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2385582 | How to solve this cubic Diophantine equation?
So I recently came across this problem: [imath]\large{x^{3}+ (x+1)^{3}+\cdots\cdots+(x+15)^{3}=y^{3}}[/imath] Find how many ordered pairs for [imath]\large{(x,y)}[/imath] satisfy the above equation. Both [imath]x,y[/imath] are positive integers. I can do it using computer program but this problem was supposed to be solved by hand and I have no idea how. How can I do it? EDIT: The other question doesn't have any answer either. | 2385215 | Find all solutions to [imath]x^3+(x+1)^3+ \dots + (x+15)^3=y^3[/imath]
Find all pairs of integers [imath](x, y)[/imath] such that [imath]x^3+(x+1)^3+ \dots + (x+15)^3=y^3[/imath] What I have tried so far: The coefficient of [imath]x^3[/imath] is [imath]16[/imath] in the left hand side. It is not useful then to trying bound LHS between, for example, [imath](ax+b)^3[/imath] and [imath](ax+c)^3[/imath] and then say that [imath]ax+b<y<ax+c[/imath]. I also tried to use modulo a prime. But it seems unlikely to bound variables this way. EDIT : Though, it can be factored as [imath](2x+15)(x^2+15x+120)=(y/2)^3[/imath]. LSH factors are almost co-prime and we can say that [imath]x^2+15x+120=3z^3[/imath] or [imath]x^2+15x+120=5z^3[/imath]. These are still too difficult to solve! Any ideas? |
2385774 | Not sure about a convergent series
I solved this series [imath]\sum_{n=2}^\infty \frac 1 {n^2\ln n}[/imath] wit Condensation Test and I got now [imath]\sum_{n=2}^\infty \frac{1}{2^nnln(ln(2))}[/imath] Can I use now a geometric series like [imath]\sum_{n=0}^\infty \frac 1 {2^n}[/imath] with comparasion test for say that [imath]\frac{1}{2^nln(ln(2))}<\frac 1 {2^n}[/imath] and get the conclusion that the series is Convergent? Thanks | 2385717 | Divergent or not series?
How tipe are this series? [imath]\sum_{n=2}^\infty \frac 1 {n^2\ln n}[/imath] But [imath]\sum_{n=2}^\infty \frac 1 {n\ln n\ln(\ln n)}[/imath] I used Condensation Test, but I stack after that. Thanks! |
1119049 | Difficulty with Jensen's Equation.
Its easy to find all continuous function [imath]f: \Bbb R \to \Bbb R [/imath] satisfing the Jensen equation [imath]f \left( \frac{x+y}{2}\right )=\frac{f(x)+f(y)}{2}[/imath] But I am finding difficulty in finding all function continuous on [imath](a,b),a,b \in \Bbb R[/imath], satisfying the Jensen equation. | 1850101 | Function that is both midpoint convex and concave
Which functions [imath]f:\mathbb{R} \to \mathbb{R}[/imath] do satisfy [imath]f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}[/imath] for all [imath]x,y \in \mathbb{R}[/imath] I think the only ones are of type [imath]f(x) = c[/imath] for some constant [imath]c\in \mathbb{R}[/imath] and the solutions of the Cauchy functional equation [imath]f(x+y) = f(x)+f(y)[/imath] and the sums and constant multiples of these functions. Are there other functions which are both midpoint convex and concave? |
2386332 | A dilogarithm identity (simplification/compaction)
I'm wondering if there is any compact expression to compute (or approximate): [imath]\operatorname{Li}_2(pe^{-\alpha})-\operatorname{Li}_2(pe^{\alpha})[/imath] or [imath]\operatorname{Re}\{\operatorname{Li}_2(pe^{-\alpha})-\operatorname{Li}_2(pe^{\alpha})\}[/imath] The problem is that I cannot use [imath]\operatorname{Li}_2(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^2}[/imath] as for my values, [imath]|z|[/imath] is NOT necessarily less than 1. I am hoping, maybe, having exponential inputs to [imath]\operatorname{Li}_2[/imath] might do some magic. Unfortunately, [imath]p[/imath] and [imath]\alpha[/imath] are pure real. A nice approximation or any piece of advice is greatly appreciated. | 2383308 | A dilogarithm identity?
I'm wondering whether there any nice identities (or relationships) that can simplify or possibly compact the following expressions: [imath]\operatorname{Li}_2(\beta e^{\alpha x}) - \operatorname{Li}_2(\beta e^{-\alpha x}) [/imath] and [imath]\operatorname{Li}_2(pe^{\alpha x}) + \operatorname{Li}_2(qe^{\alpha x}) [/imath] |
2288082 | Finding [imath]\det(I+A^{100})[/imath] where [imath]A\in M_3(R)[/imath] and eigenvalues of [imath]A[/imath] are [imath]\{-1,0,1\}[/imath]
I have a matrix [imath]A \in M_3(R)[/imath] and it is known that [imath]\sigma (A)=\{-1, 0, 1\}[/imath], where [imath]\sigma (A)[/imath] is a set of eigenvalues of matrix [imath]A[/imath]. I am now supposed to calculate [imath]\det(I + A^{100})[/imath]. I know that [imath]A^{100}[/imath] could be calculated using a diagonal matrix which has the eigenvalues of [imath]A[/imath] on it's diagonal and using matrices which are formed using the eigenvectors of [imath]A[/imath], but I am not sure how to get there. Or it might not even be the right approach. I know there is a similar question, but I don't really understand the answer given there. It's not fully explained. So if anyone could help, that would be great. Thanks | 2394219 | What is [imath]\det(I + A^{50})[/imath] if [imath]A[/imath] has eigenvalues [imath]1[/imath], [imath]-1[/imath] and [imath]0[/imath]?
Since [imath]3 \times 3[/imath] matrix [imath]A[/imath] has [imath]0[/imath] as one of its eigenvalues, we know that [imath]A[/imath] is singular. From this we can conclude that [imath]\det(A) = 0[/imath] also [imath]\mbox{tr} (A) = 0[/imath]. Further, [imath]\det(A^{50}) = 0[/imath]. Now I think that [imath]\mbox{tr} (I + A^{50}) = 3[/imath] but how can I get the determinant from here? |
2387110 | Prove that a bounded operator [imath]A[/imath] on Banach space is invertible (bijective) if [imath]\left\|I - A\right\| < 1 [/imath] holds?
I can't seem to make any progress with the following problem: If [imath] A [/imath] is a bounded operator on Banach space [imath] X [/imath] and if [imath] \lVert I - A \rVert < 1 [/imath], prove that [imath] A [/imath] is invertible/bijective. This seems to be a known result, but I wasn't able to found a source. | 1826577 | [imath]||T-I|| < 1[/imath] implies that [imath]T[/imath] is invertible.
Let [imath]B[/imath] be a banach space and [imath]T : B \to B[/imath] be a bounded linear transformation. If for identity transformation [imath]I : B \to B[/imath] , [imath]||T-I||[/imath]<1 , then [imath]T[/imath] is invertible. || || is norm of transformation. |
2387114 | Determining whether Newton-Raphon's method will converge linearly or quadraticaly
So lets say you have the following two polynomials: [imath]f_1(x)=(x-1)(x-2)^2[/imath] and [imath]f_2(x)=(x-1)^2(x-2)[/imath] And we want know whether Newton-Raphon's method will converge quadraticly or linearly for finding the root [imath]x=2[/imath]. How would I go by doing this? | 805490 | Newton's method linear convergence proof
How would you show that if f'(a)=0 then the Newton's Method is linear convergent when 1. [imath]f''(a)\neq 0[/imath] 2. [imath]f''(a)=0, f'''(a) \neq 0[/imath]? I am having some trouble getting it to the point where you can take the limit of ratio of the error terms and using L'hospital's rule to prove this. |
2031856 | Inequality with five variables. Prove that [imath]\sum\limits_{cyc}\frac{a-b}{b+c}\geq0[/imath]
Let [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], [imath]d[/imath] and [imath]e[/imath] be positive numbers. Prove that: [imath]\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq0[/imath] I tried C-S, BW and more, but without success. | 621937 | How prove this inequality [imath]\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0[/imath]
let [imath]a,b,c,d,e[/imath] are positive real numbers,show that [imath]\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0[/imath] My try: I have solved follow Four-variable inequality: let [imath]a,b,c,d[/imath] are positive real numbers,show that [imath]\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge 0[/imath] poof:since \begin{align*} &\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}=\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}-4\\ &=(a+c)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)+(b+d)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)-4 \end{align*} By Cauchy-Schwarz inequality we have [imath]\dfrac{1}{b+c}+\dfrac{1}{d+a}\ge\dfrac{4}{(b+c)+(d+a)},\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{(c+d)+(a+b)}[/imath] so we get [imath]\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge\dfrac{4(a+c)}{(b+c)+(d+a)}+\dfrac{4(b+d)}{(c+d)+(a+b)}-4=0[/imath] Equality holds for [imath]a=c[/imath] and [imath]b=d[/imath] By done! But for Five-variable inequality,I can't prove it.Thank you |
2387335 | Showing the following integral is 0.
Let [imath]f[/imath] be in [imath]L^1(\mathbb{R}). [/imath] Show that [imath]\lim_{a\to 0}\int_{\mathbb{R}}|f(x)-f(x-a)|dx=0.[/imath] As [imath]f\in L^1(\mathbb{R})[/imath], we have [imath]\int_{\mathbb{R}}|f(x)|dx<\infty. [/imath] Define [imath]f_n(x):=f(x-a_n)[/imath]. For each fixed [imath]n[/imath], [imath]f_n(x)\in L_1(\mathbb{R}).[/imath] I don't know how to solve this kinda of problems. Should I apply Fatou's lemma or the dominated convergence theorem? But I'm not evey sure if [imath]f[/imath] is bounded. I appreciate any help for this problem. I don't undstrand the proof provided in the mentioned link. I'm very new in measure theory and I think there must be some easier solution for this question. | 2065360 | Do we have [imath]\lim_{h\to 0^+}\int_0^\infty|f(t+h)-f(t)|dt=0[/imath]?
Let [imath]f:\mathbb{R^+}\to\mathbb{R}[/imath] be an integrable function ([imath]f\in L^1(\mathbb{R}^+,\mathbb{R})[/imath]). Do we have [imath]\lim_{h\to 0^+}\int_0^\infty|f(t+h)-f(t)|dt=0[/imath] ? How can we prove it ? |
2387364 | Why do [imath]2\log x[/imath] and [imath]\log x^2[/imath] look different when graphing?
I understand how both graphs are drawn, but I do not understand why you cannot just convert one into another. It feels natural to me to just convert [imath]2\log x[/imath] to [imath]\log x^2[/imath] and not have to worry about the domain restriction. | 31331 | [imath]\ln(x^2)[/imath] vs [imath]2\ln x[/imath]
These two are supposed to be equivalent because of the properties of logarithms, but the domains of [imath]\ln(x^2)[/imath] and [imath]2\ln x[/imath] seem different to me. For example, if I substitute [imath]x=-1[/imath] into the first, I get 0. But in the second, I get a non-real answer. Why is this? The domains of these functions, when graphed, seem different as well. But everything I've been taught so far, and most of the things I can find on the web, do not explain this inconsistency. Typically, I consider the property [imath]\ln(a^b) = b \ln a[/imath] to be true... I am only a senior in high school, currently in Calculus I, but hopefully the explanation won't be too outside of the realm of my understanding. But even if it is, I would still like to know the answer. Thanks in advance |
2387724 | With [imath]X[/imath] compact Hausdorff space and [imath]f:X \to X[/imath] is continuous, prove there exists a closed set [imath]C[/imath] such that [imath]f( C)=C[/imath]
Prove that if [imath]X[/imath] is a non-empty topological Hausdorff compact space and [imath]f : X\to X[/imath] is continuous, there exists a non-empty closed set [imath]C \subset X[/imath] such that [imath]f(C)=C[/imath]. My thoughts are to show that we can travel from a starting point [imath]x[/imath] through [imath]f(x),f(f(x)),...,f^n(x)[/imath] such that [imath]f^n(x)=x[/imath]. that's we found a closed set. I can't seem to find the use of [imath]f[/imath] being continuous. Thanks! | 127020 | Fixed Set Property?
As far as I know, there are fixed-point-like results for continuous functions from a convex compact subset [imath]K[/imath] of an Euclidean space to itself. I have one question in mind: Does there exist a set [imath]A\subseteq X[/imath] for which [imath]f(A)=A[/imath]? Let's say [imath]X[/imath] is a compact metric space and [imath]f[/imath] is continuous. |
2388572 | meaning of comma in bottom of binomial coefficient
I know what the binomial coefficient means in combinatorics, e.g. [imath]\binom6{2}[/imath] means "6 choose 2" i.e. how many different subsets of size 2 can there be, out of a set of 6 elements. But what does it mean when there is a comma in the bottom of the coefficient, for example [imath]\binom6{2,2}[/imath]? I saw this usage in the first answer here, and I'd never seen it before: Problem 7, Ch1 from Blitzstein and Hwang, Intro to Probability | 2312648 | What does a binomial coefficient with commas mean?
I'm reading through this paper (on Dyck Paths). Near the middle of the second page, the author states the following: Remark For the set [imath]h_n[/imath] there are [imath]{k\choose t_1, t_2, ... , t_m }[/imath] [imath]n + k \choose{k}[/imath] [imath]=[/imath] [imath] n + k\choose {n, t_1, t_2, ..., t_m}[/imath] different Dyck paths. What does [imath]{k\choose t_1, t_2, ... , t_m }[/imath] mean? I know what the binomial coefficient is, but I'm not sure how to interpret it when one of the parameters is a set. |
2388354 | What is Aut [imath]Z \oplus Z[/imath]?
What is Automorphism group of [imath]Z \oplus Z[/imath]? Please give some hints. I feel it is some matrix group looking as the group as a Z module. | 210487 | [imath]\text{Aut}(F)[/imath] is isomorphic to the multiplicative group of all [imath]n\times n[/imath] matrices over [imath]\mathbb Z[/imath]
I want to prove that: If [imath]F[/imath] is a free abelian group of rank [imath]n[/imath], then [imath]\text{Aut}(F)[/imath] is isomorphic to the multiplicative group of all [imath]n\times n[/imath] matrices over [imath]\mathbb Z[/imath] with [imath]\text{det}=\pm1[/imath]. What I have tried: Since [imath]F[/imath] is a free abelian group of rank [imath]n[/imath] so I can write [imath]F=\langle x_1,x_2,...,x_n\rangle[/imath]. If [imath]\phi\in\text{Aut}(F)[/imath] then it should map any [imath]x_i[/imath] to another element [imath]x_j[/imath] of the basis [imath]X=\{x_1,x_2,...,x_n\}[/imath]. My idea is to map any element of [imath]\text{Aut}(F)[/imath] to a suitable matrix. For example when [imath]n=2[/imath]: [imath]\phi_1:=\{x_1\to x_1, x_2\to x_2\}\Longrightarrow \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\\ \phi_2:=\{x_1\to x_2, x_2\to x_1\}\Longrightarrow \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)[/imath] Am I on a right approach? Or there is another better way? Thanks. |
2388760 | Let [imath]\{ a_{n}\}[/imath] be a positive real sequence [imath]a_{n}=\sqrt{a_{n-1}a_{n-2}}[/imath] for [imath]n≥3[/imath], then [imath]\{ a_{n}\}[/imath] converges to [imath](a_{1}a_{2}^{2})^{1/3}[/imath].
Let [imath]\{ a_{n}\}[/imath] be a sequence of positive real numbers such that [imath]a_{n}=\sqrt{a_{n-1}a_{n-2}}[/imath] for [imath]n≥3[/imath], then [imath]\{ a_{n}\}[/imath] converges to [imath](a_{1}a_{2}^{2})^{\frac{1}{3}}[/imath]. My attempt:- I multiplied the all new terms and simplify the terms based on the recursive relation [imath]a_{n}.a_{n-1}...a_{2}.a_{1}=\sqrt{a_{n-1}.a_{n-2}}.\sqrt{a_{n-2}.a_{n-3}}...a_{2}.a_{1}[/imath] Cancel the like terms, I get [imath]a_{n}\sqrt{a_{n-1}}=a_{2}\sqrt{a_{1}}[/imath]. Limit , I am getting the result. How is my steps? Does it have any mistakes? How to prove the existence?. I am not able to judge whether it is monotonically decreasing/ increasing. How to prove the sequence is bounded? | 975312 | Limit of a sequence defined by recursive relation : [imath] a_n = \sqrt{a_{n-1}a_{n-2}}[/imath]
We're given a sequence defined by the recursive relation: [imath]a_n = \sqrt{a_{n-1}a_{n-2}}[/imath] [imath]a_1[/imath] and [imath]a_2[/imath] are positive constants. We have to show the following: The sequences [imath]\{ b_n \} = \{ a_{2n-1} \}[/imath] and [imath]\{ c_n \} = \{ a_{2n} \} [/imath] are monotonic, and if one is increasing, the other is decreasing. The limit of the sequence [imath]\{ a_n \} [/imath] is [imath] \left(a_1a_2^2\right)^{\frac13} [/imath] Now, I have proved the first part. Besides that, I have also proved a few other things: If [imath]a_1 > a_2[/imath], then: a. [imath] \{ b_n \} [/imath] decreases, and [imath] \{ c_n \} [/imath] increases. b. [imath] c_n < b_n [/imath] If [imath]a_1 < a_2[/imath], we just flip [imath]\{ b_n \}[/imath] and [imath]\{c_n\}[/imath] Besides, I have also shown that both the sequences : [imath]\{b_n\}[/imath] and [imath]\{ c_n\}[/imath] have the same limit. What I don't know, is how to evaluate the limit. |
2389056 | How to evaluate this integral from -infinity to infinity (improper integral)?
I'm having difficulty evaluating this integral: [imath]\int_{-\infty}^{\infty} {x^6 e^{-x^2}} dx[/imath] All I've been able to do is separate them and evaluate them separately although I haven't been able to successfully do that: [imath]\int_{-\infty}^{0} {x^6 e^{-x^2}} dx + \int_{0}^{\infty} {x^6 e^{-x^2}} dx[/imath] I was given this hint although I can't understand how they are equal (the integral on the RHS is the Gaussian integral): [imath]\int_{0}^{\infty} {x^6 e^{-x^2}} dx = -\frac{d^3}{da^3} \int_{0}^{\infty} { e^{-ax^2}} dx [/imath] That's pretty much all I've got so far. Any help would really be appreciated! Edit: This is different from the supposed duplicate because in the other one, I was asked to make one post per question and the answer was the hint but I didn't understand how it was equal. | 2388199 | How to evaluate the following items in mathematical methods in physics?
I had taken a long break from math and physics for months due to stress and an illness so when I returned to it and tried answering problems given, I had difficulty figuring out how to do some of them all over again. Here were the items I had encountered: 1) Evaluate [imath]\sum_{n=0}^{\infty} {p^n}{sin(2nx)}[/imath] 2) Solve for x(t): [imath]\frac{dx}{dt} = x_o =-cos(t)x + t^3, x(t=0)[/imath] 3) Evaluate [imath]\int_{0}^{\pi} \frac{d\theta}{(a+cos\theta)^2}, a>1[/imath] 4) Evaluate [imath]\int_{-\infty}^{\infty} {x^6 e^{-x^2}} dx[/imath] 5) Evaluate [imath]L_n(0)[/imath] given the generating function for Laguerre polynomials is [imath]g(x,t) = \sum_{n=0}^{\infty} L_n(x) t^n = \frac {e^{xt/(1-t)}}{1-t}[/imath] 6) Solve: [imath]y'' + x^2 y'+(x-1)y = 0[/imath] -> For this, I tried the Frobenius method but didn't get far. Some weren't taught in my previous class but I was expected to already know it for my course this coming semester. I have tried searching already but I have not gotten far for these. I am not expecting all of these to be answered. I just really need help getting started again so any help for any of these would be really appreciated. You can just choose one among them. If you could perhaps give me a way I can try, step-by-step, or a really good reference for me to look at so I can self-study. |
2389075 | Are the determinants of two specific matrices equal?
