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221211 | Maximal order of an element in a symmetric group
If we let [imath]S_n[/imath] denote the symmetric group on [imath]n[/imath] letters, then any element in [imath]S_n[/imath] can be written as the product of disjoint cycles, and for [imath]k[/imath] disjoint cycles, [imath]\sigma_1,\sigma_2,\ldots,\sigma_k[/imath], we have that [imath]|\sigma_1\sigma_2\ldots\sigma_k|=\operatorname{lcm}(\sigma_1,\sigma_2,\ldots,\sigma_k)[/imath]. So to find the maximum order of an element in [imath]S_n[/imath], we need to maximize [imath]\operatorname{lcm}(\sigma_1,\sigma_2,\ldots,\sigma_k)[/imath] given that [imath]\sum_{i=1}^k{|\sigma_i|}=n[/imath]. So my question: How can we determine [imath]|\sigma_1|,|\sigma_2|,\ldots,|\sigma_k|[/imath] such that [imath]\sum_{i=1}^k{|\sigma_i|}=n[/imath] and [imath]\operatorname{lcm}(\sigma_1,\sigma_2,\ldots,\sigma_k)[/imath] is at a maximum? Example For [imath]S_{10}[/imath] we have that the maximal order of an element consists of 3 cycles of length 2,3, and 5 (or so I think) resulting in an element order of [imath]\operatorname{lcm}(2,3,5)=30[/imath]. I'm certain that the all of the magnitudes will have to be relatively prime to achieve the greatest lcm, but other than this, I don't know how to proceed. Any thoughts or references? Thanks so much. | 971173 | What is maximal possible order of an element in [imath]S_{10}[/imath] ? Why?
What is maximal possible order of an element in [imath]S_{10}[/imath]? Why? Give an example of such an element. How many elements will there be in [imath]S_{10}[/imath] of that order? |
2617337 | Finding value of [imath]\sin(2^\circ)\cdot \sin(4^\circ)\cdot\cdot \cdot \sin(88^\circ)[/imath]
Finding value of [imath]\sin(2^\circ)\cdot \sin(4^\circ)\cdot\cdot \cdot \sin(88^\circ)[/imath] Try: Assuming [imath]X=\sin(2^\circ)\cdot \sin(4^\circ)\cdot\cdot\cdot \sin(88^\circ)[/imath] and [imath]Y=\cos(2^\circ)\cdot \cos(4^\circ)\cdot\cdot\cdot \cos(88^\circ)[/imath] So [imath]2^{22}XY=\sin(4^\circ)\cdot \sin(8\circ)\cdot\cdot \sin(176^\circ)[/imath] Could some help me to solve it, thanks in advance | 812083 | Value of [imath]\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ) [/imath]
How can I calculate the value of [imath]\sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)[/imath] [imath]\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)[/imath] My solution: Let [imath]A = \sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)[/imath] [imath]B = \sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)[/imath] Now [imath] \begin{eqnarray} A\cdot B &=& (\sin 1^\circ \cdot \sin 89^\circ)\cdot (\sin2^\circ\cdot \sin 88^\circ)\cdots(\sin 44^\circ\cdot \sin 46^\circ) \cdot \sin 45^\circ \\ &=& (\sin 1^\circ\cdot \cos1^\circ)\cdot (\sin2^\circ\cdot \cos 2^\circ)\cdots(\sin 44^\circ\cdot \cos 44^\circ)\cdot \sin 45^\circ \\ &=& \frac{1}{2^{44}}\left[\sin(2^\circ)\cdot \sin (4^\circ)\cdots \sin(88^\circ) \cdot \sin (90^\circ)\right]\cdot \frac{1}{\sqrt{2}} \\ &=& \frac{1}{2^{\frac{89}{2}}}\cdot B \end{eqnarray} [/imath] Which implies that [imath]B\cdot \left(A-2^{-\frac{89}{2}}\right) = 0[/imath] So [imath]A = 2^{-\frac{89}{2}}[/imath] because [imath]B\neq 0[/imath]. But I do not understand how can I calculate the value of [imath]B[/imath]. Can we calculate these values using complex numbers? |
2617389 | radius of convergence of number of divisors
i am stuck on this question for a while and i'm not sure about it i need to calculate the radius of convergence for: [imath]\sum _{n=1}^\infty d(n)x^n[/imath] what i did: since i cannot use the root test or check the next eleent against its previous(ratio test), i think that to find the radius of convergence i need to check for the upper limit(sup). so what i did is this: if n=1 we get that its one, so while going to a very large n, i think that [imath]d(n) < n[/imath]. if this is correct, then tha radius should be: [imath]\frac{1}{R}=\limsup_{n \to \infty}d(n) ^ {\frac{1}{n}}[/imath]. however, i don't know how to calculate it. am i correct? how can i calculate the radius of convergence here? | 7111 | Radius of Convergence of a given Power series
I came across this question. Let [imath]d(n)[/imath] denote the number of divisors of [imath]n[/imath]. Let [imath]\nu(z) = \sum\limits_{n=1}^{\infty} d(n) z^{n}[/imath] Whats the radius of convergence of this power series. We also have to show that [imath]\nu(z) = \sum\limits_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}[/imath] Regarding the divisor function, we have Dirichlet's formula in hand. But will that help! |
2071459 | Prob. 6(d), Chap. 3 in Baby Rudin: Given [imath]z \in \mathbb{C}[/imath], does the series [imath]\sum \frac{1}{1+z^n}[/imath] converge or diverge?
Here's Prob. 6(d), Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: Investigate the behavior (convergence or divergence) of [imath]\sum a_n[/imath] if [imath]a_n = \frac{1}{1+z^n}[/imath] for complex values of [imath]z[/imath]. My effort: If [imath]\left\vert z \right\vert > 1[/imath], then we note that [imath] \left\vert a_n \right\vert = \frac{1}{\left\vert 1 + z^n \right\vert} \leq \frac{1}{\left\vert z \right\vert^n - 1} < \frac{1}{\left\vert z\right\vert^n},[/imath] and the series [imath]\sum_{n=0}^\infty \frac{1}{\left\vert z\right\vert^n} = \frac{1}{1- \frac{1}{\left\vert z\right\vert} } = \frac{ \left\vert z \right\vert }{ \left\vert z \right\vert - 1 } < +\infty,[/imath] which implies that our series converges if [imath]\left\vert z \right\vert > 1[/imath]. Am I right? If [imath]\vert z \vert < 1[/imath], then [imath]\lim_{n \to \infty} \vert z \vert^n = 0[/imath] and so [imath]\left\vert a_n \right\vert = \frac{1}{\left\vert 1 + z^n \right\vert} \geq \frac{1}{1 + \left\vert z \right\vert^n} \to 1 \ \mbox{ as } \ n \to \infty,[/imath] so [imath]\lim_{n\to\infty} a_n \neq 0,[/imath] which implies that our series diverges. Am I right? If [imath]\vert z \vert = 1[/imath], then we note that [imath]\left\vert a_n \right\vert = \frac{\vert z \vert^{-n}}{ \left\vert z^{-n} + 1 \right\vert} = \frac{1}{ \left\vert z^{-n} + 1 \right\vert } \geq \frac{1}{\vert z \vert^{-n} + 1 } = \frac{1}{2},[/imath] which implies that [imath]\lim_{n\to\infty} a_n \neq 0,[/imath] showing that our series diverges. Am I right? | 1257017 | Proving the convergence of [imath]\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}[/imath] for [imath]|z| > 1[/imath]
[imath]\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}[/imath], [imath]|z|>1[/imath]. There are two facts that my professor uses that I am confused about. The first is: [imath]|1+z^n| \geq ||z|^n-1|[/imath], I believe this is true for any [imath]|z|[/imath]. The other is: [imath]\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}[/imath], I believe this is also true for any [imath]|z|[/imath]. Can anyone prove these statements for me? |
2616863 | (Problem Solving) Proving [imath]\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{k+m+1}=\sum_{k=0}^{m}(-1)^k\binom{m}{k}\frac{1}{k+n+1}[/imath]
I have been given a problem as part of a practice set for a problem solving class. Here, I am asked to prove the following equality (for positive integers [imath]m[/imath] and [imath]n[/imath]): [imath]\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{k+m+1}=\sum_{k=0}^{m}(-1)^k\binom{m}{k}\frac{1}{k+n+1}[/imath] I have tried to form a counting argument,though that doesn't seem to be working too well. The theme of these problems is that they can be re-written as simpler problems, although I am just not sure how to do that here. Some other things I tried include rewriting each sum as the difference of two other sums (ie putting the positive elements into one sum and the negatives into another) but that doesn't seem to simplify it at all... Any help is much appreciated | 2610545 | Finding value of [imath]S-T[/imath] in [imath]2[/imath] binomial sum.
If [imath]\displaystyle S=\sum^{n}_{k=0}(-1)^k\frac{1}{k+m+1}\binom{n}{k}[/imath] and f [imath]\displaystyle T=\sum^{m}_{k=0}(-1)^k\frac{1}{k+n+1}\binom{m}{k}[/imath].Then [imath]S-T[/imath] is Try:[imath]S=\sum^{n}_{k=0}(-1)^k\int^{1}_{0}x^{k+n}\binom{n}{k}dx=\int^{1}_{0}x^n\sum^{n}_{k=0}(-x)^k\binom{n}{k}dx[/imath] Could some help me, thanks |
2618318 | Deriving formula for partial sum of power series
I have the sum [imath]\sum_{i=0}^n i\cdot a^i[/imath] and according to wolframalpha, the partial sum evaluates to [imath] \frac{a (n\ a^{n + 1} - (n + 1) a^n + 1)}{(a - 1)^2}.[/imath] How does one arrive at the above formula? The usual trick of expanding the sum and subtracting partial sums to arrive at a closed form doesn't seem to work here. | 2181640 | Partial sums of [imath]nx^n[/imath]
WolframAlpha claims: [imath]\sum_{n=0}^m n x^n = \frac{(m x - m - 1) x^{m + 1} + x}{(1 - x)^2} \tag{1}[/imath] I know that one can differentiate the geometric series to compute [imath](1)[/imath] when it is a series, i.e. [imath]m=\infty[/imath]. However, I'm wondering how the closed form for the partial sum is obtained. Actually, WolframAlpha gives an explicit formula for [imath]\sum_{n=0}^m n(n-1)(n-2)\cdots (n-k)x^n \tag{2}[/imath] where [imath]k[/imath] is an integer between [imath]0[/imath] and [imath]n-1[/imath], so it seems that there are some differentiation involved. But already for [imath]k=5[/imath], the formula becomes very messy. My question: How to prove [imath](1)[/imath] and how to build a similar formula for [imath](2)[/imath] assuming [imath]k[/imath] is given? |
2604430 | weak limit similiar to central limit theorem
I need to calclulate [imath]\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n \frac{S_i }{\sqrt{n}}[/imath] where [imath] S_i=\sum_{i=1}^n X_i[/imath] for i.i.d [imath]X_1,X_2,...[/imath] with Expectation 0 and variance 1. I know from the CLT that [imath]\sum_{i=1}^n \frac{X_i }{\sqrt{n}}[/imath] converges to the Standard-Normal distribution. Note that I need to calculate the weak limit. What does that even mean? Solved by JGWang | 2601463 | Calculation problem with Central limit theorem
Let [imath]X_1,X_2,\dots\,[/imath] i.i.d random variables with mean zero and variance [imath]1[/imath]. Let [imath]S_n=\sum_{i=1}^n X_i\,,n\in \mathbb N.[/imath] Compute the weak limes [imath]\lim_{n\to\infty} \frac1n \sum_{i=1}^n \frac{S_i}{\sqrt n}[/imath] Surely we will have to use the CLT. First I tried to simplify the expression, but I am not sure how to continue here. [imath]\lim_{n\to\infty} \frac1n \sum_{i=1}^n \frac{S_i}{\sqrt n}=\dots=\lim_{ n\to\infty}\frac{1}{\sqrt n} \frac{nX_1+(n-1)X_2+\dots+X_n}{n}[/imath] Edit(2) According to the comments, we have to verify Lindberg's condition (https://en.wikipedia.org/wiki/Lindeberg%27s_condition) Lindberg's condition: [imath]\lim_{n\to\infty} \frac{1}{s_n^2} \sum_{k=1}^n E[(X_k - \mu_k)^2 \mathbb 1_{\{\mid X_k - \mu_k \mid > \epsilon s_n \}}=0,\quad \text{for all $\epsilon >0$}[/imath] Here: [imath]E(S_i) {\overset{\text{[/imath]X_i[imath] i.i.d}}{=}}0[/imath] , [imath]Var(S_i) {\overset{\text{[/imath]X_i[imath] i.i.d}}{=}} \sum Var( X_i) {\overset{\text{[/imath]X_i[imath] i.i.d}}{=}} i[/imath] for all [imath]i=1,2,\dots[/imath] Furthermore [imath]s_n^2= \sum_{i=1}^n \sigma_i^2 =Var(S_1)+Var(S_2)+\dots + Var(S_n)=1+2+\dots +n=\frac{n(n+1)}{2}[/imath]. Plugging in: [imath]\lim_{n\to\infty}\frac{2}{n^2+n}\sum_{k=1}^n E(S_k)^2 1_{\{\mid S_k \mid > \epsilon {\frac{\sqrt {n^2+n}}{\sqrt 2}}\}}[/imath] Intuitively this does not seem correct to me. Furthermore I am not sure how to simplify this expression. Some help is welcome and obviously needed! |
2618990 | Evaluation of integral involving trigonometric functions
Evaluate [imath]\int_0^{\pi}\frac{x\sin x}{1 + \cos^2x}\,dx[/imath] I've tried many substitutions like [imath]u = x\sin x [/imath], [imath]u = \cos x[/imath] and [imath]u = 1+\cos^2 x[/imath] but they didn't work. | 1394173 | calculate [imath]\int_o^\pi \frac{x\sin x}{1+\cos^2x} \,dx[/imath]
Given [imath]I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx[/imath] prove without calculating [imath] \frac{\pi}{2} \le I \le \pi [/imath] calulate [imath]I[/imath] so far I know [imath]I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx \le \int\limits_0^\pi x\sin x =\pi[/imath] |
2617311 | Number theory: find [imath]a, b[/imath] such that [imath]\frac{a}{b} = b.a[/imath] in a general base [imath]\mathcal{B}\neq 10[/imath]
I was playing with numbers and thinking about this "coincidence" [imath]\frac{5}{2} = 2.5[/imath] That is, for positive [imath]a[/imath] and [imath]b[/imath] we have [imath]\frac{a}{b} = b.a[/imath] And those questions came into my mind: 1. Could we find all such integers pair [imath]a, b[/imath]? (clearly in base [imath]10[/imath]) And due to the fact that we work in base [imath]10[/imath], a more general problem popped up, that is, to write our numbers in a base [imath]\mathcal{B}\neq 10[/imath] and thence look for triplets [imath](\mathcal{B}, a, b)[/imath] such that [imath]\frac{a}{b} = b.a[/imath] When [imath]a, b[/imath] are written in base [imath]\mathcal{B}[/imath]. 2. Could we find a general formula that will produce infinitely many such integer triples? I am not really into number theory, except for few little questions, so this is more a sort of "I am asking to you experts in the field" question. If for some reason this problem is unclear or wrong or impossible, just tell me! Thank you! | 623148 | Solve [imath]\frac{b}{a}=a.b[/imath] in decimal
Are there any equations of the form [imath]\dfrac{b}{a}=a.b[/imath] other than [imath]\dfrac{5}{2}=2.5[/imath]? Denote [imath]n=\lfloor\log_{10}b\rfloor+1,[/imath] then [imath]\frac{b}{a}=a.b=a+\frac{b}{10^n},\\ b=a^2+\frac{ab}{10^n},a^2<b<a^2+a<(a+1)^2, a=\lfloor\sqrt{b}\rfloor,10^n\mid ab.[/imath] I searched [imath]b<10^5[/imath] such that [imath]10^n\mid ab[/imath] and got a table: [imath]\begin{array}{|c|c|c|}\hline b & a & d(b)=a^2+a b/10^n-b \\\hline 5 & 2 & 0 \\ 75 & 8 & -5 \\ 100 & 10 & 1 \\ 400 & 20 & 8 \\ 500 & 22 & -5 \\ 640 & 25 & 1 \\ 900 & 30 & 27 \\ 2600 & 50 & -87 \\ 5000 & 70 & -65 \\ 6500 & 80 & -48 \\ 9375 & 96 & -69 \\ 10000 & 100 & 10 \\ 25625 & 160 & 16 \\ 31250 & 176 & -219 \\ 40000 & 200 & 80 \\ 62800 & 250 & -143 \\ 65625 & 256 & 79 \\ 76000 & 275 & -166 \\ 90000 & 300 & 270 \\\hline \end{array}[/imath] [imath]\frac{b}{a}=a.b[/imath] iff [imath](b)=0[/imath]. For example, [imath]d(640)=1\neq0[/imath] and [imath]\frac{640}{25}=25.6[/imath]. Also we can search the solutions start with [imath]a,[/imath] then [imath]n=\lfloor\log_{10}a \rfloor+1[/imath] or [imath]n=\lfloor\log_{10}a\rfloor+2[/imath], [imath]b=\frac{10^n a}{10^a-n}.[/imath] I think someone has conducted research and gives a complete solution to this problem, because the problem is so interesting and natural, so I hope you can give me a reference or write an answer here, thanks in advance! |
1118030 | Polynomial equation [imath]f(x)f(2x^2)=f(2x^3+x)[/imath]
Find all polynomials [imath]f(x)[/imath], for which [imath]f(x)f(2x^2)=f(2x^3+x)[/imath]. I have no idea how to do it. | 2600903 | Solve [imath]f(x)f(2x^2) = f(2x^3+x)[/imath]
The full question is to solve [imath]f(x)f(2x^2) = f(2x^3+x)[/imath], prove there are at most one solution per degree of [imath]f(x)[/imath] and then find all the solutions. (Here [imath]f(x)[/imath] is assumed to be polynomial.) So far I have proven that [imath](x^2+1)^n[/imath] works for even degree polynomials, and I'm pretty near certain there are no odd solutions. Is the following a valid argument for there only being at most 1 solution per degree? Let [imath]K[/imath] be the degree of [imath]f(x)[/imath] then [imath]f(x)f(2x^2) = f(2x^3+x)[/imath] produces [imath]3k+1[/imath] linearly independent equations with only [imath]k+1[/imath] independent variables. Therefore for any degree [imath]K[/imath] polynomial there can be at most 1 solution. Finally, is there any elegant way to prove that no odd solutions exist? |
2619082 | Inequality concept problem
Hello I am a student and had the following problems in inequality: [imath]-2\le 1 [/imath] Squaring both sides the sign changes [imath]4\gt 1[/imath] But for [imath]-1\lt 1[/imath] Squaring the terms we get [imath]1\le 1[/imath] I am stuck here please help me to understand this concept | 2591472 | Squaring sides of an inequality
I want to know what will happen to the inequality in these cases [imath]a>x>b[/imath] If [imath]a[/imath] and [imath]b[/imath] negative When i square the both sides of it Is it going to be [imath]b^2>x^2>a^2[/imath] Ok but what would happen if [imath]a[/imath] is a positive and [imath]b[/imath] is a negative number when i square ? EXAMPLE [imath]2>x>-1[/imath] another case [imath]2>x>-4[/imath] i forgot mentioning this case ? If it is something like [imath]2>x>-2[/imath] |
2619049 | Proof of an identity in probability theory
I want to prove the following : [imath]E\lvert X \rvert^{\alpha}=\int_{0}^{\infty} P\left\{\lvert X \rvert^{\alpha}>x\right\} dx=\alpha \int_{0}^{\infty} x^{\alpha-1} P\left\{\lvert X \rvert>x\right\} dx[/imath] The first equality is easy to prove. I'm having trouble with the second one. It appears to be some differentiation technique has to be used but I don't have a clue on how to proceed. Any help would be much appreciated. | 2005549 | Generalization to higher moments of the survival function method for computing the first moment.
If [imath]X[/imath] is a nonnegative random variable, we may compute the first moment by integrating the survival function: [imath] E(X) = \int_0^\infty S_X(x) \, dx [/imath] Is there a reasonable generalization of this for computing higher moments? |
2619021 | Why is [imath]\underbrace {i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}[/imath] converging?
