## Dirichlet generating functions

Suppose $f(n)$ is a function defined for positive integers $n \geq 1$. Then we can define an infinite series $\zeta_f(s)$ as follows:

$\displaystyle \begin{array}{rcl}\zeta_f(s) &=& \displaystyle \frac{f(1)}{1^s} + \frac{f(2)}{2^s} + \frac{f(3)}{3^s} + \dots + \frac{f(n)}{n^s} + \dots \\[1em] &=& \displaystyle \sum_{n \geq 1} \frac{f(n)}{n^s} \end{array}$

(This might look a bit strange, but bear with me!) For example, suppose $f(n) = n+2$ for all $n$. Then

$\displaystyle \zeta_f(3) = \displaystyle \frac{1+2}{1^3} + \frac{2+2}{2^3} + \frac{3+2}{3^3} + \frac{4+2}{4^3} + \dots \approx 4.0490\dots$

(Note that in this case, with $s = 3$, the infinite sum converges; but often we just use $s$ as a formal placeholder and don’t particularly care about convergence. That is, $\zeta_f$ is perfectly well defined as an infinite series without worrying about the value of $s$.)

In fact $\zeta_f$ is called the Dirichlet generating function for $f$. Of course, this name should remind you of Dirichlet convolution—and it’s no coincidence; Dirichlet convolution and Dirichlet generating functions are closely related. Let’s see how.

Suppose we have two functions $f(n)$ and $g(n)$, and consider multiplying their Dirichlet generating functions (with plain old, regular multiplication):

$\displaystyle \zeta_f(s) \zeta_g(s) = \displaystyle \left( \sum_{n \geq 1} \frac{f(n)}{n^s} \right) \left( \sum_{n \geq 1} \frac{g(n)}{n^s} \right)$

We have a big (well, infinite) sum of things multiplied by another big sum. By distributivity, we’re going to get a big sum as a result, where each term of the resulting sum is the product of one thing chosen from the left-hand sum and one thing chosen from the right-hand sum. (This is just like FOIL, only way cooler and more infinite-r.) That is, the resulting sum is going to look something like this:

$\displaystyle \displaystyle \zeta_f(s) \zeta_g(s) = \dots + \frac{f(a)}{a^s} \frac{g(b)}{b^s} + \dots = \sum_{a,b \geq 1} \frac{f(a)}{a^s} \frac{g(b)}{b^s}$

with one term for every possible choice of $a$ and $b$. The $a^s b^s$ in the denominator is of course equal to $(ab)^s$, and we can collect up all the fractions with the same denominator: for each $n$, we will get a denominator of $n^s$ in each case where we pick some $a$ and $b$ whose product is $n$. So we can reorganize the terms in the above sum, grouping together all the terms where the product of $a$ and $b$ is the same, and rewrite it like this (is this starting to look familiar…?):

$\displaystyle \displaystyle \zeta_f(s) \zeta_g(s) = \sum_{n \geq 1} \sum_{ab = n} \frac{f(a) g(b)}{n^s}$

Now we can factor out the $1/n^s$ (since it doesn’t depend on $a$ or $b$), like so:

$\displaystyle \displaystyle \zeta_f(s) \zeta_g(s) = \sum_{n \geq 1} \frac{1}{n^s} \sum_{ab = n} f(a) g(b)$

But the inner sum is now just the definition of the Dirichlet convolution of $f$ and $g$! So the whole thing becomes

$\displaystyle \displaystyle \zeta_f(s) \zeta_g(s) = \sum_{n \geq 1} \frac{1}{n^s} \sum_{ab = n} f(a) g(b) = \sum_{n \geq 1} \frac{(f \ast g)(n)}{n^s}$

And finally, we note that the thing on the right is itself the Dirichlet generating function for $f \ast g$. So in summary, we have shown that

$\displaystyle \displaystyle \zeta_f(s) \zeta_g(s) = \zeta_{f \ast g}(s)$

Neato! So in some sense, Dirichlet generating functions “turn Dirichlet convolution into regular multiplication”.

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 5 Responses to Dirichlet generating functions

1. Rob says:

These functions (with z for s) are used a lot in signal processing, under the name z-transforms. https://en.m.wikipedia.org/wiki/Z-transform

• Brent says:

Hi Rob, thanks for the link! Though after reading about z-transforms I do not think they are the same. Notice that with a z-transform you have something like F(z) = sum_n … z^-n, but with Dirichlet generating functions you have F(s) = sum_n … n^-s . The base and the exponent are switched, which actually makes a very big difference.

2. Brendan says:

Does this make the Dirichlet Generating function operator a monoid homomorphism from the monoid of $(\mathbb{R} \to \mathbb{R}, \lambda f,g . \lambda x . f(x)g(x), \lambda x . 1)$ to the Dirichlet convolution monoid?

• Brent says:

That’s the right idea but I think it’s the other way around.