multiplicative function

In number theory, a multiplicative function is an arithmetic function $f \colon \mathbb{N} \to \mathbb{C}$ such that $f(1)=1$ and, for all $a,b \in \mathbb{N}$ with $\gcd(a,b)=1$ , we have $f(ab)=f(a)f(b)$ .

An arithmetic function $f(n)$ is said to be completely multiplicative if $f(1)=1$ and $f(ab)=f(a)f(b)$ holds for all positive integers $a$ and $b$ , even when they are not relatively prime. In this case, the function is a homomorphism of monoids and, because of the fundamental theorem of arithmetic, is completely determined by its restriction to prime numbers. Every completely multiplicative function is multiplicative.

Outside of number theory, the term multiplicative is usually used for all functions with the property $f(ab)=f(a)f(b)$ for all arguments $a$ and $b$ . This entry discusses number theoretic multiplicative functions.

Examples

Examples of multiplicative functions include many important functions in number theory, such as:

• $\varphi(n)$ : the Euler totient function (also denoted $\phi(n)$ ), counting the totatives of $n$ ;
• $\mu(n)$ : the Möbius function, which determines the parity of the prime factors of $n$ if $n$ is squarefree;
• $\tau(n)$ : the divisor function (also denoted $d(n)$ ), counting the positive divisors of $n$ ;
• $\sigma(n)$ : the sum of divisors function (also denoted $\sigma_1(n)$ ), summing the positive divisors of $n$ ;
• $\sigma_{k}(n)$ : the sum of the $k$ -th powers of all the positive divisors of $n$ for any complex number $k$ (typically a natural number);
• ${id}(n)$ : the identity function, defined by ${id}(n) = n$ ;
• ${id}^{k}(n)$ : the power functions, defined by ${id}^{k}(n) = n^{k}$ for any complex number $k$ (typically a natural number);
• $1(n)$ : the constant function, defined by $1(n)=1$ ;
• $\varepsilon(n)$ : the convolution identity function, defined by:
  
  where $d$ runs through the positive divisors of $n$ .

Properties

A multiplicative function is completely determined by its values at the powers of prime numbers, a consequence of the fundamental theorem of arithmetic. Thus, if $n$ is a product of powers of distinct prime numbers, say $n=p^aq^b\cdots$ , then $f(n)=f(p^a)f(q^b)\cdots$ . This property of multiplicative functions significantly reduces the need for computation, as in the following examples for $n=144=2^4 \cdot 3^2$ :

\begin{eqnarray*} \sigma(144) &=& \sigma_{1}(144)=\sigma_{1}(2^4)\sigma_{1}(3^2)= (1^1+2^1+4^1+8^1+16^1)(1^1+3^1+9^1)=31 \cdot 13=403 \\ \sigma_{2}(144) &=& \sigma_{2}(2^4)\sigma_{2}(3^2) = (1^2+2^2+4^2+8^2+16^2)(1^2+3^2+9^2)=341 \cdot 91=31031 \\ \sigma_{3}(144) &=& \sigma_{3}(2^{4})\sigma_{3}(3^{2}) = (1^3+2^3+4^3+8^3+16^3)(1^3+3^3+9^3)=4681 \cdot 757=3543517 \end{eqnarray*} Similarly, we have:

\begin{eqnarray*} \tau(144) &=& \tau(2^4)\tau(3^2)=(4+1)(2+1)=5 \cdot 3=15 \\ \varphi(144) &=& \varphi(2^4)\varphi (3^2)=2^3(2-1)3^1(3-1)=8 \cdot 1 \cdot 3 \cdot 2=48 \end{eqnarray*}

Convolution

Recall that, if $f$ and $g$ are two arithmetic functions, one defines a new arithmetic function $f*g$ , the Dirichlet convolution (or simply convolution) of $f$ and $g$ , by$$ (f*g)(n) = \sum_{d\vert n} f(d)g \left( {n\over d} \right),$$ where the sum extends over all positive divisors $d$ of $n$ . Some general properties of this operation with respect to multiplicative functions include (here the argument $n$ is omitted in all functions):

• If both $f$ and $g$ are multiplicative, then so is $f*g$ (proven here);
• $f*g = g*f$ (proven here);
• $(f*g)*h = f*(g*h)$ (proven here);
• $f*\varepsilon = \varepsilon *f = f$ (proven here);
• If $f$ is multiplicative, there exists a multiplicative function $g$ such that $f*g=\varepsilon$ (proven here). In other words, every multiplicative function has a convolution inverse that is also multiplicative.

This shows that, with respect to convolution, the multiplicative functions form an abelian group with identity element $\varepsilon$ . Relations among the multiplicative functions discussed above include:

• $\mu*1=\varepsilon$ (the Möbius inversion formula)
• $1*1=\tau$
• ${id}*1 = \sigma$
• ${id}^{k}*1 = \sigma ^{k}$
• $\phi*1 = {id}$

Given a completely multiplicative function $f$ , its convolution inverse is $f\mu$ . See this entry for a proof.