Suppose F is a field, [imath]X \in M_n(F)[/imath], [imath]XX^t = I[/imath], [imath]det(X) = 1[/imath], [imath]X = \begin{bmatrix} A & B \\ C & D \end{bmatrix}[/imath], where A and D are square matrices. Is it always true that det(A) = det(D)? This statement seems to be true for all the examples I have tried, but I do not know how to prove it in general. Any help will be appreciated. | 122639 | Sub-determinants of an orthogonal matrix
Let [imath]A[/imath] be a matrix in the special orthogonal group, [imath]A \in SO_n[/imath]. This means that [imath]A[/imath] is real, [imath]n \times n[/imath], [imath]A^t A = I[/imath] and [imath]Det(A)=1[/imath], that is, the column vectors of [imath]A[/imath] make a positively-oriented orthonormal basis for [imath]\mathbb R^n[/imath]. Decompose [imath]A[/imath] as a block matrix [imath]A = \begin{pmatrix} B & C \\ D & E\end{pmatrix}[/imath] where [imath]B[/imath] is [imath]k \times k[/imath] and [imath]E[/imath] is [imath](n-k)\times (n-k)[/imath]. I'm looking for a basic linear-algebra argument that [imath]Det(B) = Det(E)[/imath], ideally something that could be presented in a 2nd or 3rd year undergraduate course. So I do not want people to invoke anything like tensor products or differential forms. |
2389478 | Is that group abelian?
Suppose [imath]G[/imath] is a group and [imath]N[/imath] is its normal subgroup, such that every subgroup of [imath]N[/imath] is normal in [imath]G[/imath] and [imath]C_G(N)\subset N [/imath]. Is [imath]\frac{G}{N}[/imath] always abelian? I do not know such [imath]G[/imath] and [imath]N[/imath] where it is not, but neither I know general proof of this statement. Any help will be appreciated. | 1024075 | If N and every subgroup of N is normal in G then G/N is abelian .
Let [imath]N[/imath] be a normal subgroup of a group [imath]G[/imath] such that every subgroup of [imath]N[/imath] is normal in [imath]G[/imath] and [imath]C_G(N)\subseteq N [/imath]. Prove that [imath]G/N[/imath] is abelian. Here, as usual, [imath]$C_G\left(N\right)$[/imath] means the centralizer of [imath]N[/imath] in [imath]G[/imath] (i.e., those elements of [imath]$G$[/imath] that commute with everything in [imath]$N$[/imath]). I think we need to use that every subgroup of [imath]$N$[/imath] is normal in [imath]$G$[/imath] but i can't use .Please help me with Hints. |
2389523 | A Cyclic Extension of degree 4 does not contain [imath]i[/imath]
The full question is: Let [imath]K \subseteq \mathbb{C}[/imath] be a cyclic exntesion of [imath]\mathbb{Q}[/imath] of degree 4. Prove that [imath]i \notin K[/imath]. I was thinking that since [imath]K[/imath] over [imath]\mathbb{Q}[/imath] is finite (has degree 4) and is separable (since it's Galois), by the Primitive Element Theorem, [imath]K = \mathbb{Q}(\alpha)[/imath] for some [imath]\alpha \notin \mathbb{Q}[/imath]. So if I show that [imath]i \neq \alpha[/imath] then the statement holds. So suppose [imath]i = \alpha[/imath]. Then [imath]K = \mathbb{Q}(i)[/imath]. However, the degree of [imath]\mathbb{Q}(i)[/imath] over [imath]\mathbb{Q}[/imath] is not equal to 4. So [imath]i \neq \alpha[/imath]. Does this work? If not, please explain why/any ideas you may have. Thank you! | 1912901 | Galois Basic Question (Cyclic Order 4 Extension of Q cannot contain i)
Suppose [imath]K[/imath] is an extension of [imath]\mathbb{Q}[/imath] in [imath]\mathbb{C}[/imath], where [imath]Gal(K/\mathbb{Q})[/imath] is cyclic of order 4. Show that [imath]i\notin K[/imath]. ([imath]i[/imath] is the imaginary number [imath]i^2=-1[/imath].) My Galois theory is quite weak, hope someone can check if my attempt is correct. My attempt: Suppose to the contrary [imath]i\in K[/imath]. Let [imath]\sigma\in Gal(K/\mathbb{Q})[/imath]. Note that [imath]\sigma (i)\sigma(i)=\sigma(i^2)=\sigma(-1)=-1[/imath] since [imath]\sigma[/imath] fixes [imath]\mathbb{Q}[/imath]. This means that [imath]\sigma(i)=i[/imath] or [imath]\sigma(i)=-i[/imath]. The first case is ruled out since [imath]\sigma[/imath] only fixes [imath]\mathbb{Q}[/imath]. So [imath]\sigma(i)=-i[/imath]. This means that [imath]\sigma(a+bi)=a-bi[/imath], so [imath]\sigma[/imath] is effectively complex conjugation, which has order 2. Since [imath]\sigma[/imath] was arbitrary, this contradicts that [imath]Gal(K/\mathbb{Q})[/imath] has an element of order 4. Is this ok? Thanks. Update: Now I see that my argument is clearly flawed. What would be the correct proof? |
1380775 | Matrix [imath]\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}[/imath] to a large power
Compute [imath]\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}[/imath] What is the easier way to do this other than multiplying the entire thing out? Thanks | 499646 | How can I show that [imath]\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}[/imath]?
Well, the original task was to figure out what the following expression evaluates to for any [imath]n[/imath]. [imath]\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}[/imath] By trying out different values of [imath]n[/imath], I found the pattern: [imath]\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}[/imath] But I have yet to figure out how to prove it algebraically. Suggestions? |
2389987 | Find the infinite series?
Find the sum of the following infinite series [imath]\frac {1}{3!} + \frac {4}{4!} + \frac {9}{5!} + \ldots[/imath] I was trying this question ,but I couldn't get it. This is homework that my teacher has given me. I solved this question using d'Alembert's ratio test but I didn't get a finite value. If anyone could help me, that would be much appreciated. Thank you | 1094852 | Sum [imath]\sum_{n=1}^\infty \frac{n^2}{(n+2)!}[/imath]
Problem is to find sum [imath]\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots[/imath] What I knew doesn't apply on this problem Some series are telescoping, some types are solvable using binomial , both look useless here Binomial gives [imath]n (n-1) (n-2)x^3=1[/imath] [imath]n (n-1)( n-2)(n-3)x^4=4[/imath] What approach to use here? |
2390103 | Evaluating [imath]\lim_{x \to 0} \frac{1}{x}\ln(1 + x)[/imath]
[imath]\lim_{x \to 0} \frac{1}{x} \ln(1 + x) = 1 [/imath] Limit is of type [imath]+\infty \cdot 1[/imath], so must be [imath]+\infty[/imath], but answer is natural exponential to the power [imath]1/2[/imath]. | 690898 | Determine the following limit as x approaches 0: [imath]\frac{\ln(1+x)}x[/imath]
[imath]\lim_{x\to 0} \frac{\ln(1+x)}x[/imath] The process I want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule) [imath]\lim_{h \to 0} \frac{f(x+h) - f(x)}h[/imath] [imath]\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h[/imath] [imath]\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}[/imath] At this point I get confused because I know the answer is [imath]1[/imath], but I am not getting this answer through simplification of my formula. |
2390360 | Trigonometry Minimum value using A.M. and G.M.
If [imath]f(x)=(\sin x + \csc x)^2+(\cos x + \sec x)^2[/imath], then what is the minimum value of [imath]f(x)[/imath]? My Try : Now upon using Arithmetic Mean is greater than or equal to Geometric Mean I got minimum values for the following expressions as :- [imath]\sin^2x+\csc^2x=2[/imath] and also, [imath]\cos^2x+\sec^2x=2[/imath] But surprisingly, the answer I get after opening the square and using this is [imath]8[/imath] while my book shows its [imath]9[/imath]. Someone please tell me what I'm doing wrong. | 1065943 | To prove [imath](\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9[/imath]
I used the following way but got wrong answer [imath]A.M. \ge G.M.[/imath] [imath] \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}[/imath] Squaring both sides, \begin{equation*} (\sin\theta + \csc\theta )^2 \ge 4 \tag{1} \end{equation*} Similarly \begin{equation*} (\cos\theta + \sec\theta )^2 \ge 4 \tag{2} \end{equation*} Adding equation (1) and (2) \begin{equation*} (\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8 \end{equation*} What is wrong? |
2387116 | Sequence of random variables [imath]X_n[/imath] converging in probability to [imath]X[/imath] and [imath]X_n[/imath] converging in distribution to [imath]Z[/imath], then [imath]X[/imath] has same distribution as Z?
Suppose [imath]\left\{X_n\right\}_{n\in\mathbb{N}}[/imath] is a sequence of random variables on [imath](\Omega,\mathscr{F},\mathbb{P})[/imath] such that [imath]X_n[/imath] converges in probability to [imath]X[/imath]. Moreover suppose that [imath]X_n[/imath] converges in distribution to a random variable [imath]Z[/imath]. My question is does [imath]X[/imath] have the same distribution as [imath]Z[/imath]? And could you please provide an proof for it? Thanks =) | 994884 | Is the limit of a sequence of random variables unique?
If [imath]Y[/imath] and [imath]Z[/imath] are two distinct random variables with the same distribution (for example maybe [imath]Y[/imath] is constant equal to [imath]1[/imath] and [imath]Z[/imath] is equal to [imath]1[/imath] almost everywhere), then surely any sequence [imath]X_n[/imath] which converges in distribution to [imath]Y[/imath] also converges to [imath]Z[/imath]. This seems to conflict with the fact that convergence in distribution is metrizable, and therefore limits should be unique. What about convergence in probability, or almost sure convergence? |
2391034 | Find out value of [imath](3Z^{100}+\frac{2}{Z^{100}}+1)(Z^{100}+\frac{2}{Z^{100}}+3)?[/imath]
If a complex number [imath]Z[/imath] satisfy following relationship [imath](Z+\frac{1}{Z})(Z+\frac{1}{Z}+1)=1[/imath] then find out value of [imath](3Z^{100}+\frac{2}{Z^{100}}+1)(Z^{100}+\frac{2}{Z^{100}}+3)?[/imath] My Attempt let [imath]u=(Z+\frac{1}{Z})(Z+\frac{1}{Z}+1)[/imath] Then given condition become [imath](u)(u+1)=1[/imath]. [imath](3Z^{100}+\frac{2}{Z^{100}}+1)(Z^{100}+\frac{2}{Z^{100}}+3)=(2(Z^{100}+\frac{1}{Z^{100}})+Z^{100}+1)((Z^{100}+\frac{1}{Z^{100}})+\frac{1}{Z^{100}}+3)[/imath]. With this I expected to reduce it down in terms of [imath](Z+\frac{1}{Z})[/imath] and then using given condition we can find the value. But I am unable to reduce it down in terms of [imath](Z+\frac{1}{Z})[/imath]. Can anyone help? Or am I making a mistake in strategy? Thanks in advance. | 364214 | [imath]c[/imath] is a complex number that satisyfing [imath](c+\frac{1}{c}+1)(c+\frac{1}{c}) = 1[/imath]
Let [imath]c[/imath] is complex-number satisfying : [imath](c+\frac{1}{c}+1)(c+\frac{1}{c}) = 1[/imath] So, how could i get [imath](3c^{100}+\frac{2}{c^{100}}+1)(c^{100}+\frac{2}{c^{100}}+3)[/imath] ? |
2391576 | A function that maps from A to its power set.
Let [imath]A[/imath] be a non empty set and [imath]f: A \to P(A)[/imath] some function.[imath]P(A)[/imath] is the set of all the subsets of the set [imath]A[/imath]. I have to show that [imath]Z_{f} := \{x \in A | x \not\in f(x)\}[/imath] is not in the image of [imath]f[/imath]. I have no idea how to do this, please help. I guess showing that [imath]f[/imath] is not surjective would be one way, since that means it won't take the whole image space but further than that I don't understand.. Thank you in advance | 1113507 | Cantor - No set is equinumerous with its power set.
Theorem: No set is equinumerous with its power set. Proof: Let [imath]A[/imath] be a set. We want to show that if [imath]f: A \to \mathcal{P}A[/imath] (a random function) then [imath]f[/imath] is not surjective. We define the set [imath]D=\{ x \in A: x \notin f(x)\}[/imath] and obviously [imath]D \in \mathcal{P}A[/imath]. We assume that there is an [imath]a \in A[/imath] such that [imath]f(a)=D[/imath]. Then we have: [imath]a \notin D \leftrightarrow a \notin f(a) \leftrightarrow a \in D, \text{ contradiction}[/imath] Therefore for each [imath]x \in A, f(x) \notin D[/imath], i.e. [imath]f[/imath] is not surjective. Could you explain me the proof from the point where we assume that there is an [imath]a \in A[/imath] such that [imath]f(a)=D[/imath]? We have show that [imath]D \in \mathcal{P}A[/imath] and we want to show that [imath]D \notin f(A)[/imath] in order to show that [imath]\mathcal{P}A[/imath] isn't the image of [imath]f[/imath]. Why do we do it like that? Why do we assume an [imath]a \in A[/imath] such that [imath]f(a)=D[/imath]? |
2391685 | Infimum of a metric is continuous
So assume [imath](X,d)[/imath] is a metric space with [imath]A \subseteq X[/imath] as a subspace. Show that the function [imath]f: X \rightarrow \mathbb{R}[/imath] defined by [imath]f(x)=\inf\{d(x,a)|a\in A\}[/imath]is continuous. My instinct is to use the definition of continuity for metric spaces. Namesly, given any [imath]\epsilon >0[/imath], there is a [imath]\delta[/imath], such that: [imath]d(x,y) < \delta \implies d(f(x),f(y))< \epsilon[/imath], can I therefore assume that [imath]\mathbb{R}[/imath] has the usual topology or does that change the nature of what's being asked? Thanks | 1342934 | Distance of a point to a subset.
Let [imath](M,d)[/imath] be a metric space. For a subset [imath]A\subseteq M[/imath] we define the distance of a point [imath]x[/imath] to [imath]A[/imath] as [imath]\alpha_A(x):=\operatorname{dist}(x,A):=\inf_{y\in A}d(x,y)[/imath] Prove that: a) [imath]|\alpha_A(x_1)-\alpha_A(x_2)|\leq d(x_1,x_2)\text{ for all }x_1,x_2\in M[/imath]. b) [imath]\alpha_A:M\to\mathbb{R}[/imath] is continuous. c) [imath]\alpha_A(x):=0\leftrightarrow x\in\bar{A}[/imath]. d) If [imath]A[/imath] is closed and [imath]K\subseteq M[/imath] compact with [imath]A\cap K=\emptyset[/imath], then [imath]A[/imath] and [imath]K[/imath] have a positive distance, meaning [imath]\operatorname{dist}(K,A):=\inf d(x,y)>0\text{ for }x\in K,y\in A.[/imath] My ideas so far: a) I can imagine that this is true because I imagine [imath]A[/imath] being a ball and individual distance on [imath]\alpha[/imath] is the closest distance to that ball. But I don't know how to write down the proof mathematically. b) Now I'm lost here. Don't know how to approach this. c) Again, I can easily imagine that since [imath]x[/imath] is in [imath]A[/imath], meaning the closest distance to the subset is [imath]0[/imath], but I can't prove that mathematically. d) Similar to before, I have no idea how to write down the mathematical proof. But I imagine this one as a situation with two separate balls, since the intersection is the empty set. Meaning that the "gap" between them is the positive distance. Can someone show me how to write down mathematical proofs on this type of questions? |
2392289 | Determine if all [imath]n \times n[/imath] matrices such that [imath]AB = BA[/imath] for a fixed [imath]n \times n[/imath] matrix [imath]B[/imath] form a subspace of [imath]M_{nn}[/imath]
I don't know how to start this one. Can I show that [imath]A[/imath] is closed under addition and scalar multiplication by saying for following: If [imath]a, b \in \mathbb R[/imath] and [imath]A[/imath] and [imath]B[/imath] are [imath]n \times n[/imath] matrices such that [imath]AB = BA[/imath] then [imath](aA + bB)B = aAB + bBB[/imath] I'm not even sure this is legitimate what I have done so far. Some help starting would be appreciated. Why in the duplicate question in the accepted answer does @Arturo say that you also need to show that there is at least one matrix such that [imath]AB = BA[/imath]? I thought you only have to show closure under addition and scalar multiplication? As has @Samuel noted below also. | 74357 | Is the set of all [imath]n\times n[/imath] matrices, such that for a fixed matrix B AB=BA, a subspace of the vector space of all [imath]n\times n[/imath] matrices?
Is the set of all [imath]n\times n[/imath] matrices, such that for any fixed matrix B AB=BA, a subspace of the vector space of all [imath]n\times n[/imath] matrices? Alright, I understand the question and I know what I have to do, basically. I need to show that additive closure and multiplicative closure are satisfied. The problem is, I can't figure out how to do this generally. I tried playing around with [imath]2\times 2[/imath] matrices but that seemed like a dead end. Obviously two such matrices are the 0 matrix and the identity matrix, and those form a subspace, but that doesn't really tell me about all the matrices. Any ideas for how I should be tackling this? I feel like I'm not thinking generally enough. |
2359963 | Positive eigenvalues and/or positive definiteness of [imath]A^+DA[/imath]
I have stumbled across the following fact in robotics. Let [imath]A[/imath] an [imath]m\times n[/imath] matrix with [imath]rank(A)=n[/imath]. Consider also a diagonal matrix [imath]D[/imath] with positive entries. The following two claims are posed: Is [imath]A^+DA[/imath] is positive definite where [imath]A^+[/imath] is the pseudoinverse matrix of [imath]A[/imath]? Are the eigenvalues of [imath]A^+DA[/imath] positive? This is obvious if for example [imath]A[/imath] is a square matrix. Edit: I just found an example showing that claim 1. is not true in general. Specifically, for [imath]A=\left[\matrix{1 & 1\\0 & 1}\right]\qquad,\qquad D=\left[\matrix{10 & 0\\0 & 0.1}\right][/imath] then [imath]A^+DA+(A^+DA)^T=\left[\matrix{20 & 9.9\\9.9 & 0.2}\right][/imath] which is not positive definite. | 2379250 | Prove that [imath]J^+AJ[/imath] has positive eigenvalues
Let [imath]J \in \mathbb{R}[/imath] be a [imath]m \times n[/imath] matrix of rank [imath]n[/imath] and a [imath]A \in \mathbb{R}[/imath] be a [imath]m \times m[/imath] diagonal positive definite matrix. Denote by [imath]J^+[/imath] a pseudoinverse of [imath]J[/imath]. As was shown in this question [imath]J^+AJ[/imath] is not positive definite in general. However, I was wondering whether one can prove or disprove that [imath]J^+AJ[/imath] has real and positive eigenvalues. |
563569 | Dimension of local ring as vector space over [imath]\mathbb C[/imath]
I want to know what the dimension of each of the local ring [imath]\mathbb C[x,y]_p/(y^2-x^7,y^5-x^3)[/imath] is, where [imath]p\neq (0,0)[/imath] over [imath]\mathbb C[/imath]-vector space. I know the dimension of it in the origin point, but I don't know other cases. | 561351 | Krull dimension of this local ring
I want to know what the Krull dimension of this ring [imath]\mathbb C[x,y]_p/(y^2-x^7,y^5-x^3)[/imath] is, where [imath]p\neq (0,0)[/imath]. I know the dimension of it in the origin point, but I don't know other cases. |
2393153 | The series [imath]1-1+1-1+\dots[/imath]
[imath]1+x+x^2+\dots=\frac{1}{1-x}[/imath], where [imath]-1<x<1[/imath] But I saw in one video this expression is true for [imath]x=-1[/imath] too. So can I write [imath]1+x+x^2+\dots=\frac{1}{1-x}[/imath], where [imath]-1\le x<1[/imath]? | 544797 | Grandi's Series; tends to [imath]1/2[/imath], but why is this considered a valid sum?
