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Fourier (sin(2x)*x^2)

  • Thread starter pedro_bb7
  • Start date
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)
 
Can you help me, please?
 
[PLAIN]http://img826.imageshack.us/img826/6708/semttulocz.png [Broken]
 
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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

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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 bn's are correct except for b2. I get

[tex]b_2 = -\frac 1 8 +\frac {\pi^2} 3[/tex]

And remember when plotting your graph, that your formula for your original function only works on (-pi,pi), so look at the convergence there.
 
Thanks, the problem was at the B2.
Cya.
 
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: [tex]\frac {\pi^2} 3 [/tex]
 
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LCKurtz

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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

[tex]\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx[/tex]
 
That formula doesn't work for n = 2 so you have to do b2 separately. Just directly work out the integral

[tex]\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx[/tex]
Thanks, finally I got the right answer.
You helped me a lot.
 
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