Introduction: Recently I found out that [imath]i^i \approx 0.20788[/imath] has no imaginary part. I got interested and then wanted to know whether there are other [imath]n[/imath] for which [imath]\underbrace {i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}[/imath] has no inaginary part. So I wrote this python script (I'm a beginner at python, could be very bad code :) which plots what I call the [imath]i-Tower\ up\ to\ n = 100[/imath]. It looks like this: Question: Let's use this convention: [imath]{}^ni = \underbrace{i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}[/imath]. Why is the sequence [imath]{}^ni\ |\ n \in \mathbb{N}[/imath] converging? [imath]{}^{20}i \approx 0.48770 + 0.41217i[/imath] [imath]{}^{60}i \approx 0.437584 + 0.360535i[/imath] [imath]{}^{100}i \approx 0.43829+ 0.36059i[/imath] What I already noticed is that the angle between the lines you can draw from [imath]{}^ni[/imath] over [imath]{}^{n+1}i[/imath] to [imath]{}^{n+2}i[/imath] is [imath]<90°[/imath]. Is that angle equal for all [imath]n[/imath]? Is there a way to give the exact value of [imath]x = {}^ni \ [/imath]with[imath]\ {n \rightarrow \infty}[/imath]? What is the connection between [imath]i[/imath] and [imath]x[/imath]. Is [imath]x[/imath] a special complex number that is maybe already known or comes up in other places? | 1757474 | Convergence properties of [imath]z^{z^{z^{...}}}[/imath] and is it "chaotic"
[imath]\DeclareMathOperator{\Arg}{Arg}[/imath] Let [imath]z \in \mathbb{C}.[/imath] Let [imath]b = W(-\ln z)[/imath] where [imath]W[/imath] is the Lambert W Function. Define the sequence [imath]a_n[/imath] by [imath]a_0 = z[/imath] and [imath]a_{n+1} = {a_0}^{a_n}[/imath] for [imath]n \geq 1[/imath], that is to say [imath]a_n[/imath] is the sequence [imath]z, z^z, z^{z^z} ...[/imath] I am trying to classify the sequence [imath]a_n[/imath] as convergent or divergent (i.e. not convergent) according to [imath]|b|[/imath]. My approach to this involves defining [imath]2[/imath] additional sequences: [imath]b_n = \ln a_n[/imath] and [imath]c_n = b_n - b[/imath]. Then [imath]b_{n+1} = \ln \left(z^{e^{b_n}}\right) = e^{b_n} \ln z[/imath] and [imath]c_{n+1} = e^{(b + c_n)} \ln z - b= (e^b \ln z) e^{c_n} - b = b e^{c_n} - b = b (e^{c_n} - 1)[/imath]. Suppose [imath]b_n \to b \implies c_n \to 0 \implies e^{c_n} - 1 \sim c_n \implies c_{n+1} \sim bc_n[/imath]. From here there are [imath]3[/imath] cases to consider: If [imath]|b| > 1[/imath] then [imath]c_n[/imath] is increasing without bound, so it clearly cannot converge except if [imath]c_0 = 0[/imath] which would imply [imath]c_n = 0[/imath] for every [imath]n[/imath] which is not the case because [imath]b_0 =\ln z[/imath] so [imath]b_1 = e^{\ln z}\ln z \ne b_0[/imath] so [imath]c_1 \ne c_0[/imath], a contradiction [imath]\implies b_n[/imath] is divergent [imath]\implies a_n[/imath] is divergent. If [imath]|b| < 1[/imath] then [imath]c_n[/imath] is decreasing towards [imath]0 \implies b_n[/imath] is convergent [imath]\implies a_n[/imath] is convergent. If [imath]|b| = 1[/imath] then [imath]c_n \to c \ne 0 \implies b_n[/imath] is divergent [imath]\implies a_n[/imath] is divergent. I would like to give credit to user1952009. The method involving the sequences [imath]b_n[/imath] and [imath]c_n[/imath] is due to them. Edit 1: After reading Yiannis Galidakis's answer several times, I have decided it is mostly satisfactory. Having said that, I feel I must clarify my position on chaotic sequences. When I said the concept of a chaotic sequence was nonsense, that was an exaggeration. What I meant to say was the definition given in the previous post didn't really make sense to me. The one given here is detailed and precise enough that I can understand what it actually means. However, I do not believe the statements "[imath]a_n[/imath] is chaotic" and "[imath]a_n[/imath] does not converge" are incompatible. Setting aside the question of whether [imath]a_n[/imath] is chaotic or not, I feel I have given a plausibility argument (although not a rigorous proof) that [imath]a_n[/imath] is convergent if and only if [imath]c_n \to 0[/imath] which happens if and only if [imath]|b| < 1[/imath]. On a different note, I have been experimenting numerically with [imath]a_n[/imath] for at least [imath]3[/imath] years. I believe (but cannot prove) that whenever [imath]z[/imath] is not purely real and [imath]|b| = |W(-\ln z)| > 1[/imath] the sequence [imath]a_n[/imath] actually has a [imath]k[/imath]-cycle for some [imath]k \in \mathbb{N}[/imath]. It appears that [imath]k = 3[/imath] for all sufficiently large [imath]|z|[/imath] and that [imath]k \to \infty[/imath] as [imath]|b| \to 1[/imath]. Similarly, whenever the sequence "shoots off" too a neighborhood of [imath]\infty[/imath] I have found that [imath]a_n \sim[/imath] {[imath]..., \infty, 0, 1, z, z^z ...[/imath]} I will admit however that I cannot predict when the sequence will "shoot of" to [imath]\infty[/imath]. At this point I have [imath]2[/imath] main questions: is my convergence analysis correct? Are there any values of [imath]z[/imath], not purely real, such that [imath]a_n[/imath] is unbounded, i.e, the sequence "blows up" to [imath]\infty[/imath] and never "comes back down." Apologies for a somewhat lengthy post. Edit 2: In Addendum 2 in Yiannis Galidakis's answer, it is stated that if [imath]c \in \mathbb{C}[/imath] and [imath]|\Re(c)| > \exp(\exp(-1))[/imath] and [imath]\Arg{c}=2\pi\alpha[/imath] with [imath]\alpha\in\mathbb{R}\setminus\mathbb{Q}[/imath] then the iterated exponential will be unbounded. This is not quite correct. The part about irrational rotation is fine, but it is not enough to have [imath]|\Re(c)| > \exp(\exp(-1))[/imath]. Consider [imath]c = 2\exp(e\pi i/14) \approx 1.6393177+1.1457037i[/imath]. Clearly [imath]c[/imath] satisfies both conditions. However [imath]|W(-\ln c)| \approx .890512[/imath]. Hence the iterated exponential is convergent, hence it is bounded. A few other counterexamples are given by [imath]c = k\exp(e\pi i/14)[/imath] for [imath]k \in \mathbb{R}[/imath] and [imath]1.76253 \le k \le 2.34896[/imath] |
2619742 | How to find a dual basis given a basis?
So I am trying to get my head around the concept of a dual basis. I think I have an okay idea of it, but my problem lays at finding a dual basis given a basis [imath]B[/imath]. So I have given that [imath]B = ((2,1),(3,1))[/imath], which is a basis for [imath]V=\mathbb{R}^2[/imath]. Then, apparently, [imath]B^*=(-1x_1+3x_2,x_1-2x_2)[/imath] is an ordered dual basis for [imath]B[/imath]. I know that given a vector [imath]\textbf{x} \in V[/imath] we can get the function [imath]\phi_k(\textbf{x})=([\textbf{x}]_B)_k[/imath] (evaluating it's coordinates in the basis B and taking the coefficient that corresponds to the k-th position). If we then take [imath]B^*=(\phi_1,\phi_2)[/imath], we have an ordered dual basis to [imath]B[/imath]. However, I don't understand how we can compute [imath]\phi_1[/imath] in such an instance, like is done in the example I've given. Could anybody help me out? Thanks, K. Kamal | 268906 | Determine a formula for a dual basis.
Let [imath]\beta= \{ (2,1),(3,1) \} [/imath] be an ordered basis for [imath]\Bbb R^2[/imath]. Suppose that the dual basis of [imath]\beta[/imath] is given by [imath]\beta^*= \{f_1,f_2 \} [/imath] To explicitly determine a formula for [imath]f_1[/imath] we need to consider the equations [imath]1=f_1(2,1)=f_1(2e_1+e_2)=2f_1(e_1)+f_1(e_2)[/imath] [imath]0=f_1(3,1)=f_1(3e_1+e_2)=3f_1(e_1)+f_1(e_2)[/imath] Solving this equations, we obtain [imath]f_1(e_1)=-1[/imath] and [imath]f_1(e_2)=3[/imath], that is [imath]f_1(x,y)=-x+3y[/imath]. My question is why do we need to solve the above equations for 1 and 0 respectively? |
2618164 | Showing that [imath](\partial_{z}f_{n}(z_{n}))^{k} \rightarrow (\partial_{z}f(z_{o}))^{k}. [/imath] via Compact Sets and Cauchy Estimate?
In the text "Function Theory of a Complex Variable" by Robert E. Greene and Steven G. Krantz I'm having difficulty verifying my appoarch to proving to Exercise 41 on (pg.100) much of the details pertaining to the proof of [imath]\text{Proposition}\, \, \, (1.2)[/imath] can be seen in [imath]\text{Lemma (1.2)-(1.3)}.[/imath] Also what makes the question unique is the particular appoarch taken throughout [imath]\text{Lemma (1.2)-(1.3)}.[/imath] [imath](1.0)[/imath] Let [imath]f_{n}[/imath] be continuous on the open set [imath]U[/imath] and let [imath]f_{n} \rightarrow f \in U[/imath] uniformly converage on compact sets then prove that [imath]f_{n}(z_{n}) \rightarrow f(z_{o})[/imath] [imath]\text{Proposition}\, \, \, (1.2)[/imath] If the [imath]f_{n}[/imath] are also holomorphic and if [imath]0 < k \in \mathbb{Z}[/imath], then prove that in [imath](1.3)[/imath] [imath](1.3)[/imath] [imath](\partial_{z}f_{n}(z_{n}))^{k} \rightarrow (\partial_{z}f(z_{o}))^{k}.[/imath] To prove [imath](1)[/imath], one must rely on the Cauchy Inequality's and compactness arguments much of the formal developments can be seen in [imath]\text{Lemma (1.2)-(1.3)}[/imath] [imath]\text{Lemma (1.2)}[/imath] Allow [imath]\Psi[/imath] be a compact subset of [imath]U[/imath] such that there is an [imath]r >0[/imath] such that, for each [imath]z \in \Psi[/imath], the closed disc [imath]\overline{D(z,r)}[/imath] is contained in [imath]U[/imath]. Fix an r. Then the set [imath]\Psi_{r}=\overline{\Bigg(\bigcup_{z \in \Psi}D(z,r) \Bigg)}[/imath] compact subset of [imath]U[/imath]. [imath]\text{Lemma (1.3)}[/imath] [imath]\text{Theorem 1.3} \, \, \text{(The Cauchy estimate)}[/imath]. Let [imath]f: U \rightarrow \mathbb{C}[/imath] be a holomorphic function on an Open set [imath]U[/imath] such that [imath]P \in U[/imath] and assume that the closed disc [imath]\overline D(\zeta,r)[/imath], [imath]r>0[/imath] is contained in [imath]U[/imath].[imath]M=\sup_{z \in \overline D(\zeta,r)}|f(z)|[/imath] Then for [imath]k=1,2,3...[/imath] we have [imath]\bigg| \partial_{z^{k}}f(P) \bigg | \leq \frac{k!}{r^{k}}\sup_{z \in \overline D(\zeta,r)}|f(z)|.[/imath] Applying the Cauchy Inequality's our original proposition in [imath](1)[/imath] now becomes for each [imath]z \in \Psi[/imath] the following bound [imath]\left| \partial_{z^{k}} (f_{n_{1}}(z_{1})-f_{n_{2}}(z_{2})) \right|\leq \frac{k!}{r^{k}}\sup_{|\zeta -z| \leq r} |f_{n_{1}}(\zeta) - f_{n_{2}}(\zeta)| \leq \frac{k!}{r^{k}}\sup_{\zeta \in \Psi_{r}}|f_{n_{1}}(\zeta)-f_{n_{2}}(\zeta)|.[/imath] So for all [imath]z \in \Psi[/imath] we can construct the following estimate [imath]\left| \partial_{z^{k}} (f_{n_{1}}(z_{1})-f_{n_{2}}(z_{2})) \right| \leq \frac{k!}{r^{k}}\sup_{\zeta \in \Psi_{r}}|f_{n_{1}}(\zeta)-f_{n_{2}}(\zeta)|[/imath] The rhs side goes to [imath]0[/imath] as [imath]n_{1},n_{2}\rightarrow \infty[/imath], since [imath]\Psi_{r}[/imath] is compact such that [imath]{f_{n}(z_{n}})[/imath] converges uniformly on [imath]\Psi_{r}[/imath] Thus [imath](\partial / \partial_{z} f_{n}(z_{n}))[/imath] is uniformly Cauchy on [imath]\Psi_{r}[/imath] [imath]\text{Summary}[/imath] In summary construction of the Compact Subset [imath]\Psi_{r}[/imath] is to allow one at least at the global level to use compactness arguments to allow for the uniform convergence of [imath](\partial_{z}f_{n}(z_{n}))[/imath].In summary at an intuitive level for the subset [imath]\Psi_{r}[/imath] to be compact one takes [imath]\overline{\Bigg({\bigcup_{z \in \Psi}D(z,r)} \Bigg)}[/imath] to from [imath]\Psi_{r}[/imath].Which means we have a set that is not "too large" in which we can have functions defined on [imath]\Psi_{r}[/imath] for one to have Cauchy criterion for [imath]((\partial / \partial_{z}f_{n}(z_{n}) \rightarrow \partial / \partial_{z}f_{n}(z_{o}))[/imath] uniformly on [imath]\Psi_{r}[/imath]. | 1975774 | uniform convergence and function [imath]\partial^k f_n(z_n) \rightarrow \partial^k f(z_0)[/imath]
Let [imath]f_n[/imath] be continuous on the open set [imath]U[/imath] and let [imath]f_n \rightarrow f[/imath] uniformly on compact sets. and if [imath]U \supseteq \{z_n\}[/imath] and [imath]z_n \rightarrow z_0 \in U[/imath] and [imath]f_j[/imath] are holomorphic, then what i want to show is for [imath]0 < k \in \mathbb{Z}[/imath], \begin{align} \left(\frac{\partial}{\partial z}\right)^k f_n(z_n) \rightarrow \left(\frac{\partial}{\partial z}\right)^kf(z_0) \end{align} My trial is similar treatment for [imath]\lim_{n\rightarrow \infty} f_n(z_n) = f(z)[/imath]. But i am not sure about considering derivatives. |
2618248 | Higher order derivative of a composition of vector valued multivariate functions
I have this composition of twice differentiable functions : [imath] \Bbb R ^m \overset{f}{\longrightarrow} \Bbb R^n \overset{g}{\longrightarrow} \Bbb R^k[/imath] And I'm trying to get this result from a book : [imath] (g \circ f)''(x) (h,k) = (g''\circ f')(x)\big(f'(x)h,f'(x)k\big) + (g'\circ f)(x)\big(f''(x)(h,k)\big) [/imath] A poor attempt : [imath] \begin{align} (g \circ f)^{\prime\prime}(x) (h,k) & = \bigg(\big((g' \circ f)(x) f'(x)\big)(h)\bigg)'(k) \\ & = \big((g' \circ f)(x)\big)'(k) f'(x)(h) + (g' \circ f)(x)\big(f''(x)(h,k)\big) \\ & = (g'' \circ f)(x)f'(x)(k) f'(x)(h) + (g' \circ f)(x)\big(f''(x)(h,k)\big) \\ & = (g''\circ f \space)(x)\big(f'(x)k,f'(x)h\big) + (g' \circ f)(x)\big(f''(x)(h,k)\big) \\ (?) & = (g''\circ f')(x)\big(f'(x)h,f'(x)k\big) + (g'\circ f)(x)\big(f''(x)(h,k)\big) \end{align}[/imath] So It feels like it follows the rules of bilinear differentiation but then [imath]h[/imath] and [imath]k[/imath] should swap places and there's that extra [imath]f'[/imath]. What am I doing wrong. | 619778 | Second derivative of a composite function
Say, we have three Banach spaces [imath]X, Y, Z[/imath] and [imath]g:X \to Y, \ \ f:Y \to Z[/imath] are twice (Fréchet) differenciable. The question is: what is [imath](f \circ g)''[/imath]? Since [imath](f \circ g)'':X \to \mathcal{L}^2(X,Z)[/imath], I am interested in the most explicit form: [imath](f \circ g)''(a)[k, h][/imath] which would describe the entire action taking place. Let's pick some [imath]a \in X[/imath] and get started: \begin{align} \\ (f \circ g)''(a)[k, h] &= D_{k}D_{h}(f \circ g)(a) \\ &= D_{k} \bigg( D_{h}(f \circ g)(a) \bigg) \\ &= D_{k} \bigg( (f \circ g)'(a)[h] \bigg) \\ &= \bigg( D_{k} (f \circ g)'(a)\bigg) [h] \ \ \ \ \ \text{(by linearity)} \\ &= \bigg( D_{k} f'(g(a)) \circ g'(a)\bigg) [h] \ \ \ \ \ \text{(chain rule)} \\ &= \bigg( \frac{d}{dt} f'(g(a+tk)) \circ g'(a+tk)|_{t=0}\bigg) [h] \end{align} I’m really confused what to do next. It looks like we’d need to apply some sort of the product rule (but there is no product here: only compositions). And whatever I do next, I immediately lose track of how it all fits with [imath]k[/imath] and [imath]h[/imath]. Any explanations (the more detailed the better) are hugely appreciated. |
2620331 | Why quaternions is a group?
Let's consider the set consisting of [imath]8[/imath] elements [imath]\{\pm1, \pm i,\pm j, \pm k\}[/imath] with the following multiplication table: We see that [imath]1[/imath] commutes with any of the [imath]\{i,j,k\}[/imath]. For example, we want to consider the product [imath](-1)\cdot j[/imath], since [imath]i^2=-1[/imath] then we can rewrite it as: [imath](-1)\cdot j=(i\cdot i)\cdot j=i\cdot(i\cdot j)=i\cdot k=-j[/imath]. In this example and in many others in order to multiply elements we need to use associativity property. How to prove that associativity is true in this set? In similar topic I have seen approach using automorphism. However, I was not able to comprehend it. Can anyone explain it please? Would be grateful if somebody can demonstrate some elementary approach. In my opinion it is definitely important to know. | 401506 | Quaternion group associativity
Consider the eight objects [imath]\pm 1, \pm i, \pm j, \pm k[/imath] with multiplication rules: [imath]ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j,i^2=j^2=k^2=-1[/imath], where the minus signs behave as expected and [imath]1[/imath] and [imath]-1[/imath] multiply as expected. Show that these objects form a group containing exactly one involution. Well, it is easy to determine closure from the definition. [imath]1[/imath] is clearly the identity, and the inverses can also be determined [imath](i,-i),(j,-j),(k,-k)[/imath], [imath]-1[/imath] with itself, and [imath]1[/imath] with itself. The only involution is [imath]-1[/imath]. Is there an easy way to check associativity of this group? There are too many possible combinations [imath](ab)c=a(bc)[/imath] to check directly. ([imath]8^3[/imath] possible combinations) |
2615686 | Proof verification: Let [imath]p\geq 2[/imath]. If [imath]2^{p}-1[/imath] is prime, then [imath]p[/imath] must also be prime.
Can someone please verify whether my proof is logically correct? :) Let [imath]p\geq 2[/imath]. If [imath]2^{p}-1[/imath] is prime, then [imath]p[/imath] must also be prime. Proof: Let [imath]p=a\cdot b[/imath] with [imath]a>1[/imath] and [imath]b>1[/imath] being integers. Then [imath]2^{p}-1 =2^{ab}-1=(2^{a}-1)(2^{b(a-1)}+2^{b(a-2)}+\cdots +2^{a}+1) = (2^{b}-1)(2^{a(b-1)}+2^{a(b-2)}+\cdots +2^{b}+1)[/imath] where [imath]2^{a}-1>1[/imath] and [imath]2^{b}-1>1[/imath] since [imath]a>1[/imath] and [imath]b>1[/imath]. If p is composite with [imath]p=a\cdot b[/imath] and [imath]1<a<p[/imath], then [imath]2^{p}-1[/imath] is composite, since [imath]2^{ab}-1[/imath] is divisible by [imath]2^{a}-1[/imath]. [imath]\square[/imath] I don't understand why it is showing if [imath]p[/imath] is composite, then [imath]2^{p}-1[/imath] is composite though. | 920991 | Let [imath]p \geq 2[/imath]. Prove that if [imath]2^p-1[/imath] is prime then [imath]p[/imath] must also be prime.
Would the following be a valid proof? Let [imath]r[/imath] and [imath]s[/imath] be positive integers, then the polynomial [imath]x^{rs}-1=(x^r -1)(x^{s(r-1)}+x^{s(r-2)}+\cdots+x^r+1)[/imath]. So if [imath]p[/imath] is composite (say [imath]rs[/imath] with [imath]1<s<p[/imath]), then [imath]2^p-1[/imath] is also composite (because it is divisible by [imath]2^s-1[/imath]) |
2126611 | [imath]2^5 \times 9^2 =2592[/imath]
[imath]2^5 \times 9^2 =2592[/imath] I am trying to find any other number in this pattern. That is find natural numbers [imath]a[/imath] , [imath]b[/imath] , [imath]c[/imath] and [imath]d[/imath] such that [imath]a^b \times c^d = \overline{abcd} [/imath] We have [imath]a^b \times c^d = \overline{cd} + 100\overline{ab} [/imath] So [imath]a^b \times c^d - \overline{cd} = 100\overline{ab} [/imath] LHS is a multiple of [imath]100[/imath] .Any help from here will be greatly appreciated | 2505284 | A number with an interesting property. [imath]abcd=a^b c^d[/imath]
I am finding a 4-digit number [imath]abcd[/imath] (in base-10 representation) satisfying the following property. [imath]abcd=a^b c^d[/imath] I have been running my mind around this problem since a long time, but got no success. I wrote a python program for a number with this property, and got the answer to be [imath]2592[/imath] since [imath]2592=2^5\times 9^2[/imath] But I am looking for a purely mathematical way to solve this problem. Thanks! |
2621133 | Prove that, [imath]\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}[/imath]
Reading through my textbook I came across the following problem and I am looking for some help solving it. Prove that, [imath]\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}[/imath] In a previous problem I solved the following but I wasn't sure how to use it to prove the question above. Prove, [imath]\binom{n}{r}=\binom{n}{n-r}[/imath] Solution: left Side, [imath]\binom{n}{r}=\frac{n!}{(n-r)!r!}[/imath] right side, [imath]\binom{n}{n-r}=\frac{n!}{(n-(n-r))!(n-r)!} = \frac{n!}{(n-r)!r!}[/imath] Therefore completing the proof, but how do I solve my first question? thanks! | 1125923 | Proving Pascal's identity
So I came across Pascal's identity: Prove that for any fixed [imath]r\geq 1[/imath], and all [imath]n\geq r[/imath], [imath] \binom{n+1}{r}=\binom{n}{r}+\binom{n}{r-1}. [/imath] I know you can use basic algebra or even an inductive proof to prove this identity, but that seems really cumbersome. I was wondering if anyone had a "cleaner" or more elegant way of proving it. For example, I think the following would be a decent combinatorial proof. Proof: Let [imath]S[/imath] be a set with [imath]n+1[/imath] elements, and consider some fixed [imath]x\in S[/imath]. There are [imath]\binom{n+1}{r}\;\; r[/imath]-subsets of [imath]S[/imath]--count them according to whether or not they contain [imath]x[/imath]: there are [imath]\binom{n}{r}[/imath] not containing [imath]x[/imath], (each formed by choosing [imath]r[/imath] of the remaining [imath]n[/imath] elements in [imath]S\setminus\{x\}[/imath]), and there are [imath]\binom{n}{r-1}\;\; r[/imath]-sets containing [imath]x[/imath], (each formed by selecting an additional [imath]r-1[/imath] elements in [imath]S\setminus\{x\}[/imath]). Is that right? Are there any other efficient ways of doing it? |
2621277 | [imath]\int_0^1 \int_0^{1-y} \cos\Big( \frac{x-y}{x+y} \Big) \, dx dy[/imath]
Reviewing old homework sets for a class and I came across this integral: [imath]\displaystyle \int_0^1 \int_0^{1-y} \cos\Big( \frac{x-y}{x+y} \Big) \; dx dy,[/imath] which the question suggests to evaluate using a change of coordinates; however, I haven't a clue where to begin to identify a useful change of coordinates. I tried [imath]u = x-y[/imath] and [imath]v = x+y[/imath], but then wasn't sure how I'd convert the domain of integration. After that, I looked to the given limits for inspiration and noticed that [imath]0<x<1-y[/imath] could be rewritten [imath]y < x+y < 1[/imath], so tried [imath]u = x+y[/imath] and [imath]v = y[/imath], which yielded [imath]\displaystyle \int_0^1 \int_y^{1} \cos\Big( \frac{ u - 2y }{u} \Big) \; du\,dy,[/imath] but that doesn't seem any simpler than the original, to me. Any advice would be appreciated! | 2452462 | How to get the interval after change of variables?
