Question

Mini intro to Dirichlet convolutions, motivated by the Riemann Zeta Function! =D

Answer 1

First, this is the zeta function!\[\Large \zeta (s) = \frac{1}{1^s}+ \frac{1}{2^s}+ \frac{1}{3^s}+ \frac{1}{4^s}+\cdots\] we can write it more compactly as \[\Large \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\] there's another nice way to write it, by following these algebra steps \[\large \frac{1}{2^s}\zeta (s) = \frac{1}{2^s}+ \frac{1}{4^s}+ \frac{1}{6^s}+ \frac{1}{8^s}+\cdots\]\[\large \zeta(s)- \frac{1}{2^s}\zeta (s) =\left( 1-\frac{1}{2^s} \right) \zeta(s) = \frac{1}{1^s}+ \frac{1}{3^s}+ \frac{1}{5^s}+ \frac{1}{7^s}+\cdots\]\[\large \frac{1}{3^s}\left( 1-\frac{1}{2^s} \right) \zeta(s) = \frac{1}{3^s}+ \frac{1}{9^s}+ \frac{1}{15^s}+ \frac{1}{21^s}+\cdots\]\[\large \left( 1-\frac{1}{3^s} \right) \left( 1-\frac{1}{2^s} \right) \zeta(s) = \frac{1}{1^s}+ \frac{1}{5^s}+ \frac{1}{7^s}+ \frac{1}{11^s}+\cdots\] keep doing this for every prime! \[\Large \zeta(s) \prod_{i=1}^\infty \left(1 - \frac{1}{p_i^s} \right)=1\] Now we can write \[\Large \zeta(s) = \prod_{i=1}^\infty \left( \frac{1}{1-p_i^{-s} }\right)\] But this isn't directly important to understanding the Dirichlet convolution, it will be intuitively helpful I believe. The summation version of the zeta function is what we're interested in!

Answer 2

Now to the Dirichlet Convolution. All it is, is what happens to a function of n multiplied by the zeta function terms in the sum like this: \[\Large D(f;s) = \sum_{n=1}^\infty \frac{f(n)}{n^s}\] It is pretty simple to see that adding two separate series like this will just add term by term perfectly \[ D(f;s) + D(g;s) = \sum_{n=1}^\infty \frac{f(n)}{n^s}+ \sum_{n=1}^\infty \frac{g(n)}{n^s}= \sum_{n=1}^\infty \frac{f(n)+g(n)}{n^s}=D(f+g;s)\] and it's multiplication that gets interesting, as we'll see in a moment, but I'll denote it by a special symbol *. \[D(f;s)D(g;s)=D(f*g;s)\] So here's what it would look like if we expanded this multiplication out: \[\small \frac{f(1)g(1)}{1^s}+\frac{f(1)g(2)+f(2)g(1)}{2^s}+\frac{f(1)g(3)+f(3)g(1)}{3^s}+\frac{f(1)g(4)+f(2)g(2)+f(4)g(1)}{4^s}+ \cdots\] Each of the coefficients really obey this formula: \[\Large f*g = \sum_{ab=n} f(a)g(b)\] and this is the Dirichlet convolution.

Answer 3

The Dirichlet convolution takes multiplicative functions and outputs multiplicative functions, which means that it forms a group. I'll list a handful of interesting multiplicative functions: \[I(n)=\lfloor \frac{1}{n} \rfloor \\ u(n)=1 \\ N(n) = n \\ \tau(n) = \text{number of divisors of n} \\ \sigma(n) = \text{summation of divisors of n} \\ \mu(n) = \text{mobius function} \\ \phi(n) = \text{euler's totient function}\] Assume that all these only take positive values for n.

First off, I(n) is like a dirac delta, it returns 1 only when n=1 and 0 otherwise. This is an identity function of the Dirichlet convolution, and it should be obvious thinking back to the coefficients on zeta function definition that this will just leave the single term: \[D(I;s)=\sum_{n=1}^\infty \frac{I(n)}{n^s}=\frac{1}{1^s}+\frac{0}{2^s}+\frac{0}{3^s}+ \cdots\] so for any arbitrary function f(n) \[\Large f= f * I\] We can also look at a function based on its definition: \[\Large \sigma(n) = \sum_{d|n}d = \sum_{ab=n}N(a)u(b) \\ \Large \sigma = N * u\] Anyways that's it for now, if anyone wants me to clarify anything please ask, I left a lot out on purpose but tried to hit the high points.

Answer 4

Follow

Answer 5

Another important thing is that the convolution is invertible! So above we saw \[\Large \sigma = N*u\] however (to be proved shortly) the inverse of u is the mobius function! \[\Large I = \mu * u\] So we can convolve both sides by mu to get: \[\Large \sigma * \mu = N*u*\mu = N*I = N\] And did I mention that the convolution is both associative and commutative? It is fairly obvious that these are true since regular addition and multiplication we used to get the coefficients on the zeta function obey the same rules already, so it's not really that new.

Another fun fact is that all multiplicative functions obey f(1)=1.

Now for the proof I promised of this: \[\Large \mu*u = I\] So to begin let's write the summation \[\Large \sum_{ab=n}\mu(a)u(b)\] u(b)=1 always so not important, throw it away. Now what about the mobius function? Well it's multiplicative and 0 if it's a square so I'll separate it out into the product of sums. \[\Large \prod_{i=1}^k \left( \sum_{j=0}^1\mu(p_i^{j}) \right)\] So to be clear, i runs from the first prime to the kth (last) prime in the factorization of n, the product is here because it's multiplicative in each prime, so that's how I'm able to look at each prime individually. Now in this sum the exponent goes from 0 to 1, since in the factorization if it's 2 or higher the mobius function makes it 0, since it only looks at square free numbers. At this point it's arbitrary what the powers are on the prime because if it has any prime factor at all, the sum of mobius functions for any prime term will be: \[\Large \mu(p^0)+\mu(p^1)=1-1=0\] So it'll make the whole thing 0.

However if n=1 then you will simply have mu(1)=1. And this function we've just described that's 1 at n=1 and 0 otherwise is the identity function!

Answer 6

*

Answer 7

Cool stuff, keep up the good work! :D