Given a series $p_n(s)=\sum_{k=1}^n a_k $. I'd like to get $\hat{p}_n(s)=\sum_{k=1}^n \mu(k)a_k $. Think of $a_k=k^{-s}$ for example. If you let $n$ go to $\infty$, you'll see the well-known relation between $\zeta$ and $1/\zeta$.

May I apply something like a Möbius operator on $p_n(s)$, such that $\mu\circ p=\hat p$?

The wiki page on the Primon gas mentions the Fermi operator $(-1)^F$, so I thought about the following: Given a $n$-dimensional vector $\vec a=(a_k)^T$. Let $\vec e=(1,\dots,1)^T$, then $p=\vec e \cdot \vec a$ and $\hat p=\vec e \cdot (\mu \vec a)$, with $\mu$ being a diagonal matrix with entries $\mu_{kk}=\mu(k)$. Is this a finite dimensional matrix representation of the Fermi operator?

EDIT: The thing is, I don't see why there is on the one hand something like

Another inversion formula is (where we assume that the series involved are absolutely convergent): $$ g(x) = \sum_{m=1}^\infty \frac{f(mx)}{m^s}\quad\mbox{ for all } x\ge 1\quad\Longleftrightarrow\quad f(x) = \sum_{m=1}^\infty \mu(m)\frac{g(mx)}{m^s}\quad\mbox{ for all } x\ge 1. $$ (see here)

further on the other

The classic version states that if g(n) and f(n) are arithmetic functions satisfying $$ g(n)=\sum_{d\,\mid \,n}f(d)\quad\text{for every integer }n\ge 1 $$ then $$ f(n)=\sum_{d\,\mid\, n}\mu(d)g(n/d)\quad\text{for every integer }n\ge 1 $$ (see there)

and on the last hand

A related inversion formula more useful in combinatorics is as follows: suppose F(x) and G(x) are complex-valued functions defined on the interval [1,∞) such that $$ G(x) = \sum_{1 \le n \le x}F(x/n)\quad\mbox{ for all }x\ge 1 $$ then $$ F(x) = \sum_{1 \le n \le x}\mu(n)G(x/n)\quad\mbox{ for all }x\ge 1. $$

None of them fits my needs :-( I want someting like $F(x) = \sum_{1 \le n \le x}\mu(n)G(n)\quad\mbox{ for all }x\ge 1. $

EDIT: Peter Sarnak's Or Gil Kalai's work (e.g. The Depth Of The Möbius Function or The AC0 Prime Number Conjecture) shows the right kind of formulas, but I'm not sure if they are talking about what I need: My function $G(n)$ would sent the natural numbers to $be^{ia}$, with $b$ being an integer and not only to $\{-1,1\}$...

  • $\begingroup$ What do you mean by "get" in your first sentence? $\endgroup$ – Gerry Myerson Oct 7 '12 at 23:20
  • $\begingroup$ I'd like to start out with $p_n(s)$ and then do something like the Dirichlet convolution without the $d|n$ summation restriction to get $\hat p_n(s)$. BTW, it's the second sentence... $\endgroup$ – draks ... Oct 8 '12 at 5:30
  • $\begingroup$ Instead of "Dirichlet convolution" it's rather a Moebius Inversion $\mu * p$ (which is a special kind of Dirichlet convolution), but summed without the $d|n$- restriction. $\endgroup$ – draks ... Oct 8 '12 at 5:42
  • $\begingroup$ @GerryMyerson e.g. the inverse of $\zeta$ would be an example, which sums over infinitely many elements without regarding $d|n$. I'm interested in the finite case. $\endgroup$ – draks ... Oct 8 '12 at 5:49
  • $\begingroup$ "On the one hand ... on the other ... on the last hand". I see you've got three hands, lucky you! $\endgroup$ – Marc van Leeuwen Oct 9 '12 at 7:41

What you want to do goes against the nature of the Möbius function, so I don't think you are entitled to hope that anything simple and useful could be be said about the operation you describe. I'll explain a bit.

One should think of the Möbius function as a function of two integer arguments, $\mu(d,n)$, which represents an (infinite) upper unitriangular matrix $(\mu(d,n))_{d,n\geq1}$ that is the inverse of the matrix $D=([\,d\mid n\,])_{d,n\geq1}$ (the brackets being the Iverson bracket, whose value is $1$ or $0$ according as the divisibilty $d\mid n$ does or does not hold). It happens that $\mu(d,n)$ depends only the value of $n/d$ (because the divisibility interval $\{\,x\in\Bbb N: d\mid x\mid n\,\}$ is isomorphic to $\{\,x\in\Bbb N: 1\mid x\mid n/d\,\}$), so that it can be (and is) written as $\mu(n/d)$, but I consider this an almost accidental circumstance.

The natural thing to do with a matrix is multiplying a vector by it; in this case right multiplication by $D$ applied to an infinite sequence $f=(f(n))_{n\geq1}$ gives $$ f\cdot D=(g(n))_{n\geq1} \quad\text{where}\quad g(n)=\sum_{d\mid n}f(d), $$ which operation is the Dirichlet convolution by the constant sequence $(1)_{n\geq1}$, and right multiplication by $M=(\mu(d,n))_{d,n\geq1}$ is therefore the inverse operation $$ g\cdot M=(f(n))_{n\geq1} \quad\text{where}\quad f(n)=\sum_{d\mid n}g(d)\mu(d,n), $$ which operation, due to $\mu(d,n)=\mu(n/d)$, is Dirichlet convolution by the sequence $(\mu(m))_{m\geq1}$.

This gives your middle statement (except that it switches $d$ and $n/d$ which is of course harmless; I prefer to write $\mu(n/d)$ to remind of the matrix nature). For the other two statements, I can see how they follow, but I don't right know see how to get them precisely into this picture, although I get the impression that the first form is related to left-multiplication by these matrices, giving rise to infinite sums that somehow are mitigated by the factors $m^{-s}$.

In any case, what you want to do seems like doing a Hadamard product of power series defined to be mutliplicative inverses of something; not easy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.