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equidistribution of Moebius function

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• 1. Re: The distribution of Moebius(n) Posted by: Phil Carmody thefatphil@yahoo.co.uk thefatphil Date: Sun Jun 18, 2006 10:13 am (PDT) The thing that is not
Message 1 of 4 , Jun 19, 2006
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1. Re: The distribution of Moebius(n)
Posted by: "Phil Carmody" thefatphil@... thefatphil
Date: Sun Jun 18, 2006 10:13 am (PDT)

The thing that is not intuitively clear to me is that those that remain,
those
with mu=+1 and -1, should be equidistributed. If anyone has a good heuristic
argument I'm all ears.

Phil

*****

Equidistribution is related to the equidistribution of odd and even in
binomial formula

(1 - 1 )^ m = 0

Sum of positive coeficients = magnitude of sum of negative coeficients.

The square free integers are the coeficients in the formal expansion of

(1 + 2) (1 + 3)(1 +5)(1+7)...(1+p)......

= 1 + 2 + 3 + 6 + 5 + 10 + 15 + 30 +
7 + 14 + 21 + 42 + 35 + 70 + 105 + 210 + ...

The number of terms in the formal expansion of

m products ( 1 + p1)(1 + p2) (1+p3)....(1 + pm) is 2^m.

half of these terms correspond to a product of an even number of primes,
and
half of these terms correspond to a product of an odd number of primes.

Kermit < kermit@... >
• ... You ve re-ordered two subsequences of the squarefree integers, and provided a bijection between those two subsequences. Regarding the re-ordering, note
Message 2 of 4 , Jun 20, 2006
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--- Kermit Rose <kermit@...> wrote:
>> The thing that is not intuitively clear to me is that those that remain,
>> those
>> with mu=+1 and -1, should be equidistributed. If anyone has a good heuristic
>> argument I'm all ears.
>
> Equidistribution is related to the equidistribution of odd and even in
> binomial formula
>
> (1 - 1 )^ m = 0
>
> Sum of positive coeficients = magnitude of sum of negative coeficients.
>
> The square free integers are the coeficients in the formal expansion of
>
> (1 + 2) (1 + 3)(1 +5)(1+7)...(1+p)......
>
> = 1 + 2 + 3 + 6 + 5 + 10 + 15 + 30 +
> 7 + 14 + 21 + 42 + 35 + 70 + 105 + 210 + ...
>
>
> The number of terms in the formal expansion of
>
> m products ( 1 + p1)(1 + p2) (1+p3)....(1 + pm) is 2^m.
>
> half of these terms correspond to a product of an even number of primes,
> and
> half of these terms correspond to a product of an odd number of primes.

You've re-ordered two subsequences of the squarefree integers, and provided a
bijection between those two subsequences. Regarding the re-ordering, note that
the largest member of the two subsequences are different by a factor of 2.

This same type of argument can provide a bijection between the primes and the
square-free non-primes, despite the fact that the former has zero density in
the squarefree integers.

So from my PoV, it's not a good enough heuristic as it stands, sorry. It was
close, though.

Phil

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• I doubt that there is a good heuristic. As important as Moebius(n) itself is, the cumulative value is more relevant, i.e. the number you get if you add
Message 3 of 4 , Jun 22, 2006
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I doubt that there is a good heuristic.

As important as Moebius(n) itself is, the cumulative value is more relevant, i.e. the number you get if you add Moebius(1)+Moebius(2)+Moebius(3)...+Moebius(k) for some number k. This is Mertens Function M(k). Its first 10 values for arguments k=1,2,3... up to 10, are 1,0,-1,-1,-2,-1,-2,-2,-2,-1.

M(k) is a very irregular function, oscillating back and forth in a "random walk".

For arguments 1000,2000 ...up to 10,000 it has values: 2,5,-6,-9,2,0,-25,-1,1,-23.

For arguments 1,000,000, 2,000,000 to 10,000,000, it has values 212,-247,107,192,-709,257,-184,-189,-340,1037. Ignoring the signs, it's evident that the size of M(k) increases, but nothing else is clear.

Cheers

Bob

Phil Carmody <thefatphil@...> wrote:
--- Kermit Rose <kermit@...> wrote:
>> The thing that is not intuitively clear to me is that those that remain,
>> those
>> with mu=+1 and -1, should be equidistributed. If anyone has a good heuristic
>> argument I'm all ears.
>
> Equidistribution is related to the equidistribution of odd and even in
> binomial formula
>
> (1 - 1 )^ m = 0
>
> Sum of positive coeficients = magnitude of sum of negative coeficients.
>
> The square free integers are the coeficients in the formal expansion of
>
> (1 + 2) (1 + 3)(1 +5)(1+7)...(1+p)......
>
> = 1 + 2 + 3 + 6 + 5 + 10 + 15 + 30 +
> 7 + 14 + 21 + 42 + 35 + 70 + 105 + 210 + ...
>
>
> The number of terms in the formal expansion of
>
> m products ( 1 + p1)(1 + p2) (1+p3)....(1 + pm) is 2^m.
>
> half of these terms correspond to a product of an even number of primes,
> and
> half of these terms correspond to a product of an odd number of primes.

You've re-ordered two subsequences of the squarefree integers, and provided a
bijection between those two subsequences. Regarding the re-ordering, note that
the largest member of the two subsequences are different by a factor of 2.

This same type of argument can provide a bijection between the primes and the
square-free non-primes, despite the fact that the former has zero density in
the squarefree integers.

So from my PoV, it's not a good enough heuristic as it stands, sorry. It was
close, though.

Phil

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• ... Good -- elegant -- you might be right. However, I think that I can formulate a trend to look for that would imply the equidistribution of the +1 and -1
Message 4 of 4 , Jun 23, 2006
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--- Bob Gilson <bobgillson@...> wrote:
> I doubt that there is a good heuristic.

Good -- elegant -- you might be right. However, I think that I can
formulate a trend to look for that would imply the equidistribution
of the +1 and -1 values.

For squarefree numbers in the range [N-N*eps, N+N*eps), create an
expression in terms of N for the number or the proportion of them
that have f \in {1, 2, 3, ... } factors. I believe such expressions
exist, as they certainly do for f=1, the primes. Then accumulate
the terms for odd and even f's, et viola - perhaps there is an
expression which has a simple limit.

Annoyingly, I suspect each term in the expression would (very slowly)
tend (as a proportion) to zero as N increases, so one could not simply
take a finite number of terms, nor even forget any earlier terms. I
think it would be a very nasty expression indeed.

The error bounds on this expression would also be fairly hideous, and
unfortunately that might matter. Euler's gamma shows how much the
primes can connive with their non-randomness.

Phil

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