The Probabilistic argument

**zetafunction** · May4-09, 05:07 PM

the question is if we suppose that a function f(n) can take only two values +1 and -1 both with equal probability and define the summatory

$LaTeX Code: \\sum_{n=0}^{x}f(n) =A(x)$

how can one prove that $LaTeX Code: A(x)= O(x^{1/2+e})$
?? if we set A(n)=M(n) the Mertens function and since

$LaTeX Code: \\sum_{n=0}^{\\infty}(M(n)-M(n-1))n^{-s} =1/ \\zeta (s)$

then is RH true by this argument? ,

**matt grime** · May5-09, 02:34 AM

A(x) is a random variable, so you oughtn't you to be asking for something like E(A(x)), or E(A(x)^2)?

**CRGreathouse** · May5-09, 09:59 AM

Originally Posted by zetafunction

then is RH true by this argument?

Are you really arguing that the value of the Mertens function is independent of its argument? That seems like a non-starter to me.

At best you have a heuristic suggesting that RH 'should' be true, but there are plenty of those.

**gel** · May11-09, 07:21 PM

Originally Posted by zetafunction

how can one prove that $LaTeX Code: A(x)= O(x^{1/2+e})$

That is true, with probability one, if f(n) are independent. In fact, the law of the iterated logarithm says that with probability one it is of order $LaTeX Code: O(\\sqrt{x\\log\\log x})$ .

Originally Posted by zetafunction

then is RH true by this argument? ,

as CRGreathouse mentions, this does not follow because the Mertens function is deterministic and not random. It's expected to share many properties of a random sequence, but that hasn't been proven.