## Monday Math 93

Last Monday, I discussed the Dirichlet inverse. I showed that the Dirichlet inverse can be found by the recursive formula:
,
and for n>0,
.
Further, since the inverse of a multiplicative function is also multiplicative, for a multiplicative function f(n) this formula becomes
, and for p prime and k>0,
.

Now, what happens when f(n) is completely multiplicative? Then . We have , and via k=1, , we see that
,
,

and so we see that if f(n) is completely multiplicative, then f-1(n) is the multiplicative function with
.
For f(n)=1(n), this gives
, which gives , as we’ve found before.
Next, for f(n)=idx(n), we see
, which, with a little examination, we see means that , which reduces to the above for x=0. [Just as the inverse pair of 1(n) and μ(n) gives us the Möbius inversion formula, the above pair of idx(n) and idx(n)μ(n) gives a generalized Möbius inversion.]

Lastly, consider the Liouville function . This is completely multiplicative, with λ(p)=-1. Then we have , which we should recognize as .