Difference between revisions of "Multiplicative function"

(wotw)
Line 4: Line 4:
  
 
Most frequently, one deals with multiplicative functions <math>f : \mathbb{Z}_{>0} \to \mathbb{C}</math>.  These functions appear frequently in [[number theory]], especially in [[analytic number theory]].  In this case, one sometimes also defines ''weak multiplicative functions'': a function <math>f: \mathbb{Z}_{>0} \to \mathbb{C}</math> is weak multiplicative if and only if <math>f(mn) = f(m)f(n)</math> for all pairs of [[relatively prime]] [[integer]]s <math>(m, n)</math>.
 
Most frequently, one deals with multiplicative functions <math>f : \mathbb{Z}_{>0} \to \mathbb{C}</math>.  These functions appear frequently in [[number theory]], especially in [[analytic number theory]].  In this case, one sometimes also defines ''weak multiplicative functions'': a function <math>f: \mathbb{Z}_{>0} \to \mathbb{C}</math> is weak multiplicative if and only if <math>f(mn) = f(m)f(n)</math> for all pairs of [[relatively prime]] [[integer]]s <math>(m, n)</math>.
 +
 +
Let <math>f(n)</math> and <math>g(n)</math> be multiplicative in the number theoretic sense ("weak multiplicative"). Then the function of <math>n</math> defined by <cmath>\sum_{d|n} f(d) g(\frac{n}{d})</cmath> is also multiplicative.
 
{{stub}}
 
{{stub}}

Revision as of 17:44, 31 March 2008

This is an AoPSWiki Word of the Week for March 28-April 5

A multiplicative function $f : S \to T$ is a function which commutes with multiplication. That is, $S$ and $T$ must be sets with multiplication such that $f(x\cdot y) = f(x) \cdot f(y)$ for all $x, y \in S$.

Most frequently, one deals with multiplicative functions $f : \mathbb{Z}_{>0} \to \mathbb{C}$. These functions appear frequently in number theory, especially in analytic number theory. In this case, one sometimes also defines weak multiplicative functions: a function $f: \mathbb{Z}_{>0} \to \mathbb{C}$ is weak multiplicative if and only if $f(mn) = f(m)f(n)$ for all pairs of relatively prime integers $(m, n)$.

Let $f(n)$ and $g(n)$ be multiplicative in the number theoretic sense ("weak multiplicative"). Then the function of $n$ defined by \[\sum_{d|n} f(d) g(\frac{n}{d})\] is also multiplicative. This article is a stub. Help us out by expanding it.

Invalid username
Login to AoPS