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Dirichlet Inverse

  1. Jun 23, 2010 #1
    Find G in terms of F if

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

    Also what is the Dirichlet convolution inverse of log n?
  2. jcsd
  3. Jun 25, 2010 #2
    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)= [tex]\sum b(m)F(4x/m)[/tex] .Sum ranges from 1 to infinity.
    So we need F(x)= [tex]\sum\sum b(m)a(n)F(4x/nm)[/tex] . n,m>=1. a(n)=log n
    Taking mn=p, we get above sum as
    F(x)= [tex]\sum\sum b(m)a(p/m)F(4x/p)[/tex][/tex] . p>=1, m divides p.
    So b(1)=0(for p=1,2,3), b(2)= 1/log2(p=4) .
    Considering [tex]\sum b(m)a(p/m)[/tex] = 0 , p>4 , we can get all coefficients b.
  4. Jun 25, 2010 #3

    Thank you for the response.
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