Timeline for How to determine if you've discovered a new identity for a special function
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2021 at 23:54 | comment | added | Somos | By the way, $\sum_{j=0}^\infty {2j \choose j} x^j = (1-4x)^{-1/2}$ is well known and your result follows. | |
| Sep 2, 2021 at 4:18 | history | became hot network question | |||
| Sep 1, 2021 at 21:35 | comment | added | Alexandre Eremenko | If you really discovered a new identity, just contact the editors of NIST, and ask for inclusion. | |
| Sep 1, 2021 at 21:06 | vote | accept | garserdt216 | ||
| Sep 1, 2021 at 20:55 | answer | added | Carlo Beenakker | timeline score: 11 | |
| Sep 1, 2021 at 20:38 | comment | added | garserdt216 | @მამუკაჯიბლაძე Thanks! I've removed this part of the question so that more important (soft) question remains. | |
| Sep 1, 2021 at 20:37 | history | edited | garserdt216 | CC BY-SA 4.0 |
proposed unknown result is known and was thus removed. soft question remains relevant.
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| Sep 1, 2021 at 20:19 | comment | added | მამუკა ჯიბლაძე |
Sum[Binomial[2j,j](Cos[x]/2)^(2j),{j,0,\[Infinity]}] in Mathematica gives 1/Sqrt[Sin[x]^2]
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| Sep 1, 2021 at 20:11 | history | asked | garserdt216 | CC BY-SA 4.0 |