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Thread: power mean derivative

  1. #1
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    Exclamation power mean derivative

    For given positive values $\displaystyle a_{1},\ldots,a_{n}$, I have the power mean: $\displaystyle P(\alpha):=\left(\frac{1}{n}\sum_{k=1}^{n}a_{k}^{\ alpha}\right)^{\frac{1}{\alpha}}$ for which I need to find $\displaystyle \frac{d}{d\alpha}P(\alpha)$. Please Help!!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by realpart1/2 View Post
    For given positive values $\displaystyle a_{1},\ldots,a_{n}$, I have the power mean: $\displaystyle P(\alpha):=\left(\frac{1}{n}\sum_{k=1}^{n}a_{k}^{\ alpha}\right)^{\frac{1}{\alpha}}$ for which I need to find $\displaystyle \frac{d}{d\alpha}P(\alpha)$. Please Help!!!
    The simplest way to do this is probably as follows:

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

    or:

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

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

    CB
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