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Math Help - Help on Fourier Series of Sin(x/3)

  1. #1
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    Help on Fourier Series of Sin(x/3)

    THE QUESTION STATES:

    Let f be the 2 periodic function defined on [−pi,pi ) by
    f(x) = sin(x/3)

    Find the Fourier coefficients of f.

    My attempt at the question,

    After much work using trigonometry and integration by parts I have deduced

    bn = (1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

    Now I only need to put the limits in. The trouble is when i put the limits [-pi,pi) into the equation I get 0 which is not the answer!

    and the next part of the question states:

    show for all x E (-pi,pi)

    sin (x/3) = (((9)(3^.5))/pi)(((-1)^n)n)/(1-9n^2))

    I think I may need to use Dirichlets theorem however I need the answer to the part of the question before!
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  2. #2
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    Quote Originally Posted by zizou1089 View Post
    THE QUESTION STATES:

    Let f be the 2 periodic function defined on [−pi,pi ) by
    f(x) = sin(x/3)

    Find the Fourier coefficients of f.

    My attempt at the question,

    After much work using trigonometry and integration by parts I have deduced

    bn = (1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]
    This makes no sense. bn cannot depend on x. Do you mean this is the integral of sin(nx)sin(x/3) before you evaluate at the limits of integration?

    Now I only need to put the limits in. The trouble is when i put the limits [-pi,pi) into the equation I get 0 which is not the answer!
    How do you get that? When n= 1, for example, that gives
    \frac{1}{2\pi}[(3/2) sin((2/3)\pi)- (3/2)sin((4/3)\pi))= \frac{1}{2\pi}(\sqrt{3}{2}- (-\sqrt{3}{2})= \frac{\sqrt{3}}{2\pi} at x= \pi and evaluating at x= -\pi gives the negative of that so the difference is \frac{\sqrt{3}}{\pi}.

    and the next part of the question states:

    show for all x E (-pi,pi)

    sin (x/3) = (((9)(3^.5))/pi)(((-1)^n)n)/(1-9n^2))
    and this can't be true! The left side depends on x and the right side doesn't!

    I think I may need to use Dirichlets theorem however I need the answer to the part of the question before!
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  3. #3
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    For the first bit yes i meant this is the integral of sin(nx)sin(x/3) is
    (1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

    before you evaluating at the limits of integration. I just dont know how to simplify the answer after putting the limits in

    For the second part, the question only states:

    show for all x E (-pi,pi)
    .....................................infinity
    sin (x/3) = (((9)(3^.5))/pi) SUM (((-1)^n)n)/(1-9n^2))
    ......................................n=1
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    This makes no sense. bn cannot depend on x. Do you mean this is the integral of sin(nx)sin(x/3) before you evaluate at the limits of integration?


    How do you get that? When n= 1, for example, that gives
    \frac{1}{2\pi}[(3/2) sin((2/3)\pi)- (3/2)sin((4/3)\pi))= \frac{1}{2\pi}(\sqrt{3}{2}- (-\sqrt{3}{2})= \frac{\sqrt{3}}{2\pi} at x= \pi and evaluating at x= -\pi gives the negative of that so the difference is \frac{\sqrt{3}}{\pi}.


    and this can't be true! The left side depends on x and the right side doesn't!
    For the first bit yes i meant this is the integral of sin(nx)sin(x/3) is
    (1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

    before you evaluating at the limits of integration. I just dont know how to simplify the answer after putting the limits in

    For the second part, the question only states:

    show for all x E (-pi,pi)
    .....................................infinity
    sin (x/3) = (((9)(3^.5))/pi) SUM (((-1)^n)n)/(1-9n^2))
    ......................................n=1
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