Let [imath]D[/imath] be the region bounded by [imath]x+y = 1[/imath], [imath]x= 0[/imath], [imath]y = 0[/imath]. Use the result of Exercise 19 to show that [imath]\iint_D\cos\bigg(\frac{x-y}{x+y}\bigg)~dx~dy = \frac{\sin 1}{2} [/imath] and graph [imath]D[/imath] on an [imath]xy[/imath] plane and a [imath]uv[/imath] plane, with [imath]u = x-y[/imath] and [imath]v = x+y[/imath]. I got the Jacobian determinant is [imath]\dfrac{1}{2}[/imath], but how do I get the limits of integration for [imath]u[/imath] and [imath]v[/imath]? |
756510 | Convergence of [imath]\sum \frac{\sqrt{a_n}}{n^p}[/imath]
For [imath]a_n \geq 0[/imath], and [imath]\sum a_n[/imath] convergent, show that [imath]\sum \frac{\sqrt{a_n}}{n^p}[/imath] is also convergent for [imath]p > 1/2[/imath]? What bugs me more is why isn't [imath]\sum \sqrt{\frac{a_n}{n}}[/imath] convergent?? Clearly it converges for [imath]p > 1[/imath], and [imath]a_n[/imath] somehow helps out in [imath]p \in (1/2,1)[/imath]. Only hints are welcomed Thanks in advance.. | 767733 | If [imath]\sum_{n=1}^\infty a_n[/imath] converges and [imath]\sum_{n=1}^\infty \frac{\sqrt a_n}{n^p}[/imath] diverges, then p [imath]\in[/imath] {?}
Let {[imath]a_n[/imath]} be a sequence of non-negative real numbers such that the series [imath]\sum_{n=1}^\infty a_n[/imath] is convergent. If p is a real number such that the series [imath]\sum_{n=1}^\infty \frac{\sqrt a_n}{n^p}[/imath] diverges, then what can be said about the value of p? |
2621343 | Example of convex nonlinear equality constraint
I know that affine equalities such as [imath]a_1 x_1 + a_2 x_2 + a_3 x_3 = b_3[/imath] are convex. I'm wondering whether there are nonlinear equality constraints, for instance something like [imath]a_1 x_1 + a_2 x_2 + a_3 x_3 = b_3 + x^{2p},\quad p \in N[/imath] or even something like [imath]x_1 = x_2^2[/imath] Are these kinds of equality constraint also convex? Can you give me any other example showing non-affine convex functions? Thanks! | 2596938 | In convex optimization, must equality constraints be linear or affine?
In convex optimization, the feasible region is convex if equality constraints [imath]h(x)[/imath] are linear or affine, and inequality constraints [imath]g(x) \leq 0[/imath] are convex. Does this mean that if [imath]h(x)[/imath] is nonlinear, then the optimization problem is non-convex? Is there any special case where [imath]h(x)[/imath] is nonlinear and the feasible region is still convex? |
2621473 | Graph theory Orientable Graphs
I need help doing this question, and I believe a 'standard' solution would require proof by induction. Here is the question: Suppose that n ≥ 3 and a simple graph G has n vertices and at least [imath]\frac{n^2−3n+6}{2}[/imath] edges. Prove that G is orientable. | 2610990 | Prove that simple graph [imath]G[/imath] is orientable
I am a student in an introductory class for graph theory and this question is on our first homework assignment and I don't know where to start to prove the following: Suppose that [imath]n \geq 3[/imath] and a simple graph G has [imath]n[/imath] vertices and at least [imath](n^2 - 3n + 6)/2[/imath] edges. Prove that G is orientable. Thank you for your time. Any hints or help will be greatly appreciated. EDIT: A graph g is orientable if it is the underlying graph of a strongly connected digraph. A digraph is strongly connected if for all vertices [imath]v[/imath] and [imath]w[/imath] there exists a walk [imath]v-w[/imath]. |
2622609 | How many 3 digits numbers are equal to the sum of their first digit plus their second digit squared plus the third cubed?
[imath]100a + 10b + c = a + b^2 + c^3[/imath] I've tried restricting the range of the values but I couldn't get a unique solution. I am looking for a formula if there is one. I wonder which is the best way to solve this problem. | 1845102 | [imath]x^3 +y^2 +z =100z+10y+x[/imath] What is the largest and smallest integer that satisfies this equation.
[imath]x^3 + y^2 +z=\overline{zyx}[/imath], where [imath]\overline{zyx}[/imath] denotes the sequence of the digits. [imath]x^3 +y^2 +z =100z+10y+x[/imath],where [imath]x,y,z>0[/imath] The maximum value of [imath]x[/imath], [imath]y[/imath], [imath]z[/imath] individually can only be [imath]9[/imath]. [imath]\text{Maximum value}= 9^3 + 9^2 + 9 = 819[/imath] So, [imath]100z+10y+z<819[/imath] Then I'm lost here a bit. What do I do next? |
2621623 | Calculate infinite summation of sin(x)/x
How do you calculate the infinite sum [imath]\sum_{i=1}^{\infty} \frac{\sin(i)}{i}[/imath]? According to Wolfram Alpha, the value of the sum is [imath]\frac{\pi - 1}{2}[/imath], but it does not tell me the method by which it gets this result. | 829523 | Why does [imath]\sum_{k=1}^{\infty}\frac{{\sin(k)}}{k}={\frac{\pi-1}{2}}[/imath]?
Inspired by this question (and far more straightforward, I am guessing), Mathematica tells us that [imath]\sum_{k=1}^{\infty}\dfrac{{\sin(k)}}{k}[/imath] converges to [imath]\dfrac{\pi-1}{2}[/imath]. Presumably, this can be derived from the similarity of the Leibniz expansion of [imath]\pi[/imath] [imath]4\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{2k-1}[/imath]to the expansion of [imath]\sin(x)[/imath] as [imath]\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)!}x^{2k+1},[/imath] but I can't see how... Could someone please explain how [imath]\dfrac{\pi-1}{2}[/imath] is arrived at? |
1661442 | Show that [imath]f_n=(1+\frac{x}{n})^n[/imath] converge uniformly on all compact of [imath]\mathbb R[/imath]
Let [imath]f_n(x)=\left(1+\frac{x}{n}\right)^n[/imath]. Show that it converges uniformly to [imath]e^x[/imath] on every compact [imath]K\subset \mathbb R[/imath]. This is how I would like to do: Let [imath]K[/imath] a compact. Let [imath]g_n(x)=\left(1+\frac{x}{n}\right)^n-e^x[/imath]. Since [imath]g_n[/imath] is continuous on [imath]K[/imath], it take it's maximum on [imath]K[/imath]. Let denote [imath]a_n\in K[/imath] s.t. [imath]|g_n(x)|\leq g_n(a_n)[/imath] for all [imath]n[/imath]. Set [imath]b_n=g_n(a_n)[/imath]. How can I show that [imath]\lim_{n\to\infty }b_n=0\ \ ?[/imath] I tried to show by induction that [imath](b_n)[/imath] is decreasing, but it looks very complicate. And even if it converge, it looks complicate to show that it converge to [imath]0[/imath] as well. | 1462404 | Show that [imath](1+\frac{x}{n})^n e^{-x} \rightarrow 1[/imath] converges uniformly.
This is my attempt at a proof: [imath]\forall x \in \mathbb{R}, \lim_{n \to \infty}(1+\frac{x}{n})^n=\lim_{m \to 0}\frac{log(1+mx)}{m}=\lim_{m \to 0} \frac{x}{1+mx}=x \iff (1+\frac{x}{n})^n \rightarrow e^x[/imath] uniformly. It follows that [imath](1+\frac{x}{n})^n e^{-x} \rightarrow 1[/imath] uniformly. Is this sufficient? |
2622908 | Where am I making a mistake in showing that countability isn't a thing?
I am not a mathematician so I might not be entirely accurate in my mathematical depictions, but I will correct them if I can. My problem is like that: I have [imath]f_n:\mathbb R ^+\to\mathbb R ^+[/imath] with [imath]f_{n}(x)=x_{m_{n}}[/imath] For the moment I will explain the function using a [imath]n=10[/imath] as [imath]n[/imath] is the basis in which the number is written and "processed" by my function. So given the normal decimal basis I can write the function as: [imath]f(\sum c_i*10^i)=\sum c_i*10^{-1-i}[/imath]. I think the sum goes from [imath]-\infty[/imath] to [imath]\infty[/imath] for any given number as the digits that are not written are zeros. So what my function does (or at least if think it does) is basically a mirroring of digits over the decimal point (ex.: [imath]f(23.45)=54.32[/imath]). I should be easy to prove that the function is bijective as [imath]f\circ f=1_{\mathbb R ^+}[/imath]. Now given the established function my problem comes when I restrict the function to the interval [imath](0;1)[/imath] as i believe that [imath] Im_f(0;1)=\mathbb N ^* [/imath] which seems to indicate that [imath]{\mathbb R}[/imath] could be countable. EDIT: Since this was a commonly raised problem about my question I will add as many forms as I have encountered it in with what I hope are satisfying answers: [imath]\mathbb N[/imath] has "finitely many digits": Note the "finite number of digits" with [imath]k[/imath]. Take [imath]10^{k+1}[/imath]. [imath]\mathbb R^+[/imath] can not have numbers with infinite digits (to the left of the decimal point) then [imath]lim_{x\to\infty}\int log_{10}x\ dx=?[/imath] any other proof of something is finite using induction is also flawed from a simple perspective: [imath]\mathbb N[/imath] is not finite to begin with the argument about the lack of a successor and a predecessor holds no ground. If i can't add or subtract 1 from a decimal number knowing all the digits involved (at least in theory) I would have to go and start school again from the first grade. | 1729362 | Why there are real numbers with infinite digits, but no such natural numbers (or another reason why real numbers are uncountable)
This question is me trying to understand (again) why there can be no one-to-one correspondence between the sets of natural and real numbers. The source of confusion is this: if we abstract completely from the properties and operations on numbers and just think of them as strings of digits, it appears to be possible to list all the real numbers between [imath]0[/imath] and [imath]1[/imath] and all positive integers having the same digits in the same (or reverse) order, for example: [imath]0.1 \to 1[/imath] [imath]0.01 \to 10[/imath] [imath]0.0101 \to 1010[/imath] [imath]0.110101 \to 101011[/imath] I use binary expansion for simplicity. Since irrational numbers have infinite expansions, we would have to consider some kind of 'infinite natural numbers', like: [imath]101110001\dots[/imath] or [imath]\dots101110001[/imath] I think the main (and only) reason why such numbers don't exist is this requirement: We need the sets of real and natural numbers to be ordered. We need to be able to do arithmetics on real and natural numbers. Obviously, if we consider any of two types of 'infinite natural numbers' and the usual absolute value, we find that it's impossible to order them, or do arithmetics with them. The only way is to change the norm (to [imath]p-[/imath]adic norm for example). While with the real numbers it's still possible to order them, and do arithmetics with them when they have infinite amount of digits in their expansion (or at least in principle). Is my reasoning correct? Are the two requirements (ordering and arithmetics) enough to explain why natural numbers with infinite amount of digits do not exist, while real numbers do? This confusion is not restricted to me alone, see this paper for example |
2622995 | Ambiguity in Question!
I was trying to solve questions from "Mathematical Analysis" by "Apostol" and came across this question which states ""If [imath]a, b, c[/imath] and [imath]d[/imath] are rational and if [imath]x[/imath] is irrational, prove that [imath]\dfrac{\left( ax + b \right)}{\left( cx + d \right)}[/imath] is usually irrational."" Now, the problem with this is that how can we prove something is usually True? All we can do is to show some cases where this is not true. Like in this example, we can give the solution as:- The only time it will be rational is when the numerator is some integer times the denominator or vice - versa. The rest of the times, when this is not true, it will be irrational. Is this the correct way of proof or there is some other war to prove something usually true? Note: My question is not about finding the possibilities that the given expression is rational. The question is about how to prove something usually True? Because, as far as I have learnt about logic, we may prove something to be True (completely) or False (by just giving one counter example). So, here, I am asking that if such a statement comes, what does it mean to prove it? In fact, I have solved the question and its solution is written in the explaination. All I want to know is that if my solution is correct or is there a proper way to prove something usually True? | 1882242 | If a.b.c,d are rationals and x is irrational number, then prove that [imath]\frac{ax+b}{cx+d}[/imath] is usually a irrational number. When do exceptions occur?
If a.b.c,d are rationals and x is irrational number, then prove that [imath]\frac{ax+b}{cx+d}[/imath] is usually a irrational number. When do exceptions occur? This is a homework Question. i did this by assuming that given fraction is rational say [imath]\frac{p}{q}[/imath]. Now i write [imath]\frac{ax+b}{cx+d} = \frac{p}{q}[/imath]. After cross multiplying i get [imath]\frac{qb-pd}{pc-qa}=x[/imath] So this contradiction as right hand side is rational and left hand side is irrational. Now the question asks where does exceptions occur. I see that exceptions can occur when [imath]a=c=0[/imath] and [imath]a=b=c=d[/imath]. But what are other exceptions and is my working method correct? Thanks |
2622936 | Proof maximum number of equidistant points for a given dimension
The question is quite straight forward, on [imath]\mathbb{R}^2[/imath] the maximum number of equidistant distinct points that can be layed (using the euclidean distance) is 3, forming thus an equilateral triangle. On [imath]\mathbb{R}^3[/imath] the maximum number of points is 4, forming thus a tetrahedon. My question is quite simple, is the maximum number of points that can be layed in [imath]\mathbb{R}^n[/imath] equal to [imath]n+1[/imath]? | 1302395 | n points can be equidistant from each other only in dimensions [imath]\ge n-1[/imath]?
2 points are from equal distance to each other in dimensions 1,2,3,... 3 points can be equidistant from each other in 2,3,... dimensions 4 points can be equidistant from each other only in dimensions 3,4,... What is the property of number dimensions that relates the number of points that can be equidistant to all other points? |
2623219 | Formula for Curvature
How do I derive the formula for curvature through differentiation? [imath] k(t)= \dfrac{(x'y''-x''y')}{(x'^2+y'^2)^\frac32} [/imath] I know that [imath]k(t)=[/imath] the modulus of [imath] T'(s) [/imath] where [imath]T[/imath] is the unit tangent vector | 275248 | Deriving curvature formula
How do you derive the formula for unsigned curvature of a curve [imath]\gamma (t) = (x(t), y(t)[/imath] which is not necessarily parameterised by arc-length. The formula given is [imath]|\kappa (t)| = \left|\frac{x'' y' - y'' x'}{((x')^2 + (y')^2)^{3/2}}\right|[/imath] All the definitions of curvature in my notes give it in terms for an arc-length parameterisation [imath]T'(s) = \kappa(s) N(s)[/imath]. This is apparently meant to be 'easily derivable', am I missing something? |
2622296 | Cubic Equation With Complex Roots
Please Forgive me for any mistake in the proposal of the problem in advance. Please feel free to edit it. Problem Find the roots of the following cubic equation [imath]x^3-3x^2+3=0[/imath]. My Approach After removing the second degree of [imath]x[/imath] we get [imath]y^3+1-3y=0[/imath] where [imath]y=x-1[/imath]. Then Taking [imath]y=u+v[/imath] and then cubing both sides I got [imath]y^3-(u^3+v^3)-3uvy=0[/imath]. Now after equating the coefficients I got [imath]u^3+v^3=-1 , u^3v^3=1[/imath]. After This I formed the quadratic equation The quadratic equation formed is [imath]t^2+t+1=0[/imath]. The roots are [imath]\dfrac{-1+i\sqrt3}{2}[/imath] and [imath]\dfrac{-1-i\sqrt3}{2}[/imath]. After this I take [imath]r\cos\theta=\dfrac{-1}{2}[/imath] and [imath]r\sin\theta=\dfrac{\sqrt3}{2}[/imath] I got [imath]\theta=60^\circ[/imath]. However I don't know how to approach further.Please tell me if I am on the right track or not. Note:-I use Cardano's Method. Any help is welcome. Note:-This Question is different from the duplicate as this problem states the problem after we form the quadratic equation and try to get the cube roots of a complex number. | 498636 | Roots of [imath] x^3-3x+1[/imath]
How can I find the roots of [imath]x^3-3x+1[/imath] using Cardano's formula? So far, I found that [imath]x = \sqrt[3]{\dfrac{-1+\sqrt{-3}}{2}} + \sqrt[3]{\dfrac{-1-\sqrt{-3}}{2}}[/imath] [imath]x = \sqrt[3]{\dfrac{-1+i\sqrt{3}}{2}} + \sqrt[3]{\dfrac{-1-i\sqrt{3}}{2}}[/imath] [imath]x = \sqrt[3]{\frac{-1}{2}+\frac{i\sqrt{-3}}{2}} + \sqrt[3]{\frac{-1}{2}-\frac{i\sqrt{-3}}{2}}[/imath] I am now trying to express each cubic radicand in their exponential form. Euler's formula : [imath]e^{i\theta} = \cos \theta + i\sin \theta[/imath]. I have [imath]r = |z| = 1[/imath]. However, when I try to find the angle, I get [imath]\theta = \arccos \frac{-1}{2} = 2\pi/3[/imath] but at the same time [imath]\theta = \arcsin \frac{\sqrt{-3}}{2} = \pi/3[/imath] Shouldn't the two angles be the same for one radicand? Plus, once I will have expressed the two radicands in their exponential form, what is the next step? [imath]x = \sqrt[3]{e^{i2\pi/3}} + \sqrt[3]{e^{i4\pi/3}}[/imath] [imath]x = e^{i2\pi/9} + e^{i4\pi/9}[/imath] Am I on the right track? |
2623979 | Prove that the homomorphism [imath]\phi:R\to S^{-1}R[/imath] is injective if and only if [imath]S[/imath] contains no zero-divisors.
Suppose that [imath]S\subset R[/imath] is a multiplicative set in [imath]R[/imath], where [imath]R[/imath] is a commutative ring with identity [imath]1\neq 0[/imath]. Prove that the homomorphism [imath]\phi:R\to S^{-1}R[/imath] is injective if and only if [imath]S[/imath] contains no zero-divisors. I have an idea about how to approach this question, but I am not sure exactly how to structure it. For one direction, where we assume [imath]\phi[/imath] is injective, this means that [imath]\ker(\phi)=\{0\}[/imath], which I think means that there are no zero divisors in [imath]R[/imath], so there are no zero divisors in [imath]S[/imath] because it is a subset of [imath]R[/imath]? Am I headed in the right direction? Edit: I believe this question is different than the one it is marked as duplicate to, because that question uses the concept of monomorphisms which are not mentioned here. | 416999 | Rings of fractions and homomorphisms
Let the ring of fractions be denoted as [imath]S^{-1}R[/imath]. Proposition (from Robert Ash's textbook "Basic Abstract Algebra") Define [imath]f: R \rightarrow S^{-1}R[/imath] by [imath]f(a) = a/1[/imath]. Then f is a ring homomorphism. If S has no zero divisors then f is a monomorphism, and we say that R can be embedded in [imath]S^{-1}R[/imath]. In particular: (i) A commutative ring R cna be embedded in its complete (or *full) ring of fractions ([imath]S^{-1}R[/imath], where S consists of all non-divisors of zero in R). (ii) An integral domain can be embedded in its quotient field. Why does the proposition say "If S has no zero divisors..."? If S really had zero divisors, then this wouldn't even be properly defined, right? Because we would have a/0 for some a in R. |
2623568 | Why can't all pointwise continuous functions preserve Cauchy sequences?
If [imath]f[/imath] is a real-valued uniformly continuous function on [imath]A[/imath], then for every Cauchy sequence [imath](x_n)[/imath] in [imath]A[/imath], [imath](f(x_n))[/imath] is a real Cauchy sequence. But why do we need the uniform continuity of [imath]f[/imath] for this? If [imath]f[/imath] is continuous on [imath]A[/imath] then for any point [imath]c[/imath] in [imath]A[/imath], and any Cauchy sequence [imath](x_n)[/imath] converging to [imath]c[/imath], [imath](f(x_n))[/imath] converges to [imath]f(c)[/imath], thus making it a Cauchy sequence. | 2506933 | If continuity preserves convergence, and Cauchy sequences are convergent sequences, why do we need uniform continuity to preserve Cauchy sequences?
In [imath]\mathbb R[/imath], all Cauchy sequences are convergent and all convergent sequences are Cauchy. So, why isn't continuity enough to preserve Cauchy sequences? A function is continuous iff it preserves convergent sequences. A sequence is convergent iff it is Cauchy. So, why doesn't it follow that continuous functions preserve Cauchy sequences? |
2622829 | Does [imath]\sqrt{i^4}[/imath] imply +/-1 = -/+1?
