Let $\psi(n,x)$ denote the polygamma function.

In this answer Lucia gave linear relations for $\psi(m,1/3),\psi(m,1/6),\zeta(m+1)$.

The computer managed to find closed form for $\psi(2,1/6)$ and $\psi(4,1/6)$ in terms of $\zeta, \pi$ and $\sqrt{3}$. As a byproduct this give another closed form for $\zeta(3),\zeta(5)$.

Working with precision 1000 decimal digits we have:

$$\psi(2,1/6)= -2/3 \sqrt{3}(6\pi^3 + 91\sqrt{3}\zeta(3))$$ $$\psi(4,1/6)=-176 \sqrt{3}\pi^5 - 90024\zeta(5)$$ $$ \zeta(3)=-1/546\sqrt{3}(12 \pi^3 + \sqrt{3}\psi(2, 1/6)) $$

$$\zeta(5)= -2/1023 \sqrt{3} \pi^5 - 1/90024 \psi(4, 1/6) $$

Are these experimental results correct? Do they generalize for $\psi(m,1/6)$ for other $m$?

Added Looks like for all $m$, $\pi,\psi(m,1/6),\psi(m,1/3)$ are algebraically dependent and if we add $\sqrt{3}$ they are linearly dependent. Proving this and Lucia's answer will answer the question.

Adding linear dependencies.

In machine readable form:

 p_2_16= -2/3*sqrt(3)*(6*pi^3 + 91*sqrt(3)*zeta(3)) # psi(2,1/6)
 p_4_16=-176*sqrt(3)*pi^5 - 90024*zeta(5) # psi(4,1/6)

 zeta3=-1/546*sqrt(3)*(12*pi^3 + sqrt(3)*psi(2, 1/6))
 zeta5= -2/1023*sqrt(3)*pi^5 - 1/90024*psi(4, 1/6)

 print 'psi(2,1/6)',(p_2_16 - psi(2,1/6))
 print 'psi(4,1/6)',p_4_16 - psi(4,1/6)
 print 'zeta(3)',zeta3 - zeta(3)
 print 'zeta(5)',zeta5 - zeta(5)

 #linear depend.
 m= 2 ; x0=psi(2, 1/6); x1=psi(2, 1/3); x3=sqrt(3)*pi^3 ;f= 9*x0 - 63*x1 + 8*x3
 m= 3 ; x0=psi(3, 1/6); x1=psi(3, 1/3); x2=pi^4 ;f= 3*x0 - 51*x1 + 16*x2
 m= 4 ; x0=psi(4, 1/6); x1=psi(4, 1/3); x3=sqrt(3)*pi^5 ;f= 3*x0 - 93*x1 + 32*x3
 m= 5 ; x0=psi(5, 1/6); x1=psi(5, 1/3); x2=pi^6 ;f= 9*x0 - 585*x1 + 832*x2
 m= 6 ; x0=psi(6, 1/6); x1=psi(6, 1/3); x3=sqrt(3)*pi^7 ;f= 3*x0 - 381*x1 + 896*x3
 m= 7 ; x0=psi(7, 1/6); x1=psi(7, 1/3); x2=pi^8 ;f= -3*x0 + 771*x1 - 10496*x2
 m= 8 ; x0=psi(8, 1/6); x1=psi(8, 1/3); x3=sqrt(3)*pi^9 ;f= 27*x0 - 13797*x1 + 414208*x3
 m= 9 ; x0=psi(9, 1/6); x1=psi(9, 1/3); x2=pi^10 ;f= 3*x0 - 3075*x1 + 687104*x2
 m= 10 ; x0=psi(10, 1/6); x1=psi(10, 1/3); x3=sqrt(3)*pi^11 ;f= -3*x0 + 6141*x1 - 3782656*x3
 m= 11 ; x0=psi(11, 1/6); x1=psi(11, 1/3); x2=pi^12 ;f= 9*x0 - 36873*x1 + 206614528*x2
  • Where the term sqrt(3) came from? Just shooting in the dark, but could this be \psi doubling formula? – joro Oct 11 at 15:02

