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===Proof of pseudo-associative property ===
:<math>
<math> f \cdot \left(g*h \right)(n) = f(n) \cdot \sum_{d|n} g(d) h \left( \frac{n}{d} \right) =</math> ▼\begin{align}
::<math>=f \sum_{d|cdot \left(g*h \right)(n}) &= f(n) \cdot (\sum_{d|n} g(d) h \left( \frac{n}{d} \right)) =</math> \\
▲<math> f&= \ cdot \left(g*h \right)(sum_{d|n ) =} f(n) \cdot \sum_{d|n} (g(d) h \left( \frac{n}{d} \right )) = </math> \\
::<math>&= \sum_{d|n} (f(d) f \left( \frac{n}{d} \right)) \cdot (g(d) h \left( \frac{n}{d} \right)) </math>\text{ (since ''} f'' \text{ is completely multiplicative) <math>} =</math> \\
::<math>&= \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>▼
&= (f \cdot g)*(f \cdot h).
▲::<math>= \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>
\end{align}
</math>
==See also==
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