Completely multiplicative function: Difference between revisions

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Properties: clarify implicit product: pointwise multiplication, not Dirichlet convolution
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A completely multiplicative function is completely determined by its values at the prime numbers, a consequence of the [[fundamental theorem of arithmetic]]. Thus, if ''n'' is a product of powers of distinct primes, say ''n'' = ''p''<sup>''a''</sup> ''q''<sup>''b''</sup> ..., then ''f''(''n'') = ''f''(''p'')<sup>''a''</sup> ''f''(''q'')<sup>''b''</sup> ...
 
While the [[Dirichlet convolution]] of two multiplicative functions is multiplicative, the [[Dirichlet convolution]] of two completely multiplicative functions need not be completely multiplicative.
 
There are a variety of statements about a function which are equivalent to it being completely multiplicative. For example, if a function ''f'' is multiplicative then it is completely multiplicative if and only if its [[Dirichlet inverse]] is <math>\mu\cdot f</math> where <math>\mu</math> is the [[Möbius function]].<ref>Apostol, p. 36</ref>
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<math>f \cdot (g*h)=(f \cdot g)*(f \cdot h)</math>
 
where ''*'' represents the [[Dirichlet product]] and <math>\cdot</math> represents [[Hadamard product (matrices)|pointwise multiplication]].<ref>Apostol pg. 49</ref> One consequence of this is that for any completely multiplicative function ''f'' one has
 
<math>f*f = \tau \cdot f</math>