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Help with computing/understanding Fourier Sine Expansion.

  1. Mar 3, 2013 #1
    1. Find the Fourier sine expansion of [itex]\phi(x)=1[/itex]. This was posted in Calculus and Beyond thread, but I realized that this thread may be more appropriate.

    2. The attempt at a solution.
    I start with [itex]\phi(x)=A_1sin(\pi x)+A_2sin(2\pi x)+\cdots+A_nsin(n\pi x),[/itex] and then add multiply by [itex]A_msin(m\pi x)[/itex] term on each side and integrate from 0 to 1.
    So I have [itex]\int\phi(x)A_msin(m\pi x)=\int\left(A_1sin(\pi x)A_msin(m\pi x)+A_2sin(2\pi x)A_msin(m\pi x)+\cdots+A_nsin(n\pi x)A_msin(m\pi x)\right).[/itex]
    I know that due to orthogonality you can discard the terms where m is not equal to n (but I don't really understand why so if you can explain this I would appreciate it).
    Discarding those terms and using a trig relation I get, [itex]\int\phi(x)A_msin(m\pi x)=1/2\int\left(cos((m-n)\pi x)-cos((m+n)\pi x)\right).[/itex]
    I then solve the integral and try to get ##A_m=\frac{4}{\pi}\left(\frac{1}{m}\right),## but I don't know why the answer has only the odd terms.
    The book answer is [itex]1=\frac{4}{\pi}\left(sin(\pi x)+\frac{1}{3}sin(3\pi x)+\frac{1}{5}sin(5\pi x) +\cdots\right).[/itex]

    Any and all help is greatly appreciated.
     
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 6, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    The starting point is
    [tex]\phi(x) = \displaystyle \sum_{n=1}^{\infty} a_n \sin(n\pi x)[/tex]

    To work out [itex]a_n[/itex], multiply by [itex]\sin (m\pi x)[/itex] and integrate between 0 and 1:
    [tex]
    \int_0^1 \phi(x) \sin(m\pi x)\,\mathrm{d}x
    = \sum_{n=1}^\infty a_n \int_0^1 \sin(n \pi x) \sin(m\pi x)\,\mathrm{d}x
    [/tex]

    We need to calculate [itex]I = \int_0^1 \sin(n\pi x)\sin(m \pi x)\,\mathrm{d}x[/itex]. This we do by parts:
    [tex]
    I = \int_0^1 \sin(n\pi x)\sin(m \pi x)\,\mathrm{d}x \\
    = \left[ -\frac{1}{m\pi} \sin(n\pi x)\cos(m\pi x)\right]_0^1
    + \frac{n}{m}\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\mathrm{d}x \\
    = \frac{n}{m}\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\mathrm{d}x
    [/tex]
    where the last follows because [itex]\sin(0) = \sin(m\pi) = 0[/itex]. Now we can integrate by parts again:
    [tex]
    I = \frac{n}{m}\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\mathrm{d}x \\
    = \frac{n}{m}\left[ \frac{1}{m\pi} \cos(n\pi x)\sin(m \pi x)\right]_0^1
    + \frac{n^2}{m^2} \int_0^1 \sin(n\pi x)\sin(m\pi x)\,\mathrm{d}x \\
    = \frac{n^2}{m^2} I[/tex]

    So [itex](1 - n^2/m^2)I = 0[/itex]. It follows then that if [itex]n \neq m[/itex] then [itex]I = 0[/itex]. If [itex]n = m[/itex] then the integrand is [itex]\sin^2(m\pi x)[/itex] which is non-negative and not identically zero, so its integral must be strictly positive. So we have, using the identity [itex]\sin^2 \theta = \frac12 (1 - \cos 2\theta)[/itex]:
    [tex]
    \int_0^1 \phi(x) \sin(m\pi x)\,\mathrm{d}x
    = a_m \int_0^1 \sin(m \pi x) \sin(m\pi x)\,\mathrm{d}x \\
    = a_m \int_0^1 \sin^2(m \pi x)\,\mathrm{d}x \\
    = \frac12 a_m \int_0^1 1 - \cos(2m \pi x)\,\mathrm{d}x\\
    = \frac12 a_m
    [/tex]
    so that
    [tex]
    a_m = 2 \int_0^1 \phi(x) \sin(m\pi x)\,\mathrm{d}x.
    [/tex]

    [tex]a_m = 2\int_0^1 \sin(m\pi x)\,\mathrm{d}x = -\frac{2}{m\pi} \left[\cos(m\pi x)\right]_0^1 \\
    = - \frac{2}{m\pi}( \cos(m\pi) - 1) \\
    = - \frac{2}{m\pi}( (-1)^m - 1) \\
    = \frac{2}{m\pi}(1 - (-1)^m)[/tex]
    If [itex]m[/itex] is even, [itex](-1)^m = 1[/itex] so [itex]a_m = 0[/itex]. If [itex]m[/itex] is odd, [itex](-1)^m = -1[/itex] and [itex]a_m = 4/(m\pi)[/itex], which is the given answer.

     
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