A Completely Multiplicative Function

Section 6.5 A Completely Multiplicative Function

Definition 6.5.1.

A function \(f\) is called completely multiplicative provided that for all \(m, n \in\mathbb{N},\)

\begin{equation*} f(mn) = f(m)f(n). \end{equation*}

Recall from a previous lecture that the arithmetic function \(\Omega:\mathbb{N}\rightarrow\mathbb{N},\) is defined by \(\Omega(n) = k_1 +k_2 +\cdots +k_r,\) where \(n = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}\) is the prime factorization of \(n.\)

Definition 6.5.2.

The Liouville \(\lambda\)-function, \(\lambda:\mathbb{N}\rightarrow \{1,-1\},\) is defined by \(\lambda(1) = 1,\) and \(\lambda(n) = (-1)^{\Omega(n)},\) where \(n = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}\) is the prime factorization of \(n,\) and \(\Omega(n)\) is as defined previously.

Theorem 6.5.3.

The Liouville \(\lambda\)-function is completely multiplicative.

We can also use Liouville's \(\lambda\)-function to rewrite both the prime number theorem and the Riemann hypothesis.

Theorem 6.5.4. Prime Number Theorem.

We have

\begin{equation*} \lim_{n\rightarrow\infty}\frac{\lambda(1)+\lambda(2)+\cdots +\lambda(n)}{n}=0. \end{equation*}

Conjecture 6.5.5. Riemann Hypothesis.

For any \(\varepsilon\gt 0,\) we have

\begin{equation*} \lim_{n\rightarrow\infty}\frac{\lambda(1)+\lambda(2)+\cdots +\lambda(n)}{n^{\frac{1}{2}+\varepsilon}}=0. \end{equation*}