Hölder Mean for Exponent -1 is Harmonic Mean

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Theorem

Let $x_1, x_2, \ldots, x_n \in \R_{\ge 0}$ be real numbers.

For $p \in \R_{\ne 0}$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.


Then:

$\map {M_{-1} } {x_1, x_2, \ldots, x_n} = \dfrac 1 {\dfrac 1 n \paren {\dfrac 1 {x_1} + \dfrac 1 {x_2} + \cdots + \dfrac 1 {x_n} } }$

which is the harmonic mean of $x_1, x_2, \ldots, x_n$.


Proof

Recall the definition of the Hölder mean with exponent $p$:

$\ds \map {M_p} {x_1, x_2, \ldots, x_n} = \paren {\frac 1 n \sum_{k \mathop = 1}^n {x_k}^p}^{1 / p}$

Then:

\(\ds \map {M_{-1} } {x_1, x_2, \ldots, x_n}\) \(=\) \(\ds \paren {\frac 1 n \sum_{k \mathop = 1}^n {x_k}^{-1} }^{1 / -1}\)
\(\ds \) \(=\) \(\ds \paren {\frac 1 n \sum_{k \mathop = 1}^n {\dfrac 1 {x_k} } }^{-1}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {\ds \frac 1 n \sum_{k \mathop = 1}^n {\dfrac 1 {x_k} } }\) simplifying

which is the harmonic mean by definition.

$\blacksquare$


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