Fourier Series: Solving Homework Equations for f(x)

[Poirot]

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 ... = Σ^∞_j=1 1/(2j - 1)²

Homework Equations

f(x) = a₀/2 + ∑^∞_n=1 a_ncos(nx) + b_n sin(nx)

a₀ = 1/π ∫^π_-π f(x) dx
a_n = 1/π ∫^π_-π f(x) cos(nx) dx
b_n = 1/π ∫^π_-π f(x) sin(nx) dx

The Attempt at a Solution

So I've found the coefficients:
a₀ = π
a_n = -2(1 - (-1)ⁿ)/πn²
b_n = 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.

 

[stevendaryl]

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 x=\pi into your expression. (You can see that when n is even, you get 0, so you only have to consider the case for n is odd.)

 

[Samy_A]

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 ... = Σ^∞_j=1 1/(2j - 1)²

Homework Equations

f(x) = a₀/2 + ∑^∞_n=1 a_ncos(nx) + b_n sin(nx)

a₀ = 1/π ∫^π_-π f(x) dx
a_n = 1/π ∫^π_-π f(x) cos(nx) dx
b_n = 1/π ∫^π_-π f(x) sin(nx) dx

The Attempt at a Solution

So I've found the coefficients:
a₀ = π
a_n = -2(1 - (-1)ⁿ)/πn²
b_n = 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.

 

[Poirot]

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

 

[Charles Link]

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.

 

[Poirot]

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.