Prime Numbers

Date: 07/10/98 at 20:37:23
From: Mark Anderson
Subject: Prime numbers

If you multiply all the prime numbers up to N together, the limit 
appears to be exp(N) as N gets large. Is there a simple reason for 
this?

Date: 07/14/98 at 10:14:01
From: Doctor Wilkinson
Subject: Re: Prime numbers

This is a very nice observation. I believe this follows from the Prime
Number Theorem, though I have yet to prove this rigorously. 

The Prime Number Theorem states that the number of primes less than N 
is approximately N/log(N) when N is large. If you estimate that on the 
average a prime between 1 and N is about N/2, this gives as an 
estimate for the product of the primes less than N

    (N/2)^(N/log(N))

which is

    exp(log(N/2) N/log(N))

For large values of N, log(N/2) is not very different from log(N), so 
this is about exp(N).

I think these ideas could be used to prove your conjecture with a 
little work. (The Prime Number Theorem itself is quite difficult, 
however).

Thanks for a stimulating question!

- Doctor Wilkinson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 07/14/98 at 19:56:13
From: Anderson,Mark
Subject: Re: Prime numbers

Thanks very much for your reply

You must admit it is a beautifully simple, almost unbelievable result
that the geometric mean of all the primes is just e

Mark

Date: 07/14/98 at 20:46:21
From: Doctor Wilkinson
Subject: Re: Prime numbers

Careful!  That's not quite what you discovered. The geometric mean of 
the primes <= N would be the kth root of the product, where k is the 
number of primes, not the Nth root.

The result is still very pretty. After some thinking and research, I 
noticed that this result is quite well-known; in the equivalent form 
that the sum of the logs of the primes less than x is asymptotically 
equal to x, it is easily shown to be equivalent to the Prime Number 
Theorem; in fact it is often used to prove the PNT rather than the 
other way around.

Although your discovery is not original, it is still a very good 
observation, and you are to be congratulated.

- Doctor Wilkinson, The Math Forum
  http://mathforum.org/dr.math/