Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For which real numbers $x$ does the infinite sum $\sin(x)+\sin^2(2x)+\sin^3(3x)+ \cdots+\sin^n(nx)+\cdots$ converge?

How does its graph look?

share|improve this question
2  
Possible duplicate –  Envious Page Aug 26 '12 at 2:37
    
Well, at least for $\,x=2k\pi\,\,,\,\,k\in\Bbb Z\,$...:) –  DonAntonio Aug 26 '12 at 2:38
    
But not for $x = \pi q$ where $q$ is rational but not an integer. –  nayrb Aug 26 '12 at 2:39
    
@hjg I notice you are not registered. I recommend registering; it doesn't take long and helps you keep track of questions while allowing you to earn more privileges on the site. –  Alex Becker Aug 26 '12 at 5:07

2 Answers 2

up vote 1 down vote accepted

Let $x=r\pi$ and consider the following cases:

  1. $r=\frac{a}{2b}$: For infinitely many $n$ we have $n=(2m+1)b$ thus $\sin^n(nx)=\sin^n\left(\frac{(2m+1)a}{2}\pi\right)=\pm 1$ so the sequence does not go to $0$ hence the series does not converge.
  2. $r=\frac{a}{2b+1}$: For any $m$ we have $\left|nx-\frac{2m+1}{2}\pi\right|=\pi\left|\frac{2na-(2m+1)(2b+1)}{4b+2}\right|\geq \frac{\pi}{4b+2}$ and so $|\sin(nx)|$ is bounded above by $\lambda=\sin\left(\frac{2b}{2b+1}\pi\right)<1$, hence we have $$\sum\limits_{n=1}^\infty |\sin^n(nx)|<\sum\limits_{n=1}^\infty \lambda^n=\frac{\lambda}{1-\lambda}$$ so the series converges absolutely.
  3. $r$ is irrational: This is more difficult. Let $[r]$ denote the fractional part of $r$. The key is to examine terms which are nearly half-integers. Let $N=\lfloor1/[r]\rfloor$. Consider the points $[r],[2r],\ldots,[nr]$ in $[0,1)$ ordered as real numbers. It is a theorem that there exist $\alpha>\beta>\gamma$ such that the distance between any two consecutive points is one of these three, and that the number of pairs with distance equal to each of these values differs by at most $N$. In particular, there are at least $\frac{n-2N}{3}$ pairs a distance $\alpha$ apart, so $\alpha<\frac{3}{n-2N}$. Thus by the pigeonhole principle, we have some $m\leq n$ such that $\left|[mr]-\frac12\right|<\frac{3}{n-2N}$. For such $m$ we have $$|\sin(mx)|>\left|\sin\left(\frac{(2k+1)\pi}{2}-\frac{3}{n-2N}\right)\right|=\cos\left(\frac{3}{n-2N}\right)>1-\frac{9}{2(n-2N)^2}$$ thus we get $$|\sin^m(mx)|>\left(1-\frac{9}{2(n-2N)^2}\right)^m>1-\frac{9m}{2(n-2N)^2}\geq 1-\frac{9n}{2(n-2N)^2}\to 1$$ and so the sequence does not converge to $0$, hence the series does not converge.

Since this does series not converge except on a set of measure $0$, I'm not sure how to meaningfully describe its graph. However, if you want to compute its value for $r=\frac{a}{2b+1}$ you can break the series up into sums over equivalence classes $\bmod 2b+1$ using absolute convergence, which gives $$\begin{align} \sum_{n=1}^\infty \sin^n(nx)&=\sum_{i=1}^{2b+1}\sum_{j=0}^\infty(-1)^j\sin^{j(2b+1)+i}\left(\frac{a+i}{2b+1}\pi\right)\\ &=\sum_{i=1}^{2b+1}\sin^i\left(\frac{a+i}{2b+1}\pi\right)\sum_{j=0}^\infty(-1)^j\sin^{j(2b+1)}\left(\frac{a+i}{2b+1}\pi\right)\\ &=\sum_{i=1}^{2b+1}\sin^i\left(\frac{a+i}{2b+1}\pi\right)\frac{1}{1+\sin^{(2b+1)}\left(\frac{a+i}{2b+1}\pi\right)}\\ \end{align}$$ which, while not the prettiest, is at least a finite sum.

share|improve this answer

Hint: Let $x=r \pi$.

If $r$ is irrational, then the set $nx -2k\pi$ is dense in $\mathbb{R}$. It is easy to get from here that $\lim_n \sin(nx)$ does not exist....

If $r=\frac{m}{k}$ with $gcd(m,k)=1$ and $\lim_n \sin(n\frac{m}{k} \pi)=0$ what can you say about $k$?

share|improve this answer
    
The OP edited the question, so unfortunately this no longer answers it. –  Alex Becker Aug 26 '12 at 5:08
    
At least this time I didn't get any negative votes for this :) –  N. S. Aug 26 '12 at 12:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.