N-th root of the arithmetic mean of the given numbers raised to the power n
Plot of several generalized means
M
p
(
1
,
x
)
{\displaystyle M_{p}(1,x)}
.
In mathematics , generalized means (or power mean or Hölder mean from Otto Hölder )[1] are a family of functions for aggregating sets of numbers. These include as special cases the Pythagorean means (arithmetic , geometric , and harmonic means ).
Definition [ edit ]
If p is a non-zero real number , and
x
1
,
…
,
x
n
{\displaystyle x_{1},\dots ,x_{n}}
are positive real numbers, then the generalized mean or power mean with exponent p of these positive real numbers is[2] [3]
M
p
(
x
1
,
…
,
x
n
)
=
(
1
n
∑
i
=
1
n
x
i
p
)
1
/
p
.
{\displaystyle M_{p}(x_{1},\dots ,x_{n})=\left({\frac {1}{n}}\sum _{i=1}^{n}x_{i}^{p}\right)^{{1}/{p}}.}
(See p -norm ). For 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
,
…
,
x
n
)
=
(
∏
i
=
1
n
x
i
)
1
/
n
.
{\displaystyle M_{0}(x_{1},\dots ,x_{n})=\left(\prod _{i=1}^{n}x_{i}\right)^{1/n}.}
Furthermore, for a sequence of positive weights wi we define the weighted power mean as[2]
M
p
(
x
1
,
…
,
x
n
)
=
(
∑
i
=
1
n
w
i
x
i
p
∑
i
=
1
n
w
i
)
1
/
p
{\displaystyle M_{p}(x_{1},\dots ,x_{n})=\left({\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}}{\sum _{i=1}^{n}w_{i}}}\right)^{{1}/{p}}}
and when
p = 0, it is equal to the
weighted geometric mean :
M
0
(
x
1
,
…
,
x
n
)
=
(
∏
i
=
1
n
x
i
w
i
)
1
/
∑
i
=
1
n
w
i
.
{\displaystyle M_{0}(x_{1},\dots ,x_{n})=\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)^{1/\sum _{i=1}^{n}w_{i}}.}
The unweighted means correspond to setting all wi = 1/n .
Special cases [ edit ]
A few particular values of p yield special cases with their own names:[4]
minimum
M
−
∞
(
x
1
,
…
,
x
n
)
=
lim
p
→
−
∞
M
p
(
x
1
,
…
,
x
n
)
=
min
{
x
1
,
…
,
x
n
}
{\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}\}}
A visual depiction of some of the specified cases for n = 2 with a = x 1 = M ∞ and b = x 2 = M −∞ : harmonic mean, H = M −1 (a , b ) ,
geometric mean, G = M 0 (a , b )
arithmetic mean, A = M 1 (a , b )
quadratic mean, Q = M 2 (a , b )
harmonic mean
M
−
1
(
x
1
,
…
,
x
n
)
=
n
1
x
1
+
⋯
+
1
x
n
{\displaystyle M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}}
geometric mean
M
0
(
x
1
,
…
,
x
n
)
=
lim
p
→
0
M
p
(
x
1
,
…
,
x
n
)
=
x
1
⋅
⋯
⋅
x
n
n
{\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}}}}
arithmetic mean
M
1
(
x
1
,
…
,
x
n
)
=
x
1
+
⋯
+
x
n
n
{\displaystyle M_{1}(x_{1},\dots ,x_{n})={\frac {x_{1}+\dots +x_{n}}{n}}}
root mean square or quadratic mean[5] [6]
M
2
(
x
1
,
…
,
x
n
)
=
x
1
2
+
⋯
+
x
n
2
n
{\displaystyle M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}}
cubic mean
M
3
(
x
1
,
…
,
x
n
)
=
x
1
3
+
⋯
+
x
n
3
n
3
{\displaystyle M_{3}(x_{1},\dots ,x_{n})={\sqrt[{3}]{\frac {x_{1}^{3}+\dots +x_{n}^{3}}{n}}}}
maximum
M
+
∞
(
x
1
,
…
,
x
n
)
=
lim
p
→
∞
M
p
(
x
1
,
…
,
x
n
)
=
max
{
x
1
,
…
,
x
n
}
{\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}\}}
Proof of
lim
p
→
0
M
p
=
M
0
{\textstyle \lim _{p\to 0}M_{p}=M_{0}}
(geometric mean)
For the purpose of the proof, we will assume without loss of generality that
w
i
∈
[
0
,
1
]
{\displaystyle w_{i}\in [0,1]}
and
∑
i
=
1
n
w
i
=
1.
