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 .


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2 Responses to “Monday Math 93”

  1. Monday Math 94 « Twisted One 151’s Weblog Says:

    […] here I showed that the Dirichlet series generating function for |μ| is . I also demonstrated here that |μ| and the Liouville function lambda(n) are Dirichlet inverses, so we can see that […]

  2. Monday Math 95 « Twisted One 151’s Weblog Says:

    […] is also zero in that case. Now, for n>1 with prime factorization , we have , with . I showed here that the Dirichlet inverse f-1(n) of a completely multiplicative function f(n) is the […]

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