Value of Fourier Series from Two Equations

[Incand]

Homework Statement

Derive
##\sum_{n=1}^\infty \frac{1}{n^2+b^2} = \frac{\pi}{2b}\coth b\pi - \frac{1}{2b^2}##

from either
##e^{b\theta} = \frac{e^{2\pi b}-1}{2\pi} \sum_{-\infty}^\infty \frac{e^{in\theta}}{b-in}##
for ##0 < \theta < 2\pi##.

or
##e^{b\theta} = \frac{\sinh b\pi}{\pi}\sum_{-\infty}^\infty \frac{(-1)^n}{b-in}e^{in\theta}##
for ##-\pi < \theta < \pi##.

Homework Equations

N/A

The Attempt at a Solution

To get the series to be from ##1\to \infty## it seems a good idea to set ##\theta=\pi## which means we have to use the first equation. We get
##e^{b\pi } = \frac{e^{2\pi b}-1}{2\pi}\sum_{-\infty}^\infty \frac{e^{in\pi}}{b-in} =\frac{e^{2\pi b}-1}{2\pi} \sum_{-\infty}^\infty \frac{\cos (n\pi)(b+in)}{b^2+n^2} ##
from that all ##\sin n\pi = 0 \; \forall n \in Z##. Moving thing around and realising that ##\cos (n\pi) = (-1)^n## we get
##\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} = \sum_{-\infty}^\infty \frac{(-1)^n(b+in)}{b^2+n^2} = b\sum_{-\infty}^\infty \frac{(-1)^n}{b^2+n^2}##
since the terms with ##in## all cancel our and are zero for ##n=0##.

We realize that every term with ##n## is equal to the term with ##-n## and we get
##\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} =\frac{1}{b}+ 2b\sum_1^\infty \frac{1}{n^2+b^2}##
where we get ##\frac{1}{b}## from ##n=0##. The sum of the series is then
##\sum_1^\infty \frac{1}{n^2+b^2} = \frac{\pi e^{b\pi }}{b(e^{2\pi b}-1)}-\frac{1}{2b^2}##
I can't get this to contain
##\coth b\pi = \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}}##.

 

[Dr. Courtney]

How can you say the relevant equations are not applicable?

 

[Incand]

How can you say the relevant equations are not applicable?

They're not? The exercise give a warning "be careful" that may allude to that they or one of them may not be applicable. I thought since choosing ##\theta = \pi## seemed like a good idea to me and since ##0 < \pi < 2\pi## the first equation should be applicable?

Edit: I guess you meant why i wrote N/A. I just put them in the formulation of the exercise since they were that way in the book. I could've put them there instead ofcouse.

 

[vela]

$$e^{b\theta} = \frac{e^{2\pi b}-1}{2\pi} \sum_{-\infty}^\infty \frac{e^{in\theta}}{b-in}$$ for ##0 < \theta < 2\pi## or
$$e^{b\theta} = \frac{\sinh b\pi}{\pi}\sum_{-\infty}^\infty \frac{(-1)^n}{b-in}e^{in\theta}$$ for ##-\pi < \theta < \pi##.

I don't see how both of these can be right because
$$\frac{e^{2\pi b}-1}{2\pi} = \frac{e^{\pi b}(e^{\pi b}-e^{-\pi b})}{2\pi} = \frac{e^{\pi b}}{\pi} \sinh \pi b,$$ so for ##0 < \theta < \pi##, you'd have
$$e^{\pi b}\sum_{n=-\infty}^\infty \frac{e^{in\theta}}{b-in} = \sum_{n=-\infty}^\infty (-1)^n\frac{e^{in\theta}}{b-in}$$ Maybe that's true, but it doesn't look right to me. Consider the case when ##b=0##, for instance.

To get the series to be from ##1\to \infty##, it seems a good idea to set ##\theta=\pi##, which means we have to use the first equation. We get
$$e^{b\pi } = \frac{e^{2\pi b}-1}{2\pi}\sum_{-\infty}^\infty \frac{e^{in\pi}}{b-in} =\frac{e^{2\pi b}-1}{2\pi} \sum_{-\infty}^\infty \frac{\cos (n\pi)(b+in)}{b^2+n^2}$$
from that all ##\sin n\pi = 0 \; \forall n \in Z##. Moving thing around and realising that ##\cos (n\pi) = (-1)^n##, we get
$$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} = \sum_{-\infty}^\infty \frac{(-1)^n(b+in)}{b^2+n^2} = b\sum_{-\infty}^\infty \frac{(-1)^n}{b^2+n^2}$$ since the terms with ##in## all cancel our and are zero for ##n=0##.