I know that the minus sign of -r, where r is real, can never be moved under a radical sign. But there are no minus signs present when the sqrt of [imath]i^4[/imath] is taken in the following manner [imath]1 = i^4[/imath] [imath]\sqrt{1} = \sqrt{i^4}[/imath] [imath]+/-1 = +/-i^2[/imath] [imath]+/-1 = -/+1[/imath] I realize it might be argued that i also can never be moved under the radical sign. But [imath]\sqrt{i}[/imath] does have an open form solution which is [imath]+/-[(1+i)/\sqrt2][/imath]. Repeating the above calculation with this open form I find that the associative law of multiplication no longer holds and all sorts of contradictions ensue. It might also be argued that [imath]i^4[/imath] factors into -1 x -1 and therefore cannot be placed under the radical sign. But that is also true of any positive real number. Any ideas for a consistent solution to the above derivation? Thanks. | 362014 | What is the square root of [imath]i^4[/imath]?
What is the [imath]\sqrt{i^4}[/imath]? [imath]i^4[/imath] = [imath](i^2)^2[/imath] So is [imath]\sqrt{i^4}[/imath] = [imath]\sqrt{(i^2)^2}[/imath] = [imath]i^2[/imath] = [imath]-1[/imath]? Or is [imath]\sqrt{i^4}[/imath] = [imath]\sqrt{1}[/imath] = [imath]1[/imath]? When I plug it into my TI-89 Titanium, I get [imath]1[/imath]. Edit: I now see why [imath]i^4 \neq -1[/imath]. [imath]\sqrt{x^2} = \left|{x}\right| [/imath] So[imath]\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1[/imath] |
1686533 | Why can we act like functions are totally ordered by their orders?
For simplicity, consider only functions from [imath]\Bbb N[/imath] to [imath]\Bbb R^{>0}[/imath]. Let [imath]f\preceq g[/imath] if there is an [imath]A>0[/imath] such that for all sufficiently large [imath]n[/imath], [imath]f(n)\le A g(n)[/imath]. We normally would write this [imath]f\in O(g)[/imath], but I want to emphasize the order properties in this post. This is a preorder on the space of functions, so we can quotient out by the equivalence class [imath]f\sim g\iff (f\preceq g)\land (f\succeq g)[/imath] to get a partial order. It is also a lattice, with [imath]f\lor g\sim \max(f,g)\sim f+g[/imath]. It even has an abelian group structure (multiplication of functions) that respects the order, with [imath]f\preceq g\iff f/g\preceq 1[/imath]. But it is not a total order; any function both unbounded above and below, such as [imath]f(n)=\begin{cases}n&n\mbox{ even}\\1/n&n\mbox{ odd}\end{cases}[/imath], is incomparable to [imath]1[/imath]. In fact, this kind of "oscillation" between different orders is necessary for incomparability to occur, although it may still occur in monotonic functions, simply by multiplying by a sufficiently fast-growing function, such as [imath]n^{2n}[/imath] (that is, [imath]n^{2n}[/imath] is incomparable with [imath]n^{2n}f(n)[/imath] even though [imath]n^{2n}f(n)[/imath] is monotonic). My question is, why is it that we can basically ignore this fact completely when discussing orders of common functions? Given an arbitrary function, we seem to have some method for choosing "simpler" functions which are drawn from some chain [imath]S[/imath]. So for example the function [imath]f(n)[/imath] above would generally only be compared as [imath]1/n\preceq f\preceq n[/imath], where [imath]1/n[/imath] and [imath]n[/imath] are members of this [imath]S[/imath]. But what is [imath]S[/imath]? Desirable properties for [imath]S[/imath]: [imath]S[/imath] is a chain, i.e. [imath]f\preceq g[/imath] or [imath]f\succeq g[/imath] when [imath]f,g\in S[/imath] [imath]n\in S[/imath] [imath]S[/imath] is a subgroup and a sublattice, i.e. [imath]f,g\in S\implies f/g,f+g\in S[/imath] [imath]S[/imath] is closed under [imath]\log[/imath] [imath]S[/imath] is closed under [imath]\exp[/imath] (note: [imath]\exp[/imath] is not well-defined on equivalence classes) Does such an [imath]S[/imath] satisfying the properties actually exist? The toughest restriction is (5); most of the others play well with the total order constraint. I think this covers all the functions I have seen in an asymptotic expression, not counting multivariate expressions like [imath]O(mn)[/imath] or composite expressions like [imath]e^{n^2+O(n)}[/imath]. Exceptions include [imath]\log^*(n)[/imath] and fast growing functions like the Ackermann function. | 2387792 | Natural class of functions whose [imath]\mathcal O[/imath]-sets are linearly ordered
In doing complexity analysis of algorithms, we often restrict to their asymptotic complexity. A function [imath]g: \mathbb N \to \mathbb R_+[/imath] is asymptotically bounded by [imath]f: \mathbb N \to \mathbb R_+[/imath] if there are constants [imath]c \in \mathbb R_+, N \in \mathbb N[/imath] such that for each [imath]n \geq N[/imath], [imath] g(n) \leq c\cdot f(n). [/imath] The set of all such [imath]g[/imath] is referred to as [imath]\mathcal O(f)[/imath]. By letting [imath]g \preceq f[/imath] if [imath]\mathcal O(g) \subseteq \mathcal O(f)[/imath], we obtain a preordering on the set of functions [imath]\mathbb N \to \mathbb R_+[/imath]. Examples of functions that typically come up as complexities of algorithms are: [imath]\tag{*} \log(\log(n)), \log(n),\log(n)^2, \sqrt{n}, n, n\log(n),n\sqrt{n},n^3, 2^n, 3^n, 2^{2^n}. [/imath] Interestingly, these functions are all linearly ordered by [imath]\preceq[/imath]. Note that it is not hard to find functions which are not linearly ordered by [imath]\preceq[/imath], like [imath]|\sin(n)|[/imath] and [imath]|\cos(n)|[/imath], and with some fiddling it would not be too hard to make examples of strictly increasing (or even convex) functions which are not linearly ordered by [imath]\preceq[/imath]. However, these examples all seem artificial, and in particular I would be surprised if they ever came up as complexities of algorithms. Can we explicitly give a nice class of functions that includes at least all the functions listed in (*), and for which [imath]\preceq[/imath] is a linear ordering? Is this class also closed under some operations, like multiplication, exponentiation by a constant, or taking logarithms? And, perhaps, could we even justify why any algorithm is likely to have a complexity in this class? |
2623994 | Let [imath]A,B\in M_n(\mathbb{R})[/imath], Prove [imath]\det\begin{pmatrix} A & -B\\ B & A \end{pmatrix}=|\det(A+iB)|^2[/imath]
Let [imath]A,B\in \mathcal M_n(\mathbb{R})[/imath], Prove [imath]\det\begin{pmatrix} A & -B\\ B & A \end{pmatrix}=\bigg|\det(A+iB)\bigg|^2[/imath] My work: We know: [imath]|\det(A+iB)|^2=\sqrt{(\det(A+iB))^2}^2=\det(A+iB)^2=\det(A+iB)\det(A+iB)=\det((A+iB)(A+iB))=\det(A^2+AiB+iBA+i^2B^2)=\det(A^2+AiB+iBA-B^2)[/imath] Here I'm stuck. Can someone help me? | 725852 | Proving [imath]\det \left( \begin{smallmatrix} A & -B \\ B & A \end{smallmatrix} \right) =|\det(A+iB)|^2[/imath]
The complex general linear group is a subgroup of the group of real matrices of twice the dimension and with positive determinant. Let us decompose complex matrices [imath]M[/imath] as [imath]M=A+iB[/imath], where [imath]A,B[/imath] are real matrices. Now consider the correspondence [imath]f(A+iB)=\begin{pmatrix} A & -B \\ B & A\end{pmatrix}.[/imath] If [imath]\det f(M)=|\det M|^2[/imath] for square matrices, then we would have [imath]GL(n,\mathbb C)\subseteq GL_+(2n,\mathbb R)[/imath] with the identification [imath]M\to f(M)[/imath], which is an injective homomorphism. In other words, the complex general linear group would be a subgroup of the group of real matrices of twice the dimension and with positive determinant. How is [imath]\det f(M)=|\det M|^2[/imath]? |
2000378 | Calculate [imath]\lim\limits_{n \to \infty} \frac1n\cdot\log\left(3^\frac{n}{1} + 3^\frac{n}{2} + \dots + 3^\frac{n}{n}\right)[/imath]
Calculate [imath]L = \lim\limits_{n \to \infty} \frac1n\cdot\log\left(3^\frac{n}{1} + 3^\frac{n}{2} + \dots + 3^\frac{n}{n}\right)[/imath] I tried putting [imath]\frac1n[/imath] as a power of the logarithm and taking it out of the limit, so I got [imath] L = \log\lim\limits_{n \to \infty} \left(3^\frac{n}{1} + 3^\frac{n}{2} + \dots + 3^\frac{n}{n}\right)^\frac1n [/imath] At this point I thought of the fact that [imath]\lim\limits_{n \to \infty} \sqrt[n]{a_1^n+a_2^n+\dots+a_k^n} = max\{a_1, a_2, \dots,a_k\}[/imath] but this won't be of any use here, I guess. How can I calculate this limit, please? | 412765 | Evaluating [imath]\lim\limits_{n\to\infty}(a_1^n+\dots+a_k^n)^{1\over n}[/imath] where [imath]a_1 \ge \cdots\ge a_k \ge 0[/imath]
Need to find [imath]\lim\limits_{n\to\infty}(a_1^n+\dots+a_k^n)^{1\over n}[/imath] Where [imath]a_1\ge\dots\ge a_k\ge 0[/imath] I thought about Cauchy Theorem on limit [imath]\lim\limits_{n\to\infty}\dfrac{a_1+\dots+a_n}{n}=\lim a_n[/imath] and something like what happen if all [imath]a_i=0[/imath] or [imath]a_1=\dots=a_k[/imath], but may be something I am thinking wrong? Maybe it is too simple but I am not getting it; please help. |
2624498 | Evaluate [imath]\lim_{n \rightarrow\infty} \sqrt[n]{3^{n} +5^{n}}[/imath]
Evaluate [imath]\lim_{n \rightarrow\infty} \sqrt[n]{3^{n} +5^{n}}[/imath] Attempt: The only sort of manipulation that has come to mind is: [imath]e^{\frac{1}{n}ln(e^{n\ln(3)} + e^{n\ln(5)})}[/imath] So what is the trick to successfully evaluate this? | 1521019 | Find [imath]\lim_{n\to \infty}\sqrt[n]{3^n+5^n}[/imath]
[imath]\lim_{n\to \infty}\sqrt[n]{3^n+5^n}[/imath] I'm stuck. Tried to apply the squeeze theorem, but found only the [imath]\le[/imath] side, which is [imath]\sqrt[n]{5^n}[/imath], approaching 5. How do I proceed from there? |
2624598 | If we have a set of n linearly independent vectors in [imath]R^n[/imath], does that mean they span [imath]R^n[/imath]?
I know that they will form a basis, but will they span the entire [imath]R^n[/imath]? | 978327 | Does a set of [imath]n[/imath] independent vectors in [imath]R^{n}[/imath] always span [imath]R^{n}[/imath]?
Does a set of [imath]n[/imath] independent vectors in [imath]R^{n}[/imath] always span [imath]R^{n}[/imath]? If so, what's the reasoning behind this? |
1313285 | Show determinant of [imath]\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \det{A}\cdot \det{D}[/imath]
Let [imath]A \in \mathbb{R}^{n, n}[/imath], [imath]B \in \mathbb{R}^{n, m}[/imath], [imath]C \in \mathbb{R}^{m, n}[/imath] and [imath]D \in \mathbb{R}^{m, m}[/imath] be matrices. Now, I have seen on Wikipedia the explanation of why determinant of [imath]\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \det{A}\cdot \det{D}[/imath], but I still did not get it. Specifically, the explanation is: This can be seen ... from a decomposition like: [imath]\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \left[\begin{matrix} A & 0 \\ C & I_{m}\end{matrix}\right]\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right][/imath] I understood that the equation is true from the standard rules of matrix-matrix multiplication, but it is still not too clear why this should prove what we want to prove or show. If [imath]A[/imath], [imath]B[/imath], [imath]C[/imath] and [imath]D[/imath] were regular reals (and [imath]I_{i}[/imath] was [imath]1[/imath]), then the equation and the explanation would be obvious, because of the standard rules of calculating determinants... But in this case, I cannot understand why the equation shows that the final determinant is [imath]\det{A} \cdot \det{D}[/imath] Those 2 matrices [imath]\left[\begin{matrix} A & 0 \\ C & I_{m}\end{matrix}\right][/imath] and [imath]\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right][/imath] basically could not be triangular or diagonal matrices, from my understanding... | 266068 | The determinant of block triangular matrix as product of determinants of diagonal blocks
I am given the following partitioned, upper-triangular matrix: [imath] \begin{bmatrix} A_1 &* &* &* &* &* \\ 0& A_2 &* &* &* &* \\ .& 0& A_3 &* &* &* \\ .& 0& 0 &... &* &. \\ .& 0& 0& 0& ... &. \\ 0& .& ...& 0&0 & A_m \end{bmatrix} [/imath] where each of the [imath]A_i[/imath] are block-matrices. I have to prove that the determinant of this general matrix is: [imath] \prod_{n=1}^m \det A_i. [/imath] We already proved in class that it is true for [imath] m=2 [/imath] and we can use it in our proof. Thanks for your answers! |
2624800 | [imath]a_{n+1}-a_{n}\ge\frac{1}{n}[/imath]. Prove: [imath]\lim\limits_{n \to \infty} a_n= \infty[/imath]
[imath]\{a_n\}[/imath] a sequence such that for all [imath]n[/imath]: [imath]a_{n+1}-a_{n}\ge\frac{1}{n}[/imath] Prove: [imath]\lim\limits_{n \to \infty} a_n= \infty[/imath] I have figured already that [imath]a_{n+1}-a_{1}\ge\sum_{k=1}^n \frac{1}{k}[/imath]. Am I on the right track? | 1547303 | Let [imath]a_n[/imath] be a sequence such that: [imath] a_{n+1}-a_n \ge \frac{1}{n}[/imath]. prove that [imath]\lim_\limits{n \to \infty} a_n = \infty[/imath].
Let [imath]a_n[/imath] be a sequence such that: [imath] a_{n+1}-a_n \ge \frac{1}{n}[/imath]. prove that [imath]\lim_\limits{n \to \infty} a_n = \infty[/imath]. SOLUTION: I want to prove this by disproving Cauchy's test. which means, I want to prove that: There is an [imath]\epsilon>0[/imath] , such that for each [imath]N[/imath], there is [imath]n>N[/imath] and [imath]p \in[/imath] [imath]\mathbb N[/imath] such that:[imath]|a_{n+p}-a_n| \ge \epsilon[/imath]. but for some reason I couldn't find that [imath]\epsilon[/imath]. the "[imath]\frac{1}{n}[/imath]" made my job harder. To conclude: I want to disprove that [imath]a_n[/imath] converges by disproving that its Cauchy's and from [imath] a_{n+1}-a_n \ge \frac{1}{n}[/imath] we get that the sequence is monotonic increasing, thus, this will prove that: [imath]\lim_\limits{n \to \infty} a_n = \infty[/imath]. Am I on the right track? are all the conclusions that I've reached are right? How do I really complete this? any kind of help would be appreciated. |
2624897 | Ways to show [imath]\left (\frac{n^2}{2^n} \right)[/imath] is converging to zero
I still have problems understanding how to prove that a sequence converges, stuck with [imath]a_n=\left(\frac{n^2}{2^n} \right)[/imath] I know that [imath]\lim_{n\to\infty} a_n=0[/imath] but don't know how to proceed. I tried replacing [imath]2^n[/imath] with [imath]\sum_{k = 0}^n { \binom{n}{k}}[/imath] to get [imath]\left(\frac{n^2}{\sum_{k = 0}^n \left(\frac{n!}{k!(n-k)!} \right)} \right)[/imath]. Next I want to manipulate the term somehow (don't understand how this could be done) to get [imath]n[/imath] instead of [imath]n^2[/imath] as numerator to show that the denominator is always bigger than [imath]n^2[/imath] starting from some [imath]n_0[/imath]. How would I then get the value of [imath]\epsilon[/imath]? Is this a valid approach? How can I extrace a [imath]n[/imath] from the sum? Also different approaches are appriciated! @edit forgot to include that I want to show that this sequence converges to [imath]0[/imath], so showing convergence is not enough. In the linked dublicate it's written as fact if [imath]a_n > 0[/imath] and the ratio test [imath]< 1[/imath]. Why is this true? | 1003016 | How to prove that [imath]\lim_{n\rightarrow \infty}\frac {n^{2}}{2^{n}}=0[/imath] & [imath]\lim_{n\rightarrow \infty}\frac {n^{2}}{n!}=0[/imath]?
Prove that: [imath]\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0[/imath] I think I need to show [imath]2^n \geq n^3[/imath] [imath]\forall n \geq10[/imath]. Is it true? What about: [imath]\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0[/imath] What to do with [imath]n![/imath]? |
2625114 | find the number of analytic functions which vanish only on [imath]S[/imath]?
Let [imath]S = \{0\}\cup\{1\}\cup \{\frac{1}{4n+7} : n =1,2,\ldots\}[/imath]. How to find the number of analytic functions which vanish only on [imath]S[/imath]? as i just modify the given question Finding the number of analytic functions which vanish only on a given set. Options are a: [imath]\infty[/imath] b: [imath]0[/imath] c: [imath]1[/imath] d: [imath]2[/imath] i think answer will be 1 is its corrects or not ? Pliz tell me | 144496 | Finding the number of analytic functions which vanish only on a given set.
Let [imath]S = \{0\}\cup \{\frac{1}{4n+7} : n =1,2\ldots\}[/imath]. How to find the number of analytic functions which vanish only on [imath]S[/imath]? Options are a: [imath]\infty[/imath] b: [imath]0[/imath] c: [imath]1[/imath] d: [imath]2[/imath] |
2625234 | Proof request of [imath]p(A) < 1 \space \text{iff} \space \lim_{k \to \infty} A^k = 0[/imath]
I'm seeking a proof to the following lemma that I found being stated without being proven at my Numerical Linear Algebra course textbook, which involves a correlation between the spectral radius of a matrix [imath]A[/imath], [imath]p(A)[/imath] and the limit of the power of [imath]A[/imath] : Let [imath]p(A)[/imath] denote the spectral radius of the matrix [imath]A[/imath]. Then, it is [imath]p(A) <1[/imath] if and only if [imath]\lim_{k \to \infty} A^k = 0[/imath]. | 1231866 | How to prove that the matrix [imath]A^k[/imath] approaches [imath]0[/imath] as [imath]k[/imath] approaches infinity
First of all, what does it mean to say an eigenvalue is "less than unity"? I'm not exactly sure what this means. Secondly, how do I show that [imath]\lim_{k\to\infty} A^k=0[/imath] given that all eigenvalues of [imath]A[/imath] is less than unity? |
1066614 | How to prove [imath]\hat f[/imath] is uniformly continuous in [imath]R^n[/imath]?
Let the Fourier transform be defined by [imath]\hat f(\xi)=\int_{R^n}f(x)e^{-ix\xi}dx[/imath]. Suppose [imath]f\in L^1(R^n)[/imath]. How to prove [imath]\hat f[/imath] is uniformly continuous in [imath]R^n[/imath]? | 68642 | Fourier transform is uniformly continuous
I am trying to prove the following statement: If [imath]f \in L^1[/imath], then [imath]\hat f[/imath] is uniformly continuous. The argument given is as follows : [imath]|\hat f (\xi +h )-\hat f (\xi)| = \left| \int f(x) (e^{-2 \pi i x \cdot (\xi+h)}- e^{-2 \pi i x \cdot (\xi)})\mathrm dx \right| \leq 2 \|f\|_{L^1}[/imath] Now I suppose we have to use the Dominated Convergence Theorem, but I am unable to see to what sequence of functions we apply the theorem to. Any help is greatly appreciated. |
2624667 | How to prove that this representation is irreducible?
Consider the [imath]3[/imath] dimensional Lie algebra [imath]L[/imath] of differential operators [imath]x=u\frac{\partial{ }}{\partial{v}},\text{ } y=v\frac{\partial{ }}{\partial{u}}, \text{ }h=u\frac{\partial{ }}{\partial{u}}-v\frac{\partial{ }}{\partial{v}}[/imath] on the space of smooth complex vaued functions in [imath]u[/imath] and [imath]v[/imath]. (1) Prove that they satisfy satisfy [imath][xy]=h, \text{ }[hx]=2x, \text{ }[hy]=-2y[/imath] (2) Prove using the killing form that [imath]L[/imath] is semisimple. Since [imath]L[/imath] is [imath]3[/imath] dimensional, this forces [imath]L[/imath] to be simple. (3) Prove that the set of homogeneous polynomials of degree [imath]n[/imath] in [imath]u[/imath] and [imath]v[/imath] is a irreducible representation space of [imath]L[/imath]. I have proved the first two parts. But I am unable to prove the last part. Any ideas? | 2232436 | Representation of [imath]\text{sl}(2,\mathbb{C})[/imath]
Consider the Lie algebra [imath]\text{sl}(2,\mathbb{C})[/imath] (commutator as Lie bracket) with its standard basis consisting of [imath]e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.[/imath] Define the map [imath]R: \text{sl}(2,\mathbb{C})\rightarrow \text{End}(\mathbb{C}[x,y])[/imath] on the basis as [imath]h\mapsto x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y},~e\mapsto x\frac{\partial}{\partial y},~f\mapsto y\frac{\partial}{\partial x}.[/imath] I want to show that this is a Lie algebra homomorphism where [imath]\text{End}(\mathbb{C}[x,y])[/imath] is equipped with the commutator as Lie bracket. For that I will use the relations [imath][e,f]=h,~[h,e]=2e,~[h,f]=-2f[/imath]. [imath]R[/imath] is of course [imath]\mathbb{C}[/imath]-linear since it is defined as a linear extension. In order to show [imath]R([A,B])=[R(A),R(B)][/imath] for alle [imath]A,B\in\text{sl}(2,\mathbb{C})[/imath], it is enough to show this on the basis. But here is my issue: For example [imath]R([e,f])=R(h)=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}\neq 0=x\frac{\partial}{\partial y}y\frac{\partial}{\partial x}-y\frac{\partial}{\partial x}x\frac{\partial}{\partial y}=[R(e),R(f)][/imath]. What am I doing wrong? |
2625227 | Arranging five letters of the word MISSISSIPPI
I am looking for some help with the following two questions, I have included my explanation for part a but I am not sure if it is correct or how to do part b, thanks! Question a) how many ways are there to arrange the letters in MISSISSIPPI? Explanation 1) The word MISSISSIPPI has [imath]11[/imath] letters, not all of them distinct. There are [imath]\dbinom{11}{4}[/imath] ways to choose the slots where the four S's will go. For each of these ways, there are [imath]\dbinom{7}{4}[/imath] ways to decide where the four I's will go. There are [imath]\dbinom{3}{2}[/imath] ways to decide where the P's will go. That leaves [imath]1[/imath] gaps, and [imath]1[/imath] singleton letters, which can be arranged in [imath]1![/imath] ways, for a total of [imath]\binom{11}{4}\binom{7}{4}\binom{3}{2}1!.[/imath] Explanation 2) In general if you have n objects with [imath]{r_1}[/imath] objects of one kind, [imath]{r_2}[/imath] objects of another,...,and [imath]{r_k}[/imath] objects of the kth kind, they can be arranged in [imath]\frac{n!}{(r_1)!(r_2)!...(r_k)!}[/imath] Therefore, [imath]\frac{11!}{4!4!2!}[/imath] Question b) How many arrangements are possible using [imath]5[/imath] letters from MISSISSIPPI? | 1855369 | MISSISSIPPI problem
How many arrangement of the letters in MISSISSIPPI have at least 2 adjacent S's? I was thinking that I can glue two of the S's together, so there will be 9 letters plus the special letter SS, and the number of arrangements will be [imath]10!/(2!2!4!)[/imath], this is not the correct answer, but what is wrong with my reasoning. (the correct answer is [imath]11!/(4!4!2!)-7!/(4!2!)\binom{8}{4}[/imath], I understand the solution) |
2625332 | Consider the sequence [imath]x_0 =\cos 0[/imath], [imath]x_{n+1} =\cos x_n[/imath], how do I show it converges?
