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[parent] formula for the convolution inverse of a completely multiplicative function (Corollary)
Corollary   If $ f$ is a completely multiplicative function, then its convolution inverse is $ f\mu$, where $ \mu$ denotes the Möbius function.
Proof. Recall the Möbius inversion formula $ 1*\mu = \varepsilon$, where $ \varepsilon$ denotes the convolution identity function. Thus, $ f(1*\mu) = f\varepsilon$. Since pointwise multiplication of a completely multiplicative function distributes over convolution, $ (f \cdot 1)*(f\mu)=f\varepsilon$. Note that, for all natural numbers $ n$, $ f(n)1(n)=f(n) \cdot 1=f(n)$ and $ f(n)\varepsilon(n)=\varepsilon(n)$. Thus, $ f*(f \mu)=\varepsilon$. It follows that $ f\mu$ is the convolution inverse of $ f$. $ \qedsymbol$



"formula for the convolution inverse of a completely multiplicative function" is owned by Wkbj79.
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See Also: criterion for a multiplicative function to be completely multiplicative


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completely multiplicative functions whose convolution inverses are completely multiplicative (Corollary) by Wkbj79
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Cross-references: natural numbers, convolution identity function, Möbius inversion, Möbius function, convolution inverse, completely multiplicative function
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This is version 2 of formula for the convolution inverse of a completely multiplicative function, born on 2007-04-14, modified 2007-04-15.
Object id is 9181, canonical name is FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction.
Accessed 645 times total.

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AMS MSC11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas)

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