Here is what I do: Take the logarithm: [imath]\begin{align} \lim\limits_{n\to\infty}\log \prod\limits_{1\leq k\leq n}\left( 1+\frac{k}{n} \right)^{\frac{1}{k}}&=\lim\limits_{n\to\infty}\sum_{k=1}^{n}\frac{\log (1+\frac{k}{n})}{\frac{k}{n}}\frac{1}{n}\\ &=\int_{0}^{1}\frac{\log (1+x)}{x}dx \end{align}[/imath] Then I am stuck. How to integrate [imath] \int_{0}^{1}\frac{\log (1+x)}{x}dx [/imath]? The question is from: (8) of https://math.uchicago.edu/~min/GRE/files/week4.pdf Edit: Different approaches are very welcome! The hint says 'estimate from above and below'. Maybe someone can provide a solution without integration? | 183551 | Compute [imath]\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{{1}/{2}}\cdots\left(1+\frac{n}{n}\right)^{{1}/{n}}[/imath]
Compute the limit: [imath]\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots\left(1+\frac{n}{n}\right)^{\frac{1}{n}}[/imath] |
2964071 | Explicit solution of [imath]XA=BX[/imath]
Consider the equation [imath]XA=BX,[/imath] where [imath]X,A,B[/imath] could be matrices or (pseudo) differential operators. How to represent [imath]X[/imath] using [imath]A,B[/imath], that is [imath]X=f(A,B),[/imath] where [imath]f[/imath] is a known function. | 2189116 | Solution to matrix equation PX=YP
Consider three [imath]n \times n[/imath] real valued matrices, [imath]P[/imath], [imath]X[/imath], and [imath]Y[/imath] and the following equation: [imath]PX=YP[/imath], where [imath]P[/imath] is the unknown term. Under what conditions is this equation solvable? Is it there a closed-form solution? Your help is greatly appreciated. |
2968141 | How to calculate the summation of [imath]n \cdot 2^n[/imath]?
So I know that you can take the derivative of this and multiply by x and do integrals or something like that. However, I am just wondering if there is a way to come to the summation of the series without doing derivatives or integrals. Is there a simple way or would that way just be guess and check? | 2906492 | Solving summations of form [imath]\sum_{i=0}^{\infty} i^kx^i[/imath] without Differentiation and Integration
The actual sum I'm looking to solve is [imath]\sum x^i(i^3-(i-1)^3)=(1-x)\sum x^ii^3[/imath] But I'd rather do a general answer first. I know the method of differentiation and integration but I'd like to do it without that. I remember seeing a method which reduced the power of [imath]i[/imath] on Brilliant but I can't find it anymore. |
2967952 | How to prove [imath]x^5 + x^4 + x^3 + x^2 + 1 = 0[/imath] has no rational roots?
I know I can set [imath]x = \frac{p}{q}[/imath] and then multiply everything and then show there is a contradiction but is there an easier way to do this? like by some theorem? | 1051759 | Prove [imath]x^5 + x^4 + x^3 + x^2 + 1 = 0[/imath] has no rational solution
Prove [imath]x^5 + x^4 + x^3 + x^2 + 1 = 0[/imath] has no rational solution I want to prove it by Proof by Contradiction, but I am not sure how to proceed with the proof. |
2968157 | Convexity of a function over matrices
How to show that [imath]f(X)=tr(X^{-1})[/imath] is convex over the set of symmetric real positive definite matrices by showing that [imath]g(t)=f(X+tV)[/imath] is convex for any such [imath]X[/imath] and [imath]V[/imath]? Clearly we need to show that [imath]g(t)\le (1-t)g(0)+tg(1)[/imath] but I'm stuck to find a simplified version of [imath]tr((X+tV)^{-1})[/imath]. Thank you for your help! | 297635 | Is the trace of inverse matrix convex?
Hi I would like to know whether the trace of the inverse of a symmetric positive definite matrix [imath]\mathrm{trace}(S^{-1})[/imath] is convex. Actually I know that the trace of a symmetric positive definite matrix [imath]S\in M_{m,m}[/imath] is convex since we can find [imath]B\in M_{n,m}[/imath] such that [imath]S=B^T\times B[/imath] then we can write the trace as the sum of scalar quadratic forms, i.e. [imath]\mathrm{trace}(S)=\mathrm{trace}(B^T\times B)=\sum_{j=1}^mb_j^T\times b_j[/imath] where [imath]b_j[/imath] is the [imath]j^{th}[/imath] column of [imath]B[/imath]. for instance if we have [imath]trace([\begin{array}{cc} 1 & 2 \\ 3 & 4 \\ \end{array}] \times [\begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array}])= [\begin{array}{cc} 1 & 2 \\ \end{array}]\times [\begin{array}{c} 1 \\ 2 \\ \end{array}]+ [\begin{array}{cc} 3 & 4 \\ \end{array}]\times [\begin{array}{c} 3 \\ 4 \\ \end{array}]=30[/imath] And so I wonder if [imath]\mathrm{trace}(S^{-1})[/imath] is convex too.. |
2961466 | Lie group embeddings [imath]SU(5) \supset SU(3) \times SU(2) \times U(1)?[/imath]
Does the special unitary Lie group [imath]SU(5)[/imath] contains [imath]SU(3) \times SU(2) \times U(1)[/imath] as a subgroup? Can one show which of the following embeddings are possible rigorously: [imath]SU(5) \supset SU(3) \times SU(2) \times U(1)?[/imath] [imath]SU(5) \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_2}?[/imath] [imath]SU(5) \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3}?[/imath] [imath]SU(5) \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3 \times \mathbb{Z}_2}?[/imath] Here [imath]\mathbb{Z}_n[/imath] means the cyclic group of order [imath]n[/imath]. it looks to me that the first one is impossible, can we prove it is a no go? (p.s. [imath]SO(10)[/imath] and [imath]SU(5)[/imath] are different groups - how can the two questions be duplicate to each other, without knowing the way of embedding?) | 2961471 | Lie group embeddings [imath]SO(10) \supset SU(3) \times SU(2) \times U(1)?[/imath]
Let [imath]G[/imath] be [imath]SO(10)[/imath] or [imath]Spin(10)[/imath]. Does either of them [imath]G=SO(10)[/imath] or [imath]G=Spin(10)[/imath] contain [imath]SU(3) \times SU(2) \times U(1)[/imath] as a subgroup? Can one show which of the following embeddings are possible rigorously: [imath]G\supset SU(3) \times SU(2) \times U(1)?[/imath] [imath]G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_2}?[/imath] [imath]G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3}?[/imath] [imath]G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3 \times \mathbb{Z}_2}?[/imath] Here [imath]\mathbb{Z}_n[/imath] means the cyclic group of order [imath]n[/imath]. it looks to me that the first one is impossible, can we prove it is a no go? for [imath]G=SO(10)[/imath] or [imath]G=Spin(10)[/imath]? |
2968110 | Minimal prime of a ring is contraction of some prime ideal of any extension of the ring.
Let [imath]S[/imath] be a commutative ring with [imath]1[/imath] and let [imath]R[/imath] be any subring of [imath]S.[/imath] Let [imath]p[/imath] be a minimal prime of [imath]R.[/imath] Then how can I show that there is a prime ideal [imath]q[/imath] of [imath]S[/imath] whose contraction is [imath]p,[/imath] i.e., [imath]q \cap R = p.[/imath] What I am trying to show is the necessary and sufficient condition that [imath]p=p^{ec}.[/imath] Clearly [imath]p \subset p^{ec}.[/imath] But the other way I couldn't complete. I need some help. | 935987 | A property of minimal prime ideal
Let [imath]R[/imath] commutative ring with unity, [imath]S\subseteq R[/imath] subring, [imath]p[/imath] minimal prime ideal of [imath]S[/imath]. Show there exists a minimal prime ideal [imath]q[/imath] in [imath]R[/imath] with the property that the contraction [imath]q^c=q\cap S=p[/imath]. First of all, I am not sure whether the minimality of [imath]q[/imath] refers to the prime ideals in [imath]R[/imath] or to the prime ideals with such contraction property, if the latter is the case, then we only need to show the existence of such prime ideal, minimal one would be given by Zorn's lemma. Second, if we drop the minimality condition, then the proposition clearly doesn't hold ([imath]\mathbb{Z}\subseteq\mathbb{Q}[/imath] for instance), so this condition must be crucial here. |
2969115 | is the sequence [imath]\{a_n\}[/imath] is strictly increasing or strictly decreasing?
