most recent 30 from http://mathoverflow.net 2013-06-18T05:27:28Z http://mathoverflow.net/feeds/question/98566 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98566/asymptotics-of-the-n-th-prime-using-the-gamma-function Nilotpal Sinha 2012-06-01T11:47:53Z 2012-06-01T16:34:25Z

<p>In the paper <a href="http://rgmia.org/papers/v8n2/eepnt.pdf" rel="nofollow">http://rgmia.org/papers/v8n2/eepnt.pdf</a>, the author proves that proves an explicit inequality on prime numbers using the gamma function and as a corollary, he showed that.</p> <p>$$ p_n = n \frac{\Gamma'(n)}{\Gamma(n)} + o(n \ln n). $$</p> <p>I obtained a stronger form of this result namely</p> <p>$$ p_n = n \ln \frac{\Gamma'(n)}{\Gamma(n-1)} + O\Big(\frac{n\ln\ln n}{\ln n}\Big). $$</p> <p>The gamma function seems to beautifully approximate $p_n$. To get the same error term using the regular Cipolla's asymptotic expansion of the $p_n$ we would need three terms.</p> <p>Can someone explain why the gamma function approximated the n-th prime so nicely? Is this a coincidence or is there some underlying phenomenon governing this result that can shed some new light distribution of prime numbers. </p>

http://mathoverflow.net/questions/98566/asymptotics-of-the-n-th-prime-using-the-gamma-function/98578#98578 juan 2012-06-01T15:09:12Z 2012-06-01T16:34:25Z

<p>The asymptotic expansion of Cipolla starts $$p_n=n\log n+n\log\log n-n+n\frac{\log\log n}{\log n}+O(n(\log\log n/\log n)^2)$$ So the given approximations have errors $$p_n=n\frac{\Gamma'(n)}{\Gamma(n)}+\Theta(n\log\log n)$$ and $$p_n=n\log\frac{\Gamma'(n)}{\Gamma(n-1)}+\Theta(n).$$ I would not say these are good approximations with so big errors. </p> <p>The inverse function of the log integral function $\text{li}^{-1}(x)$ has error $$p_n= \text{li}^{-1}(n) +O(n \exp(-c\sqrt{\log n})$$ which assumming Riemann hypothesis can be reduced to $$|p_n-\text{li}^{-1}(n)|\le \pi^{-1} \sqrt{n}(\log n)^{\frac52}\qquad n>11.$$ (see arXiv:1203.5413)</p>