Challenge 1: Multiple Zeta Values

  • #1
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99

Main Question or Discussion Point

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]
 

Answers and Replies

  • #2
1,796
53
[tex] \zeta(2,2) = \sum_{m =1}^{\infty} \sum_{n = 1}^{m-1} \frac{1}{ m^2 n^2} [/tex].
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?
 
  • #3
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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)
 
  • #4
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
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.
 
  • #5
CompuChip
Science Advisor
Homework Helper
4,302
47
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.
Be careful. Last time I heard someone say that it took about 3 centuries before they got it right.
 
  • #6
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
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.
 
  • #7
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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.
 
  • #8
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
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.
 
  • #9
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
jbunniii, the relation [itex] \zeta(4) = 4\zeta(3,1)[/itex] might be a red herring....
 
  • #10
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
jbunniii, the relation [itex] \zeta(4) = 4\zeta(3,1)[/itex] might be a red herring....
Well, that's an intriguing remark... OK, I'll try to think about this a bit more tonight.
 
  • #11
612
23
jbunniii, the relation [itex] \zeta(4) = 4\zeta(3,1)[/itex] might be a red herring....
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.
 
  • #12
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
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.
 
  • #13
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
Oops, deleted my last message due to some indexing errors. I'll save it and post an edited version later.
 
  • #14
612
23
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:
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.
 
  • #15
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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:
  • #16
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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.
 
  • #17
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
There still exists a completely elementary three line proof waiting to be found!
 
  • #18
2,111
17
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.
 

Related Threads on Challenge 1: Multiple Zeta Values

Replies
1
Views
1K
Replies
28
Views
4K
Replies
8
Views
2K
Replies
4
Views
826
  • Last Post
Replies
8
Views
1K
Replies
8
Views
2K
Replies
26
Views
3K
Replies
49
Views
4K
  • Last Post
Replies
3
Views
591
Replies
5
Views
1K
Top