Section 19.2 Conjectures and Proofs
Remark 19.2.1.
Don't read this section until you have tried some of the exploration in the previous section!
Sage note 19.2.2. Syntax for sigma.
Here is the syntax for doing this in Sage. However, for this function it is better to try it out by hand first!
xxxxxxxxxx
sigma(12,1),sigma(12,0)
If you do not put the second argument in, Sage just computes
xxxxxxxxxx
sigma(12)
if is prime. if is a prime power. is in fact multiplicative for
for a prime power. is odd precisely if is a perfect square.
Subsection 19.2.1 Prime powers
Again, usually one will have discovered various formulas that are special cases of the following, among others. It's surprisingly easy to find the patterns!Fact 19.2.3.
If
Proof.
There isn't much to prove here, once discovered. Both formulas come from the same fundamental observation.
All possible divisors of a prime power must have only that prime as divisors, by the Fundamental Theorem of Arithmetic. So, these divisors are just other (smaller) powers of that prime.
There are exactly \(e+1\) of these divisors, and these divisors are the ones summed up in the \(\sigma_1\) formula.
The fraction formula for \(\sigma_1\) is just the usual geometric summation formula familiar from precalculus, or perhaps calculus.
Subsection 19.2.2 Multiplicativity
It's a bit harder to prove the following. See Definition 18.1.2 to remind yourself of the definition of multiplicative.Fact 19.2.4.
For any
Theorem 19.2.5.
If we factor
then we have formulas
Principle 19.2.6.
In the long run, it is better to prove general results for sums of arithmetic functions than to do each one by itself.
Subsection 19.2.3 A very powerful lemma
LetTheorem 19.2.7.
If
then
Proof.
We follow here [E.2.1]. Let \(m\) and \(n\) be coprime; we are interested in \(f(mn)\text{.}\)
Basically, this all boils down to asking what the divisors of \(mn\) look like. Any divisor of \(mn\) must be the product of some divisor \(a\) of \(m\) and some divisor \(b\) of \(n\text{.}\)
The previous observation is just about multiplication and divisibility, not even coprimeness. But that guarantees that \(a\) and \(b\) are coprime as well, given that \(m\) and \(n\) are. So each divisor \(d\mid mn\) gives us a (unique) pair of (coprime) divisors \(a\) and \(b\) of \(m\) and \(n\text{.}\)
Instead of summing over all divisors of \(mn\text{,}\) we can instead sum over each divisor of \(n\) for each divisor of \(m\text{.}\) In symbols,
Now we can use all the facts we have at hand (coprimeness, multiplicativity, etc.) to finish it off.
Corollary 19.2.8.
Since
is also multiplicative.
Definition 19.2.9.
Let us set the following two arithmetic functions:
is the unit function is the identity function