Verify this result when ## n=10 ##

[Math100]

Homework Statement

Prove that $\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d})$, and verify this result when $n=10$.

Relevant Equations

None.

Proof:

Observe that
\begin{align*}
\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})&=(N^{2}\ast u)\ast (N^{2}\ast u) \text{ since } \sigma_{2}(n)=\sum_{d\mid n}d^{2}\cdot 1=N^{2}\ast u)\\
&=(u\ast N^{2})\ast (N^{2}\ast u)\\
&=u\ast (N^{2}\ast N^{2})\ast u\\
&=u\ast (N^{2}\sigma_{0})\ast u \text{ since } N^{2}\ast N^{2}=\sum_{d\mid n}d^{2}(\frac{n^{2}}{d^{2}})=n^{2}\sum_{d\mid n}1=N^{2}\sigma_{0} \\
&=N^{2}\sigma_{0}\ast (u\ast u) \text{ since } u\ast u=\sigma_{0} \\
&=N^{2}\sigma_{0}\ast \sigma_{0}\\
&=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}).\\
\end{align*}
Therefore, $\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d})$.
Now we will verify this result when $n=10$.
Consider the divisors $d\in\left \{ 1, 2, 5, 10 \right \}$.
Observe that
\begin{align*}
\sum_{d\mid 10}\sigma_{2}(d)\sigma_{2}(\frac{10}{d})&\stackrel{?}{=}\sum_{d\mid 10}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{10}{d})
\end{align*}
\begin{align*}
&\sigma_{2}(1)\sigma_{2}(\frac{10}{1})+\sigma_{2}(2)\sigma_{2}(\frac{10}{2})+\sigma_{2}(5)\sigma_{2}(\frac{10}{5})+\sigma_{2}(10)\sigma_{2}(\frac{10}{10})\\
&\stackrel{?}{=}1^{2}\sigma_{0}(1)\sigma_{0}(\frac{10}{1})+2^{2}\sigma_{0}(2)\sigma_{0}(\frac{10}{2})+5^{2}\sigma_{0}(5)\sigma_{0}(\frac{10}{5})+10^{2}\sigma_{0}(10)\sigma_{0}(\frac{10}{10}).
\end{align*}

 

[Math100]

I was thinking $\sigma_{0}(1)=1, \sigma_{0}(2)=2, \sigma_{0}(5)=2, \sigma_{0}(10)=4$. But what should $\sigma_{2}(1), \sigma_{2}(2), \sigma_{2}(5), \sigma_{2}(10)$ be?

 

[fresh_42]

Homework Statement:: Prove that $\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d})$, and verify this result when $n=10$.
Relevant Equations:: None.

Proof:

Observe that
\begin{align*}
\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})&=(N^{2}\ast u)\ast (N^{2}\ast u) \text{ since } \sigma_{2}(n)=\sum_{d\mid n}d^{2}\cdot 1=N^{2}\ast u)\\
&=(u\ast N^{2})\ast (N^{2}\ast u)\\
&=u\ast (N^{2}\ast N^{2})\ast u\\
&=u\ast (N^{2}\sigma_{0})\ast u \text{ since } N^{2}\ast N^{2}=\sum_{d\mid n}d^{2}(\frac{n^{2}}{d^{2}})=n^{2}\sum_{d\mid n}1=N^{2}\sigma_{0} \\
&=N^{2}\sigma_{0}\ast (u\ast u) \text{ since } u\ast u=\sigma_{0} \\
&=N^{2}\sigma_{0}\ast \sigma_{0}\\
&=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d}).\\
\end{align*}
Therefore, $\sum_{d\mid n}\sigma_{2}(d)\sigma_{2}(\frac{n}{d})=\sum_{d\mid n}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{n}{d})$.
Now we will verify this result when $n=10$.
Consider the divisors $d\in\left \{ 1, 2, 5, 10 \right \}$.
Observe that
\begin{align*}
\sum_{d\mid 10}\sigma_{2}(d)\sigma_{2}(\frac{10}{d})&\stackrel{?}{=}\sum_{d\mid 10}d^{2}\sigma_{0}(d)\sigma_{0}(\frac{10}{d})
\end{align*}
\begin{align*}
&\sigma_{2}(1)\sigma_{2}(\frac{10}{1})+\sigma_{2}(2)\sigma_{2}(\frac{10}{2})+\sigma_{2}(5)\sigma_{2}(\frac{10}{5})+\sigma_{2}(10)\sigma_{2}(\frac{10}{10})\\
&\stackrel{?}{=}1^{2}\sigma_{0}(1)\sigma_{0}(\frac{10}{1})+2^{2}\sigma_{0}(2)\sigma_{0}(\frac{10}{2})+5^{2}\sigma_{0}(5)\sigma_{0}(\frac{10}{5})+10^{2}\sigma_{0}(10)\sigma_{0}(\frac{10}{10}).
\end{align*}

I corrected the LaTeX. I cannot check the proof since I still don't know what $N$ and $u$ are.

The equation where I added a question mark needs to be checked, i.e. you have to calculate the sigma function values and add them. We have
$$
\sigma_k(n) =\sum_{d|n}d^k
$$
This means that $\sigma_0$ is the number of divisors, e.g. $\sigma_0(10)= |\{1,2,5,10\}|=4.$ For $\sigma_2$ we have to calculate e.g.
$$
\sigma_2(10) =\sum_{d|10}d^2=\sum_{d\in \{1,2,5,10\}} d^2 =1^2+2^2+5^2+10^2=130
$$

Or you look up the values in the table here:
https://de.wikipedia.org/wiki/Teilerfunktion

 

[Math100]

I corrected the LaTeX. I cannot check the proof since I still don't know what $N$ and $u$ are.

The equation where I added a question mark needs to be checked, i.e. you have to calculate the sigma function values and add them. We have
$$
\sigma_k(n) =\sum_{d|n}d^k
$$
This means that $\sigma_0$ is the number of divisors, e.g. $\sigma_0(10)= |\{1,2,5,10\}|=4.$ For $\sigma_2$ we have to calculate e.g.
$$
\sigma_2(10) =\sum_{d|10}d^2=\sum_{d\in \{1,2,5,10\}} d^2 =1^2+2^2+5^2+10^2=130
$$

Or you look up the values in the table here:
https://de.wikipedia.org/wiki/Teilerfunktion

It's okay, you don't necessarily need to check the proof. I just wanted to know how to compute those values at the end, the verification part.

 

[fresh_42]

It's okay, you don't necessarily need to check the proof. I just wanted to know how to compute those values at the end, the verification part.

The formula looks like a convolution, a variation of
$$
(f\ast g)(n)=\sum_{k\in D} f(k) g(n/k).
$$
Would be nice to understand the details, i.e. to see the proof why the divisor formulas can be written that way.

 

[fresh_42]

I think I have found it in a book. Möbius function seems to be the key word.

 

[Math100]

I think I have found it in a book. Möbius function seems to be the key word.

What book?

 

[fresh_42]

What book?

A book on number theory. It's been a long time since I last time had a look.

 

[Math100]

A book on number theory. It's been a long time since I last time had a look.

I see. And the results when $n=10$ match each other in the equation. It's $130+130+130+130=4+16+100+400=520$.