Timeline for Is the following recursion formula for $\zeta(2n)$ known?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 23, 2019 at 12:06 | vote | accept | CommunityBot | ||
| Jan 22, 2019 at 11:22 | answer | added | David Chow | timeline score: 15 | |
| Jan 20, 2019 at 13:56 | comment | added | Carlo Beenakker | @GeraldEdgar --- the familiar recursion formula for Bernoulli numbers (due to Euler) gives a bilinear recursion, see mathoverflow.net/q/306898/11260 --- the OP has a linear recursion, which is somewhat special. | |
| Jan 20, 2019 at 13:37 | comment | added | Gerald Edgar | If you write $\zeta(2n)$ in terms of Bernoulli numbers, and use the known recursion for Bernoulli numbers, what do you get? | |
| Jan 20, 2019 at 12:24 | answer | added | Carlo Beenakker | timeline score: 8 | |
| Jan 20, 2019 at 12:23 | comment | added | LeechLattice | The value of the Riemann ζ function on even positive integers can be reduced to Bernoulli numbers. | |
| Jan 20, 2019 at 12:05 | review | First posts | |||
| Jan 20, 2019 at 14:17 | |||||
| Jan 20, 2019 at 12:03 | history | asked | user134766 | CC BY-SA 4.0 |