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Differentiation of Arithmetic Functions 📂Number Theory

Differentiation of Arithmetic Functions

Definition 1

The derivative or differential ff ' of an arithmetic function ff is defined as follows. f(n):=f(n)logn,nN f ' (n) := f(n) \log n \qquad , n \in \mathbb{N}

Basic Properties

  • [1] Sum rule of derivatives: (f+g)=f+g(f+g)' = f '+g'
  • [2] Product rule of derivatives: (fg)=fg+fg\left( f \ast g \right)' = f '\ast g + f \ast g'
  • [3] Quotient rule of derivatives: If f(1)0f(1) \ne 0, then (f1)=f (f f)1\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 Λ\Lambda, the following equation holds. Λ u=u \Lambda \ast\ u = u ' The fact that u(n)=1u(n) = 1 is a widely used unit function, shows the endless possibilities for application.

[1]

(f+g)(n)=[f(n)+g(n)]logn=f(n)logn+g(n)logn=f(n)+g(n) (f+g)' (n) = \left[ f(n)+g(n) \right] \log n = f(n) \log n + g(n) \log n = f '(n) +g ' (n)

[2]

Since logn=logd+log(n/d)\log n = \log d + \log (n/d), (fg)(n)=dnf(d)g(nd)logn=dnf(d)g(nd)logd+dnf(d)g(nd)lognd=fg+fg \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*}

[3]

The existence of the Dirichlet inverse is assumed for f(1)0f(1) \ne 0. Meanwhile, the derivative of the Dirichlet identity II is I=0I’ = 0, and because I=f f1I = f \ast\ f^{-1}, 0=(ff1)=ff1+f(f1) 0 = (f*f^{-1})' = f '*f^{-1} + f \ast (f^{-1})' therefore, we obtain f(f1)=ff1 f \ast (f^{-1})' = - f '*f^{-1}, and by multiplying both sides by f1f^{-1}, (f1)=f (f f)1 \left( f^{-1} \right)' = - f ' \ast\ (f \ast\ f)^{-1}


  1. Apostol. (1976). Introduction to Analytic Number Theory: p45. ↩︎