Grandi's series, [imath]1+1-1+1-1+1-1+...[/imath] can be expressed as the below: [imath]\sum_{n=0}^\infty(-1)^n[/imath] Two valid sums that make sense to me are [imath]1[/imath], and [imath]0[/imath], depending on how you approach the series. [imath](1+1)-(1+1)-(1+1)-...=0[/imath], and [imath]1+(1-1)+(1-1)+(1-1)+...=1[/imath]. There is consensus, however, that the actual sum is [imath]\frac{1}{2}[/imath]. Why? I understand the approach of finding partial means of the series, and they do indeed tend to [imath]\frac{1}{2}[/imath], but it seems unintuitive to assert that the sum is neither [imath]1[/imath] or [imath]0[/imath]. A more convincing method I found was assuming the series is [imath]S[/imath], then shifting it such that [imath]S-1 = S[/imath], then through algebra finding [imath]S = \frac{1}{2}[/imath], but again, it seems more intuitive answer is either [imath]0[/imath] or [imath]1[/imath]. I say this strictly because adding and subtracting integers should equal an integer, never a fraction. Is this a characteristic of infinite series, which is not specific to Grandi's series? |
2392765 | Show that [imath] \left( 1 + \frac{x}{n}\right)^n[/imath] is uniformly convergent on [imath]S=[0,1][/imath].
Show that [imath] \left( 1 + \frac{x}{n}\right)^n[/imath] is uniformly convergent on [imath]S=[0,1][/imath]. Given [imath]f_n(x)=\left( 1 + \frac{x}{n}\right)^n[/imath] is a sequence of bounded function on [imath][0,1][/imath] and [imath]f:S \rightarrow \mathbb R[/imath] a bounded function , then [imath]f_n(x)[/imath] converges uniformly to [imath]f[/imath] iff [imath]\lim\limits_{n \rightarrow \infty} ||f_n - f|| = \lim\limits_{n \rightarrow \infty}\left(\sup |f_n(x) - f(x)| \right)= 0[/imath] As [imath]f(x) = \lim\limits_{n \rightarrow \infty} \left( 1 + \frac{x}{n}\right)^n = e^x[/imath] We have [imath]\begin{split} \lim\limits_{n \rightarrow \infty} \left\|\left( 1 + \frac{x}{n} \right)^n - e^x\right\| &= \lim\limits_{n \rightarrow \infty} \left|\sup_{x \in S}\left[\left( 1 + \frac{x}{n}\right)^n - e^x \right ]\right|\\ &= \lim\limits_{n \rightarrow \infty} \left|\left( 1 + \frac{1}{n} \right)^n - e^1 \right|\\ &= \lim\limits_{n \rightarrow \infty} |e-e| \\ &= 0 \end{split} [/imath] I am new to sequence. Is this appropriate to show convergence? Going back to the definition, How can I show that: "[imath]f_n(x)=\left( 1 + \frac{x}{n}\right)^n[/imath] is a sequence of bounded function on [imath][0,1][/imath]"? [imath]f:S \rightarrow \mathbb R[/imath] a bounded function? | 2036694 | Show that [imath](1+\frac{x}{n})^n \rightarrow e^x[/imath] uniformly on any bounded interval of the real line.
Show that [imath](1+\frac{x}{n})^n \rightarrow e^x[/imath] uniformly on any bounded interval of the real line. I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling quite a lot. My efforts so far have concentrated on trying to find a sequences, [imath]{a_n}[/imath] which tends to zero, such that [imath]|(1+\frac{x}{n})^n -e^x |\leq a_n[/imath] for all [imath]n[/imath]. But I have been unsuccessful thus far. All help is greatly appreciated. |
2393512 | Let [imath]\phi:G\rightarrow G[/imath] be a group homomorphism. Prove that there exists [imath]n[/imath] such that [imath]G\cong ker(\phi^n)\bigoplus \phi^n(G)[/imath]
Let [imath]G[/imath] be a finite abelian group. Let [imath]\phi:G\rightarrow G[/imath] be a group homomrphism. Prove that there exists [imath]n[/imath] such that [imath]G\cong ker(\phi^n)\bigoplus \phi^n(G)[/imath]. Any ideas? | 1079124 | Show that there is a positive integer [imath]n[/imath] such that [imath]G \cong \ker(\phi^{n}) \times \phi^{n}(G)[/imath]
Let [imath]G[/imath] be a finite abelian group and let [imath]\phi: G \rightarrow G[/imath] be a group homomorphism. I am trying to show that there is a positive integer [imath]n[/imath] such that [imath]G \cong \ker(\phi^{n}) \times \phi^{n}(G)[/imath]. I know that since [imath]G[/imath] is abelian we have that [imath]\ker(\phi^{k})[/imath] and [imath]\phi^{k}(G)[/imath] are normal subgroups of [imath]G[/imath] for any [imath]k[/imath] I also suspect we have the following towers which stabilize at [imath]m[/imath] and [imath]m'[/imath] [imath]\phi(G) \unrhd \phi^{2}(G) \unrhd \cdots \unrhd \phi^{m}(G)=\phi^{m+1}(G)=\cdots[/imath] [imath]\ker(\phi) \unlhd \ker^{2}(\phi) \unlhd \cdots \unlhd \ker^{m'}(\phi)=\ker^{m'+1}(\phi)=\cdots[/imath] I know that if given a group of the form [imath]HK[/imath] where [imath]H[/imath] and [imath]K[/imath] are normal subgroups of [imath]HK[/imath] and [imath]H\cap K=1[/imath] then [imath]H \times K \cong HK[/imath]. I want to apply this to this situation but I am unable to show how write [imath]G[/imath] as a product [imath]HK[/imath] Resolution One of the comments directed me to the following article: http://en.wikipedia.org/wiki/Fitting_lemma The last part of which answers the question. Below is what is says: Choose [imath]n=\max(m,m')[/imath], then we have for [imath]x \in \ker^{n}(\phi) \cap \phi^{n}(x)[/imath], this means that [imath]x=\phi^{n}(y)[/imath] for some [imath]y \in G[/imath]. This gives: [imath]0=\phi^{n}(x)=\phi^{2n}(y)[/imath] which means that [imath]y \in \ker^{2n}\phi=\ker^{n}{\phi}[/imath]. and then we have that [imath]0=x=\phi^{n}(y)[/imath]. The answer below kindly points out that every element [imath]x[/imath] is contained in one of the cosets of [imath]G/\ker^{n}(\phi)[/imath], this means that [imath]x=k+g[/imath] for some [imath]k \in \ker^{n}(\phi)[/imath] and [imath]g \in G[/imath]. We also have that [imath]g=\phi^{n}(h)[/imath] for some [imath]h \in G[/imath]. Writing [imath]x=k+\phi^{n}(h)[/imath] essentially shows that [imath]G=\ker(\phi^{n}) + \phi^{n}(G)[/imath]. Using the fact above, the claim follows. |
2393753 | Cutting pizza with seven knife strokes
What is the maximum number of pieces that a pizza can be cut into by [imath]7[/imath] knife strokes? | 1657081 | Maximum number of pieces of pizza when making 7 cuts
If we have a circular pizza then the maximum number of pieces we can get by making [imath]7[/imath] cuts in it? The fact that I know the solution only got me the way to find it but it was like a kid trying to make all those cuts on a piece of a paper. is there some logical way of getting the answer? |
2393669 | Let [imath](X,\mathscr U)[/imath] be a uniform space. Then why does [imath]\mathcal N_x=\{U[x]\ |\ U\in\mathscr U\}[/imath] form a neighborhood base for [imath]x\in X[/imath]?
Let [imath](X,\mathscr U)[/imath] be a uniform space, i.e. [imath]X[/imath] is a set and [imath]\mathscr U\subset 2^{X\times X}[/imath] is a filter such that [imath]\Delta\subset U[/imath] for all [imath]U\in\mathscr U[/imath] [imath]U\circ V,U^{-1}\in\mathscr U[/imath] for all [imath]U,V\in\mathscr U[/imath] For all [imath]U\in\mathscr U[/imath], there exists some [imath]V\in\mathscr U[/imath] such that [imath]V\circ V\subset U[/imath] where we define [imath]U^{-1}=\{(y,x)\ |\ (x,y)\in U\}[/imath] and [imath]U\circ V = \{(x,z)\ |\ (x,y)\in U,(y,z)\in V\hbox{ for some }y\in X\}.[/imath] Furthermore we define [imath]U[x]=\{ y\in X\ |\ (x,y)\in X\}.[/imath] Meanwhile, for any set [imath]X[/imath] and family of filters [imath](\mathcal N_x)_{x\in X}[/imath] in [imath]X[/imath], we say that [imath](\mathcal N_x)_{x\in X}[/imath] forms a neighborhood filter base if [imath]x\in N[/imath] for all [imath]N\in\mathcal N_x[/imath] For all [imath]N\in\mathcal N_x[/imath], there exists some [imath]M\subset N[/imath], [imath]M\in\mathcal N_x[/imath] such that [imath]y\in M[/imath] implies that [imath]M\in\mathcal N_y[/imath]. A neighborhood filter base thus defines a topology [imath]\tau[/imath] on [imath]X[/imath]. Note that we still have a topology [imath]\tau[/imath] on [imath]X[/imath] if we remove condition 2, but in that case, [imath]\mathcal N_x[/imath] might not be the neighborhood filter associated with [imath]\tau[/imath] at [imath]x[/imath]. Now, if we set [imath]\mathcal N_x=\{U[x]\ |\ U\in\mathscr U\}[/imath] for all [imath]x\in X[/imath], then it is clear that [imath]\mathcal N_x[/imath] is a filter satisfying condition 1, but I have yet to succeed in proving condition 2. I suspect I must use conditions 2 and 3 for the definition of a uniform space, but I don't see how. | 2391994 | Neighbourhood filter of a uniform space
The Wikipedia page for Uniform space includes the following section: Every uniform space [imath]V[/imath] becomes a topological space by defining a subset [imath]O[/imath] of [imath]X[/imath] to be open if and only if for every [imath]x[/imath] in [imath]O[/imath] there exists an entourage [imath]V[/imath] such that [imath]V[x][/imath] is a subset of [imath]O[/imath]. In this topology, the neighbourhood filter of a point [imath]x[/imath] is [imath]\{V[x] : V∈Φ\}[/imath]. This can be proved with a recursive use of the existence of a "half-size" entourage. Compared to a general topological space the existence of the uniform structure makes possible the comparison of sizes of neighbourhoods: [imath]V[x][/imath] and [imath]V[y][/imath] are considered to be of the "same size". While I have worked through and proved that the topology described is in fact a topology, I'm stuck on how it can be proved that [imath]\{V[x] : V∈Φ\}[/imath] defines the neighbourhood filter of [imath]x[/imath], as I don't understand what the hint there is trying to tell me. I want to work through the full proof myself, but are there any extra hints to push me in the right direction? |
2394088 | Is it true that [imath]|a-b|[/imath]
Let [imath]a[/imath],[imath]b \in \mathbb{R}[/imath] be two real numbers and let [imath]t \in \mathbb{R}_+[/imath] a real positive number. I would like to know if is it true that [imath] |a-b|<t \Longrightarrow |a|<|b|+t [/imath] Thanks. | 2220089 | Show that [imath]|a-b|<\epsilon \Rightarrow |a|<|b|+\epsilon[/imath]
For any [imath]\epsilon>0[/imath] and [imath]a,b\in\mathbb{R}[/imath] show that [imath]|a-b|<\epsilon\Rightarrow |a|<|b|+\epsilon[/imath] In some notes I found the following property [imath]|a|-|b|\leq ||a|-|b||\leq|a-b|[/imath] So [imath]|a-b|<\epsilon\Rightarrow |a|-|b|<\epsilon\Rightarrow |a|<|b|+\epsilon[/imath] I have two doubts 1) Is it right? 2)In triangular inequality I have that [imath]|a+b|\leq |a|+|b|[/imath] [imath]|a+(-b)|\leq |a|+|-b|\leq|a|+|b|[/imath] Is wrong do that? |
2394269 | [imath]f^{'}(z)[/imath] exists if and only if [imath]\dfrac{df}{d\overline{z}} =0[/imath]
[imath]f^{'}(z)[/imath] exists if and only if [imath]\dfrac{df}{d\overline{z}} = \dfrac{df}{dx}\dfrac{dx}{d\overline{z}} + \dfrac{df}{dy}\dfrac{dy}{d\overline{z}} = \dfrac{1}{2}\dfrac{df}{dx}-\dfrac{1}{2i}\dfrac{df}{dy}=0[/imath] My professor briefly mentioned this in his notes, however, i tried to read up more on this but cannot find a convincing proof or the intuition behind this. Is it related to the Cauchy Riemann equations? And if let's say there is a question as such: Is the function [imath]f(z) = z + \overline{z}[/imath] differentiable? Instead of using the Cauchy Riemann, i can use the above formula and say that since [imath]\dfrac{df}{d\overline{z}} = 0 + 1 = 1 \neq 0[/imath], it follows that [imath]f[/imath] is not differentiable anywhere? Here i treated [imath]z[/imath] as a constant? Please help! Thanks | 1631088 | Show that a function satisfies the Cauchy-Riemann function if and only if [imath]\frac{df}{d\overline{z}}=0[/imath]
The full question is "Define the operator [imath]\frac{d}{d \overline{z}} = \frac{1}{2}(\frac{d}{dx}+i\frac{d}{dy})[/imath]. Show that a function [imath]f(z)=u(x,y)+iv(x,y)[/imath] satisfies the Cauchy-Riemann equation if and only if [imath]\frac{df}{d\overline{z}}=0[/imath]." I have already figured out the [imath]\rightarrow[/imath] direction and only need help with the [imath]\leftarrow[/imath] direction. So far I have: Let [imath]\frac{df}{d\overline{z}}=0[/imath] and [imath]f(z)=u(x,y)+iv(x,y)[/imath]. Then [imath]0=\frac{df}{d\overline{z}}[/imath] [imath]=\frac{1}{2}\frac{d}{dx}(u+iv)+\frac{i}{2}\frac{d}{dy}(u+iv)[/imath] [imath]=\frac{1}{2}(u_x-v_y)+\frac{i}{2}(u_y+v_x)[/imath] And this is where I get stuck. I am not sure where to go from there so any help would be much appreciated. (Also feel free to fix my formatting I know it is atrocious." |
1531811 | [imath]G[/imath] is a group with [imath]|G| = n[/imath] and [imath]p[/imath] is the smallest prime dividing [imath]|G|[/imath] , then any subgroup of index [imath]p[/imath] is normal.
What we do in this proof is : We say let [imath]G[/imath] acting on the left cosets of [imath]H[/imath] in [imath]G[/imath] where [imath]H[/imath] is a subgroup of [imath]G[/imath] , with [imath] | G : H | = p[/imath] We consider [imath]K[/imath] as the kernel of the action such that [imath] | H : K | = k[/imath], Then we make a statement that [imath] G/K [/imath] is isomorphic to a "subgroup" of [imath]S_p[/imath] , I know the first theorem of Isomorphism is applied here , but I am a little confused , According to the first isomorphism theorem , [imath] G/K [/imath] should be isomorphic to [imath]S_p[/imath] and hence , [imath] |G/K| = |S_p|[/imath] , i.e [imath] |G/K| = p![/imath] .. From where does that "subgroup" term comes from ? I know , only that "subgroup" helps in completing the proof as , that implies [imath]|G/K|[/imath] divides [imath]p![/imath] , but the actual statement of the theorem confuses me , could anyone help me with this? | 1825321 | If a subgroup has smallest prime index, then it is normal
Assume that [imath]G[/imath] is finite with [imath]p[/imath] the smallest prime dividing its order. Suppose [imath]H < G[/imath] with [imath][G:H]=p[/imath]. Prove that [imath]H \lhd G[/imath]. I've seen this question a few times on here but all the proofs I saw appeal to the First Isomorphism Theorem. I had old class notes showing this proved solely with group actions and orbits but I could only make out some of it so I recreated it from scratch. Can you guys check this for me? Let [imath]H[/imath] act on the set [imath]G/H[/imath] by left multiplication. Consider the orbit of [imath]H[/imath] itself. [imath]\mathcal{O}_H=\{hH \mid h\in H\}[/imath] and [imath]hH = H[/imath], thus [imath]\mathcal{O}_H[/imath] contains one coset, namely, [imath]H[/imath]. Let [imath]x\in G - H[/imath]. What can be said about [imath]\mathcal{O}_{xH}[/imath]? This action of left multiplication partitions [imath]G/H[/imath] into disjoint orbits of cosets. Because [imath]|G/H| = p[/imath], the sum of the sizes of the orbits is equal to [imath]p[/imath]. We already know that [imath]|\mathcal{O}_H| = 1[/imath] therefore [imath]|\mathcal{O}_{xH}| \leq p-1[/imath]. The Orbit-Stabilizer theorem gives us [imath]|\mathcal{O}_{xH}| |Stab_{xH}| = |H|[/imath]. We see that [imath]|\mathcal{O}_{xH}|[/imath] divides [imath]|H|[/imath] and hence divides [imath]|G|[/imath]. But since [imath]p[/imath] was the smallest prime dividing the order of [imath]G[/imath], [imath]|\mathcal{O}_{xH}| = 1[/imath]. The immediate consequence is that [imath]|Stab_{xH}| = |H|[/imath]. As [imath]Stab_{xH} \leq H[/imath], we have [imath]Stab_{xH} = H[/imath]. Therefore for every [imath]h \in H[/imath], [imath]hxH=xH[/imath]. This implies [imath]x^{-1}hx\in H[/imath] and thus [imath]x^{-1}Hx = H[/imath] for all [imath]x \in G[/imath]. Therefore, [imath]H \lhd G[/imath]. |
2391660 | Inequality involving the pseudo inverse [imath]A^+[/imath]
Say I have a matrix of the form [imath] A = \begin{bmatrix} A_1 \\ A_2 \end{bmatrix} [/imath] where [imath]A_1[/imath] is a [imath]n \times n[/imath] matrix of full rank, and [imath]A_2[/imath] is a [imath](m-n) \times n[/imath] matrix (arbitrary). I would like to show that [imath]||A^+||_2\leq||A_1^{-1}||_2[/imath] where [imath]A^{+}[/imath] is the pseudo inverse, defined [imath](A^*A)^{-1}A^*[/imath]. I have said that, [imath]A^+ = (A_1^{*}A_1 + A_2^{*}A_2 )^{-1}[A_1^* A_2^*][/imath] lets look at the first part of this matrix [imath](A_1^{*}A_1 + A_2^{*}A_2 )^{-1}A_1^* = [A_1 + A_1^{-*}A_2^*A_2]^{-1}[/imath] this is as far as I can go with regards to manipulating this matrix as [imath]A_2[/imath] is not square it does not have inverse. I am a bit unsure how to progress... | 2393912 | Pseudoinverse and norm
Suppose the m * n matrix A =[imath] \begin{bmatrix} A_1 \\A_2 \end{bmatrix}[/imath] where A1 is a nonsingular square matrix (n*n) and A2 is an arbitrary matrix. Why is [imath]||A^{+}||_2 \leq ||A_1^{-1}||_2[/imath] ? ([imath]A^{+}[/imath] denotes the psuedoinverse) |
2394230 | Expressions of Inverse Hyperbolic functions
I have been trying to understand hyperbolic functions for some time now. I have a problem concerning the expressions of inverse hyperbolic functions. The text( G.B. Thomas ) mentions nothing about their expressions. While plotting the Hyperbolic Sine was easy, plotting its inverse was not. By definition we have, [imath]\sinh(x) = \frac{e^x - e^{-x}}{2}[/imath] I plotted its graph below- The dotted lines are the function's asymptotes. I then looked at the graph of inverse hyperbolic Sine and decided to obtain its expression. I tried by finding the inverses of the asymptotes because I knew that the inverse function would get close to them eventually. I plotted a graph. I was happy. But not for long. It turned out that I cannot have an algebraic combination of these two asymptotes. This was my second approach, to be honest. A holy approach to finding inverses is, express [imath]x[/imath] in terms of [imath]y[/imath]. I got stuck at the very beginning and could never proceed. Is there a way around this? If there indeed is a way to express [imath]x[/imath] in terms of [imath]y[/imath], could anyone give me a few hints to do it? If there isn't, how do we obtain an expression for the inverse hyperbolic Sine? | 268591 | How to derive inverse hyperbolic trigonometric functions
[imath]e^{i\theta}=\cos\theta + i\sin \theta[/imath] [imath]e^{i\sin^{-1}x}=\cos(\sin^{-1}x)+i\sin(\sin^{-1}x)[/imath] [imath]i\sin^{-1}x=\ln|\sqrt{1-x^2} + ix|[/imath] [imath]\sin^{-1}x=-i\ln|\sqrt{1-x^2} + ix|[/imath] Now from here I'm kind of lost, since it seems like this should be the definition, but when I look it up, the definition of inverse hyperbolic sine is: [imath]\sinh^{-1}x=\ln(\sqrt{1+x^2} + x)[/imath] So although they're very similar, I guess I just don't know how to handle the logarithm and anything to the ith power or drop off the absolute value. |
2393599 | Spanning Trees and Graphs
Suppose [imath]Z_1[/imath] and [imath]Z_2[/imath] are two different spanning trees of a simple connected undirected graph [imath]H[/imath]. Show that if [imath]a_1[/imath] is an edge in [imath]Z_1[/imath] that is not in [imath]Z_2[/imath], then there is an edge [imath]a_2[/imath] in [imath]Z_2[/imath] that is not in [imath]Z_1[/imath] such that the graph ([imath]Z_1[/imath] − [imath]a_1[/imath]) + [imath]a_2[/imath] (obtained from [imath]Z_1[/imath] on replacing [imath]a_1[/imath] by [imath]a_2[/imath]) is also a spanning tree of [imath]H[/imath]. This is a practice problem assigned in the book, but the answer provided is a little confusing. Could someone break this down a little better? Answer provided: Essentially, it breaks it down through the associative property of addition between the two spanning trees. Since the two spanning trees are of a simple connected graph, subtracting one edge from a tree and adding an edge that is not in the previous tree will produce a spanning tree that is still representative of the graph H. | 2392698 | Spanning Trees: Prove if [imath]e_1[/imath] is an edge in [imath]X[/imath] that isn't in [imath]Y[/imath], then there exists an edge ...