Given [imath]a_n = (1 + \frac{1}{n})^{n+1}[/imath] then the sequence [imath]\{a_n\}[/imath] is strictly increasing or strictly decreasing ? My attempts : [imath]\sqrt [n+1] e > 1 + \frac{1}{n}[/imath] that is [imath]e >(1 + \frac{1}{n})^\frac{1}{n+1}[/imath] so [imath]\{a_n\}[/imath] is strictly increasing Pliz verified Am i right /wrong ? Any hints/solution will be appreciated thanks u | 926076 | Monotonicity of the sequences [imath]\left(1+\frac1n\right)^n[/imath], [imath]\left(1-\frac1n\right)^n[/imath] and [imath]\left(1+\frac1n\right)^{n+1}[/imath]
I am working on the following sequences. [imath]x_n=\left(1+\frac1n\right)^n \qquad z_n=\left(1-\frac1n\right)^n \qquad y_n=\left(1+\frac1n\right)^{n+1}[/imath] I am trying to prove that [imath]x_n[/imath] and [imath]z_n[/imath] are increasing sequences and [imath]y_n[/imath] is a decreasing sequence. I proved that [imath]x_n[/imath] is increasing using the binomial theorem: as [imath]n[/imath] increases, the number of terms in my sum increases. Also, for [imath]k\ge1[/imath] and [imath]n\ge k[/imath] the [imath](k+1)th[/imath] term in our sum is [imath]\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\left(\frac{1}{n}\right)^k[/imath] which can be rewritten as [imath]\frac{1}{k!}\left[\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}{n}\right)\right]\qquad(1)[/imath] Here, the product in [imath](1)[/imath] increases as [imath]n[/imath] increases. I am in trouble proving that [imath]z_n[/imath] and [imath]y_n[/imath] are decreasing. Could you suggest me an approach to this problem please? Thank you. Edit Since I obtained an answer for [imath]y_n[/imath], I kept working on [imath]z_n[/imath]. I took the continuous function [imath]z(x)=(1-\frac1x)^x=\exp \left\{x\log(1-\frac1x)\right\}[/imath]. Then I derived it to obtain [imath]z'(x)=\exp\left\{x\log(1-\frac1x)\right\}\left[\log(1-\frac1x)+\frac{x}{(x-1)/x}\cdot\frac{1}{x^2}\right][/imath] But [imath]z'(x)=\exp\left\{x\log(1-\frac1x)\right\}\left[\log(1-\frac1x)+\frac{1}{x-1}\right][/imath] [imath]e^a[/imath] is always positive. Hence the sign of [imath]z'(x)[/imath] depends on what we have between square brackets, which is greater than zero (I don't know how to prove it; I used a plot) for [imath]x\ge1[/imath]. Hence [imath]z_n[/imath] is increasing. |
2970149 | Prove that if [imath]f(x) > 0 \; \forall \;x \in (a, b)[/imath], then [imath]\int_a^b{f(x) \; \mathrm{d}x} > 0[/imath].
Question is in the title. It appears obvious at first glance, but I've got no idea how to prove it. Edit: I am referring to the Riemann integral. | 1693153 | Show that the integral of a positive function is positive
Suppose [imath]f:[0,1] \to (0,\infty)[/imath] is a Riemann integrable function. Prove that the integral of the function from [imath]0[/imath] to [imath]1[/imath] is strictly positive. I have been trying to do this for awhile but I can't seem to get it. Here is my thought process: If the function is Riemann integrable, then its set of discontinuities has measure zero (Lebesgue). I am not sure how to connect this to the integral being positive |
2970197 | Is the function [imath]f(x) = e^x \cos (1/x)[/imath] is uniformly continuous on [imath]A = (0,1)[/imath].
The question and its answer is given below: Determine whether t the function [imath]f(x) = e^x \cos (1/x)[/imath] is uniformly continuous on [imath]A = (0,1)[/imath]. but I am wondering why the [imath]n[/imath] is chosen like this, and if it should be [imath]N[/imath]? and why [imath]1/n\pi[/imath] is [imath]\leq \delta?[/imath] could anyone explain this for me please? | 2970165 | Determine whether the function [imath]f(x) = e^x \cos (1/x)[/imath] is uniformly continuous on [imath]A = (0,1)[/imath].
Determine whether the function [imath]f(x) = e^x \cos (1/x)[/imath] is uniformly continuous on [imath]A = (0,1)[/imath]. I see that the function [imath]f(x) = e^x \cos (1/x)[/imath] is uniformly continuous on [imath]A = (0,1)[/imath]. Am I correct? My answer: since the function [imath]\cos (1/x)[/imath] is continuous on [imath]\mathbb{R} - \{0\}[/imath] so it is continuous on any closed interval not containing [imath]0[/imath] say [imath][a,b][/imath], [imath]a\neq 0[/imath]. Since [imath]e^x[/imath] is continuous for all [imath]x[/imath] then we can say that [imath]f(x) = e^x \cos (1/x)[/imath] is continuous on any interval [imath][a,b][/imath] such that [imath]a\neq 0[/imath]. So [imath]f[/imath] is continuous on any subset of [imath][a,b][/imath] so it is uniformly continuous using the theorem that: A continuous function on an interval [imath][a,b][/imath] is uniformly continuous . Is my solution correct? If not could you please provide me with the correct solution? |
2970840 | Find optimal estimator for exponential distributions.
Consider [imath]X_{1} \dots X_{n}[/imath] are i.r.v with [imath]Exp(\theta)[/imath] ([imath]p_{\theta} = \theta e^{-\theta x}[/imath]). We want to find optimal estimator of [imath]\theta[/imath] (i.e. unbiased estimator with minimal variance in unbiased estimator class). Our teacher tells us that it can be done with using sufficient and completeness statistics. It's easy to determine that [imath]X_{(1)}[/imath] is satisfy this properties. Now we want to make it unbiased. We need to find some [imath]f(x)[/imath], such as : [imath]\mathbb{E}(f(X_{(1)}) = \theta[/imath]. That is a problem moment. Actually I'm thinking that there is no such functions. I've tried to consider some easy examples : [imath]f(x) = x^{m}, f(x) = \log{x}[/imath] and [imath]f(x) = e^{x}[/imath] but it give's me only [imath]\theta^{a}[/imath] ,where [imath]a \ne 1[/imath] or functions like [imath]\frac{\theta^{m}}{\theta^{2} + const}[/imath] . Any hint's will be good . | 2034206 | Finding UMVUE of [imath]\theta[/imath] when the underlying distribution is exponential distribution
Hi I'm solving some exercise problems in my text : "A Course in Mathematical Statistics". I'm in the chapter "Point estimation" now, and I want to find a UMVUE of [imath]\theta[/imath] where [imath]X_1 ,...,X_n[/imath] are i.i.d random variables with the p.d.f [imath]$f(x; \theta)=\theta e^{-\theta x}, x\gt0$[/imath]. I know that [imath]E(X_i)=1/\theta,[/imath] for each [imath]i[/imath], and also have that [imath]\bar{X}[/imath] (or equivalently [imath]\sum_1^n X_i[/imath]) is a complete sufficient statistic for [imath]\theta[/imath]. But I cannot go any further here. Somebody can help me? |
2971078 | Proof that [imath]\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}[/imath]
I saw in this paper the following identity: [imath]\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}[/imath] I have a pervious post on an integral quite closely related to this identity, but I still do not know how to derive/prove the identity. I'm really not that good at combinatorics or evaluating series, so I don't know how to start, which is why I don't have any attempts to show you. Please explain your steps thoroughly. Edit: I know there other posts on this series, but I did not get from them the proof I was satisfied by. | 2956365 | How to apply induction to this formula?
I want to justificate following equation: [imath]\sum_{k=0}^n \frac{(-1)^k}{k!(n-k)!}\frac{1}{2k+1} = \frac{2^n}{(2n+1)!!}[/imath] I calculated the both sides for [imath]n[/imath] from 1 to 10 and it was true. How the mathematical induction can be applied to this equation? Or is there other way to justificate it? |
2970603 | Integration of [imath]\frac{1}{\log(x)}[/imath]?
