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Math Help - fourier series representation

  1. #1
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    fourier series representation

    f(x) = cosax between pi and -pi and a is a constant

    can find the an coefficent to be (sin(n+a)x)/(n+a) + (sin(n-a)x)/(n-a) but cant get any further.
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  2. #2
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    Quote Originally Posted by dannie View Post
    f(x) = cosax between pi and -pi and a is a constant

    can find the an coefficent to be (sin(n+a)x)/(n+a) + (sin(n-a)x)/(n-a) but cant get any further.
    This makes no sense. The coefficient can't be a function of x. Did you forget to evaluate at \pi and -\pi?

    cos(ax) is an even function- it will have no "sin(nx)" terms in it. The coefficient of cos(nx) will be
    \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx=  \frac{1}{2\pi}\left(\int_{-\pi}^\pi cos((n+a)x)dx+ \int_{-\pi}^\pi cos(n- a)x dx\right)
    for n> 0.

    The n= 0 term is \frac{1}{2\pi}\int_{-pi}^{\pi} cos(ax)dx.

    As I said above, all coefficients of "sin(nx)" will be 0.
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    This makes no sense. The coefficient can't be a function of x. Did you forget to evaluate at \pi and -\pi?

    cos(ax) is an even function- it will have no "sin(nx)" terms in it. The coefficient of cos(nx) will be
    \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx=  \frac{1}{2\pi}\left(\int_{-\pi}^\pi cos((n+a)x)dx+ \int_{-\pi}^\pi cos(n- a)x dx\right)
    for n> 0.

    The n= 0 term is \frac{1}{2\pi}\int_{-pi}^{\pi} cos(ax)dx.

    As I said above, all coefficients of "sin(nx)" will be 0.
    You can also use by-parts here.

    C_n= \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx

     U = cos(ax) and  dV = cos(nx)dx

     du = -asin(ax) and  V= \frac{ sin(nx) }{n}

     \frac{1}{\pi}\int_{-\pi}^{\pi} cos(ax) cos(nx) dx = \frac{a}{\pi} [ \frac{ sin(nx)cos(ax) }{n} + \frac{a}{n} \int_{-\pi}^{\pi} sin(ax)sin(nx) dx]

    \frac{a^2}{n \pi} [ \int_{-\pi}^{\pi} sin(ax)sin(nx) dx]

     U = sin(ax) and  dV = sin(nx) dx

     du = acos(ax) and  V= - \frac{ cos(nx) }{n}

    \frac{a^2}{n \pi} [ \int_{-\pi}^{\pi} sin(ax)sin(nx) dx] = \frac{a^2}{n \pi} [ -\frac{ cos(nx)sin(ax) }{n} + \frac{1}{n} \int_{-\pi}^{\pi} cos(ax)cos(nx)dx ]

    <br />
C_n = \frac{a^2}{n \pi} [ -( \frac{ (-1)^n sin(a \pi ) }{n} - \frac{ (-1)^n sin(- a \pi ) }{n} ) + \frac{ \pi }{n} C_n]

    <br />
C_n = \frac{a^2}{n^2 \pi} [ -2 (-1)^n sin(a \pi ) + \pi C_n]

     C_n (1 - \frac{a^2}{n^2} ) = -2 (-1)^n sin(a \pi )

     C_n = - \frac{ 2 (-1)^n sin(a \pi ) } { (1 - \frac{a^2}{n^2 } ) }

    I'm fairly certain I've missed something or forgot something major (spidy sense is tingling...).
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