Editing Additive function (section)

== Summatory functions ==

Given an additive function <math>f</math>, let its summatory function be defined by <math display="inline">\mathcal{M}_f(x) := \sum_{n \leq x} f(n)</math>. The average of <math>f</math> is given exactly as
<math display=block>\mathcal{M}_f(x) = \sum_{p^{\alpha} \leq x} f(p^{\alpha}) \left(\left\lfloor \frac{x}{p^{\alpha}} \right\rfloor - \left\lfloor \frac{x}{p^{\alpha+1}} \right\rfloor\right).</math>

The summatory functions over <math>f</math> can be expanded as <math>\mathcal{M}_f(x) = x E(x) + O(\sqrt{x} \cdot D(x))</math> where
<math display=block>\begin{align} 
E(x) & = \sum_{p^{\alpha} \leq x} f(p^{\alpha}) p^{-\alpha} (1-p^{-1}) \\ 
D^2(x) & = \sum_{p^{\alpha} \leq x} |f(p^{\alpha})|^2 p^{-\alpha}. 
\end{align}</math>

The average of the function <math>f^2</math> is also expressed by these functions as
<math display=block>\mathcal{M}_{f^2}(x) = x E^2(x) + O(x D^2(x)).</math>

There is always an absolute constant <math>C_f > 0</math> such that for all [[natural number]]s <math>x \geq 1</math>,
<math display=block>\sum_{n \leq x} |f(n) - E(x)|^2 \leq C_f \cdot x D^2(x).</math>

Let
<math display=block>\nu(x; z) := \frac{1}{x} \#\!\left\{n \leq x: \frac{f(n)-A(x)}{B(x)} \leq z\right\}\!.</math>

Suppose that <math>f</math> is an additive function with <math>-1 \leq f(p^{\alpha}) = f(p) \leq 1</math> 
such that as <math>x \rightarrow \infty</math>,
<math display=block>B(x) = \sum_{p \leq x} f^2(p) / p \rightarrow \infty.</math>

Then <math>\nu(x; z) \sim G(z)</math> where <math>G(z)</math> is the [[normal distribution|Gaussian distribution function]]
<math display=block>G(z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{z} e^{-t^2/2} dt.</math>

Examples of this result related to the [[prime omega function]] and the numbers of prime divisors of shifted primes include the following for fixed <math>z \in \R</math> where the relations hold for <math>x \gg 1</math>:
<math display=block>\#\{n \leq x: \omega(n) - \log\log x \leq z (\log\log x)^{1/2}\} \sim x G(z),</math> 
<math display=block>\#\{p \leq x: \omega(p+1) - \log\log x \leq z (\log\log x)^{1/2}\} \sim \pi(x) G(z).</math>