How to integrate [imath]\frac{1}{\log(x)}[/imath]? I have tried integration by parts, but it is a never ending series with no specific general term. PS: It is indefinite integration. | 714020 | Integration of [imath]\frac{1}{\log(x)}[/imath]
Please integrate [imath]\frac{1}{\log(x)}[/imath] and illustrate the steps of your method of integration. I have already tried integration by parts following the ILATE rule and otherwise. Eventually it forms a loop that takes me back to the direct or substituted term where I started the process of integration. |
2967487 | [imath]H^*(\mathbb{G}(k,n))[/imath] free with basis the partitions that index the Schubert varieties
I have to proof the fact that [imath]H^*(\mathbb{G}(k,n))[/imath] is, as an abelian group, free with basis the partitions that index the Schubert varieties, but I'm having trouble doing it myself. Denote [imath]\sigma_{\lambda}=[X_\lambda][/imath] the Schubert class associated to a partition [imath]\lambda \subset m\times n[/imath]. For all partitions [imath]\lambda[/imath] contained in an [imath]m\times n[/imath] rectangle, the Schubert class [imath]\sigma_{\lambda}[/imath] is an element of [imath]H^{2|\lambda|}(\mathbb{G}(m,m+n))[/imath], and we have the decomposition [imath]H^*(\mathbb{G}(m,m+n))=\bigoplus_{\lambda \subset m\times n} \mathbb{Z}\sigma_\lambda[/imath] Thanks in advance Edit: The question linked does not have an answer and the notes contain a dead link. Moreover the question asks for an example while I need a more general proof or at least an outline for it. Edit 2: How can I use the fact that [imath]\Omega_\lambda \simeq \mathbb{C}^{mn-|\lambda|}[/imath] and [imath]\mathbb{G}(m,n)[/imath] is the disjoint union of Schubert cells? | 65209 | How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles?
Let [imath]G(r, n)[/imath] be the Grassmannian of the set of all [imath]r[/imath]-planes in a [imath]n[/imath]-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, [imath]G(2,4)[/imath] and show that it has a basis consisting of the equivalent classes represented by Schubert cycles. |
2971768 | Sum of infinite telescoping series [imath]\sum_{r=2}^\infty \frac{1}{r^2-1}[/imath]?
How do I find sum of [imath]\sum_{r=2}^\infty \frac{1}{r^2-1}[/imath]? The answer given in my book is 3/4. I can decompose the general term into [imath](\frac{1}{r-1}-\frac{1}{r+1})[/imath] multiplied by 1/2 but since it is infinite I don't know what to do next. How should I do this? | 1698963 | find [imath]\sum_{k=2}^{\infty}\frac{1}{k^2-1}[/imath]
[imath]\sum_{k=2}^{\infty}\frac{1}{k^2-1}[/imath] I found that: [imath]\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}[/imath] and [imath]\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}[/imath] is a telescopic series so we need [imath]lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}[/imath] but the answer is [imath]\frac{3}{4}[/imath] |
2971609 | If [imath]E[/imath] is measurable and [imath]0, then for any \epsilon \in (0,1) there is an open interval I s.t m(E \cap I)>(1-\epsilon)m(I)[/imath]
Prove that is [imath]E[/imath] is measurable and [imath]0<m(E)<\infty[/imath], then for any [imath]0<\epsilon<1[/imath] there is an open interval [imath]I[/imath] such that [imath]m(E \cap I)>(1-\epsilon)m(I)[/imath] How would I solve this without using The Lebesgue density theorem | 2896620 | Let [imath]\lambda(A)[/imath] be the Lebesgue measure of [imath]A[/imath]. There exists an interval [imath]I[/imath] such that [imath]\lambda(E \cap I) / \lambda(I) < 1-\epsilon[/imath]
(Not mentioned in title but [imath]\epsilon[/imath] is a number greater than [imath]0[/imath] and [imath]E[/imath] a Lebesgue measurable subset of [imath]\mathbb{R}^n[/imath].) I know a question equivalent to this one has been asked (here). But it was not answered. Also my thinking is a bit different. So I would love to get some help on this problem. So far I've got: [imath]\frac{\lambda(E \cap I)}{\lambda(I)} > 1- \epsilon \iff \frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)} < \epsilon[/imath] And also, a theorem states that, for all [imath]\epsilon[/imath] there exists a closed set [imath]F\subset E[/imath] such that [imath]\lambda(E\setminus F) < \epsilon[/imath] so it would sufice to prove that there exists an interval [imath]I[/imath] such that [imath]\frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)}<\lambda(E\setminus F).[/imath] I'm not sure if this helps me much but any help is appreciated. What would you do? |
2972544 | Set of symmetric matrices is closed
Show that the following set is closed. [imath] S_n = \{ A \;\epsilon \; \mathbb R^{n^2} | \text{A is symmetric} \} \,\subset\, \mathbb R ^{n^2}[/imath] We are given a hint to use this theorem: Let f be a function from [imath]\mathbb R^n \, to \,\mathbb R^m[/imath] Then, f conts [imath]\iff f^{-1}(F) [/imath] is closed for each closed set F in [imath]\mathbb R^m[/imath]. Not sure where to even begin. Thanks! | 1457591 | Open and closed set in n by n matrices space.
This problem seems to be easy yet I have no idea to deal with. If [imath]\mathbb{M} ^n[/imath] is the set of all real square matrices of order n, identified here with [imath]\mathbb{R} ^{n^2}[/imath] equipped with its usual Euclidean norm which induce an usual topology on the space [imath]\mathbb{M}^n[/imath]. Show that: The set of all invertible real matrices of order n is open in [imath]\mathbb{M} ^n[/imath] with respect to the Euclidean norm topology; The set [imath]\mathbb{S} ^n[/imath] which consists of all real symmetric matrices of order n is closed in [imath]\mathbb{M} ^n[/imath] with respect to the Euclidean norm topology. For question #1, by the hint which is given by @Brent Kerby, that [imath]det:\mathbb{M}^n \rightarrow \mathbb{R}[/imath] is continuous. |
2972852 | Product of binomials
I am trying to compute this quantity. It seems we have to use Stirling’s formula but I don’t really succeed... Although trying to compute the quantity with a large n it seems to converge to [imath]e^{1/2}[/imath]. Thank you for you help | 2936916 | Evaluating [imath]\lim_{n \to \infty}\left[(\mathrm{^nC_o ^nC_1....^nC_n} )^{\frac{1}{n(n+1)}}\right][/imath]
[imath]\lim_{n \to \infty}\left[(\mathrm{^nC_o ^nC_1...^nC_n} )^{\dfrac{1}{n(n+1)}}\right][/imath] is equal to: a) [imath]e[/imath] b) [imath]2e[/imath] c) [imath]\sqrt e[/imath] d) [imath]e^2[/imath] Though it looks really innocent at first sight, it certainly isn't. Attempt: It's infinity^infinity form. I had tried taking the product raised to the power [imath]1/(n(n+1))[/imath] as function [imath]f(n)[/imath] Then I took logarithm of both sides to see if things simplifying. Even after factoring out the extra factorials it wasn't easy. Note that [imath]\mathrm{^nC_x} = \dbinom{n}{x}[/imath] |
2971980 | Show that if [imath]0 it follows that \lim_{n\to\infty}b^n=0[/imath]
Show that if [imath]0<b<1[/imath] it follows that [imath]\lim_{n\to\infty}b^n=0[/imath] I have no idea how to express [imath]N[/imath] in terms of [imath]\varepsilon[/imath]. I tried using logarithms but I don't see how to find [imath]N[/imath] from this. | 1063774 | Limit of the geometric sequence $\{r^n\}$, with $|r| < 1$, is $0$?
Prove that the [imath]\lim_{n\to \infty} r^n = 0[/imath] for [imath]|r|\lt 1[/imath]. I can't think of a sequence to compare this to that'll work. L'Hopital's rule doesn't apply. I know there's some simple way of doing this, but it just isn't coming to me. :( |
2973128 | Log Fibonacci = Theta
I'm trying to prove that [imath]\log F_n = Θ(n)[/imath] and where [imath]F_n = F_{n-1} + F_{n-2}[/imath] [imath]F_1 = 1[/imath], [imath]F_0 = 0[/imath] There's already a thread about this question, but the accepted answer doesn't explain a certain calculation in detail, and I was hoping someone here understands and could explain it to me! I don't understand the part [imath]\log F_{n+1}=\log(F_n+F_{n-1})=\log F_n+\log\left(1+{F_{n-1}\over F_n}\right)[/imath] More precisely I don't understand how [imath]log(F_{n-1}) = \log\left(1+{F_{n-1}\over F_n}\right)[/imath] I hope someone can explain how this is possible. | 2971350 | Show that log [imath]Fib_{n}[/imath] is [imath]\theta(n)[/imath]
I need to show log [imath]Fib_{n}[/imath] is [imath]\theta(n)[/imath] by the Fibonacci numbers defined as [imath] F_n=F_{n-1}+F_{n-2}[/imath] for [imath] n \geq 2 [/imath] [imath] F_{0} = 0 [/imath] and [imath] F_{1} = 1 [/imath] I'm not sure how to approach this. I can see it grows exponentially as I've shown a basecase for [imath]F_{6}[/imath]. Basecase for [imath]F_{6}[/imath]: [imath] F_{2} = F_{1} + F_{0} = 1 + 0 [/imath] [imath] F_{3} = F_{2} + F_{1} = 1 + 1 [/imath] [imath] F_{4} = F_{3} + F_{2} = 2 + 1 [/imath] [imath] F_{5} = F_{4} + F_{3} = 3 + 2 [/imath] [imath] F_{6} = F_{5} + F_{4} = 5 + 3 [/imath] But how I prove it's true for log [imath]Fib_{n}[/imath] is [imath]\theta(n)[/imath] I don't know. Hope someone can help! |
2973751 | [imath]E[/imath] be an algebraic extension of [imath]F[/imath]. Suppose [imath]K[/imath] is an integral domain s.t. [imath]F \leq K \leq E[/imath]. Show that [imath]K[/imath] is a field.
