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The computer found this, but couldn't prove it.

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

With precision 500 decimal digits we have:

$$ \pi^2 = \frac{1}{4}(15 \psi(1, \frac13) - 3 \psi(1, \frac16)) $$

Is it true?

In machine readable form:

 pi^2 == 1/4*(15*psi(1, 1/3) - 3*psi(1, 1/6))
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  • $\begingroup$ Both sides are periods, so maybe someday the computer will also get an algorithm to prove it or disprove it. $\endgroup$
    – user44143
    Commented Oct 10, 2018 at 16:12
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    $\begingroup$ Boils down easily to $\zeta(2)$ calculation. $\endgroup$
    – Lucia
    Commented Oct 10, 2018 at 16:16
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    $\begingroup$ I have the following similar conjectures: $$\psi\left(2,\frac16\right)-9\psi\left(2,\frac13\right)-52\zeta(3)=0$$ and $$\psi\left(3,\frac16\right)-17\psi\left(3,\frac13\right)+480\zeta(4)=0.$$ $\endgroup$
    – Zhi-Wei Sun
    Commented Oct 10, 2018 at 17:14
  • $\begingroup$ More conjectural formula: $$\psi\left(4,\frac16\right)-33\psi\left(4,\frac13\right)-5808\zeta(5)=0.$$ $\endgroup$
    – Zhi-Wei Sun
    Commented Oct 10, 2018 at 17:25
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    $\begingroup$ Include notation ... $\psi(m,x) = \psi^{(m)}(x)$ the $m$th derivative of $\psi$, and the digamma function $\psi(x)$ is $\Gamma'(x)/\Gamma(x)$. $\endgroup$
    – Gerald Edgar
    Commented Oct 10, 2018 at 17:42

1 Answer 1

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Note that $$ \psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}. $$ Therefore $$ \psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. $$ Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is \begin{align*} &(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\ &= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). \end{align*} We also have $$ \psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). $$ From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$: namely, $$ \psi(m,1/6) = (2^{m+1}+1) \psi(m,1/3)+(-1)^m m! (3^{m+1}-1) \zeta(m+1). $$

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  • $\begingroup$ Thank you. Is there simpler closed form for psi(1,1/3) or psi(1,1/6)?. psi(1,1/4) is directly related to catalan and pi. $\endgroup$
    – joro
    Commented Oct 11, 2018 at 5:47
  • $\begingroup$ I managed to express psi(2,1/6),psi(4,1/6) in terms of zeta, pi,sqrt(3) only: mathoverflow.net/questions/312550/… $\endgroup$
    – joro
    Commented Oct 11, 2018 at 8:22
  • $\begingroup$ @joro We have for example $\psi(1,1/3)=2\pi^2/3+9L(\chi,2)/2$ where $\chi$ is the non-trivial character mod 3, see my answer to the question in your last comment. I know of no closed form for $L(\chi,2)$ itself and I don't expect there is one (which is not trivially equivalent) but these are definitely interesting numbers. $\endgroup$
    – François Brunault
    Commented Oct 11, 2018 at 14:18

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