# Power mean

• August 29th 2008, 08:37 PM
particlejohn
Power mean
From the power mean inequality $P_r = \left(\frac{a_{1}^{r} + \ldots + a_{n}^{r} }{n}\right)^{1/r}$ why does $\lim_{r \to 0} P_{r} = (a_{1}a_{2} \cdots a_{n})^{1/n}$?
• August 29th 2008, 09:12 PM
mr fantastic
Quote:

Originally Posted by particlejohn
From the power mean inequality $P_r = \left(\frac{a_{1}^{r} + \ldots + a_{n}^{r} }{n}\right)^{1/r}$ why does $\lim_{r \to 0} P_{r} = (a_{1}a_{2} \cdots a_{n})^{1/n}$?

There's a proof in this link: Generalized mean - Wikipedia, the free encyclopedia
• August 30th 2008, 05:16 AM
PaulRS
First: $
\mathop {\lim }\limits_{r \to 0} \log \left( {P_r } \right) = \mathop {\lim }\limits_{r \to 0} \tfrac{1}
{r} \cdot \log \left( {\tfrac{{\sum\nolimits_{k = 1}^n {a_k ^r } }}
{n}} \right)
$

Now we have: $
\log \left( {\tfrac{{\sum\nolimits_{k = 1}^n {a_k ^r } }}
{n}} \right)\mathop \sim \limits_{r \to 0} \tfrac{{\sum\nolimits_{k = 1}^n {a_k ^r } }}
{n} - 1 = \tfrac{{\sum\nolimits_{k = 1}^n {\left( {a_k ^r - 1} \right)} }}
{n}
$
(since what's inside the logarithm tends to 1)

So: $
\mathop {\lim }\limits_{r \to 0} \log \left( {P_r } \right) = \mathop {\lim }\limits_{r \to 0} \tfrac{1}
{r} \cdot \tfrac{{\sum\nolimits_{k = 1}^n {\left( {a_k ^r - 1} \right)} }}
{n} = \tfrac{1}
{n} \cdot \sum\nolimits_{k = 1}^n {\mathop {\lim }\limits_{r \to 0} \tfrac{1}
{r} \cdot \left( {a_k ^r - 1} \right)}
$
(by the linearity of the limit- all of those are defined-)

$
\mathop {\lim }\limits_{r \to 0} \tfrac{1}
{r} \cdot \left( {a_k ^r - 1} \right) = \mathop {\lim }\limits_{r \to 0} \tfrac{1}
{r} \cdot \left( {e^{r \cdot \log \left( {a_k } \right)} - 1} \right) = \log \left( {a_k } \right)
$

So: $
\mathop {\lim }\limits_{r \to 0} \log \left( {P_r } \right) = \tfrac{1}
{n} \cdot \sum\nolimits_{k = 1}^n {\log \left( {a_k } \right)}
$
and the rest follows easily by the properties of the logarithm and the continuity of the exponential function.