Difference between revisions of "Mobius inversion formula"

(created page (please expand))
 
m (Cosmetic edit for parentheses)
Line 1: Line 1:
Suppose that <math>f</math> and <math>g</math> are [[function|functions]] from the [[natural number|natural numbers]] to the [[real number|real numbers]] such that <math>f(n) = \sum_{d|n}g(d)</math>. Then we can express <math>g</math> in terms of <math>f</math> as <math>g(n) = \sum_{d|n} \mu(\frac{n}{d})f(d)</math> where <math>\mu</math> is the [[Mobius function]]. This formula is useful in [[number theory]].
+
Suppose that <math>f</math> and <math>g</math> are [[function|functions]] from the [[natural number|natural numbers]] to the [[real number|real numbers]] such that <math>f(n) = \sum_{d|n}g(d)</math>. Then we can express <math>g</math> in terms of <math>f</math> as <math>g(n) = \sum_{d|n} \mu\left(\frac{n}{d}\right)f(d)</math> where <math>\mu</math> is the [[Mobius function]]. This formula is useful in [[number theory]].
  
 
{{stub}}[[Category:Number Theory]]
 
{{stub}}[[Category:Number Theory]]

Revision as of 20:54, 13 March 2022

Suppose that $f$ and $g$ are functions from the natural numbers to the real numbers such that $f(n) = \sum_{d|n}g(d)$. Then we can express $g$ in terms of $f$ as $g(n) = \sum_{d|n} \mu\left(\frac{n}{d}\right)f(d)$ where $\mu$ is the Mobius function. This formula is useful in number theory.

This article is a stub. Help us out by expanding it.