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Challenge 1: Multiple Zeta Values

  1. Sep 26, 2013 #1

    Office_Shredder

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    A multiple zeta value is defined as
    [tex] \zeta(s_1,...,s_k) = \sum_{n_1 > n_2 ... > n_k > 0} \frac{1}{n_1^{s_1} n_2^{s_2}...n_k^{s_k}} [/tex].
    For example,
    [tex] \zeta(4) = \sum_{n = 1}^{\infty} \frac{1}{n^4} [/tex]
    and
    [tex] \zeta(2,2) = \sum_{m =1}^{\infty} \sum_{n = 1}^{m-1} \frac{1}{ m^2 n^2} [/tex].

    Prove the following relationship:
    [tex] \zeta(2)^2 = 4 \zeta(3,1) + 2 \zeta(2,2) [/tex]
     
  2. jcsd
  3. Sep 26, 2013 #2
    This looks like when m=1, then n goes from 1 to 0 causing the denominator to be zero. Am I mis-interpreting your double sum and if so could you clarify please?
     
  4. Sep 26, 2013 #3

    Office_Shredder

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    The intention is that there are no entries in the sum when m = 1 (if you check the general definition the smallest value n1 can be is 2 if k=2)
     
  5. Sep 26, 2013 #4

    jbunniii

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    I hope I'm not violating the rules by posting a partial solution. The FAQ is silent on the subject.

    First we note that, by symmetry of ##m## and ##n##,
    $$\sum_{m=1}^{\infty}\sum_{n > m} \frac{1}{m^2 n^2} = \sum_{m=1}^{\infty}\sum_{n < m} =
    \frac{1}{m^2 n^2} = \zeta(2,2)$$
    Therefore, we may write
    $$\begin{align}
    \zeta(2)^2 &= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2 n^2} \\
    &= \sum_{m=1}^{\infty} \sum_{n<m} \frac{1}{m^2 n^2} + \sum_{m=1}^{\infty} \sum_{n>m} \frac{1}{m^2 n^2} + \sum_{n=1}^{\infty} \frac{1}{n^4} \\
    &= 2\zeta(2,2) + \zeta(4)\end{align}$$
    So the problem reduces to showing that ##\zeta(4) = 4\zeta(3,1)##. I have a proof for this but the margin is too small to contain it. I'll see if I can find a shorter one.
     
  6. Sep 27, 2013 #5

    CompuChip

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    Be careful. Last time I heard someone say that it took about 3 centuries before they got it right.
     
  7. Sep 27, 2013 #6

    jbunniii

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    I've been playing around with the usual trick of expressing the series in terms of a power series so calculus techniques can be applied. Formally (without justifying the interchanges of sum and integral), we have:
    $$\begin{align}
    \zeta(4) &= \sum_{n=1}^{\infty}\frac{1}{n^4}\\
    &= \left. \sum_{n=1}^{\infty}\frac{x^n}{n^4}\right|_{x=1} \\
    &= \sum_{n=1}^{\infty} \frac{1}{n^3} \left.\left(\frac{x^n}{n}\right)\right|_{x=1} \\
    &= \sum_{n=1}^{\infty} \frac{1}{n^3} \int_{0}^{1} x^{n-1} dx \\
    &= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{x^{n-1}}{n^3} dx\\
    &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^n}{n^3} dx
    \end{align}$$
    Now we can use almost the same trick with ##\sum x^n / n^3## as we did with ##\sum x^n / n^4##. (Unfortunately we'll have a couple more iterations of this.)
    $$\begin{align}
    &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{1}{n^2} \left(\frac{x^n}{n}\right) dx\\
    &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{1}{n^2} \left(\int_{0}^{x} y^{n-1}dy\right) dx\\
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \sum_{n=1}^{\infty} \frac{y^{n-1}}{n^2} dy dx \\
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \sum_{n=1}^{\infty} \frac{y^n}{n^2} dy dx \\
    \end{align}$$
    And here we go again...
    $$\begin{align}
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{y^n}{n}\right) dy dx\\
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \sum_{n=1}^{\infty} \frac{1}{n} \left(\int_{0}^{y}z^{n-1}dz\right) dy dx \\
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \sum_{n=1}^{\infty} \frac{z^{n-1}}{n} dz dy dx \\
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \sum_{n=1}^{\infty} \frac{z^n}{n} dz dy dx \\
    \end{align}$$
    And one last time...
    $$\begin{align}
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \sum_{n=1}^{\infty} \int_{0}^{z} w^{n-1} dw dz dy dx \\
    &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \int_{0}^{z} \sum_{n=1}^{\infty} w^{n-1} dw dz dy dx\\
    \end{align}$$
    Now ##\sum_{n=1}^{\infty}w^{n-1} = 1/(1-w)## for ##|w| < 1##, so the above reduces to
    $$\begin{align}
    \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \int_{0}^{z} \frac{1}{1-w} dw dz dy dx\\
    \end{align}$$
    Unfortunately after all this work, it's still not clear whether it will get us anywhere. But the next step would be to start manipulating ##\zeta(3,1)## in a similar way and see if we can get a match. Even then, we won't be done, as all the integral-sum interchanges will have to be justified. I think this will be no big deal as we can invoke the uniform convergence of power series.
     
