Show that this series sums to value shown using Fourier technique

[bossman007]

Homework Statement

Use the Fourier series technique to show that the following series sums to :
1+\frac{1}{3^2}+\frac{1}{5^2}+...=\frac{\pi^2}{8}

Homework Equations

The Attempt at a Solution

Don't know what the first few steps are...but I assume that I need to first express the sum as \frac{1}{(2n-1)^2}, but have no idea where to go from there. I know that Fourier series involve finding the coefficients for the a_0, a_n, and b_n.

 

[bossman007]

it seems the most difficult thing is picking the f(x) function

 

[vela]

You need to be familiar with the Fourier series for common functions. I'd look for one that has only odd harmonics.

 

[bossman007]

would you be referring to a square wave? if so, I'm having trouble setting it up in relation to this series or whatever I have to do

 

[vela]

The square wave doesn't quite do it because you need n² in the denominator whereas the Fourier coefficients for the square wave are proportional to 1/n, but it's a good place to start.

What can you do to the series to get another factor of n in the denominator?

 

[bossman007]

multiply the series by the integral of (cos(n*pi*x)? because taking the integral will yield a 1/n multiplied by the series? , I am still confused what to do after I have that

 

[bossman007]

If that's completely wrong i apologize, I've had my head buried in Fourier series all day and I'm probably mixing stuff up :/

 

[vela]

I'm not sure exactly what you mean, but I think you're on the right track. Can you show what you mean using math starting with the Fourier series for the square wave?

 

[bossman007]

My classmates led me to believe that i should choose the function for a square wave :

f(x) = -1 when -pi < x < 0
1 when 0 < x < pi

doing this I get a_0=0

and b_n = \frac{1}{\pi}\int^\pi_0 sin(n*\pi*x)\,dx , but I don't see how this helps. I know the fact cos(n\pi)=(-1)^n and sin(n\pi)=0 pop up in a lot of these problems we have been doing, and I don't see how the square wave b_n coefficient I wrote out makes and sense for this series

 

[vela]

b_n = \frac{1}{\pi}\int^\pi_0 sin(n*\pi*x)\,dx

You don't want the factor of $\pi$ in the argument of the sine. So what did you get when you did the integral? What happens when n is even? What happens when n is odd?

 

[bossman007]

when n is even or odd sin dissapears. doing the integral without the factor of pi in the argument of sine yields:

\frac{1}{n\pi}-\frac{cos(n\pi}{n\pi}=\frac{1}{n\pi}-\frac{-1^n}{n\pi}

 

[vela]

From your final result, if n is even, you have $\cos n\pi = 1$, so b_n=0. If n is odd, you get $b_n = \frac{2}{n\pi}$, right? So the first few terms of the series are
$$\frac{2}{\pi}\left(\frac{1}{1}\sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots \right)$$ Now what do you get if you integrate that?

 

[bossman007]

IT GIVES ME MY ORIGINAL SERIES :D, thank u so much, but not sure what I need to do after that to prove it sums to pi^2/8

 

[vela]

You have
$$f(x) = \frac{2}{\pi}\left(\frac{1}{1}\sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots \right)$$ You integrated the righthand side, so you have to integrate the lefthand side. Then picking an appropriate value of x, you'll get the result you desire.

 

[bossman007]

many thanks,

little confused on the integral on the left hand and right hand side should be

I have "integral of f(x)" = (2[-cos(x)-(1/(3^2))cos(3x)-1/(5^2))cos(5x)]) / pi

I can't find a value of x that makes it pi^2/8

I know if x=pi, the RHS will yield my initial series multipled by the 2 out front of the brackets, but I'm tripping myself up over the left hand side and what to integrate

I'm also confused because the RHS is infinite series and not sure what to do

 

[vela]

You wrote what f(x) is above, in post 9. You can either use definite integrals with the limits conveniently chosen, or you can use indefinite integrals and deal with the constants of integration.

Also, I just noticed you're missing a factor of 2 in your formula for b_n.

 

[bossman007]

I ended up getting a close answer to pi^2/8

still not totally sure what my Fourier formula should look like with everything plugged int