Homogeneous Functions
We start with equivalence of two integrals  . I assume here
that p is a real nonzero number. For p = 1, the second integral reduces to the first and, therefore,
is a clear generalization. However, once we know how to compute the more special integral, we may, using a standard substitution, compute the more general one.
Indeed, let u = x/p. Then successively x = pu, dx = pdu, and
Since
this leads to
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The situation, where one parameter appears superfluous, is quite common to the class of homogeneous functions.
Definition
A function f(x, y) such that, for t>0, f(tx, ty) = taf(x, y), is called (positively) homogeneous of order (or degree) a.
Examples
- f(x, y) = x2 + y2 is homogeneous of order 2. Indeed, f(tx,ty) = (tx)2 + (ty)2 = t2(x2 + y2) = t2f(x,y).
- f(x, y) = (x2 + y2)/x is homogeneous of order 1
 is also homogeneous of order 1
- f(x, y) = (x2 - y2)/(x2 + y2) has order 0.
Functions in examples 1 and 3 are also symmetric because they do not change when one swaps values of x and y.
The famous Binomial Theorem (discovered and repeatedly used but not proved by I.Newton) may be
written in two equivalent forms:
All terms in the latter are homogeneous of order a.
The notion of homogeneity extends to functions of more than 2 variables. For example, all kinds of means are symmetric and naturally homogeneous of order 1. For N variables,
- Arithmetic mean, a(x, y, z, ...) = (x + y + z + ...)/N
- Geometric mean, g(x, y, z, ...) = (x·y·z· ...)1/N
- Harmonic mean, h(x, y, z, ...) = N/(1/x + 1/y + 1/z + ...)
A famous inequality relates arithmetic and geometric means of nonnegative numbers:
a(x, y, z, ...) g(x, y, z, ...). If we can prove this inequality
for powers of 2, N = 2n, then from a previous result it will follow
for all integer N. Let's use induction.
Let n = 1. The inequality is then equivalent to (x + y)24xy which is true because, by rearranging terms, it reduces to (x - y)20 which is obviously true.
For the inductive step, assume the inequality has been proven for N = 2k, and let there be given N = 2k+1. Note that N = 2k+1 = 2·2k = 2M. Thus we may split the given set of N numbers into two
groups of M elements each. Let these be x, y, z, ... and u, v, w, ... We have
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| a(x, y, z, ..., u, v, w, ...) | = (x + y + z + ... + u + v + w + ...)/N |
| | = [(x + y + z + ...)/M + (u + v + w + ...)/M]/2 |
| | = (a(x, y, z, ...) + a(u, v, w, ...))/2 |
| | = a(a(x, y, z, ...), a(u, v, w, ...)). |
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Quite similarly g(x, y, z, ..., u, v, w, ...) = g(g(x, y, z, ...),g(u, v, w, ...)). Since, by the inductive assumption,
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a(x, y, z, ...)g(x, y, z, ...) and a(u, v, w, ...)g(u, v, w, ...)
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we finally have
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| a(x, y, z, ..., u, v, w,...) | = a(a(x, y, z, ...), a(u, v, w, ...)) |
| | a(g(x, y, z, ...), g(u, v, w, ...)) |
| | g(g(x, y, z, ...), g(u, v, w, ...)) |
| | = g(x, y, z, ..., u, v, w, ...). |
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Q.E.D.
References
- G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Cambridge University Press (2nd edition) 1988.
- G. Polya, Mathematical Discovery, John Wiley & Sons, 1981
Copyright © 1996-2008 Alexander Bogomolny
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