I know that this sequence is bounded and also that if it converges, then it should converge at [imath]x[/imath] such that [imath]x = \cos x[/imath]. Assuming I am not allowed to use calculator (and hence not able to solve [imath]x = \cos x[/imath] numerically), how would I prove the convergence? | 2164009 | Is [imath]x_{n+1}=\cos x_n[/imath] convergent?
If [imath]x\in \Bbb R[/imath] ; then is the sequence [imath]\{a_n\}[/imath] where [imath]a_1=x[/imath]; [imath]a_{n+1}=\cos (a_n)[/imath] convergent? Obviously [imath]|a_n|\le1[/imath] and hence [imath](a_n)[/imath] is bounded. Also [imath]f(x)=\cos x[/imath] is decreasing for [imath]x>0[/imath]. But here [imath]x\in \Bbb R[/imath] .How to proceed here?Please help. |
2624959 | Differentiating [imath]\int_x^{x+1/x}e^{t^2}dt[/imath]
This is problem from here, but I didn't receive any answers [imath]\lim_{x\to \infty}e^{-x^2}\int_x^{x+1/x}e^{t^2}dt[/imath]. I need to differentiate [imath]\int_x^{x+1/x}e^{t^2}dt[/imath] and I got [imath](1- 1/x^2) e^{(x+1/x)^2}-e^{x^2}[/imath] Am I right? I am asking because Spivak Calculus 1994 in chapter 18 problem 31, Calculus 1994 got another answer. He got [imath]e^{(x+1/x)^2}-e^{x^2}[/imath] I think he is missing derivative of [imath]x+1/x[/imath] | 2624929 | [imath]\lim_{x\to \infty}e^{-x^2}\int_x^{x+1/x}e^{t^2}dt[/imath]
[imath]\lim_{x\to \infty}e^{-x^2}\int_x^{x+1/x}e^{t^2}dt[/imath] This is problem from Spivak calculus 1994, chapter 18, 31. The book gives following solution: simply apply Lhopitatal rule, but what puzzles me is the way numerator is differentiated. So they got [imath]e^{(x+1/x)^2}-e^{x^2}\over 2xe^{x^2}[/imath] But I think it should be [imath](1- 1/x^2) e^{(x+1/x)^2}-e^{x^2}\over 2xe^{x^2}[/imath] . I think they are missing derivative of [imath]x+1/x[/imath]. Am I wright or wrong? P.S Could anyone please help me editing this question? |
2626075 | Orthogonal coordinate frame
On a Riemannian manifold [imath](M,g)[/imath], there always exists local orthonormal frame [imath]\{E_i\}[/imath] with respect to the metric [imath]g[/imath]. But there does not necessarily exist orthonormal coordinate frame [imath]\{ \frac{\partial}{\partial x^i} \}[/imath] in a small neighborhood around a point [imath]p[/imath], where [imath]\{x^i:U \rightarrow \mathbb{R}[/imath] } are coordinate maps. My question is the follow: instead of local orthonormal coordinate frame, is it possible to find local orthogonal coordinate frame? i.e. with respect to some coordinate system, the metric g can be written locally as [imath]g= g_{ii} {dx^{i}}dx^i[/imath] For surface, I know this is true because of something stronger, the existence of isothermal coordinate. I wonder whether this is possible for higher dimension. | 2037169 | Existence of orthogonal coordinates on a Riemannian manifold
This is probably a very naive question, but so far I could not find an answer: Let [imath](M,g)[/imath] be a Riemannian manifold. Can we always find "orthogonal coordinates" locally? More precisely, I am asking if for every [imath]p \in M[/imath] there exists a neighbourhood [imath]U[/imath] and a diffeomorphism [imath]\phi:\mathbb{R}^n \to U[/imath], such that [imath]g_{ij}=g(d\phi(e_i),d\phi(e_j))=0[/imath] for [imath]i \neq j[/imath]. Clarification: Note that I want [imath]g_{ij}=0[/imath] on all [imath]U[/imath], not just at [imath]p[/imath]. Also, I allow [imath]g_{ii} \neq g_{jj}[/imath] for [imath]i \neq j[/imath] (the special case where [imath]g_{ii}[/imath] is independent of [imath]i[/imath] is called isothermal coordinates-and corresponds to conformal flatness of [imath]U[/imath]). Of course, this is weaker than requiring [imath]M[/imath] to be conformally flat, since a (linear) map which maps an orthogonal basis to an orthogonal basis does not need to be conformal. |
2626366 | Proving that a three complex numbers form an equilateral triangle with vertices on unit circle
While solving questions from Mathematial Analysis by Apostol, I came across this question:- Q. Given three complex numbers [imath]z_1[/imath], [imath]z_2[/imath] and [imath]z_3[/imath] such that [imath]\left| z_1 \right| = \left| z_2 \right| = \left| z_3 \right| = 1[/imath] and [imath]z_1 + z_2 + z_3 = 0[/imath]. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with the center at origin. Unfortunately, I could not get much ideas to solve this questions. However, from the given data, I inferred that the numbers are on the unit circle and hence, the vertices of the triangle are on the unit circle. Also, I figured out that the centroid of this triangle is at origin. Now, if somehow we prove that the orthocenter of this triangle is also at origin, we will prove that the triangle is equilateral. But, I could not find a possible way to prove it by just using the information given in the question. Another thought that I came across was to show that between any pair of the complex numbers (given) the difference in their argument, as measured from the positive X - axis, is [imath]\dfrac{2\pi}{3}[/imath] or [imath]120^o[/imath]. If we prove this, we will prove that the triangle is equilateral. Again, there are no further ideas to implement this idea as well for the final proof! Help will be appreciated! | 201100 | Vertices of equilateral triangle inscribed in the unit circle
Prove that if [imath]z_{1}+z_{2}+z_{3}=0[/imath] and [imath]|z_{1}|=|z_{2}|=|z_{3}|=1[/imath] then the points [imath]z_{1},z_{2},z_{3}[/imath] are the vertices of an equilateral triangle inscribed in the unit circle [imath]|z|=1[/imath]. My idea was the following: since the sum of the numbers has to be [imath]0[/imath] and they have equal modulus, the interior angles between them should be equal to [imath]120º[/imath]. Later I could use the Inscribed Angles Theorem to prove that [imath]\arg \left(\frac{z_{j}-z_{i}}{z_{j}-z_{k}}\right)=\frac{1}{2}\arg\left(\frac{z_{i}}{z_{k}}\right)[/imath] and since all [imath]\arg(z_{i}/z_{k})=120º[/imath], all angles would be equal. So the triangle formed would be equilateral. I am wondering if this is correct and if would there be a more organized way to prove this. |
1901349 | Riemannian metric on Lie Groups and [imath]S^n[/imath]
I am studying Riemannian Geometry by myself and I am following the textbook Riemannian Geometry by Manfredo Perdigao do Carmo. The Riemannian metric on a Lie group G is given by defining it on the tangent space of the neutral element e and it is given by: [imath]\langle u,v\rangle_x = \langle (dL_{x^{-1}})_x(u),(dL_{x^{-1}})_x(v)\rangle_e\ ,\text{ where } x \in G, u,v \in T_xG.[/imath] I have verified all the properties of a Riemannian metric except for the smoothness. Since a Riemannian metric is a 2-tensor and not a differential form, I am not sure how to understand it's pullback (by [imath]L_{x^{-1}})[/imath]. Can somebody explain how the metric is varying smoothly due to the smoothness of [imath]L_{x^{-1}}[/imath]. Thanks in advance. Edit: I also want to know how to find an explicit formula for a Riemannian metric on [imath]S^n[/imath] which is induced by the pullback of the inclusion map from [imath]\Bbb R^n[/imath] to [imath]S^n[/imath]. | 1326735 | Proving smoothness of left-invariant metric on a Lie Group
Assume [imath]G[/imath] is a Lie group. The standard construction of a left invariant metric on [imath]G[/imath] goes as follows: Take an arbitrary inner product [imath]\langle,\rangle_e[/imath] on [imath]T_eG[/imath] and define [imath]\langle u , v\rangle_x = \langle (dL_{x^{-1}})_xu , (dL_{x^{-1}})_xv\rangle_e [/imath] , where for every [imath]g\in G[/imath] [imath]L_g[/imath]:[imath]G\rightarrow G[/imath] is left multiplication by [imath]g[/imath]. How to prove this metric is smooth? (There are several different equivalent criterions,I tried to prove for instance that for any two smooth vector fields X,Y on [imath]G[/imath] [imath]\langle X,Y \rangle[/imath] is a smooth function but failed. |
2626224 | How to evaluate the integral[imath]\int_0^1 \ln r \ln(1-r)[/imath]?
Who can show that [imath]\int_0^1 \ln r \ln(1-r)=2-\frac{\pi^2}{6}[/imath] without using strange functions such as [imath]Li[/imath] | 650745 | Integrating [imath]\ln(x)\times\ln(1-x)[/imath]
Is there a way I can derive the value of the integral [imath] \int_0^1 \ln(x)\ln(1-x)dx[/imath] using the fact that [imath]\displaystyle\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}[/imath] ? (the actual value of the integral is [imath]2-\frac{\pi^2}{6}[/imath]) Thanks in advance |
2612366 | Integral sequence limit: [imath]\lim_{n\to+\infty}\int_{\mathbb{R}} \sin(nx)h(x) \, dx[/imath]
I am to computing this limit: [imath]\lim_{n\to+\infty}\int_{\mathbb{R}} \sin(nx)h(x) \,, dx[/imath] where [imath]h(x)[/imath] stays in [imath]L^1(\mathbb{R})[/imath]. Is it correct this method? I choose a sequence of functions [imath]h_n(x)\in C_c^\infty[/imath] which converges to [imath]h[/imath] in [imath]L^1[/imath] sense and I have [imath]\int_{\mathbb{R}} \sin(nx)h_n(x) \, dx = \left.-\frac{\cos(nx)} n h_n(x)\right|_{-\infty}^{+\infty} + \int_{\mathbb{R}} \frac{\cos(nx)}{n}h_n'(x) \, dx[/imath] so [imath]\left|\int_{\mathbb{R}} \sin(nx)h_n(x) \, dx\right|\le \frac Cn\to0.[/imath] In this way I have [imath]\|\sin(nx)h(x)\|_{L^1}\le\| \sin(nx)h(x)-\sin(nx)h_n(x)\|_{L^1}+\| \sin(nx) h_n(x) \|_{L^1} \to 0[/imath] | 1350584 | Question about the Riemann-Lebesgue Lemma proof
Ok, so one of the formulations of the Riemann-Lebesgue Lemma says: [imath] f\in L^1(\mathbb{R}) \implies \hat{f}(\omega)\to 0\;\mbox{ when } \;|\omega|\to\infty.[/imath] I get all the steps of the proof, except the one which says that if [imath]f[/imath] is an arbitrary integrable function, it may be approximated in the [imath]L^1[/imath] norm by a compactly supported smooth function [imath]g\in L^1[/imath]. My question is: Why? How do you know that for all [imath]\epsilon>0[/imath], and for all arbitrary function [imath]f\in L^1[/imath] exists a compactly supported smooth function [imath]g[/imath] so that [imath]||f-g||_{L^1}<\epsilon[/imath]? The "proof" can be found in many places, but this step can be found in Wikipedia https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma |
2624359 | Rudin chap 3 exercise 7.
The question is basically to prove that convergence of [imath]\sum{a_n}[/imath] implies convergence of [imath]\sum {\frac{\sqrt a_n}n}[/imath] with [imath]a_n>0.[/imath] Basically what I tried to show is that: Since [imath]\sum{a_n}[/imath] converges, then [imath]\lim a_n= 0[/imath] when [imath]n \rightarrow \infty[/imath] , and so [imath]\sqrt a_n \rightarrow 1[/imath] when [imath]n \rightarrow \infty,[/imath] which means it is bounded. On the other hand, [imath]\frac1n \rightarrow 0[/imath] as [imath]n \rightarrow \infty[/imath] and is always decreasing. Per theorem 3.42 in Rudin, that means [imath]\sum{\frac{\sqrt a_n}n}[/imath] also converges. Is that correct or am I missing something? The solutions on the net appear to differ. | 1378353 | Is this correct for Rudin exercise 3.7? Prove the series is convergent
This is Baby Rudin exercise 7 of Chapter 3. Prove that the convergence of [imath]\sum{a_n}[/imath] implies the convergence of [imath]\sum{\sqrt {a_k} \over k}[/imath] if [imath]a_n \ge 0[/imath]. Proof: I will attempt to show that the cauchy criterion is met so [imath]{\sum_{k=n}^m {\sqrt {a_k} \over k}} \le \epsilon [/imath] for all [imath]n,m \le N_1[/imath]. Im disregarding the absolute value since its all positive terms. Since [imath]\sum{a_n}[/imath] converges, we know that the [imath]\lim \limits_{n \to \infty} a_n =0[/imath] So there exists some [imath]N_2[/imath] such that for all [imath]n \ge N_2[/imath], [imath] a_n \lt \ {{(m-n)}^2k^2(\epsilon)^2} [/imath] I am again disregarding the absolute values because its always positive Let N=max{[imath]N_1,N_2[/imath]} then for all [imath]n,m \ge N[/imath] [imath]\sum_{k=n}^m {\sqrt {a_k} \over k} \le \sum_{k=n}^m {\sqrt {{(m-n)}^2k^2(\epsilon)^2} \over k} \le \sum_{k=n}^m{(m-n)(\epsilon)} = \epsilon [/imath] Maybe this is just crazy, should I just use Cauchy-Schwarz inequality and forget about this or could I possibly make this method work? |
2626658 | Prove that if [imath]G[/imath] is a group with [imath](ab)^{2}=a^{2}b^{2}[/imath] for all [imath]a,b\in G[/imath], then [imath]G[/imath] is abelian.
Prove that if [imath]G[/imath] is a group with [imath](ab)^{2}=a^{2}b^{2}[/imath] for all [imath]a,b\in G[/imath], then [imath]G[/imath] is abelian. I am not sure how to prove this. This is my work so far: Proof so far: [imath]G[/imath] is abelian iff [imath]ab=ba[/imath]. If [imath](ab)^{2}=a^{2}b^{2}[/imath] for all [imath]a,b\in G[/imath], then [imath](ae)^{2}=a^{2}e^{2}=a^{2}[/imath] or [imath]a=a[/imath] for all [imath]a\in G[/imath]. I am not sure whether I am approaching this correctly at this point, since it looks a little silly to me. If this is the right way, I am not sure how to continue. | 1624012 | If [imath]G[/imath] is a group show that if [imath](a \cdot b)^2 = a^2 \cdot b^2[/imath] then [imath]G[/imath] must be abelian.
If [imath]G[/imath] is a group show that if [imath](a \cdot b)^2 = a^2 \cdot b^2[/imath] then [imath]G[/imath] must be abelian. [imath]\begin{aligned}(a \cdot b)^2 = a^2 \cdot b^2 & \iff (a\cdot b)\cdot(a \cdot b) = (a \cdot a)\cdot (b \cdot b) \\& \iff a \cdot (b \cdot (a \cdot b)) =a(a \cdot (b\cdot b)) \\& \iff (a^{-1} \cdot a) \cdot (b \cdot (a \cdot b)) =(a^{-1}\cdot a)(a \cdot (b\cdot b))\\& \iff (b \cdot (a \cdot b)) =(a \cdot (b\cdot b)) \\& \iff (b \cdot a) \cdot b =(a \cdot b)\cdot b\\& \iff (b \cdot a) \cdot (b \cdot b^{-1}) =(a \cdot b)\cdot (b \cdot b^{-1}) \\& \iff b \cdot a = a \cdot b\end{aligned}[/imath] Thus [imath]G[/imath] must be abelian. Is this right? Is there less clunky way to write it if it's? |
2626524 | Find the limit of [imath]a_{n+1}=\cos(a_n)[/imath]
Let [imath]\beta\in[-1,1][/imath] and let [imath]a_n[/imath] defined by [imath] \begin{cases} a_1=\beta, \\ a_{n+1}=\cos(a_n) \end{cases}[/imath] Let [imath]c\in \mathbb{R}[/imath], s.a [imath]\cos(c)=c[/imath]. Prove that: [imath]\lim\limits_{n\to\infty}a_n=c[/imath] My work so far: I know that [imath]c\in(0,1)[/imath]. [imath]|\cos(a_n)-c|=|\cos(a_n)-\cos(c)|=|-2\sin(\frac{a_n+c}{2})\sin(\frac{a_n-c}{2})|\le\frac{1}{2}|a_n+c||a_n-c|[/imath] But I don't know how to continue from here. | 876819 | If $a_{n+1}=\cos(a_n)$ for [imath]n\ge0[/imath] and [imath]a_0 \in [-\pi/2,\pi/2][/imath], find [imath]\lim_{n \to \infty}a_n[/imath] if it exists
Let [imath]a_{n+1}=\cos(a_n)[/imath] for [imath]n\ge0[/imath] and [imath]a_0 \in [0,\pi/2][/imath] Find [imath]\lim_{n \to \infty}a_n[/imath] if it exists. I drew some sketches and it does seem like the limit exists, it's probably [imath]x[/imath] such that [imath]\cos(x)=x[/imath] I have no idea how to go about solving this, hints would really be appreciated. Thank you for your time! |
2626069 | Do discontinuities only exist strictly inside the domain of a function?
For example the graph [imath]f(x) = 1/x[/imath] approaches [imath]\infty[/imath] at [imath]x=0[/imath] but we would not say this is an infinite discontinuity, just an asymptote, correct? Unless we specifically said the domain of the function included [imath]x=0[/imath]? Or must domains by definition exclude discontinuities and undefined points? Do discontinuities only exist if we can "split" the domain into two new non-empty intervals that each exclude the discontinuity? | 1087623 | Is function [imath]f:\mathbb C-\{0\}\rightarrow\mathbb C[/imath] prescribed by [imath]z\rightarrow \large \frac{1}{z}[/imath] by definition discontinuous at [imath]0[/imath]?
