convolution inverses for arithmetic functions


An arithmetic functionMathworldPlanetmath f has a convolution inverse if and only if f(1)0.


If f has a convolution inverse g, then f*g=ε, where ε denotes the convolution identity function. Thus, 1=ε(1)=(f*g)(1)=f(1)g(1), and it follows that f(1)0.

Conversely, if f(1)0, then an arithmetic function g must be constructed such that (f*g)(n)=ε(n) for all n. This will be done by inductionMathworldPlanetmath on n.

Since f(1)0, we have that 1f(1). Define g(1)=1f(1).

Now let k with k>1 and g(1),,g(k-1) be such that (f*g)(n)=ε(n) for all n with n<k. Define

g(k)=-1f(1)d|k and d<kf(kd)g(d).


(f*g)(k) =d|kf(kd)g(d)
=f(1)g(k)+d|k and d<kf(kd)g(d)
=f(1)(-1f(1)d|k and d<kf(kd)g(d))+d|k and d<kf(kd)g(d)

In the entry titled arithmetic functions form a ring, it is proven that convolution is associative and commutativePlanetmathPlanetmathPlanetmath. Thus, G={f:|f(1)0} is an abelian group under convolution. The set of all multiplicative functions is a subgroupMathworldPlanetmathPlanetmath of G.

Title convolution inverses for arithmetic functions
Canonical name ConvolutionInversesForArithmeticFunctions
Date of creation 2013-03-22 15:58:32
Last modified on 2013-03-22 15:58:32
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 27
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 11A25
Related topic ArithmeticFunction
Related topic MultiplicativeFunction
Related topic ArithmeticFunctionsFormARing
Related topic ElementaryResultsAboutMultiplicativeFunctionsAndConvolution