[imath]E[/imath] be an algebraic extension of [imath]F[/imath]. Suppose [imath]K[/imath] is an integral domain s.t. [imath]F \leq K \leq E[/imath]. Show that [imath]K[/imath] is a field. I've been struggling with this one for awhile now! Anyone got any insights? Thanks! | 2223928 | Algebraic field extension and intermediate integral domain
Let [imath]F/K[/imath] be an algebraic field extension. Suppose [imath]D[/imath] is an integral domain with [imath]K\subset D\subset F[/imath]. Show that [imath]D[/imath] is a field. \\\\\My idea was wrong, so I have to delete them. \\\\\\ Thanks! |
2974630 | Given, [imath]f(x)=f(2x)[/imath] evaluate [imath]\int_{-1}^1{f(f(x))}dx[/imath]
Let [imath]f:\mathbb{R}\to \mathbb{R}[/imath] be a continuous function and [imath]f(x)=f(2x)[/imath] is true [imath]\forall{x\in{\mathbb{R}}}[/imath]. If [imath]f(1)=3[/imath], then the value of [imath]\int_{-1}^{1}f(f(x))dx[/imath] equals. I don't have any idea how to proceed. Any hints would be helpful. Thank you. | 1682803 | If [imath]f[/imath] is continuous with [imath]f(x) = f(2x),f(1) = 3[/imath], then what is [imath] \int_{-1}^{1}f(f(x))\,dx[/imath]?
If [imath]f(x)[/imath] is a continuous function such that [imath]f(x) = f(2x)[/imath] and [imath]f(1) = 3\;,[/imath] Then [imath]\displaystyle \int_{-1}^{1}f(f(x))\,dx[/imath] [imath]\bf{My\; Try::}[/imath] Here [imath]-\infty <x<\infty[/imath] and Given [imath]f(x) = f(2x)[/imath] So Using Recursively [imath]\displaystyle f(x) = f(2^1x)=f(2^2x)=..........=\lim_{n\rightarrow \infty}f\left(2^{n-1}x\right)[/imath] OR we can write it as [imath]f(x) = f\left(\frac{x}{2}\right)=f\left(\frac{x}{2^2}\right)=........=\lim_{n\rightarrow \infty}f\left(\frac{x}{2^{n-1}}\right)[/imath] Now How can i prove that [imath]f(x)[/imath] is a constant function. Help me Thanks |
2974924 | Are [imath]\frac {\mathbb R[x]} {(x^2 -1)}[/imath] and [imath]\frac {\mathbb R[x]}{(x^2)}[/imath] Isomorphic?
Are [imath]\frac {\mathbb R[x]} {(x^2 -1)}[/imath] and [imath]\frac {\mathbb R[x]}{(x^2)}[/imath] Isomorphic? Can anyone please give me a hint? My attempt: I can find the elements of two rings. But I have no clue to prove or disprove that statement. | 2894580 | Describe [imath]\mathbb{R}[x]/(x^{2}+ax+b)[/imath] where [imath]a,b \in \mathbb{R}[/imath]
So obviously using quadractic formula, we have three cases for the roots which depends on the discriminant. I am not sure if I am right on this but, Case 1: [imath]a^{2}-4b>0[/imath]. Then we have two distinct real roots so this factors as [imath]\mathbb{R}[x]/(x-\alpha) \oplus \mathbb{R}[x]/(x-\beta).[/imath] (My guess here is since [imath]\alpha[/imath] and [imath]\beta[/imath] are different, these two ideals are comaximal so I can apply the Chinese Remainder Theorem). But this is isomorphic to [imath]\mathbb{R} \oplus \mathbb{R}.[/imath] Case 2: [imath]a^{2}-4b < 0[/imath]. In this case, we have a complex root so we have an extension of degree [imath]2[/imath]. But obviously whatever root we have, this is contained in [imath]\mathbb{R}[i]\cong \mathbb{C}[/imath] so we must have this is [imath]\mathbb{C}[/imath] by the tower law since [imath]\mathbb{R}[i][/imath] is also a degree [imath]2[/imath] extension of [imath]\mathbb{R}[/imath]. Case 3: [imath]a^{2}-4b=0[/imath]. Then we have a real root with multiplicity [imath]2[/imath]. I am not sure what this is as if we have the root is [imath]\alpha[/imath], we get [imath]\mathbb{R}[x]/(x-\alpha)^{2}[/imath] Since these roots are the same, [imath](x-\alpha)[/imath] and [imath](x-\alpha)[/imath] are not comaximal I believe so I don't know what this ring is. |
2974405 | Conjunctions of disjunctions equivalent to disjunctions of conjunctions?
For finite sets [imath]I[/imath], [imath]J[/imath] and formulas [imath]A_{i, j}[/imath] is this equivalence true? [imath]\bigwedge_{i\in I} \bigvee_{j\in J}A_{i, j} \equiv \bigvee_{j\in J}\bigwedge_{i\in I}A_{i, j} [/imath] I tried it out with some examples and it seemed correct. How would prove that? | 2506455 | True or false? [imath]\bigwedge_{i \in I} \bigvee_{j \in J}a_{i,j} \equiv \bigvee_{j \in J} \bigwedge_{i \in I}a_{i,j}[/imath]
Let [imath]I = \left\{i_1,...,i_m\right\}[/imath] and [imath]J = \left\{j_1,...,j_n\right\}[/imath] be finite sets, and for every [imath]i \in I[/imath] and [imath]j \in J[/imath], there is a formula [imath]a_{i,j}[/imath] given. Do you always have that [imath]\bigwedge_{i \in I} \bigvee_{j \in J}a_{i,j} \equiv \bigvee_{j \in J} \bigwedge_{i \in I}a_{i,j}[/imath] In my other question I ask how can write as example [imath]\bigwedge_{i=1}^{2} \bigvee_{j=1}^{2}a_{ij}[/imath] You write it like this: [imath] (a_{11} \lor a_{12} \lor a_{13} \lor a_{14}) \land (a_{21} \lor a_{22} \lor a_{23} \lor a_{24})[/imath] Now when we swap the symbol (look at right side of equivalence symbol) we have: [imath] (a_{11} \land a_{12} \land a_{13} \land a_{14}) \lor (a_{21} \land a_{22} \land a_{23} \land a_{24})[/imath] From this example we see that not same because sign is all opposite, never the same.. And also the index [imath]i,j[/imath] can be different, then also no same. That why this is false? |
2975173 | Let [imath]F[/imath] be a finite field with [imath]p^n[/imath] elements where char[imath](F)=p[/imath]
I faced the following question while prove the proposition "the group of automorphisms of a finite field is cyclic"- Let [imath]F[/imath] be a finite field with [imath]p^n[/imath] elements where [imath]\operatorname{char}(F)=p[/imath]. Then how to show that any automorphism of [imath]F[/imath] fixes [imath]F_p[/imath] pointwise i.e. if [imath]f\in \operatorname{Aut}(F)[/imath] then [imath]f(x)=x \forall x\in F_p[/imath] where [imath]F_p\simeq \Bbb{Z}_p[/imath]. Can anybody clear up query? Thanks for assistance in advance. | 1758671 | How would you show that field automorphisms fix prime subfields?
Suppose [imath]K[/imath] is a prime subfield of [imath]E[/imath], then if [imath]\phi[/imath] is an automorphism from [imath]E[/imath] to [imath]E[/imath], we have for all [imath]x \in K[/imath], [imath]\phi(x) = x[/imath]. I feel like this is just the definition of a field automorphism, but my book says this should be proven as an exercise. |
2973432 | Shortest distance between a point and a parabola.
Find the shortest distance from the point [imath](1, 0)[/imath] to the parabola [imath]y^2 = 4x[/imath]. As- Let [imath]D^2=(x-1)^2 + y^2[/imath]. Now reduce this to one variable by putting [imath]y^2 = 4x[/imath] to get [imath]D^2 = (x-1)^2+ 4x[/imath]. We are working with [imath]D^2[/imath] to make the calculations easier. Clearly [imath]\frac{d}{dx}D^2 = 2(x-1)+4 = 0 \implies x = -1[/imath] At [imath]x = -1[/imath], we have [imath]\frac{d^2}{dx^2}D^2 = 2 < 0[/imath] which shows that [imath]D^2[/imath] is minimum. At this [imath]x = -1[/imath]. But this is not a point on parabola. Am i doing some mistake or we have to follow some other method to get the minima? | 2963331 | Point on parabola which is at shortest distance?
