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Math Help - Basel problem application

  1. #1
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    Basel problem application

    Prove that \frac{\pi^2}{8} = \sum \frac{1}{(2n-1)^2}

    How do I go about starting this, knowing basel's problem?

    Basel problem - Wikipedia, the free encyclopedia
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  2. #2
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    Quote Originally Posted by Lord Darkin View Post
    Prove that \frac{\pi^2}{8} = \sum \frac{1}{(2n-1)^2}

    How do I go about starting this, knowing basel's problem?

    Basel problem - Wikipedia, the free encyclopedia
    Start by noting that  \sum_{n=1}^{+ \infty} \frac{1}{n^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2} + \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2}.
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    Okay. I'm not sure how you got that, though.

    Also, I'm not sure how to proceed even with that.

    Am I supposed to use a taylor series anywhere during this problem? I've looked at Wiki's page and it says that sinx/x was a proof used to find pi^2/6
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Lord Darkin View Post
    Okay. I'm not sure how you got that, though.

    Also, I'm not sure how to proceed even with that.

    Am I supposed to use a taylor series anywhere during this problem? I've looked at Wiki's page and it says that sinx/x was a proof used to find pi^2/6
    As Mr. Fantastic Suggested he is breaking the sum up into the even and odd parts.

    for example

    \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}  {4^2}+\frac{1}{5^2}+...=\left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{2^2}+\frac{1}{4^2}+...\right)

    As another hint factor a 2^2 out of the even terms on the right hand side and compare the right and left hand sides of the equality.
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post

    \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}  {4^2}+\frac{1}{5^2}+...=\left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{2^2}+\frac{1}{4^2}+...\right)

    As another hint factor a 2^2 out of the even terms on the right hand side and compare the right and left hand sides of the equality.
    \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}  {4^2}+\frac{1}{5^2} + ... = \left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2

    Then that gives me

    \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}  {8^2}+\frac{1}{10^2}+...= +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2

    So ... I have now factored out the 2 squared. But that just leaves me with the same thing I had ... ?
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  6. #6
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    Quote Originally Posted by Lord Darkin View Post
    \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}  {4^2}+\frac{1}{5^2} + ... = \left( \frac{1}{1^2}+\frac{1}{3^2}+...\right) +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2

    Then that gives me

    \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}  {8^2}+\frac{1}{10^2}+...= +\left( \frac{1}{4^2}+\frac{1}{6^2}+...\right) * 2^2

    So ... I have now factored out the 2 squared. But that just leaves me with the same thing I had ... ?
    Please go back and read the replies you got again. In particular, my first reply shows exactly what to do.
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  7. #7
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    \sum_{n=1}^{+ \infty} \frac{1}{n^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2} + \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2}

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

    ( \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}  {4^2}+\frac{1}{5^2} + ... ) - ( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}  {8^2}+\frac{1}{10^2} + ... ) = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}

    ( \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}  {7^2}+\frac{1}{9^2} + ... ) = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2}

    Sorry about this I tried looking at Mr fantastic posts but I still cannot deciper it. I know that I'm close but I can't get it.
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  8. #8
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    Quote Originally Posted by Lord Darkin View Post
    \sum_{n=1}^{+ \infty} \frac{1}{n^2} = \sum_{n=1}^{+ \infty} \frac{1}{(2n-1)^2} + \sum_{n=1}^{+ \infty} \frac{1}{(2n)^2}

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

    [snip]
    Therefore:

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

    and I hope it's clear what happens next.
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