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How can I prove the following using integration and elementary functions?

Prove that:

$$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2}$$

$0 < \theta < 2\pi$

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Minus one for title and for no shown effort. –  k.stm May 7 '13 at 14:04
    
I changed the title and I will try to show some effort. –  please delete me May 7 '13 at 14:05
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I know that it is a simple Fourier series called sawtooth function but I want to do it using integration and elementary functions. –  please delete me May 7 '13 at 14:07
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Fourier series are much about integration, and every function here is elementary. So if you want specifically to avoid Fourier theory, it would be good to make that clear in your question. –  1015 May 7 '13 at 14:11
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Is this a highschool question? How did you come accross it? –  1015 May 7 '13 at 14:20

3 Answers 3

up vote 11 down vote accepted

Let, $$S_1 = \sum_{n=1}^{\infty}\frac{\cos n\theta}{n}\\ S_2 = \sum_{n=1}^{\infty}\frac{\sin n\theta}{n}$$

Then $$S_1 + iS_2 = \sum_{n=1}^{\infty}\frac{\cos(n\theta)+i\sin(n\theta)}{n}=\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}$$

Now, from the Taylor expansion, $\ln (1+x) = x -\frac{x^2}{2}+\frac{x^3}{3} ...$ $$\implies -\ln(1-x) = x+ \frac{x^2}{2}+\frac{x^3}{3} ... = \sum_{n=1}^{\infty}\frac{x^n}{n}$$ $$\begin{align} \therefore S_1+iS_2 &= -\ln(1-e^{i\theta}) \\&=-\ln(1-\cos\theta-i\sin \theta) \\ &=-\ln(2\sin^2\theta/2 - 2i\sin(\theta/2)\cos(\theta/2)) \\ &=-\ln(2\sin\theta/2)-\ln(\sin\theta/2-i\cos\theta/2) \\ &=-\ln(2\sin\theta/2)+\ln(\sin\theta/2+i\cos\theta/2) \\ &=-\ln(2\sin\theta/2)+\ln(e^{i(\pi/2-\theta/2)}) \\ &=-\ln(2\sin\theta/2)+i(\pi/2-\theta/2) \end{align} $$

Taking the imaginary part of both sides, $$S_2 = \frac{\pi}{2} - \frac{\theta}{2}$$

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You can do it a little more directly by using $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ –  Glen O May 7 '13 at 14:38
    
Yes, I'm not very sure how rigorous this is. With the complex logarithm and the Taylor expansion of natural logarithm, I too doubt this is high school level, but it was the simplest I could get it. –  Milind May 7 '13 at 14:39
    
Note in particular that you are on the boundary of the disk of convergence. Where convergence is the tricky part. –  1015 May 7 '13 at 14:40
    
Thanks a lot for the answer, it doesn't use integration though. –  please delete me May 7 '13 at 14:42
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@AlexanderJones: It can be converted to integration by integrating $\frac{1}{1-x}$... –  Aryabhata May 7 '13 at 14:45

In this answer, I show

$$ \begin{align} \sum_{k=1}^\infty\frac{\sin(2kx)}{k} &=\sum_{k=1}^\infty\frac{e^{i2kx}-e^{-i2kx}}{2ik}\\ &=\frac1{2i}\left(-\log(1-e^{i2x})+\log(1-e^{-i2x})\right)\\ &=\frac1{2i}\log(-e^{-i2x})\\[4pt] &=\frac\pi2-x\quad\text{for }x\in\left(0,\pi\right) \end{align} $$

which is, in essence, milind's answer. However, the question asks about integration. This sounds as if the question is asking to find the Fourier Series of $f(\theta)=\frac\pi2-\frac\theta2$. First, note that $f(\theta)$ is odd; that is, $$ \begin{align} f(2\pi-\theta) &=\frac\pi2-\frac{2\pi-\theta}2\\ &=\frac\theta2-\frac\pi2\\ &=-\left(\frac\pi2-\frac\theta2\right)\\[6pt] &=-f(\theta)\tag{1} \end{align} $$ Equation $(1)$ implies that $$ \begin{align} \color{#00A000}{\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta} &=\int_0^{2\pi}f(2\pi-\theta)\cos(n 2\pi-n\theta)\,\mathrm{d}\theta\\ &=-\color{#00A000}{\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta}\\[6pt] &=0\tag{2} \end{align} $$ because $\color{#00A000}{x}=-\color{#00A000}{x}\implies\color{#00A000}{x}=0$.

