1
$\begingroup$

I searched and wasn't able to find a question similar enough to mine. Here's the problem:

Show that $\sigma=c_1*c_1$, where $c_1$ is the constant function $1$.

Here is my attempt. My argument makes sense to me, but it seems kind of short. I'm wondering if there is anything I need to add to my argument or if I'm sort of using circular reasoning.

The operation $*$ is the convolution product defined as: \begin{equation} \begin{split} (c_1 * c_1)(n) &= \sum_{d \mid n}c_1(d)c_1\left(\frac nd \right) \\ \end{split} \end{equation}

The formula above is clearly true if $n=1$. Assume that $n > 1$ and write $n=p_1^{e_1} \dots p_s^{e_s}$. In this sum, all terms, $c_1(d)c_1(\frac nd)$, are equal to $1$ for all $d \mid n$. So we will end up multiplying $1$ by the number of divisors, which is exactly $\sigma(n)$.

As always, thank you all for your help.

$\endgroup$
  • $\begingroup$ You've written the question backwards, first introducing $\sigma$ and $*$, and only later telling people what you mean by those symbols. Also, $\sigma(n)$ is generally used for the sum of the divisors, not the number of divisors. $\endgroup$ – Gerry Myerson Apr 23 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.