Got numerical support that for odd $n$, $\zeta(n)$ might be expressed in terms of the derivatives of $\zeta(\frac12)$.

Based on More Zeta Functions for the Riemann Zeros, Andre Voros, p.12, Table 3:

Conjecture: For odd $n$, $$\zeta(n) = \left(\frac{2}{(n-1)!} (\log(|\zeta|)^{(n)} (\frac12) - 2^n \beta(n))\right)/(2^n-1)$$

$\beta(n)$ is Dirichlet beta function and it is a rational multiple of $\pi^n$ for odd $n$. The derivative can be expressed in terms of $\zeta(\frac12),\zeta^{(k)}(\frac12)$

For $n=3$ get numerical support for:

$$\zeta(3) = (-\zeta'''(\frac12)/|\zeta(\frac12)| -3 \zeta''(\frac12) \zeta'(\frac12)/|\zeta(\frac12)|^2 -2 \zeta'(\frac12)^3/|\zeta(\frac12)|^3- \pi^3 / 4)/7 $$

The last equality holds with precision $10^4$ decimal digits. One can eliminate the first derivative since there is closed form for $\zeta'(\frac12)/\zeta(\frac12)$

Is this result true?

sage/mpmath code in case of typos of the latex.

#run in sage

import mpmath
from mpmath import mpf

def zeta3test():



    # eliminate the first derivative

    rh1= -z3/z12a -3*z2*z1/z12a**2 -2* z1**3/z12a**3

    #print 'rh',mpmath.chop(rh0-rh1)
    #rhs= (mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - mpmath.pi^3 / 4 )/(7)
    rhs= (rh1 - mpmath.pi**3 / 4 )/(7)
    print mpmath.chop(zeta3-rhs)

def conjecture1(n):

    voros, p. 12
    assert n%2==1
    a1= mpmath.zeta(n)
    a2= (2/factorial(n-1) * mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - 2**(n) * dirichletbeta(n))/(2**n-1)
    print mpmath.chop(a1-a2)

def dirichletbeta(s):
    dirichlet beta
    return 4**(-s) * (mpmath.hurwitz(s,1/4)-mpmath.hurwitz(s,3/4))
  • $\begingroup$ What do you mean with "the last equality holds to precision 10^-4"? If the result was true, shouldn't the equality hold up to any precision? $\endgroup$ – Matthias Ludewig May 5 '13 at 10:39
  • $\begingroup$ @Kofi: I mean experimentally with precision $10^4$ decimal digits the equality is true. Probably will edit the question. If it is true, it will hold with any precision of course, but experimental results don't prove it. btw, I don't mean the low 10^-4. $\endgroup$ – joro May 5 '13 at 10:55
  • $\begingroup$ I deleted some (now irrelevant) comments, because they were causing typesetting issues. They are preserved at tea.mathoverflow.net/discussion/1589/cleanup $\endgroup$ – Scott Morrison May 8 '13 at 0:44

In fact for odd $n\ge3$ we have

$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|_{s=\frac12}= \frac{(n-1)!}{2}\Bigl(2^n L(n,\chi)+(2^n-1)\zeta(n)\Bigr)$$

The proof (due to Voros) is the following: It is well known that $$\frac{\zeta'(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum_\rho\Bigl(\frac{1}{s-\rho}-\frac{1}{\rho}\Bigr) +\sum_{n=1}^\infty \Bigl(\frac{1}{s+2n}-\frac{1}{2n}\Bigr)+B$$

Differentiating this we get $$\frac{d^{n-1}}{ds^{n-1}}\frac{\zeta'(s)}{\zeta(s)}=\frac{(-1)^{n-2} (n-1)!}{(s-1)^{n}}-\sum_{\rho} \frac{(n-1)!}{(\rho-s)^{n}}+\sum_{k=1}^\infty \frac{(-1)^{n-1} (n-1)!}{(s+2k)^{n}}.$$ Now take $s=\frac12$

We get


$$(n-1)! 2^{n}- \sum_{\rho}\frac{(n-1)!}{(\rho-\frac12)^{n}}+\sum_{k=1}^\infty \frac{(-1)^{n-1} (n-1)! 2^{n}} {(4k+1)^{n}}.$$

When $n$ is odd the sum on the non trivial zeros is $0$ by symmetry. There is only the question of recognizing the sum.

Assuming $n$ even we have


$$=2+2\sum_{k=1}^\infty \frac{1}{(4k+1)^{n}}$$

It is easily seen that $$2\sum_{k=1}^\infty \frac{1}{(4k+1)^{s}}=-2+L(s,\chi)+(1-2^{-s})\zeta(s)$$ from which the result follows.

  • $\begingroup$ Thank you Juan. Got similar results for zeta(3) and derivatives of Dirichlet beta at 1/2. $\endgroup$ – joro Jun 4 '13 at 6:23

Looks like this is proved by Andre Voros, "Zeta Functions over Zeros of Zeta Functions",p. 69, eq. 7.49:

$$ (\log{|\zeta|})^{(n)} (\frac12) = \frac12 (n-1)![(2^n-1)\zeta(n) + 2^n\beta(n)],\qquad n > 1 , \qquad (7.49) $$

From which the conjecture follows.


This is just partial numerics, but the following Mathematica code strengthens the conjecture, with 500 decimal places:

Here is code for those that want to perform numerics in Mathematica:

prec = 500; (* Precision of calculations. *)
lhs = N[Zeta[3], prec]
rhs = N[(-Zeta'''[1/2]/Abs[Zeta[1/2]] -
 3 Zeta''[1/2] Zeta'[1/2]/Abs[Zeta[1/2]]^2 -
 2 Zeta'[1/2]^3/Abs[Zeta[1/2]]^3 - Pi^3/4)/7, prec]
lhs == rhs (* Gives true *)
  • 2
    $\begingroup$ In the question, the result is checked to $10^4$ digits, which seems more precision that 500! $\endgroup$ – M P May 5 '13 at 14:29
  • $\begingroup$ Oh, read it as "to $10^{-4}$ precision". Silly me... $\endgroup$ – Per Alexandersson May 5 '13 at 15:01
  • $\begingroup$ Comparing with exact equality doesn't appear very good, since the last few digits may differ by instability. $\endgroup$ – joro Aug 19 '14 at 14:45

On of methods for value of an infinite sum like $\zeta (3)$ is using following formula which by this trick we can write an infinite sum to a faster sum which by summing first terms of it we can find decimal digits .

$\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$ and by taking $f(n)=\frac{1}{n^3}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

But the second method which is more welcomed for number theorist which working on $\zeta (3)$ in recent decade is generationg function method. In fact Bernoulli numbers which is very important in generating function method is less functional for $\zeta (3)$ . So by this reason Kaneko defined a new generating function which was more applicable for finding $\zeta (3)$ up to now.

${Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}$

where Li is the polylogarithm. The $B_{n}^{(1)} $are the usual Bernoulli numbers.

Kaneko also gave following combinatorial formula:


You can follow following paper. So you can write Euler-Maclaurin formula with respect to $B_n^{(k)}$ and get more desired results for decimal digits of $\zeta (3)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.