# derivation of geometric mean as the limit of the power mean

Fix $x_{1},x_{2},\ldots,x_{n}\in\mathbb{R}^{+}$. Then let

 $\mu(r):=\left(\frac{x_{1}^{r}+\cdots+x_{n}^{r}}{n}\right)^{1/r}.$

For $r\neq 0$, by definition $\mu(r)$ is the $r$th power mean  of the $x_{i}$. It is also clear that $\mu(r)$ is a differentiable function for $r\neq 0$. What is $\lim_{r\to 0}\mu(r)$?

We will first calculate $\lim_{r\to 0}\log\mu(r)$ using l’Hôpital’s rule (http://planetmath.org/LHpitalsRule).

 $\displaystyle\lim_{r\to 0}\log\mu(r)$ $\displaystyle=\lim_{r\to 0}\frac{\log\left(\frac{x_{1}^{r}+\cdots+x_{n}^{r}}{n% }\right)}{r}$ $\displaystyle=\lim_{r\to 0}\frac{\left(\frac{x_{1}^{r}\log x_{1}+\cdots+x_{n}^% {r}\log x_{n}}{n}\right)}{\left(\frac{x_{1}^{r}+\cdots+x_{n}^{r}}{n}\right)}$ $\displaystyle=\lim_{r\to 0}\frac{x_{1}^{r}\log x_{1}+\cdots+x_{n}^{r}\log x_{n% }}{x_{1}^{r}+\cdots+x_{n}^{r}}$ $\displaystyle=\frac{\log x_{1}+\cdots+\log x_{n}}{n}$ $\displaystyle=\log\sqrt[n]{x_{1}\cdots x_{n}}.$

It follows immediately that

 $\lim_{r\to 0}\left(\frac{x_{1}^{r}+\cdots+x_{n}^{r}}{n}\right)^{1/r}=\sqrt[n]{% x_{1}\cdots x_{n}}.$
 Title derivation of geometric mean  as the limit of the power mean Canonical name DerivationOfGeometricMeanAsTheLimitOfThePowerMean Date of creation 2013-03-22 14:17:13 Last modified on 2013-03-22 14:17:13 Owner Mathprof (13753) Last modified by Mathprof (13753) Numerical id 8 Author Mathprof (13753) Entry type Derivation Classification msc 26D15 Related topic LHpitalsRule Related topic PowerMean Related topic WeightedPowerMean Related topic ArithmeticGeometricMeansInequality Related topic ArithmeticMean Related topic GeometricMean Related topic DerivationOfZerothWeightedPowerMean