Timeline for Curious infinite product, convergence, connection to prime numbers
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2023 at 13:40 | vote | accept | Vincent Granville | ||
| Jan 25, 2023 at 14:12 | comment | added | Michael Rieck | Vincent, perhaps this will be helpful: Let $F(z) = \prod_{n=1}^\infty (1-\frac{z^2}{n^2})^{d(n)} = \sin^2(\pi z) f(z) / [ (\pi z)^2 (1-z^2) ]$. Then, $F'(z) / F(z) = \frac{d}{dz} \sum_{n=1}^\infty d(n) \ln(1-\frac{z^2}{n^2}) = 2 z \sum_{n=1}^\infty \frac{d(n)}{z^2-n^2}$. | |
| Jan 24, 2023 at 18:14 | comment | added | Vincent Granville | I haven't double-checked your formula, I assume it is correct; I expect it to be something like that. Thank you for your answer! But I have a more difficult time dealing with the asymptotic formula, valid for primes it seems, probably for many non-integer numbers, but of course not for composite integer numbers. Hoping to get some help with that. | |
| S Jan 24, 2023 at 14:19 | review | First answers | |||
| Jan 24, 2023 at 14:25 | |||||
| S Jan 24, 2023 at 14:19 | history | edited | Michael Rieck | CC BY-SA 4.0 |
edited body
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| Jan 24, 2023 at 14:16 | comment | added | Michael Rieck | The doubly infinite product just equals $\prod_{n=1}^\infty ( 1 - \frac{x^2}{n^2})^{d(n)}$ where $d(n)$ is the number of positive divisors of $n$, of course. | |
| Jan 24, 2023 at 13:43 | history | edited | Alex M. | CC BY-SA 4.0 |
MathJaxed
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| Jan 24, 2023 at 13:30 | review | Low quality posts | |||
| Jan 24, 2023 at 13:43 | |||||
| Jan 24, 2023 at 13:19 | comment | added | Vincent Granville | Writing $\sin(\pi x)/(\pi x)$ as an infinite product, and grouping each factor $1/[1-(x/k)^2]$ with each factor $\sin(\pi x/k)/(\pi x /k)$, may help understand what is going on. | |
| S Jan 24, 2023 at 13:12 | review | First answers | |||
| Jan 24, 2023 at 13:22 | |||||
| S Jan 24, 2023 at 13:12 | history | answered | Michael Rieck | CC BY-SA 4.0 |