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HELP: Fourier series Problem

  1. May 21, 2006 #1
    Hi

    Given the function [tex]f(t) = t^2[/tex], were [tex]t \in ]- \pi, pi[[/tex], and is continious find the fourier series for f(t). [tex]L = 2 \pi[/tex].

    Then

    [tex] A_0 = \frac{1}{2 \pi} \int \limit_{-\pi} ^{\pi} t^2 dt = \frac{\pi ^2}{3}[/tex]

    [tex]A_n = \int \limit_{-\pi} ^{\pi} t^2 \cdot cos(\matrm{n} \pi \mathrm{t}) dt[/tex]

    [tex]A_n = \int \limit_{-\pi} ^{\pi} u^2 \cdot cos(u) du[/tex], where u = n \pi t,

    The new limit gives:

    [tex]A_n = \int \limit_{0} ^{2 \pi^2 n} u^2 \cdot cos(u) du[/tex]

    [tex]A_n = [u^2 \cdot sin(u) - 2sin(u) + 2u sin(u)]_{0} ^{2 \pi ^ 2 n}[/tex]

    Then

    [tex]B_n = \int \limit_{0} ^{2n \pi^2} u^2 \cdot sin(u) du[/tex]

    which gives

    [tex]B_n = (2-4n^2 * \pi ^4 * cos(2n * \pi ^2) + 4n *sin (2n * \pi ^2) \pi ^2 -2[/tex]

    Therefore the Fourier series for f(t) is:

    [tex]\frac{\pi ^ 2}{3} + \sum \limit_{n=1} ^{\infty} A_n cos(nt) + B_n sin(nt)[/tex]

    Could someone please me if my calculations are correct?

    Sincerley

    Yours
    Hummingbird
     
    Last edited: May 21, 2006
  2. jcsd
  3. May 21, 2006 #2

    benorin

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    you may check here that the Fourier coefficients ought to be

    [tex]A_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t) dt[/tex]

    [tex]A_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos (nt) dt[/tex]

    and

    [tex]B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin (nt) dt[/tex]

    so that

    for [tex]f(t)=t^2[/tex] we have

    [tex]A_0=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2 dt = \frac{2}{\pi}\int_{0}^{\pi}t^2 dt = \frac{2\pi ^3}{3}[/tex]

    where the second move is from the integral of an even function over a symmetric interval (i.e. [-a,a]) is twice the integral over [0,a], also note that the same applies to An, hence

    [tex]A_n=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2 \cos (nt) dt = \frac{2}{\pi}\int_{0}^{\pi}t^2 \cos (nt) dt[/tex]

    integrate by parts twice to get EDIT: forgot to multiply by the [tex]\frac{1}{\pi}[/tex]!

    [tex]A_n=\frac{2}{\pi}\int_{0}^{\pi}t^2 \cos (nt) dt = \frac{2}{\pi}\left[ \frac{2t^2}{n}\sin (nt) + \frac{4t}{n^2}\cos (nt)-\frac{4}{n^3}\sin (nt)\right]_{t=0}^{\pi} = \frac{1}{\pi}\left( \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )\right) [/tex]
    [tex] = \frac{2\pi }{n}\sin (n\pi ) + \frac{4}{n^2}\cos (n\pi )-\frac{4}{\pi n^3}\sin (n\pi )[/tex]

    know also that [tex]t^2\sin (nt)[/tex] is an odd function, and that the integral of an odd function over a symmetric interval (i.e. [-a,a]) is zero, hence

    [tex]B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2\sin (nt) dt=0[/tex]
     
    Last edited: May 21, 2006
  4. May 21, 2006 #3

    benorin

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  5. May 21, 2006 #4
    Hello ben,

    I guess it mixed up the details from a textbook example. Sorry,

    Then the fourier series is expressed:

    [tex]\frac{2\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )[/tex] ?

    Sincerely
    Hummingbird25

     
  6. May 21, 2006 #5

    benorin

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    Rather it is expressed:

    [tex]\frac{1}{2}A_0+\sum_{n=1}^{\infty} A_n\cos (nt)+ \sum_{n=1}^{\infty} B_n\sin (nt)=\frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \left( \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )\right) \cos (nt)[/tex]
     
  7. May 21, 2006 #6
    Okay I get that,

    By the way,

    f(t) has continious deriatives, and is periodic [tex]2 \pi[/tex]

    Then the Fourier series of f(t) converge to f(t) Uniformly on [tex]]-\pi, \pi[[/tex] ??

    Or am I missing a condition?

    Sincerely

    Hummingbird

     
  8. May 21, 2006 #7

    benorin

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    I forgot to distribute the [tex]\frac{1}{\pi}[/tex] in the calculation of A_n in my first post, I fixed it: look for the EDIT,

    very important simplifications: [tex]\sin (n\pi ) = 0\mbox{ for }n=1,2,3,...[/tex]

    and

    [tex]\cos (n\pi ) = (-1)^{n}\mbox{ for }n=1,2,3,...[/tex]

    and hence [tex]A_n = \frac{2\pi }{n}\sin (n\pi ) + \frac{4}{n^2}\cos (n\pi )-\frac{4}{\pi n^3}\sin (n\pi ) = \frac{4}{n^2}(-1)^{n}[/tex]

    so the series becomes

    [tex]\frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \left( \frac{2\pi }{n}\sin (n\pi ) + \frac{4}{n^2}\cos (n\pi )-\frac{4}{\pi n^3}\sin (n\pi )\right) \cos (nt ) = \frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} (-1)^{n}\frac{4}{n^2}\cos (nt )[/tex]

    that is

    [tex]\boxed{t^2 \sim \frac{\pi ^3}{3} + 4\sum \limit_{n=1} ^{\infty} (-1)^{n}\frac{\cos (nt)}{n^2} = \frac{\pi ^3}{3} -4\left( \frac{\cos (t)}{1^2}-\frac{\cos (2t)}{2^2}+\frac{\cos (3t)}{3^2}-\mdots \right)}[/tex]​
     
  9. May 21, 2006 #8

    benorin

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    Uniform convergence can be proved by the Weierstrass M-test, just note that

    [tex] \left| (-1)^{n}\frac{\cos (nt)}{n^2} \right| \leq \frac{1}{n^2}=M_n[/tex] for all n

    and [tex]\sum \limit_{n=1} ^{\infty}M_n=\sum \limit_{n=1} ^{\infty}\frac{1}{n^2} [/tex] converges, so the given series converges uniformly on [tex] \left[ -\pi ,\pi \right] [/tex] by the Weierstrass M-test.
     
  10. May 21, 2006 #9
    Okay thank You then I only have one final question.

    Given the two series

    [tex]\sum_{k=1} ^{\infty} \frac{1}{k^4}[/tex]

    and

    [tex]\sum_{k=1} ^{\infty} (-1)^{k-1} \frac{1}{k^2}[/tex]

    I need to find the sum of these two.

    In series number 1:

    I can see that by the test of comparison, that it converges

    [tex]\frac{1}{k^{2+t}} < \frac{1}{k^{2}}[/tex]

    But what is the next step in finding the sum here?

    In series two:

    What do I here? Do I test for convergens ?

    Sincerely
    Hummingbird
     
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