Say [imath]X[/imath] and [imath]Y[/imath] are two different spanning trees of a simple and connected and undirected graph [imath]G[/imath]. Prove that if [imath]e_1[/imath] is an edge in [imath]X[/imath] that isn't in [imath]Y[/imath], then there exists an edge [imath]e_2[/imath] in [imath]Y[/imath] that isn't in [imath]X[/imath]... so the graph [imath](X − e_1) + e_2[/imath] (obtained from [imath]X[/imath] on replacing [imath]e_1[/imath] by [imath]e_2[/imath]) is also a spanning tree of the original graph, [imath]G[/imath]. |
2395270 | [imath]f(x)=\frac{x^2 -4}{x-2}[/imath], Can [imath]x=2[/imath] or not?
Given: [imath]f(x)=\frac{x^2-4}{x-2}[/imath] First thoughts would be than [imath]x\ne 2[/imath] But since [imath]f(x)[/imath] can be simplified to: [imath]f(x)=x+2[/imath] It now seems than [imath]x[/imath] can be a solution to [imath]f(x)[/imath]. If I graph this function I get the line [imath]f(x)=x+2[/imath], but it that with the point [imath](2,4)[/imath] not included? My book gives an answer where [imath]x\ne 2[/imath] | 1141492 | Graphing [imath]y = \frac{x^2-4}{x-2}[/imath]
I'm trying to graph the function [imath]y = \frac{x^2-4}{x-2}[/imath]. I factored the numerator into [imath](x+2)(x-2)[/imath], and then canceled the common factor (x-2), leaving the function as [imath]y = x + 2[/imath]. Is it acceptable to graph the original function as [imath]y = x + 2[/imath], a straight line with domain and range all real numbers? Or do you have to restrict the domain as in the original function? Thanks, Sean |
2393111 | Induction (Geometric sequence)
There are two one-letter words in English (“I” and “a”), and according to http://www.scrabble.org.au/words/twos.htm there are 122 two-letter words. Let [imath]A_n[/imath] be the number of strings of n letters that may be formed from some sequence of one- and two-letter words, by concatenating them all together. Find [imath]a_3[/imath] Give an expression for an in terms of [imath]a_{n-1}[/imath] and [imath]a_{n-2}[/imath], that works for all n ≥ 3. So far a1 makes sense to just be "I" and "a" so a total of 2. a2 would be the 122 two letter strings plus a combination of adding the one letter words (i and a) giving ai, ia, ii and aa. This would give a total of 126 for [imath]a_2[/imath]. doing the combinations for [imath]a_3[/imath] gives 495 but that doesn't seem right as there is no evident patter to use the formula [imath]a_n=a_1r^{n-1}[/imath]. | 2397858 | Solving Problem by Induction
I've been stuck on the following puzzle: There are [imath]2[/imath] one-letter words in English ("I" and "a"), and there are [imath]124[/imath] two-letter words. Let [imath]a_n[/imath] be the number of strings of [imath]n[/imath] letter that may be formed from some sequence of one- and two-letter words, by concatenating them all together. Prove, by induction on [imath]n[/imath], that [imath]a_n \le 13^n \ \forall n[/imath]. I would appreciate it if people could please take the time to explain this. |
2395498 | Show that [imath]\frac{d}{dx}T_n(x)=nU_{n-1}(x)[/imath]
[imath]T_n(x)[/imath] is a polynomial of degree [imath]n[/imath] such that [imath]T_n(x)=\cos(n\cos^{-1}(x))[/imath] (i) Show that [imath]T_2(x)=2x^2-1[/imath] and [imath]T_3(x)=4x^3-3x[/imath] (ii) Show that [imath]T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)[/imath] [imath]U_n(x)[/imath] is an [imath]n[/imath] degree polynomial defined by [imath]U_n(x)=\frac{\sin((n+1)\theta)}{\sin\theta}[/imath] where [imath]x=\cos\theta[/imath]. (iii) Show that [imath]\frac{d}{dx}T_n(x)=nU_{n-1}(x)[/imath] (iv) Show that [imath]\frac{d}{dx}U_{n-1}(x)=\frac{xU_{n-1}(x)-nT_n(x)}{1-x^2}[/imath] I'm having a lot of problems with part (iii). I typed out part (ii) because I thought perhaps the result from part (ii) maybe be cleverly used to solve part (iii). Differentiating [imath]T_n(x)[/imath] would yield an algebraic expression in [imath]x[/imath]. On the other hand, [imath]nU_{n-1}(x)[/imath] is in [imath]\theta[/imath] unless we explicitly express [imath]U_n(x)[/imath] in [imath]x[/imath] which would be very troublesome because there is no simple way of expressing [imath]\sin n\theta[/imath] in terms of [imath]\cos\theta[/imath] and therefore [imath]x[/imath]. Some help would be of great help thank you! [I don't need help with part (i) and (ii)] | 1463422 | relation between first kind Chebyshev poly and second kind Chebyshev poly
How do you prove following relation between Chebyshev poly of first kind and Chebyshev poly of second kind: [imath]dT_n(x)/dx=nU_{n-1}(x)[/imath] |
2395637 | Help on proof of convergent sequence without limit point
Please kindly help with this problem on converging sequence. Question Determine whether the sequence [imath]\bigl(\sqrt{4n^2+n}-2n\bigr)_{n\in\mathbb N}[/imath] converges and if it converges, guess its limits and prove your guess. Now what I don't really understand is to proof the convergence of the sequence, because it has no limit point, so how do we prove it is convergent, and show me your guess. | 483481 | How would you prove that [imath]\lim\limits_{n\to\infty}(\sqrt{4n^2+n}-2n)=\frac14[/imath]?
[imath]\lim_{n\rightarrow \infty}\left[{\sqrt{4n^2+n}-2n}\right]=\frac{1}{4}[/imath] I am trying to use the definition of the limit but have no idea how to simplify the expression with radical! ---edit--- so by the definition, [imath]\forall n > N \rightarrow |a_n - a | < \epsilon, \text{where}\, \ a_n=\sqrt{4n^2+n}-2n\ \, \text{and}\,\ a = \frac{1}{4}[/imath] so after multiply the conjugate and negate [imath]\frac{1}{4}[/imath], I get [imath]\frac{2n-\sqrt{4n^2+n}}{4(\sqrt{4n^2+n}+2n)}[/imath] Since there is still radical in the numerator, I think I have multiply the conjugate again...right?? And then find some formula that is greater!!??? confuse me this real analysis!! |
2395906 | Question: How to prove this inequality
I have to prove this inequality. [imath]\sqrt[n]{n} - 1 \lt \sqrt{\dfrac 2n} ~~\mbox{for}~~ n \gt 2[/imath] So far I reached this result: [imath]n^{\dfrac {n+2}{n}} - n^{\dfrac {n+1}{n}} + n - n^{\dfrac {n+1}{n}} \lt 2[/imath] | 471730 | An [imath]n[/imath]th root inequality: [imath]\sqrt[n]{n} < 1 + \sqrt{2/n}[/imath]
Prove that for any positive integer [imath]n[/imath], [imath]n^{1/n} < 1 + \sqrt{\frac{2}{n}}.[/imath] This due to Victor Linis, Eureka, Vol. 2, No. 2, February 1976, p. 29. Hint: Use the binomial theorem. |
2352691 | Proof that [imath]W^{\perp\perp}=W[/imath] (in a finite dimensional vector space)
Let [imath]V[/imath] be a finite dimensional vector space over the field [imath]K[/imath], with a non-degenerate scalar product. Let [imath]W[/imath] be a subspace. Show that [imath]W^{\perp\perp}=W[/imath]. I have no idea how to solve this one. Since I am self-studying I have no way to know if I get this right, that is the reason I ask for a resolution. Questions: Can someone provide me a resolution? Thanks in advance! | 1690797 | How to show that [imath](W^\bot)^\bot=W[/imath] (in a finite dimensional vector space)
I need to prove that if [imath]V[/imath] is a finite dimensional vector space over a field K with a non-degenerate inner-product and [imath]W\subset V[/imath] is a subspace of V, then: [imath] (W^\bot)^\bot=W [/imath] Here is my approach: If [imath]\langle\cdot,\cdot\rangle[/imath] is the non-degenerate inner product of [imath]V[/imath] and [imath]B={w_1, ... , w_n}[/imath] is a base of [imath]V[/imath] where [imath]{w_1, ... , w_r}[/imath] is a base of [imath]W[/imath] then I showed that [imath] \langle u,v\rangle=[u]^T_BA[v]_B [/imath] for a symmetric, invertible matrix [imath]A\in\mathbb{R}^{n\times n}[/imath]. Then [imath]W^\bot[/imath] is the solution space of [imath]A_rx=0[/imath] where [imath]A_r\in\mathbb{R}^{r\times n}[/imath] is the matrix of the first [imath]r[/imath] lines of [imath]A[/imath]. Is all this true? I tried to exploit this but wasn't able to do so. How to proceed further? |
2396135 | A relation between gcd and lcm
Let [imath]a,b,c[/imath] be non-zero integers . Let [imath](x,y)[/imath] denote the gcd and [imath][x,y][/imath] denote the lcm of two non-zero integers . Then is it true that [imath] (a,[b,c])=[(a,b),(a,c)][/imath] ? | 147973 | [imath]\gcd(a,\operatorname{lcm}(b,c))=\operatorname{lcm}(\gcd(a,b),\gcd(a,c))[/imath]
I am trying to show that [imath]\gcd(a,\operatorname{lcm}(b,c))=\operatorname{lcm}(\gcd(a,b),\gcd(a,c))[/imath]. If [imath]d=\gcd(a,\operatorname{lcm}(b,c))[/imath], then [imath]d=ax-my[/imath], where [imath]m=\operatorname{lcm}(b,c)[/imath]. So [imath]d=ax-b(ky)[/imath] and [imath]d=ax-c(ly)[/imath], which means that [imath]\gcd(a,b)|d[/imath] and [imath]\gcd(a,c)|d[/imath], or [imath]d[/imath] is a common multiple. But I wasn't able to show that it is the smallest common multiple. |
2394785 | The Fox and the Duck in a Square Pond
A duck is in the center of a square pond and cannot fly. A fox is in the corner of the pond and cannot swim. If the duck can swim with speed 1, and both play optimally, what is the minimal fox movement speed [imath]v[/imath] such that the fox can prevent the duck from escaping the pond? I expect there is a closed form solution, and here is my thinking about how to solve it. The combined positions of the duck and the fox form a point in three dimensional space. At the critical speed [imath]v[/imath], there should be a critical set of positions such that the duck can approach arbitrarily close to the set, but if the duck reaches just beyond it, the duck can escape. Analysis of the puzzle, or perhaps numerical simulation of the perfect play, should reveal the geometry of this set, after which the problem may become straightforward. For the circular pond (of radius [imath]r[/imath]), the solution is well-known: Near the critical speed, the duck stays opposite of the fox until the duck gets to distance [imath]r/v[/imath] from the center, and then the duck continues on the straight path (tangent to the radius [imath]r/v[/imath] circle) to the edge of the pond, just barely escaping at [imath]v=4.603...[/imath]. For the square pond problem, the lack of rotational symmetry makes the game state three-dimensional and thus harder. Update: See this question for the generalization to a regular polygon and some good but (at least as of this writing) suboptimal strategies. I am looking either for an exact solution, or an accurate numerical simulation of the perfect play, including a graph showing the perfect play at the critical speed (minus epsilon, thus allowing the duck to escape). Despite the simplicity of the problem, the continuous nature of the play makes efficient accurate simulation tricky. | 1762665 | Can the boy escape the teacher for a regular [imath]n[/imath]-gon?
This is related to Prove that the boy cannot escape the teacher Suppose there is a boy in the center of a regular [imath]n[/imath]-gon. The teacher is on the edge of the [imath]n[/imath]-gon (but cannot leave the edge) and wants to capture the boy. At the beginning he is on a vertex. The teacher is [imath]v(n)[/imath] times faster than the boy. Which is the maximum [imath]v(n)[/imath] such that the boy can escape? (By escaping means he reaches the edge of the [imath]n[/imath]-gon and the teacher is not there) From the linked question we know [imath]3 \le v(4) < 6[/imath], and for [imath]n= \infty[/imath] (a circle) then I know a strategy such that [imath]v(\infty) = \pi + 1[/imath] suffice; I don't know if this is optimal. I also put convergence in the tags because my wild hypothesis is that the maximum [imath]v(n)[/imath] will converge to some value and it would be interesting to know which one! Any cool way to solve this? |
263896 | Nilradical of polynomial ring
Let [imath]R[/imath] be a commutative ring. The nilradical [imath]\text{nil}(R)[/imath] is the set of all nilpotent elements, and it is the intersection of all the prime ideals of [imath]R[/imath]. Is the following true in the polynomial ring? [imath]\text{nil}(R[X])=(\text{nil}(R))[X][/imath] Any polynomial in the right hand side has all its coefficients nilpotent in [imath]R[/imath], hence also in [imath]R[X][/imath], and therefore it is nilpotent in [imath]R[X][/imath] as the sum of nilpotent elements. The other inclusion is not clear to me. Is there a counterexample? | 2669954 | Nilpotent Polynomial
Let [imath]R[/imath] be commutative ring [imath]R[/imath] with a multiplicative identity [imath]1[/imath]. Suppose [imath]p(x)=a_{n}x^{n}+\cdots a_{1}x+a_{0} [/imath] is a polynomial in [imath]R[x][/imath]. Prove that each [imath]a_0,a_{1},\dots,a_n[/imath] is a nilpotent element in [imath]R[/imath] if and only if [imath]p(x)[/imath] is nilpotent in [imath]R[x][/imath]. Here is my attempt at the forward direction. [imath](\Rightarrow)[/imath] Suppose each [imath]a_0,a_{1},\dots,a_n[/imath] is nilpotent in [imath]R[/imath]. Then [imath]a_{i}^{n_{i}}=0[/imath] for each [imath]a_{i}[/imath] and for some positive integer [imath]n_{i}[/imath]. Let [imath]N=\prod\limits_{i=0}^{n}n_{i}[/imath]. Then [imath][p(x)]^{N}=0[/imath] which shows that [imath]p(x)[/imath] is nilpotent in [imath]R[x][/imath]. I think this is correct, but I'm not positive. I couldn't find any concrete examples of a polynomial ring which contains more than one nilpotent element. For the [imath](\Leftarrow)[/imath] direction, I was thinking that I should use induction on the degree of [imath]p(x)[/imath]. If this is the wrong approach, then I could use some advice on how to prove this. |
2396690 | Is there any positive function [imath]f[/imath] such that [imath]f(x)f(y)\leq |x-y|[/imath] for every [imath]x[/imath] rational and [imath]y[/imath] irrational?
We are given a function [imath]f[/imath] from reals to positive reals (not including [imath]0)[/imath] satisfying [imath]f(x)f(y)\leq |x-y|[/imath] for every rational number [imath]x[/imath] and irrational number [imath]y[/imath]. Does this function exist? If [imath]f[/imath] is continuous, it is easy to show that [imath]f[/imath] must be a constant zero function (So it does not exist, in this case). Otherwise, for an irrational number [imath]y[/imath] it can be shown that for any sequence [imath]x_n[/imath] of rationals converging to [imath]y[/imath], [imath]f(x_n)[/imath] converges to zero and the same is true for rational number [imath]x[/imath] and the sequence of irrationals [imath]y_n[/imath]. Does this argument give us any information about [imath]f[/imath], and does it say whether [imath]f[/imath] exists at all? | 1694434 | Is there a positive function [imath]f[/imath] on real line such that [imath]f(x)f(y)\le|x-y|, \forall x\in \mathbb Q , \forall y \in \mathbb R \setminus \mathbb Q[/imath]?