{\displaystyle \sum _{i=1}^{n}w_{i}=1.}
We can rewrite the definition of
M
p
{\displaystyle M_{p}}
using the exponential function as
M
p
(
x
1
,
…
,
x
n
)
=
exp
(
ln
[
(
∑
i
=
1
n
w
i
x
i
p
)
1
/
p
]
)
=
exp
(
ln
(
∑
i
=
1
n
w
i
x
i
p
)
p
)
{\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. We assume that
p
∈
R
{\displaystyle p\in \mathbb {R} }
but p ≠ 0 , and that the sum of wi is equal to 1 (without loss in generality);[7] Differentiating the numerator and denominator with respect to p , we have
lim
p
→
0
ln
(
∑
i
=
1
n
w
i
x
i
p
)
p
=
lim
p
→
0
∑
i
=
1
n
w
i
x
i
p
ln
x
i
∑
j
=
1
n
w
j
x
j
p
1
=
lim
p
→
0
∑
i
=
1
n
w
i
x
i
p
ln
x
i
∑
j
=
1
n
w
j
x
j
p
=
∑
i
=
1
n
w
i
ln
x
i
∑
j
=
1
n
w
j
=
∑
i
=
1
n
w
i
ln
x
i
=
ln
(
∏
i
=
1
n
x
i
w
i
)
{\displaystyle {\begin{aligned}\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 _{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{aligned}}}
By the continuity of the exponential function, we can substitute back into the above relation to obtain
lim
p
→
0
M
p
(
x
1
,
…
,
x
n
)
=
exp
(
ln
(
∏
i
=
1
n
x
i
w
i
)
)
=
∏
i
=
1
n
x
i
w
i
=
M
0
(
x
1
,
…
,
x
n
)
{\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.
[2]
Proof of
lim
p
→
∞
M
p
=
M
∞
{\textstyle \lim _{p\to \infty }M_{p}=M_{\infty }}
and
lim
p
→
−
∞
M
p
=
M
−
∞
{\textstyle \lim _{p\to -\infty }M_{p}=M_{-\infty }}
Assume (possibly after relabeling and combining terms together) that
x
1
≥
⋯
≥
x
n
{\displaystyle x_{1}\geq \dots \geq x_{n}}
. Then
lim
p
→
∞
M
p
(
x
1
,
…
,
x
n
)
=
lim
p
→
∞
(
∑
i
=
1
n
w
i
x
i
p
)
1
/
p
=
x
1
lim
p
→
∞
(
∑
i
=
1
n
w
i
(
x
i
x
1
)
p
)
1
/
p
=
x
1
=
M
∞
(
x
1
,
…
,
x
n
)
.
{\displaystyle {\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}=M_{\infty }(x_{1},\dots ,x_{n}).\end{aligned}}}
The formula for
M
−
∞
{\displaystyle M_{-\infty }}
follows from
M
−
∞
(
x
1
,
…
,
x
n
)
=
1
M
∞
(
1
/
x
1
,
…
,
1
/
x
n
)
=
x
n
.
{\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})={\frac {1}{M_{\infty }(1/x_{1},\dots ,1/x_{n})}}=x_{n}.}
Properties [ edit ]
Let
x
1
,
…
,
x
n
{\displaystyle x_{1},\dots ,x_{n}}
be a sequence of positive real numbers, then the following properties hold:[1]
min
(
x
1
,
…
,
x
n
)
≤
M
p
(
x
1
,
…
,
x
n
)
≤
max
(
x
1
,
…
,
x
n
)
{\displaystyle \min(x_{1},\dots ,x_{n})\leq M_{p}(x_{1},\dots ,x_{n})\leq \max(x_{1},\dots ,x_{n})}
.Each generalized mean always lies between the smallest and largest of the x values.
M
p
(
x
1
,
…
,
x
n
)
=
M
p
(
P
(
x
1
,
…
,
x
n
)
)
{\displaystyle M_{p}(x_{1},\dots ,x_{n})=M_{p}(P(x_{1},\dots ,x_{n}))}
, where
P
{\displaystyle P}
is a permutation operator.Each generalized mean is a symmetric function of its arguments; permuting the arguments of a generalized mean does not change its value.
M
p
(
b
x
1
,
…
,
b
x
n
)
=
b
⋅
M
p
(
x
1
,
…
,
x
n
)
{\displaystyle M_{p}(bx_{1},\dots ,bx_{n})=b\cdot M_{p}(x_{1},\dots ,x_{n})}
.Like most
means , the generalized mean is a
homogeneous function of its arguments
x 1 , ..., xn . That is, if
b is a positive real number, then the generalized mean with exponent
p of the numbers
b
⋅
x
1
,
…
,
b
⋅
x
n
{\displaystyle b\cdot x_{1},\dots ,b\cdot x_{n}}
is equal to
b times the generalized mean of the numbers
x 1 , ..., xn .
M
p
(
x
1
,
…
,
x
n
⋅
k
)
=
M
p
[
M
p
(
x
1
,
…
,
x
k
)
,
M
p
(
x
k
+
1
,
…
,
x
2
⋅
k
)
,
…
,
M
p
(
x
(
n
−
1
)
⋅
k
+
1
,
…
,
x
n
⋅
k
)
]
{\displaystyle M_{p}(x_{1},\dots ,x_{n\cdot k})=M_{p}\left[M_{p}(x_{1},\dots ,x_{k}),M_{p}(x_{k+1},\dots ,x_{2\cdot k}),\dots ,M_{p}(x_{(n-1)\cdot k+1},\dots ,x_{n\cdot k})\right]}
.
Generalized mean inequality [ edit ]
Geometric proof without words that max (a ,b ) > root mean square (RMS ) or quadratic mean (QM ) > arithmetic mean (AM ) > geometric mean (GM ) > harmonic mean (HM ) > min (a ,b ) of two distinct positive numbers a and b [note 1]
In general, if p < q , then
M
p
(
x
1
,
…
,
x
n
)
≤
M
q
(
x
1
,
…
,
x
n
)
{\displaystyle M_{p}(x_{1},\dots ,x_{n})\leq M_{q}(x_{1},\dots ,x_{n})}
and the two means are equal if and only if
x 1 = x 2 = ... = xn .
The inequality is true for real values of p and q , as well as positive and negative infinity values.
It follows from the fact that, for all real p ,
∂
∂
p
M
p
(
x
1
,
…
,
x
n
)
≥
0
{\displaystyle {\frac {\partial }{\partial p}}M_{p}(x_{1},\dots ,x_{n})\geq 0}
which can be proved using
Jensen's inequality .
In particular, for p in {−1, 0, 1} , the generalized mean inequality implies the Pythagorean means inequality as well as the inequality of arithmetic and geometric means .
Proof of the weighted inequality [ edit ]
We will prove the weighted power mean inequality. For the purpose of the proof we will assume the following without loss of generality:
w
i
∈
[
0
,
1
]
∑
i
=
1
n
w
i
=
1
{\displaystyle {\begin{aligned}w_{i}\in [0,1]\\\sum _{i=1}^{n}w_{i}=1\end{aligned}}}
The proof for unweighted power means can be easily obtained by substituting wi = 1/n .
Equivalence of inequalities between means of opposite signs [ edit ]
Suppose an average between power means with exponents p and q holds:
(
∑
i
=
1
n
w
i
x
i
p
)
1
/
p
≥
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
{\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}
applying this, then:
(
∑
i
=
1
n
w
i
x
i
p
)
1
/
p
≥
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
{\displaystyle \left(\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{p}}}\right)^{1/p}\geq \left(\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{q}}}\right)^{1/q}}
We raise both sides to the power of −1 (strictly decreasing function in positive reals):
(
∑
i
=
1
n
w
i
x
i
−
p
)
−
1
/
p
=
(
1
∑
i
=
1
n
w
i
1
x
i
p
)
1
/
p
≤
(
1
∑
i
=
1
n
w
i
1
x
i
q
)
1
/
q
=
(
∑
i
=
1
n
w
i
x
i
−
q
)
−
1
/
q
{\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{-p}\right)^{-1/p}=\left({\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{p}}}}}\right)^{1/p}\leq \left({\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{q}}}}}\right)^{1/q}=\left(\sum _{i=1}^{n}w_{i}x_{i}^{-q}\right)^{-1/q}}
We get the inequality for means with exponents −p and −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.
Geometric mean [ edit ]
For any q > 0 and non-negative weights summing to 1, the following inequality holds:
(
∑
i
=
1
n
w
i
x
i
−
q
)
−
1
/
q
≤
∏
i
=
1
n
x
i
w
i
≤
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
.
{\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{-q}\right)^{-1/q}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.}
The proof follows from Jensen's inequality , making use of the fact the logarithm is concave:
log
∏
i
=
1
n
x
i
w
i
=
∑
i
=
1
n
w
i
log
x
i
≤
log
∑
i
=
1
n
w
i
x
i
.
{\displaystyle \log \prod _{i=1}^{n}x_{i}^{w_{i}}=\sum _{i=1}^{n}w_{i}\log x_{i}\leq \log \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
∏
i
=
1
n
x
i
w
i
≤
∑
i
=
1
n
w
i
x
i
.
{\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}.}
Taking q -th powers of the xi yields
∏
i
=
1
n
x
i
q
⋅
w
i
≤
∑
i
=
1
n
w
i
x
i
q
∏
i
=
1
n
x
i
w
i
≤
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
.
