Fourier Series for sin(2x)*x^2: Finding the Correct Solution

[pedro_bb7]

Hello, I got a problem using Fourier series.

f(x) = sin(2x)*(x^2) (-pi < x < pi)

I got this answer, although it seems to be wrong.

f(x) = (-16/9)*sin(x) + (-1/8)*sin(2x) + sum((2*(-1)^n/(2-n)^2 - 2*(-1)^n/(2+n)^2)*sin(nx) , n, 3, 100)

 

[pedro_bb7]

Can you help me, please?

 

[pedro_bb7]

[PLAIN]http://img826.imageshack.us/img826/6708/semttulocz.png

 

[least_action]

Can you help me, please?

What part are you having trouble with?

 

[pedro_bb7]

Its hard to explain.
I did this exercise a lot of times, every time I get the same answer.
I can post the whole exercise to help me find what is wrong.
That Bn is kind hard to get right.

 

[LCKurtz]

Its hard to explain.
I did this exercise a lot of times, every time I get the same answer.
I can post the whole exercise to help me find what is wrong.
That Bn is kind hard to get right.

I think your b_n's are correct except for b₂. I get

b_2 = -\frac 1 8 +\frac {\pi^2} 3

And remember when plotting your graph, that your formula for your original function only works on (-pi,pi), so look at the convergence there.

 

[pedro_bb7]

Thanks, the problem was at the B2.
Cya.

 

[pedro_bb7]

I got this:
Bn = (2*(-1)^n)/(2-n)^2 - (2*(-2)^n)/(2+n)^2
B2 = (2/(2-2)^2) - 1/8

What did you make to get that: \frac {\pi^2} 3

 

[LCKurtz]

I got this:
Bn = (2*(-1)^n)/(2-n)^2 - (2*(-2)^n)/(2+n)^2
B2 = (2/(2-2)^2) - 1/8
What did you make to get that: (pi^2)/3

That formula doesn't work for n = 2 so you have to do b₂ separately. Just directly work out the integral

\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx

 

[pedro_bb7]

That formula doesn't work for n = 2 so you have to do b₂ separately. Just directly work out the integral

\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx

Thanks, finally I got the right answer.
You helped me a lot.