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Fourier Series

  • Thread starter Poirot
  • Start date
  • #1
94
2

Homework Statement


The following function is periodic between -π and π:

f(x) = |x|
Find the Coefficients of the Fourier series and, by examining the Fourier series at x=π or otherwise, determine:
1 + 1/32 + 1/52 + 1/72 ... = Σj=1 1/(2j - 1)2

Homework Equations


f(x) = a0/2 + ∑n=1 ancos(nx) + bn sin(nx)

a0 = 1/π ∫π f(x) dx
an = 1/π ∫π f(x) cos(nx) dx
bn = 1/π ∫π f(x) sin(nx) dx

The Attempt at a Solution


So I've found the coefficients:
a0 = π
an = -2(1 - (-1)n)/πn2
bn = 0 (as even function)

and so f(x) = π/2 + ∑-2(1 - (-1)n)/πn2 cos(nx)

I don't know how to do the last bit however, I don't really understand if I'm meant to come out with a number or something...
Any help would be greatly appreciated thank you.
 

Answers and Replies

  • #2
stevendaryl
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Science Advisor
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and so f(x) = π/2 + ∑-2(1 - (-1)n)/πn2 cos(nx)

I don't know how to do the last bit however, I don't really understand if I'm meant to come out with a number or something...
Any help would be greatly appreciated thank you.
Yes. Just plug [itex]x=\pi[/itex] into your expression. (You can see that when [itex]n[/itex] is even, you get 0, so you only have to consider the case for [itex]n[/itex] is odd.)
 
  • #3
Samy_A
Science Advisor
Homework Helper
1,241
510

Homework Statement


The following function is periodic between -π and π:

f(x) = |x|
Find the Coefficients of the Fourier series and, by examining the Fourier series at x=π or otherwise, determine:
1 + 1/32 + 1/52 + 1/72 ... = Σj=1 1/(2j - 1)2

Homework Equations


f(x) = a0/2 + ∑n=1 ancos(nx) + bn sin(nx)

a0 = 1/π ∫π f(x) dx
an = 1/π ∫π f(x) cos(nx) dx
bn = 1/π ∫π f(x) sin(nx) dx

The Attempt at a Solution


So I've found the coefficients:
a0 = π
an = -2(1 - (-1)n)/πn2
bn = 0 (as even function)

and so f(x) = π/2 + ∑-2(1 - (-1)n)/πn2 cos(nx)

I don't know how to do the last bit however, I don't really understand if I'm meant to come out with a number or something...
Any help would be greatly appreciated thank you.
Yes, they expect a value for ##\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)²}##.

You have computed the Fourier series, and found ##\displaystyle |x|=\frac{\pi}{2}- \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(1-(-1)^n)}{n²}\cos(nx)##.

What you have to do next is actually stated in the question: "examining the Fourier series at x=π". In other words, set x=π in your Fourier series.
 
Last edited:
  • #4
94
2
Ok, Yeah I've done that and get

π/2 +4/π∑1/n^2 (From n=1 to ∞ for n odd only)

And can't see how to get rid of the π/2 or 4/π

Thanks
 
  • #5
Charles Link
Homework Helper
Insights Author
Gold Member
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1,926
Ok, Yeah I've done that and get

π/2 +4/π∑1/n^2 (From n=1 to ∞ for n odd only)

And can't see how to get rid of the π/2 or 4/π

Thanks
Take another look at it: You have ## \pi=\pi/2+(4/\pi)(...) ## The rest is algebra.
 
  • #6
94
2
Take another look at it: You have ## \pi=\pi/2+(4/\pi)(...) ## The rest is algebra.
Thank you for your help, after doing a few more questions I finally understood the point of the question and I got an answer of π2/8.
 

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