Is function [imath]f:\mathbb C-\{0\}\rightarrow\mathbb C[/imath] prescribed by [imath]z\rightarrow \large{\frac{1}{z}}[/imath] by definition discontinuous at [imath]0[/imath]? Personally I would say: "no". In my view a function can only be (dis)continuous at [imath]z[/imath] if [imath]z[/imath] belongs to its domain. But I have heard other sounds, that made me curious. This question was inspired by comments/answers on this question. |
2625383 | Choosing correct contour for [imath]\int_{-\infty}^{\infty}\frac{x-\sin x}{x^3}\,dx[/imath]
I am trying to compute the integral[imath]\int_{-\infty}^{\infty}\frac{x-\sin x}{x^3}\,dx[/imath] Let [imath]f(z)=\dfrac{z-\sin z}{z^3}[/imath]. I tried to integrate this along a semi-circular contour in the upper half plane, taking the radius to [imath]\infty[/imath], but the [imath]\sin z[/imath] term explodes along the circular arc. I then thought about a rectangular contour, but then the vertical contours on the sides do not give a very nice expression. (By the way, [imath]f(z)[/imath] has no singularities.) What contour should I choose? How can you decide this will work when approaching the problem to start with? | 2395247 | Contour method to show that [imath]\int_0^\infty\frac{x-\sin x}{x^3} \, dx=\frac\pi4[/imath]
Show that [imath]\int_{0}^{\infty} \frac{x - \sin(x)}{x^3} \, dx = \frac{\pi}{4}[/imath] My attempt is as follows: Let [imath]f(z) = \frac{z - i e^{iz}}{z^3}[/imath] and consider the contour on [imath][\epsilon, R][/imath] followed by a semicircular arc in the counter clockwise direction, then on [imath][-R, -\epsilon][/imath], then the semicircular clockwise contour avoiding the origin. We have, then, that [imath]0 = \int_{\Gamma} f(z) dz = \int_{[\epsilon, R]} f(t) dt + \int_{C_R}f(Re^{it})Rie^{it}dt + \int_{[-R, -\epsilon]}f(t)dt + \int_{C_{\epsilon}}f(\epsilon e^{-it})\epsilon i e^{-it} dt[/imath] Then the first and third integrals ( [imath]I_1[/imath] and [imath]I_3[/imath]) combine so that [imath]I_1 + I_3 = 2\int_{\epsilon}^R \frac{t - \sin{t}}{t^3}\,dt[/imath] Further, [imath]|I_{C_R}| \leq \int_0^\pi \left|\frac{Re^{it} - ie^{-R\sin{t}}e^{iRcos{t}}}{R^2 e^{2it}} \right|dt \rightarrow 0 \text{ as } R\rightarrow \infty[/imath] (I've omitted the details, it isn't too bad to bound) However, I'm having trouble computing the limit [imath]\lim_{\epsilon \rightarrow 0}\int_{C_{\epsilon}} f(\epsilon e^{-it})\epsilon i e^{-it} dt[/imath] No matter which way I look at it, it seems like this limit does not exist. Perhaps I'm seeing something wrong or have I chosen a bad [imath]f(z)[/imath]? |
2625949 | A fly flying to [imath]300[/imath] points, but each time it increases its flying distance by [imath]1.[/imath]
There are 300 points around circle. A fly sits in one of the points. It starts moving clockwise. At first the fly moves to the next point. Then it skips one point. Then it flies skipping 2 points, then 3 points and so on. Will the fly eventually visit all the points? I am not sure where to start. I have tried to shown it for a smaller case with the number of points being 6, but it doesn't seem to work. How can I generalize it? | 2272691 | Jumping frog problem
Consider [imath]n \ (>1)[/imath] lotus leaves placed around a circle. A frog jumps from one leaf to another in the following manner. It starts from some selected leaf. From there, it skips one leaf in clockwise direction and jumps to the next one. Then it skips exactly [imath]2[/imath] leaves in clockwise direction and jumps to the next one. Then the frog skips [imath]3[/imath] in clockwise direction and so on. Suppose it turns out that if frog continues this way all leaves are visited by the frog. Show that [imath]n[/imath] cannot be odd. My attempt, Let frog be initially placed at leaf named [imath]0[/imath], and name the remaining leaves [imath]1[/imath], [imath]2[/imath] , [imath]3[/imath]... [imath]n - 1[/imath] in clockwise direction. After [imath]k[/imath] moves the frog will be on label number [imath]0+(1+1)+(2+1)+\ ...\ + (k+1) = k (k+3)/2[/imath]. Thus the question reduces to proving that there does not exist any [imath]i[/imath] such that [imath]n = 2i+1[/imath] and the congruence stated below is satisfied for all [imath]t[/imath] for some value of [imath]k[/imath]: [imath]k(k+3)/2 \equiv t \,\pmod n, \;\; 0\leq t \lt n[/imath] |
298832 | If [imath]\lim_{x \to +\infty} f'(x) = L[/imath], then [imath]\lim_{x \to \infty} \frac {f(x)}{x} = L[/imath]
I'm trying to solve this question: Let [imath]f:[0,+\infty) \to \mathbb{R}[/imath] be derivable and [imath]\lim_{x \to +\infty} f'(x) = L[/imath], then [imath]\lim_{x\to \infty}\frac {f(x)}{x}=L[/imath]. I'm trying to solve this question using l'Hôpital rule, but I couldn't use it, because I don't know if [imath]\lim_{x\to +\infty} f(x)=+\infty[/imath]. I need help. Thanks a lot. | 1097366 | [imath]f'(t)\rightarrow b[/imath] as [imath]t\rightarrow +\infty[/imath] [imath]\Rightarrow f (t)/t\rightarrow b [/imath] using Mean Value Theorem
Suppose that [imath]f [/imath] is differentiable on [imath](0,\infty) [/imath] and [imath]f'(t) \rightarrow b [/imath] as [imath]t\rightarrow + \infty[/imath] . Show that [imath]f (t)/t\rightarrow b [/imath]. I want to use mean value theorem here, but I can't figure out where. When [imath] f (0) \ne 0[/imath], I can't make [imath]f'(c)=\frac {f (t)-f (a)}{t-a} [/imath] become [imath]f (t)/t [/imath]. So now I am just not really sure what to do. |
2626181 | Algebra - Prove this is a Ring
Let [imath]S[/imath] be a nonempty set. Prove that the power set [imath]P(S)[/imath] whose elements are all subsets of [imath]S[/imath], forms ring under the following operations: \begin{align*} a + b &= (a \cup b) \setminus (a \cap b)\\ a \cdot b &= a \cap b \end{align*} For proving it's an abelian group first, I've gotten that the identity is [imath]\emptyset[/imath]. \begin{align*} a + \emptyset &= (a \cup \emptyset) \setminus (a \cap \emptyset)\\ &= a \setminus \emptyset\\ &= a \end{align*} Each element is it's own inverse. \begin{align*} a + a &= (a \cup a) \setminus (a \cap a)\\ &= a \setminus a\\ &= \emptyset \end{align*} It's certainly commutative. \begin{align*} a + b &= (a \cup b) \setminus (a \cap b)\\ &= (b \cup a) \setminus (b \cap a)\\ &= b + a \end{align*} But I can't get associative. That is, [imath](a + b) + c = a + (b + c)[/imath]. Anyone want to aid? The properties for multiplication aren't too bad either. The only one that is throwingme a curve ball is the left/right distribution. That is, [imath]a(b + c) = ab + ac[/imath] and [imath](a + b)c = ac + bc[/imath]. While this questions is asked elsewhere on this site and has answers to most parts in proving that it is a ring, it does not explain or show the associativity portion for addition in detail. That is [imath](a + b) + c = a + (b + c)[/imath] is left very vague. | 1986529 | Power Set of [imath]X[/imath] is a Ring with Symmetric Difference, and Intersection
I'm studying for an abstract algebra exam and one of the review questions was this: Let [imath]X[/imath] be a set, and [imath]\mathcal P(X)[/imath] be the power set of [imath]X[/imath]. Consider the operations [imath]\Delta[/imath] = symmetric difference (a.k.a. "XOR"), and [imath]\bigcap[/imath] = intersection. a) Does [imath]\Delta[/imath] and [imath]\bigcap[/imath] make [imath]P(X)[/imath] into a ring? b) If so, is it a ring with unity? c) Is the ring commutative? d) Is it a field? For parts a) and b), I think it does form a ring with unity, but I'm not quite sure how to get started on proving it. For part c), it is a commutative ring since [imath]\mathcal P(X)[/imath] is closed under symmetric difference and intersection, right? Not even sure how to get started on d). I'd really like to understand this question fully, so any kind of input would be tremendously helpful. Thank you! |
2626688 | Fixed point of a map [imath]\mathbb R^n \rightarrow \mathbb R^n[/imath]
Today I cam across the following exercise: exercise Let [imath]\alpha: \mathbb R^n \rightarrow \mathbb R^n[/imath] be a map with [imath]\alpha^p=1[/imath] for some prime [imath]p[/imath]. Show that [imath]\alpha[/imath] has a fixed point. origin I found the exercise somewhere in my notes. The setting was roughly orientation and the orientation bundle. approach For my approach, of course I assumed that [imath]\alpha[/imath] has no fixed point and just as in the Brouwer fixed point theorem I obtained a map [imath]\mathbb R^n \rightarrow \mathbb S^{n-1}[/imath]. But now I don't know how to continue. I especially have trouble using the assumption that [imath]\alpha^p=1[/imath]. Any hint or help for this exercise is welcome! Sincerely Slin Edit This indeed is a duplicate of this The answer uses some more advanced theory. I am interested how to do it the way it is suggested in Bredon's "Topology and Geometry". I cannot make the connection between [imath]\mathbb R^n[/imath] and coverings. | 269486 | Can a finite group act freely (as homeomorphisms) on [imath]\mathbb R^n[/imath]
I am asking if whether or not a finite group acts freely (as homeomorphisms) on [imath]\mathbb R^n[/imath]. To answer in the negative, it suffices to show: for any homeomorphism [imath]f[/imath] such that [imath]f^d=\text{id}_{\mathbb R^n}[/imath], then [imath]f[/imath] has a fixed point. I am looking for a complete self-contained answer. |
2615740 | Proof verification: Let [imath]a,b\in \mathbb{Z}_{0}[/imath]. If there exists integers [imath]r[/imath] and [imath]s[/imath] such that [imath]ar+bs=1[/imath], show that [imath]gcd(a,b)=1[/imath].
Can someone please verify whether my proof is logically correct? :) Let [imath]a,b\in \mathbb{Z}_{0}[/imath]. If there exists integers [imath]r[/imath] and [imath]s[/imath] such that [imath]ar+bs=1[/imath], show that [imath]gcd(a,b)=1[/imath]. Proof: Assume that [imath]a[/imath] and [imath]b[/imath] are not relatively prime. Then there exists an integer [imath]k>1[/imath] such that [imath]k|a[/imath] and [imath]k|b[/imath]. Then [imath]k|ar+bs[/imath]. Then [imath]k|1[/imath], which forms a contradiction ([imath]k>1[/imath] so [imath]k[/imath] does not divide [imath]1[/imath]) by assuming [imath]a[/imath] and [imath]b[/imath] are not relatively prime. [imath]\square[/imath] | 925807 | If [imath]ar + bs =1[/imath], then [imath]\gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1[/imath]
Here's the question: Let [imath]a[/imath] and [imath]b[/imath] be integers such that [imath]\gcd(a,b) = 1[/imath]. Let [imath]r[/imath] and [imath]s[/imath] be integers such that [imath]ar + bs =1.[/imath] Prove that [imath]\gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1[/imath]. I was stuck how to solve this problem. My first instinct is to do a proof by contradiction, that is, assume the [imath]\gcd(a,s) > 1[/imath] - therefore there exists a [imath]d > 1[/imath] such that [imath]a = dn[/imath] for an integer [imath]n[/imath] and [imath]s = dm[/imath] for an integer [imath]m[/imath]. However, I don't know where to go from here (or even if this is the correct route). I would really appreciate some help - thank you in advance for everything! Thanks! |
2627616 | Proving that [imath]a_{n} \rightarrow L[/imath] given that [imath]a_{n}+ 2a_{n+1} \rightarrow 3L[/imath] .
Let [imath](a_{n})_{n=1}^{\infty}[/imath] be a bounded sequence, and suppose that the sequence [imath](a_{n}+ 2a_{n+1})_{n=1}^{\infty}[/imath] converges to [imath]3L \in R[/imath]. How can I prove that [imath](a_{n})_{n=1}^{\infty}[/imath] converges to [imath]L[/imath] ? How can I go about proving this? I tried using the definition of a limit for the 2 sequences, and tried iterating the inequalities, but I did not seem to be getting anywhere. Could someone provide a proof for this problem, I am really trying my best to understand a proof for this problem. Also, is there any way to prove the above without, assuming that [imath](a_{n})_{n=1}^{\infty}[/imath] is bounded. Really would appreciate your help on this problem. Thank You! | 1823080 | With [imath]y_n[/imath] a sequence of real numbers, prove that if [imath]y_n=x_{n-1}+2x_{n}[/imath] converges then [imath]x_n[/imath] also converges
Let [imath]y_n[/imath] be a sequence of real numbers. Prove that if [imath]y_n=x_{n-1}+2x_{n}[/imath] converges then [imath]x_n[/imath] also converges. Let us suppose that [imath]y_{n}[/imath] goes to a limit [imath]L[/imath]. Then for all [imath]\varepsilon >0[/imath], for sufficiently large [imath]n[/imath], [imath]|y_n -L| < \varepsilon[/imath] . But [imath]|y_n - L| < 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|[/imath] But then how to proceed? |
2008656 | Find and prove a limit of a sequence
[imath]\lim_{x \to \infty }\sqrt[x]{a^{x}+b^{x}+c^{x}} [/imath]; [imath]a,b,c\in \mathbb{R}[/imath]. I need to find and prove a limit of this sequence. I know that for example the limit of [imath]a^{x}[/imath] is [imath]\infty[/imath] for [imath]a>1[/imath]. And the limit of [imath]\sqrt[x]{a^{x}+b^{x}+c^{x}}[/imath] should be equal [imath]\sqrt[x]{\lim_{x \to \infty }{a^{x}+b^{x}+c^{x}}}[/imath]. Am I right? But I have no idea what to do next. | 582669 | Find [imath]L=\lim \limits_{n\to \infty}\sqrt[n]{x^n+y^n+z^n}[/imath]
Find the limit following: [imath]L=\lim \limits_{n\to \infty}\sqrt[n]{x^n+y^n+z^n}[/imath] With [imath]x,\: y\: z\in R[/imath] P.S I think this limit result is [imath]L=max\left\{x,\: y\: z \right\}[/imath]. But i'm not find it, so expect people to help me find out the results by some solution Extend this limit: [imath]S=\lim \limits_{n\to \infty}\sqrt[n]{\sum_{i}^{m}a_{i}}[/imath] with [imath]a_i\in R,\: i=1,\: m[/imath] |
2627558 | Find the domain of composite function
[imath]f(x) = \dfrac x {(x-3)}[/imath]; [imath]g(x)= \dfrac {-7}{(x+7)}[/imath] I found The domain of [imath](f \circ g)[/imath] is [imath](- \infty, -28/3) \cup (-28/3 , -7) \cup (-7, \infty)[/imath] which is correct. Can someone please tell me the domain of [imath](g \circ f)(x)[/imath] ? Show work if possible please. Thanks! | 2627493 | Find the domain of composition of two functions
Let [imath]f(x)= \frac{x}{x-9} \text{ and } g(x) = \frac{-6}{x+5} [/imath] Find the domain of [imath]f \circ g.[/imath] Please show the steps and substitution/algebra to the problem. Thank you. |
2626403 | Let [imath]f: \mathbb{R} \to \mathbb{R}[/imath] and [imath]\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}[/imath]
Let [imath]f: \mathbb{R} \to \mathbb{R}[/imath] and [imath]\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}[/imath] then find the : [imath]\lim_{x \to \infty} f(x)=?[/imath] My Try : [imath]f^2(x+b)+\frac{1}{4}-f(x+b)=f(x)-f^2(x)[/imath] and let [imath]b=0[/imath] then [imath]f^2(x)+\frac{1}{4}-f(x)=f(x)-f^2(x)[/imath] so [imath]b \neq 0[/imath] now what do i do ? | 1147980 | Periodic Function With A Given Functional Equation
Let [imath]f[/imath] be a real valued function defined for all real numbers [imath]x[/imath] such that for some positive constant '[imath]a[/imath]' the equation [imath]f(x+a)=1/2+\sqrt{f(x)-(f(x))^2}[/imath] holds for all [imath]x[/imath]. Then prove that [imath]f[/imath] is periodic |
652581 | Showing [imath]\frac{x}{1+x}<\log(1+x) for all x>0 using the mean value theorem[/imath]
I want to show that [imath]\frac{x}{1+x}<\log(1+x)<x[/imath] for all [imath]x>0[/imath] using the mean value theorem. I tried to prove the two inequalities separately. [imath]\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0[/imath] Let [imath]f(x) = \frac{x}{1+x} -\log(1+x).[/imath] Since [imath]f(0)=0[/imath] and [imath]f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}<0[/imath] for all [imath]x > 0[/imath], [imath]f(x)<0[/imath] for all [imath]x>0[/imath]. Is this correct so far? I go on with the second part: Let [imath]f(x) = \log(x+1)[/imath]. Choose [imath]a=0[/imath] and [imath]x>0[/imath] so that there is, according to the mean value theorem, an [imath]x_0[/imath] between [imath]a[/imath] and [imath]x[/imath] with [imath]f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}[/imath]. Since [imath]x_0>0 \Rightarrow \frac{1}{x_0+1}<1.[/imath] [imath]\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)[/imath] | 1376462 | how to prove that [imath]\ln(1+x)< x[/imath]
I want to prove that: [imath]\ln(x+1)< x[/imath]. My idea is to define: [imath]f(x) = \ln(x+1) - x[/imath], so: [imath]f'(x) = \dfrac1{1+x} - 1 = \dfrac{-x}{1+x} < 0, \text{ for }x >0[/imath]. Which leads to [imath]f(x)<f(0)[/imath], so [imath]\ln (x+1)-x<0[/imath]. Is that a valid proof? Any other ideas? Thanks. |
2628164 | Find number of functions
Find the number of functions [imath]f: \{1,2,3,\dots,1999\}\to\{2000,2001,2002,2003\}[/imath] satisfying the condition that [imath]f(1)+f(2)+f(3)+\dots+f(1999)[/imath] is odd. Upon first thought my try was that for any function [imath]2000p+2001q+2002r+2003s[/imath] is odd where [imath]p, q, r, s[/imath] are natural numbers which also satisfy [imath]p+q+r+s=1999[/imath] and hence [imath]q, s[/imath] must be odd. Using generating functions I found the answer but then the thought stuck me that I am asked to find the number of functions possible. So I would like to know how would one do such questions using the mapping technique or any other combinatorial method. | 2017208 | Determine the number of functions [imath]f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}[/imath]satisfying the condition that [imath]f(1)+f(2)+...f(1999)[/imath] is odd.
Problems:Determine the number of functions [imath]f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}[/imath] satisfying the condition that [imath]f(1)+f(2)+...f(1999)[/imath] is odd. My Attempt: Each integer in domain has [imath]4[/imath] choices and therefore the total number of functions is [imath]f[/imath] is [imath]4^{1999}.[/imath] Since there are an equal number of functions that yield odd or even result, we can directly write that the number of functions satisfying the above condition to be [imath]2*4^{1998}.[/imath] I am unsure about this last claim and would like to prove it (if it is true.)Also is the answer to this problem correct? Edit [imath]1[/imath]: I tried to proceed further in the following manner: For a given mapping [imath]f[/imath] let the number of integers assigned to [imath]2000,2001,2002[/imath] amd [imath]2003[/imath] be [imath]p,q,r[/imath] and [imath]s[/imath] respectively, where [imath]p,q,r,s\in \mathbb{Z}^+[/imath]. Therefore the sum in question can be written as [imath]f(1)+f(2)+...+f(1999)=2000p+2001q+2002r+2003s.[/imath] Clearly the sum will be odd iff [imath]q+s[/imath] is odd. Also note that [imath]p+q+r+s=1999\Rightarrow p+r[/imath] is even. On the other hand if we let [imath]q+s[/imath] as even then [imath]p+r[/imath] must be even. Now the number of values of [imath]p,q,r[/imath] and [imath]s[/imath] for which the above two lemmas hold is the same and therefore if we let [imath]f[/imath] to be the function which maps elements form [imath]\{1,2,3,...,1999\}[/imath] to [imath]\{2000,2001,2002,2003\}[/imath] such that [imath]q+s[/imath] is even and [imath]g[/imath] to be the function hich maps elements form [imath]\{1,2,3,...,1999\}[/imath] to [imath]\{2000,2001,2002,2003\}[/imath] such that [imath]p+q[/imath] is even we will get a bijection.(Maybe!) |
2629250 | To prove a sequence is convergent
If a sequence [imath]\{x_n\}_{n=1}^{\infty}[/imath] of real numbers satisfies [imath]lim_{n\rightarrow\infty} n\Arrowvert x_n-x_{n+1}\Arrowvert=0[/imath],can we prove that it is a convergent sequence? I have tried to prove it is a Cauchy sequence but it seems hard to deal with the part of [imath]\frac{1}{n}[/imath]. | 2098853 | How to give a example that [imath]\lim\limits_{n\to\infty}n|a_{n}-a_{n+1}|=0[/imath] and {[imath]a_{n}[/imath]} divergent
I have tried to link with the series {[imath]\sum \frac{1}{n}[/imath]} but it does not work. |
2629243 | If [imath]Y_1,...,Y_n[/imath] are a finite collection of compact subsets of [imath]X[/imath], then their union [imath]Y_1\cup....\cup Y_n[/imath] are also compact.
If [imath]Y_1,...,Y_n[/imath] are a finite collection of compact subsets of [imath]X[/imath], then their union [imath]Y_1\cup....\cup Y_n[/imath] are also compact. Proof: Assume that [imath]Y_1,...Y_n[/imath] are a finite collection of compact subsets of [imath]X[/imath] WTS that [imath]Y_1\cup....\cup Y_n[/imath] are compact In order for [imath]Y_1 \cup...\cup Y_n[/imath] to be compact, their union must be closed and bounded Lets first prove that the union of [imath]Y_1\cup...\cup Y_n[/imath] is closed I will do this using a previous proposition which I learned Prop: If [imath]Y_1,...Y_n[/imath] are are finite collection of of closed sets in X, then [imath]Y_1\cup...\cup Y_n[/imath] is also closed (I have the proof of this proposition) By our assumption [imath]Y_1,...,Y_n[/imath] are a finite collection of compact subsets of [imath]X[/imath]. And since they are compact, then they must be closed Now I have to prove that the union of [imath]Y_1\cup...\cup Y_n[/imath] is bounded By our assumption [imath]Y_1,...,Y_n[/imath] are a finite collection of compact subsets of [imath]X[/imath]. And since they are compact, then they must be bounded Since [imath]Y_1,...,Y_n[/imath] are bounded subsets of X then by definition [imath]\exists B(x,r)[/imath] in X which contains [imath]Y_1,...,Y_n[/imath] If [imath]\exists B(x,r)[/imath] in X which contains [imath]Y_1,...,Y_n[/imath] then this means that this ball [imath]B(x,r)[/imath] will also contain [imath]Y_1\cup...\cup Y_n[/imath] (Intuitively, this seems true to me, but I am not sure if it is and how to show this) Can I please get some help in proving this? | 1973340 | Prove that finite unions of compact sets are compact
The question is: let [imath]F_1, ... F_n[/imath] be compact subsets of X. Show that [imath]\cup^{N}_{n=1} F_n[/imath] is compact. know that a set [imath] F \subset X[/imath] is compact if every open cover [imath]\mathcal {G}[/imath] of F contains a finite subcover [imath]\mathcal {H}[/imath]. Intuitively, I think that since [imath]F_1, .. F_n[/imath] are compact, they contain finite subcovers and so [imath]\cup^{N}_{n=1} F_n[/imath] at most has finite subcovers. I'm just stuck at trying to formally prove this out. Help would be much appreciated! |
2629335 | Defining an iterative method.