Now the question is $$ \begin{align} \int_0^{2\pi}f(\theta)\sin(n\theta)\,\mathrm{d}\theta &=\int_0^{2\pi}\left(\frac\pi2-\frac\theta2\right)\sin(n\theta)\,\mathrm{d}\theta\\ &=-\frac1n\int_0^{2\pi}\left(\frac\pi2-\frac\theta2\right)\,\mathrm{d}\cos(n\theta)\\ &=\left.-\frac1n\left(\frac\pi2-\frac\theta2\right)\cos(n\theta)\right]_0^{2\pi}\\ &\hphantom{=\,}+\frac1n\int_0^{2\pi}\cos(n\theta)\,\mathrm{d}\left(\frac\pi2-\frac\theta2\right)\\[4pt] &=\frac\pi{n}\tag{3} \end{align} $$ $(3)$ says that the Fourier series for $f(\theta)$ on $(0,2\pi)$ is $$ f(\theta)=\sum_{n=1}^\infty\frac{\sin(n\theta)}{n}\tag{4} $$

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Detail: the Fourier series converges to $(f(\theta^-)+f(\theta^+))/2$. That's $f(\theta)$ on $(0,2\pi)$, but $0$ at $0$ and $2\pi$. –  1015 May 7 '13 at 18:01
    
@julien: indeed, but the question only considers $0\lt\theta\lt2\pi$. –  robjohn May 7 '13 at 18:03
    
Yes, +1. I guess the simplest justification for pointwise convergence would be that $f$ is piecewise $C^1$ (here meaning continuously differentiable except at two points where right and left derivatives exist), which yields Dirichlet. –  1015 May 7 '13 at 18:11
    
@julien: Since $0\equiv2\pi$ on $\mathbb{R}/(2\pi\mathbb{Z})$ there is only one point of discontinuity, or are you talking about something else? –  robjohn May 8 '13 at 17:36
    
No, you're right. I somehow meant $0$ and $2\pi$, not considering the quotient. But that does not make much sense. –  1015 May 8 '13 at 17:39

$$ I=\sum_{n=1}^\infty \frac{\sin(n\theta)}{n} $$ Now, \begin{align} \frac{dI}{d\theta} &= \sum_{n=1}^\infty \cos(n\theta)\\ \frac{dI}{d\theta}\cos(\theta)&=\sum_{n=1}^\infty \cos(n\theta)\cos(\theta)\\ &=\sum_{n=1}^\infty \frac{\cos((n-1)\theta)+\cos((n+1)\theta)}2\\ &=\sum_{n=0}^\infty \frac{\cos(n\theta)}{2}+\sum_{n=2}^\infty\frac{\cos(n\theta)}{2}\\ &=\frac{1}2+\sum_{n=1}^\infty \frac{\cos(n\theta)}{2}+\sum_{n=1}^\infty\frac{\cos(n\theta)}{2}-\frac{\cos(\theta)}2\\ &=\frac{1-\cos(\theta)}{2}+\frac{dI}{d\theta}\\ \frac{dI}{d\theta}(\cos(\theta)-1) &= -\frac{\cos(\theta)-1}{2}\\ \frac{dI}{d\theta} &= -\frac12 \end{align} Noting that $$ I(\pi) = \sum_{n=1}^\infty \frac{\sin(n\pi)}n = 0 $$ we integrate around $\theta=\pi$ to get $$ I = -\frac\theta2 + \frac\pi2 = \frac\pi2-\frac\theta2 $$ Note that this doesn't strictly require that the $\cos$ sum converges, as we may alter the summation process to obtain convergence. What is important is which terms may be extracted for the purposes of the integration.


There are two obvious ways to handle the nonconvergent nature of the sum for $\frac{dI}{d\theta}$.

Option 1: use $$I = \sum_{n=1}^\infty \frac{\sin(n\theta)}n z^n$$ and then take the limit as $z\to1^{-}$. For any $|z|<1$, the sum in the derivative will converge, and in the limit it will be $\frac12$.

Option 2: Change the order of summation. Let $$ I = \sum_{n=1}^\infty \frac{\sin(n\theta)}n \sum_{k=1}^\infty 2^{-k} $$ which is the same as multiplying by $1$. Now change the order of summation to $n+k=m$ first, as $$ I = \sum_{m=2}^\infty \sum_{k=1}^{m-1} \frac{\sin((m-k)\theta)}{m-k}2^{-k} $$ When summed in this order, the derivative converges.

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I think it should be $I(\pi) = 0$, not $\pi/2$. –  Milind May 7 '13 at 15:41
    
@milind - good point. A minor typo (repeated when I said "around $\theta=\pi/2$"). Now fixed. –  Glen O May 7 '13 at 15:42
    
Your answer is a manipulation of a series which converges nowhere. It would be great if you explained what you mean by altering the summation process. –  1015 May 7 '13 at 18:16
    
@julien: I've edited in an explanation. –  Glen O May 7 '13 at 20:36
    
You realize that when there is no absolute convergence, all these manipulations are not "obvious", right? –  1015 May 7 '13 at 20:47

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