Does Randomness in Function f(n) Impact the Riemann Hypothesis?

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Discussion Overview

The discussion centers on the relationship between a function f(n) that takes values +1 and -1 with equal probability and its impact on the Riemann Hypothesis (RH). Participants explore the summatory function A(x) defined as the sum of f(n) over a range, questioning how to prove its growth rate and whether this has implications for RH.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if A(x) can be shown to equal O(x^{1/2+e}), it might imply the truth of RH, referencing the Mertens function.
  • Another participant suggests that A(x) should be treated as a random variable, raising the need to consider expected values like E(A(x)) or E(A(x)^2).
  • A participant challenges the idea that the Mertens function's value is independent of its argument, arguing that this undermines the initial claim.
  • It is noted that A(x) is expected to be of order O(√(x log log x)) with probability one if f(n) are independent, but this does not directly support the truth of RH.
  • Another participant emphasizes that while the Mertens function may exhibit properties similar to a random sequence, it is deterministic and this distinction is crucial.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the properties of A(x) and the Mertens function, with no consensus reached on whether the arguments presented support the truth of the Riemann Hypothesis.

Contextual Notes

Participants highlight the need for careful consideration of the definitions and assumptions involved, particularly regarding the deterministic nature of the Mertens function versus the randomness of f(n).

zetafunction
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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

\sum_{n=0}^{x}f(n) =A(x)

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

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

then is RH true by this argument? ,
 
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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)?
 
zetafunction said:
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.
 
zetafunction said:
how can one prove that 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 O(\sqrt{x\log\log x}).

zetafunction said:
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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