Does there exist a function [imath]f:\mathbb R \to (0,\infty)[/imath] such that [imath]f(x)f(y)\le|x-y|, \forall x\in \mathbb Q , \forall y \in \mathbb R \setminus \mathbb Q[/imath] ? |
2396558 | Prove that [imath]f(\overline A) \subseteq \overline{f(A)}[/imath]
Let [imath]f : X \longrightarrow Y[/imath] be continuous, and let [imath]A \subseteq X[/imath]. Show that [imath]f(\overline A) \subseteq \overline{f(A)}[/imath]. My attempt Let [imath]y \in f(\overline A)[/imath]. Then there exists [imath]x \in \overline A[/imath] such that [imath]f(x) = y[/imath]. Now let us take an open neighbourhood [imath]V[/imath] of [imath]y[/imath] in [imath]Y[/imath] arbitrarily. If we can show that [imath]V \cap f(A) \neq \emptyset[/imath] then our purpose will be served. Now since [imath]f(x) = y \in V[/imath], we have [imath]x \in f^{-1} (V)[/imath]. Since [imath]x \in \overline A[/imath] and [imath]f^{-1} (V)[/imath] is open in [imath]X[/imath] (since [imath]f[/imath] is continuous), we have that [imath]f^{-1} (V) \cap A \neq \emptyset[/imath]. Let [imath]z \in f^{-1} (V) \cap A[/imath]. Then [imath]f(z) \in V \cap f(A)[/imath], which proves that [imath]V \cap f(A) \neq \emptyset[/imath] and we are done. Is my reasoning correct at all? Please verify it. | 2133249 | proof [imath]f(\overline{A})\subseteq \overline{f(A)} \Leftrightarrow f[/imath] continuous
It'd be great if someone checked the proof I did for the following problem: [imath]f:X\longrightarrow Y[/imath], [imath]f(\overline{A})\subseteq \overline{f(A)},\forall A\subseteq X \Leftrightarrow f[/imath] continuous proof: Suppose [imath]f(\overline{A})\subseteq \overline{f(A)}[/imath] for any [imath]A\subseteq X[/imath] . Let [imath]C\subseteq Y[/imath] be closed and define [imath]A=f^{-1}(C)[/imath]. [imath]f(\overline{A})\subseteq \overline{f(A)}\subseteq\overline{C}=C[/imath], therefore [imath]\overline{A}\subseteq f^{-1}(C)=A[/imath], so [imath]A[/imath] is closed. *Conversely, suppose [imath]f[/imath] is continuous. Let [imath]p\in\overline{A}[/imath] and let [imath]V[/imath] be an open neighborhood of [imath]f(p)[/imath]. Since [imath]f[/imath] is continuous, [imath]f^{-1}(V)[/imath] is open. On the other hand [imath]p\in f^{-1}(V)[/imath], but since [imath]p\in\overline{A}[/imath] we have that [imath]f^{-1}(V)[/imath] contains points of [imath]A[/imath]. Therefore [imath]V[/imath] contains points of [imath]f(A)[/imath]. Since [imath]V[/imath] is an arbitrary neighborhood of [imath]f(p)[/imath], we have [imath]f(p)\in \overline{f(A)}[/imath]. So finally, [imath]f(\overline{A})\subseteq\overline{f(A)}[/imath]. Thank you very much! |
2396619 | Branch cut for multi valued complex functions
This question is about the so called 'branch cut' for defining multivalued function continuously on complex plane.For example to define inverse of the function [imath]f(z)=z^2[/imath] we consider the domain of the inverse function as the slit plane [imath]C\setminus(\infty,0][/imath]. The case is also similar when defining Log function. So my question, is there any sort of 'rule of thumb' which guarantees that removing this sort slit assures you that the domain is right for defining inverse function? Also why we are removing straight lines always? Would we get same results if we have removed any arbitary type of curves from C? | 245579 | How does a branch cut define a branch?
I am studying complex analysis and I have problem understanding the concept of branch cut. The lecturer draw this as some curve that starts from a point and goes on and on in some direction (for example, something like [imath]y=x[/imath] for [imath]x\geq0[/imath] , but it doesn't have to be straight). The definition given in the lecture is A branch cut is a curve that is being presented in order to define a branch and then he added a note Points on the branch cut are singular. Can someone please explain how does such a curve define a branch of a function ? As I understand it, if [imath]f(z)=u(r,\theta)+iv(r,\theta)[/imath] then we want [imath]\alpha<\theta\leq\alpha+2\pi[/imath] for some real [imath]\alpha[/imath]. How does a curve define this [imath]\alpha[/imath] ? Why are the points on a branch cut singular? Are we also assuming something about the function that we are trying to define a branch of it? |
2397426 | Second fundamental form of an implicit surface
I'm interested in expressions for the second fundamental form of an implicit surface [imath]f(x,y,z) = 0[/imath], in terms of the first and second derivatives [imath]f_x, \ldots, f_{yz}[/imath]. I am not looking for the expressions of a function in Monge form, [imath]z = h(x,y)[/imath]. That is I don't want to assume that I can rewrite the expression locally in the form. I have access only to [imath]f[/imath] and its derivatives. As a concrete example, suppose I have a function [imath]x^2+y^2+z^2 - 1 = 0[/imath], and I want to compute the coefficients of the second fundamental form [imath]L, M, N[/imath] in terms of the available derivatives, without rewriting [imath]f[/imath] in parts such as [imath]z = \sqrt{1-x^2-y^2}[/imath]. | 111577 | About the second fundamental form
Let [imath]U\subset\mathbb R^3[/imath] be an open set, and [imath]f:U\to \mathbb R[/imath] be a smooth function. Suppose that the level set [imath]S=f^{-1}(\{0\})[/imath] is non-empty, and that at each [imath]p\in S,[/imath] the gradient [imath]\overrightarrow \nabla f(p)[/imath] is not the zero vector. Then [imath]S[/imath] is a smooth two-dimensional surface in [imath]U[/imath], and [imath]p\mapsto \overrightarrow \eta(p)=\frac{1}{||\overrightarrow \nabla f(p)||}\overrightarrow \nabla f(p)[/imath] defines a smooth unit-length normal vector field along [imath]S[/imath]. At each [imath]x\in U,[/imath] write [imath]H(f)_{(x)}[/imath] for the [imath]3\times 3[/imath] Hessian matrix specified by [imath](H(f)_{(x)})_{ij}=\frac{\partial^2f}{\partial x_i\partial x_j}(x).[/imath] Show that, at each [imath]p\in S[/imath], the second fundamental form [imath]II_p: T_p(s)\times T_p(s)\to \mathbb R[/imath] is the symmetric bilinear map [imath]II_p(\overrightarrow v,\overrightarrow w)=\frac{-1}{||\overrightarrow \nabla f(p)||}\overrightarrow v\cdot H(f)_{(p)}\overrightarrow w,[/imath]for all [imath]\overrightarrow v ,\overrightarrow w \in T_p(s)[/imath]. (Here, we view the tagent space [imath]T_p(S)[/imath] as the two-dimensional subspace [imath](span\{ {\overrightarrow \eta(p)}\})^{\bot}[/imath] of [imath]\mathbb R^3[/imath]. Edit: Actually my question is why the second fundamental form under the usual definition can be written in this way. Definition: The quadratic form [imath]II_p[/imath], defined in [imath]T_p(S)[/imath] by [imath]II_p(v)=-<d N_p(v),v>[/imath] is called the second fundamental form of [imath]S[/imath] at [imath]p[/imath], where [imath]dN_p:T_p(S)\to T_p(S)[/imath] is the differential of the Gauss map. Hopefully, I express this problem explicitly. I was just wondering how to prove this statement. I took a diffrential geometry class last semester, and when I organized my notes this morning, I found this statement, but there was no proof... Looking forward to an understandable explaination. Thanks in advance. Edit 2:Furthermore, show that, at each point [imath]p\in S[/imath], the expression [imath]\phi_p(z)=det\pmatrix{-H(f)_{(p)}-zI_{3\times 3} & \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)& 0}[/imath] (the underlying matrix here is [imath]4\times 4[/imath]) defines a second-degree polynomial whose roots [imath]\lambda_1[/imath] and [imath]\lambda_2[/imath] are [imath]||\overrightarrow \nabla f(p)||k_1[/imath] and [imath]||\overrightarrow \nabla f(p)||k_2[/imath], where [imath]k_1[/imath] and [imath]k_2[/imath] are the principal curvatures of [imath]S[/imath] at [imath]p[/imath]. Also, if a non-zero vector [imath]\pmatrix {\overrightarrow v \\c}[/imath] lies in the kernel of the [imath]4\times 4[/imath] matrix [imath]\pmatrix{-H(f)_{(p)}-\lambda_jI_{3\times 3} & \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)& 0},[/imath] then [imath]\vec v[/imath] is a non-zero element of [imath]T_p(S)[/imath] and lies in the "principal direction" corresponding to [imath]K_j[/imath]. |
2397294 | Why is [imath]\sqrt{ab} = \sqrt{a} \sqrt{b} [/imath] where a and b are positive real numbers?
That's the question. I try to explain this to myself I can't find any good resources. Note: I have researched for about 15 mins and yet haven't found the answer. I do apologize for any inconveniences. | 1274936 | Proving [imath]\sqrt{ab} = \sqrt a\sqrt b[/imath]
I am currently in high school and we are studying radicals. I had asked my math teacher why [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] (for all a,b>0) and he tries to prove it by arguing that [imath]a^{1/2}*b^{1/2}=(ab)^{1/2}[/imath] (an exponent law). However, I find this proof problematic since [imath]x^{1/2}[/imath] is simply defined as [imath]\sqrt{x}[/imath], so the reasoning is circular. My view is that [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] because once we square both sides we get [imath]ab=ab[/imath]. Since we're obviously referring to the positive root, and the function [imath]f(x) = \sqrt{x}[/imath] is injective, it necessarily follows that the original expressions [imath]\sqrt{ab}[/imath] and [imath]\sqrt{a}\sqrt{b}[/imath] are equivalent because for injective functions it is not possible to map distinct elements in the domain to the same element in the range ([imath]ab[/imath]). Hence they are equivalent expressions. My question therefore is, is my proof valid and/or rigorous (I find it convincing but maybe it's wrong; I just want to be clear) and secondly was my teacher's proof correct? |
2396921 | Roulette Probabilty
Question: A roulette wheel has 21 red, 21 black numbers and 4 zeros, A player places [imath]20 bets on red numbers. If the roulette wheel spins a red number the player gets his [/imath]20 back and wins another [imath]20. If the roulette does not spin a red number the player loses his bet. the player sits down with [/imath]40 and keeps playing until he has no money. How many spins can he expect to be able to play until he has no money? This problem has got me in a bit of a pickle. I know I have to use the binomial theory but I am getting confused on what you would consider the number of successful trials. Eg. to go bust after 2 spins you would have to land on a colour other then red twice. This probability is 23/44 and each event is independent. For 4 spins there are 2 ways to go bust. For 6, there are 4 ways to go bust and so one and so forth. I am not sure if I am heading the right path but the probability of going bust after 4 attempts say is: Pr(x=4) = 2!/(2-4)!*4! * (21/44)^4!(23/44)^(4-2)... Am I heading down the right path? | 2397097 | Roulette wheel probability question
A roulette wheel has [imath]21[/imath] red numbers, [imath]21[/imath] black numbers, and [imath]4[/imath] zeros. A player places [imath]20[/imath] dollar bets on the red numbers. If the roulette spins a red number, the player get his [imath]20[/imath] dollars back, and wins another [imath]20[/imath] dollars. If the roulette does not spin a red number, the player loses his bet. The player sits down with [imath]40[/imath] dollars, and keeps on playing until he has no more money. How many spins can he expect to be able to play until he has no more money? |
2397894 | Prove that: [imath]\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\leq \frac{a+b+c}{2}[/imath]
Prove that: [imath]\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\leq \frac{a+b+c}{2} : (a, b, c) \in \mathbb{R}^+[/imath] | 1026954 | Prove that [imath]\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{a+b+c}{2}[/imath]
If [imath]$a,b,c$[/imath] are positive real numbers, not all equal, prove that [imath]$$\dfrac{bc}{b+c}+\dfrac{ca}{c+a}+\dfrac{ab}{a+b}<\dfrac{a+b+c}{2}$$[/imath] Progress: [imath]\frac{bc}{b+c}-\frac{b+c}{2}=[/imath] |
2399269 | Prove that for every integer [imath]n \geq 0[/imath], the number [imath]4^{2n+1}+3^{n+2}[/imath] is a multiple of 13.
Prove that for every integer [imath]n \geq 0[/imath], the number [imath]4^{2n+1}+3^{n+2}[/imath] is a multiple of 13. | 859523 | [imath]13\mid4^{2n+1}+3^{n+2}[/imath]
How can I prove that [imath]4^{2n+1}+3^{n+2}[/imath] is always divisible by 13? |
2399409 | Find the maximum value of [imath]a^2b^3[/imath] where [imath]a,b[/imath] are positive real numbers satisfying [imath]a+b=10[/imath]
Find the greatest value of [imath]a^2b^3[/imath] where [imath]a,b[/imath] are positive real numbers satisfying [imath]a+b=10[/imath].Determine the values of [imath]a,b[/imath] for which the greatest value is attained. It is my question.I continuously tried to use weighted A.M-G.M. inequality, but unfortunately I found no way.I don't think it can be done using Tchebycheff's inequality or Cauchy-Schwartz's theorem. Please give me any hint for doing that.I also failed to solve another similar problem (but reversed. "Find [imath]\min(3x+2y)[/imath], where [imath]x^2y^3[/imath]=48). If possible then please give me hint in that also. Thank you. | 2049448 | Find the greatest value of [imath]a^2\cdot b^3[/imath] where [imath]a+b=10[/imath]
Find the greatest value of [imath]a^2\cdot b^3[/imath] where [imath]a,b[/imath] are positive real numbers satisfying [imath]a+b=10[/imath]. Determine the values of [imath]a,b[/imath] for which the greatest value is attained. At first I tried to find the value by putting [imath]a=10-b[/imath], I found a expression, but I got not way to do anything. If we apply A.M-G.M. inequality, I think we found the minimum value or something else. It is a pre-Olympiad level problem. So I need some answers or some good hints. |
2400380 | Complex number inequality on the unit disk
This was posted on another website, but there are no responses so far: Let [imath]a,b[/imath] and [imath]c[/imath] be complex numbers such that [imath]|az^2+bz+c| \le 1[/imath] for [imath]z[/imath] in the closed unit disk. Show that [imath]|bc| \le \frac{3\sqrt{3}}{16}[/imath]. | 781820 | How find the maximum value of [imath]|bc|[/imath]
Question: Given complex numbers [imath]a,b,c[/imath], we have that [imath]|az^2 + bz +c| \leq 1[/imath] holds true for any complex number [imath]z, |z| \leq 1[/imath]. Find the maximum value of [imath]|bc|[/imath] It is said this is answer is [imath]|bc|\le \dfrac{3\sqrt{3}}{16}[/imath] My idea: let [imath]z=1[/imath],then [imath]|a+b+c|\le 1[/imath] let [imath]z=-1[/imath],then [imath]|a-b+c|\le 1[/imath] let [imath]z=0[/imath], then [imath]|c|\le 1[/imath] let [imath]|\theta|=1[/imath],then we have [imath]|a(z/\theta)^2+b(z/\theta)+c|\le 1\Longrightarrow |az^2+b\theta z+c\theta^2|\le 1[/imath] and only this can't solve this problem,Thank you |
2400607 | Evaluate the surface area of the cone delimited by the plane
Surface is the cone [imath]z^2=x^2+y^2[/imath] delimited by the plane [imath]z=0[/imath] and [imath]x+2z=3[/imath]. Well, the first thing I thought is to devide it in two parts, the one that is a simple cone and the other part. But I'm a bit lost of how do I find the proper parametrization for this surface. | 681102 | Please tell me how to evaluate this integral.
Compute the area of that portion of the conical surface [imath]x^2 + y^2 = z^2[/imath] which lies between the two planes [imath]z = 0[/imath] and [imath]x + 2z = 3[/imath]. Ans:[imath]2\pi\sqrt{6}[/imath]. Thank you. |
250254 | Can distance between two closed sets be zero?
Is given metric space [imath](M, d)[/imath]. Let [imath]A\cap B = \emptyset; \,\,\text{dist}(A,B):=\inf\{d(x,y):x\in A, y\in B\}[/imath]. [imath]A, B[/imath] are both closed sets. Is it possible that [imath]\text{dist}(A,B)=0[/imath]? The first thought comes into mind is that obviously [imath]\text{dist}(A,B)>0[/imath], but possibly there are some tricky [imath]d[/imath] and [imath]A, B[/imath] so that it's untrue. Thanks in advance! | 420122 | Find two closed subsets or real numbers such that [imath]d(A,B)=0[/imath] but [imath]A\cap B=\varnothing[/imath]
Here is my problem: Find two closed subsets or real numbers such that [imath]d(A,B)=0[/imath] but [imath]A\cap B=\varnothing[/imath]. I tried to use the definition of being close for subsets like intervals but I couldn't find any closed sets. Any hint? Thank you. |
2400577 | Does [imath]\lim_{x\to 0}\frac{1}{x}[/imath] exist?
Consider the following function: [imath]\displaystyle f(x) = \frac{x^2-4}{x-2}[/imath] and now its limit: [imath]\displaystyle \lim \atop x \to 2[/imath] [imath]\displaystyle \frac{x^2-4}{x-2}[/imath] Upon evaluating, this limit becomes: [imath]\displaystyle \lim \atop x\to 2[/imath] [imath]\displaystyle x+2 = 4[/imath] The limit therefore exists since it meets each of the following criteria: The limit approaches a finite value ([imath]4[/imath]), the limits on either side of [imath]x=2[/imath] correspond and the limit approaches a particular value ([imath]4[/imath]) With this in mind, does that mean that [imath]\displaystyle \lim_{x\to 0}\frac{1}{x}[/imath] does not exist? Seeing as it does not approach a finite value? | 851498 | Proof that the limit of [imath]\frac{1}{x}[/imath] as [imath]x[/imath] approaches [imath]0[/imath] does not exist
Hello I was hoping that someone might be able to verify that the following proof that [imath]\lim_{x\to 0} {1\over x}[/imath] does not exist is correct. First assume that [imath]\lim_{x\to 0} {1\over x} = L[/imath]. This means that for every [imath]\epsilon > 0[/imath] there is a [imath]\delta > 0[/imath] such that for all [imath]x[/imath] if [imath]0 < \lvert x\rvert < \delta[/imath] then [imath]\lvert{1\over x} - L \rvert <\epsilon.[/imath] Now let [imath]\epsilon=1[/imath] and choose [imath]x < \min(1,\delta)[/imath]. Therefore there is a [imath]\delta>0[/imath] such that if [imath]0 < \lvert x\rvert < \delta[/imath] then [imath]\lvert{1\over x} - L \rvert <\epsilon[/imath]. It follows that [imath]\lvert{1\over x}\rvert < 1+ \lvert L\rvert[/imath]. However clearly [imath]\lvert{1\over x}\rvert>1[/imath] and therefore a contradiction has been reached and there is no number [imath]L[/imath] such that [imath]\lim_{x\to 0} {1\over x} = L[/imath]. |
2400587 | Show that the product of all group elements is [imath]e[/imath] or [imath]u[/imath] if [imath]G[/imath] has exactly one involution [imath]u[/imath]
Let [imath]G[/imath] be an finite abelian group, define [imath]s= \Pi_{g\in G} g[/imath], show that [imath]s = e[/imath] unless [imath]G[/imath] has exactly one involution [imath]u[/imath], in this case [imath]s=u[/imath]. I have the start of a proof, but I get stuck at a point. Proof Write [imath]G[/imath] as [imath]G = \{e\} \cup I \cup \{g_1, g_1^{-1}, \ldots, g_n, g_n^{-1}\}[/imath] where [imath]I[/imath] is a set of the all the groups involutions. Since [imath]G[/imath] is abelian [imath]\Pi_{g\in \{g_1, g_1^{-1}, \ldots\}} g = e[/imath] we just need to examine [imath]\Pi_{u\in I} u[/imath] Now: If [imath]|I|=0[/imath] then the theorem holds, If [imath]|I| = 1[/imath] then [imath]I=\{u\}[/imath] and [imath]s= u[/imath], the theorem holds If [imath]|I| = 2[/imath], requires closer examination: Let [imath]I = \{u_1,u_2\}[/imath] be the set of all involutions (not equal to [imath]e[/imath]), then [imath]u_1u_2[/imath] is also an involution (since [imath]G[/imath] is abelian). If [imath]u_1u_2 =e[/imath] then [imath]u_1=u_2[/imath], contradiction, if [imath]u_1u_2 = u_1[/imath] then [imath]u_2=e[/imath], contradiction. [imath]u_1 u_2=u_2[/imath] results in the same contradiction. This means that the case [imath]|I|=2[/imath] is not possible. the case [imath]|I|=3[/imath] is possible, let [imath]I = \{u_1,u_2,u_3\}[/imath] then it's easy to prove [imath]u_3 = u_1u_2[/imath]. The theorem holds. How should I continue? I've tried splitting [imath]I[/imath] as [imath]I=\{u_i: u_i \not = u_ju_k (\text{for a certain }j\not = k)\} \cup \{u_i = u_ju_k (\text{for a certain} j \not = k)\}[/imath] but I still get stuck. Also the case [imath]|I|= 4[/imath] seems contradictory once again.. | 2550052 | Let [imath]G[/imath] be a finite abelian group with exactly one element of order [imath]2[/imath] denoted by [imath]\alpha[/imath]. Prove that [imath]\prod_{g \in G} g = \alpha[/imath].
