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The problem

Can be $\zeta(3)$ written as $\alpha\pi^\beta$, where ($\alpha,\beta \in \mathbb{C}$), $\beta \ne 0$ and $\alpha$ doesn't depend of $\pi$ (like $\sqrt2$, for example)?

Details

Several $\zeta$ values are connected with $\pi$, like:

$\zeta$(2)=$\pi^2/6$

$\zeta$(4)=$\pi^4/90$

$\zeta$(6)=$\pi^6/945$

...

and so on for all even numbers.

See this mathworld link to more details: Riemann Zeta Function

So the question is, could $\zeta(3)$ be written as:

$$\zeta(3)=\alpha\pi^\beta$$ $$\alpha,\beta \in \mathbb{C}$$ $$\beta \ne 0$$ $$\alpha \text{ not dependent of } \pi$$

See $\alpha$ not essencially belongs $\mathbb{Q}$ and $\alpha,\beta$ could be real numbers too.

When I wrote $\alpha$ is not dependent of $\pi$ it's a strange and a hard thing to be defined, but maybe $\alpha$ can be written using $e$ or $\gamma$ or $\sqrt2$ or some other constant.

Edit:

Maybe this still a open question. If

$ \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!}$

in $-4 \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!}\pi^3$ be of the form $\frac{\delta}{\pi^3}$ with $\delta$ not dependent of $\pi$

and $- 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}$ not dependent of $\pi$ too, this question still hard and open.

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See Alex Bartel's answer below; it is not expected that $\zeta(3)$ will be related to $\pi$ in a meaningful way! –  Matt E Apr 28 '11 at 1:42
    
@Matt E : what is meant by meaningful? is 3 more meaning full than $\zeta(3)$ ? –  Arjang Apr 28 '11 at 2:08
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@Arjang: Dear Arjang, What I mean is that $\zeta(3)$ is not expected to be a rational multiple of $\pi$, and indeed is expected to be algebraically independent of $\pi$. In fact, all the positive odd zeta values are expected to be algebraically independent of one another, and of $\pi$. So they should satisfy no algebraic relations among themselves; this is what I intended by saying that there is no meaningful relationship. Regards, –  Matt E Apr 28 '11 at 2:19
    
@Arjang: PS. I'm not sure what you mean by "is $3$ more meaningful then $\zeta(3)$". –  Matt E Apr 28 '11 at 2:20
    
Hi @Matt, your comment about algebraic independence is very interesting, +1 to that! Regards –  Arjang Apr 28 '11 at 2:37

5 Answers 5

up vote 35 down vote accepted

The question is whether or not $\zeta(3)$ is connected with $\pi$. The answer is yes. Moreover, $\zeta(3) = \beta \pi^{\alpha}$ for some complex $\alpha, \beta$. Take $\alpha = 3$ and $\beta = 0.0387682...$. It is not known whether $\beta = 0.0387682...$ is algebraic or transcendental. It is known, however, that $\zeta(3)$ is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. If $\alpha, \beta > 0$ such that $\alpha \beta = \pi^{2}$, then for each non-negative integer $n$, \begin{align} \alpha^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \alpha} - 1} \right) & = (- \beta)^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \beta} - 1} \right) - \end{align} \begin{align} \qquad 2^{2n} \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n - 2k + 2)!} \alpha^{n - k + 1} \beta^{k}. \end{align} where $B_n$ is the $n^{\text{th}}$-Bernoulli number.

For odd positive integer $n$, we take $\alpha = \beta = \pi$, \begin{align} \zeta(2n+1) = -2^{2n} \left( \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n - 2k + 2)!} \right) \pi^{2n+1} - 2 \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 \pi k} - 1}. \end{align}

In particular, for $n = 1$, \begin{align} \zeta(3) = -4 \left( \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!} \right) \pi^{3} - 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}. \end{align}

Observe that the coefficient of $\pi^{3}$ is rational, however, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that $\frac{\zeta(3)}{\pi^{3}}$ is transcendental.

Update: Recently, Takaaki Musha claims to have proved that $\frac{\zeta(2n+1)}{(2 \pi)^{2n+1}}$ is irrational for positive $n \geq 1$. However, some objection has since been raised (read comments below).

