Timeline for maybe Faulhaber polynomial $S_{k}(x)=0$ have only rational roots $0,-\frac{1}{2},-1$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2023 at 0:45 | vote | accept | math110 | ||
| S Apr 23, 2023 at 11:08 | history | bounty ended | CommunityBot | ||
| S Apr 23, 2023 at 11:08 | history | notice removed | CommunityBot | ||
| Apr 19, 2023 at 9:09 | answer | added | Karl Fabian | timeline score: 5 | |
| Apr 15, 2023 at 9:34 | history | edited | math110 | CC BY-SA 4.0 |
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| S Apr 15, 2023 at 9:30 | history | bounty started | math110 | ||
| S Apr 15, 2023 at 9:30 | history | notice added | math110 | Authoritative reference needed | |
| Apr 13, 2023 at 15:05 | comment | added | Todd Trimble | Something looks wrong with $S_7$: the top coefficient should be $1/8$, whereas you have $3/90$. | |
| Apr 13, 2023 at 12:23 | comment | added | Gerald Edgar | To investigate known results, look for "Bernoulli polynomial". If $B_k(x)$ is the Bernoulli poynomial of degree $k$, then $S_{k-1}(x) = (-1)^{k}B_{k}(-x)/k$ except for the constant term . | |
| Apr 13, 2023 at 1:45 | history | edited | math110 | CC BY-SA 4.0 |
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| Apr 13, 2023 at 1:15 | history | edited | math110 | CC BY-SA 4.0 |
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| Apr 13, 2023 at 1:02 | history | asked | math110 | CC BY-SA 4.0 |