Editing Möbius inversion formula (section)

==Proofs of generalizations==

The first generalization can be proved as follows.  We use [[Iverson's convention]] that [condition] is the indicator function of the condition, being 1 if the condition is true and 0 if false.  We use the result that
:<math>\sum_{d|n}\mu(d)=\varepsilon (n),</math>
that is, <math> 1 * \mu = \varepsilon</math>, where <math>\varepsilon</math> is the [[unit function]].

We have the following:
:<math>\begin{align}
\sum_{1\le n\le x}\mu(n)g\left(\frac{x}{n}\right)
 &= \sum_{1\le n\le x} \mu(n) \sum_{1\le m\le \frac{x}{n}} f\left(\frac{x}{mn}\right)\\
 &= \sum_{1\le n\le x} \mu(n) \sum_{1\le m\le \frac{x}{n}} \sum_{1\le r\le x} [r=mn] f\left(\frac{x}{r}\right)\\
 &= \sum_{1\le r\le x} f\left(\frac{x}{r}\right) \sum_{1\le n\le x} \mu(n) \sum_{1\le m\le \frac{x}{n}} \left[m=\frac{r}{n}\right] \qquad\text{rearranging the summation order}\\
 &= \sum_{1\le r\le x} f\left(\frac{x}{r}\right) \sum_{n|r} \mu(n) \\
 &= \sum_{1\le r\le x} f\left(\frac{x}{r}\right) \varepsilon (r) \\
 &= f(x) \qquad\text{since } \varepsilon (r)=0\text{ except when }r=1
\end{align}</math>

The proof in the more general case where {{math|''α''(''n'')}} replaces 1 is essentially identical, as is the second generalisation.