Let [imath]G[/imath] be a finite abelian group with exactly one element of order [imath]2[/imath] denoted by [imath]\alpha[/imath]. Prove that [imath]\prod_{g \in G} g = \alpha[/imath] Okay so I was given two hints for this problem Hint 1 : If [imath]g \in G[/imath] doesn't have order [imath]2[/imath], then [imath]g \neq g^{-1}[/imath] Hint 2: If you write out [imath]0+1+2+3+4+50+1+2+3+4+5 \mod 6[/imath], why does it come out [imath]3[/imath], other than computing brute force? I'm not sure at all how to use hint 1. The answer for Hint 2 is that you can pair up [imath]5+1=6[/imath] and [imath]4+2 = 6[/imath] and be left with [imath]3[/imath]. In that hint we are looking at the additive group [imath]\mathbb{Z}/6\mathbb{Z}[/imath], and we have [imath]|5| = |1| = 6[/imath] and [imath]|4| = |2| = 3[/imath], so this hint would suggest looking at group operating each two elements of [imath]G[/imath] which have the same order with each other. But the problem is, is that even though [imath]G[/imath] may be a finite group, some [imath]g \in G[/imath] may have [imath]|g| = \infty[/imath], and furthermore every [imath]g \in G[/imath] may have a unique order, (please correct me if I'm wrong). So I'm not sure how I could use it to prove [imath]\prod_{g \in G} g = \alpha[/imath] At this point all I can say about [imath]G[/imath], is that by Lagrange's Theorem, [imath]|G| = 2k[/imath] for some [imath]k \in \mathbb{Z}^+[/imath], and that (trivially) letting [imath]\alpha \in G[/imath] be the element of order [imath]2[/imath], we have [imath]\alpha = \alpha^{-1}[/imath]. I'm assuming there's some algebraic trick that I'm missing here, because I'm sure all I'd have to do is just write out [imath]\prod_{g \in G} g = g_1 \cdot g_2 \cdot \ ... \ \cdot g_{2k}[/imath] and find some way to simplify this expression. EDIT: Thanks to the comments and answers below I realized that since [imath]G[/imath] is abelian, and for any [imath]g \in G[/imath], we have [imath]g^{-1} \in G[/imath], so [imath]\prod_{g \in G} g = g_1 \cdot (g_1^{-1}) \cdot ... \cdot g_i \cdot(g_i^{-1}) \cdot ... \cdot (g_k) \cdot (g_k^{-1})[/imath] where for some [imath]i \in \{1, ..., k\}[/imath] we have [imath]g_i = \alpha[/imath], but now I don't see why [imath]\prod_{g \in G} g \neq e[/imath] where [imath]e[/imath] is the identity of [imath]G[/imath]? |
2400897 | Show that [imath]a^4+b^4\geq \frac18[/imath] given that [imath]a+b=1[/imath]
Show that [imath]a^4+b^4\geq \frac18[/imath] given that [imath]a+b=1[/imath] [imath]b=1-a\Rightarrow a^4+b^4=2a^4-4a^3+6a^2-4a+1[/imath] If we try to find the minimum of a one-variable function,we must solve a 3rd degree equation,on the other hand making perfect squares seems somewhat difficult! Please help. | 2095390 | For [imath]x+y=1[/imath], show that [imath]x^4+y^4\ge \frac{1}{8}[/imath]
As in the title. Let [imath]x,y[/imath] be two real numbers such that [imath]x+y=1[/imath]. Prove that [imath]x^4+y^4\ge \frac{1}{8}[/imath]. Any hints? Basically, the only method I am aware of is plugging [imath]y=1-x[/imath] into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality. |
2400197 | How to find recurrence of prime numbers with periodic function
The aim is to find a recurrence of prime numbers. So we'll take the trigonometric functions as they're standard periodic functions, like [imath]\sin(x)[/imath]: we'll take as an argument of [imath]\sin(x)[/imath] the angle found before: [imath]y=\sin(πx/n) \mid n ∈ ℕ, n>1.[/imath] Fixed [imath]n[/imath] the function gives as result 0 exactly when [imath]x=n[/imath] for obvious reasons and the other zeros are given for every k*n, where [imath]k=1,2,3,...[/imath] Then take [imath]n_1 ∈ ℕ, n_1 > 1[/imath], for all [imath]n ≠ n_1[/imath] and [imath] n_1 ∉ [/imath] set of all zeros of the harmonics, [imath]n_1[/imath] is prime because there is only one harmonic that has the first zero in [imath]x=n_1[/imath]. This harmonic is called prime harmonic. The set of all zeros of prime harmonics are ℕ/{1} then we can build a function that preserves these zeros. The best method is the multiplication. If the first prime harmonic has n=2, the next number without the zero is 3. Then take the second harmonic that has n=3, the next number without zero is 5 and so on. With this method we build gradually a function [imath]p(x)[/imath]: [imath]p(x) = \prod_{n=2}^\infty \sin (πx/n)[/imath] where [imath]x, n ∈ ℕ[/imath] for all [imath]n \in \sin (πx/n).[/imath] This function [imath](p(X))[/imath] is called prime function. | 776997 | Formula for prime counting function
I saw this formula on this paper page 2 [imath]\pi (n)=\sum_{j=2}^{n}\frac{\sin^{2}\left(\pi \frac{(j-1)!^{2}}{j}\right)}{\sin^{2}(\frac{\pi }{j})}[/imath] Where [imath]\pi(n)[/imath] is the prime counting function. Is this true? How to prove it? |
2400942 | Find sum of [imath]n[/imath]-terms of series
I want to find the sum of a series for which the [imath]n^{th}[/imath] term is given by [imath]T_n=n(n+1)(-1)^{n+1}[/imath] Basically if we find the sum of [imath]n[/imath] terms the series turns out to be like this: [imath]1 \times 2 -2 \times 3 +3 \times 4 -4 \times 5 +5 \times 6\ldots[/imath] | 2400151 | summation of multiplying series with negative terms
i know the summation of [imath]1 \times 2 + 2\times 3 + 3\times 4 = \frac{k(k+1)(k+2)}3[/imath] but this one has negative terms. so we need to find below [imath]\sum_{i=1}^n(-1)^{i+1} (i(i+1))[/imath] e.g. [imath]1\times 2 - 2\times 3 + 3\times 4 ....[/imath] if you can share the derivation then it would be very great. |
2401281 | Show that if [imath]a+b+c=0[/imath], [imath]2(a^4 + b^4+ c^4)[/imath] is a perfect square
Show that for [imath]\{a,b,c\}\subset\Bbb Z[/imath] if [imath]a+b+c=0[/imath] then [imath]2(a^4 + b^4+ c^4)[/imath] is a perfect square. This question is from a math olympiad contest. I started developing the expression [imath](a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2[/imath] but was not able to find any useful direction after that. Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question. | 2161515 | Show that [imath]a+b+c=0[/imath] implies that [imath]32(a^4+b^4+c^4)[/imath] is a perfect square.
There are given integers [imath]a, b, c[/imath] satysfaying [imath]a+b+c=0[/imath]. Show that [imath]32(a^4+b^4+c^4)[/imath] is a perfect square. EDIT: I found solution by symmetric polynomials, which is posted below. |
2381988 | Find [imath](a^2+b^2),[/imath] where [imath](a+b)=\dfrac{a}{b}+\dfrac{b}{a}[/imath]
The question is the same .Find [imath]a^2+b^2[/imath]. I think we have to find [imath]a[/imath] and [imath]b[/imath] firstly. Given that [imath]a[/imath] and [imath]b[/imath] are integers. | 2399350 | Can sum of a rational number and its reciprocal be an integer?
Can sum of a rational number and its reciprocal be an integer? My brother asked me this question and I was unable to answer it. The only trivial solutions which I could think of are [imath]1[/imath] and [imath]-1[/imath]. As to what I tried, I am afraid not much. I have never tried to solve such a question, and if someone could point me in the right direction, maybe I could complete it on my own. Please don't misunderstand my question. I am looking for a rational number [imath]r[/imath] where [imath]r + \frac{1}{r}[/imath] is an integer. |
2402701 | Construct a continuous function such that [imath]\int_{-\infty}^\infty|f(x)|\,dx<\infty[/imath] but [imath]\lim_{x\to \infty}|f(x)|[/imath] does not exists.
I have to construct a continuous function such that [imath]\int_{-\infty}^\infty |f(x)| \, dx<\infty[/imath] but [imath]\lim_{x\to \infty}|f(x)|[/imath] does not exists. I have already known one messy example that deals with lines and minimum distance. I just want to see different examples. I know we can construct a triangle with fixed height and decreasing base such that the area get's smaller and the integral is like the geometric series. However, the fixed height makes the limit inexistent. I just dont know how to describe that properly. Thanks. | 469184 | Give an example of a continuous function [imath]f : [0, ∞) \mapsto [0, ∞)[/imath] such that [imath]\int_{0}^{\infty}f(x)dx[/imath] exists but [imath]f[/imath] is unbounded.
Give an example of a continuous function [imath]f : [0, ∞) \mapsto [0, ∞)[/imath] such that [imath]\int_{0}^{\infty}f(x)dx[/imath] exists but [imath]f[/imath] is unbounded. I have been thinking about this. And I have come to the conclusion that I will need to construct a function, [imath]f[/imath], such that [imath]f[/imath] is a sequence of triangles of increasing height, but decreasing base. I obviously need [imath]f[/imath] such that both the height of the triangles, and sum of the bases tends to infinity. But I also need that the [imath]\sum (\text{height} \times \text{base}) \leq \infty [/imath] |
2402514 | help proving an binomial identity about catalan numbers
I need help proving the following identity: [imath]\sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k} \binom{2n-2k}{n-k} = \binom{2n+1}{n}.[/imath] It has to do with the Catalan numbers and Dyck walks. Notice that [imath]\frac{1}{k+1} \binom{2k}{k}[/imath] is the Catalan number [imath]C_k[/imath]. | 65944 | Proving this identity [imath]\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}[/imath] using lattice paths
How can I prove the identity [imath]\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}[/imath]? I have to prove it using lattice paths, it should be related to Catalan numbers The [imath]n[/imath]th Catalan number [imath]C_n[/imath] counts the number of monotonic paths along the edges of a grid with [imath]n\times n[/imath] square cells, which do not pass above the diagonal. See for example this link For example [imath]\frac{1}{k}\binom{2k-2}{k-1}[/imath] is exactly [imath]C_{k-1}[/imath], and the other terms can also be expressed in terms of the Catalan numbers. The second part of the exercise ask to prove the recurrence formula [imath]C_n=\sum_{k=1}^n C_{k-1}C_{n-k}[/imath] using similar reasoning (i.e. lattice paths). So we can't use this formula to prove the first. Could you help me please? |
2402938 | Asymptotics of sum without Euler-Maclaurin
I want to show that [imath]\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n n^{1/k} = 2[/imath] I found the limit with Euler-Maclaurin but this was messy involving integral [imath]\int_1^nn^{1/x}[/imath]dx. What is an easier way using squeeze or other technique? | 307028 | Methods to find [imath]\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^nn^{1/k} [/imath]
What would you suggest here? [imath]\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k} [/imath] |
2402741 | If the scalar product is non degenerate then the metric matrix is invertible and symmetric : why?
I would like to know how to prove that if if the scalar product is non degenerate then the metric matrix is invertible and symmetric. It is probably trivial, probably not but I really don't know how to prove it. They define a non degenerate scalar product, a scalar product verifying : [imath]\forall w ~ \langle v |w\rangle =0 \Rightarrow v=0 [/imath] | 342267 | Non-degenerate bilinear forms and invertible matrices
I have the following question: Show that a symmetric bilinear form is nondegenerate if and only if its matrix in a basis is invertible Ok so both directions "if non-degenerate then the matrix is invertible" and "if matrix is invertible then the form is nondegenerate" have to be proven for this. For the first direction. If the bilinear form is non-degenerate its null space is [imath]\left\{0\right\}[/imath], so for every [imath]v \neq 0[/imath] there exists a [imath]v'[/imath] such that [imath]\langle v, v' \rangle \neq 0[/imath] and so there are no zero eigenvalues, I'm pretty sure this somehow implies the determinant of the matrix is also non-zero, as it is the product of the eigenvalues and hence invertible (but not sure how to show this). For the other direction, if the matrix is invertible, then the determinant is non-zero and, if my previous assumption is correct, this implies there are no zero eigenvectors, which implies the null space is [imath]\left\{0\right\}[/imath] and hence non-degenerate? Thanks for any guidance, the notes I have on this topic are pretty poor so I'm having a hard time fully understanding everything. Thanks again. |
2402811 | Find the maximum value of [imath]x_1x_2+x_2x_3+\cdots+x_nx_1[/imath]
Find the maximum value of [imath]x_1x_2+x_2x_3+\cdots+x_nx_1[/imath] if [imath]x_1+x_2+\cdots+x_n=0[/imath] and [imath]x_1^2+x_2^2+...+x_n^2=1[/imath]. I can get by Cauchy-Schwarz that [imath]x_1x_2+x_2x_3+\cdots+x_nx_1 \leqslant 1[/imath], but can expression [imath]x_1x_2+x_2x_3+\cdots+x_nx_1[/imath] be smaller? | 2213960 | Optimising [imath]x_1x_2+x_2x_3+\cdots+x_nx_1[/imath] given certain constraints
To seek the maximum value of [imath]S=x_1x_2+x_2x_3+\cdots+x_nx_1[/imath] on this domain: [imath]x_1+x_2+\cdots+x_n=0[/imath] and [imath]x_1^2+x_2^2+\cdots+x_n^2=1[/imath]. I have made some trivial observations: 1) [imath]S\in[-1,1)[/imath] by the rearrangement inequality. 2) We can make [imath]S[/imath] arbitrarily close to [imath]1[/imath] by increasing [imath]n[/imath]. 3) An equivalent problem is to minimise [imath](x_1-x_2)^2+\cdots+(x_n-x_1)^2[/imath]. But does the maximum have a meaningful closed form for each [imath]n[/imath]? |
2403368 | Compactness of a subset of [imath]\mathbb{Q}[/imath]
I tried to solve this exercise taken from the text Manetti - Topologia, pg. 81. Let's consider the metric space [imath](\mathbb{Q},d_e)[/imath], with [imath]d_e[/imath] the standard euclidian distance. Show that [imath]K=[/imath]{[imath]x \in \mathbb{Q}: 0 \leq x \leq \sqrt2[/imath]} is closed, bounded, but not compact. It's a bounded set because [imath]sup_{x,y \in K}[/imath] [imath]d_e(x,y)=\sqrt2[/imath]. It's closed because the complementar set is [imath]C_\mathbb{Q}(K)=(-\infty,0) \cup (\sqrt2,+\infty)[/imath], which is an open set since for every [imath]x[/imath] there is an open set [imath]B[/imath]: [imath]x \in B_\epsilon(x)=[/imath] {[imath]y \in \mathbb{Q}: |x-y| <\epsilon[/imath]} [imath]\subset A[/imath]. I think it's not compact because if I have an open set [imath]U_\alpha[/imath] which covers [imath]\sqrt2[/imath], then there's an open ball [imath]B_(\sqrt2)[/imath] and so there's a [imath]x \in \mathbb{Q}[/imath] such that [imath]x=\sqrt2[/imath], which is a contradiction. Is it right? | 1038094 | Is the subset [imath][0, \sqrt2] ∩\mathbb{Q} ⊂ \mathbb{Q}[/imath] closed, bounded, compact?
Letting [imath]\mathbb{Q}[/imath] be equipped with the Euclidean metric. What I can work out is that it is bounded as its contained in the closed ball of radius [imath]{\sqrt2}/{2}[/imath] centred at [imath]{\sqrt2}/{2} [/imath]. Its not compact as it can be expressed as union of the two disjoint open sets [imath][0,{\sqrt2}/{2}) [/imath]and[imath] ({\sqrt2}/{2}, \sqrt2)[/imath] (though I'm not sure if this makes it not compact in [imath]\mathbb{Q}[/imath] or just [imath]\mathbb{R}[/imath]). And its not closed as the sequence of truncations of [imath]\sqrt2[/imath] in [imath]\mathbb{Q}[/imath] converges to [imath]\sqrt2[/imath] |
2403203 | Can [imath]f[/imath] ever has uncountable number of Local maxima or minima?
This question suddenly came to my mind (May be very silly question...) Suppose [imath]f[/imath] is a continuous function over [imath]\mathbb{R}[/imath] and twice differentiable. Can [imath]f[/imath] ever has uncountable number of Local maxima or minima? (Twice differentiablity can be ignored for finitely many points). I was thinking of counter-example (obviously), but could not find one.(I am really really bad at finding counter-example actually!) I thought about the example [imath]x^2\sin\Big(\dfrac{1}{x}\Big)[/imath] when [imath]x\neq 0[/imath], and [imath]0[/imath], when [imath]x=0[/imath], but this function is not twice differentiable everywhere. Added: Note: Twice differentiablity can be ignored for finitely many points (Because I think if I impose this condition the problem will be interesting...) | 1339170 | Does there exist a continuous function from [0,1] to R that has uncountably many local maxima?
Does there exist a continuous function from [imath][0,1][/imath] to [imath]R[/imath] that has uncountably many strict local maxima? |
2403608 | Solving the Unknown in the Figure
I was asked to solve for the [imath]\theta[/imath] shown in the figure below. My work: The [imath]\Delta FAB[/imath] is an equilateral triangle, having interior angles of [imath]60^o.[/imath] I don't think [imath]\Delta HIG[/imath] and [imath]\Delta DEC[/imath] are right triangles. So far, that's all I know. I'm confused on how to get [imath]\theta.[/imath] How do you get the [imath]\theta[/imath] above? | 1434534 | [imath]ABCD[/imath] is a square and [imath]AEB[/imath] is an equilateral triangle. Find [imath]\angle DEC[/imath].
I found this question in an exam question paper. But I was stumped by it. It seemed that all the answers were correct. I tried: Let [imath]\angle ADE=x[/imath] and [imath]\angle EDC=y[/imath]. [imath]\angle DAB=90^{\circ}=\angle DAE+\angle EAB=30^{\circ}+60^{\circ}[/imath]. [imath]\text{ref}\angle AEB=300^{\circ}[/imath]. And we know that [imath]x+y=90^{\circ}[/imath] But now I am stuck. How to proceed? |
2403653 | Is there any formula for this sum of power of positive integers?
