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I was recently told that the Möbius Inversion Formula can be applied to the Chebyshev Function.

Let $\vartheta(x)$,$\psi(x)$ be the first and second Chebyshev functions so that:

$$\vartheta(x) = \sum_{p\le{x}}\log p$$

$$\psi(x) = \sum_{n=1}^{\infty}\vartheta(\sqrt[n]{x})$$

Then applying the Möbius Inversion Formula, we get:

$$\vartheta(x) = \sum_{k=1}^{\infty}\mu(k)\psi(\sqrt[k]{x}) = \psi(x) - \psi(\sqrt{x}) -\psi(\sqrt[3]{x}) -\psi(\sqrt[5]{x}) + \psi(\sqrt[6]{x}) + \ldots$$

As I understand it, Möbius Inversion Formula can only be applied to functions of the following form:

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

With the inversion being of the form:

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

So, I'm not clear how the inversion formula can be applied to the Chebyshev functions.

If someone could help me to understand why the inversion formula can be applied in this situation, that will really help.



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up vote 7 down vote accepted

I spent an embarrassingly long time thinking this through, trying to do all-too-clever things (as it turns out) before this appeared for me. But here's how this works.

$$ \begin{align} \sum_{k \geq 1} \mu(k) \psi(x^{1/k}) &= \sum_{k \geq 1} \mu(k) \sum_{l \geq 1} \vartheta (x^{1 / kl}) \\ &= \sum_{n \geq 1} \sum_{k \mid n}\mu(k)\vartheta(x^{1/n}) \\ &= \sum_{n \geq 1} \left( \delta_{1,n}\right)\vartheta(x^{1/n}) \\ &= \vartheta(x) \end{align}$$

This shows that we always have that $$F(x) = \sum_{k \geq 1} G(x^{1/k}) \iff G(x) = \sum_{k \geq 1}\mu(k)F(x^{1/k})$$

or, more generally (but with no extra difficulty), if $\alpha(n)$ is arithmetic with Dirichlet convolution inverse $\alpha^{-1}(n)$, $$F(x) = \sum_{k \geq 1} \alpha(k)G(x^{1/k}) \iff G(x) = \sum_{k \geq 1}\alpha^{-1}(k)F(x^{1/k})$$

So in fact we have another form of Möbius Inversion, relying on nothing that we haven't seen before.

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What's worse is that I now recognize that I've seen this before, too. (I think) – mixedmath Apr 23 '13 at 6:05
Thanks very much. This is a nice example of how to reason with the Mobius Inversion Formula! :-) – Larry Freeman Apr 23 '13 at 6:23
I don't know if this is the place where you saw it before, but I found the same proof with explanation about $\delta_{1,n}$, here. See page 2. – Larry Freeman May 7 '13 at 15:42

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