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Sum of 1/(n^2) as n goes to infinity

  1. Oct 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove Ʃ1/(n^2) as n goes to infinity = (∏^2)/8


    2. Relevant equations



    3. The attempt at a solution
    No idea how to start. Pls guide.

    Thanks
     
  2. jcsd
  3. Oct 12, 2013 #2

    Ray Vickson

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    You need to show your work, first. Anyway, what is the context? Is this a problem in a course? If so, what is the course subject (analytic function theory, differential equations, Fourier analysis...)?
     
  4. Oct 12, 2013 #3

    arildno

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    You won't be able to prove it, since the result is wrong. The sum converges to [itex]\frac{\pi^{2}}{6}[/itex] instead
     
  5. Oct 12, 2013 #4

    I got that from a manual solution, I don't know where should I ask my question so I put it here .
    This answer is for quantum mechanics problem. I don't think that is Fourier series.
     

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  6. Oct 12, 2013 #5

    arildno

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    Well, in that thumbnail, it isn't 1/n^2 for n the integers, is it?
     
  7. Oct 12, 2013 #6

    berkeman

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    Check your PMs -- you must show your work to receive tutorial help (and please do not use txt speak here like "pls")
     
  8. Oct 12, 2013 #7

    Ray Vickson

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    You gave us the wrong problem; it should be
    [tex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}.[/tex]
    I would be willing to bet this can be obtained from an appropriate Fourier series, but I will leave the fun of discovery to you.
     
  9. Oct 13, 2013 #8
    Sorry for my mistakes. Thanks for replying.
    Do I need to know f(x)= ? for solving this problem? Like f(x)=x for showing 1 - (1/3) + (1/5)....= ∏/4 .as shown below
     

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  10. Oct 13, 2013 #9

    Dick

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    Use the same function, but instead of evaluating the series at a point, use Parseval's identity.
     
  11. Oct 13, 2013 #10
    f(x)= x .Range? -∏ <x< ∏ ?
    What do you mean by evaluating the series at a point? Fourier series evaluating at a point
    Complex fouries series and Parseval's theorem ?
    Why do we need two types?
     
  12. Oct 13, 2013 #11

    Dick

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    Yes, exactly the same function, exactly the same series. Did you look up Parseval's theorem? Evaluating the series gives you things like 1/n, Parseval's gives you things like 1/n^2. Just try it.
     
  13. Oct 13, 2013 #12
    I got the answer by using parseval's theorem when f(x)=x over the interval -∏<x<∏.
    Thanks

    But when do I know I need to use complex Fourier series and parseval's theorem? Is that because there is a square? (2m+1)^2
    [tex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}.[/tex]

    Can I say Fourier series is used when my conjugate ,g(x) is not complex. Complex conjugate of g(x) = g(x), Eg. ∫g(x)f(x)dx= c
    But when g(x) is complex , complex conjugate of g(x) is not equal to g(x),then I should use complex Fourier series?
    This is my own conclusion after reading . Please Correct if it's wrong.

    I am still don't understand what do you mean evaluate the series at a point, is that meant f(x)= x , the x should not be a point, so I have to integrate by using parseval's , but how do you know we can't treat it as a point?
     
  14. Oct 13, 2013 #13

    Dick

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    Yes, the square is a hint you might need Parseval. But you didn't really finish the proof you presented in post 8. Could you do that? Then you might see why evaluation of the Fourier series at a point can give you the sum of some series.
     
  15. Oct 13, 2013 #14

    Ray Vickson

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    You ask "...how do you know...?" You don't! You can try 9 methods that lead to failure, while the 10th method works. You may generate dozens of pages of scrap paper full of failed attempts, and you may spend many hours on "wasted" efforts (not really wasted, though---they teach you something). Be assured, all the posters here that are professors, etc., have gone through this type of experience many times in the past.

    However, there are some lessons you can take with you. Often (not always!) when we have sums involving π or π2, etc. on the right, it will be a result of either evaluating some definite integral in two ways---once using some method that gives the right-hand-side directly, and the other by integrating a Taylor series term-by-term, for example; or it may come somehow from Fourier series. You just get to "guess" these things after gaining experience through working lots of problems. Maybe you don't know this stuff going into a course, but hopefully you do (at least a bit) when you have finished the course! That's called education.
     
  16. Oct 13, 2013 #15
    Thank you guys.



    Understood~
     
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