# Series expansion of Coth(pi)

• Ratpigeon
In summary, the Fourier series expansion of e^x is Sinh(\pi)/\pi(1+\sum 2(-1)^m/(1+n^2) (cos(mx)-n sin(nx)). But this is wrong because of the one out the front - I know that needs to stay there, which means I somehow need to stick a -sinh(pi) into the front of my Fourier series of e^pi. I would say that it would be my term for n= zero, but I already took care of that one, which was sinh(pi)/pi... So, I need to somehow sub in e^-pi for n=0 in order for my equation to

## Homework Statement

show coth($\pi$)=1/$\pi$ (1+2 $\sum$$\infty$n=1 (1/(1+n^2)

## Homework Equations

The Fourier expansion of e^x is
Sinh($\pi$)/$\pi$(1+$\sum$ 2(-1)^m/(1+n^2) (cos(mx)-n sin(nx)

## The Attempt at a Solution

I subbed in
Coth(Pi)=1+e^-pi/sinh(pi) =1+1/$\pi$ (1+2 $\sum$$\infty$n=1 (1/(1+n^2)

But this is wrong because of the one out the front - I know that needs to stay there, which means I somehow need to stick a -sinh(pi) into the front of my Fourier series of e^pi I would say that it would be my term for n= zero, but I already took care of that one, which was sinh(pi)/pi...

Ratpigeon said:

## Homework Statement

show coth($\pi$)=1/$\pi$ (1+2 $\sum$$\infty$n=1 (1/(1+n^2)

## Homework Equations

The Fourier expansion of e^x is
Sinh($\pi$)/$\pi$(1+$\sum$ 2(-1)^m/(1+n^2) (cos(mx)-n sin(nx)
You're missing closing parentheses and have two index variables, m and n, so I'm not sure what you meant to write exactly. Did you mean
$$\frac{\sinh \pi}{\pi}\left(1+\sum_{n=1}^\infty \frac{2(-1)^n}{1+n^2}(\cos nx - n\sin nx)\right)?$$

## The Attempt at a Solution

I subbed in
Coth(Pi)=1+e^-pi/sinh(pi) =1+1/$\pi$ (1+2 $\sum$$\infty$n=1 (1/(1+n^2)

But this is wrong because of the one out the front - I know that needs to stay there, which means I somehow need to stick a -sinh(pi) into the front of my Fourier series of e^pi I would say that it would be my term for n= zero, but I already took care of that one, which was sinh(pi)/pi...
How'd you come up with ##\coth \pi = 1+\frac{e^{-\pi}}{\sinh \pi}##?

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Coth($\pi$) =(e$\pi$+e-$\pi$)/(e$\pi$-e-$\pi$)

which equals
1+2e-$\pi$/(e$\pi$-e-$\pi$)
which equals
1+2e^pi/2sinh(pi)

Ratpigeon said:
The Fourier expansion of e^x is
Sinh($\pi$)/$\pi$(1+$\sum$ 2(-1)^m/(1+n^2) (cos(mx)-n sin(nx)

As vela pointed out, this makes no sense. What is the actual Fourier series expansion of $e^x$?

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It's the one that vela used - I usually use n's but the lecturer uses m's so it's a bit messed up, and they are both standing for the same thing. And the e^pi is a typo - I have it right in my working...
I've found another page that uses this as the Fourier expansion of e^x, so I'm fairly sure that it's right (after making all the indeces the same)

Hint: What does your Fourier series converge to when you set ##x=\pi##? Keep in mind there's a discontinuity there.

It'll be... halfway between the limit of the Fourier series at Pi and -pi?
Which means...
1/2(e^pi+e^-pi)=Sn(+/-pi)?

I don't know what you mean by "Sn(+/-pi)", but yeah, it'll be the average.

And that let's me calculate e^pi+e^-pi from my Fourier series at Pi; but how do I get e^pi-e^-pi for the denominator then?

Here's another hint: What's ##\cosh \pi## equal to?

wait, no, I got it. The denominator is 2 sinh(pi) and the numerator is twice the Fourier series at Pi - which is sinH(pi) multiplied by the series I want.
Thanks!
:)