I'm learning about iterative methods to solve the system [imath]Au = f[/imath], and I came across the following in my book: To construct an iterative scheme, we assume that a non-singular matrix [imath]M[/imath] exists and define the matrix [imath]N[/imath] as [imath]N = M - A[/imath]. We can then write [imath]A = M - N[/imath] The linear system [imath]Au = f[/imath] can then be written as [imath]Mu = Nu + f[/imath]. By multiplying to the left and right with [imath]M^{-1}[/imath] we can define an iterative scheme [imath]\begin{split}u^{k+1} &= M^{-1}Nu^k + M^{-1}f\\&=M^{-1}(M-A)u^k + M^{-1}f\\&=u^k + M^{-1}(f - Au^k)\\&=u^k + M^{-1}r^k\end{split}[/imath] Question: This all makes a lot of sense to me, except for one thing; why is it allowed to write [imath]u^{k+1} = M^{-1}Nu^k + M^{-1}f[/imath] with the superscripts [imath]k+1[/imath] and [imath]k[/imath] over [imath]u[/imath]? The construction of an iterative method goes from [imath]Mu = Nu + f[/imath] to [imath]u^{k+1} = M^{-1}Nu^k + M^{-1}f[/imath] and I don't understand how. Thanks in advance! | 2623208 | Understanding iterative methods for solving [imath]Ax=b[/imath] and why they are iterative
Firstly let me preface by saying I wasn't sure if this is the correct site to ask the question. I am currently trying to gain a better understanding of where iterative methods to solve systems such as [imath]Ax=b[/imath] come from and I am missing what I think is quite a fundamental understanding. The question is fairly simple. Often we want to solve the aforementioned system but it isn't practical for reasons of stability etc to compute an inverse of [imath]A[/imath] and directly solve [imath]x=A^{-1}b[/imath]. So instead we aim to solve it using an iterative method by splitting [imath]A[/imath] into parts. For example if we perform the following splitting of [imath]A[/imath], [imath]A=S-T[/imath], we could write [imath] Ax = b\\ Sx_{n+1}=Tx_n+b. [/imath] What I want to know is why does this now become an iterative system, other than the fact I have written it as such. For me personally I would not come up with the step to split [imath]A[/imath] into [imath]S[/imath] and [imath]T[/imath] and then re-write iteratively. I might split it and end up with a line such as [imath]Sx=Tx+b[/imath] but I wouldn't make the jump myself to make it [imath]iterative[/imath]. I know how to apply the methods and I get that by doing so we slowly improve our guess for [imath]x[/imath] each time until we converge to a solution. But why? I am just making it iterative because I have been taught that's what to do. I hope I managed to express what I don't get clearly but if I haven't please let me know. |
2629315 | [imath]\{x: 2x^2\cos(1/x) \le 1\} \cup \{0\}[/imath] is complete.
Prove or disprove that [imath] \{x: 2x^2\cos(1/x) \le 1\} \cup \{0\}\ \text{is complete}. [/imath] Define [imath]f(x)=\begin{cases} 2x^2\cos\left(\frac{1}{x}\right), & x\neq 0\\ 0, & x=0 \end{cases}[/imath] Clearly, [imath]f[/imath] is continuous and [imath]T=f^{-1}\big((-\infty,1]\big)[/imath] which is closed and hence [imath]T[/imath] is complete. Am I right? | 498312 | whether they are complete metric space
[imath]S=\{x\in\mathbb{R}: 2x^2\cos{1\over x}=1\}[/imath] [imath]T=\{x\in\mathbb{R}:2x^2\cos{1\over x}\le 1\}\cup\{0\}[/imath] I need to tell whether they are complete metric space under usual metric. As they are closed subsets of complete metric space [imath]\mathbb{R}[/imath], so they are complete. Am I right? Edit: [imath]S=f^{-1}\{1\},T=\{0\}\cup f^{-1}(-\infty,1][/imath] [imath]f[/imath] is continuous. |
2629372 | A problem of Number Theory
Find a positive integer [imath]n[/imath] such that [imath]n/2[/imath] is a square, [imath]n/3[/imath] is a cube and [imath]n/5[/imath] is a fifth power. Above problem was given in the text “An Introduction to the Theory of Numbers” by “Ivan Niven and Zuckerman". I tried to generalize the problem statement as: Find a positive integer [imath]n[/imath] such that [imath]n/2[/imath] is a square, [imath]n/3[/imath] is a cube, [imath]n/5[/imath] is a fifth power, [imath]n/7[/imath] is a seventh power ... [imath]n/p_{k}[/imath] is a [imath]p_{k}[/imath]-th power. I have found the solution for this and trying to generalize it more. Now, How can I know that this work is authentic i.e. it has not been done by anyone earlier? Any sort of help is welcomed. | 420054 | There are infinitely many [imath]N[/imath] such that [imath]\frac{N}{2}[/imath] is a perfect square, [imath]\frac{N}{3}[/imath] is a perfect cube, and [imath]\frac{N}{5}[/imath] is a perfect fifth power
Show that there are infinitely many [imath]N[/imath] such that [imath]\frac{N}{2}[/imath] is a perfect square, [imath]\frac{N}{3}[/imath] is a perfect cube, and [imath]\frac{N}{5}[/imath] is a perfect fifth power. A hint is given with this question, which is: [imath]7^{30}[/imath] is a perfect square, a perfect cube and a perfect fifth power. Can anyone solve this? Thanks! |
2630100 | Inclusion and Element of in Set theory
Not a duplicate. This deals with sets that are elements of other sets. In other words, sets of sets. The other question deals with elements of sets and subsets of sets. The key difference in here is that the sets are the elements. Besides the other question only asks if [imath]A \subset A[/imath] which is trivial. When it comes to sets of sets, [imath] \subset[/imath] and [imath]\in[/imath] always confuse me in set theory. Could someone explain in detail with an example? I greatly appreciate it. In other words, when do we say [imath]\mathbb{A} \in \mathbb{B}[/imath] and when do we say [imath]\mathbb{A} \subset \mathbb{B}[/imath] (besides the definition that every element in [imath]\mathbb{A}[/imath] ought to be in [imath]\mathbb{B}[/imath]) Thank you | 131309 | Set theory: difference between belong/contained and includes/subset?
This is a total noob question. I am reading Naive Set Theory by Paul R. Halmos, and I'm having difficulty to understand something which seems to be trivial. In the first chapter he writes: If [imath]x[/imath] belongs to [imath]A[/imath] ([imath]x[/imath] is an element of [imath]A[/imath], [imath]x[/imath] is contained in [imath]A[/imath]), we shall write [imath]x\in A[/imath] I understand this. Then, he write: If [imath]A[/imath] and [imath]B[/imath] are sets and if every element of [imath]A[/imath] is an element of [imath]B[/imath], we say that [imath]A[/imath] is a subset of [imath]B[/imath], or [imath]B[/imath] includes [imath]A[/imath], and we write: [imath]A \subset B[/imath] I understand this too. Then he says: The working of the definition implies that each set must be considered to be included in itself ([imath]A \subset A[/imath]); this fact is described by saying that set inclusion is reflexive. I understand this too. But then: Observe that belonging ([imath]\in[/imath]) and inclusion ([imath]\subset[/imath]) are conceptually very different things indeed. One important difference has already manifested itself above: inclusion is always reflexive, whereas it is not at all clear that belonging is ever reflexive. That is: [imath]A \subset A[/imath] is always true; is [imath]A\in A[/imath] ever true? It is certainly not true of any reasonable set that anyone has ever seen. And this is where I don't think I understand anything. There is not more elaboration on this point in the text. I tried to skip this but it seems it is quite fundamental for understanding what follows in the book. Could someone explain what is meant here? |
2630757 | Is [imath]f(x)=\sum_{n=1}^{\infty} \frac{1}{n}\sin\left(\frac{x}{n}\right)[/imath] bounded?
It's relatively simple to prove that this sum converges by using that [imath]|\sin(t)|\leq|t|[/imath] to get that [imath]|f(x)|<\frac{\pi^2 x}{6}[/imath], but this doesn't give boundedness. I've tried using Abel summation and the approximation [imath]\lfloor t \rfloor \approx t[/imath], but the maximum possible error term is linear in [imath]x[/imath] ([imath]2x/\pi[/imath] if I remember correctly), and I've tried to use the Chinese Remainder Theorem to construct values of [imath]x=\frac{\pi}{2} N[/imath] for which the [imath]\sin(x/n)\approx 1[/imath] for small values of [imath]n[/imath], but I haven't been able to get anywhere with that as the behavior of numbers that are close to [imath]n\bmod 4n[/imath] for lots of [imath]n[/imath] is not very nice. My hunch says that there should be values of [imath]x[/imath] for which "the stars align" and the first few terms are close to [imath]1/n[/imath], where "few" is enough for [imath]f(x)[/imath] to get arbitrarily big, but I'm not sure if there's a way to rigorize this via a probabilistic argument. For completeness, here's a bit of what I've done so far: Say it were possible for [imath]\sin(x/n)[/imath] to equal [imath]1[/imath] for every [imath]1\leq n\leq K[/imath]. Then [imath]f(x) = H_K+\sum_{n=K+1}^{\lfloor x\rfloor} \frac{\sin(x/n)}{n} + \sum_{n=\lfloor x\rfloor+1}^{\infty} \frac{\sin(x/n)}{n}.[/imath] Using for the first sum that [imath]-1\leq \sin(x)[/imath] and for the second sum that [imath]t\sin(1)\leq \sin(t)[/imath] for [imath]0\leq t\leq 1[/imath], [imath]f(x)\geq H_K - \left(H_{\lfloor x\rfloor}-H_{K+1}\right)+x\sin(1)\zeta\left(2,\lfloor x\rfloor+1\right),[/imath] where [imath]\zeta(s,q)[/imath] is the Hurwitz Zeta Function. It shouldn't be too hard to prove that [imath]\zeta(2,q)>1/q[/imath] (although I don't know of a proof offhand), so this implies [imath]f(x)\geq H_K - \left(H_{\lfloor x\rfloor}-H_{K+1}\right)+\frac{\sin(1)x}{\lfloor x\rfloor+1}.[/imath] If [imath]K[/imath] and [imath]x[/imath] are both *large*, this is asymptotically about [imath]\ln\left(\frac{K(K+1)}{x}\right)+C[/imath] for some constant [imath]C[/imath]. Thus, if we could make [imath]K[/imath] grow a bit faster than [imath]\sqrt{x}[/imath], we would be able to prove that [imath]f[/imath] is unbounded. Unfortunately, [imath]\sin(x/n)=1[/imath] for all [imath]1\leq n\leq K[/imath] is not possible for [imath]K\geq 2[/imath] for modular reasons (specifically numbers that are [imath]1\bmod 4[/imath] are not also [imath]2\bmod 8[/imath]). So, we would need to find a threshold [imath]\mu[/imath] as well as a [imath]K[/imath] for which [imath]\sin(x/n)>\mu[/imath] for all [imath]1\leq n\leq K[/imath], and then weaken our bounds accordingly. Or, we could try to use a better bound than [imath]-1[/imath] on [imath]\sin[/imath] for the first sum. I'm not really sure. | 2539942 | Proof of the unbounded-ness of [imath]\sum_{n\geq 1}\frac{1}{n}\sin\frac{x}{n}[/imath]
For any [imath]x\in\mathbb{R}[/imath], the series [imath] \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) [/imath] is trivially absolutely convergent. It defines a function [imath]f(x)[/imath] and I would like to show that [imath]f(x)[/imath] is unbounded over [imath]\mathbb{R}[/imath]. Here there are my thoughts/attempts: [imath](\mathcal{L} f)(s) = \sum_{n\geq 1}\frac{1}{1+n^2 s^2} = \frac{-s+\pi\coth\frac{\pi}{s}}{2s}=\sum_{m\geq 1}\frac{(-1)^{m+1}\,\zeta(2m)}{s^{2m}}[/imath] is a function with no secrets. It behaves like [imath]\frac{\pi}{2s}[/imath] in a right neighbourhood of the origin, like [imath]\frac{\pi^2}{6s^2}[/imath] in a left neighbourhood of [imath]+\infty[/imath]. The origin is an essential singularity and there are simple poles at each [imath]s[/imath] of the form [imath]\pm\frac{i}{m}[/imath] with [imath]m\in\mathbb{N}^+[/imath]. These facts do not seem to rule out the possibility that [imath]f[/imath] is bounded; For any [imath]N\in\mathbb{N}^+[/imath] there clearly is some [imath]x\ll e^N[/imath] such that [imath]\sin(x),\sin\left(\frac{x}{2}\right),\ldots,\sin\left(\frac{x}{N}\right)[/imath] are all positive and large enough, making a partial sum of [imath] \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) [/imath] pretty close to [imath]C\log N[/imath]. On the other hand I do not see an effective way for controlling [imath] \sum_{n>N}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) [/imath] - maybe by summation by parts, by exploiting the bounded-ness of the sine integral function? Some probabilistic argument might be effective. For any [imath]n\geq 3[/imath] we may define [imath]E_n[/imath] as the set of [imath]x\in\mathbb{R}^+[/imath] such that [imath]\sin\left(\frac{x}{n}\right)\geq \frac{1}{\log n}[/imath]. The density of any [imath]E_n[/imath] in [imath]\mathbb{R}^+[/imath] is close to [imath]\frac{1}{2}[/imath], so by a Borel-Cantellish argument it looks reasonable that the set of points such that [imath]|f(x)|\geq \frac{\log x}{100}[/imath] is unbounded, but how to make it really rigorous? To compute [imath]\lim_{x\to x_0}f(x)[/imath] through convolutions with approximate identities seems doable but not really appealing. |
2631163 | show that every irreducible polynomial on [imath]\mathbb{F}_q[x] [/imath] is separable
I have already been able to show that if the irreducible polynomial [imath] f [/imath] has degree n, then its roots are [imath] a, a^q \ldots, a^{q^{n-1}}[/imath], where [imath] a [/imath] is a roots of [imath] f [/imath] to some extent. How can I show that they are all different? | 1351176 | Every irreducible polynomial f over perfect field F is separable
Every irreducible polynomial [imath]f[/imath] over a perfect field [imath]F[/imath] is separable. Can you check my proof? Assume that [imath]f[/imath] is irreducible but inseparable. So we have [imath]f=\sum_i h_ix^i[/imath] and [imath]f^p=\sum_i h_i^px^{ip}[/imath]. Now I use Frobenius mapping [imath]\phi (f)=f^p=(\sum_i h_ix^i)^p=\sum_i h_i^px^{ip}[/imath] so it is reducible. Contradiction. Should I also show for char(F)=0? |
2631602 | Prove that [imath]n^2 -1[/imath] is divisible by [imath]8[/imath], for every odd integer [imath]n[/imath].
Prove that [imath]n^2 - 1[/imath] is divisible by [imath]8[/imath], for every odd integer [imath]n[/imath]. I tried to factorize and find something but still have no idea. | 199185 | Proof problem: Show that [imath]n^2-1[/imath] is divisible by [imath]8[/imath], if [imath]n[/imath] is an odd positive integer.
Show that [imath]n^2-1[/imath] is divisible by [imath]8[/imath], if [imath]n[/imath] is an odd positive integer. Please help me to prove whether this statement is true or false. |
1055844 | How can we identify the quotient group [imath]GL(n,\mathbb{R})/SL(n,\mathbb{R})[/imath]
How can we identify the quotient group [imath]GL(n,\mathbb{R})/SL(n,\mathbb{R})[/imath] I am using First isomorphisam theorem,but i am not getting the solution how can we find out the right hand side set of the mapping and kernel | 361612 | [imath]GL(n,\mathbb R)/SL(n,\mathbb R)[/imath] is isomophic to [imath]\mathbb R^\times[/imath]
Let [imath]G[/imath] be the group of all [imath]n\times n[/imath] matrices with real entries that are invertible. The operations of the group are matrix multiplication, matrix inversion, and identity matrix. Let [imath]S[/imath] be the set of [imath]n\times n[/imath] matrices with real entries and with determinant [imath]1[/imath]. Let [imath]R^\times[/imath] denote the group of nonzero numbers with the operations of multiplication, multiplicative inverse, and [imath]1[/imath]. Prove that [imath]G/S \cong R^{\times}[/imath]. |
2617684 | A Gift Problem for the Year 2018
We had this problem in exam class yesterday on Combinatoric and it was supposed to be the new year gift from our teacher. The exercise was entitled A Gift Problem for the Year 2018 Problem: The numbers [imath]1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{2018} [/imath] are written on the blackboards. John chooses any two numbers say [imath]x[/imath] and [imath]y[/imath] erases them and writes the number [imath]x+y+xy[/imath]. He continues to do so until there is only one number left on the board. What are the possible value of the final number? I understood the problem as follows for instance if John take [imath]x=1[/imath] and [imath]y=\frac{1}{2}[/imath] then [imath]x+y+xy =2[/imath] and the new list becomes [imath]2,\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{2018} [/imath] continuing like this and so on..... Please bear with me that I do not want to propose my solution since I fell like it was wrong and I don't want to fail the exam before the result get out. but by the way I found, [imath]2017[/imath], [imath]2018[/imath] and [imath]2019[/imath] but I am still suspicious. You may help is you have an idea. | 431520 | The final number after [imath]999[/imath] operations.
I wanted to know, let the numbers [imath]1,\frac12,\frac13,\dots,\frac1{1000}[/imath] be written on a blackboard. One may delete two arbitrary numbers [imath]a[/imath] and [imath]b[/imath] and write [imath]a+b+ab[/imath] instead. After [imath]999[/imath] such operations only one number is left. What is this final number. I tried, let [imath]*[/imath] be an operation [imath]a*b = (1+a)(1+b) - 1[/imath], and [imath](a*b)*c = (1+a)(1+b)(1+c) -1[/imath]. by induction we can show that for [imath]N[/imath] such no.s we have [imath](1+a)(1+b)(1+c)\dots(1+d) - 1[/imath], where [imath]n\{a,b,c,\dots,d\} = N[/imath]. In the question we have numbers [imath]1,\frac12,\frac13,\dots,\frac1{1000}[/imath]. plugging in we get the final number [imath]1000[/imath], but the answer given is [imath]100[/imath]. What am i doing wrong? Any help appreciated. Thanks. |
2631947 | Module over integral domain, "Rank-nullity theorem", Exact Sequence
I have read questions related to [imath]\mathbb{Z}[/imath]-module and exact sequence of it. I have a question about generalising this result (or finding counter example) Let [imath]R[/imath] be an integral domain and [imath]0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0[/imath] be an exact sequence of finitely generated [imath]R[/imath]-module. Then, is [imath]rank(M)=rank(M_1)+rank(M_2)[/imath] true? (Definition of rank: [imath]rank(M)[/imath] denotes the cardinality of the largest linearly independent subset of [imath]M[/imath]) I have read statements online and I found that "flat modules" satisfy this result (though I have not learnt tahings about maps and tensor product so I cannot effectively understand all proofs), but there are no general result about integral domains. Hence, I would like to ask this question here and seek for help. Thank you for your kind help! (I have stuck in this for half a week and I am not making any progress. The current state of my work is that: I have (kind of?) proved that [imath]Rank(M)\geq Rank(M_1)+Rank(M_2)[/imath] by putting their linear independent sets together and prove that it is still linearly independent. I have proved that the statement is true if [imath]Rank(M)-Rank(M_1)=0[/imath] or [imath]1[/imath]. The proof of [imath]0[/imath] is by torsion module, and the proof of [imath]1[/imath] is proved by hardcoding (but it cannot be generalised to [imath]2[/imath] or larger. ) | 1641562 | Exact Sequences of Modules and Rank
Suppose that: [imath]0 \rightarrow M_{1} \rightarrow M \rightarrow M_{2} \rightarrow 0[/imath] Is an exact sequence of [imath]R[/imath]-modules (for [imath]R[/imath] commutative integral domain). Show that [imath]\mathrm{rk}(M) = \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})[/imath] I have shown the inequality [imath]\mathrm{rk}(M) \geq \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})[/imath], however for the opposite direction I'm lost. I tried to take [imath]N[/imath] as a linear independent subset and tried to decompose it into [imath]N_{1}[/imath] (the elements in [imath]\mathrm{Ker}(g)[/imath]) and [imath]N_{2}[/imath] (the elements not in [imath]\mathrm{Ker}(g)[/imath]), but I haven't got far, since [imath]N_{1}[/imath] gives us a linearly independent set in [imath]M_{1}[/imath], however [imath]N_{2}[/imath] can give us a linearly dependent set in [imath]M_{2}[/imath]. We define [imath]\mathrm{rk}(M)[/imath] in this case as the supremum over all cardinalities of the linearly independent subsets of [imath]M[/imath] (where the only finite linear combination equaling [imath]0[/imath] implies trivial coefficients for [imath]r \in R[/imath]). Would this be the proper way to show the result? I think I'm missing something important. |
2516670 | Inverse Laplace Transform of f(s)=[imath]\frac{1}{s\sinh^2{c\sqrt{s}}}[/imath]
Please help guys.. I have to calculate inverse laplace of f(s)=[imath]\frac{1}{s\sinh^2{c\sqrt{s}}}[/imath], where s is Laplacian variable. By residue theorem, the function has one residue at s=0 (solving this is easy) and infinitely many at [imath]c\sqrt{s} = n\pi i[/imath] which I have some problems because it is order of 2! How can I solve the second part of the problem? Thanks in advance. | 2442326 | Inverse Laplace transform of [imath]\frac{1}{s\sinh{c\sqrt{s}}}[/imath]
I have some questions. I have to take the inverse Laplace transform of the given function below by residue theorem: [imath]\frac{1}{s\sinh{c\sqrt{s}}}[/imath] where s is Laplacian variable and c is a real constant. Actually it has a very crowded numerator which I do not need to define. So far I know that it has a simple pole at s = zero and infinitely many poles at c√s = nπi Questions: 1- What is the order of the poles at nπi? Is it 1/2? 2- At zero, what is the order? (Because to residue at zero, s also contributes) 3- What is the way of solving this question at nπi by residue theorem (without defining sinh function as exponential function)? Thanks for any help. |
2632425 | Show that [imath]|f(0)| \leq \frac{1}{4}[/imath] for every holomorphic function [imath]f:\Bbb D \to \Bbb D[/imath] such that [imath]|f(\frac{1}{2})|+|f(-\frac{1}{2})|=0[/imath]
Let [imath]\Bbb D= \{z : |z|<1\}[/imath] be the unit disc. Let [imath]f:\Bbb D \to \Bbb D[/imath] be a holomorphic function such that [imath]|f(\frac{1}{2})|+|f(-\frac{1}{2})|=0[/imath]. Prove: [imath]|f(0)| \leq \frac{1}{4}[/imath]. The only thing I thought about was using Schwarz lemma, but I'm not sure how. I thought about defining [imath]g(z)=\frac{f(z+\frac{1}{2})+f(z-\frac{1}{2})}{2}[/imath], but now [imath]g[/imath] is not defined in all [imath]\Bbb D[/imath], and I'm pretty much stuck. | 1177230 | If [imath]f \in \operatorname{Hol}(D)[/imath], [imath]f(\frac{1}{2}) + f(-\frac{1}{2}) = 0[/imath], prove that [imath]|f(0)| \leq \frac{1}{4}[/imath]
If [imath]f \in \operatorname{Hol}(D),f(\frac{1}{2}) + f(-\frac{1}{2}) = 0[/imath], prove that [imath]|f(0)| \leq \frac{1}{4}[/imath] [imath]D = \{ z \in \mathbb{C} : |z| < 1 \} [/imath] My thoughts so far: Let's say [imath]f(0) = a[/imath]. Define [imath]g = \frac{z-a}{1-\bar{a}z}[/imath] and [imath]h(z) = (g(f(z))[/imath] Now all the conditions for the Shwarz lemma are met, and I can conclude that [imath]|h(\frac{1}{2})| \leq \frac{1}{2}[/imath] and [imath]|h(-\frac{1}{2})| \leq \frac{1}{2}[/imath]. The idea would then be to multiply the two inequalities together and try to somehow separate [imath]a[/imath], but the algebra gets really messy and I feel like I'm doing something wrong. Any help would be appreciated! |
2631463 | [imath]\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor[/imath],
Sorry about the title i didn't know what to call this other than "analytic number theory" I am asked to show that [imath]\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor[/imath], where the sum is over all d for which ever prime factor is less or equal to [imath]\sqrt x [/imath] The furthest right hand part of the equation looks like a mobius inversion but im not sure how to do it over root x and the rest i am completely lost on. | 914994 | Legendre's formula
Legendre's formula counts the number of positive integers less than or equal to a number [imath]x[/imath] which are not divisible by any of the first [imath]a[/imath] primes: [imath]\begin{align} &\phi(x,a)=\lfloor x \rfloor-\sum_{p_i\le a}\left\lfloor \dfrac{ x }{(p_i)}\right\rfloor+\sum_{p_i<p_j\le a}\left\lfloor\dfrac{ x}{(p_ip_j)}\right\rfloor-\sum_{p_i<p_j<p_k\le a}\left\lfloor \dfrac{x}{(p_ip_jp_k)}\right\rfloor+\dots \end{align}[/imath] then [imath]\pi(x)=\phi\bigl(x,\pi (\sqrt{x})\bigr)+\pi (\sqrt{x})-1[/imath] as given here. My question is, why does [imath]\bigl\lfloor\phi(x,\lfloor\sqrt{x}\rfloor)+\sqrt{x}-1\bigr\rfloor[/imath] also equal [imath]\pi(x)?[/imath] |
2633081 | Evaluating [imath]\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}[/imath]
Evaluate: [imath]\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}[/imath] I have been trying to solve this for [imath]15[/imath] minutes but sin(sin(x)) part has me stuck. My attempt: I tried multiplying with [imath]x[/imath] inside the [imath]\sin[/imath] as [imath]\sin{(\frac{x\sin{x}}{x})}[/imath]. No leads. | 437926 | A limit problem [imath]\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}[/imath]
This is a problem from "A Course of Pure Mathematics" by G H Hardy. Find the limit [imath]\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}[/imath] I had solved it long back (solution presented in my blog here) but I had to use the L'Hospital's Rule (another alternative is Taylor's series). This problem is given in an introductory chapter on limits and the concept of Taylor series or L'Hospital's rule is provided in a later chapter in the same book. So I am damn sure that there is a mechanism to evaluate this limit by simpler methods involving basic algebraic and trigonometric manipulations and use of limit [imath]\lim_{x \to 0}\frac{\sin x}{x} = 1[/imath] but I have not been able to find such a solution till now. If someone has any ideas in this direction please help me out. PS: The answer is [imath]1/18[/imath] and can be easily verified by a calculator by putting [imath]x = 0.01[/imath] |
2633204 | Is the limit of the empty diagram considered a product?
