Editing Arithmetic function (section)

== Dirichlet convolution ==
Given an arithmetic function ''a''(''n''), let ''F''<sub>''a''</sub>(''s''), for complex ''s'', be the function defined by the corresponding [[Dirichlet series]] (where it [[Convergent series|converges]]):<ref>Hardy & Wright, § 17.6, show how the theory of generating functions can be constructed in a purely formal manner with no attention paid to convergence.</ref>
<math display="block"> F_a(s) := \sum_{n=1}^\infty \frac{a(n)}{n^s} .</math>
''F''<sub>''a''</sub>(''s'') is called a [[generating function]] of ''a''(''n''). The simplest such series, corresponding to the constant function ''a''(''n'') = 1 for all ''n'', is ''ζ''(''s'') the [[Riemann zeta function]].

The generating function of the Möbius function is the inverse of the zeta function:
<math display="block">\zeta(s)\,\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=1, \;\;\Re s >1.</math>

Consider two arithmetic functions ''a'' and ''b'' and their respective generating functions ''F''<sub>''a''</sub>(''s'') and ''F''<sub>''b''</sub>(''s''). The product ''F''<sub>''a''</sub>(''s'')''F''<sub>''b''</sub>(''s'') can be computed as follows:
<math display="block"> F_a(s)F_b(s) = \left( \sum_{m=1}^{\infty}\frac{a(m)}{m^s} \right)\left( \sum_{n=1}^{\infty}\frac{b(n)}{n^s} \right) . </math>

It is a straightforward exercise to show that if ''c''(''n'') is defined by
<math display="block"> c(n) := \sum_{ij = n} a(i)b(j) = \sum_{i\mid n}a(i)b\left(\frac{n}{i}\right) , </math>
then <math display="block">F_c(s) = F_a(s) F_b(s).</math>

This function ''c'' is called the [[Dirichlet convolution]] of ''a'' and ''b'', and is denoted by <math>a*b</math>.

A particularly important case is convolution with the constant function ''a''(''n'') = 1 for all ''n'', corresponding to multiplying the generating function by the zeta function:
<math display="block">g(n) = \sum_{d \mid n}f(d).</math>

Multiplying by the inverse of the zeta function gives the [[Möbius inversion]] formula:
<math display="block">f(n) = \sum_{d\mid n}\mu\left(\frac{n}{d}\right)g(d).</math>

If ''f'' is multiplicative, then so is ''g''. If ''f'' is completely multiplicative, then ''g'' is multiplicative, but may or may not be completely multiplicative.