The Probabilistic argument

  • #1
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Main Question or Discussion Point

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

[tex] \sum_{n=0}^{x}f(n) =A(x) [/tex]

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

[tex] \sum_{n=0}^{\infty}(M(n)-M(n-1))n^{-s} =1/ \zeta (s) [/tex]

then is RH true by this argument? ,
 
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Answers and Replies

  • #2
matt grime
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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)?
 
  • #3
CRGreathouse
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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.
 
  • #4
gel
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how can one prove that [tex]A(x)= O(x^{1/2+e}) [/tex]
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 [itex]O(\sqrt{x\log\log x})[/itex].

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.
 

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