Motivation for the Derivative of an Arithmetic Function

289 Views Asked by At

I'm currently reading through Apostol's Introduction to Analytic Number Theory. In Chapter 2, he discusses arithmetic functions (i.e. functions $f: \mathbb{N} \to \mathbb{C}$) under the Dirichlet Convolution $$ (f \ast g)(n) = \sum_{d \mid n} f(d)g(\frac{n}{d}).$$ He defines the derivative of an arithmetic function by setting $f'(n) = f(n)\log(n)$, goes on to prove that it has "nice" properties, e.g. linearity and the product rule. What I'm struggling to understand is why we chose this definition. Is there a motivation behind this definition, or are we making it because it happens to give us the structure we want?

1

There are 1 best solutions below

0
On BEST ANSWER

Consider the following Dirichlet series:


$$F(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\ n^{-s}\right)\tag{1}$$

$$G(s)=-F'(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\ \log(n)\ n^{-s}\right)\tag{2}$$


For $a(n)=1$, $F(s)=\zeta(s)$ and $G(s)=-F'(s)=-\zeta'(s)$


As another example, for $a(n)=\left(\left\{\begin{array}{cc} 0 & n=1 \\ \frac{\Lambda (n)}{\log (n)} & n>1 \\ \end{array}\right.\right)$, $F(s)=\log\zeta (s)$ and $G(s)=-F'(s)=-\frac{\partial\ \log\zeta(s)}{\partial s}=-\frac{\zeta'(s)}{\zeta(s)}$.