6
$\begingroup$

I would appreciate help as to how to apply the Möbius inversion theorem to prime counting $\Pi (x)$ and $\pi (x)$, where:

$$\Pi(x) := \sum_{n = 1}^\infty \frac{1}{n} \pi(x^{1/n})$$

and $\pi (x) = \sum_{n = 1}^\infty \frac{\mu (n)}{n} \Pi(x^{1/n})$.

What I am having difficulty with, to begin with, is putting the $\Pi (x)$ relation in the form that makes the Möbius inversion theorem applicable:

$$f(n) = \sum_{d\mid n} g(d)$$

(I hope if I can see that, then I can figure out the $g(n)$ inverse.)

=====================

Edwards, "Riemann's Zeta Function" page 34 has a nice algorithm replacing $f(x)$ with $f(x) - (1/p)f(x^{1/p})$ on each side of the $\Pi (x)$ equation above which I tried. Of course it works.

I am also wondering if this works for all Möbius inversion applications. And also how this ties in with basic relations of the Möbius inversion theorem as stated above.

EDIT: In that I am asking two questions in one, perhaps any responders might like to give separate answers so that I may fully acknowledge their kind efforts.

Thanks very much.

$\endgroup$
  • $\begingroup$ But I think Edwards actually does that calculation in detail on p. 34. Then at 217-18 he has a discussion of the inversion generally--its derivation. This corresponds to section 2.7 in Apostol's Intro to Analytic Number Theory. A more general (and relevant) statement of inversion is $G(x) =\sum_{n\leq x}\alpha(n)F(x/n) \leftrightarrow F(x)=\sum \mu(n)\alpha(n)G(x/n).$ $\endgroup$ – daniel Aug 12 '13 at 1:25
  • $\begingroup$ @daniel Thanks for your comment. I see the relevance of the more general formula (also saw it on wikipedia). Ironically, I'm sure it's available somewhere, but the discussions I have seen (including a glimpse at Apostol at Amazon, deal with the relations as I posted them rather than the general statement. That, in fact, is really my first question. Regards, $\endgroup$ – user12802 Aug 12 '13 at 1:53
6
$\begingroup$

Let $F(x)$ and $G(x)$ (here) be real-valued functions defined on the positive real axis such that $F(x)$ and $G(x)$ are $0$ for $0< x< 1.$

Let $\alpha(n) = 1/n.$ This function is called completely multiplicative since for all m,n

$$\alpha(m)\cdot\alpha(n) = \alpha(mn),$$ which is true since

$\frac{1}{n}\cdot\frac{1}{m} = \frac{1}{mn}.$

The Dirichlet inverse $\alpha^{-1}(n) $ of a completely multiplicative function $\alpha(n) $ is

$\alpha^{-1}(n) = \mu(n)\alpha(n),~~~ n\geq 1.$

Given that $\alpha^{-1}(n) $ exists,

$$ G(x) = \sum_{n\leq x}\alpha(n)F(x/n) \leftrightarrow F(x) = \sum_{n\leq x}\alpha^{-1}(n)G(x/n) = \mu(n)\alpha(n)G(x/n)$$

and conversely. This is the generalized Möbius inversion formula.

The application to Edwards' inversion is straightforward taking $G(x) = \Pi(x), F(x) = \pi(x)$ and $\alpha(n) = 1/n $ as the completely multiplicative function required for the application of the general inversion theorem.

Writing out the terms of Edwards' inversion gives some intuition about how $\mu(n)$ works here. It is surely possible to work backwards and relate the simpler statements of the inversion formula to the general one, though that is a separate question and I have to leave off here.

This material is covered generally in Apostol's Intro. to Analytic Number Theory pp. 30-40.

$\endgroup$

Your Answer

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