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Infinite Summations

  1. Feb 5, 2006 #1

    eep

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    I'm having a lot of trouble with infinite summations. For example, I need to sum the following quantity:

    [tex]\sum_{n=1,3,5...}^{\infty}\frac{1}{n^2}[/tex]

    I tried to find the result on the internet but to no avail. I really have no idea where to begin when evaluating infinite sums, so if someone could please recommend a good book or point me to a good resource, I'd greatly appreciate it.
     
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  3. Feb 5, 2006 #2

    AKG

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    Observe that:

    [tex]\sum _{n\ odd} \frac{1}{n^2} = \left(\sum _{n \in \mathbb{N}}\frac{1}{n^2}\right) - \left(\sum _{n\ even}\frac{1}{n^2}\right)[/tex]

    [tex] = \left(\sum _{n \in \mathbb{N}}\frac{1}{n^2}\right) - \left(\sum _{n = 2,4,6,\dots}^{\infty}\frac{1}{n^2}\right)[/tex]

    [tex] = \left(\sum _{n \in \mathbb{N}}\frac{1}{n^2}\right) - \left(\sum _{m = 1,2,3,\dots}^{\infty}\frac{1}{(2m)^2}\right)[/tex]

    [tex] = \left(\sum _{n \in \mathbb{N}}\frac{1}{n^2}\right) - \left[\frac{1}{4}\left(\sum _{m = 1,2,3,\dots}^{\infty}\frac{1}{m^2}\right)\right][/tex]

    [tex] = \frac{3}{4}\left(\sum _{n \in \mathbb{N}}\frac{1}{n^2}\right)[/tex]

    I'm not sure what that sum is. We all learn that it converges, I don't recall if we ever learn how to compute it. It's 1.6 something or 1.5 something.
     
    Last edited: Feb 5, 2006
  4. Feb 6, 2006 #3

    quasar987

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    In the context of which class is this question asked? If you're allowed to use Fourier series, than it can be done. Otherwise, I don't recall of a way either.
     
  5. Feb 6, 2006 #4

    eep

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    Fourier series would be fair game, I suppose. This is from a Griffiths' Quantum Mechanics problem in which given [itex]\psi(x,0)[/itex] you're supposed to find [itex]\psi(x,t)[/itex] using fourier series. The problem then asks me for the expected energy, which left me with the infinte sum from above.
     
  6. Feb 6, 2006 #5

    quasar987

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    K, then here's a way:

    Develop the function f(x) = x defined on [0,pi] in a serie of cosines (i.e. consider f as a mere restriction of an even function f*, and expand f* in a Fourier serie as you normally would)

    The serie will be summed over odd integers and 0 only and the coefficients of that serie will be a_0 = 1 and a_n = -4/(pi² n²). Now set x = 0. All the cos(nx) terms become 1 and you can solve for

    [tex]\sum_{n=1,3,5...}^{\infty}\frac{1}{n^2}[/tex]

    And also find

    [tex]\sum_{n \in \mathbb{N}}^{\infty}\frac{1}{n^2}[/tex]

    following AKG'S procedure if you desire.

    Funny how the odd part of the serie makes for a whooping 3/4 of its sum and the even part only 1/4.

    I find

    [tex]\sum_{n=1,3,5...}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{4} \Rightarrow \sum_{n \in \mathbb{N}}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{3}[/tex]

    But this does not seem right, as after 17 terms, it can be established that the 2 first digits of the sum are 1.5 < pi²/3 :(
     
    Last edited: Feb 6, 2006
  7. Feb 6, 2006 #6

    arildno

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    Eeh???
    [tex]\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}[/tex]
     
  8. Feb 6, 2006 #7

    Hurkyl

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    A friend of mine had a novel way of doing these -- get a numerical estimate, and then try multiplying or dividing the results by appropriate powers of pi. (He had ways of telling which powers were good) When you get one that looks like a rational number, you have good odds that's your sum!
     
  9. Feb 6, 2006 #8

    quasar987

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    Well indeed, but can you see the flaw in my reasoning? The values a_0 = 1 and a_n = -4/(pi² n²) are taken straight from the solution manual of my analysis textbook.
     
  10. Feb 6, 2006 #9

    AKG

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    Hurkyl, how would that work exactly? Given your initial guess, you're going to change it by at least a factor of [itex]\pi[/itex]. This method is only good if your initial guess is thus off by at least a factor of [itex]\pi[/itex], i.e. your initial guess has to be way off. Am I missing something?
     
  11. Feb 6, 2006 #10

    Hurkyl

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    You add up the first two terms of your sum and get, say, 1.6448. You suspect there's a pi^2 factor, because it's a sum of inverse squares, so you divide by pi^2 leaving 0.16665. That smells like 1/6, so you decree that the sum is pi^2 / 6.
     
  12. Feb 6, 2006 #11

    AKG

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    I see. I tried it with 1/n³, and summing the first 10 terms gives 1.197532... Dividing by [itex]\pi ^3[/itex] gives 0.03862... which is 1/25.89182... As more terms are added, the estimate will increase, so the result of dividing by [itex]\pi ^3[/itex] will increase so expressing it in the form 1/x will make the x decrease. So if anything, the 25.89182... will have to decrease to 25, which seems to drastic a drop. Maybe it's not [itex]\pi ^3/x[/itex] but [itex]y\pi ^3/x[/itex] where y and x are co-prime, y is not 1. Or it may not be [itex]\pi ^3[/itex] at all. However, I tried it with a few other powers (1st, 2nd, 4th, and 6th) and none of them give anything that looks like 1/x in the end, for some integer x (i.e. the x that comes out is far away from an integer).

    I tried it with [itex]1/n^4[/itex], and dividing by [itex]\pi ^4[/itex] it seems that maybe the answer should be:

    [tex]\frac{\pi ^4}{90}[/tex]

    Does anyone know if this is right? Or even better, does anyone know an online resource where the values of various sums are given?
     
  13. Feb 6, 2006 #12

    Hurkyl

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  14. Feb 6, 2006 #13

    quasar987

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    It helps to know that zeta(p) is of the form pi^p/something only if p is even. That way you know the trick of your friend will only work for p even. That would explain your problem with zeta(3) AKG. But I like the trick too. It can be useful in an exam.
     
  15. Feb 7, 2006 #14

    AKG

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    Thanks Hurkyl.
     
  16. Feb 7, 2006 #15

    matt grime

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    Not really, the odd terms contain 1, which is distinctly larger than any of the other entries, and takes up a signficant part of pi^2/6 on its own (about 2/3 of it). Correspondingly the odd terms will be even more significant if we were to look at other higher (lower) powers of 1/n (n)
     
  17. Feb 7, 2006 #16

    shmoe

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    Check how your text defines the fourier series and coefficients. There's probably a 1/2 term in front of the a_0 term in the series.
     
  18. Feb 7, 2006 #17

    quasar987

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    Ah! Yes, I had forgotten about that! =)
     
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