If p is a non-zero real number, and are positive real numbers, then the generalized mean or power mean with exponent p of these positive real numbers is[2][3]
(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):
Furthermore, for a sequence of positive weights wi we define the weighted power mean as[2]
and when p = 0, it is equal to the weighted geometric mean:
The unweighted means correspond to setting all wi = 1.
For the purpose of the proof, we will assume without loss of generality that
and
We can rewrite the definition of using the exponential function as
In the limit p → 0, we can apply L'Hôpital's rule to the argument of the exponential function. We assume that 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
By the continuity of the exponential function, we can substitute back into the above relation to obtain
as desired.[2]
Proof of and
Assume (possibly after relabeling and combining terms together) that . Then
Let be a sequence of positive real numbers, then the following properties hold:[1]
.
Each generalized mean always lies between the smallest and largest of the x values.
, where 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.
.
Like most means, the generalized mean is a homogeneous function of its arguments x1, ..., xn. That is, if b is a positive real number, then the generalized mean with exponent p of the numbers is equal to b times the generalized mean of the numbers x1, ..., xn.
.
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.
Suppose an average between power means with exponents p and q holds:
applying this, then:
We raise both sides to the power of −1 (strictly decreasing function in positive reals):
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.
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
Taking q-th powers of the xi yields
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:
Of course, taking each side to the power of a negative number -1/q swaps the direction of the inequality.
We are to prove that for any p < q the following inequality holds:
if p is negative, and q is positive, the inequality is equivalent to the one proved above:
The proof for positive p and q is as follows: Define the following function: f : R → R. f is a power function, so it does have a second derivative:
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:
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:
Using the previously shown equivalence we can prove the inequality for negative p and q by replacing them with −q and −p, respectively.
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]
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.
^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.
^ abSýkora, Stanislav (2009). "Mathematical means and averages: basic properties". Stan's Library. III. Castano Primo, Italy. doi:10.3247/SL3Math09.001.
^ abcP. S. Bullen: Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177