We realize that every term with ##n## is equal to the term with ##-n##, and we get
$$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} =\frac{1}{b}+ 2b\sum_1^\infty \frac{1}{n^2+b^2}$$ where we get ##\frac{1}{b}## from ##n=0##.

You should have
$$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} =\frac{1}{b}+ 2b\sum_{n=1}^\infty \frac{(-1)^n}{n^2+b^2}.$$ You changed the form of the summand when you dropped the alternating signs.

The sum of the series is then
$$\sum_1^\infty \frac{1}{n^2+b^2} = \frac{\pi e^{b\pi }}{b(e^{2\pi b}-1)}-\frac{1}{2b^2}.$$ I can't get this to contain ##\coth b\pi = \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}}##.

 

[Incand]

I don't see how both of these can be right because
$$\frac{e^{2\pi b}-1}{2\pi} = \frac{e^{\pi b}(e^{\pi b}-e^{-\pi b})}{2\pi} = \frac{e^{\pi b}}{\pi} \sinh \pi b,$$ so for ##0 < \theta < \pi##, you'd have
$$e^{\pi b}\sum_{n=-\infty}^\infty \frac{e^{in\theta}}{b-in} = \sum_{n=-\infty}^\infty (-1)^n\frac{e^{in\theta}}{b-in}$$ Maybe that's true, but it doesn't look right to me. Consider the case when ##b=0##, for instance.

I'm pretty sure they're both right. I checked the book that I wrote them correctly (Folland, Fourier analysis and it's applications). It doesn't say that b is restricted in anyway but if we choose ##b=0## in the first equation we end up with ##1 =0## so perhaps ##b \ne 0##. I think you end up with a special case when deriving the coefficients for ##b=0##. Something along the lines of ##c_n = \frac{1}{2\pi} \int_{-\pi}^\pi e^{inx}dx = \frac{-e^{-in\pi}+e^{in\pi}}{2in\pi} = \frac{1}{2\pi} \sin n\pi##. (complex sine function). So you get other ##c_n## when ##b = 0## but I only did this real quick so may be wrong. Actually thinking about this made me come up with the solution to the original problem, cheers!

$$\frac{2\pi e^{b\pi }}{e^{2\pi b}-1} =\frac{1}{b}+ 2b\sum_{n=1}^\infty \frac{(-1)^n}{n^2+b^2}.$$ You changed the form of the summand when you dropped the alternating signs.

Thanks! You(we) actually proved another execise in the book by mistake!

I think I got it now, posting the solution if anyone is interested. Since the functions are discontinuous I can't input ##0## or ##\pi## in the wrong equation to cancel our the alternatiing ##(-1)^n## and I have to use the convergence theorem:
If ##f## is ##2\pi##-periodic and piecewise smooth we have (limit of partial sum)
##\lim_{N \to \infty} S_N^f (\theta ) = \frac{1}{2} \left[ f(\theta - ) + f(\theta +) \right]##
So If we pick ##\theta = 0## in the first equation we end up with
##\frac{e^{b2\pi}+e^{0}}{2} = \frac{e^{2\pi b}-1}{2\pi } \sum_{-\infty}^\infty \frac{b+in}{b^2+n^2}##
Rearanging we get
##b\sum_{-\infty}^\infty \frac{1}{n^2+b^2} = \frac{4\pi (e^{\pi b}+1)}{e^{2\pi b}-1}##
And summing from ##1 \to \infty## instead we get
##\frac{1}{b} + 2b\sum_1^\infty \frac{1}{n^2+b^2} = \frac{\pi (e^{2\pi b}+1)}{e^{2\pi b}-1}##
and we get
##\sum_1^\infty \frac{1}{n^2+b^2} = -\frac{1}{2b^2} + \frac{\pi (e^{2\pi b}+1)}{2b(e^{2\pi b}-1)} = -\frac{1}{2b^2} + \frac{\pi}{2b} \coth b\pi##

 

[vela]

I'm pretty sure they're both right. I checked the book that I wrote them correctly (Folland, Fourier analysis and it's applications). It doesn't say that b is restricted in anyway but if we choose ##b=0## in the first equation we end up with ##1 =0## so perhaps ##b \ne 0##.

##b=0## was a poor choice because that would make ##\sinh \pi b=0##, so I effectively divided by 0. I suspect with the first one, you end up having to take a limit as ##b\to 0## to evaluate the righthand side. While ##\sinh \pi b \to 0##, the series probably diverges in such a way that the product is goes to 1.