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Math Help - Another cant get this to equal this :)

  1. #1
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    Another cant get this to equal this :)

    So after the half range fourier series of sinx i have got the left hand side.

    But my teachers answer has got this on the right how do i get it to equal this?



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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    So after the half range fourier series of sinx i have got the left hand side.

    But my teachers answer has got this on the right how do i get it to equal this?



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    Look at n even and odd separately.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Look at n even and odd separately.
    Ive done that but i cant get the answer i think my arithmetic is wrong.
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by adam_leeds View Post
    Ive done that but i cant get the answer i think my arithmetic is wrong.
    hmmm...

     \frac{1}{ \pi } [ \frac{1}{n+1} (-1)^n - \frac{1}{n-1} (-1)^n + \frac{1}{n+1} - \frac{1}{n-1} ]

    Lets work with the inside for now,

    \frac{1}{n+1} (-1)^n - \frac{1}{n-1} (-1)^n + \frac{1}{n+1} - \frac{1}{n-1}

    Group terms

     \frac{1}{n+1} ( 1 +(-1)^n) - \frac{1}{n-1} ( 1 + (-1)^n)

    Note that if  n = odd then the above must equal 0. So n must be even, which means

     \frac{1}{n+1} ( 1 +(-1)^n) - \frac{1}{n-1} ( 1 + (-1)^n) \to \frac{2}{n+1} - \frac{2}{n-1}

    This gives us

     \frac{ 2(n-1) -2(n+1) }{(n+1)(n-1)}

     - \frac{ 4} {(n+1)(n-1)}

    So my answer will be,

     - \frac{4}{ \pi} \frac{ 1} {(n+1)(n-1)}

    Is this the same as theres? Lets change it around

    If you sub in  n = 2 into their equation and mine you arrive at  - \frac{4}{3 \pi }

    So I think they are equal. let's equate them

     - \frac{4}{ \pi} \frac{ 1} {(n+1)(n-1)} = -\frac{2}{ \pi } \frac{ (-1)^n +1 } {n^2 - 1}

     2 \frac{ 1} {n^2 - 1} = \frac{ (-1)^n + 1 }{n^2 - 1}

    Note how these are the same if n=even!

    However, if n=odd the right side is equal to 0, which is the same result as I achieved, however i had the stipulation before making my equation that n cannot be odd.

    Thus, left side = right side!
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  5. #5
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    Quote Originally Posted by adam_leeds View Post
    Ive done that but i cant get the answer i think my arithmetic is wrong.
    \frac{1}{n+1}-\frac{1}{n-1} = \frac{n-1 - (n+1)}{n^2-1} = \frac{-2}{n^2-1}

    So when n is even you have two of these inside the brackets, making overall \frac{-4}{\pi(n^2-1)} and when n is odd you have a positive and negative of these canceling out to 0.
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  6. #6
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    You two are legends
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