# formula for the convolution inverse of a completely multiplicative function

###### Corollary 1.

If $f$ is a completely multiplicative function^{}, then its convolution inverse is $f\mathit{}\mu $, where $\mu $ denotes the Möbius function^{}.

###### Proof.

Recall the Möbius inversion formula^{} $1*\mu =\epsilon $, where $\epsilon $ denotes the convolution identity function. Thus, $f(1*\mu )=f\epsilon $. Since pointwise multiplication^{} of a completely multiplicative function distributes over convolution (http://planetmath.org/PropertyOfCompletelyMultiplicativeFunctions), $(f\cdot 1)*(f\mu )=f\epsilon $. Note that, for all natural numbers^{} $n$, $f(n)1(n)=f(n)\cdot 1=f(n)$ and $f(n)\epsilon (n)=\epsilon (n)$. Thus, $f*(f\mu )=\epsilon $. It follows that $f\mu $ is the convolution inverse of $f$.
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Title | formula for the convolution inverse of a completely multiplicative function |
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Canonical name | FormulaForTheConvolutionInverseOfACompletelyMultiplicativeFunction |

Date of creation | 2013-03-22 16:55:09 |

Last modified on | 2013-03-22 16:55:09 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 5 |

Author | Wkbj79 (1863) |

Entry type | Corollary |

Classification | msc 11A25 |

Related topic | CriterionForAMultiplicativeFunctionToBeCompletelyMultiplicative |