Timeline for Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function?
Current License: CC BY-SA 3.0
14 events
when toggle format | what | by | license | comment | |
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Jun 1, 2014 at 21:38 | answer | added | ACL | timeline score: 8 | |
Jun 1, 2014 at 19:47 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 50 characters in body; edited title
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Jun 1, 2014 at 12:23 | answer | added | Testcase | timeline score: 33 | |
May 31, 2014 at 22:19 | vote | accept | William Chang | ||
May 30, 2014 at 19:06 | answer | added | David E Speyer | timeline score: 60 | |
Apr 23, 2014 at 19:07 | vote | accept | William Chang | ||
May 31, 2014 at 22:19 | |||||
Apr 23, 2014 at 15:08 | answer | added | so-called friend Don | timeline score: 41 | |
S Apr 23, 2014 at 13:40 | history | suggested | GNiklasch | CC BY-SA 3.0 |
rearranged first chain of equalities and added missing factor in one denominator
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Apr 23, 2014 at 13:18 | review | Suggested edits | |||
S Apr 23, 2014 at 13:40 | |||||
Apr 23, 2014 at 13:00 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
reworded a little to remove "infinite primes"
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Apr 23, 2014 at 11:48 | comment | added | KConrad | The product seems "simple" only because the expression was rigged to remove the powers of $\pi$ in $\zeta(2)$ and $\zeta(4)$. Also, your factorization with terms $(p-1)/(p\pm i)$ is into two products that are each divergent, though that wouldn't stop Ramanujan. | |
Apr 23, 2014 at 11:27 | comment | added | Dietrich Burde | The result is attributed to Ramanujan (1913-1914), who may have used other methods (like hypergeometric series, continued fractions etc.). A reference is Le Lionnais, F. Les nombres remarquables. Paris: Hermann, p. 46, 1983. | |
Apr 23, 2014 at 6:54 | review | First posts | |||
Apr 23, 2014 at 7:32 | |||||
Apr 23, 2014 at 6:37 | history | asked | William Chang | CC BY-SA 3.0 |