Completely multiplicative function: Difference between revisions

Revision as of 19:31, 2 May 2009

Definition

Suppose $f$ is an arithmetic function. In other words, $f$ is a function from the natural numbers to some commutative unital ring $R$ . Then, $f$ is termed completely multiplicative if it is a monoid homomorphism from the multiplicative monoid of natural numbers to the multiplicative monoid of $R$ . In other words, $f$ satisfies the following two conditions:

$f(1)=1$ .
$f(mn)=f(m)f(n)$ for all natural numbers $m,n\in \mathbb {N}$ . Here, the multiplication on the left happens in $\mathbb {N}$ while the multiplication on the right happens in $R$ .

Facts

Determined by values at primes

A completely multiplicative function is completely determined by the values it takes at all prime numbers. Further, there is no interdependence between the values taken at prime numbers. In other words, completely multiplicative functions correspond to infinite sequences of elements of the ring: given a sequence, the corresponding completely multiplicative function is the function sending the $k^{th}$ prime to the $k^{th}$ element of the sequence.

Dirichlet series

The Dirichlet series of a completely multiplicative function can be expressed particularly nicely in terms of the values at primes. Formally, if $f$ is completely multiplicative, then:

$\sum _{n\in \mathbb {N} }{\frac {f(n)}{n^{s}}}=\prod _{p\in \mathbb {P} }{\frac {1}{1-f(p)p^{-s}}}$ ,

where $\mathbb {P}$ is the set of primes.

Relation with other properties

Weaker properties

Effect of operations

Dirichlet product

The Dirichlet product of completely multiplicative functions need not be completely multiplicative. However, it is a multiplicative function. Further information: Multiplicative functions form a group under Dirichlet product

Pointwise product

A pointwise product of completely multiplicative functions is completely multiplicative.

Composition

If $g$ is a completely multiplicative function with the property that $g$ takes only positive integer values, and $f$ is a completely multiplicative function, then $f\circ g$ makes sense and is also completely multiplicative.

@@ Line 7: / Line 7: @@
 * <math>f(1) = 1</math>.
 * <math>f(mn) = f(m)f(n)</math> for all natural numbers <math>m,n \in \mathbb{N}</math>. Here, the multiplication on the left happens in <math>\mathbb{N}</math> while the multiplication on the right happens in <math>R</math>.
+==Facts==
+===Determined by values at primes===
+A completely multiplicative function is ''completely'' determined by the values it takes at all [[prime number]]s. Further, there is no interdependence between the values taken at prime numbers. In other words, completely multiplicative functions correspond to infinite sequences of elements of the ring: given a sequence, the corresponding completely multiplicative function is the function sending the <math>k^{th}</math> prime to the <math>k^{th}</math> element of the sequence.
+===Dirichlet series===
+The Dirichlet series of a completely multiplicative function can be expressed particularly nicely in terms of the values at primes. Formally, if <math>f</math> is completely multiplicative, then:
+<math>\sum_{n \in \mathbb{N}} \frac{f(n)}{n^s} = \prod_{p \in \mathbb{P}} \frac{1}{1 - f(p)p^{-s}}</math>,
+where <math>\mathbb{P}</math> is the set of primes.
 ==Relation with other properties==
@@ Line 14: / Line 28: @@
 * [[Stronger than::Multiplicative function]]
 * [[Stronger than::Divisibility-preserving function]]
+==Effect of operations==
+===Dirichlet product===
+The [[Dirichlet product]] of completely multiplicative functions need not be completely multiplicative. However, it is a [[multiplicative function]]. {{further|[[Multiplicative functions form a group under Dirichlet product]]}}
+===Pointwise product===
+A pointwise product of completely multiplicative functions is completely multiplicative.
+===Composition===
+If <math>g</math> is a completely multiplicative function with the property that <math>g</math> takes only positive integer values, and <math>f</math> is a completely multiplicative function, then <math>f \circ g</math> makes sense and is also completely multiplicative.