I wonder if there is any formula for this sum. [imath]k^\gamma+(k-1)^\gamma+\cdots+1^\gamma,[/imath] where [imath]k[/imath] is positive integer and [imath]\gamma\in(0,1)[/imath]. And how about [imath]\gamma<0[/imath]? Or is there any known asymptotic when [imath]k[/imath] tends to infinity? | 63986 | Asymptotic behaviour of sums of consecutive powers
Let [imath]S_k(n)[/imath], for [imath]k = 0, 1, 2, \ldots[/imath], be defined as follows [imath]S_k(n) = \sum_{i=1}^n \ i^k[/imath] For fixed (small) [imath]k[/imath], you can determine a nice formula in terms of [imath]n[/imath] for this, which you can then prove using e.g. induction. For small [imath]k[/imath] we for example get [imath]\begin{align} S_0(n) &= n\\ S_1(n) &= \frac{1}{2}n^2 + \frac{1}{2}n \\ S_2(n) &= \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \\ S_3(n) &= \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 \\ S_4(n) &= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3 - \frac{1}{30}n \end{align}[/imath] The coefficients of these polynomials are related to the Bernoulli-numbers, and getting arbitrary coefficients in these polynomials (i.e. the coefficient of [imath]n^m[/imath] in [imath]S_k(n)[/imath] for large [imath]k,m[/imath]) is not so easy. However, th first two coefficients follow a simple pattern: the coefficient of [imath]n^{k+1}[/imath] is [imath]\frac{1}{k+1}[/imath] and the coefficient of [imath]n^k[/imath] (for [imath]k > 0[/imath]) is always [imath]\frac{1}{2}[/imath]. My main question now is: How can we prove that [imath]S_k(n) = \frac{1}{k+1}n^{k+1} + \frac{1}{2}n^k + O(n^{k-1})[/imath] for [imath]k > 0[/imath]? The first coefficient can be explained intuitively, as [imath]S_k(n) = \sum_{i=1}^n \ i^k \approx \int_{i=1}^n i^k di \approx \frac{n^{k+1}}{k+1}[/imath] Maybe you could make this more rigorous, but I don't see how you will get the term [imath]\frac{1}{2}n^k[/imath] with this. Also, while the coefficient of [imath]n^{k+1}[/imath] can be explained intuitively, it's not clear to me why the coefficient of [imath]n^k[/imath] is [imath]\frac{1}{2}[/imath], and why this one is fixed while e.g. the coefficient of [imath]n^{k-1}[/imath] is different for different [imath]k[/imath]. If someone could explain that, that would be appreciated as well. Thanks. |
1007977 | Proof that [imath] 3 > (1+\frac{1}{n})^n \geq 2[/imath]
I am studying computer science in first term, and i got a task that i was not able to solve for a long time now. I have to prove that [imath] 3 > (1+\frac{1}{n})^n>=2[/imath] for every [imath]n \in \mathbb{N}[/imath] [imath](1+\frac{1}{n})^n>=2[/imath] Can be proved easily with Bernoullis Inequality: [imath](1+x)^n>=1+x*n[/imath] [imath](2= 1+\frac{1}{n}*n)[/imath] Thats cool. But how do i prove that it is smaller than 3? I thought of the following. Using bionomical theorem we can write the above term as: [imath]\sum_{k=0}^n\binom{n}{k}*\frac{1}{n^k}[/imath] If we write the first to sums for k= 0 and k= 1 seperately, we can see that they both equal 1. [imath]\binom{n}{0}*\frac{1}{n^0}[/imath] And [imath]\binom{n}{1}*\frac{1}{n^1}[/imath] Both equal one, am i correct. So the sum [imath]\sum_{k=2}^N\binom{n}{k}*\frac{1}{n^k}[/imath] (Without the first two sums) has to be >1. But thats where my problem is. I mean i already proved that Each of the sums is smaller than 1 but that doesnt prove anything, right? [imath]>1+>1\neq >1[/imath] if you know what i mean. The whole thing has to be proofed to be smaller than 1. How can i see that. P.S: The answer is NOT that the above term wont get bigger as Eulers Number, as by definition. I mean this statement is correct, but it would not solve the question. I really have to, and want, to show it on my own. Any help is much apprechiated, been thinking about this for so many hours now. | 200141 | Inequality [imath](1+\frac1k)^k \leq 3[/imath]
How can I elegantly show that: [imath](1 + \frac{1}{k})^k \leq 3[/imath] For instance I could use the fact that this is an increasing function and then take [imath]\lim_{ k\to \infty}[/imath] and say that it equals [imath]e[/imath] and therefore is always less than [imath]3[/imath] Is this sufficient? What is a better wording than "increasing function" |
2404139 | When is [imath]\sqrt[3]{a+\sqrt b}+\sqrt[3]{a-\sqrt b}[/imath] an integer?
I saw a Youtube video in which it was shown that [imath](7+50^{1/2})^{1/3}+(7-50^{1/2})^{1/3}=2[/imath] Since there are multiple values we can choose for the [imath]3[/imath]rd root of a number, it would also make more sense to declare the value of this expression to be one of [imath]2, 1 + \sqrt{-6},[/imath] or [imath]1 - \sqrt{-6}[/imath] We may examine this more generally. If we declare [imath]x[/imath] such that [imath]x=(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}[/imath] [imath]\text{(supposing } a \text{ and } b \text{ to be integers here)}[/imath] one can show that [imath]x^3+3(b-a^2)^{1/3}x-2a=0[/imath] Which indeed has [imath]3[/imath] roots. We now ask For what integer values of [imath]a[/imath] and [imath]b[/imath] is this polynomial solved by an integer? I attempted this by assuming that [imath]n[/imath] is a root of the polynomial. We then have [imath]x^3+3(b-a^2)^{1/3}x-2a[/imath] [imath]||[/imath] [imath](x-n)(x^2+cx+d)[/imath] [imath]||[/imath] [imath]x^3+(c-n)x^2+(d-nc)x-nd[/imath] Since [imath](c-n)x^2=0[/imath] we conclude that [imath]c=n[/imath] and we have [imath]x^3+3(b-a^2)^{1/3}x-2a=x^3+(d-c^2)x-cd[/imath] And - to continue our chain of conclusions - we conclude that [imath]3(b-a^2)^{1/3}=d-c^2 \quad\text{and}\quad 2a=cd[/imath] At this point I tried creating a single equation and got [imath]108b=4d^3+15c^2d^2+12c^4d-c^6[/imath] This is as far as I went. | 374619 | Is [imath]\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n[/imath], [imath](p,q,n)\in\mathbb{N} ^3[/imath] solvable?
In this recent answer to this question by Eesu, Vladimir Reshetnikov proved that [imath] \begin{equation} \left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1} \end{equation} [/imath] I would like to know if this result can be generalized to other triples of natural numbers. Question. What are the solutions of the following equation? [imath] \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left( p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right) \in \mathbb{N} ^{3}.\tag{2} \end{equation} [/imath] For [imath](1)[/imath] we could write [imath]26+15\sqrt{3}[/imath] in the form [imath](a+b\sqrt{3})^{3}[/imath] [imath] 26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \sqrt{3} [/imath] and solve the system [imath] \left\{ \begin{array}{c} a^{3}+9ab^{2}=26 \\ a^{2}b+b^{3}=5. \end{array} \right. [/imath] A solution is [imath](a,b)=(2,1)[/imath]. Hence [imath]26+15\sqrt{3}=(2+\sqrt{3})^3 [/imath]. Using the same method to [imath]26-15\sqrt{3}[/imath], we find [imath]26-15\sqrt{3}=(2-\sqrt{3})^3 [/imath], thus proving [imath](1)[/imath]. For [imath](2)[/imath] the very same idea yields [imath] p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3} \sqrt{3} [/imath] and [imath] \left\{ \begin{array}{c} a^{3}+9ab^{2}=p \\ 3( a^{2}b+b^{3}) =q. \end{array} \right. \tag{4} [/imath] I tried to solve this system for [imath]a,b[/imath] but since the solution is of the form [imath] (a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5} [/imath] where [imath]x[/imath] satisfies the cubic equation [imath] 64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6} [/imath] would be very difficult to succeed, using this naive approach. Is this problem solvable, at least partially? Is [imath]\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n[/imath], [imath](p,q,n)\in\mathbb{N} ^3[/imath] solvable? |
2404869 | Two polynomials sharing two values
Let [imath]f, g \colon \mathbb C \to \mathbb C[/imath] be two nonconstant polynomial functions such that [imath]f(x) = 1 \iff g(x)=1[/imath] and [imath]f(x) = 0 \iff g(x)=0.[/imath] Does it imply that [imath]f \equiv g[/imath], that is, they come from the same polynomial? This is certainly not true for real polynomials, as an example of [imath]g(x) = f(x)^3 = x^3[/imath] shows. If possible, I would like to know about generalisations to fields other than [imath]\mathbb C[/imath] - I couldn't find any reference in literature. | 677815 | Do two distinct level sets determine a non-constant complex polynomial?
Let [imath]f[/imath] and [imath]g[/imath] be non-constant complex polynomials in one variable. Let [imath]a\neq b[/imath] be complex numbers and suppose [imath]f^{-1}(a)=g^{-1}(a)[/imath] and [imath]f^{-1}(b)=g^{-1}(b)[/imath]. Does this imply [imath]f=g[/imath]? If we think of entire functions instead of polynomials, the answer is negative: take [imath]e^{-z}[/imath] and [imath]e^{z}[/imath] and they share the same level sets for 0 and 1. More generally, Nevanlinna's 5-values theorem says that 5 level sets completely determine a non-constant meromorphic function. Can we lower this number when dealing with polynomials? |
2405053 | Proof that the sum of the even side and the hypotenuse of a coprime (and positive) Pythagorean triple is a square number
I wish to prove that for any coprime (and positive) Pythagorean triple [imath](x,y,z)[/imath], where [imath]x[/imath] is even, [imath]x+z[/imath] is always a square number. By inspection, this appears true; [imath](3,4,5)[/imath] gives [imath]9[/imath], [imath](5,12,13)[/imath] gives [imath]25[/imath], [imath](8,15,17)[/imath] gives [imath]25[/imath], [imath](7,24,25)[/imath] gives [imath]49[/imath] et cetera. Yet I am stuck on the proof. I do not know the original source of the problem. | 395543 | Formulas for calculating pythagorean triples
I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is: [imath]\begin{align} &a = m^2-n^2 \\ &b = 2mn\\ &c = m^2+n^2 \end{align}[/imath] With numerous parameters. So are there other formulas/methods? |
2405185 | The answer to the equation [imath]\log_3x+\log_2x+\log_4x=1[/imath]
The answer to the equation : [imath]\log_3x+\log_2x+\log_4x=1[/imath] my try : [imath]\log_3x+\log_2x+\log_4x=1\\ \log_2x+\log_2\sqrt{x}+\log_{3}x=1\\\log_2x\sqrt{x}=\log_{3}{\dfrac{3}{x}}[/imath] now ? | 2397733 | Is there an analytic solution for the equation [imath]\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1[/imath]?
I am looking for a close form solution for below equation. [imath]\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1.[/imath] I solve it by graphing, but I don't know is there a way to find [imath]x[/imath] analytically ? |
2404834 | Show subgroups of an abelian group isomorphic to each other
I am working on this problem and got stuck: Let [imath]G[/imath] be a group with three normal subgroups [imath]N_1, N_2, N_3[/imath]. Suppose whenever [imath]i[/imath] and [imath]j[/imath] are distinct in [imath]\{1,2,3\}[/imath], [imath]G = N_i N_j[/imath], and [imath]N_i \cap N_j = \{e\}[/imath]. Show that [imath]N_1 \simeq N_2 \simeq N_3[/imath]. I have shown that [imath]G[/imath] is in fact abelian, hence [imath]N_i[/imath]'s are abelian subgroups. In order to show that they are isomorphic to each other, i tried to define an isomorphism between [imath]N_i[/imath] and [imath]N_j[/imath] but I can't come to the conclusion. Is my approach not feasible? Should I try another way? | 294487 | [imath]N_1,N_2,N_3 \unlhd G, N_i\cap N_j =\{e\}, G = N_iN_j[/imath]. Want to show that [imath]G[/imath] is abelian, [imath]N_i[/imath] are isomorphic.
The following is a problem from the Berkeley Problems book. Let [imath]G[/imath] be a group with three normal subgroups [imath]N_1[/imath] , [imath]N_2[/imath], and [imath]N_3[/imath]. Suppose [imath]N_i \cap N_j = \{ e\}[/imath] and [imath]N_iN_j = G[/imath] for all [imath](i,j )[/imath] with [imath]i \ne j[/imath] . Show that [imath]G[/imath] is abelian and [imath]N_i[/imath], is isomorphic to [imath]N_j[/imath] for all [imath]i, j[/imath] . Attempt: I have not got very far. The only thing that comes to my mind is that as [imath]|HK|=\frac{|H||K|}{|H \cap K|}[/imath], [imath]|N_1|=|N_2|=|N_3|=m[/imath], and order of [imath]|G|=m^2[/imath]. I have made other futile attempts of trying to show m prime, finding an isomorphism between the [imath]N's[/imath]. I think the fact that the equation [imath]N_i N_j=G[/imath], could be seen as, left coset of [imath]N_j[/imath] with the multiplying elements restricted to only [imath]N_i[/imath] could provide me an answer, but I haven't got anywhere with this. Thanks in advance. |
2405154 | Why is [imath](x_i-\bar{x})^2[/imath] in the formula for variance as opposed to [imath]|x_i-\bar{x}|[/imath]?
For a data set [imath]\bigcup_{i=1}^nx_i[/imath] of [imath]n[/imath] values with mean [imath]\bar{x}[/imath] the variance is defined as [imath]\sigma^2=\frac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n}[/imath] My textbook says that “the square ensures that each term in the sum is positive, which is why the sum turns out not to be zero.” However wouldn’t [imath]\sigma^2=\frac{\displaystyle\sum_{i=1}^n|x_i-\bar{x}|}{n}[/imath] also prevent a sum of zero? | 717339 | Why is variance squared?
The mean absolute deviation is: [imath]\dfrac{\sum_{i=1}^{n}|x_i-\bar x|}{n}[/imath] The variance is: [imath]\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}[/imath] So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a positive value, so the sum won't be zero, but absolute value also gives a positive value. Why isn't it [imath]|x_i-\bar x|^2[/imath], then? Squaring just enlarges, why do we need to do this? A similar question is here, but mine is a little different. Thanks. |
2402308 | Infinite ring is an integral domain if the quotient with each non-trivial ideal is finite
Suppose [imath]R[/imath] is a commutative ring with identity of infinite cardinality. If for every non-zero ideal [imath]I[/imath] of [imath]R[/imath] we have [imath]|\frac{R}{I}|<\infty[/imath] then [imath]R[/imath] is an integral domain. How do I go on proving this? | 126206 | Prove that an infinite ring with finite quotient rings is an integral domain
How can we show that if [imath]R[/imath] is an infinite commutative ring and [imath]R/I[/imath] is finite for every nonzero [imath]I \unlhd R[/imath], then [imath]R[/imath] is an integral domain? I tried proceeding by contradiction: assume [imath]a[/imath],[imath]b[/imath] [imath]\in R \backslash \{0\}[/imath] and [imath]ab=0[/imath]; then [imath]R/(a)[/imath] and [imath]R/(b)[/imath] must be finite, say [imath]R/(a)=\{k_i + (a) : 1 \leq i \leq m\}[/imath] and [imath]R/(b)=\{l_j + (b) : 1 \leq j \leq n\}[/imath]. Does this mean [imath]R[/imath] must be finite? Or what about using the fact that [imath]R/(a,b)[/imath] finite? Thanks for any help with this! |
2405419 | [imath] (\tan\frac{A}{2})^2 +(\tan \frac{B}{2})^2 + (\tan \frac{C}{2})^2 \ge 1[/imath]
Prove that [imath] (\tan\frac{A}{2})^2 + (\tan \frac{B}{2})^2 + (\tan \frac{C}{2})^2 \ge 1[/imath] when [imath]A,B,C[/imath] are the angles in a triangle. I am trying to solve it using Jensen's Equality but not getting any desired result. Is there any other methodology of solving this problem. | 337628 | Trigonometric Functions And Identities Question .
If [imath]A[/imath] , [imath]B[/imath] and [imath]C[/imath] are the angles of a triangle , I have to show that : [imath] \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 [/imath] . I had only arrived at [imath]A+B+C = \pi [/imath] , thus [imath]\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} [/imath] What to do next? |
2405325 | Proof of liminf subset limsup
I am sure this is fairly simple, but I am just not getting it. I am not sure how to show that [imath]\liminf_n→_∞ A_n \ [/imath] ⊆ [imath]\limsup_n→_∞ A_n \ [/imath] I guess I am not understanding lim inf and lim sup properly. | 213327 | Proofs with limit superior and limit inferior: [imath]\liminf a_n \leq \limsup a_n[/imath]
I am stuck on proofs with subsequences. I do not really have a strategy or starting point with subsequences. NOTE: subsequential limits are limits of subsequences Prove: [imath]a_n[/imath] is bounded [imath]\implies \liminf a_n \leq \limsup a_n[/imath] Proof: Let [imath]a_n[/imath] be a bounded sequence. That is, [imath]\forall_n(a_n \leq A)[/imath]. If [imath]a_n[/imath] converges then [imath]\liminf a_n = \lim a_n = \limsup a_n[/imath] and we are done. Otherwise [imath]a_n[/imath] has a set of subsequential limits we need to show [imath]\liminf a_n \leq \limsup a_n[/imath]: This is where I am stuck... |
2402963 | Power Sum and Negative Zeta
Power sum is given by [imath] 1^m + \cdots + n^m = \frac1{m+1} \sum_{k=0}^m (-1)^k \binom{m+1}kB_k n^{m+1-k}[/imath] and negative zeta values are given by [imath]\zeta(-m) = (-1)^n \frac{B_{m+1}}{m+1}[/imath] But heuristically, [imath]\zeta(-m) = 1^m + 2^m + \cdots [/imath]. So it feels as if taking [imath]\lim_{n \rightarrow \infty}[/imath] in the power sum formula should give us the negative zeta value. All this is obviously mathematically not rigorous ([imath]\zeta(s)[/imath] with [imath]\text{Re}(s)\le 1[/imath] must be calculated using reflection formula and taking such limit is impossible). But to me, the occurrence of Bernoulli number in both expressions seems too suspicious for a coincidence. Is it really just a coincidence? | 2252550 | Integrating the formula for the sum of the first [imath]n[/imath] natural numbers
I was messing around with some math formulas today and came up with a result that I found pretty neat, and I would appreciate it if anyone could explain it to me. The formula for an infinite arithmetic sum is [imath]\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},[/imath] so if you want to find the sum of the natural numbers from [imath]1[/imath] to [imath]n[/imath], this equation becomes [imath]\frac{n^2+n}{2},[/imath] and the roots of this quadratic are at [imath]n=-1[/imath] and [imath]0[/imath]. What I find really interesting is that [imath]\int_{-1}^0 \frac{n^2+n}{2}dn=-\frac{1}{12}[/imath] There are a lot of people who claim that the sum of all natural numbers is [imath]-\frac{1}{12}[/imath], so I was wondering if this result is a complete coincidence or if there's something else to glean from it. |
2405270 | Finding the matrix derivative of [imath]X^{-1}[/imath] with respect to [imath]X[/imath]
Assume [imath]X \in \mathbb{R^{n \times n}}[/imath]. I could not found particular formula to calculate the Derivative of [imath]X^{-1}[/imath] with respect to [imath]X[/imath], but I found a formula related to inverse of matrix as follows: (1)[imath]\frac{\partial}{\partial X} (a^TX^{-1}b) = -X^{-T}ab^TX^{-T} \quad a, b \in \mathbb{R}^n[/imath] Can anyone give an insight on how derive a formula for derivative of [imath]X^{-1}[/imath] or formula (1) please? Thank you in advance. ====== This post shared and discussed the same topic and I was asked if the current post is redundant. I believe the way the problem stated and discussed in these two posts is different. Specifically, I was trying to learn a simple approach for finding the derivative of a matrix expression that contains inverse of a matrix. I believe the detailed answer and discussion in the post is helpful to other learners like me(with an elementary calculus and matrix understanding). | 190424 | How to evaluate the derivatives of matrix inverse?
