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Power Means Inequality (the geometric part)

  1. Nov 20, 2007 #1
    Hi everyone! So I'm trying to learn more about inequalities and the one I'm starting with is the power means inequality. But it all seems pretty intuitive except how they define the n=0 power mean (i.e. the geometric mean). I read that it's actually the limit as n->0, but I don't see why that's true. Any help? Thanks!
     
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  3. Nov 20, 2007 #2

    Gib Z

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    Could you perhaps quote the 'power means' inequality? I may have seen it before under a different name, or under no name.
     
  4. Nov 20, 2007 #3

    CRGreathouse

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    You're perhaps asking about
    [tex](a^x+b^x)^{1/x}[/tex]
     
  5. Nov 21, 2007 #4
    So the equality is that given [tex] r > s [/tex] and [tex] x_1, x_2, \ldots x_n \epsilon \; \Re [/tex] where [tex] s \geq 0[/tex] if any of the [tex] x_1, x_2, \ldots x_n = 0[/tex]. Then:
    [tex] P(r) \geq P(s) [/tex]
    where [tex] P(r) = \left \{ {\begin{array}{*{20}c}
    {(\frac{x_1^r + x_2^r \ldots + x_n^r}{n})^{1/r},} & {r \neq 0} \\
    {\sqrt[n]{x_1 x_2 \ldots x_n},} & {r = 0} \\
    \end{array}} \right.
    [/tex]
    And supposedly if we take the limit of P(r) to 0 we get P(0) but I don't see how that's true.

    On another note, how do I make my fraction look bigger. It looks really small. Thanks!
     
  6. Nov 21, 2007 #5
    In the limit, you have an indeterminate form like [tex]1^\infty[/tex]. The standard way to deal with that kind of indeterminate form is to take the log and then use L'Hopital's Rule. If you'd like me to work it out for you, let me know, but I think you can probably get it yourself from there.
     
  7. Nov 21, 2007 #6
    Success! Thanks!
     
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