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.
$\blacksquare$
Sources
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Generalized Mean: $3.1.20$