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Series expansion of Coth(pi)

  • Thread starter Ratpigeon
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  • #1
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Homework Statement



show coth([itex]\pi[/itex])=1/[itex]\pi[/itex] (1+2 [itex]\sum[/itex][itex]\infty[/itex]n=1 (1/(1+n^2)

Homework Equations


The fourier expansion of e^x is
Sinh([itex]\pi[/itex])/[itex]\pi[/itex](1+[itex]\sum[/itex] 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/[itex]\pi[/itex] (1+2 [itex]\sum[/itex][itex]\infty[/itex]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...
Please help - I've spent about fifteen pages on the algebra for this and I'm getting really sick of the stupid question...
 

Answers and Replies

  • #2
vela
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Homework Statement



show coth([itex]\pi[/itex])=1/[itex]\pi[/itex] (1+2 [itex]\sum[/itex][itex]\infty[/itex]n=1 (1/(1+n^2)

Homework Equations


The fourier expansion of e^x is
Sinh([itex]\pi[/itex])/[itex]\pi[/itex](1+[itex]\sum[/itex] 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/[itex]\pi[/itex] (1+2 [itex]\sum[/itex][itex]\infty[/itex]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...
Please help - I've spent about fifteen pages on the algebra for this and I'm getting really sick of the stupid question...
How'd you come up with ##\coth \pi = 1+\frac{e^{-\pi}}{\sinh \pi}##?
 
Last edited:
  • #3
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Coth([itex]\pi[/itex]) =(e[itex]\pi[/itex]+e-[itex]\pi[/itex])/(e[itex]\pi[/itex]-e-[itex]\pi[/itex])

which equals
1+2e-[itex]\pi[/itex]/(e[itex]\pi[/itex]-e-[itex]\pi[/itex])
which equals
1+2e^pi/2sinh(pi)
 
  • #4
gabbagabbahey
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The fourier expansion of e^x is
Sinh([itex]\pi[/itex])/[itex]\pi[/itex](1+[itex]\sum[/itex] 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 [itex]e^x[/itex]?
 
Last edited:
  • #5
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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)
 
  • #6
vela
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Hint: What does your Fourier series converge to when you set ##x=\pi##? Keep in mind there's a discontinuity there.
 
  • #7
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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)?
 
  • #8
vela
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I don't know what you mean by "Sn(+/-pi)", but yeah, it'll be the average.
 
  • #9
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And that lets 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?
 
  • #10
vela
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Here's another hint: What's ##\cosh \pi## equal to?
 
  • #11
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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!
:)
 

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