# zeta(3) in terms of derivatives of zeta at 1/2 and pi

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
mpmath.mp.pretty=True

def zeta3test():

n=3
mpmath.mp.dps=10^3
zeta3=mpmath.zeta(3)

Pi=mpmath.pi
gamma=mpmath.euler
z12=mpmath.zeta(1/mpmath.mpf(2))
z1=mpmath.zeta(1/mpmath.mpf(2),derivative=1)
z2=mpmath.zeta(1/mpmath.mpf(2),derivative=2)
z3=mpmath.zeta(1/mpmath.mpf(2),derivative=3)

# eliminate the first derivative

#rh0=1/32*(72*Pi*mpmath.log(2)*mpmath.log(Pi)*z12+144*gamma*mpmath.log(2)*mpmath.log(Pi)*z12+72*mpmath.log(2)*z12*gamma*Pi+24*mpmath.log(Pi)*z12*gamma*Pi-144*z2*mpmath.log(2)-48*z2*mpmath.log(Pi)+216*mpmath.log(2)^3*z12+8*mpmath.log(Pi)^3*z12-48*z2*gamma-24*z2*Pi+8*z12*gamma^3+z12*Pi^3+72*mpmath.log(2)*z12*gamma^2+18*mpmath.log(2)*z12*Pi^2+24*mpmath.log(Pi)*z12*gamma^2+6*mpmath.log(Pi)*z12*Pi^2+216*mpmath.log(2)^2*mpmath.log(Pi)*z12+72*mpmath.log(2)*mpmath.log(Pi)^2*z12+216*gamma*mpmath.log(2)^2*z12+24*gamma*mpmath.log(Pi)^2*z12+108*Pi*mpmath.log(2)^2*z12+12*Pi*mpmath.log(Pi)^2*z12+32*z3+12*z12*gamma^2*Pi+6*z12*gamma*Pi^2)/z12
z12a=mpmath.fabs(z12)
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))

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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? –  Kofi May 5 '13 at 10:39
@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. –  joro May 5 '13 at 10:55
I deleted some (now irrelevant) comments, because they were causing typesetting issues. They are preserved at tea.mathoverflow.net/discussion/1589/cleanup –  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

$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|_{s=\frac12}=$$

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

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

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

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

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.

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Thank you Juan. Got similar results for zeta(3) and derivatives of Dirichlet beta at 1/2. –  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.

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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 *)

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In the question, the result is checked to $10^4$ digits, which seems more precision that 500! –  M P May 5 '13 at 14:29
Oh, read it as "to $10^{-4}$ precision". Silly me... –  Per Alexandersson May 5 '13 at 15:01
Thank you for verifying Per. –  joro May 6 '13 at 8:38
Comparing with exact equality doesn't appear very good, since the last few digits may differ by instability. –  joro Aug 19 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:

$B_{n}^{(-k)}=\sum_{m=0}^{n}(-1)^{m+n}m!S(n,m)(m+1)^{k},$

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 $\zeta (3)$.

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