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Math Help - Fourier series representation

  1. #1
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    Fourier series representation

    okay, so i was given the piecewise function f(t)= k if -pi/2< x< pi/2
    0 if pi/2 < x < 3pi/2

    i computed the series to be

    Sf(t)= \frac{k}{2}  + \sum_{n = 1}^\infty\frac{2k}{(2n-1)\pi  } {(-1)}^{ n+1} \cos((2n-1)t)

    which i think is right,

    but now i need to show that \frac{\pi}{4 } = [1-1/3+1/5-1/7+........]

    i know it is probably pretty simple but iam a bit stuck so what steps should i take to show this.
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  2. #2
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    Quote Originally Posted by olski1 View Post
    okay, so i was given the piecewise function f(t)= k if -pi/2< x< pi/2
    0 if pi/2 < x < 3pi/2

    i computed the series to be

    Sf(t)= \frac{k}{2}  + \sum_{n = 1}^\infty\frac{2k}{(2n-1)\pi  } {(-1)}^{ n+1} \cos((2n-1)t)

    which i think is right,

    but now i need to show that \frac{\pi}{4 } = [1-1/3+1/5-1/7+........]

    i know it is probably pretty simple but iam a bit stuck so what steps should i take to show this.
    Try integrating from -pi to pi.

    CB
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  3. #3
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    Alternatively, evaluate it at t=0.
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  4. #4
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    If you let t=0, you can arrange it so you end up with pi/4 in the LHS and corresponding infinite series in the RHS.
    Last edited by Mondreus; May 9th 2011 at 03:50 AM. Reason: Ninja'd
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  5. #5
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    why t = 0?
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  6. #6
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    okay so let me see if i have it right,

    i set my fourier series of f(t) = k, set t =0 because it is within the domain of k
    and do some re arranging and get the sum = pi/4.

    and i got the correct answer.

    is that correct? and is that why i choose t=0?
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  7. #7
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    Yes, it's a common trick to evaluate infinite series using the fourier representation of a function and for a certain t. In this case we chose t=0 because it makes the cos term become 1 for all values of n. Another choice could have been 2pi, but then the LHS wouldn't be defined.
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