Complex Fourier Series of f: Evaluating Parseval's Relation

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Homework Help Overview

The discussion revolves around finding the complex Fourier series of the function f(t) = e^{-t} defined on the interval [-π, π) and extended to be 2π-periodic. Participants are also exploring the application of Parseval's relation to evaluate a specific infinite sum.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the Fourier coefficients and the simplification of the resulting expressions. Questions arise about the application of Parseval's relation and how to evaluate the infinite sum based on the results obtained.

Discussion Status

There is an ongoing exploration of the mathematical expressions involved, with some participants providing insights and identities that may assist in simplification. However, there is no explicit consensus on the evaluation of the sum or the application of Parseval's relation to this specific problem.

Contextual Notes

Participants mention potential issues with algebraic manipulation and the clarity of the problem setup, indicating that there may be some confusion or missing information regarding the integral and the sum involved.

AkilMAI
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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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You can use

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

Not sure if it would really help much though.
 
maybe I did something wrong with the integral
 
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].
 
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?
 
What does Parseval's relation tell you?
 
that the integral of the square of a function is equal to the sum of the square of its transform...?
 
So what do you get when you apply it to this particular problem?
 
[tex]\frac{ \frac{\pi^{2}}{4}*cosh(iK)^{2}}{k^{2} +1}=\frac{1}{2\pi}*(e^{-2\pi} - e^{2\pi})[/tex]
 
Last edited:
  • #10
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
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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