Completely multiplicative function: Difference between revisions

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where ''*'' represents the [[Dirichlet product]] and <math>\cdot</math> represents 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>

which deduced from the latter/above for [both] <math>g = h = 1</math>, where <math>1(n) = 1</math> is well-known [[multiplicative function#Examples|constant function]].

Here <math> \tau</math> is the [[divisor function]].

===Proof of pseudo-associative property ===

<math> f \cdot \left(g*h \right)(n) = f(n) \sum_{d|n} g(d) h \left( \frac{n}{d} \right) =</math>

::<math> f \cdot \left(g*h \right)(n) = \sum_{d|n} f(n) g(d) h \left( \frac{n}{d} \right) </math>

::<math> f \cdot \left(g*h \right)(n) = \sum_{d|n} (f(d) f \left( \frac{n}{d} \right)) \cdot g(d) h \left( \frac{n}{d} \right) </math> (since ''f'' is completely multiplicative)

::<math> f \cdot \left(g*h \right)(n) = \sum_{d|n} (f(d) g(d)) \cdot (f \left( \frac{n}{d} \right) h \left( \frac{n}{d} \right)) = (f \cdot g)*(f \cdot h).</math>

<math> f \cdot \left(g*h \right)(n) = (f \cdot g)*(f \cdot h).</math>

==See also==