Cliff Taubes wrote in his differential geometry book that: We now calculate the directional derivatives of the map [imath]M\rightarrow M^{-1}[/imath] Let [imath]\alpha\in M(n,\mathbb{R})[/imath] denote any given matrix. Then the directional derivatives of the coordinates of the map [imath]M\rightarrow M^{-1}[/imath] in the drection [imath]\alpha[/imath] are the entries of the matrix [imath]-M^{-1}\alpha M^{-1}[/imath] Consider, for example, the coordinate given by the [imath](i,j)[/imath]th entry, [imath](M^{-1})_{ij}[/imath]. The directional derivative in the drection [imath]\alpha[/imath] of this function on [imath]GL(n,\mathbb{R})[/imath] is [imath]-(M^{-1}\alpha M^{-1})_{ij}[/imath] In particular, the partial derivative of the function [imath]M\rightarrow (M^{-1})_{ij}[/imath] with respect to the coordinate [imath]M_{rs}[/imath] is [imath]-(M^{-1})_{ir}(M^{-1})_{sj}[/imath]. I am wondering why this is true. He did not give any deduction of this formula, and all the formulas I know for matrix inverse does not generate anything similar to his result. So I venture to ask. |
2405612 | Finding [imath]O(ab)[/imath] when [imath]O(a)[/imath] and [imath]O(b)[/imath] are given in non abelian Group
Let [imath]G[/imath] be a non abelian group. If [imath] a[/imath] and [imath]b[/imath] [imath]\in[/imath] [imath]G[/imath] and [imath]O(a) = 3[/imath] and [imath]O(b) = 4[/imath]. Then what is the order of the element [imath]ab?[/imath] Now Since [imath]O(a)= 3[/imath] and [imath]O(b)= 4[/imath] , [imath]O(G)[/imath] must be of the form [imath]12.n[/imath] for some [imath]n \in N[/imath]. therefore [imath]O(ab)[/imath] should divide [imath]12.n.[/imath] I dont know how to proceed from here...please give me a hint to solve this problem [imath]a) 6 \quad b) 12 \quad c)[/imath] of [imath]12k[/imath] for some [imath]k \in R \quad d)[/imath] need not be finite | 1178038 | Let [imath]G[/imath] be a nonabelian group. Let [imath]a \in G[/imath] have order [imath]4[/imath] and let [imath]b \in G[/imath] have order [imath]3[/imath]. Then the order of the element [imath]ab[/imath] in [imath]G[/imath]
Let [imath]G[/imath] be a nonabelian group. Let [imath]a \in G[/imath] have order [imath]4[/imath] and let [imath]b \in G[/imath] have order [imath]3[/imath]. Then the order of the element [imath]ab[/imath] in [imath]G[/imath] : A. is [imath]6[/imath]. B. is [imath]12[/imath] C. is of the form [imath]12k, k \geq 2[/imath] D. need not be finite' Attempt: I think the correct option should be [imath]D.[/imath] need not be finite. Since, [imath]G[/imath] is non abelian, A,B,C shouldn't be true. I just am not able to find a suitable example for D. Am I correct in my inference? Thank you for your help in this regard. |
2379458 | Is [imath]\sum_{n\ge0}\frac1{2^{2^n}}[/imath] rational?
How might I determine whether [imath]\sum_{n\ge0}\frac1{2^{2^n}}[/imath] is rational? I have seen in the answers to this question that the sum converges and it was also mentioned that this is called Kempner's number. Another question mentions this under the name Fredholm number. | 788772 | How to sum this series to infinity: [imath]\sum_{n=0}^{\infty} \frac1{2^{2^n}}[/imath]
How to sum the series: [imath]\sum _{ n=0 }^{ n=\infty }{ \frac { 1 }{ { 2 }^{ { 2 }^{ n } } } }[/imath] PS: Just a hint would suffice. |
2405673 | Maximum value of [imath]\sin(n)[/imath] for integer [imath]n[/imath]
We know that for any integer [imath]n[/imath], [imath]\sin(n)[/imath] can never equal to one. So what is the maximum value of [imath]sin(n)[/imath]? How close can the sine of an integer get to unity? I'll be really interested to know if such a value exists. | 1481458 | Does [imath]\sin n[/imath] have a maximum value for natural number [imath]n[/imath]?
In formal, does there exist [imath]k\in\mathbb{N}[/imath] such that [imath]\sin n\leq\sin k[/imath] for all [imath]n\in\mathbb{N}[/imath]? |
1931193 | Convergence of series based on iteratively applying the logarithm function
Context From the Riemann criterion, we know that [imath]\sum \frac 1{n^s}[/imath] converges when [imath]s>1[/imath]. Continuing in the same path, we know thanks to Bertrand series that [imath]\sum \frac 1{n\ln (n)^s}[/imath] converges when [imath]s>1[/imath]. And so on, i.e. if we note [imath]\ln^{(k)}[/imath] the [imath]k[/imath]-th iteration of the logarithm: [imath]\ln^{(k)}:=\underbrace{\ln\circ\ln\circ\cdots\circ\ln}_{k \text{ times}}[/imath] then for all [imath]\ell[/imath] [imath]\sum \left(\ln^{(\ell)}(n)^s\prod_{k=0}^{\ell-1} \ln^{(k)}(n)\right)^{-1}[/imath] converges when [imath]s>1[/imath]. The problem So I was wondering what happens at the limit. Does it go infinity or does it converge ? Let's consider the following series: [imath]\mathscr S:=\sum_{n=1}^\infty \left(\prod_{\substack{k=0 \\ \ln^{(k)}(n)\geqslant 1}}^{\infty} \ln^{(k)}(n)\right)^{-1},[/imath] where the apparently infinite product is in fact finite for all [imath]k[/imath]. Basically, we take as much logarithm as we can so it doesn't get smaller than [imath]1[/imath]. Concretely We have [imath]\ln^{(0)}(1)=1\geqslant 1[/imath] and [imath]\ln^{(1)}(1)=0< 1[/imath] so we stop there. And [imath]\ln^{(0)}(2)=2\geqslant 1[/imath] and [imath]\ln^{(1)}(2)\approx 0.69< 1[/imath] so we stop there. And [imath]\ln^{(1)}(3)\approx 1.1\geqslant 1[/imath] and [imath]\ln^{(2)}(3)\approx 0.1< 1[/imath] so we stop there. [imath]\vdots[/imath] And [imath]\ln^{(1)}(15)\approx 2.7\geqslant 1[/imath] and [imath]\ln^{(2)}(15)\approx 0.996< 1[/imath] so we stop there. And [imath]\ln^{(2)}(16)\approx 1.02\geqslant 1[/imath] and [imath]\ln^{(3)}(16)\approx 0.02< 1[/imath] so we stop there. [imath]\vdots[/imath] So it starts like this: [imath]\mathscr S=\frac 11+\frac 1{2}+\frac 1{3\ln (3)}\ldots+\frac 1{15\ln (15)}+\frac 1{16\ln (16)\ln\ln (16)}+\ldots.[/imath] The question Does [imath]\mathscr S[/imath] converge ? What could work We can prove Riemann criterion and Bertrand's one using the Cauchy condensation test: [imath]\sum f(n)<\infty \iff \sum 2^nf(2^n)<\infty.[/imath] But I didn't get anywhere. We can also notice that since we want [imath]n[/imath] such that [imath]1\leqslant \log^{(k)}(n)<e[/imath] we want [imath]n[/imath] sucht that [imath]e^{(k)}\leqslant n < e^{(k+1)}.[/imath] So we can rewrite the series: [imath]\mathscr S:=\sum_{k=0}^\infty \sum_{n=[e^{(k)}]+1}^{[e^{(k+1)}]}\left(\prod_{\ell=0}^{k} \ln^{(\ell)}(n)\right)^{-1}.[/imath] Therefore, if we take [imath]n\in \{[e^{(k)}]+1, \ldots, [e^{(k+1)}]\}[/imath] we have: [imath]\sum_{n=[e^{(k)}]+1}^{[e^{(k+1)}]}\left(\prod_{\substack{\ell=0 \\ \ln^{(k)}(n)\geqslant 1}}^{k} \ln^{(\ell)}(n)\right)^{-1}\leqslant \frac{e^{(k+1)}-e^{(k)}}{\displaystyle\prod_{\ell=0}^{k} \ln^{(\ell)}(e^{(k+1)})}=\frac{e^{(k+1)}-e^{(k)}}{\displaystyle\prod_{\ell=0}^{k} e^{(\ell)}}.[/imath] Now we only need to understand whether or not [imath]\sum_{k=0}^{\infty} (e^{(k+1)}-e^{(k)})\left(\prod_{\ell=0}^{k} e^{(\ell)}\right)^{-1}[/imath] converges. | 1764465 | Does the series converge
We know that all series of the following form diverge: \begin{equation} S_k = \sum_{n=\left\lceil \mathrm{e}^k \right\rceil}^\infty \frac{1}{n (\ln n) (\ln \ln n)\dots(\ln^k n)} \end{equation} where the notation [imath]\ln^k[/imath] is function composition. Additionally, this series is interesting because for large [imath]k[/imath], it diverges very slowly. Then does the following series converge? \begin{equation} S = \sum_{n=1}^\infty \frac{1}{f(n)} \end{equation} where [imath]f(n)[/imath] is defined recursively by the monotonic function \begin{equation} f(n) = \begin{cases} n & \text{if }n \le \mathrm{e} \\ n \cdot f{\left(\ln n\right)} & \text{otherwise} \end{cases} \end{equation} The first few terms of this series are \begin{equation} 1 + \frac{1}{2} + \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} + \dots + \frac{1}{16 (\ln 16) (\ln \ln 16)} + \dots \end{equation} and the maximum power of [imath]\ln[/imath] in the denominator gradually increases. This series should eventually grow slower than all [imath]S_k[/imath] (it is in some kind of sense a limit) because the terms will eventually drop below those of [imath]S_k[/imath] for any particular [imath]k[/imath]. But at the same time it seems to grow faster than all series of the form \begin{equation} T_k = \sum_{n=\left\lceil \mathrm{e}^k \right\rceil}^\infty \frac{1}{n (\ln n) (\ln \ln n)\dots(\ln^{k-1} n){(\ln^k n)}^2} \end{equation} which are known to converge, albeit also very slowly. |
2406057 | Let [imath]G=\{x,y│x^4=y^4=e,xyxy^{-1}=e\}.[/imath]
Let [imath]G=\{x,y│x^4=y^4=e,xyxy^{-1}=e\}[/imath] 1.Show that [imath]|G|\leq 16[/imath]. 2.Suppose that [imath]|G|=16[/imath] a.Find [imath]Z(G)[/imath]. b.Find a familiar group that is isomorphic to [imath]G/〈y^2 〉[/imath] and show that these are isomorphic. I am having trouble with part 2b. So far, I have Assume [imath]|G|=16[/imath] then [imath]|G|/<y^2> = 8[/imath] => [imath]|G|/<y^2> = |D_4|[/imath] I'm not sure which direction to head in after this. Any feedback is appreciated. | 1339383 | Question on Groups [imath]G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle[/imath]
Studying for an exam, a review question... Given [imath]G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle[/imath]. Show [imath]|G|\leq16[/imath]. For this, I want to consider that [imath]x^3=x^{-1}[/imath] and [imath]y^3=y^{-1}[/imath] based on our assumptions. I am a little lost as to how to put the second part, [imath]xyxy^{-1}=e[/imath] to show there are no more than 16 elements. If [imath]|G|=16[/imath], find the center of the group and find a group that is isomorphic to [imath]G/\langle y^2\rangle[/imath]. I'm pretty sure the group is centerless. |
2405352 | Trigonometry pre calculus level good question
Here is a trigonometry problem. Given [imath]\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m[/imath] Show that [imath]m^2+m\cos(\alpha) = 2.[/imath] I tried to convert [imath]\sin^3(x)[/imath] into [imath]\sin(3x)[/imath] and similarly to cosine term, but couldn't get the answer. Please tell how to proceed further and any other way to solve it. | 2405291 | Trigonometry pre calculus level question
Here is a trigonometry problem. Given [imath]\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m[/imath] Show that [imath]m^2+m\cos(\theta) = 2.[/imath] I tried to convert [imath]\sin^3(x)[/imath] into [imath]\sin(3x)[/imath] and similarly to cosine term, but couldn't get the answer. Please tell how to proceed further and any other way to solve it. |
2406118 | For every analytic function [imath]f[/imath] on [imath]G[/imath] (simply connected,open) s.t [imath]f(z) \neq 0, \forall z \in G[/imath], then [imath]\exists[/imath] g analytic in [imath]G[/imath] s.t [imath]g^2 = f[/imath]
I have to show that for every analytic function [imath]f[/imath] on [imath]G \subseteq \mathbb{C}[/imath] (Simply connected open set) such that [imath]f(z) \neq 0, \forall z \in G[/imath], then [imath]\exists[/imath] a function g analytic in [imath]G[/imath] such that [imath]g^2 = f[/imath]. Also, I need to give an example of an open set [imath]G[/imath] which is not simply connected and an analytic function [imath]f[/imath] on [imath]G[/imath] with [imath]f(z) \neq 0, \forall z \in G[/imath] such that [imath]f \neq g^2[/imath] for every analytic function [imath]g \in G[/imath]. I do not see how to approach this problem. If I get the first part, maybe the example will be easier. | 2384369 | Does there exist a holomorphic function [imath]\varphi(z)[/imath] such that [imath]\varphi^2(z) = \psi(z)[/imath]
[imath]\psi(z) \neq 0[/imath] is a holomorphic function on [imath]D \subseteq \mathbb{C}[/imath]. Does there exist a holomorphic function [imath]\varphi(z)[/imath] on [imath]D[/imath] such that [imath]\varphi^2(z) = \psi(z)[/imath]? Here [imath]D[/imath] can be a single-connected region or a multi-connected region. I think for single-connected region, it is true, but I can not prove. |
2405623 | Null-homotopic closed curve in a Jordan domain
If every Jordan curve [imath]J \subset D[/imath] is null-homotopic in a domain [imath]D \subset \mathbb{R}^2[/imath], then is it clear that every closed curve [imath]C \subset D[/imath] is also null-homotopic in [imath]D[/imath]? Apparently the set [imath]\mathbb{R}^2 \setminus C[/imath] is an union of countably many Jordan domains [imath]D_i[/imath], but it's not clear to me that the result for every [imath]\partial D_i[/imath] is enough. What is the right approach? EDIT: Before you mofos shut down my question with silly down votes I ask to check out the following link http://faculty.up.edu/wootton/Complex/Chapter8.pdf There this question is answered with seemingly simple compactness argument. It would be nice if someone of you can explain why those argument are enough? | 1465907 | Existence of a simple closed curve which is not null-homotopic
Problem. Assume that [imath]U[/imath] is an open and connected subset of [imath]\mathbb R^2[/imath], and [imath]\gamma :[0,1]\to U[/imath] is a closed curve, which is not null-homotopic in [imath]U[/imath] and not necessarily simple closed. Show that there exists a simple closed curve, which is not null-homotopic in [imath]U[/imath]. I am sure that this is a standard problem, and I am looking for an elegant, if possible, solution or a reference. I have in mind the following sketch of proof: a. Construct a polygonal closed curve [imath]\tilde\gamma[/imath] which is homotopic with [imath]\gamma[/imath] in [imath]U[/imath]. b. Construct [imath]\tilde\gamma[/imath] so that it intersects itself in finitely many points, and not whole segments. c. Use induction on the number of self–intersections of [imath]\tilde\gamma[/imath], in order to show that [imath]\tilde\gamma[/imath] is written as a product (in the fundamental group's sense) of simple closed polygonal curves. |
2405500 | Prove that if square root of [imath]{m\over n}[/imath] is rational, then [imath]m[/imath] and [imath]n[/imath] are both perfect squares
Let [imath]m[/imath] and [imath]n[/imath] be positive integers with no common factor. I have to prove that if square root of [imath]{m\over n}[/imath] is rational, then [imath]m[/imath] and [imath]n[/imath] are both perfect squares, that is to say there exist integers [imath]p[/imath] and [imath]q[/imath] such that [imath]m = p^2[/imath] and [imath]n = q^2[/imath]. Can anyone help me with this?? | 1721457 | Square Root of Rational Number [imath]\frac{A}{B}[/imath]
Here's the question: Let [imath]x=\frac{A}{B}[/imath] be a positive rational number in lowers terms (i.e., [imath]A, B\in\mathbb{N}[/imath] and [imath]hcf(A,B)=1[/imath]). Prove that [imath]\sqrt{x}[/imath] is rational if and only if [imath]A[/imath] and [imath]B[/imath] are both perfect squares. (Remember that your proof should work for cases like [imath]A=40[/imath], where [imath]A[/imath] is not a perfect square but has a factor that is a perfect square.) I know prime factorization is involved but that's basically it. Proving that [imath]\sqrt{y}[/imath] (for some number y) is rational if and only if [imath]y[/imath] is a perfect square I can do. Applying that that to the fraction mentioned above? Not so much. Any and all help would be appreciated. Thank you. |
2406340 | Intersection of two hyperbola tangent
Tangents are drawn from the point ([imath]\alpha[/imath],[imath]\beta[/imath]) to the hyperbola [imath]3x^2 - 2y^2 = 6 [/imath] and are inclined at angles A and B to the x -axis. If [imath]\tan A. \tan B = 2[/imath], prove that [imath]\beta^2[/imath]=2[imath]\alpha^2[/imath] - 7. I tried the following concept,The points on Hyperbola from where tangent are drawn are [imath](\sqrt2*(\sec A),\sqrt3*(\tan A))[/imath] & [imath](\sqrt2*(\sec B),\sqrt3*(\tan B))[/imath] The two tangents are [imath]\frac{\sec A}{\sqrt2}[/imath]-[imath]\frac{\tan A}{\sqrt3}[/imath]=1 and [imath]\frac{\sec B}{\sqrt2}[/imath]-[imath]\frac{\tan B}{\sqrt3}=1[/imath], I tried entering ([imath]\alpha[/imath],[imath]\beta[/imath]) and using condition [imath]\tan A. \tan B = 2[/imath] but not able to get the result [imath]\beta^2[/imath]=2[imath]\alpha^2[/imath] - 7. | 1605641 | Tangents are drawn from the point [imath](\alpha,\beta)[/imath] to the hyperbola [imath]3x^2-2y^2=6[/imath] and are inclined at angles [imath]\theta[/imath] and [imath]\phi[/imath] to the [imath]x-[/imath]axis.
Tangents are drawn from the point [imath](\alpha,\beta)[/imath] to the hyperbola [imath]3x^2-2y^2=6[/imath] and are inclined at angles [imath]\theta[/imath] and [imath]\phi[/imath] to the [imath]x-[/imath]axis.If [imath]\tan\theta.\tan\phi=2,[/imath] prove that [imath]\beta^2=2\alpha^2-7[/imath] The equation of the pair of tangents from the point [imath](\alpha,\beta)[/imath] to the hyperbola [imath]3x^2-2y^2=6[/imath] is given by [imath](3x\alpha-2y\beta-6)^2=(3x^2-2y^2-6)(3\alpha^2-2\beta^2-6)[/imath] When i simplify this expression,i get [imath](18+6\beta^2)x^2+(6\alpha^2-12)y^2-12xy\alpha\beta+24y\beta-36x\alpha+18\alpha^2-12\beta^2=0[/imath] This is an equation of pair of lines and the angle between pair of lines [imath]ax^2+2hxy+by^2=0[/imath] is given by the formula [imath]\frac{2\sqrt{h^2-ab}}{a+b}[/imath] So [imath]\tan(\theta-\phi)=\frac{2\sqrt{36\alpha^2\beta^2-(18+6\beta^2)(6\alpha^2-12)}}{18+6\beta^2+6\alpha^2-12}[/imath] [imath]\frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}[/imath] [imath]\frac{\tan\theta-\tan\phi}{3}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}[/imath] I am stuck here.I do not know,how to prove it further. |
2406949 | limit of Recurrence relation
Let [imath]c>0[/imath]. Let a sequence [imath](a_n)[/imath]: [imath]a_1=1[/imath], [imath]a_{n+1}=\frac{1}{2}(a_{n}+\frac{c}{a_{n}})[/imath]. Calculate [imath]\lim_{n \mapsto\infty }a_n[/imath] So, my solution is: I can show in induction that [imath]a_n[/imath] is determinated for all [imath]n[/imath]. I showed in induction that [imath](a_n)[/imath] is trupped between [imath]1[/imath] and [imath]c[/imath], (by seperating the cases of [imath]c>1[/imath] and [imath]c<1[/imath]). I guess I need to separate for the 2 above cases and show that the sequence is monotonous and then I can conclude that the sequence converges, and show that the limit is [imath]\sqrt c[/imath]. But I cannot know for sure that there's exist a limit to the sequence. | 2033109 | Prove that the sequence [imath]a_{n+1} =\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)[/imath] is convergent and find its limit
Let [imath]c>0[/imath], [imath]a_{1} = 1[/imath], and [imath]a_{n+1} =\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)[/imath] I need to: Show that [imath]a_{n}[/imath] is defined for every [imath]n\geq 1[/imath] Show that this sequence is convergent. Find its limit. I proved the first part by showing by induction that this sequence is positive for every [imath]n[/imath]. To show that this sequence is convergent I'm thinking of showing that this sequence is a Cauchy series, yet can't figure out how. For the third part I'm clueless at the moment. |
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