Editing Generalized mean (section)

== Special cases ==
A few particular values of {{mvar|p}} yield special cases with their own names:<ref name="mw">{{MathWorld|title=Power Mean|urlname=PowerMean}} (retrieved 2019-08-17)</ref>
;[[minimum]] :<math>M_{-\infty}(x_1,\dots,x_n) = \lim_{p\to-\infty} M_p(x_1,\dots,x_n) = \min \{x_1,\dots,x_n\}</math>
;[[Image:MathematicalMeans.svg|thumb|A visual depiction of some of the specified cases for {{math|1=''n'' = 2}} with {{math|1=''a'' = ''x''{{sub|1}} = ''M''{{sub|∞}}}} and {{math|1=''b'' = ''x''{{sub|2}} = ''M''{{sub|−∞}}}}: {{legend|magenta|harmonic mean, {{math|''H'' {{=}} ''M''{{sub|−1}}(''a'', ''b'')}},}} {{legend|blue|geometric mean, {{math|''G'' {{=}} ''M''{{sub|0}}(''a'', ''b'')}}}} {{legend|red|arithmetic mean, {{math|''A'' {{=}} ''M''{{sub|1}}(''a'', ''b'')}}}} {{legend|lime|quadratic mean, {{math|''Q'' {{=}} ''M''{{sub|2}}(''a'', ''b'')}}}}]][[harmonic mean]] :<math>M_{-1}(x_1,\dots,x_n) = \frac{n}{\frac{1}{x_1}+\dots+\frac{1}{x_n}}</math>
;[[geometric mean]] <math>M_0(x_1,\dots,x_n) = \lim_{p\to0} M_p(x_1,\dots,x_n) = \sqrt[n]{x_1\cdot\dots\cdot x_n}</math>
;[[arithmetic mean]] :<math>M_1(x_1,\dots,x_n) = \frac{x_1 + \dots + x_n}{n}</math>
;[[root mean square]]{{anchor|Quadratic}}<br/>or quadratic mean<ref>{{cite book |last1=Thompson |first1=Sylvanus P. |title=Calculus Made Easy |date=1965 |publisher=Macmillan International Higher Education |isbn=9781349004874 |page=185 |url=https://books.google.com/books?id=6VJdDwAAQBAJ&pg=PA185 |access-date=5 July 2020 }}{{Dead link|date=May 2024 |bot=InternetArchiveBot |fix-attempted=yes }}</ref><ref>{{cite book |last1=Jones |first1=Alan R. |title=Probability, Statistics and Other Frightening Stuff |date=2018 |publisher=Routledge |isbn=9781351661386 |page=48 |url=https://books.google.com/books?id=OvtsDwAAQBAJ&pg=PA48 |access-date=5 July 2020}}</ref> :<math>M_2(x_1,\dots,x_n) = \sqrt{\frac{x_1^2 + \dots + x_n^2}{n}}</math>
;[[cubic mean]] :<math>M_3(x_1,\dots,x_n) = \sqrt[3]{\frac{x_1^3 + \dots + x_n^3}{n}}</math>
;[[maximum]] :<math>M_{+\infty}(x_1,\dots,x_n) = \lim_{p\to\infty} M_p(x_1,\dots,x_n) = \max \{x_1,\dots,x_n\}</math>

{{Math proof|title=Proof of <math display="inline"> \lim_{p \to 0} M_p = M_0 </math> (geometric mean)|proof=For the purpose of the proof, we will assume without loss of generality that 
<math display="block"> w_i \in [0,1] </math>
and 
<math display="block"> \sum_{i=1}^n w_i = 1. </math>

We can rewrite the definition of <math>M_p</math> using the exponential function as

<math display=block>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) }</math>

In the limit {{math|''p'' → 0}}, we can apply [[L'Hôpital's rule]] to the argument of the exponential function. We assume that <math>p \isin \mathbb{R}</math> but {{math|''p'' ≠ 0}}, and that the sum of {{mvar|w<sub>i</sub>}} is equal to 1 (without loss in generality);<ref>{{Cite book |title=Handbook of Means and Their Inequalities (Mathematics and Its Applications)}}</ref> Differentiating the numerator and denominator with respect to {{mvar|p}}, we have
<math display=block>\begin{align}
 \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} \\
 &= \frac{\sum_{i=1}^n w_i \ln{x_i}}{\sum_{j=1}^n w_j} \\
 &= \sum_{i=1}^n w_i \ln{x_i} \\
 &= \ln{\left(\prod_{i=1}^n x_i^{w_i} \right)}
\end{align}</math>

By the continuity of the exponential function, we can substitute back into the above relation to obtain
<math display=block>\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)</math>
as desired.<ref name="Bullen1">P. S. Bullen: ''Handbook of Means and Their Inequalities''. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177</ref>}}

{{Proof|title= Proof of <math display="inline">\lim_{p \to \infty} M_p = M_\infty</math> and <math display="inline">\lim_{p \to -\infty} M_p = M_{-\infty}</math> |proof=
Assume (possibly after relabeling and combining terms together) that <math>x_1 \geq \dots \geq x_n</math>. Then

<math display=block>\begin{align}
 \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).
\end{align}</math>

The formula for <math>M_{-\infty}</math> follows from 
<math display="block">M_{-\infty} (x_1,\dots,x_n) = \frac{1}{M_\infty (1/x_1,\dots,1/x_n)} = x_n.</math>
}}