सामान्यीकृत माध्य

यदि p एक अशून्य वास्तविक संख्या है, तथा ${\displaystyle x_{1},\dots ,x_{n}}$ धनात्मक वास्तविक संख्याएँ हैं, तो इन संख्याओं का सामान्यीकृत माध्य (generalized mean) या p घात वाला घात माध्य (power mean) निम्नलिखित है-

${\displaystyle M_{p}(x_{1},\dots ,x_{n})=\left({\frac {1}{n}}\sum _{i=1}^{n}x_{i}^{p}\right)^{\frac {1}{p}}.}$

विशिष्ट स्थितियाँ

A visual depiction of some of the specified cases for n = 2 with a = x₁ = M_+∞, b = x₂ = M_−∞, ██ harmonic mean, H = M₋₁(a, b), ██ geometric mean, G = M₀(a, b) ██ arithmetic mean, A = M₁(a, b), and ██ quadratic mean, Q = M₂(a, b).

${\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})=\lim _{p\to -\infty }M_{p}(x_{1},\dots ,x_{n})=\min\{x_{1},\dots ,x_{n}\}}$	निम्निष्ट
${\displaystyle M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}}$	हरात्मक माध्य (harmonic mean)
${\displaystyle M_{0}(x_{1},\dots ,x_{n})=\lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})={\sqrt[{n}]{x_{1}\cdot \dots \cdot x_{n}}}}$	गुणोत्तर माध्य (geometric mean)
${\displaystyle M_{1}(x_{1},\dots ,x_{n})={\frac {x_{1}+\dots +x_{n}}{n}}}$	समान्तर माध्य (arithmetic mean)
${\displaystyle M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}}$	वर्ग माध्य (quadratic mean]])
${\displaystyle M_{3}(x_{1},\dots ,x_{n})={\sqrt[{3}]{\frac {x_{1}^{3}+\dots +x_{n}^{3}}{n}}}}$	घन माध्य (cubic mean)
${\displaystyle M_{+\infty }(x_{1},\dots ,x_{n})=\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\max\{x_{1},\dots ,x_{n}\}}$	उचिष्ट (maximum)

Proof of ${\displaystyle \textstyle \lim _{p\to 0}M_{p}=M_{0}}$ (geometric mean)

We can rewrite the definition of Mp using the exponential function ${\displaystyle M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left[\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\right]}\right)}=\exp {\left({\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}\right)}}$ 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 ${\displaystyle \lim _{p\to 0}{\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}=\lim _{p\to 0}{\frac {\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}{1}}=\lim _{p\to 0}{\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}=\sum _{i=1}^{n}w_{i}\ln {x_{i}}=\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}}$ By the continuity of the exponential function, we can substitute back into the above relation to obtain ${\displaystyle \lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}\right)}=\prod _{i=1}^{n}x_{i}^{w_{i}}=M_{0}(x_{1},\dots ,x_{n})}$ as desired.

Proof of ${\displaystyle \textstyle \lim _{p\to \infty }M_{p}=M_{\infty }}$ and ${\displaystyle \textstyle \lim _{p\to -\infty }M_{p}=M_{-\infty }}$

Assume (possibly after relabeling and combining terms together) that ${\displaystyle x_{1}\geq \dots \geq x_{n}}$. Then ${\displaystyle \lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\lim _{p\to \infty }\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}=x_{1}\lim _{p\to \infty }\left(\sum _{i=1}^{n}w_{i}\left({\frac {x_{i}}{x_{1}}}\right)^{p}\right)^{1/p}=x_{1}=M_{\infty }(x_{1},\dots ,x_{n}).}$ The formula for ${\displaystyle M_{-\infty }}$ follows from ${\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})={\frac {1}{M_{\infty }(1/x_{1},\dots ,1/x_{n})}}.}$

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• हरात्मक माध्य
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