- 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@... > - --- Kermit Rose <kermit@...> wrote:
>> The thing that is not intuitively clear to me is that those that remain,

You've re-ordered two subsequences of the squarefree integers, and provided a

>> 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.

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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http://mail.yahoo.com - 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,

You've re-ordered two subsequences of the squarefree integers, and provided a

>> 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.

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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[Non-text portions of this message have been removed] - --- 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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