Solving a Differential Equation with a Fourier Series

[VinnyCee]

I am not sure I am doing this correctly, so here it is.

Problem:

Find the Fourier Series

f(s)\,=\,\left\{\begin{array}{ccc}x^2&-\pi\,<\,x\,<\,0\\0 &0\,<x\,<\,\pi \\}\end{array}\right

Answer(supposedly):

a_0\,=\,\frac{\pi^3}{3}

a_n\,=\,-\frac{2}{n^2}

b_n\,=\,\frac{4\,-\,n^2\,\pi^2}{n^3\,\pi}

Does that look right? If so, where do I gofrom here?

f(s)\,\approx\,\frac{\pi^3}{3}\,+\,\sum_{n\,=\,1}^{\infty}\,\left[-\frac{2}{n^2}\,cos(n\,x)\,+\,\frac{4\,-\,n^2\,\pi^2}{n^3\,\pi}\,sin(n\,x)\right]

Is this the final answer?

 

[VinnyCee]

Can anyone type this into mathematica?

Please double check

 

[VinnyCee]

Here is how I got the a0, an and bn factors

a_0\,=\,\frac{1}{\pi}\,\int_{-\pi}^{0}\,x^2\,dx\,=\,\frac{\pi^2}{3}

a_n\,=\,\frac{1}{\pi}\,\int_{-\pi}^{0}\,x^2\,cos\,(n\,x)\,dx

a_n\,=\,\frac{1}{\pi}\,\left[\frac{x^2\,sin\,(n\,x)}{n}\,+\,\frac{2\,x\,cos\,(n\,x)}{n^2}\,-\,\frac{2\,sin\,(n\,x)}{n^3}\right]_{-\pi}^{0}

Right? Then:

a_n\,=\,-\frac{2}{n^2}

Am I on the right track here? or am i completely missing something?

 

[VinnyCee]

Or is this how to solve a_n?

a_n\,=\,\frac{1}{\pi}\,\int_{-\pi}^{0}\,x^2\,cos\,(n\,x)\,dx\,=\,\frac{1}{\pi}\,\left[\frac{n^2\,x^2\,sin\,(n\,x)\,-\,2\,sin\,(n\,x)\,+\,2\,n\,x\,cos\,(n\,x)}{n^3}\right]_{-\pi}^{0}

a_n\,=\,\frac{1}{\pi}\,\left[\frac{(-\pi)^2\,n^2\,sin\,(-\pi\,n)\,-\,2\,sin\,(-\pi\,n)\,-\,2\,\pi\,n\,cos\,(-\pi\,n)}{n^3}\right]\,=\,\frac{1}{\pi}\,\left[\frac{2\,\pi\,n\,cos\,(-\pi\,n)}{n^3}\right]

a_n\,=\,\frac{1}{\pi}\,\left[\frac{-2\,\pi\,n\,(-1)^n}{n^3}\right]\,=\,\frac{-2\,n\,(-1)^n}{n^3}

Which one is correct?

 

[OlderDan]

I think I see several sign mistakes, with the last sign being wrong. Why not reduce the last expression n/n^3??

 

[Justin Lazear]

You need to revisit your b_n with your newly refound knowledge of the behavior of sine and cosine.

The sign of your a_n term is indeed incorrect.

Why do I let myself get dragged into other people's problems? *sighs* Back to finals! Work, damn me!

--J

 

[VinnyCee]

Is this the correct an then?

Is the term below right for an?

a_n\,=\,\frac{1}{\pi}\,\left[\frac{-2\,\pi\,n\,(-1)^n}{n^3}\right]\,=\,\frac{-2\,n\,(-1)^n}{n^3}\,=\,-\frac{2\,(-1)^n}{n^2}

Thanks again!

 

[Justin Lazear]

You got an extra negative. \cos{n\pi} = (-1)^n, which we can verify by noting that when n = 0, n\pi = 0, so \cos{n\pi} = 1. Additionally, when n = 1, we just have cosine of pi, which is -1. So we must have (-1)ⁿ instead of (-1)^n-1.

--J

 

[VinnyCee]

So, the an is:

a_n\,=\,\frac{2\,(-1)^n}{n^2}

Right?

 

[Justin Lazear]

Looks good. Now go have another look at those b coefficients.

--J

Just one final left... Yay me!

 

[VinnyCee]

Now for the bn!

b_n\,=\,\frac{1}{\pi}\,\int_{-\pi}^{0}\,x^2\,sin\,(n\,x)\,dx\,=\,\frac{1}{\pi}\,\left[\frac{-n^2\,x^2\,cos\,(n\,x)\,+\,2\,cos\,(n\,x)\,2\,n\,x\,sin\,(n\,x)}{n^3}\right]_{-\pi}^{0}

b_n\,=\,\frac{1}{\pi}\,\left[\frac{2}{n^3}\,+\,\frac{\pi^2\,(-1)^n}{n}\,-\,\frac{2\,(-1)^n}{n^3}\right]

b_n\,=\,\frac{2}{\pi\,n^3}\,+\,\frac{\pi\,(-1)^n}{n}\,-\,\frac{2\,(-1)^n}{\pi\,n^3}

Does that look right also?

Good luck on the final! I feel for you man.

 

[Justin Lazear]

Looks like you got it.

And just in case you were interested, here's some plots of the Fourier series truncated to 5, 15, 50, and 500 terms. Can't even tell it's not the function itself in the 500 one, can ya'?

Good job.

--J

 

[VinnyCee]

Awesome plots.

Thanks a lot for the help.