Can Fourier Series Help Solve this Summation Question?

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SUMMARY

The discussion centers on using Fourier series to demonstrate the convergence of the series \(\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}\) to \(\frac{\pi^2}{8}\). The Fourier series provided is \(F.S f(x) = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n=1,3}^{\infty} \frac{1}{n^2} \cos(nx)\). Participants suggest evaluating the series at specific values of \(x\), particularly \(x = \frac{\pi}{2}\) and \(x = 0\), to simplify the cosine terms and derive the desired result. The key takeaway is that choosing \(x = 0\) is essential for proving the convergence to \(\frac{\pi^2}{8}\).

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koolrizi
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Hey everyone,
I got the following Fourier series

F.S f(x)= (pi/2) - (4/pi) \sumn=1,3.. to infinity (1/n^2 cos (nx))

l= pi

After deriving it the question now is how can i use it to show

\sum n=1 to infinity (1/(2n-1)^2= 1+ 1/3^2 + 1/5^2 +... = pi^2/8

I think I am not sure what I have to do here.

Thanks
 
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Usually the idea is to find some value of x that makes your Fourier series into the series in n that you want. If you choose x=pi/2 for example, cos(nx) will vanish for odd n. Then you'd plug your choice of x into the function f(x), for which you presumably have a closed form (but you haven't told us what it is).
 
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Can you think of a value for x such that the cosine becomes zero for all even n, and non-zero (for example 1) for all odd n? Then calculate f(x) for this x.
 
f(x)= |x| over (-pi,pi]
-x over -pi<x<= 0 and x over 0<x<=pi
 
If i use x=pi/2 in Fourier series cos nx vanishes and i have 1/n^2 which for odd numbers is 1/(2n-1)^2

I can get that. how would i prove the part series converges to pi^2/8

Thanks
 
Forgot to mention the question. question was to use Fourier series and x=0 to prove it. so i cannot use x= pi/2 can i?
 
The x=pi/2 was just an example. It seems like the question even gives you the value of x you should use, so just plug it into the series and the function and you have your answer.
 

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