Partial sum of Fourier Coefficients

In summary: I'm pretty sure that the series wouldn't converge at x=pi because that is where the discontinuity is.
  • #1
8
0

Homework Statement



f (x) = 0 -pi<x<0

x^2 0<x<pi

Find the Fourier series and use it to show that

(pi^2)/6=1+1/2^2+1/3^2+...

Homework Equations




N/A

The Attempt at a Solution



I was able to find the Fourier series and my answer matched with the back of the book. But I don't understand how I am supposed to use it to prove the expansion of pi^2/6 . I tried plugging in (pi^2)/6 for f(x) but that didn't work out.


Thanks!
 
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  • #2
What do you find to be the Fourier series for f(x)?
 
  • #3
Just to echo nicksauce, if you found the Fourier series as you said, what is it?
 
  • #4
I didn't want to type out the Fourier series since I am really new to latex but here it is

f(x) =pi^2/6 +[tex]\Sigma[/tex] { (2(-1)^n/n^2) cos nx + ((-1)^(n+1) pi/n + 2/(pi * n^3) [(-1)^n-1] sin nx }

I hope that made sense and that I didn't make any typos.
 
  • #5
P.S: and n goes from 1 to infinity
 
  • #6
Thanks. Since the mean value of the function over the interval is pi^2/6, I was hoping I could just say put x=0. But you've got that alternating sign and factor of 2 in the cosine part. Are you sure of the series? Otherwise, I'll have to work it out. Didn't want to do that.
 
  • #7
I am pretty sure about the series. I verified with the back of the book. We were also supposed to show pi^2/12=1-1/2^2+ etc. I plugged in x=o and was able to prove it.

For the pi^2/6, I tried setting x to be pi/2 so that the cosine term goes away but that didn't work out.
 
  • #8
It looks to me like you should be putting x=pi. That turns cos(nx) into (-1)^n makes the sin(nx) vanish. I'm trying to check your series. I get something a lot like it. But I can't seem to get all the parts quite right. Guess I'm not very good at this...
 
  • #9
Doh. I've been being stupid. Regarded as a periodic function f(x) is discontinuous at x=pi. The series doesn't converge to f(pi). It converges to (f(pi)+f(-pi))/2. Or pi^2/2. Do you know why? Now try x=pi in your series.
 

1. What is the partial sum of Fourier coefficients?

The partial sum of Fourier coefficients is a mathematical concept used in Fourier analysis to approximate a periodic function by a finite sum of trigonometric functions. It is also known as the truncated Fourier series.

2. How is the partial sum of Fourier coefficients calculated?

The partial sum of Fourier coefficients is calculated by taking a finite number of terms from the Fourier series expansion of a periodic function. It involves finding the coefficients that best fit the given function and adding them up to form the partial sum.

3. What is the significance of the partial sum of Fourier coefficients?

The partial sum of Fourier coefficients allows us to approximate a function with a finite number of terms rather than an infinite series. This makes it a useful tool for analyzing and understanding the behavior of periodic functions.

4. How does the number of terms in the partial sum of Fourier coefficients affect the accuracy of the approximation?

The more terms we include in the partial sum, the more accurate the approximation will be. However, as the number of terms increases, the computation becomes more complex and may not always be feasible. Therefore, a trade-off between accuracy and complexity must be considered.

5. Can the partial sum of Fourier coefficients be used for non-periodic functions?

No, the partial sum of Fourier coefficients is only applicable to periodic functions. For non-periodic functions, other methods such as the Fourier transform must be used for approximation and analysis.

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