Timeline for Square root of prime numbers
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 6, 2022 at 19:40 | answer | added | Christophe Leuridan | timeline score: 7 | |
| May 6, 2022 at 9:06 | comment | added | Andrea Marino | For antiderivative I mean $G_n(t)=\frac{(x_0+t\sqrt{S}) ^{n+1}}{\sqrt{S}(n+1)}$. This has the property that $G_n'(t) =F_n(t) $. It seems like you can express the numerator as something like $F_{2^n}(1) +F_{2^n}(-1) $ and the denominator as $G_{2^n}(1) +G_{2^n}(-1) $. | |
| May 5, 2022 at 23:53 | comment | added | Gerry Myerson | @Andrea, what does "its antiderivative in $\pm1$" mean? | |
| May 5, 2022 at 21:01 | comment | added | Andrea Marino | If you call $F_n(t) = (x_0+t \sqrt{S}) ^n$, it seems like both the numerator and the denominator can be expressed in terms of $F_n $ and its antiderivative in $\pm 1$. Is this the way you found the formula? | |
| May 5, 2022 at 19:52 | history | edited | Glorfindel | CC BY-SA 4.0 |
added 3 characters in body
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| May 5, 2022 at 18:28 | comment | added | Will Sawin | Surely this works for any number $S$, not just prime numbers? Could it be a version of Newton's method? | |
| S May 5, 2022 at 18:19 | history | suggested | Steven Clark | CC BY-SA 4.0 |
Converted linked image to visible image.
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| May 5, 2022 at 16:13 | comment | added | TMM | Someone might already be aware of the result if there was a natural way how this result would come up, or if the result somehow had "practical" significance. Maybe it would help to add some context why the given formula is special or useful, and how you derived it. | |
| May 5, 2022 at 15:27 | review | Suggested edits | |||
| S May 5, 2022 at 18:19 | |||||
| S May 5, 2022 at 14:56 | review | First questions | |||
| May 5, 2022 at 19:52 | |||||
| S May 5, 2022 at 14:56 | history | asked | Salomon S Mizrahi | CC BY-SA 4.0 |