Dirichlet Inverse: Find G in Terms of F

[Charles49]

Find G in terms of F if

<br /> F(x)=\sum_{n=1}^{\infty}G\left(\frac{x}{n}\right)\log n<br />

Also what is the Dirichlet convolution inverse of log n?

 

[01030312]

Clearly log 1 is 0. So it has no dirichlet inverse. But I think, there can be an easy way to get around this and invert the function G(x).
Write G(x)= \sum b(m)F(4x/m) .Sum ranges from 1 to infinity.
So we need F(x)= \sum\sum b(m)a(n)F(4x/nm) . n,m>=1. a(n)=log n
Taking mn=p, we get above sum as
F(x)= \sum\sum b(m)a(p/m)F(4x/p)[/tex] . p>=1, m divides p.
So b(1)=0(for p=1,2,3), b(2)= 1/log2(p=4) .
Considering \sum b(m)a(p/m) = 0 , p>4 , we can get all coefficients b.

 

[Charles49]

01030312,

Thank you for the response.