Asymptotic formula for Mertens function

  • Thread starter keebs
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I've been thinking about this for a while and I just wanted somebody to show me where my proof becomes faulty. This was my attempt to find an asymptotic formula for Mertens' function (the sum of the Mobius function). Oh, and if you aren't clear of my reasoning behind something, just ask...and in case it isn't obvious, I used pn to denote the nth prime number. Sorry about the cheap scan...I don't know how to use Latex...

http://home.comcast.net/~shane_stephenson/mertens1.jpg [Broken]
 
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shmoe

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You found the sum for the mobius function on [tex]\mathbb{P}_n[/tex]. This is distinctly different than the sum for the mobius function on {1,..,n}, you are seriously mucking with the order of summation. That is there's no reason to expect

[tex]\lim_{n\rightarrow\infty}\sum_{m\in\mathbb{P}_n}\mu(m)=\lim_{n\rightarrow\infty}\sum_{1\leq m\leq n}\mu(m)[/tex]

since if you scrub off the limits you don't have equality in general. Even if you allow the n's to go to infinity seperately on each side of the equation, the fact that you've changed the order of summation means you can't expect equality- if the limit on the right did exist, it would have to be conditionally convergent. This limit of course can't exist, since the terms themselves don't go to zero.

You do know the conlcusion is false in any case right? It's known that [tex]M(x)>x^{1/2}[/tex] for some arbitrarily large values of x. Likewise for [tex]M(x)<-x^{1/2}[/tex]. It flops back and forth pretty good, the limit on the right not only fails to exist but does so in a bad way.

Something is funny with your notation by the way. It looks like you're allowing 0 for exponents in [tex]\mathbb{P}_n[/tex]? [tex]\mathbb{N}[/tex] usually denotes {1,2, ...} in number theory. If you are allowing 0 exponents, then 1 is in this set. If you're not allowing zero exponents, then there's even less reason to expect the limit equation above to be true- in this case [tex]\mathbb{P}_n[/tex] does not pick up all natural numbers.
 
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That is there's no reason to expect
[tex]\lim_{n\rightarrow\infty}\sum_{m\in\mathbb{P}_n}\m u(m)=\lim_{n\rightarrow\infty}\sum_{1\leq m\leq n}\mu(m)[/tex]
since if you scrub off the limits you don't have equality in general.
Yeah, but when n approaches infinity Pn approaches the set of all natural numbers because Pn is the set of all numbers generated by all primes up to n, so when we include all of the primes it should generate all of the natural numbers.

This limit of course can't exist, since the terms themselves don't go to zero.
Why does the terms have to go to zero in order for the function to converge? I never thought of that...

You do know the conlcusion is false in any case right? It's known that [tex]M(x)>x^{1/2}[/tex] for some arbitrarily large values of x. [tex]M(x)<-x^{1/2}[/tex]. It flops back and forth pretty good, the limit on the right not only fails to exist but does so in a bad way.
Yeah, I've read that lim sup M(x)>1.6x1/2, but that just means that it is above 1.6x1/2 at some point, but it does not necessarily mean that it stays like that, right?

Something is funny with your notation by the way. It looks like you're allowing 0 for exponents in [tex]\mathbb{P}_n[/tex]? [tex]\mathbb{N}[/tex] usually denotes {1,2, ...} in number theory. If you are allowing 0 exponents, then 1 is in this set. If you're not allowing zero exponents, then there's even less reason to expect the limit equation above to be true- in this case [tex]\mathbb{P}_n[/tex] does not pick up all natural numbers.
Yeah, I didn't think that all the way through. Thanks for the help.
 

shmoe

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keebs said:
Yeah, but when n approaches infinity Pn approaches the set of all natural numbers because Pn is the set of all numbers generated by all primes up to n, so when we include all of the primes it should generate all of the natural numbers.
You agree that you've changed the order? With an infinite sum, unlike a finite one, the order of summation does matter. It's built into the notation that you are summing over an ordered set. If you rearrange the terms there's no general guarantee what will happen. Specifically if you have a conditionally convergent series you can rearrange the terms to get anything you like (see a recent post for an example).

keebs said:
Why does the terms have to go to zero in order for the function to converge? I never thought of that...
This is a standard divergence test you see in calculus. You might want to give a stab at proving it before looking it up (I can provide the details if you wish).

keebs said:
Yeah, I've read that lim sup M(x)>1.6x1/2, but that just means that it is above 1.6x1/2 at some point, but it does not necessarily mean that it stays like that, right?


('tis 1.06 you mean not 1.6 yes?)

It means that you can find an x as large as you like with [tex]M(x)>x^{1/2}[/tex]. So no it doesn't necessarily stay there (in fact it can't given the corresponding result for lim inf), but you know that it will eventually return. You can find* some sequence [tex]a_n[/tex] where [tex]a_n\rightarrow\infty[/tex] as [tex]n\rightarrow\infty[/tex] and [tex]M(a_n)>a_n^{1/2}[/tex] for all n. In otherwords M(x) takes on arbitrarily large positive values (and negative using the lim inf version).


*I'm being liberal here. You can't actually find it as in write down some explicit sequence, but you know one exists. I don't think anyone can even give you a valid [tex]a_1[/tex] yet to start off the sequence, but this is only a matter of time and increased computational power (may be *alot* of time) before this happens.

keebs said:
Yeah, I didn't think that all the way through. Thanks for the help.
No problem. It was actually refreshing that you didn't claim to have proven the limit was zero but instead asked for a critique of your 'proof'.

As an aside, latex is well worth learning. these forums are a good intro for basics, if you don't know already you can click on the generated images to see how it was produced so you've plenty of examples.
 
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19
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You agree that you've changed the order? With an infinite sum, unlike a finite one, the order of summation does matter. It's built into the notation that you are summing over an ordered set. If you rearrange the terms there's no general guarantee what will happen. Specifically if you have a conditionally convergent series you can rearrange the terms to get anything you like (see a recent post for an example).
Hmmm...interesting. I didn't know that, thanks.

This is a standard divergence test you see in calculus. You might want to give a stab at proving it before looking it up (I can provide the details if you wish).
It makes sense because if it doesn't converge to 0 then "towards the end" (I guess you could say that) of the summation you'd just be adding values very close to a certain constant (or adding diverging terms) over and over and over again, but when you originally said that I was thinking about a function that oscillates around 0 and all of the terms except a few end up cancelling out. Now that I think about it though, if it keeps oscillating back and forth around 0 your sum will just keep oscillating as well, so it's sort of like trying to take the limit of the sin function as it approaches infinity.

('tis 1.06 you mean not 1.6 yes?)
Yeah...my mistake.

As an aside, latex is well worth learning. these forums are a good intro for basics, if you don't know already you can click on the generated images to see how it was produced so you've plenty of examples.
Yeah, I figured that out when I was trying to quote you.

Anyways, thanks again.
 

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