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The Probabilistic argument |
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| May4-09, 04:07 PM | #1 |
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The Probabilistic argument
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? , |
| May5-09, 01:34 AM | #2 |
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Recognitions:
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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)?
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| May5-09, 08:59 AM | #3 |
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Recognitions:
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At best you have a heuristic suggesting that RH 'should' be true, but there are plenty of those. |
| May11-09, 06:21 PM | #4 |
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The Probabilistic argument |
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