[math-fun] 2 surprising approximations and 1 exact formula for pi.
Hello everybody, these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13 = 119.00000000000000000000000000000009593745851025547335588584913... the precision is 31 digits. Another one is infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 7 = 10.0000000000000001901617678886626755843593058554453334802548978434061099438\ The precision here is 15 digits. For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course). in general, these sums will be near an integer if the argument is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends a little bit the known formula of Ramanujan/Berndt/etc. A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results. Because, I do have another one which is EXACT, namely for pi: In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc. here is the EXACT formula with fractional arguments mixed with integer exponents. /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------| | / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 / /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------| | / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 // /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------) - 1|| \ n = 1 \ 5 // \ n = 1 \ 5 //
evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\ 62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds.
This is trivial, ? I do not see how. If someone has a piece of information on why this exist, I would be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before, have a good day, Simon Plouffe
I find that the sum is near an integers for 7,11,13,17, and 29, but not for 163. The integer parts appear in A089034 a(n)={p^4 - 1}/240, where p=prime(n+3). terms: 10, 61, 119, 348, 543, 1166, Pari code: ----------------- default(realprecision,1000); default(format,"g.50"); { for (k=1, 1000, s = sum(n=1,100, n^3/(exp(2*Pi*n/k)-1)); if ( abs(s-round(s))<0.001, print(k,": ---- "); print(s); c = contfrac(s); print(vector(20,n,c[n])); ); ); } Output: ----------------- 7: ---- 10.000000000000000190161767888662674830486095865010 [10, 5258680601799450, 4, 1, 10, 1, 1, 1, 2, 1, 4, 2, 1, 1, 1, 3, 1, 3, 1, 1] 11: ---- 60.999999999999999999783017186376408591503435429769 [60, 1, 4608659936241490513, 5, 1, 6, 1, 7, 10, 1, 127, 1, 1, 3, 2, 1, 3, 9, 1, 1] 13: ---- 118.99999999999999822060853662753902153501763247951 [118, 1, 561989882824721, 38, 1, 1, 27, 1, 9, 1, 1, 1, 3, 1, 1, 2, 5, 2, 1, 2] 17: ---- 347.99999999978099522542236078594550239032616790505 [347, 1, 4566110495, 1, 1, 2, 1, 1, 1, 3, 1, 5, 1, 1, 1, 5, 1, 1, 2, 3] 19: ---- 542.99999998765415727636451319177064725216471202466 [542, 1, 80998924, 1, 9, 2, 7, 3, 4, 1, 4, 101, 2, 3, 1, 1, 4, 1, 2, 1] 23: ---- 1165.9999950595915588813284161938468112041252275357 [1165, 1, 202411, 2, 2, 2, 2, 2, 7, 1, 2, 3, 1, 3, 1, 1, 22, 2, 1, 1] ----------------- Remove the if() statement to observe nearness to rationals for many k.
Hello, abouthe argument 2*Pi*n/163, for 500 digits of precision I get ; 0.294129900000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000\ 0000000000000000000000001153610141122787532496931945\ 7 677118639979465080614285218174 10 which is almost 2941299. Are you certain that you have enough terms in the sum ?, those series converge slowly, n has to be 32000 to get 522 digits of precision. Unless there is somthing I am missing ? très bonne journée, Simon Plouffe
Ouch, yes! Replace the finite sum by s = suminf(n=1, n^3/(exp(2*Pi*n/k)-1)); Now I get close to integer results apparently for all k (except the initial 1) in sequence A007775 "Not divisible by 2, 3 or 5." An interesting observation of yours! * Simon Plouffe <simon.plouffe@gmail.com> [Jan 22. 2011 14:28]:
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Unless there is somthing I am missing ? très bonne journée,
Simon Plouffe
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