# Butis really so easy? (Möbius inversion)

1. May 8, 2007

### tpm

But..is really so easy? (Möbius inversion), let be F(x) and G(x) functions

$$F(x)= G(ax)+G(2ax)+G(3ax)+...............$$

for n=,1,2,3,4,5,............ a is a fixed real number. then

$$G(ax)= \sum_{n=1}^{\infty}\mu (x) F(nx)$$

is seems too easy for me, to be true.

2. May 8, 2007

### matt grime

Why don't you try it? (You might want to make sure you state it properly.)