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Convergence to pi^2/6

  1. Feb 24, 2005 #1
    How does
    Sum (n^(-2)=(pi^2)/6

    Please tell me if this has been posted before (afraid :redface: )
    (in that case, i'll see the other post)
    Last edited: Feb 24, 2005
  2. jcsd
  3. Feb 24, 2005 #2
    Take f(x)=x. Then the Fourier coeffcients of f are [itex]a_n=0[/itex] and [itex]b_n=\frac{2}{n}(-1)^{n+1}[/itex]. Parseval's theorem says that:

    [tex]\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left(a_k^2+b_k^2)[/tex]

    Since the [itex]a_n[/itex] terms are all zero, this reduces to:

    [tex]\frac{1}{\pi}\int_{-\pi}^\pi x^2\:dx=4\sum_{n=1}^\infty\frac{1}{n^2}[/tex]

    The integral is easy enough to solve, and the left hand side reduces to [itex]2\pi^2/3[/itex]. Dividing both sides by four gives us:

  4. Feb 24, 2005 #3
    I see clearly now!--thanks

    (Will no one answer my "digit-factorial question" thread :frown: )
    Last edited: Feb 24, 2005
  5. Feb 24, 2005 #4


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    And let's not forget Euler's original method.Combining the series he found for [itex] \frac{\sin x}{x} [/itex] and the one from Taylor expansion,he was able to prove it...

  6. Feb 24, 2005 #5

    matt grime

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    And there's another way:


    Evalutate that as x and y both go from 0 to 1. Do it using a substitution, and then do it by replacing the fraction inside with its series expansion and ignore the convergence issues to rearrange sum and integral.
  7. Sep 1, 2011 #6
    So this is parsevals theorem?

    [tex] \frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2) [/tex]

  8. Sep 1, 2011 #7


    Staff: Mentor

    How about this?

    [tex] \frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty( a_k^2+b_k^2) [/tex]
  9. Sep 10, 2012 #8
    hello! i am new to this forum and it looks like a very nice place!

    sorry for my english, i hope everyone can understand it..

    sorry for spamming in this thread but it looks like it is the most close to what i need.

    i think i understood the answer master_coda gave but i don't understand why he choose f(x)=x..

    for instance in my exercise i am asked to verify this equation


    what f(x) should i choose for the calculation?
  10. Sep 10, 2012 #9

    I'm answering this, in spite of being an intent of "kidnapping" a thread because

    (1) it is, perhaps unwillingly, very close to the OP, and more important

    (2) This is a newcomer so he/she doesn't know (but now you do!).

    Check the following:

    $$\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\Longleftrightarrow \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

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