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本帖最后由 hbghlyj 于 2022-10-15 12:03 编辑 Wikipedia – Generalized mean
Proof of \(\lim_{p \to 0} M_p = M_0\) (geometric mean)
We can rewrite the definition of \(M_p\) using the exponential function
\[M_p(x_1,\dots,x_n) = \exp{\left( \ln{\left[\left(\sum_{i=1}^n w_ix_{i}^p \right)^{1/p}\right]} \right) } = \exp{\left( \frac{\ln{\left(\sum_{i=1}^n w_ix_{i}^p \right)}}{p} \right) }\]
In the limit \(p→ 0\), we can apply L'Hôpital's rule to the argument of the exponential function. We assume that $p∈\Bbb R$ but $p ≠ 0$, and that the sum of $w_i$ is equal to 1 (without loss in generality); Differentiating the numerator and denominator with respect to $p$, we have
\begin{aligned}
\lim_{p \to 0} \frac{\ln{\left(\sum_{i=1}^n w_ix_{i}^p \right)}}{p} &= \lim_{p \to 0} \frac{\frac{\sum_{i=1}^n w_i x_i^p \ln{x_i}}{\sum_{j=1}^n w_j x_j^p}}{1} \\
&= \lim_{p \to 0} \frac{\sum_{i=1}^n w_i x_i^p \ln{x_i}}{\sum_{j=1}^n w_j x_j^p} \\
&= \sum_{i=1}^n \frac{ w_i \ln{x_i}}{ \lim_{p \to 0} \sum_{j=1}^n w_j \left( \frac{x_j}{x_i} \right)^p} \\
&= \sum_{i=1}^n w_i \ln{x_i} \\
&= \ln{\left(\prod_{i=1}^n x_i^{w_i} \right)}
\end{aligned}By the continuity of the exponential function, we can substitute back into the above relation to obtain
\[\lim_{p \to 0} M_p(x_1,\dots,x_n) = \exp{\left( \ln{\left(\prod_{i=1}^n x_i^{w_i} \right)} \right)} = \prod_{i=1}^n x_i^{w_i} = M_0(x_1,\dots,x_n)\]
as desired.
Proof of \(\lim_{p \to \infty} M_p = M_\infty\) and \(\lim_{p \to -\infty} M_p = M_{-\infty}\)
Assume (possibly after relabeling and combining terms together) that \(x_1 \geq \dots \geq x_n\). Then
\begin{aligned}
\lim_{p \to \infty} M_p(x_1,\dots,x_n) &= \lim_{p \to \infty} \left( \sum_{i=1}^n w_i x_i^p \right)^{1/p} \\
&= x_1 \lim_{p \to \infty} \left( \sum_{i=1}^n w_i \left( \frac{x_i}{x_1} \right)^p \right)^{1/p} \\
&= x_1 = \max (x_1,\dots,x_n).
\end{aligned}The formula for \(M_{-\infty}\) follows from \(M_{-\infty} (x_1,\dots,x_n) = \frac{1}{M_\infty (1/x_1,\dots,1/x_n)} = x_n.\) |
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