Find a point on parabola [imath]y^2 =4x[/imath] which is at shortest distance from [imath](1,0)[/imath]. Answer given is [imath](0,0)[/imath] But I am getting imaginary values. How do i get [imath](0,0)[/imath] |
2975421 | Need further explanation for this problem
I have recently asked this problem: What is the probability that [imath]m[/imath] is the largest number drawn? I don't understand how one arrive at the questions posted by lulu and greedoid. I guess I don't understand what the problem asks for. Could you provide a fuller explanation to these answers? | 2975112 | What is the probability that [imath]m[/imath] is the largest number drawn?
A box contains [imath]n[/imath] identical balls numbered [imath]1[/imath] through [imath]n[/imath]. Suppose [imath]k[/imath] balls are drawn in succession. (a) What is the probability that m is the largest number drawn? (b) What is the probability that the largest number drawn is less than or equal to [imath]m[/imath]? I don't know how to solve this problem? Could you help me? For (a), I only know that the denominator will be [imath]\binom{n}{k}[/imath]. How to find the numerator for both cases? |
2975110 | [imath]\langle x,y \rangle + \langle y,x \rangle = 2\operatorname{Re}\langle x,y \rangle[/imath]
In an inner product space, [imath]\langle x,y \rangle + \langle y,x \rangle = 2\operatorname{Re}\langle x,y \rangle[/imath] and [imath]\langle x,-y \rangle + \langle -y,x \rangle = -2\operatorname{Re}\langle y,x \rangle[/imath] Why are these two identities true? What is the significance of writing Re instead of leaving the sum as [imath]\langle x,y \rangle + \langle y,x \rangle[/imath]? | 5179 | Is complex conjugation needed for valid inner product?
What are the benefits of using a conjugate linear inner product in a complex vector space vs a simple linear inner product? That is, why do we demand that [imath](y,x) = \overline{(x,y)}[/imath] as opposed to [imath](y,x)=(x,y)[/imath]? Of course, this ensures that [imath](x,x)[/imath] is real and thus makes an easy definition of norm, but is that necessary? |
2975324 | All Möbius transformations that take the unit disk onto itself
I wish to prove that all Möbius transformation raking the unit disk into itself are of the form [imath]k\frac{z-l}{1-z\bar{l}}[/imath] where [imath]|k| = 1[/imath]. More specifically, I ask, in addition to the main question above, the maximum modulus principle implies that the analytic continuation of such mobius transform has to carry the unit circle to itself. I realize that numerous posts like this and this already seems to have an answer here, but I couldn't find a satisfactory answer yet. Letting [imath]T(z)[/imath] be an arbitrary Möbius transformation that preserves the unit disk, and writing [imath]T(z) = \frac{az+b}{cz+d}[/imath] with (without loss of generality) [imath]a=0[/imath] or [imath]a=1[/imath], we must somehow show that [imath]T[/imath] is of the form specified above... It seems like we have to use the maximum modulus principle, yet all the answers from other posts did not elaborate on how this would be used. Please note that my question only asks for one direction and not the other, and does NOT ask to show that all conformal automorphism of the unit disk are of this form. | 209308 | Can we characterize the Möbius transformations that maps the unit disk into itself?
The Möbius transformations are the maps of the form [imath] f(z)= \frac{az+b}{cz+d}.[/imath] Can we characterize the Möbius transformations that map the unit disk [imath]\{z\in \mathbb C: |z| <1\}[/imath] into itself? |
2975734 | Show that the limit exists or does not exist
[imath]\lim_{(x, y) \to (0,0)} \frac{5x^2}{x^2 + y^2}[/imath] let [imath]y = 0[/imath] [imath]\lim_{x \to 0} \frac{5x^2}{x^2} = 5[/imath] let [imath]y = x[/imath] [imath]\lim_{x \to 0} \frac{5x^2}{2x^2} = \frac{5}{2}[/imath] Since different values the limit does not exist. Would this be right? | 2961513 | Show that the limit does not exist [imath]\lim_{(x, y) \to (0,0)}\frac{5x^2}{x^2 + y^2}[/imath]
Show that the limit does not exist [imath]\lim_{(x, y) \to (0,0)}\frac{5x^2}{x^2 + y^2}[/imath] attempt: let [imath]y = 0[/imath] [imath]\lim_{x \to 0} \frac{5x^2}{x^2 + 0^2} = 5[/imath] let [imath]x = 0[/imath] [imath]\lim_{y \to 0} \frac{5(0)^2}{y^2} = 0[/imath] [imath]5 \neq 0[/imath], therefore two different values, limit does not exist right? |
2838680 | Showing that a function has the intermediate value property
Here's the question: Let [imath]f(x)=\sin \left( \frac{1}{x} \right)[/imath] for [imath]x\ne 0[/imath] and let [imath]f(0)=0[/imath]. Show that [imath]f[/imath] is discontinuous on [imath]\mathbb{R}[/imath] and still has the intermediate value property on [imath]\mathbb{R}[/imath]. It's easy to show that [imath]f[/imath] is discontinuous at [imath]x=0[/imath]. Now, I go on to prove that [imath]f[/imath] has the intermediate value property. Let [imath]a,b \in \mathbb{R}[/imath] with [imath]a<b[/imath]. Assume that [imath]y[/imath] be some real such that [imath]f(a) < y < f(b)[/imath]. If [imath]y=0[/imath] we are done for [imath]f(0)=y[/imath]. If [imath]y\ne 0[/imath], I was hoping to use the continuity of sine on [imath]\mathbb{R}[/imath] but nothing seems work out. Hints would be appreciated. I do not need full solution. | 432647 | Showing Intermediate Value Property Holds
Let [imath]f(x)=\begin{cases} \sin \tfrac 1 x &\text{if $x\ne 0$}\\ 0 & \text{if $x = 0$.}\end{cases}[/imath] I have to show that [imath]f[/imath] has the intermediate value property. That is, for any [imath]a < b[/imath], if [imath]y[/imath] is any real number such that [imath]f(a) < y< f(b)[/imath] or [imath]f(a)>y> f(b)[/imath], then there exists a [imath]c \in (a,b)[/imath] such that [imath]f(c)=y[/imath]. I feel like I kind of know how to go about completing this. I just am curious as to if I have to create a bound such as letting [imath]a = -1[/imath] and [imath]b = 1[/imath], or keep [imath]a[/imath] and [imath]b[/imath] both arbitrary. |
2975516 | [imath]Γ \models α [/imath]if and only if [imath]Γ ∪ \{(¬α)\}[/imath] is not satisfiable.
