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    It's really confusing me. I thought when the function is even then only the bn terms remain? When I sub in f(-x) here I get f(x) in both cases. I tried therefore to find the b[SUB]n[SUB] term but I got zero:

    What are my limits? I thought the usual integral is between -pi and pi and multiplied by 1/pi. Here I split it up into 2 integrals: The first integrand was -x2cos(nx) multiplied by 1/pi and integrated between -pi and 0. The second integrand was x2cos(nx) multiplied by 1/pi and integrated between 0 and pi. Is this wrong?
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    The function is odd.

    Therefore the  a_n vanish. Moreover  f(0) = 0 so  a_0 = 0 . Thus you have

     f(x) = \sum_{n = 1}^{+\infty} b_n \sin(nx) .

    For the limits you can take any interval of length  2\pi since the function is  2\pi -periodic.  [-\pi,\pi] is fine. Therefore

     b_n = \int_{-\pi}^\pi f(x) \sin(nx)dx .

    Your splitting of the integral is fine...but the functions you're integrating are  -x^2 \sin(nx) and  x^2 \sin(nx) (respectively) not cos.

    Edit:updated link
    Edit again: still doesn't work...draw your own picture.
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    (Original post by jb444)
    The function is odd.

    Therefore the  a_n vanish. Moreover  f(0) = 0 so  a_0 = 0 . Thus you have

     f(x) = \sum_{n = 1}^{+\infty} b_n \sin(nx) .

    For the limits you can take any interval of length  2\pi since the function is  2\pi -periodic.  [-\pi,\pi] is fine. Therefore

     b_n = \int_{-\pi}^\pi f(x) \sin(nx)dx .

    Your splitting of the integral is fine...but the functions you're integrating are  -x^2 \sin(nx) and  x^2 \sin(nx) (respectively) not cos.

    Edit:updated link
    Edit again: still doesn't work...draw your own picture.
    Thanks, the link worked for me and I understand the rest of your post but I still don't get the graph. It says -x^2 for x less than 0. Surely it should be below the x-axis? also i worked out a0 and got 2/3 pi^2 but i thought it was 0? i followed all the normal rules for calculating a0
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    Yes, it should be below the x axis for $x < 0$ (I deleted the link since it showed the wrong graph, wolfram doesn't load quite the same thing if you copy the URL it seems). It's this extended periodically

    Name:  Sans_titre.jpg
Views: 42
Size:  13.6 KB

    To see that  a_0 = 0 you have that

    \int_{-\pi}^\pi f(x) dx = \int_{-\pi}^0 -x^2 dx + \int_{0}^\pi x^2 dx

    so it'll cancel. Alternatively you can note that  f(0) = 0 and  \sin(n\cdot 0) = 0 for each  n , so substituting all this into the Fourier series equation you must have that  a_0 = 0 as well.

    Yet another edit: that's still the wrong graph, it should be sgn(x) x^2, obviously.
 
 
 
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