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Math Help - Difficulty finding how to state the sum of a fourier series

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    Difficulty finding how to state the sum of a fourier series

    I have been asked to find the sum of the attached fourier series when t = -3, t = 0 and t =3 but the handout doesnt tell me how I go about doing it and I cant find it on the net. Can someone show me how to find the first one with working for:

    -1/2 + 1/pi |sum symbol here|1/n (1-(-1)^n)sin(n x pi x t/2)

    Sorry about the way I've put it, the jpeg wont upload and I dont know how to use the math tags.

    or can someone direct me in the right direction where I can find it.
    Last edited by hunterage2000; February 26th 2011 at 05:41 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by hunterage2000 View Post
    I have been asked to find the sum of the attached fourier series when t = -3, t = 0 and t =3 but the handout doesnt tell me how I go about doing it and I cant find it on the net. Can someone show me how to find the first one with working for:

    -1/2 + 1/pi |sum symbol here|1/n (1-(-1)^n)sin(n x pi x t/2)

    Sorry about the way I've put it, the jpeg wont upload and I dont know how to use the math tags.

    or can someone direct me in the right direction where I can find it.
    t = 0 is easy...-1/2.

    For t = 3 (t = -3 is similar), start by looking at your coefficient in the sum:
    \displaystyle ( 1 - (-1)^n) = 0 for n even and \displaystyle ( 1 - (-1)^n) = 2 for n odd. So, letting n = 2y - 1:
    \displaystyle f(3) = -\frac{1}{2} + \frac{2}{\pi} \sum_{y = 1}^{\infty}\frac{sin \left ( \frac{3(2y -1) \pi}{2} \right )}{2y - 1}

    Now look at the sine:
    \displaystyle sin \left ( \frac{3(2y - 1) \pi}{2} \right ) = sin ( 3 \pi y ) cos \left ( \frac{3 \pi}{2} \right ) - sin \left ( \frac{3 \pi}{2} \right ) cos( 3 \pi y) = cos ( 3 \pi y )

    and since cos ( 3 \pi y ) = -1 for odd y and cos ( 3 \pi y ) = 1 for even y we get
    \displaystyle f(3) = -\frac{1}{2} + \frac{2}{\pi} \left [ -\frac{1}{1} + \frac{1}{3} - \frac{1}{5} + ~...~ \right ]

    Can you sum this?

    -Dan
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    not sure what you have done their. What do you mean by coefficient in the sum and where has n = 2y - 1 come from? Surely their is a simplfied explanation for this.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by hunterage2000 View Post
    not sure what you have done their. What do you mean by coefficient in the sum and where has n = 2y - 1 come from? Surely their is a simplfied explanation for this.
    This is proabaly as simple as it's going to get. This sum is (probably) impossible to simplify for arbitrary values of t. The tricks I used are only good for integer values of t. (And possibly 1/2 integer values of t as well.)

    The coefficient is \displaystyle \frac{1 - (-1)^n}{n}. I looked at just the numerator of this.

    After noting that the series does not involve a value of n that is even, I used n = 2y - 1. 2y - 1 = 1, 3, 5, 7, etc, for y going from 1 to infinity.

    -Dan
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