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    Hi, I have a problem with a Fourier series question and would like some assistance please:

    The question is: find the Fourier series of x^2 in the range of -1<=x<=1.

    Where f(x) = 0.5A0 + SUM OF(An cos(pi*n*x) + Bn sin(pi*n*x))

    I think I have found the Fourier series correctly, with An = 4Cos(pi*n)/(pi^2*n^2) = 4(-1)^n/(pi^2*n^2), A0 = 2/3 and Bn = 0.

    So the fourier series equation becomes: f(x) = 1/3 + 4/pi^2 * SUM of((-1)^n*cos(pi*n*x)/(n^2))

    The next question is to hence show that the answer to the SUM of ( 1 - 1/2^2 + 1/3^2 - 1/4^2 +.....) = Pi^2/12

    If my fourier series equation is correct, then that sum can only be obtained when x=1/n
    (not sure if you can even do that). But then I get an answer of -(pi)^2/12 as n -> infinity.

    Can anyone tell me if I have gone totally wrong or something I'm missing?

    Thank you.
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    (Original post by dada55)
    Hi, I have a problem with a Fourier series question and would like some assistance please:

    The question is: find the Fourier series of x^2 in the range of -1<=x<=1.

    Where f(x) = 0.5A0 + SUM OF(An cos(pi*n*x) + Bn sin(pi*n*x))

    I think I have found the Fourier series correctly, with An = 4Cos(pi*n)/(pi^2*n^2) = 4(-1)^n/(pi^2*n^2), A0 = 2/3 and Bn = 0.

    So the fourier series equation becomes: f(x) = 1/3 + 4/pi^2 * SUM of((-1)^n*cos(pi*n*x)/(n^2))

    The next question is to hence show that the answer to the SUM of ( 1 - 1/2^2 + 1/3^2 - 1/4^2 +.....) = Pi^2/12

    If my fourier series equation is correct, then that sum can only be obtained when x=1/n
    (not sure if you can even do that). But then I get an answer of -(pi)^2/12 as n -> infinity.

    Can anyone tell me if I have gone totally wrong or something I'm missing?

    Thank you.
    It has to be a silly sign error somewhere ...
    I cannot possibly see it unless I do the question or post some workings
    (no hieroglyphics please)
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    (Original post by dada55)
    Hi, I have a problem with a Fourier series question and would like some assistance please:

    The question is: find the Fourier series of x^2 in the range of -1<=x<=1.

    Where f(x) = 0.5A0 + SUM OF(An cos(pi*n*x) + Bn sin(pi*n*x))

    I think I have found the Fourier series correctly, with An = 4Cos(pi*n)/(pi^2*n^2) = 4(-1)^n/(pi^2*n^2), A0 = 2/3 and Bn = 0.

    So the fourier series equation becomes: f(x) = 1/3 + 4/pi^2 * SUM of((-1)^n*cos(pi*n*x)/(n^2))

    The next question is to hence show that the answer to the SUM of ( 1 - 1/2^2 + 1/3^2 - 1/4^2 +.....) = Pi^2/12

    If my fourier series equation is correct, then that sum can only be obtained when x=1/n
    (not sure if you can even do that). But then I get an answer of -(pi)^2/12 as n -> infinity.

    Can anyone tell me if I have gone totally wrong or something I'm missing?

    Thank you.
    I'm a bit rusty on this so I hope I don't get this wrong.

    You have x^2=\dfrac{1}{3}+\dfrac{4}{\pi^2  }\sum...

    Rearranging you get

    \dfrac{\pi^2}{12}=\sum \left( \dfrac{(-1)^n\cos(n \pi x)/(3x^2-1)}{n^2}\right)

    Now ask yourself what value of x makes this work?
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    (Original post by TeeEm)
    It has to be a silly sign error somewhere ...
    I cannot possibly see it unless I do the question or post some workings
    (no hieroglyphics please)

    I'm not sure, I just re did it and still get to the same point. Unless I am making the same mistake. I have attached the question and my working out.

    (Original post by BuryMathsTutor)
    I'm a bit rusty on this so I hope I don't get this wrong.

    You have x^2=\dfrac{1}{3}+\dfrac{4}{\pi^2  }\sum...

    Rearranging you get

    \dfrac{\pi^2}{12}=\sum \left( \dfrac{(-1)^n\cos(n \pi x)/(3x^2-1)}{n^2}\right)

    Now ask yourself what value of x makes this work?
    But if my equation up to that point is correct, then surely the only value of x that should work is if x=1/n so that the n=1 value is positive and then they continue to alternate signs. I did try something simmilar to what you did and got an answer of x = +- ROOT(6) over 3.

    edit: for some reason my working out doesn't want to attach at the right rotation: http://tinypic.com/view.php?pic=205s...8#.VK_6a3uoP5k


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    Name:  Fourier Series Q.jpg
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    (Original post by dada55)
    I'm not sure, I just re did it and still get to the same point. Unless I am making the same mistake. I have attached the question and my working out.



    But if my equation up to that point is correct, then surely the only value of x that should work is if x=1/n so that the n=1 value is positive and then they continue to alternate signs.

    I did try something simmilar to what you did and got an answer of x = +- ROOT(6) over 3.
    Name:  20150109_153920.jpg
Views: 73
Size:  505.5 KB
    Name:  Fourier Series Q.jpg
Views: 79
Size:  99.7 KB
    all correct so far

    I can tell what you cannot see!!!

    after you let x = 0 in both sides and you rearrange, the resulting sum is not the one you are after but the negative of that

    write out a few terms to see it

    your sum is pi squared over 12
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    (Original post by TeeEm)
    all correct so far

    I can tell what you cannot see!!!

    after you let x = 0 in both sides and you rearrange, the resulting sum is not the one you are after but the negative of that

    write out a few terms to see it

    your sum is pi squared over 12
    Ah yeah, I see that now. Thank you.

    I guess that means that you cannot do the substitution x=1/n. I did think that you couldn't, but I was stuck in the idea that I had to get the signs of the SUM correct. In reality, -1* the answer for the same sum but with signs swapped is correct.
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    (Original post by dada55)
    Ah yeah, I see that now. Thank you.

    I guess that means that you cannot do the substitution x=1/n. I did think that you couldn't, but I was stuck in the idea that I had to get the signs of the SUM correct. In reality, -1* the answer for the same sum but with signs swapped is correct.
    the substitutions in these types of questions will usually be 0, pi/2, pi and so on
    it is just that the sum is -1+1/4-1/9+1/16 ... and so on
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    (Original post by dada55)
    Ah yeah, I see that now. Thank you.

    I guess that means that you cannot do the substitution x=1/n. I did think that you couldn't, but I was stuck in the idea that I had to get the signs of the SUM correct. In reality, -1* the answer for the same sum but with signs swapped is correct.
    I added your series to my: Fourier series
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    (Original post by BuryMathsTutor)
    I added your series to my: Fourier series
    what program are you using?
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    (Original post by TeeEm)
    what program are you using?
    http://www.graphmatica.com/
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    thanks
 
 
 
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