7
$\begingroup$

Given the 2 functions $ g(x)= \sum_{n=1}^{\infty}f\left(\frac{x}n\right)\log(n)\;, $ how can I use Möbius inversion to recover $f$ from $g$?? I believe that

$ f(x)= \sum_{n=1}^{\infty}\mu (n)g\left(\frac{x}n\right)\log(n)\;. $ Here 'mu' is the Möbius function.

2 Answers 2

5

Suppose $f(x)=x^{10}$. Then $g(x)=\sum_1^{\infty}x^{10}n^{-10}\log n=Cx^{10}$ where $C=\sum_1^{\infty}n^{-10}\log n$ is a very small positive constant. Then $\sum_1^{\infty}\mu(n)g(x/n)\log n=Cx^{10}\sum_1^{\infty}\mu(n)n^{-10}\log n=CDx^{10}$ where $D=\sum_1^{\infty}\mu(n)n^{-10}\log n$ is a very small constant. We can't have $CD=1$, so we can't have $f(x)=\sum_1^{\infty}\mu(n)g(x/n)\log n$

  • 0
    You're right; looks like I was confusing the order of differentiation and taking the reciprocal in my Dirichlet series thinking.2012-03-21