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1  
I apologize for being a little tardy for this nice discussion. ;) –  user02138 May 9 '11 at 14:37
    
No no user02138. Thanks a lot about this explanation. Now I will try search more things about that infinity sum. –  GarouDan May 10 '11 at 15:22
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@GarouDan: Alfred van der Poorten's A Proof that Euler missed ..., section 2 ift.uni.wroc.pl/~mwolf/Poorten_MI_195_0.pdf , wrote that Elementary evaluation of ζ(2n). Maths Mag. 48 (1945), pp. 148--153 by Bruce C. Berndt, "gives many others formulas and detailed references". –  Américo Tavares May 11 '11 at 10:41
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Do you have any more information on the claim in your update? I can't seem to find a preprint or anything other than just generic information on Takaaki Musha. –  Daenerys Naharis Nov 10 '11 at 20:58
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An anonymous suggested edit to this answer read: "I (F.M.S. Lima) have found an error in the proof of Musha's 'theorem' 1, in the above mentioned article (published in "JP Journal of Algebra, Number Theory and Applications"). The error occurs at the end of p.45, where the possibility of existence of a matrix A with det(A)=0 was missed. I've sent a message alerting Musha of this error(in 03/11/2012)." I rejected this because it was more properly a comment than an edit, and have thus reproduced it as such. –  anon Mar 11 '12 at 4:54

Any complex number can be written as $\alpha \pi^\beta$ if you make no assumptions on $\alpha$ and $\beta$. You can even take $\beta = 0$.

It is however conjectured that $\zeta(2n+1)/\pi^{2n+1}$ is irrational for all $n \geq 1$. This conjecture is wide open.

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1  
Does this conjecture just come from the fact that for a "generic" arbitrary real number $x$, $x/\pi^{2n+1}$ will be irrational? Or is there some more specific reason to suspect the number is not rational? –  Jason DeVito Apr 27 '11 at 17:05
    
I edited the question... i think is more precisely now. I am not searching for a rational number, maybe could be a irrational...a strange irrational. Maybe $\alpha$ uses $\gamma$ (constant). –  GarouDan Apr 27 '11 at 19:08
    
@Jason: see the MO question in Alex Bartel's answer. There is a web of conjectures here related (as I understand it) to the theory of motives, but it seems to be, as Alex says, pretty technical stuff. –  Qiaochu Yuan Apr 27 '11 at 19:10
    
@Garou: Even with that relaxed criterion, I believe your question remains unsolved. The only progress I remember being made with Apery's constant, for instance, was it being proven irrational. –  J. M. Apr 27 '11 at 19:11
8  
@Jason: Dear Jason, There are specific reasons, related to the geometry behind the period integrals that compute these zeta values, for conjecturing this irrationality. So it is not just conjectured on the basis of a genericity argument. In fact, the algebraic relations between the zeta values (or rather, the absence of such --- see my comments below the question) should reflect the highly structured geometry (the geometry of mixed Tate motives) that underlies the period integrals giving rise to the zeta values. Best wishes, –  Matt E Apr 28 '11 at 2:49

See this question at MO. The short answer is that we don't know whether $\zeta(3)/\pi^3$ is rational (I assume that that's what you meant), but nobody seriously believes it is. Indeed, it shouldn't be, instead it should be connected to a certain higher regulator (google for Borel regulator and/or Lichtenbaum conjecture if you want to know more, but beware that it's pretty technical stuff).

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I'm hoping it'll turn out to be related somehow to OEIS A104007

$\zeta(3)=c \cdot \pi^3/60$

$\zeta(5)=c \cdot \pi^5/378$

$\zeta(7)=c \cdot \pi^7/2700$

where $c$ is either a constant or a function of $n$. The sequence 6,60,90,378,945,2700 are the denominators of coefficients in the expansion of $\frac{x^2}{(1-e^{-2x})^{2}}$.

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3  
By numerics alone you can rule out the possibility that such a constant $c$ exists. –  Gerben Apr 27 '11 at 14:32
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...and as a function of $n$, $c(n)$ exists (trivially), but may not take on values only in $\mathbb{Q}$. So, the current statement in the answer appears vacuous. –  cardinal Apr 27 '11 at 14:46
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As is stated in the comments to the MO question, this is not actually a natural conjecture to make. There are good reasons one should expect the zeta function to behave very differently at the odd and the even integers. –  Qiaochu Yuan Apr 28 '11 at 0:58
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@Mitch: actually, my understanding is that it's not natural to expect a simple expression in terms of known constants. Again, there are deep conjectures about what should happen, but I'm not at all familiar with them. –  Qiaochu Yuan Apr 28 '11 at 1:15
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Dear Mitch, Although there are not many theorems proved about $\zeta$ at odd integers, there is a well-developed theoretical picture which is expected to explain these values. Unlike in the case of positive even integers, it is not expected that the values at the positive odd integers should be related to one another, or to $\pi$. Regards, –  Matt E Apr 28 '11 at 2:16

I'll answer your modified question, as explained in your comment:

I am not searching for a rational number, maybe could be a irrational...a strange irrational. Maybe α uses γ (constant).

It's known through LLL testing that there is no simple form of $\zeta(3)$ involving low powers of $\pi$, certain other constants like $\gamma,$ and simple (small denominator) rationals. It's not expected that such a form exists at all, but of course this is not known (and probably won't be any time soon).

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