power mean derivative

realpart1/2 · #1 $Old$ July 8th, 2009, 09:24 PM

For given positive values $a_{1},\ldots,a_{n}$ , I have the power mean: $P(\alpha):=\left(\frac{1}{n}\sum_{k=1}^{n}a_{k}^{\alpha}\right)^{\frac{1}{\alpha}}$ for which I need to find $\frac{d}{d\alpha}P(\alpha)$ . Please Help!!!

CaptainBlack · #2 $Old$ July 9th, 2009, 01:01 AM

Quote:

Originally Posted by realpart1/2 $View Post$

For given positive values $a_{1},\ldots,a_{n}$ , I have the power mean: $P(\alpha):=\left(\frac{1}{n}\sum_{k=1}^{n}a_{k}^{\alpha}\right)^{\frac{1}{\alpha}}$ for which I need to find $\frac{d}{d\alpha}P(\alpha)$ . Please Help!!!

The simplest way to do this is probably as follows:

$[P(\alpha)]^{\alpha}=\frac{1}{n}\sum a_i^{\alpha}$

or:

$e^{\alpha \ln(P(\alpha))}=\frac{1}{n}\sum e^{\alpha\ln(a_i)}$

Now differentiate this with respect to $\alpha$ and rearrange and simplify.

CB