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weighted power mean (Definition)

If $ w_1,w_2,\ldots,w_n$ are positive real numbers such that $ w_1+w_2+\cdots+w_n=1$, we define the $ r$-th weighted power mean of the $ x_i$ as:

$\displaystyle M_w^r(x_1,x_2,\ldots,x_n)=\left({w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r}\right)^{1/r}.$

When all the $ w_i=\frac{1}{n}$ we get the standard power mean. The weighted power mean is a continuous function of $ r$, and taking limit when $ r\to0$ gives us

$\displaystyle M_w^0=x_1^{w_1}x_2^{w_2}\cdots w_n^{w_n}.$

We can weighted use power means to generalize the power means inequality: If $ w$ is a set of weights, and if $ r<s$ then

$\displaystyle M_w^r \leq M_w^s.$



"weighted power mean" is owned by drini. [ owner history (1) ]
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See Also: arithmetic-geometric-harmonic means inequality, arithmetic mean, geometric mean, harmonic mean, power mean, proof of arithmetic-geometric-harmonic means inequality, root-mean-square, proof of general means inequality, derivation of geometric mean as the limit of the power mean


Attachments:
derivation of zeroth weighted power mean (Derivation) by pbruin
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Cross-references: weights, power means inequality, limit, continuous function, power mean, real numbers, positive
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This is version 7 of weighted power mean, born on 2001-10-17, modified 2005-01-30.
Object id is 267, canonical name is WeightedPowerMean.
Accessed 7411 times total.

Classification:
AMS MSC26B99 (Real functions :: Functions of several variables :: Miscellaneous)

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Correct? by Patak on 2005-01-30 15:38:19
This inequality becomes an equality if and only if x1 = x2 =... = xn
Hence there must be =<
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