Prime NumbersDate: 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/ |
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