Prime Number Theorem w/o Complex Analysis - MathOverflow

Prime Number Theorem w/o Complex Analysis

2011-04-04T20:50:21Z

I know about

"Simple analytic proof of the prime number theorem" Newman, 1980

However, is there a proof of the Prime Number Theorem without the use of complex analysis? (Real analysis is fine).

Thanks!

Answer by Charles Matthews for Prime Number Theorem w/o Complex Analysis

2011-04-04T21:12:58Z

http://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf explains the classic proof in context (there is what amounts to a priority dispute).

Answer by Gerry Myerson for Prime Number Theorem w/o Complex Analysis

2011-04-05T00:06:39Z

Another exposition of an elementary proof (that is, a proof not using complex analysis) is in Gerald Tenenbaum and Michel Mendes France, The Prime Numbers and Their Distribution, which is Volume 6 of the Student Mathematical Library, published by the American Mathematical Society. The proof they give is due to Daboussi, from 1984.

Answer by M Mueger for Prime Number Theorem w/o Complex Analysis

2011-04-06T17:38:38Z

A nice exposition of an Erdos/Selberg-type elementary proof is given by Levinson in Amer. Math. Monthly 76 (1969) 225–245.

The proof by Daboussi as written up by Tenenbaum and Mendes-France was already mentioned.

Yet another one is due to Hildebrand in Mathematika 33 (1986) 23–30.

Answer by Emerton for Prime Number Theorem w/o Complex Analysis

2011-04-06T18:51:30Z

There is a terrific exposition of the elementary proof by Terry Tao, available as the file prime.dvi here. A more traditional exposition is available in Edwards's book Riemann's zeta function.

Answer by Richard Dore for Prime Number Theorem w/o Complex Analysis

2011-07-08T17:36:12Z

If you just want $\pi(n) = \Omega \left( \frac{n}{\log n} \right)$, good enough for many applications, here is a quick proof: The highest power of a prime $p$ dividing $2n \choose n$ is at most $2n$ -- you get at most one more factor of $p$ in the numerator than denominator for each power $p^i \leq 2n$. This tells you that ${2n \choose n} \leq (2n)^{\pi(2n)}$. So $\pi(2n) \geq \frac{\log_2 {2n \choose n}}{\log_2 (2n)} \geq \frac{n}{\log_2 (2n)}$.