- #1

Math100

- 773

- 219

- 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}) ##.