  8. Sep 29, 2013 #7

    Office_Shredder

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    That's a really nice calculation jbunniii - I thought I'd just give you a bit of encouragement by saying that you can get similar formulas for the other zeta values.
     
  9. Sep 29, 2013 #8

    jbunniii

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    I've spent some time obtaining a similar expression for ##\zeta(3,1)## but I haven't worked out how to manipulate it to the expression for ##\zeta(4)## with an extra factor of 4. I got tired of that after a while and tried a different approach using summation by parts, but that hasn't worked out yet either. Will keep playing with it as time allows.
     
  10. Oct 1, 2013 #9

    Office_Shredder

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    jbunniii, the relation [itex] \zeta(4) = 4\zeta(3,1)[/itex] might be a red herring....
     
  11. Oct 1, 2013 #10

    jbunniii

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    Well, that's an intriguing remark... OK, I'll try to think about this a bit more tonight.
     
  12. Oct 1, 2013 #11
    I don't think so. I'll start from the beginning.

    I remember, from a kick I had in complex analysis, the beautiful formula $$\zeta(m,n)+\zeta(n,m)=\zeta(m)\zeta(n)-\zeta(m+n).$$

    Letting ##n=m=2## gives ##2\zeta(2,2)=\zeta(2)^2-\zeta(4)##.

    After about 2 hours of getting nowhere with this, I broke down and asked the internet what it knew about multiple zeta functions. I found this source as a reference. I have used two of its theorems in the following.

    To start, I use the Sum Theorem to obtain ##\zeta(4)=\zeta(3,1)+\zeta(2,2)##. Then, I use the pretty little identity shown below the Derivation Theorem to show ##3\zeta(3,1)=\zeta(2,2)##. Plugging this into the above formula, we get ##\zeta(4)=\zeta(3,1)+3\zeta(3,1)=4\zeta(3,1)##.

    Thus, ##\zeta(2)^2=2\zeta(2,2)+4\zeta(3,1)##. Halmos.
     
  13. Oct 1, 2013 #12

    jbunniii

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    Mandelbroth - It seems like you're bringing in some pretty specialized machinery for this. Are you prepared to prove your Sum Theorem and your pretty little identity? :biggrin:

    Also, I proved the ##\zeta(m,n)+\zeta(n,m)=\zeta(m)\zeta(n)-\zeta(m+n)## identity for the special case ##m=n=2## in post #4. No complex analysis required.

    I've assumed thus far that there must be a relatively elementary proof that doesn't require developing an extensive amount of theory about the zeta function. Slicing and dicing the sums the right way, most likely. Office_Shredder's most recent hint suggests that the way I sliced it in the first post might not be the right way, so I'm going back to the beginning to try again.
     
  14. Oct 1, 2013 #13

    jbunniii

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    Oops, deleted my last message due to some indexing errors. I'll save it and post an edited version later.
     
  15. Oct 1, 2013 #14
    There are proofs in the references given in the link. :tongue:

    I agree that there is, probably, a more elementary method. I'll keep working on it after I finish my Spanish homework, and I'll see where we are in the morning.
     
  16. Oct 1, 2013 #15

    Office_Shredder

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    The intent of the question was that it was supposed to be self-contained as far as necessary knowledge of multiple zeta values (and there are multiple ways to solve the problem like this that I know of, have no fear!)

    I will admit that was a solid combination of googling and applying of theorems.
     
    Last edited: Oct 1, 2013
  17. Oct 9, 2013 #16

    Office_Shredder

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    Well, since there hasn't been any posts here in a while I thought I would give a small prod by noting that another way of writing, for example, [itex] \zeta(2,2) [/itex] is
    [tex] \zeta(2,2) = \sum_{m,n=1}^{\infty} \frac{1}{m^2(m+n)^2} [/tex]
    and there are similar expressions for the other multiple zeta values that should be obvious. Maybe this form is a bit easier to work with for what you guys are trying.
     
  18. Nov 17, 2013 #17

    Office_Shredder

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    There still exists a completely elementary three line proof waiting to be found!
     
  19. May 28, 2014 #18
    We follow Office_Shredder's latest advice:

    [tex]
    \zeta(2,2) = \sum_{m=2}^{\infty} \sum_{n=1}^{m-1} \frac{1}{m^2n^2}
    = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^2n^2}
    = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^2n^2}
    [/tex]

    Similarly:

    [tex]
    \zeta(3,1) = \sum_{m=2}^{\infty} \sum_{n=1}^{m-1} \frac{1}{m^3n}
    = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^3n}
    = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^3n}
    [/tex]

    Our goal is to prove that

    [tex]
    \sum_{n,m=1}^{\infty} \frac{1}{n^2m^2} = \sum_{n,m=1}^{\infty}
    \Big(\frac{4}{(n+m)^3n} + \frac{2}{(n+m)^2n^2}\Big)
    [/tex]

    With finite amount of work we get

    [tex]
    \frac{4}{(n+m)^3n} + \frac{2}{(n+m)^2n^2} - \frac{1}{n^2m^2} = \cdots
    = \frac{(m-n)^3}{(n+m)^3n^2m^2}
    [/tex]

    The sum over this is zero due to the antisymmetricity with respect to [itex]n[/itex] and [itex]m[/itex]. In other words, the sums over the regions [itex]n>m[/itex] and [itex]n<m[/itex] cancel each other.
     
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