सामान्यीकृत माध्य

यदि p एक अशून्य वास्तविक संख्या है, तथा $x_1,\dots ,x_n$ धनात्मक वास्तविक संख्याएँ हैं, तो इन संख्याओं का सामान्यीकृत माध्य (generalized mean) या p घात वाला घात माध्य (power mean) निम्नलिखित है-

$M_p(x_1,\dots ,x_n)={\left(\frac{1}{n}\sum _{i=1}^n{x}_i^p\right)}^{\frac{1}{p}}.$

विशिष्ट स्थितियाँ

A visual depiction of some of the specified cases for साँचा:Math with साँचा:Math, साँचा:Math, साँचा:Legend साँचा:Legend साँचा:Legend साँचा:Legend

$M_{-\mathrm{\infty }}(x_1,\dots ,x_n)=\underset{p\to -\mathrm{\infty }}{\lim }{M}_p(x_1,\dots ,x_n)=\min \{x_1,\dots ,x_n\}$	निम्निष्ट
$M_{-1}(x_1,\dots ,x_n)=\frac{n}{\frac{1}{x_1}+\dots +\frac{1}{x_n}}$	हरात्मक माध्य (harmonic mean)
$M_0(x_1,\dots ,x_n)=\underset{p\to 0}{\lim }{M}_p(x_1,\dots ,x_n)=\sqrt[n]{x_1\cdot \dots \cdot x_n}$	गुणोत्तर माध्य (geometric mean)
$M_1(x_1,\dots ,x_n)=\frac{x_1+\dots +x_n}{n}$	समान्तर माध्य (arithmetic mean)
$M_2(x_1,\dots ,x_n)=\sqrt{\frac{x_1^2+\dots +x_n^2}{n}}$	वर्ग माध्य (quadratic mean]])साँचा:Anchor
$M_3(x_1,\dots ,x_n)=\sqrt[3]{\frac{x_1^3+\dots +x_n^3}{n}}$	घन माध्य (cubic mean)
$M_{+\mathrm{\infty }}(x_1,\dots ,x_n)=\underset{p\to \mathrm{\infty }}{\lim }{M}_p(x_1,\dots ,x_n)=\max \{x_1,\dots ,x_n\}$	उचिष्ट (maximum)

Proof of $\underset{p\to 0}{\lim }{M}_p=M_0$ (geometric mean)

We can rewrite the definition of Mp using the exponential function $M_p(x_1,\dots ,x_n)=\exp (\ln \left[{\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}\right])=\exp \left(\frac{\ln \left(\sum _{i=1}^n{w}_i{x}_i^p\right)}{p}\right)$ In the limit p → 0, we can apply L'Hôpital's rule to the argument of the exponential function. Differentiating the numerator and denominator with respect to p, we have $\underset{p\to 0}{\lim }\frac{\ln \left(\sum _{i=1}^n{w}_i{x}_i^p\right)}{p}=\underset{p\to 0}{\lim }\frac{\frac{\sum _{i=1}^n{w}_i{x}_i^p\ln x_i}{\sum _{i=1}^n{w}_i{x}_i^p}}{1}=\underset{p\to 0}{\lim }\frac{\sum _{i=1}^n{w}_i{x}_i^p\ln x_i}{\sum _{i=1}^n{w}_i{x}_i^p}=\sum _{i=1}^n{w}_i\ln x_i=\ln \left(\prod _{i=1}^n{x}_i^{w_i}\right)$ By the continuity of the exponential function, we can substitute back into the above relation to obtain $\underset{p\to 0}{\lim }{M}_p(x_1,\dots ,x_n)=\exp (\ln \left(\prod _{i=1}^n{x}_i^{w_i}\right))=\prod _{i=1}^n{x}_i^{w_i}=M_0(x_1,\dots ,x_n)$ as desired.

Proof of $\underset{p\to \mathrm{\infty }}{\lim }{M}_p=M_{\mathrm{\infty }}$ and $\underset{p\to -\mathrm{\infty }}{\lim }{M}_p=M_{-\mathrm{\infty }}$

Assume (possibly after relabeling and combining terms together) that $x_1\ge \dots \ge x_n$. Then $\underset{p\to \mathrm{\infty }}{\lim }{M}_p(x_1,\dots ,x_n)=\underset{p\to \mathrm{\infty }}{\lim }{\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}=x_1\underset{p\to \mathrm{\infty }}{\lim }{\left(\sum _{i=1}^n{w}_i{\left(\frac{x_i}{x_1}\right)}^p\right)}^{1/p}=x_1=M_{\mathrm{\infty }}(x_1,\dots ,x_n).$ The formula for $M_{-\mathrm{\infty }}$ follows from $M_{-\mathrm{\infty }}(x_1,\dots ,x_n)=\frac{1}{M_{\mathrm{\infty }}(1/x_1,\dots ,1/x_n)}.$

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