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.