Differentiation of Arithmetic Functions
Definition 1
The derivative or differential $f '$ of an arithmetic function $f$ is defined as follows. $$ f ' (n) := f(n) \log n \qquad , n \in \mathbb{N} $$
Basic Properties
- [1] Sum rule of derivatives: $(f+g)' = f '+g'$
- [2] Product rule of derivatives: $\left( f \ast g \right)' = f '\ast g + f \ast g'$
- [3] Quotient rule of derivatives: If $f(1) \ne 0$, then $\left( f^{-1} \right)' = - f ' \ast\ (f \ast\ f)^{-1}$
Explanation
Arithmetic functions, conceptually, are just sequences, so it is not possible to define differentiation, which is often explained as a rate of change, for them in the usual sense. However, one can define differentiation in analytic number theory simply by multiplying the original function by a logarithm. Such differentiation has conceptually no significant meaning, but formally, one can see that it is very similar to the original differentiation.
In particular, for the Mangoldt function $\Lambda$, the following equation holds. $$ \Lambda \ast\ u = u ' $$ The fact that $u(n) = 1$ is a widely used unit function, shows the endless possibilities for application.
[1]
$$ (f+g)' (n) = \left[ f(n)+g(n) \right] \log n = f(n) \log n + g(n) \log n = f '(n) +g ' (n) $$
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[2]
Since $\log n = \log d + \log (n/d)$, $$ \begin{align*} \left( f \ast g \right)' (n) =& \sum_{d \mid n} f(d) g \left( {{ n } \over { d }} \right) \log n \\ =& \sum_{d \mid n} f(d) g \left( {{ n } \over { d }} \right) \log d + \sum_{d \mid n} f(d) g \left( {{ n } \over { d }} \right) \log {{ n } \over { d }} \\ =& f '\ast g + f \ast g' \end{align*} $$
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[3]
The existence of the Dirichlet inverse is assumed for $f(1) \ne 0$. Meanwhile, the derivative of the Dirichlet identity $I$ is $I’ = 0$, and because $I = f \ast\ f^{-1}$, $$ 0 = (f*f^{-1})' = f '*f^{-1} + f \ast (f^{-1})' $$ therefore, we obtain $ f \ast (f^{-1})' = - f '*f^{-1}$, and by multiplying both sides by $f^{-1}$, $$ \left( f^{-1} \right)' = - f ' \ast\ (f \ast\ f)^{-1} $$
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Apostol. (1976). Introduction to Analytic Number Theory: p45. ↩︎