Let [imath]\Gamma[/imath] be some set of well formed formulas and let [imath]\alpha \in \mathrm{WFF}[/imath]. Prove the following statement: [imath]\Gamma \models \alpha[/imath] if and only if [imath]\Gamma \cup \{(\lnot\alpha)\}[/imath] is not satisfiable. So far I have done this: edit: whatever I did was wrong, so can you tell me what could be done? | 975470 | Show that [imath]\Gamma \cup \{\neg \phi\}[/imath] is satisfiable if and only if [imath]\Gamma\not \models \phi[/imath]
Let [imath]\Gamma[/imath] be a set of formulas and [imath]\phi[/imath] be a formula. Show that [imath]\Gamma \cup \{\neg \phi\}[/imath] is satisfiable if and only if [imath]\Gamma\not \models \phi[/imath]. This seemed pretty obvious but I wanted to see if my proof made sense: Proof: [imath](\Rightarrow)[/imath] To derive for a contradiction, suppose that: [imath]\Gamma \models \phi[/imath]. That means for all truth assignments [imath]v[/imath], for [imath]\gamma \in \Gamma[/imath], if [imath]v(\gamma) = T[/imath], then [imath]v(\phi) = T[/imath]. But this contradicts our assumption that [imath]\Gamma \cup \{\neg \phi \}[/imath] is satisfiable by the fact that [imath]v(\phi) = T[/imath] cannot happen so: [imath]\Gamma \not \models \phi[/imath] . [imath](\Leftarrow)[/imath] So by the definition of [imath]\Gamma \not \models \phi[/imath], we have that there is some truth assignment [imath]v[/imath] which satisfies [imath]\Gamma[/imath] but does not satisfy [imath]\phi[/imath]. So that means [imath]v(\phi) = F[/imath] since [imath]v(\phi) \not = T[/imath] which implies that [imath]v(\neg \phi) = T[/imath] which means [imath]v[/imath] satisfies [imath]\Gamma \cup \{\neg \phi\}[/imath]. I feel like I'm missing something in the forward direction, but at the same time... It looks pretty trivial as well. Am I missing anything crucial? Thank you! |
2976144 | [imath]\sum_{i=0}^{n} (-1)^i l(H_i)=\sum_{i=0}^{n} (-1)^i l(G_i)[/imath]
Let [imath]0 \overset{d_{n+1}}{\rightarrow} G_n \overset{d_n}{\rightarrow} G_{n-1}\overset{d_{n-1}}{\rightarrow}\ldots\overset{d_2}{\rightarrow}G_1 \overset{d_1}{\rightarrow} 0[/imath] be a sequence of modules and homomorphisms over the commuative ring [imath]R[/imath] such that [imath]d_i \circ d_{i+1}=0[/imath] for all [imath]i=1,..,n-1[/imath], and suppose that [imath]G_i[/imath] has finite length for all [imath]i[/imath]. Set [imath]H_i=\ker d_i / \text{im} \, d_{i+1}[/imath]. Show that [imath]H_i[/imath] has finite length for all [imath]i[/imath] and that [imath]\sum_{i=0}^{n} (-1)^i l(H_i)=\sum_{i=0}^{n} (-1)^i l(G_i)[/imath] It's obvious to see that [imath]H_i[/imath] has finite length by the exact sequence [imath]0 \to Im \; d_{i+1}\to Ker \; d_i \to Ker \; d_i / Im \; d_{i+1}\to 0[/imath] But I can indicate that [imath]\sum_{i=0}^{n} (-1)^i l(H_i)=\sum_{i=0}^{n} (-1)^i l(G_i)[/imath]. Can anyone help me? | 2807224 | Show that the Euler characteristic of a chain complex is equal to the Euler characteristic of its homology
Let [imath]C_*[/imath] be a chain complex such that each [imath]C_i[/imath] is a torsion-free, finite-range abelian group with [imath]C_i=0[/imath] for all [imath]i<0[/imath]. Suppose that [imath]C_i=0[/imath] for all [imath]i[/imath] is large enough. The Euler characteristic of [imath]C_*[/imath] chain complex is defined as [imath]\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(C_i)[/imath] Prove that [imath]\chi(C_*)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))[/imath] I have to prove that [imath]\sum_{i\geq 0}(-1)^iRank(C_i)=\sum_{i\geq 0}(-1)^iRank(H_i(C_*))[/imath], I think that one way to do this is by showing that [imath]Rank(C_i)=Rank(H_i(C_*))[/imath] but I do not know if this is true in general, in a nutshell, I want to show that the cardinality of the base of any [imath]C_i[/imath] is the same as the cardinality of the basis of the corresponding homology, how can I do this? Thank you |
2976302 | How many integer solutions of [imath]x_1[/imath] + [imath]x_2[/imath] + [imath]x_3[/imath] + [imath]x_4[/imath] = 28 are there with
How many integer solutions of [imath]x_1[/imath] + [imath]x_2[/imath] + [imath]x_3[/imath] + [imath]x_4[/imath] = 28 are there with (a) 0 ≤ [imath]x_i[/imath] ≤ 12? (b) −10 ≤ [imath]x_i[/imath] ≤ 20? (c) 0 ≤ [imath]x_i[/imath], [imath]x_1[/imath] ≤ 6, [imath]x_2[/imath] ≤ 10, [imath]x_3[/imath] ≤ 15, [imath]x_4[/imath] ≤ 21? I have tried. a) To count the compliment value, so (31 choose 3) - (15 choose 3) b) Not quite sure where to begin c) (31 choose 3)-(24 choose 3)-(18 choose 3)- (13 choose 3)- (9 choose 3) | 2976442 | Combinatorics Question Asking How Many Integer Solutions Given Different Restrictions
How many integer solutions of [imath]x_1 + x_2 + x_3 + x_4 = 28[/imath] are there with: (a) [imath]0 ≤ x_i ≤ 12[/imath]? (b) [imath]−10 ≤ x_i ≤ 20[/imath]? (c) [imath]0 ≤ x_i, x_1 ≤ 6, x_2 ≤ 10, x_3 ≤ 15, x_4 ≤ 21[/imath]? \ \ My attempts (I was really only able to solve part (a) but I would appreciate if you could check my work for this one as well): (a) First I took the total number of integer solutions with no restrictions: There would be [imath]{28+3\choose 3}[/imath] = [imath]{31\choose 3}[/imath] total solutions with no restrictions. Then I used the inclusion exclusion principle and knew I would have to subtract [P(A) + P(B) + P(C) + P(D)] where A --> [imath]x_1[/imath] is greater than or equal to 13, ..., D--> [imath]x_4[/imath] is greater than or equal to 13. The probabilities of all these cases would equal [imath]{15+3\choose 3}[/imath] = [imath]{18\choose 3}[/imath] because once we assign 13 to one of the [imath]x_i[/imath] there would be [imath]28 - 13 = 15[/imath] remaining to distribute among the [imath]x_i[/imath] values. We would also need to choose which of the [imath]x_i[/imath] to give the 13 to so we would multiply the [imath]{18\choose 3}[/imath] by [imath]{4\choose 1}[/imath] = [imath]{4\choose 1}{18\choose 3}[/imath] We would then have to add back, since we "oversubtracted", the case in which there is an intersection between two of the [imath]x_i[/imath] or in other words two of the [imath]x_i[/imath] receive 13 or more. P[imath](A\bigcap B)[/imath] = [imath]28-2(13) = 2[/imath] remaining to distribute among the [imath]x_i[/imath] values. So we would also have to choose which of the [imath]x_i[/imath] to give the 13 to so we would have [imath]{4\choose 2}[/imath] ways to choose them. So there would be [imath]{4\choose 2}{2+3\choose 3}[/imath] = [imath]{4\choose 2}{5\choose 3}[/imath] solutions in this case. Thus the number of solutions would be: [imath]{31\choose 3} - {4\choose 1}{18\choose 3} + {4\choose 2}{5\choose 3}[/imath]. I'm not sure if this is correct and I'm really stuck on b and c so I would appreciate any help or feedback. Thank you! |
2976692 | How to show eigenvalues of [imath]X^T X[/imath] are all non-negative
How to show eigenvalues of [imath]X^T X[/imath] are all non-negative? | 363955 | Prove all eigenvalues of [imath]A^*A[/imath] are non-negative
Let [imath]A[/imath] be an [imath]m\times n[/imath] matrix. Prove that all eigenvalues of [imath]A^*A[/imath] are non-negative. |
2974330 | Finding the joint density of two independent exponential distributions
Let [imath]X_1,X_2[/imath] be independent random variables each having a exponential distribution with mean [imath]λ = 1[/imath]. (a) Find the joint density of [imath]Y_1 = X_1[/imath] and [imath]Y_2 = X_1 + X_2[/imath]. (b) Get the marginal density of [imath]f_{Y_1}(y_1)[/imath] and [imath]f_{Y_2}(y_2)[/imath]. These are what I've found so far. [imath]f_{X_1}(x_1)=e^{-x_1}, f_{X_2}(x_2)=e^{-x_2}[/imath] [imath]f_{X_1,X_2}(x_1, x_2)=f_{X_1}(x_1)f_{X_2}(x_2)=e^{-x_1-x_2}[/imath] Also, since [imath]Y_1=X1[/imath], [imath]f_{Y_1}(y_1)=e^{-y_1}[/imath] right? Similarly, since [imath]Y_2=X_1+X_2[/imath], [imath]f_{Y_2}(y_2)=e^{-y_2}[/imath] However, is it guaranteed that [imath]Y_1[/imath] and [imath]Y_2[/imath] are independent? or else how can I solve these problems?? Oh, and I haven't learned the Jacobian so it would be grateful if you avoid using it. | 2972227 | Joint density for exponential distribution
Let [imath]X_1[/imath] and [imath]X_2[/imath] be independent random variables each having a exponential distribution with mean [imath]\lambda = 1[/imath]. (a) Find the joint density of [imath]Y_1 = X_1[/imath] and [imath]Y_2 = X_1 + X_2[/imath]. (b) Get the marginal density of [imath]f_1(y_1)[/imath] and [imath]f_2(y_2)[/imath]. [imath]f(x_1)=e^{-x_1}[/imath], [imath]f(x_2)=e^{-x_2}[/imath] Since [imath]X_1[/imath] and [imath]X_2[/imath] are independent, [imath]f(x_1, x_2)=f(x_1)f(x_2)=e^{-x_1-x_2}[/imath] However, how could I find the joint density? I mean, it is quite easy to find marginal ones from the joint one, but how can I do the opposite? |
2976957 | [imath]a|b^2[/imath] , [imath]b^2|a^3[/imath] , [imath]a^3|b^4[/imath] , [imath]b^4| a^5[/imath]... then...
