Timeline for $\pi$ in terms of polygamma
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2018 at 8:23 | comment | added | joro | Related, found potential closed form for psi(2,1/6) and psi(4,1/6): mathoverflow.net/questions/312550/… | |
| Oct 11, 2018 at 8:20 | vote | accept | joro | ||
| Oct 10, 2018 at 21:29 | comment | added | François Brunault | @GeraldEdgar In general $\psi (1,a/N) $ can be expressed in terms of Dirichlet L-values $L (\chi,2) $ for some characters $\chi $ modulo $N $. Taking the linear combination we are interested in, the odd characters cancel, leaving only the even ones for which the L-value is proportional to $\zeta (2) $. | |
| Oct 10, 2018 at 18:05 | answer | added | Lucia | timeline score: 14 | |
| Oct 10, 2018 at 17:44 | comment | added | Gerald Edgar | For investigating $\pi^2 = \frac{1}{4}(15 \psi(1, \frac13) - 3 \psi(1, \frac16))$, perhaps first we should ask whether $\psi(1, \frac13)$ and $\psi(1, \frac16))$ have closed forms. | |
| Oct 10, 2018 at 17:42 | comment | added | Gerald Edgar | Include notation ... $\psi(m,x) = \psi^{(m)}(x)$ the $m$th derivative of $\psi$, and the digamma function $\psi(x)$ is $\Gamma'(x)/\Gamma(x)$. | |
| Oct 10, 2018 at 17:38 | comment | added | Zhi-Wei Sun | In general, I conjecture that for each positive integer $m$ there are integers $a_m$ and $b_m$ such that $$\psi\left(m,\frac16\right)=a_m\psi\left(m,\frac13\right)+b_m\zeta(m+1).$$ | |
| Oct 10, 2018 at 17:34 | comment | added | Zhi-Wei Sun | $$\psi\left(5,\frac16\right)-65\psi\left(5,\frac13\right)+87360\zeta(6)=0,$$ $$\psi\left(6,\frac16\right)-129\psi\left(6,\frac13\right)-1573920\zeta(7)=0.$$ | |
| Oct 10, 2018 at 17:25 | comment | added | Zhi-Wei Sun | More conjectural formula: $$\psi\left(4,\frac16\right)-33\psi\left(4,\frac13\right)-5808\zeta(5)=0.$$ | |
| Oct 10, 2018 at 17:14 | comment | added | Zhi-Wei Sun | I have the following similar conjectures: $$\psi\left(2,\frac16\right)-9\psi\left(2,\frac13\right)-52\zeta(3)=0$$ and $$\psi\left(3,\frac16\right)-17\psi\left(3,\frac13\right)+480\zeta(4)=0.$$ | |
| Oct 10, 2018 at 16:16 | comment | added | Lucia | Boils down easily to $\zeta(2)$ calculation. | |
| Oct 10, 2018 at 16:12 | comment | added | user44143 | Both sides are periods, so maybe someday the computer will also get an algorithm to prove it or disprove it. | |
| Oct 10, 2018 at 14:30 | history | asked | joro | CC BY-SA 4.0 |