{\displaystyle {\begin{aligned}&\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{aligned}}}
Thus, we are done for the inequality with positive q ; the case for negatives is identical but for the swapped signs in the last step:
∏
i
=
1
n
x
i
−
q
⋅
w
i
≤
∑
i
=
1
n
w
i
x
i
−
q
.
{\displaystyle \prod _{i=1}^{n}x_{i}^{-q{\cdot }w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}^{-q}.}
Of course, taking each side to the power of a negative number -1/q swaps the direction of the inequality.
∏
i
=
1
n
x
i
w
i
≥
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
.
{\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.}
Inequality between any two power means [ edit ]
We are to prove that for any p < q the following inequality holds:
(
∑
i
=
1
n
w
i
x
i
p
)
1
/
p
≤
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
{\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}
if
p is negative, and
q is positive, the inequality is equivalent to the one proved above:
(
∑
i
=
1
n
w
i
x
i
p
)
1
/
p
≤
∏
i
=
1
n
x
i
w
i
≤
(
∑
i
=
1
n
w
i
x
i
q
)
1
/
q
{\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}
The proof for positive p and q is as follows: Define the following function: f : R + → R +
f
(
x
)
=
x
q
p
{\displaystyle f(x)=x^{\frac {q}{p}}}
. f is a power function, so it does have a second derivative:
f
″
(
x
)
=
(
q
p
)
(
q
p
−
1
)
x
q
p
−
2
{\displaystyle f''(x)=\left({\frac {q}{p}}\right)\left({\frac {q}{p}}-1\right)x^{{\frac {q}{p}}-2}}
which is strictly positive within the domain of
f , since
q > p , so we know
f is convex.
Using this, and the Jensen's inequality we get:
f
(
∑
i
=
1
n
w
i
x
i
p
)
≤
∑
i
=
1
n
w
i
f
(
x
i
p
)
(
∑
i
=
1
n
w
i
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{\displaystyle {\begin{aligned}f\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)&\leq \sum _{i=1}^{n}w_{i}f(x_{i}^{p})\\[3pt]\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{q/p}&\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}\end{aligned}}}
after raising both side to the power of
1/q (an increasing function, since
1/q is positive) we get the inequality which was to be proven:
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{\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}
Using the previously shown equivalence we can prove the inequality for negative p and q by replacing them with −q and −p , respectively.
Generalized f -mean [ edit ]
The power mean could be generalized further to the generalized f -mean :
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{\displaystyle M_{f}(x_{1},\dots ,x_{n})=f^{-1}\left({{\frac {1}{n}}\cdot \sum _{i=1}^{n}{f(x_{i})}}\right)}
This covers the geometric mean without using a limit with f (x ) = log(x ) . The power mean is obtained for f (x ) = xp . Properties of these means are studied in de Carvalho (2016).[3]
Applications [ edit ]
Signal processing [ edit ]
A power mean serves a non-linear moving average which is shifted towards small signal values for small p and emphasizes big signal values for big 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 )
See also [ edit ]
^ If AC = a and BC = b . OC = AM of a and b , and radius r = QO = OG. Using Pythagoras' theorem , QC² = QO² + OC² ∴ QC = √QO² + OC² = QM . Using Pythagoras' theorem, OC² = OG² + GC² ∴ GC = √OC² − OG² = GM . Using similar triangles , HC / GC = GC / OC ∴ HC = GC² / OC = HM .
References [ edit ]
^ a b Sýkora, Stanislav (2009). "Mathematical means and averages: basic properties". Stan's Library . III . Castano Primo, Italy: Stan's Library. doi :10.3247/SL3Math09.001 .
^ a b c P. S. Bullen: Handbook of Means and Their Inequalities . Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177
^ a b de Carvalho, Miguel (2016). "Mean, what do you Mean?" . The American Statistician . 70 (3): 764‒776. doi :10.1080/00031305.2016.1148632 . hdl :20.500.11820/fd7a8991-69a4-4fe5-876f-abcd2957a88c .
^ Weisstein, Eric W. "Power Mean" . MathWorld . (retrieved 2019-08-17)
^ Thompson, Sylvanus P. (1965). Calculus Made Easy . Macmillan International Higher Education. p. 185. ISBN 9781349004874 . Retrieved 5 July 2020 . [permanent dead link ]
^ Jones, Alan R. (2018). Probability, Statistics and Other Frightening Stuff . Routledge. p. 48. ISBN 9781351661386 . Retrieved 5 July 2020 .
^ Handbook of Means and Their Inequalities (Mathematics and Its Applications) .
Further reading [ edit ]
Bullen, P. S. (2003). "Chapter III - The Power Means". Handbook of Means and Their Inequalities . Dordrecht, Netherlands: Kluwer. pp. 175–265.
External links [ edit ]