Let [imath]a[/imath] and [imath]b[/imath] be positive interger such that [imath]a|b^2[/imath] , [imath]b^2|a^3[/imath] , [imath]a^3|b^4[/imath] , [imath]b^4| a^5[/imath] ... Prove that [imath]a=b[/imath]. Here how I tried to do it, I think it isn't the best way to approach it but anyway I failed in proving it: EDIT: Let [imath]a = p_1 ^{a_1} p_2 ^{a_2} ... p_k ^{a_k}[/imath] and [imath]b= p_1 ^{b_1} p_2 ^{b_2} ... p_k ^{b_k}[/imath] and [imath]a_i, b_i \geq 0[/imath] for [imath]i[/imath], [imath] 0 \leq i \leq k[/imath]. Then for a specific prime [imath]p_i[/imath], [imath]a_i \leq 2b_i[/imath], [imath]2b_i \leq 3a_i[/imath] , [imath]3a_i \leq 4b_i[/imath] and [imath]4b_i \leq 5a_i[/imath].... I tried to get some information from that but didn't get anything. This question has been marked as a duplicate but it is in fact different, or maybe that's what I can directly deduce from the conditions. Just look carefully at the titles to realise that they are different. | 123584 | Let [imath]a,b \in {\mathbb{Z_+}}[/imath] such that [imath]a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots[/imath], Prove [imath]a=b[/imath]
Let [imath]a,b \in {\mathbb{Z_+}}[/imath] such that [imath]a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots[/imath] Prove [imath]a=b[/imath] |
2976190 | (Self-Study) UMVUE of the mean of a normal distribution
Let [imath]X_1,...,X_n[/imath] be a random sample from normal(θ,1). Is there an UMVUE of [imath]θ^2[/imath] here? [imath]X^2-1[/imath] is an unbiased estimator of [imath]θ^2[/imath]. First thing that came to my mind is to use Lehmann-Scheffe Theorem. But the problem is that, according to my calculations, n(θ,1) doesn't have a complete statistic since, even though [imath]\overline{X}=\frac{\sum_{n=1}^{n}X_i}{n}[/imath] is a minimal sufficient statistic for θ (easy to show), it is not complete (again easy to show). Thus there is simply no complete statistic for n(θ,1). Another way to prove it may use the condition that an UMVUE must be uncorrelated with all unbiased estimators of 0 (and vice-versa). But I don't have a candidate for UMVUE at all. Can anybody give a clue about this question? Thanks! | 1486975 | UMVUE for [imath]\theta^2[/imath]
Let [imath]X_1,...X_n[/imath] be a random sample with distribution [imath]\text{Normal}(\theta,1)[/imath]. Find the UMVUE for [imath]\theta^2[/imath] What I´ve done so far: I have already shown that [imath]T=\sum_{i=1}^nX_i[/imath] is a complete sufficient statistic for [imath]\theta[/imath]. Let [imath]\widehat{\theta^2}=\bar {X}^2-\frac{1}{n}[/imath] be an estimator for [imath]\theta^2[/imath] [imath]E[\widehat{\theta^2}]=E[\bar {X}^2-\frac{1}{n}]=E[\bar{X}^2]-E^2[\bar X]+E^2[\bar X]+\frac{1}{n}=Var(\bar X)+E^2[\bar X]+\frac{1}{n}[/imath] I know that [imath]\bar X[/imath] has distribution [imath]\text{Normal}(\theta, \frac{1}{n})[/imath] It follows that :[imath]E[\widehat{\theta^2}]=\theta^2[/imath] Hence [imath]\widehat{\theta^2}=\bar {X}^2-\frac{1}{n}[/imath] is unbiased estimator Know I have to compute [imath]g(T)=E[\widehat{\theta^2}|T][/imath] and by Lehmann-Scheffé this will be the UMVUE for [imath]\theta^2[/imath] but the problem is how can I compute [imath]E[\widehat{\theta^2}|T]=E[\bar {X}^2-\frac{1}{n}|\sum_{i=1}^nX_i][/imath] I know I need to find the joint density of [imath]\bar {X}^2-\frac{1}{n}[/imath] and [imath]\sum_{i=1}^nX_i[/imath] but is there an easy way to do it? Or is there another way to find the UMVUE? I would really appreciate if you can help me with this problem |
2976788 | How to inverse [imath](I + \alpha M)[/imath] for all [imath]\alpha[/imath]
I am looking for a way to solve the following equation: [imath](I + \alpha M)X=F[/imath] [imath]\forall \alpha \in R[/imath] (the real domain), with [imath]M[/imath] a square complex matrix without any particular properties, [imath]I[/imath] the identity matrix, [imath]F[/imath] a complex vector and [imath]X[/imath] the unknown. Thank you, | 17776 | Inverse of the sum of matrices
I have two square matrices - [imath]A[/imath] and [imath]B[/imath]. [imath]A^{-1}[/imath] is known and I want to calculate [imath](A+B)^{-1}[/imath]. Are there theorems that help with calculating the inverse of the sum of matrices? In general case [imath]B^{-1}[/imath] is not known, but if it is necessary then it can be assumed that [imath]B^{-1}[/imath] is also known. |
2180431 | How to show that [imath]\sqrt{p\over p+1}[/imath] is irrational?
I want to prove that [imath]\sqrt{p\over p+1}[/imath] is irrational where [imath]p[/imath] is any prime number. I have been thinking about this problem for several hours without any progress. To show that [imath]\sqrt{p\over p+1}[/imath] is irrational, I assume that it is rational. So there are integers [imath]m[/imath] and [imath]n[/imath] (which have no common divisors) such that [imath]\sqrt{p\over p+1}=\frac{m}{n}[/imath]. By squaring and rearranging, I get [imath]pn^2=(p+1)m^2[/imath] but I am not sure how to proceed from here. I can say that [imath](p+1)m^2[/imath] is divisible by [imath]p[/imath] and then I do not know what to do! | 2734338 | Prove that [imath]\sqrt{\frac{n}{n+1}}\notin \mathbb{Q}[/imath]
I would like to show that [imath]\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin \mathbb{Q}[/imath] I'm interested in more ways of proofing this. My method : suppose that [imath]\sqrt{\frac{n}{n+1}}\in \mathbb{Q}[/imath] then there exist [imath](p,q)\in\mathbb{Z}\times \mathbb{N}^*[/imath] such that [imath]\sqrt{\frac{n}{n+1}}=\frac{p}{q}[/imath] thus [imath]\dfrac{n}{n+1}=\dfrac{p^2}{q^2} \implies nq^2=(n+1)p^2 \implies n(q^2-p^2)=p^2[/imath] since [imath]p\neq q\implies p^2\neq q^2[/imath] then [imath]n=\dfrac{p^2}{(q^2-p^2)}[/imath] since [imath]n\in \mathbb{N}^*[/imath] then [imath]n\in \mathbb{Q}[/imath] I'm stuck here and I would like to see different ways to prove [imath]\sqrt{\frac{n}{n+1}}\notin \mathbb{Q}[/imath] |
2976680 | Is it possible to cut the unit square in triangles of the same area such that one of the vertices has irrational coordinates?
This may be an easy question, although it stumped some of us over lunch. Consider the unit square and cut it up in several triangles, such that all the triangles have the same area. Is it possible that one triangle has a vertex with irrational coordinates (at least the abscissa or the ordinate, not necessarily both)? This question was prompted by the theorem that in the cutting above, the number of triangles has to be even. The proof (that was inspired by this talk by Jim Fowler) uses that there exists an extension of the 2-adic valuation on [imath]\mathbb{R}[/imath], which follows from the axiom of choice. So we're wondering if it's necessary. | 1280185 | Is there an equidissection of a unit square involving irrational coordinates?