It is a theorem on the stacks project that if a category [imath]\mathcal{C}[/imath] has finite products and equalizers then it has a terminal object, this theorem is at https://stacks.math.columbia.edu/tag/04AS. But the category [imath]\mathbb{Z}[/imath] considered as a totally ordered set has all finite products which are given by taking the minimal element, and it vacuously have all equalizers, but has no final object. What is going wrong here? | 1991522 | Terminal objects as "nullary" products
I read something weird in my category theory book (Awodey p 47). " Observe also that a terminal object is a nullary product, that is, a product of no objects: Given no objects, there is an object [imath]1[/imath] with no maps, and given any other object [imath]X[/imath] and no maps, there is a unique arrow: [imath]!:X\to 1[/imath] making nothing further commute." Could anyone give a hint about what this means? I mean "given no objects, there is an object..?" Thank you |
2632237 | Implicit differentiation. Confusing assumption.
I've encountered this example of implicit differentiation. The problem asks us to find the derivative of [imath]y^2 + x^2 = 4[/imath] How can they assume that y is a function x when it clearly is not? If we isolate y, we get: [imath]y = \pm \sqrt{4-x^2}[/imath]. So then how can we assumt that y is a function of x? On a high level, I'm still a bit confused. So I guess implicit differentiation is useful when we have relations between y and x that aren't functions. That's why the [imath]\frac{dy}{dx}[/imath] is in term of 2 variables. [imath]\frac{dy}{dx}[/imath] has two values depending on the value of x and the value of y since at x, there are 2 values of y. Is that right? | 936206 | Why does implicit differentiation work on non-functions?
I've been reading Keisler's Calculus book, and there's an example where he does implicit differentiation on the equation: [imath]x^2+y^2=1[/imath] which yields:[imath]\frac{dy}{dx}=-\frac{x}{y}[/imath] I understand the technique of implicit differentiation, and I understand that it's just an application of the chain rule when [imath]y[/imath] is a function of [imath]x[/imath], or [imath]x[/imath] and [imath]y[/imath] are both functions of some third variable. What I don't understand is how you can get the derivative of a non-function. In the case above, [imath]y[/imath] is not a function of [imath]x[/imath], and as far as I know, looking at the definition of the derivative, derivatives only seem to be defined for functions. Sorry for the noob question, but what am I missing here? |
2621624 | How the sequence [imath]S_n=\frac{1}{2}({S_{n-1}+S_{n-2}}), \forall n>2,[/imath] converge?
If [imath]S_n=\frac{1}{2}({S_{n-1}+S_{n-2}}), \forall n>2,[/imath] then show that [imath]{S_n}[/imath] converges. I have proved that sequence converges and I understood all the steps except the last one. First, I assumed that the [imath]S_1<S_2[/imath] which leads to a sequence such that it has two subsequences. One of even order and another of odd order. The odd ordered subsequence is increasing and bounded above by [imath]S_2[/imath] and hence it is convergent and similarly the even ordered subsequence also converges. Also, both subsequence converge to the same limit. We can check this by simple substitution of two different expression in the above sequence and then both expression will turn out to be same. I understood all the steps of the book but then the following expression,I couldn't understand: [imath]S_{k}+ \frac {1}{2}S_{k-1}=\frac{1}{2}(S_1+2S_2).[/imath] I couldn't not understand,how the book came to the above mentioned step. Any help or hint would be great. Thanks in advance | 2559687 | Show that the sequence [imath]a_1=1[/imath], [imath]a_2=2[/imath], [imath]a_{n+2} = (a_{n+1}+a_n)/2[/imath] converges by showing it is Cauchy
Show that the sequence [imath]a_1=1[/imath], [imath]a_2=2[/imath], [imath]a_{n+2} = (a_{n+1}+a_n)/2[/imath] converges by showing it is Cauchy. My work : Need to show that for every [imath]\epsilon \gt 0[/imath] there exist [imath]N[/imath] such that [imath]n,m\ge N \implies | a_n - a_m| \lt\epsilon[/imath]. [imath]|a_n-a_m| = \dfrac{1}{2}|(a_{n-1} + a_{n-2}) - ( a_{m-1} + a_{m-2})|[/imath] I feel triangle inequality might be helpful here, but really not sure how to link it to the [imath]\epsilon[/imath]. Appreciate any help... |
2633952 | Convergence of [imath]\sum\limits_{n=2}^{\infty}\frac{(n^3+1)^\frac13-n}{\log n}[/imath]
To show that [imath]\sum\limits_{n=2}^{\infty}\frac{(n^3+1)^\frac13-n}{\log n}[/imath] is convergent My attempt: Expanding the first term using binomial series and simplifying, I tried comparison test with [imath]b_{n}=\frac{1}{n^2}[/imath] [imath]\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\log n}\right)n^2 [/imath] [imath]=\lim_{n \to \infty}\frac{n\left(\frac13\frac{1 }{n^{3}}-\frac19\frac{1 }{n^{6}}+\dots\right)n^2}{\log n}[/imath] I could simplify this using binomial theorem but I don't know how to deal with the [imath]\log[/imath] term in the denominator | 2519958 | Show that the series [imath]\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}[/imath] converges
Show that the series [imath]\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}[/imath] converges. I showed it using Abel's theorem and limit comparison test. Any other simpler method? |
2633043 | number of integral solutions of the equation
Find the number of integral solutions of [imath]\large 2x + y + z = 20[/imath] with [imath]x, y, z \ge 0[/imath] ? I saw a method of generating function but i donot have knowledge about generating function any alternative way to solve this , answer = 121 | 2017905 | Find the number of integral solutions of [imath]2x + y + z = 20[/imath]?
Find the number of integral solutions of [imath]2x + y + z = 20[/imath] where [imath]x,y,z \geq 0[/imath] ? Can I solve it by using stars and bars method of combinatorics? |
766323 | Proof that laplace's equation is rotationally invariant using chain rule
Suppose [imath](x, y)[/imath] and [imath](p, q)[/imath] are coordinates in the plane related by rotation around a fixed point [imath](a, b)[/imath], as follows: [imath]\begin{bmatrix} p\\ q\end{bmatrix} = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix} \begin{bmatrix} x-a \\ y-b \end{bmatrix}[/imath] where [imath]t[/imath] is the rotation angle. Applying the chain rule show that [imath]u(p(x,y), q(x,y)) [/imath] satisfies [imath]u_{xx}+u_{yy}=0[/imath] iff [imath]u_{pp}+u_{qq}= 0[/imath]. Where would I use chain rule in this problem? I am kind of confused because [imath]u[/imath], [imath]p[/imath], and [imath]q[/imath] are all functions of two variables. | 970892 | Show Laplace operator is rotationally invariant
I'm trying to show the Laplace operator is rotationally invariant. Essentially this boils down to showing [imath]\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}[/imath] where [imath]u = x \cos \theta + y \sin \theta[/imath] [imath]v = -x \sin \theta + y \cos \theta[/imath] I think I'm on the right track by noting that [imath]\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial u}\left(\frac{\partial f}{\partial x}\right)\frac{\partial u}{\partial x} + \frac{\partial}{\partial v}\left(\frac{\partial f}{\partial x}\right)\frac{\partial v}{\partial x}[/imath] but I'm having difficulty reaching an end game where I show [imath]\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2}({\sin}^2 \theta + {\cos}^2 \theta) + \frac{\partial^2 f}{\partial v^2}({\sin}^2 \theta + {\cos}^2 \theta)[/imath] |
2634612 | If [imath]d_1 =1, d_2, \cdots, d_k = n[/imath] are positive divisors of natural [imath]n[/imath], show [imath](d_1d_2 \cdots d_k)^2 = n^k[/imath].
If [imath]d_1 =1, d_2, \cdots, d_k = n[/imath] are the positive divisors of the natural number [imath]n[/imath], show that [imath](d_ld_2 \cdots d_k)^2 = n^k[/imath]. Seems a confusing question with no place to start with, please help. I see no approach except induction to work. Apart from induction, if some way to bring up a proof, then would be quite fine. | 1126001 | Product of Divisors of some [imath]n[/imath] proof
The function [imath]d(n)[/imath] gives the number of positive divisors of [imath]n[/imath], including n itself. So for example, [imath]d(25) = 3[/imath], because [imath]25[/imath] has three divisors: [imath]1[/imath], [imath]5[/imath], and [imath]25[/imath]. So how do I prove that the product of all of the positive divisors of [imath]n[/imath] (including [imath]n[/imath] itself) is [imath]n^{\frac{d(n)}{2}}[/imath]. For example, the divisors of [imath]12[/imath] are [imath]1[/imath], [imath]2[/imath], [imath]3[/imath], [imath]4[/imath], [imath]6[/imath], and [imath]12[/imath]. [imath]d(12)[/imath] is [imath]6[/imath], and [imath]1 · 2 · 3 · 4 · 6 · 12 = 1728 = 12^3 = 12^{\frac{6}{2}} = 12^{\frac{d(n)}{2}}[/imath] |
572433 | prove that: [imath]\lim_{x \to \infty} [f(x+1)-f(x)] = 0[/imath] just by using definitions of limit and definition of derivative.
Let f(x) be a differentiable function such that that [imath]\lim_{x\to \infty} f '(x) = 0[/imath]. I have to prove that: [imath]\lim_{x\to \infty} [f(x+1)-f(x)] = 0 [/imath] just by using definitions of limit and definition of derivative. I have no idea how to begin this...any hints? i found some posts similar to this but i need a more specific explanation.. any help ? I would be grateful. EDIT may i use MVT in some specific space? EDIT2 : im yet confused. is it better to use MVT insted of intermediate value theorem? EDIT3 : still cant get to a conclusion ..... | 179067 | Prove that: [imath]\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' [/imath]
I conjecture that in some specific conditions a differentiating function gives the following equality: [imath]\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' [/imath] However, I'm not sure yet what exactly those conditions are in order to precisely know where I may apply this rule or not. If you wanna take a look over my posted problem here you'll immediately notice that this rule applies for that case. I really appreciate if you help me clarify this. |
2634652 | Continuity and Compactness, Is this mapping onto? (Rudin)
Here the point [imath]p=(1,0) \in Y[/imath] but there is no [imath]t \in X[/imath] s.t. [imath]f(t)=p[/imath], is it still correct to say that [imath]f[/imath] is mapping of [imath]X[/imath] onto [imath]Y[/imath]? Also, can we say the reason why [imath]f^{-1}[/imath] fails to be continuous is that simply [imath]X[/imath] has no such [imath]t[/imath] so that the point [imath]p[/imath] is not mapped to anything? | 2096563 | Example 4.21 in Baby Rudin: Why is the inverse of this function not continuous at this point?
Here's Theorem 4.17 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: Suppose [imath]f[/imath] is a continuous 1-1 mapping of a compact metric space [imath]X[/imath] onto a metric space [imath]Y[/imath]. Then the inverse mapping [imath]f^{-1}[/imath] defined on [imath]Y[/imath] by [imath] f^{-1}( f(x) ) = x \ \ \ \ \ \ \ ( x \in X ) [/imath] is a continuous mapping of [imath]Y[/imath] onto [imath]X[/imath]. Now in Example 4.21 Rudin illustrates that compactness of [imath]X[/imath] is essential in Theorem 4.17. So here is Example 4.21: Let [imath]X[/imath] be the half open interval [imath][0, 2 \pi )[/imath] on the real line, and let [imath]\mathbf{f}[/imath] be the mapping of [imath]X[/imath] onto the circle [imath]Y[/imath] consisting of all points whose distance from the origin is [imath]1[/imath], given by [imath] \mathbf{f}(t) = ( \cos t, \sin t ) \ \ \ \ \ \ ( 0 \leq t < 2 \pi). [/imath] Then this function [imath]\mathbf{f}[/imath] is of course continuous and injective. But Rudin asserts that the inverse function is discontinuous at the point [imath](1, 0) = \mathbf{f}(0)[/imath]. How to demonstrate this fact rigorously? So far in the book, Rudin has not discussed the functions [imath]\cos[/imath] and [imath]\sin[/imath]. So can we come up with an example that uses only what Rudin has discussed? |
2626949 | Find [imath]\lim_{n\to \infty} \big[\frac{1}{1\cdot3}+\frac{1}{2\cdot5}+...+\frac{1}{n\cdot(2n+1)}\big][/imath]
Find [imath]\displaystyle\lim_{n\to \infty} \bigg[\frac{1}{1\cdot3}+\frac{1}{2\cdot5}+...+\frac{1}{n\cdot(2n+1)}\bigg][/imath] . This is a similar question as I've asked an hour ago . Find [imath]\lim_{n\to \infty} (1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})[/imath] But here, I am not able to rearrange terms to use the fact that [imath]\gamma_{n}=\sum_{n=1}^n \frac{1}{n} -\log(n)[/imath] . Any ideas ? Or Is there any other way to find limit? | 1334870 | How to find sum of the infinite series [imath]\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}[/imath]
[imath]\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots [/imath] How to find sum of this series? I tried this: its [imath]$n$[/imath]th term will be = [imath]\frac{1}{n}-\frac{2}{2n+1}[/imath]; after that I am not able to solve this. |
2635388 | Is there an entire function satisfying [imath]|f(z)|=|z|+1[/imath] for every [imath]|z|\geq2017[/imath]?
Is there a holomorphic function [imath]f \in Hol(\Bbb C)[/imath] such that [imath]|f(z)|=|z|+1[/imath] for every [imath]|z|\geq2017[/imath] ? I tried defining [imath]g(z)=\frac{1}{f(z)}[/imath]. Such [imath]g[/imath] is clearly holomorphic and bounded for [imath]|z| \geq 2017[/imath]. I want to show that [imath]g[/imath] is holomorphic in [imath]|z| \leq 2017[/imath] and then conclude that [imath]g[/imath] is constant. For that I want to show that [imath]f[/imath] has no zeros in [imath]|z| \leq 2017[/imath]. I tried to prove it by Rouche theorem, but didn't succeed. Any ideas? Edit: the post similar to mine doesn't solve my problem, because the ending of the proof is not clear. Where is the contradiction? | 2144104 | Does there exists an entire function such that for each [imath]|z|>C[/imath] for some constant [imath]C[/imath], [imath]|f(z)|=|z|+1[/imath]?
Does there exists an entire function such that for each [imath]|z|>C[/imath] for some constant [imath]C[/imath], [imath]|f(z)|=|z|+1[/imath]? Can I assume that since [imath]f[/imath] has a pole in infinity, [imath]Imf[/imath] or [imath]Ref[/imath] must have a pole at infinity too? If so, I can use this fact to define an exponent function of [imath]Imf[/imath] or [imath]Ref[/imath] (the one that goes to infinity) or their minus, and show that function is a bounded entire function, and therefore constant. Which gives a contradiction to the conditions about [imath]f[/imath]. |
2635460 | What is the average value of the determinant of matrices in this set?
State true or false. Let [imath]n \geq 2[/imath] be a natural number. Let [imath]S[/imath] be the set of all [imath]n \times n[/imath] real matrices whose entries are only [imath]0, 1[/imath] or [imath]2[/imath]. Then the average of determinants of matrices in [imath]S[/imath] is greater than or equal to [imath]1[/imath]. This is how I think the answer should go, but I am not fully convinced. False. The average value is zero. There are only finitely many matrices in [imath]S[/imath], and we can exhibit a one-to-one correspondence between the matrices with non-zero determinants as follows: given [imath]A \in S[/imath] with positive determinant, interchanging the first two columns of [imath]A[/imath] gives us a matrix in [imath]S[/imath] with negative determinant. As this function has an inverse (itself) it is bijective. Thus, the average value of the determinant of matrices in [imath]S[/imath] is zero. | 2055403 | The average determinant of all integer matrices with coefficients [imath]0,1,2[/imath]
Let [imath]S[/imath] denote the set of [imath]A \in M(n,\mathbb R)[/imath] such that every entry of [imath]A[/imath] is either of [imath]0[/imath], [imath]1[/imath] or [imath]2[/imath], then is it true that [imath]\sum_{A \in S} \det A \ge 3^{n^2}\ ?[/imath] |
2635592 | Find the cardinality of this set: [imath]\{x \subseteq \mathbb R : |x| < \aleph_o\}[/imath]
Find the cardinality of set [imath]A[/imath]: [imath]A = \{x \subseteq \mathbb R : |x| < \aleph_o \}[/imath] I have thought about it for a while and I've come to the conclusion that the cardinality of this set will be at least continuum, because every real number singleton is in this set. However, the only upper-bound of this cardinaltiy that I can think of is [imath]\mathbb R^{\mathbb R}[/imath] whose cardinality is more than continuum. Any suggestions and hints would be most appreciated. | 2057826 | Cardinality of the Set of all finite subset of [imath]\mathbb{R}[/imath]
Find the Cardinality of the set of all finite subsets of [imath]\mathbb{R}[/imath]. I have proved that the set of all finite subsets of [imath]\mathbb{N}[/imath] is countable . But I cannot find the cardinality of the set in case of [imath]\mathbb{R}[/imath] . My Attempt: First I have considered the set [imath]A_k=\{\{a_1, a_2, ..., a_k\}| a_i \in \mathbb{R} \ and\ a_1<...<a_k \}[/imath] Then [imath]S=The set \ of all\ finite \ subsets\ of\ \mathbb{R}=\bigcup_{n=1}^{\infty}A_{n}[/imath] and Now in the case of [imath]\mathbb{N}[/imath] I could show that this [imath]A_k[/imath] is countable and consequently the set of all finite subset of [imath]\mathbb{N}[/imath] is countable. But in this case I cannot say anything like this...In this case the set will be of the cardinality as that of [imath]\mathbb{R}[/imath] (I think) ..But cannot prove it... Please Help. |
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