 #1
Math100
 611
 171
 Homework Statement:

Prove that if ## f ## is any arithmetical function, then ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.
[Hint: if you have difficulty with this, write out both sides in the case ## n=10 ##.]
 Relevant Equations:
 None.
Proof:
Let ## n=10 ##.
Observe that
\begin{align*}
&\sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d})\\
&\sum_{d\mid 10}f(d)=\sum_{d\mid 10}f(\frac{10}{d})\\
&f(1)+f(2)+f(5)+f(10)=f(\frac{10}{1})+f(\frac{10}{2})+f(\frac{10}{5})+f(\frac{10}{10})\\
&f(1)+f(2)+f(5)+f(10)=f(10)+f(5)+f(2)+f(1).\\
\end{align*}
Therefore, ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.
Let ## n=10 ##.
Observe that
\begin{align*}
&\sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d})\\
&\sum_{d\mid 10}f(d)=\sum_{d\mid 10}f(\frac{10}{d})\\
&f(1)+f(2)+f(5)+f(10)=f(\frac{10}{1})+f(\frac{10}{2})+f(\frac{10}{5})+f(\frac{10}{10})\\
&f(1)+f(2)+f(5)+f(10)=f(10)+f(5)+f(2)+f(1).\\
\end{align*}
Therefore, ## \sum_{d\mid n}f(d)=\sum_{d\mid n}f(\frac{n}{d}) ##.