An equidissection of a square is a dissection into non-overlapping triangles of equal area. Monsky's theorem from 1970 states that if a square is equidissected into [imath]n[/imath] triangles, then [imath]n[/imath] is even. In 1968, John Thomas proved the following weaker statement: there is no equidissection of a unit square into an odd number of triangles whose vertices are rational numbers with odd denominators. I wonder to which extent the result of Thomas fails to cover all cases. It is easy to construct an equidissection where the coordinates are rational numbers with even denominators. My question is: Is there an equidissection of the unit square containing a triangle with at least one irrational coordinate? Dissecting the square into [imath]6[/imath] or less triangles seems to always yield rational coordinates. |
2977316 | About [imath]L \vDash GCH[/imath]
In the notes I'm following to understand the proof that [imath]L \vDash GCH[/imath], there is this crucial following passage: "Let [imath]A \subseteq X[/imath]. There is some [imath]\alpha[/imath] such that [imath]A ∈ L_\alpha[/imath] , and it suffices to show that the least such [imath]\alpha[/imath] is strictly smaller than [imath]\kappa^+[/imath] . Because then [imath]\mathcal{P}(X) \subseteq L_{\kappa^+}[/imath]" It is probably very simple, but why every [imath]A \in L_\alpha[/imath] where [imath]\alpha < \kappa^+[/imath] implies that [imath]P(\kappa) \subseteq L_{\kappa^+}[/imath]? My doubt is because [imath]sup\{\alpha: \exists A \subseteq \kappa \; A \in L_\alpha\}[/imath] is not necessarily less than [imath]\kappa^+[/imath] if the [imath]\alpha[/imath] are. Does [imath]L[/imath] "see" [imath]|L_\alpha| = |\alpha|[/imath]? | 264015 | Question about the proof of [imath]GCH[/imath] holds in [imath]\mathbf L[/imath]
I have a question about the proof of the following: (Lemma 37) Assume [imath]\mathbf V = \mathbf L[/imath], and let [imath]\kappa[/imath] be a cardinal. Then [imath]\mathcal P (\kappa ) \subseteq L_{\kappa^+}[/imath]. Assume we have proved the following: (Claim 38) There exists a finite subset [imath]T[/imath] of [imath]ZF + (\mathbf V = \mathbf L)[/imath] such that whenever [imath]x[/imath] is transitive and [imath]\langle x, \overline{\in} \rangle \models T[/imath] then there exists [imath]\gamma \in \mathbf{ON}[/imath] such that [imath]x = L_\gamma[/imath]. (Lemma 35) Let [imath]\alpha \in \mathbf{ON}[/imath] and [imath]x, a_0, \dots , a_{k-1} \in L_\alpha[/imath] and let [imath]\varphi(z, a_0, \dots ,a_{k-1})[/imath] be a formula of [imath]L_S[/imath]. Then there exists a [imath]\beta > \alpha[/imath] such that for every [imath]z \in x[/imath]: [imath] \langle L_\beta , \overline{\in} \rangle \models \varphi [z, a_0 , \dots, a_{k-1}] \hspace{0.5cm} \text{ iff } \hspace{0.5cm} \mathbf L \models \varphi [z, a_0 , \dots, a_{k-1}] [/imath] The proof in the book starts as follows: Suppose [imath]\mathbf V = \mathbf L[/imath] holds, let [imath]\kappa[/imath] be as in the assumptions of lemma 37 and suppose [imath]\kappa \in \mathcal P ( \kappa )[/imath]. Fix an ordinal [imath]\alpha > \kappa[/imath] such that [imath]y \in L_\alpha[/imath]. Since the theory [imath]T[/imath] of claim 38 is finite we can form the conjunction [imath]\varphi[/imath] of all sentences in [imath]T[/imath]. Question: But what is [imath]T[/imath]? I know that it's the theory from claim 38 but it's not clear to me how we can use an unknown theory to prove lemma 37. And why do we take the conjunction over [imath]T[/imath]? Then the proof proceeds: By lemma 35, Question: How does [imath]\gamma < \kappa^+[/imath] follow from [imath](2)[/imath]? Many thanks for your help. |
2977314 | Sum to n terms of the harmonic series
We know that [imath]\sum_{k=1}^{\infty}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots[/imath] diverges. But for any natural number [imath]n[/imath], [imath]\sum_{k=1}^{n}\frac{1}{k}[/imath] is finite. The question is; how to compute [imath]\sum_{k=1}^{n}\frac{1}{k}[/imath] for sum natural number [imath]n[/imath], explicitly? Say [imath]n=50[/imath], how to compute [imath]\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{49}+\frac{1}{50}[/imath], explicitly? | 439177 | Summing Finitely Many Terms of Harmonic Series: [imath]\sum_{k=a}^{b} \frac{1}{k}[/imath]
How do I calculate sum of a finite harmonic series of the form : [imath]\sum_{k=a}^{b} \frac{1}{k} = \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} +\cdots \frac{1}{b}.[/imath] Is there a general formula for this? How can we approach this if not? |
1504485 | The sum of non real roots of the polynomial equation [imath]x^3+3x^2+3x+3=0[/imath]
Problem : The sum of non real roots of the polynomial equation [imath]x^3+3x^2+3x+3=0[/imath] (a) equals 0 (b) lies between 0 and 1 (c)lies between -1 and 0 (d) has absolute value bigger than 1 My approach : The discriminant of cubic equation [imath]ax^3+bx^2+cx+d=0[/imath] is given by [imath]\Delta = 18abcd -4b^3d +b^2c^2 -4ac^3 -27a^2d^2[/imath] [imath]\Delta = -108 < 0 [/imath] Therefore the equation has one real and two non real roots. But how to find the roots of this equation not getting any idea please help . thanks | 623890 | What is the sum of non-real roots of the polynomial equation?
What is the sum of non-real roots of the polynomial equation [imath]X^{3}+3X^{2}+3X+3=0[/imath] ? |
2974499 | Equation in variable [imath](x,y)[/imath] represent curve in Cartesian plane
If [imath]x,y \in \mathbb{R}.[/imath] Then the equation [imath]3x^4-2(19y+8)x^2+361y^2+2(100+y^4)+64=2(190y+2y^2)[/imath] represent in rectangular cartesian system Options [imath](a)[/imath] Circle [imath]\;\;\;\;\;(b)[/imath] Parabola [imath]\;\;\;\;\;\; (c)[/imath] Ellipse [imath]\;\;(d)[/imath] Hyperbola Try: From [imath]3x^4-2(19y+8)x^2++2y^4+357y^2-380y+264=0[/imath] For real roots, its discriminant always [imath]\geq 0[/imath] [imath]4(19y+8)^2-4\cdot 3 \cdot (2y^4+357y^2-380y+264)\geq 0[/imath] [imath]361y^2+64+304y-6y^4-1125y^2+1140y-792\geq 0[/imath] So [imath]6y^4+764y^2-1444y+728\leq 0[/imath] I am struck at that point. did not how to solve further [imath]\bf{Added}:[/imath] I have seems that it can be convert into sum of square of numbers like [imath]()^2+()^2+()^2+\cdots =0[/imath] Could some help me to convert it , thanks | 2976143 | Integer ordered pair of [imath](x,y)[/imath] in complex algebraic equation
If [imath]x,y \in \mathbb{Z}.[/imath] Then the ordered pair of [imath](x,y)[/imath] for which [imath]3x^4-2(19y+8)x^2+361y^2+2(100+y^4)+64=2(190y+2y^2)[/imath] Try: From [imath]3x^4-2(19y+8)x^2++2y^4+357y^2-380y+264=0[/imath] For real roots, its discriminant always [imath]\geq 0[/imath] [imath]4(19y+8)^2-4\cdot 3 \cdot (2y^4+357y^2-380y+264)\geq 0[/imath] [imath]361y^2+64+304y-6y^4-1125y^2+1140y-792\geq 0[/imath] So [imath]6y^4+764y^2-1444y+728\leq 0[/imath] I am struck at that point. did not how to solve further |
2977745 | Interesting limit
Let [imath]a_n[/imath] a sequence of real numbers such that [imath]\lim_{n\to +\infty}a_n=A[/imath]. Is it true that [imath]\lim_{n\to +\infty}\Big(1+\frac{a_n}{n}\Big)^n=e^A?[/imath] | 2969851 | Prove the limit
Assume that [imath]x_n[/imath] is a sequence such that [imath]x_n[/imath] converge to a constant [imath]x[/imath]. I would like to prove that [imath]\left( 1+ x_n/n \right)^n \rightarrow e^x[/imath] Obvious if [imath]x_n = x[/imath] the result would be given. The extra sequence complicates things and I would like a hint or solution on how to proof this. |
2977966 | proving the inequality [imath]\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\ge 1[/imath]
I wish to prove the following inequality: [imath]\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\ge 1[/imath] I tried to use different ideas, I tried using squeeze theorem or assuming [imath]m\gt n[/imath] and working my way from there, but I did not manage to prove this inequality. Any suggestions? | 2972318 | how do I prove this inequallity?
[imath]m,n[/imath] are integers bigger than 0. I need to prove: [imath]\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geq 1[/imath] I tried to multiply both sides by the common factor and raise both sides to the power of [imath]m*n[/imath] but it did not work for me, and I have no other idea how to proceed. I don't want the full solution, but just a hint. Thank you. |
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