Timeline for Bernoulli sum meets golden number
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 12, 2021 at 13:40 | vote | accept | T. Amdeberhan | ||
| Jun 16, 2017 at 23:30 | vote | accept | T. Amdeberhan | ||
| Aug 12, 2021 at 13:40 | |||||
| Jun 16, 2017 at 17:27 | comment | added | T. Amdeberhan | Yes, it is cool. Thanks for the kind words too. | |
| Jun 16, 2017 at 17:10 | comment | added | efs | I doubt it is related. I just rememberd this because of the title. Beautiful identity bt the way. | |
| Jun 16, 2017 at 17:07 | comment | added | T. Amdeberhan | Yes, that is rather curious. I'm not sure of its immediate impact here. | |
| Jun 16, 2017 at 16:57 | comment | added | efs | Just a curiosity. All Bernoulli polyomials of odd degree have the 0, 1 and 1/2 as "trivial" roots. The only known Bernoulli polyomial of odd degree that has "non-trivial" roots is $B_{11}(x)$, whose "non-trivial" roots are the golden ratio and its conjugate. | |
| Jun 16, 2017 at 1:38 | answer | added | Cherng-tiao Perng | timeline score: 4 | |
| Jun 15, 2017 at 22:31 | answer | added | Suvrit | timeline score: 6 | |
| Jun 15, 2017 at 17:13 | comment | added | T. Amdeberhan | @Nemo: Thanks, there was a typo which is corrected now. | |
| Jun 15, 2017 at 17:12 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 190 characters in body
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| Jun 15, 2017 at 11:30 | answer | added | Henri Cohen | timeline score: 9 | |
| Jun 15, 2017 at 10:39 | answer | added | Henri Cohen | timeline score: 4 | |
| S Jun 15, 2017 at 10:18 | history | suggested | Martin Sleziak |
added (binomial-coefficients) and (bernoulli-numbers)
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| Jun 15, 2017 at 10:10 | review | Suggested edits | |||
| S Jun 15, 2017 at 10:18 | |||||
| Jun 15, 2017 at 10:06 | comment | added | Nemo | @T.Amdeberhan , are there any typos in this question? Did you check this result numerically? | |
| Jun 15, 2017 at 9:06 | answer | added | Nemo | timeline score: 25 | |
| Jun 15, 2017 at 2:20 | comment | added | KConrad | The right side is $-(2/\sqrt{5})\log((1+\sqrt{5})/2)$, which is $-{\rm Res}_{s=1}\zeta_{\mathbf Q(\sqrt 5)}(s)$. On the left side, $-B_{j+1}/(j+1) = -\zeta(-j)$. Not sure what you can do with that, but suggests you should negate both sides. | |
| Jun 15, 2017 at 0:59 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |