generalized mean#Quadratic

If {{mvar|p}} is a non-zero real number, and $x\_1,\; \backslash dots,\; x\_n$ are positive real numbers, then the generalized mean or power mean with exponent {{mvar|p}} of these positive real numbers is{{cite journal|last=de Carvalho|first=Miguel|title=Mean, what do you Mean?|journal=The American Statistician|year=2016|volume=70|issue=3|pages=764‒776|doi=10.1080/00031305.2016.1148632|url=https://zenodo.org/record/895400|hdl=20.500.11820/fd7a8991-69a4-4fe5-876f-abcd2957a88c|hdl-access=free}}

$$M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash left(\; \backslash frac\{1\}\{n\}\; \backslash sum\_\{i=1\}^n\; x\_i^p\; \backslash right)^\{\{1\}/\{p\}\}\; .$$

(See Norm (mathematics)#p-norm). For {{math|1=p = 0}} we set it equal to the geometric mean (which is the limit of means with exponents approaching zero, as proved below):

$$M\_0(x\_1,\; \backslash dots,\; x\_n)\; =\; \backslash left(\backslash prod\_\{i=1\}^n\; x\_i\backslash right)^\{1/n\}\; .$$

Furthermore, for a sequence of positive weights {{mvar|w_i}} we define the weighted power mean as

$$M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash left(\backslash frac\{\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^p\}\{\backslash sum\_\{i=1\}^n\; w\_i\}\; \backslash right)^\{\{1\}/\{p\}\}$$

and when {{math|1=p = 0}}, it is equal to the weighted geometric mean:

$$M\_0(x\_1,\backslash dots,x\_n)\; =\; \backslash left(\backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\backslash right)^\{1\; /\; \backslash sum\_\{i=1\}^n\; w\_i\}\; .$$

The unweighted means correspond to setting all {{math|1=w_i = 1}}.

A few particular values of {{mvar|p}} yield special cases with their own names:{{MathWorld|title=Power Mean|urlname=PowerMean}} (retrieved 2019-08-17)

;minimum :$M\_\{-\backslash infty\}(x\_1,\backslash dots,x\_n)\; =\; \backslash lim\_\{p\backslash to-\backslash infty\}\; M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash min\; \backslash \{x\_1,\backslash dots,x\_n\backslash \}$

;Image:MathematicalMeans.svgharmonic mean :$M\_\{-1\}(x\_1,\backslash dots,x\_n)\; =\; \backslash frac\{n\}\{\backslash frac\{1\}\{x\_1\}+\backslash dots+\backslash frac\{1\}\{x\_n\}\}$

;geometric mean $M\_0(x\_1,\backslash dots,x\_n)\; =\; \backslash lim\_\{p\backslash to0\}\; M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash sqrt[n]\{x\_1\backslash cdot\backslash dots\backslash cdot\; x\_n\}$

;arithmetic mean :$M\_1(x\_1,\backslash dots,x\_n)\; =\; \backslash frac\{x\_1\; +\; \backslash dots\; +\; x\_n\}\{n\}$

;root mean square{{anchor|Quadratic}}
or quadratic mean{{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 }}{{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}} :$M\_2(x\_1,\backslash dots,x\_n)\; =\; \backslash sqrt\{\backslash frac\{x\_1^2\; +\; \backslash dots\; +\; x\_n^2\}\{n\}\}$

;cubic mean :$M\_3(x\_1,\backslash dots,x\_n)\; =\; \backslash sqrt[3]\{\backslash frac\{x\_1^3\; +\; \backslash dots\; +\; x\_n^3\}\{n\}\}$

;maximum :$M\_\{+\backslash infty\}(x\_1,\backslash dots,x\_n)\; =\; \backslash lim\_\{p\backslash to\backslash infty\}\; M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash max\; \backslash \{x\_1,\backslash dots,x\_n\backslash \}$

{{Math proof|title=Proof of $\backslash lim\_\{p\; \backslash to\; 0\}\; M\_p\; =\; M\_0$ (geometric mean)|proof=For the purpose of the proof, we will assume without loss of generality that

$$w\_i\; \backslash in\; [0,1]$$

and

$$\backslash sum\_\{i=1\}^n\; w\_i\; =\; 1.$$

We can rewrite the definition of $M\_p$ using the exponential function as

$$M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash exp\{\backslash left(\; \backslash ln\{\backslash left[\backslash left(\backslash sum\_\{i=1\}^n\; w\_ix\_\{i\}^p\; \backslash right)^\{1/p\}\backslash right]\}\; \backslash right)\; \}\; =\; \backslash exp\{\backslash left(\; \backslash frac\{\backslash ln\{\backslash left(\backslash sum\_\{i=1\}^n\; w\_ix\_\{i\}^p\; \backslash right)\}\}\{p\}\; \backslash right)\; \}$$

In the limit {{math|p → 0}}, we can apply L'Hôpital's rule to the argument of the exponential function. We assume that $p\; \backslash isin\; \backslash mathbb\{R\}$ but {{math|p ≠ 0}}, and that the sum of {{mvar|w_i}} is equal to 1 (without loss in generality);{{Cite book |title=Handbook of Means and Their Inequalities (Mathematics and Its Applications)}} Differentiating the numerator and denominator with respect to {{mvar|p}}, we have

$$\backslash begin\{align\}$$

\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}

By the continuity of the exponential function, we can substitute back into the above relation to obtain

$$\backslash lim\_\{p\; \backslash to\; 0\}\; M\_p(x\_1,\backslash dots,x\_n)\; =\; \backslash exp\{\backslash left(\; \backslash ln\{\backslash left(\backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; \backslash right)\}\; \backslash right)\}\; =\; \backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; =\; M\_0(x\_1,\backslash dots,x\_n)$$

as desired.P. S. Bullen: Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177}}

{{Proof|title= Proof of $\backslash lim\_\{p\; \backslash to\; \backslash infty\}\; M\_p\; =\; M\_\backslash infty$ and $\backslash lim\_\{p\; \backslash to\; -\backslash infty\}\; M\_p\; =\; M\_\{-\backslash infty\}$ |proof=

Assume (possibly after relabeling and combining terms together) that $x\_1\; \backslash geq\; \backslash dots\; \backslash geq\; x\_n$. Then

$$\backslash begin\{align\}$$

\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}

The formula for $M\_\{-\backslash infty\}$ follows from

$$M\_\{-\backslash infty\}\; (x\_1,\backslash dots,x\_n)\; =\; \backslash frac\{1\}\{M\_\backslash infty\; (1/x\_1,\backslash dots,1/x\_n)\}\; =\; x\_n.$$

}}

Let $x\_1,\; \backslash dots,\; x\_n$ be a sequence of positive real numbers, then the following properties hold:{{cite journal|last=Sýkora|first=Stanislav|year=2009|title=Mathematical means and averages: basic properties|journal=Stan's Library |location=Castano Primo, Italy|volume=III |doi=10.3247/SL3Math09.001 }}

1. $\backslash min(x\_1,\; \backslash dots,\; x\_n)\; \backslash le\; M\_p(x\_1,\; \backslash dots,\; x\_n)\; \backslash le\; \backslash max(x\_1,\; \backslash dots,\; x\_n)$.{{block indent|left=1|text= Each generalized mean always lies between the smallest and largest of the {{mvar|x}} values.}}
   1. $M\_p(x\_1,\; \backslash dots,\; x\_n)\; =\; M\_p(P(x\_1,\; \backslash dots,\; x\_n))$, where $P$ is a permutation operator.{{block indent|left=1|text= Each generalized mean is a symmetric function of its arguments; permuting the arguments of a generalized mean does not change its value.}}
      1. $M\_p(b\; x\_1,\; \backslash dots,\; b\; x\_n)\; =\; b\; \backslash cdot\; M\_p(x\_1,\; \backslash dots,\; x\_n)$.{{block indent|left=1|text= Like most means, the generalized mean is a homogeneous function of its arguments {{math|x₁, ..., x_n}}. That is, if {{mvar|b}} is a positive real number, then the generalized mean with exponent {{mvar|p}} of the numbers $b\backslash cdot\; x\_1,\backslash dots,\; b\backslash cdot\; x\_n$ is equal to {{mvar|b}} times the generalized mean of the numbers {{math|x₁, ..., x_n}}.}}
         1. $M\_p(x\_1,\; \backslash dots,\; x\_\{n\; \backslash cdot\; k\})\; =\; M\_p\backslash left[M\_p(x\_1,\; \backslash dots,\; x\_\{k\}),\; M\_p(x\_\{k\; +\; 1\},\; \backslash dots,\; x\_\{2\; \backslash cdot\; k\}),\; \backslash dots,\; M\_p(x\_\{(n\; -\; 1)\; \backslash cdot\; k\; +\; 1\},\; \backslash dots,\; x\_\{n\; \backslash cdot\; k\})\backslash right]$.{{block indent|left=1|text= Like the quasi-arithmetic means, the computation of the mean can be split into computations of equal sized sub-blocks. This enables use of a divide and conquer algorithm to calculate the means, when desirable.}}

{{QM_AM_GM_HM_inequality_visual_proof.svg}}

In general, if {{math|p < q}}, then

$$M\_p(x\_1,\; \backslash dots,\; x\_n)\; \backslash le\; M\_q(x\_1,\; \backslash dots,\; x\_n)$$

and the two means are equal if and only if {{math|1= x₁ = x₂ = ... = x_n}}.

The inequality is true for real values of {{mvar|p}} and {{mvar|q}}, as well as positive and negative infinity values.

It follows from the fact that, for all real {{mvar|p}},

$$\backslash frac\{\backslash partial\}\{\backslash partial\; p\}M\_p(x\_1,\; \backslash dots,\; x\_n)\; \backslash geq\; 0$$

which can be proved using Jensen's inequality.

In particular, for {{mvar|p}} in {{math|{−1, 0, 1}}}, the generalized mean inequality implies the Pythagorean means inequality as well as the inequality of arithmetic and geometric means.

We will prove the weighted power mean inequality. For the purpose of the proof we will assume the following without loss of generality:

$$\backslash begin\{align\}$$

w_i \in [0, 1] \\

\sum_{i=1}^nw_i = 1

\end{align}

The proof for unweighted power means can be easily obtained by substituting {{math|1= w_i = 1/n}}.

Suppose an average between power means with exponents {{mvar|p}} and {{mvar|q}} holds:

$$\backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^p\backslash right)^\{1/p\}\; \backslash geq\; \backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^q\backslash right)^\{1/q\}$$

applying this, then:

$$\backslash left(\backslash sum\_\{i=1\}^n\backslash frac\{w\_i\}\{x\_i^p\}\backslash right)^\{1/p\}\; \backslash geq\; \backslash left(\backslash sum\_\{i=1\}^n\backslash frac\{w\_i\}\{x\_i^q\}\backslash right)^\{1/q\}$$

We raise both sides to the power of −1 (strictly decreasing function in positive reals):

$$\backslash left(\backslash sum\_\{i=1\}^nw\_ix\_i^\{-p\}\backslash right)^\{-1/p\}$$

= \left(\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^p}}\right)^{1/p}

\leq \left(\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^q}}\right)^{1/q}

= \left(\sum_{i=1}^nw_ix_i^{-q}\right)^{-1/q}

We get the inequality for means with exponents {{math|−p}} and {{math|−q}}, and we can use the same reasoning backwards, thus proving the inequalities to be equivalent, which will be used in some of the later proofs.

For any {{math|q > 0}} and non-negative weights summing to 1, the following inequality holds:

$$\backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^\{-q\}\backslash right)^\{-1/q\}\; \backslash leq\; \backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; \backslash leq\; \backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^q\backslash right)^\{1/q\}.$$

The proof follows from Jensen's inequality, making use of the fact the logarithm is concave:

$$\backslash log\; \backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; =\; \backslash sum\_\{i=1\}^n\; w\_i\backslash log\; x\_i\; \backslash leq\; \backslash log\; \backslash sum\_\{i=1\}^n\; w\_i\; x\_i.$$

By applying the exponential function to both sides and observing that as a strictly increasing function it preserves the sign of the inequality, we get

$$\backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; \backslash leq\; \backslash sum\_\{i=1\}^n\; w\_i\; x\_i.$$

Taking {{mvar|q}}-th powers of the {{mvar|x_i}} yields

$$\backslash begin\{align\}$$

&\prod_{i=1}^n x_i^{q{\cdot}w_i} \leq \sum_{i=1}^n w_i x_i^q \\

&\prod_{i=1}^n x_i^{w_i} \leq \left(\sum_{i=1}^n w_i x_i^q\right)^{1/q}.\end{align}

Thus, we are done for the inequality with positive {{mvar|q}}; the case for negatives is identical but for the swapped signs in the last step:

$$\backslash prod\_\{i=1\}^n\; x\_i^\{-q\{\backslash cdot\}w\_i\}\; \backslash leq\; \backslash sum\_\{i=1\}^n\; w\_i\; x\_i^\{-q\}.$$

Of course, taking each side to the power of a negative number {{math|-1/q}} swaps the direction of the inequality.

$$\backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; \backslash geq\; \backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^\{-q\}\backslash right)^\{-1/q\}.$$

We are to prove that for any {{math|p < q}} the following inequality holds:

$$\backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^p\backslash right)^\{1/p\}\; \backslash leq\; \backslash left(\backslash sum\_\{i=1\}^nw\_ix\_i^q\backslash right)^\{1/q\}$$

if {{mvar|p}} is negative, and {{mvar|q}} is positive, the inequality is equivalent to the one proved above:

$$\backslash left(\backslash sum\_\{i=1\}^nw\_i\; x\_i^p\backslash right)^\{1/p\}\; \backslash leq\; \backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; \backslash leq\; \backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^q\backslash right)^\{1/q\}$$

The proof for positive {{mvar|p}} and {{mvar|q}} is as follows: Define the following function: {{math|f : R₊ → R₊}} $f(x)=x^\{\backslash frac\{q\}\{p\}\}$. {{mvar|f}} is a power function, so it does have a second derivative:

$$f\text{'}\text{'}(x)\; =\; \backslash left(\backslash frac\{q\}\{p\}\; \backslash right)\; \backslash left(\; \backslash frac\{q\}\{p\}-1\; \backslash right)x^\{\backslash frac\{q\}\{p\}-2\}$$

which is strictly positive within the domain of {{mvar|f}}, since {{math|q > p}}, so we know {{mvar|f}} is convex.

Using this, and the Jensen's inequality we get:

$$\backslash begin\{align\}$$

f \left( \sum_{i=1}^nw_ix_i^p \right) &\leq \sum_{i=1}^nw_if(x_i^p) \\[3pt]

\left(\sum_{i=1}^n w_i x_i^p\right)^{q/p} &\leq \sum_{i=1}^nw_ix_i^q

\end{align}

after raising both side to the power of {{math|1/q}} (an increasing function, since {{math|1/q}} is positive) we get the inequality which was to be proven:

$$\backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^p\backslash right)^\{1/p\}\; \backslash leq\; \backslash left(\backslash sum\_\{i=1\}^n\; w\_i\; x\_i^q\backslash right)^\{1/q\}$$

Using the previously shown equivalence we can prove the inequality for negative {{mvar|p}} and {{mvar|q}} by replacing them with {{mvar|−q}} and {{mvar|−p}}, respectively.

{{Main|Generalized f-mean|l1=Generalized {{mvar|f}}-mean}}

The power mean could be generalized further to the generalized f-mean:

$$M\_f(x\_1,\backslash dots,x\_n)\; =\; f^\{-1\}\; \backslash left(\{\backslash frac\{1\}\{n\}\backslash cdot\backslash sum\_\{i=1\}^n\{f(x\_i)\}\}\backslash right)$$

This covers the geometric mean without using a limit with {{math|1= f(x) {{=}} log(x)}}. The power mean is obtained for {{mvar|1= f(x) {{=}} x^p}}. Properties of these means are studied in de Carvalho (2016).

A power mean serves a non-linear moving average which is shifted towards small signal values for small {{mvar|p}} and emphasizes big signal values for big {{mvar|p}}. Given an efficient implementation of a moving arithmetic mean called smooth one can implement a moving power mean according to the following Haskell code.

powerSmooth :: Floating a => ([a] -> [a]) -> a -> [a] -> [a]

powerSmooth smooth p = map (** recip p) . smooth . map (**p)

• For big {{mvar|p}} it can serve as an envelope detector on a rectified signal.
• For small {{mvar|p}} it can serve as a baseline detector on a mass spectrum.

{{cols|colwidth=26em}}

• Arithmetic–geometric mean
• Average
• Heronian mean
• Inequality of arithmetic and geometric means
• Lehmer mean – also a mean related to powers
• Minkowski distance
• Quasi-arithmetic mean – another name for the generalized f-mean mentioned above
• Root mean square

{{colend}}

{{notelist}}

{{reflist|group=note}}

{{reflist}}

• {{cite book|first1=P. S. |last1=Bullen|title=Handbook of Means and Their Inequalities|location=Dordrecht, Netherlands|publisher=Kluwer|year=2003|chapter=Chapter III - The Power Means|pages=175–265}}

• [http://mathworld.wolfram.com/PowerMean.html Power mean at MathWorld]
• [http://people.revoledu.com/kardi/tutorial/BasicMath/Average/Generalized%20mean.html Examples of Generalized Mean]
• A [https://planetmath.org/ProofOfGeneralMeansInequality proof of the Generalized Mean] on PlanetMath

Category:Means

Category:Inequalities (mathematics)

Category:Articles with example Haskell code