These relations are well-understood and can be proved rigorously. Recall that the polygamma function $\psi(n,x)$ can be expressed in terms of the Hurwitz zeta function (see here): $$\psi(m,x)=(-1)^{m+1} m! \cdot \zeta(m+1,x) \quad \textrm{where} \quad \zeta(m+1,x)=\sum_{n=0}^{\infty} \frac{1}{(x+n)^{m+1}}.$$ In particular for $x=a/N$ with $0<a<N$ and $(a,N)=1$, we get $$\psi(m,a/N)=(-1)^{m+1} m! \cdot \zeta(m+1,a/N)=(-N)^{m+1} m! \sum_{\substack{n \geq 1 \\ n \equiv a (N)}} \frac{1}{n^{m+1}}.$$ These series is a particular case of the Dirichlet series $\sum_{n \geq 1} \theta(n)/n^s$ where $\theta : \mathbb{Z} \to \mathbb{C}$ is $N$-periodic. In our case $\theta$ is the characteristic function $\theta_a$ of $\bar{a} \in \mathbb{Z}/N\mathbb{Z}$. Now you can decompose $\theta_a$ as a linear combination of Dirichlet characters modulo $N$, which gives $$\psi(m,a/N)=\frac{(-N)^{m+1} m!}{\varphi(N)} \sum_{\chi \textrm{ mod } N} \overline{\chi}(a) L(\chi,m+1).$$ The artihmetic nature of the special value $L(\chi,m+1)$ is very different according to the parity of $\chi$. If $\chi$ and $m+1$ have the same parity, that is $\chi(-1)=(-1)^{m+1}$, then $L(\chi,m+1)$ is a critical value in the sense of Deligne (the $\Gamma$-factor of $L(\chi,s)$ has no pole at $s=-m$) and $L(\chi,m+1)$ is an algebraic multiple of $\pi^{m+1}$. The expression is explicit in terms of generalized Bernoulli numbers. The factor $\sqrt{3}$ that you found out is essentially the Gauss sum of the non-trivial character modulo $3$.

On the other hand, if $\chi$ and $m+1$ have opposite parity, then $L(\chi,m+1)$ is non-critical, the simplest example being $\zeta(3)$. In this case no expression in terms of usual special functions is known, but there does exist a nice theory about these $L$-values: they are related to the Borel regulator on the $K$-group $K_{2m+1}(F_\chi)$, where $F_\chi \subset \mathbb{Q}(\zeta_N)$ is the abelian number field cut out by the kernel of $\chi$. See for example this article by Zagier.

Of course, it may happen that suitable linear combinations of $\theta_a$'s are even or odd, which explains the formulas involving only a power of $\pi$.

  • Thanks. Is it known for which $x$ we have zeta(3) = f(psi(2,x),simpler terms)? – joro Oct 11 at 14:20
  • @joro If you mean $\zeta(3)$ expressed in terms of a single $\psi(2,a/N)$ and other simpler terms, then I guess a necessary condition is that the group $(\mathbb{Z}/N\mathbb{Z})^\times/\pm 1$ is trivial, so $N=2,3,4,6$. – François Brunault Oct 11 at 14:44
  • Thanks, this is exactly what I mean. This generalizes to higher zeta? The case N=4 is known to Wolfram Alpha, but N=6 appears unknown to it. Where the term sqrt(3) came from? – joro Oct 11 at 14:47
  • @joro Yes, this generalizes to zeta at any odd integer. The term $\pi^3/\sqrt{3}$ in $\psi(2,1/3)$ and $\psi(2,1/6)$ comes from the non-trivial character $\chi$ mod 3 (or 6). It is known in general that $L(\chi,1-m) \in \mathbb{Q}(\chi) = \mathbb{Q}$ here. Now the functional equation gives $L(\chi,m) = (\textrm{some factor}) (2\pi i)^m/G(\bar{\chi}) L(\chi,1-m)$ where $G$ is the Gauss sum. This explains the $\sqrt{3}$. – François Brunault Oct 11 at 15:40
  • @joro In a more elementary way, you can avoid the functional equation and prove directly these identities by considering the Fourier expansion of the (1-periodic version of the) Bernoulli polynomials, and use the fact that the discrete Fourier transform of $\chi$ as a function on $\mathbb{Z}/N\mathbb{Z}$ is $G(\chi) \bar{\chi}$ if $\chi$ is primitive. In this way you get $L(\chi,m) \sim (2\pi i)^m/G(\bar{\chi})$ if $\chi(-1)=(-1)^m$. – François Brunault Oct 11 at 16:00

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