tag:blogger.com,1999:blog-264226589944705290.post7029508140381811782..comments2023-11-05T03:45:25.001-08:00Comments on God Plays Dice: Heuristic derivation of the Prime Number TheoremMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-264226589944705290.post-24930299237337464502010-01-26T20:08:31.446-08:002010-01-26T20:08:31.446-08:00hydrants coated touch isolation dysphagia grunge i...hydrants coated touch isolation dysphagia grunge iiianupama resignation gabon channelled stationary <br />servimundos melifermulyAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-12832994954500089332010-01-25T07:50:38.873-08:002010-01-25T07:50:38.873-08:00pawan drop ping webct exciting btinternet booths m...pawan drop ping webct exciting btinternet booths mcgill expert hallsumit verkhovna <br />servimundos melifermulyAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-79296098051952019322008-12-03T20:30:00.000-08:002008-12-03T20:30:00.000-08:00In the same vein, what about the Legendre's conjec...In the same vein, what about the Legendre's conjecture, saying that there is at least one prime between n^2 and (n+1)^2? Among 2n+1 consecutive numbers, 1/2 are odd, among these 2/3 aren't divisible by 3, etc., (p-1)/p aren't divisible by p. Taking all the primes up to n+1 and multiplying all these factors together, and then by 2n+1, we get the value more than 2, so the Legendre's conjecture is plausible.mishahttps://www.blogger.com/profile/01166708933155105921noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-673607203258761592008-12-03T20:27:00.000-08:002008-12-03T20:27:00.000-08:00This comment has been removed by the author.mishahttps://www.blogger.com/profile/01166708933155105921noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-27488428631637190462008-11-25T09:14:00.000-08:002008-11-25T09:14:00.000-08:00Interesting! Do you know the following heuristic a...Interesting! <BR/><BR/>Do you know the following heuristic argument, which tells you that the constant C should be zero?<BR/> <BR/>The zeta function is defined by <BR/><BR/>zeta(s) = sum_n 1/n^s<BR/><BR/>and obeys the Euler product formula <BR/><BR/>zeta(s) = <BR/>prod_p 1/(1-p^{-s}).<BR/><BR/> Taking the logarithmic derivative of the latter formula, we get<BR/><BR/> zeta'(s)/zeta(s) = <BR/>(sum_p (log p)p^{-s} )<BR/>+O(1)<BR/><BR/>where O(1) is an analytic function of s that remains bounded as s -> 1.<BR/><BR/>Now, by approximation by an integral, <BR/><BR/>zeta(s) = 1/(s-1)+O(1) <BR/><BR/>so zeta'(s)/zeta(s)=<BR/>1/(s-1)+O(1) <BR/><BR/>and we deduce that <BR/><BR/>sum_p log p/p^s =<BR/>sum_n 1/n^s + O(1) (*)<BR/><BR/>Approximating both sides of equation (*) by integrals, we have<BR/><BR/>int P(x) log x dx/x^s = int dx x^{-s}+O(1).<BR/><BR/>Now, that tells you P(x) = 1/log(X) + e(X) where e is an error term small enough that <BR/><BR/>int e(x) log x dx/x^s<BR/><BR/>remains bounded as s -> 1. <BR/><BR/>Intuitively, this shows that <BR/><BR/>e(x) = o(1/(log x)^2).<BR/><BR/>If you can make that rigorous, you have a new proof of the prime number theorem. But this really can be turned into a proof that, if e(X) = A/(log x)^2 <BR/> + o(1/(log x)^2)<BR/><BR/>then A is zero. In particular, <BR/><BR/>1/(C + log X) = <BR/>1/log X - C/(log X)^2 + o(1/(log x)^2)<BR/><BR/>so, if your heuristic is correct, then C=0.<BR/><BR/>You can push this argument further by taking d derivatives of equation (*) with respect to s. This gives<BR/><BR/>sum (log p)^{d+1}/p^s =<BR/>sum (log n)^d/n^s+O(1).<BR/><BR/>(Here we need to know that O(1) is analytic in s, so that we know each derivatives stay bounded as well.)<BR/><BR/>Running through the same approximation by integrals heuristic, you get that <BR/><BR/>int e(x) (log x)^{d+1} <BR/> *dx/x^s<BR/><BR/>remains bounded as s -> 1 for any d. Heuristically, this tells you that <BR/><BR/>e(x) = <BR/>o(1/(log x)^{d+1})<BR/><BR/>and rigorously it tells you that you can't have<BR/><BR/>e(X) = A/(log x)^{d+1}<BR/> + o(1/(log x)^{d+1})Anonymousnoreply@blogger.com