derivation of zeroth weighted power mean

Let $x_1,x_2,\mathrm{\dots },x_n$ be positive real numbers, and let$w_1,w_2,\mathrm{\dots },w_n$ be positive real numbers such that$w_1+w_2+\mathrm{\cdots }+w_n=1$. For $r\ne 0$, the $r$-th weighted power meanof $x_1,x_2,\mathrm{\dots },x_n$ is

$$M_w^r(x_1,x_2,\mathrm{\dots },x_n)={(w_1{x}_1^r+w_2{x}_2^r+\mathrm{\cdots }+w_n{x}_n^r)}^{1/r}.$$

Using the Taylor series expansion $e^t=1+t+𝒪(t^2)$, where $𝒪(t^2)$is Landau notation for terms of order $t^2$ and higher, we canwrite $x_i^r$ as

$$x_i^r=e^{r\log x_i}=1+r\log x_i+𝒪(r^2).$$

By substituting this into the definition of $M_w^r$, we get

$M_w^r(x_1,x_2,\mathrm{\dots },x_n)$$=$${\left[w_1(1+r\log x_1)+\mathrm{\cdots }+w_n(1+r\log x_n)+𝒪(r^2)\right]}^{1/r}$

$=$${\left[1+r(w_1\log x_1+\mathrm{\cdots }+w_n\log x_n)+𝒪(r^2)\right]}^{1/r}$

$=$${\left[1+r\log (x_1^{w_1}{x}_2^{w_2}\mathrm{\cdots }{x}_n^{w_n})+𝒪(r^2)\right]}^{1/r}$

$=$$\exp \left\{{\displaystyle \frac{1}{r}}\log \left[1+r\log (x_1^{w_1}{x}_2^{w_2}\mathrm{\cdots }{x}_n^{w_n})+𝒪(r^2)\right]\right\}.$

Again using a Taylor series, this time $\log (1+t)=t+𝒪(t^2)$, we get

$M_w^r(x_1,x_2,\mathrm{\dots },x_n)$$=$$\exp \left\{{\displaystyle \frac{1}{r}}\left[r\log (x_1^{w_1}{x}_2^{w_2}\mathrm{\cdots }{x}_n^{w_n})+𝒪(r^2)\right]\right\}$

$=$$\exp \left[\log (x_1^{w_1}{x}_2^{w_2}\mathrm{\cdots }{x}_n^{w_n})+𝒪(r)\right].$

Taking the limit $r\to 0$, we find

$M_w^0(x_1,x_2,\mathrm{\dots },x_n)$$=$$\exp \left[\log (x_1^{w_1}{x}_2^{w_2}\mathrm{\cdots }{x}_n^{w_n})\right]$

$=$$x_1^{w_1}{x}_2^{w_2}\mathrm{\cdots }{x}_n^{w_n}.$

In particular, if we choose all the weights to be $\frac{1}{n}$,

$$M^0(x_1,x_2,\mathrm{\dots },x_n)=\sqrt[n]{x_1{x}_2\mathrm{\cdots }{x}_n},$$

the geometric mean of $x_1,x_2,\mathrm{\dots },x_n$.