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Math Help - Fourier Series

  1. #1
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    Fourier Series

    Can someone help me with these two Fourier Series problems please? Also, can you explain how you did the problems because I dont understand how to do Fourier Series ...Thanks in advance!

    #1 Use the Fourier Series
    f(x) = {0, -pi < x < 0; x^2, 0 < x < pi

    to show that
    1 - 1/4 + 1/9 - 1/16 +.... = (pi^2)/12


    #2 Establish the result in the problem below, where m and n are positive integers. (Hint: sinA sinB = (1/2) [cos(A-B) - cos(A+B)].)
    L
    ∫ sin((n pi x)/L) sin((m pi x)/L) dx = {0, m≠n; L, m=n
    -L
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by cherry106
    Can someone help me with these two Fourier Series problems please? Also, can you explain how you did the problems because I dont understand how to do Fourier Series ...Thanks in advance!

    #1 Use the Fourier Series
    f(x) = {0, -pi < x < 0; x^2, 0 < x < pi

    to show that
    1 - 1/4 + 1/9 - 1/16 +.... = (pi^2)/12
    The Fourier series representation of f(x) is:

    <br />
f(x)=a_0+\sum_1^{\infty}a_n \cos(nx) +b_n \sin(nx)<br />
,

    where:

    <br />
a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) dx =\frac{1}{2\pi}\int_0^{\pi} x^2 dx=\frac{\pi^2}{6}<br />

    <br />
a_n=\frac{1}{\pi}\int_{-pi}^{\pi}f(x) \cos(nx) dx=\frac{1}{\pi}\int_{0}^{\pi}x^2 \cos(nx) dx=\frac{(-1)^n 2}{n^2}<br />

    and

    <br />
b_n=\frac{1}{\pi}\int_{-pi}^{\pi}f(x) \sin(nx) dx=\frac{1}{\pi}\int_{0}^{\pi}x^2 \sin(nx) dx<br />

    So:

    <br />
f(0)=a_0+2\sum_1^{\infty} b_n=\frac{\pi^2}{6}+2\sum_1^{\infty} \frac{(-1)^n}{n^2}=0<br />
.

    Rearranging gives:

    <br />
1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+...=\frac{\pi^2}{12}<br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by cherry106
    #2 Establish the result in the problem below, where m and n are positive integers. (Hint: sinA sinB = (1/2) [cos(A-B) - cos(A+B)].)
    L
    ∫ sin((n pi x)/L) sin((m pi x)/L) dx = {0, m≠n; L, m=n
    -L
    <br />
\int_L^L \sin(n \pi x/L) \sin(m \pi x/L) dx= \frac{1}{2} \int_L^L [\cos((n-m) \pi x/L) -\cos((n+m) \pi x/L)]dx<br />

    Now if n \ne m the integral on the RHS is over an integer number of cycles of each of the cosines
    which sum to the integrand and so is zero.

    When n=m the integand on the RHS is one plus a cosine whose period
    is a sub-multiple of the interval of integration and so the integral is L/2 and when
    this is multipled by two gives the intregral on the LHS is L

    RonL
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  4. #4
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    Show that,
    1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+...=\frac{  \pi^2}{8}
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