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===Proof of pseudo-associative property ===
<math> f \cdot \left(g*h \right)(n) = f(n) \cdot \sum_{d|n} g(d)
::<math>= \sum_{d|n} f(n) \cdot (g(d)
::<math>= \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>=</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>
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