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*)=*id*_{x}(*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 *id*_{x}(*n*) and *id*_{x}(*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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Tags: DIrichlet Inverse, Math, Monday Math, Multiplicative Function

This entry was posted on October 19, 2009 at 1:25 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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