Looking for help understanding the Möbius Inversion Formula

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.

Thanks,

-Larry

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.