📂Number Theory

Divisor Function in Analytic Number Theory

Definition ¹

For $\alpha \in \mathbb{C}$, the following $\sigma_{\alpha} : \mathbb{N} \to \mathbb{C}$ is defined as a divisor function. $$ \sigma_{\alpha} (n) := \sum_{d \mid n} d^{\alpha} $$

Basic Properties

• [1] Multiplicativity: For all $m, n \in \mathbb{N}$ that satisfy $\gcd (m,n) = 1$, $\sigma_{\alpha} (mn) = \sigma_{\alpha} (m) \sigma_{\alpha} (n)$
• [2]: For a prime $p$ and natural number $a$, $$ \sigma_{\alpha} \left( p^{a} \right) = \begin{cases} a +1 & , \alpha = 0 \\ {{ p^{\alpha (a+1)} - 1 } \over { p^{\alpha} - 1 }} &,\alpha \ne 0 \end{cases} $$

Explanation

Especially

• If $\alpha = 0$, it can also be represented by the function $d := \sigma_{0}$ that indicates the number of divisors.
• If $\alpha = 1$, it becomes the sigma function of elementary number theory $\sigma := \sigma_{1}$.

Proof

[1]

Dirichlet product and multiplicative property: If $f$ and $g$ are multiplicative functions, then $f \ast\ g$ is also a multiplicative function.

Let’s define the unit function $u$ and the power function $N^{\alpha}$ as follows. $$ u(n) := 1 \\ N^{\alpha} (n) := n^{\alpha} $$ Since $u$ and $N^{\alpha}$ are multiplicative functions, their convolution $$ \left( N^{\alpha} \ast\ u \right)(n) = \sum_{d \mid n} N^{\alpha} (d) u \left( {{ d } \over { n }} \right) = \sum_{d \mid n} d^{\alpha} = \sigma_{\alpha} (n) $$ must also be a multiplicative function.

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[2]

Since the divisors of $p^{a}$ are $1 , p , \cdots ,p^{a}$, $$ \sigma_{\alpha} ( n) = 1 + p^{\alpha} + \cdots + p^{a\alpha} $$ if $\alpha = 0$, $$ \sigma_{\alpha} ( n) = \underbrace{1 + 1 + \cdots + 1}_{a+1} = a + 1 $$ if $\alpha \ne 0$, according to the geometric series formula, $$ \sigma_{\alpha} ( n) = {{ p^{\alpha (a+1)} - 1 } \over { p^{\alpha} - 1 }} $$

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