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Math Help - Pi Identity

  1. #1
    Senior Member DivideBy0's Avatar
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    Pi Identity

    In the 18th Century, Euler proved the remarkable fact that

    1/1^2 + 1/2^2 + 1/3^2+ 1/4^2 + 1/5^2 ... = pi^2/6

    Use this to determine the value of

    1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 ...


    Please help me on this... thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by DivideBy0 View Post
    In the 18th Century, Euler proved the remarkable fact that

    1/1^2 + 1/2^2 + 1/3^2+ 1/4^2 + 1/5^2 ... = pi^2/6

    Use this to determine the value of

    1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 ...


    Please help me on this... thanks.
    1/1^2 + 1/2^2 + 1/3^2+ 1/4^2 + 1/5^2 ... =

    ....[1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 ...] +

    .............[1/2^2 + 1/4^2 + 1/6^2 + 1/8^2 + 1/10^2 ...] =

    ....[1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 ...] +

    .............. [1/[1/4][1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 ...]

    So we have:

    1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 ... = [3/4][1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 ...]

    ................. = [3/4][pi^2/6] = pi^2/8

    RonL
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  3. #3
    Senior Member DivideBy0's Avatar
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    I don't understand the step where you take a quarter of the reciprocated even squares.

    How did you come to the conclusion that

    [1/2^2 + 1/4^2 + 1/6^2 + 1/8^2 + 1/10^2] = 1/4[1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2]

    Can you take maybe a smaller example to show me how this works? Thanks.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by DivideBy0 View Post
    I don't understand the step where you take a quarter of the reciprocated even squares.

    How did you come to the conclusion that

    [1/2^2 + 1/4^2 + 1/6^2 + 1/8^2 + 1/10^2] = 1/4[1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2]

    Can you take maybe a smaller example to show me how this works? Thanks.
    The n-th even number (counting 2 as the first) is 2n, and:

    1/(2n)^2 = (1/2^2)(1/n^2) = (1/4)(1/n^2)

    RonL
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