Fourier series

  • Thread starter AkilMAI
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  • #1
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Define [tex]f(t)=e^{-t}[/tex] ont he interval [tex][-\pi,\pi)[/tex],and extend f to [itex]2\pi[/itex]-periodic.Find the complex
Fourier series of f.Then, apply Parseval's relation to f to evaluate
[tex]\sum^{\infty}_0 \frac{1}{1+k^{2}}[/tex]

For the first part when I calculate c_k [tex]\frac{1}{2\pi}\ \int^\pi_{-\pi} e^{-ikt-t}dt[/tex].....I get the following [tex]\frac{-e^{-(ik+1)\pi} + e^{(ik+1)\pi}}{2\pi(ik+1)}[/tex]...is there any way to simplify it?Also for the second part,how can I apply Parseval's relation to evaluate the sum?
Thanks in advance
 

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  • #2
vela
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You can use

[tex]\sinh x = \frac{e^x-e^{-x}}{2}[/tex]

Not sure if it would really help much though.
 
  • #3
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maybe I did something wrong with the integral
 
  • #4
vela
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You have

[tex]\frac{-e^{-(ik+1)\pi} + e^{(ik+1)\pi}}{2\pi(ik+1)} = \frac{\sinh \pi(1+ik)}{\pi(1+ik)}[/tex]

Use the identity [itex]\sinh(a+b) = \sinh a \cosh b + \cosh a \sinh b[/itex].
 
  • #5
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thanks vela.... plugging that result in the Parseval’s relation provides me with another result...the questions is how can I use it to evaluate the sum from above?
 
  • #6
vela
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What does Parseval's relation tell you?
 
  • #7
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that the integral of the square of a function is equal to the sum of the square of its transform...?
 
  • #8
vela
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So what do you get when you apply it to this particular problem?
 
  • #9
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[tex]\frac{ \frac{\pi^{2}}{4}*cosh(iK)^{2}}{k^{2} +1}=\frac{1}{2\pi}*(e^{-2\pi} - e^{2\pi}) [/tex]
 
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  • #10
vela
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Check your algebra. You should have a factor of [itex]\sinh \pi[/itex] in there, and you can simplify [itex]\cosh ik[/itex]. Also, where's the sum?
 
  • #11
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sorry I do have a pi in there ,latex typo also is the forum,the loading time is very high...yes the sum is in the l.h.s.....
 

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