# Homogeneous Functions

We start with equivalence of two integrals $\displaystyle\int\frac{dx}{1+x^2}$ and $\displaystyle\int\frac{dx}{p^{2}+x^2}.$ 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

$\displaystyle\int\frac{dx}{p^{2}+x^2}=\frac{1}{p^2}\int\frac{p\space du}{1+u^2}.$

Since

$\displaystyle\int\frac{dx}{1+x^2}=\arctan(x)+C$

where $C$ is an arbitrary constant, this leads to

$\displaystyle\int\frac{dx}{p^{2}+x^2}=\frac{1}{p}\arctan(\frac{x}{p})+C.$

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 $t\gt 0,$ $f(tx, ty) = t^{a}f(x, y),$ is called (positively) homogeneous of order (or degree) $a.$

## Examples

$f(x, y) = x^{2} + y^{2}$ is homogeneous of order 2. Indeed, $f(tx,ty) = (tx)^{2} + (ty)^{2} = t^{2}(x^{2} + y^{2}) = t^{2}f(x,y).$

$f(x, y) = (x^{2} + y^{2})/x$ is homogeneous of order $1$

$f(x, y) = \sqrt{x^{2} + y^{2}}$ is also homogeneous of order $1$

$f(x, y) = (x^{2} - y^{2})/(x^{2} + y^{2})$ 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

$\displaystyle (1+x)^{a}=1+\frac{a}{1}x+\frac{a(a-1)}{1\cdot 2}x^{2}+\frac{a(a-1)(a-2)}{1\cdot 2\cdot 3}x^{3}+\ldots$

$\displaystyle (x+y)^{a}=x^{a}+\frac{a}{1}x^{a-1}y+\frac{a(a-1)}{1\cdot 2}x^{a-2}y^{2}+\frac{a(a-1)(a-2)}{1\cdot 2\cdot 3}x^{a-3}y^{3}+\ldots$

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, $g(x, y, z, ...) = (x\cdot y\cdot z\cdot ...)^{1/N}$

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 $2,$ $N = 2^{n},$ 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 $(x + y)^{2} \ge 4xy$ which is true because, by rearranging terms, it reduces to $(x - y)^{2} \ge 0$ which is obviously true.

For the inductive step, assume the inequality has been proven for $N = 2^{k},$ and let there be given $N = 2^{k+1}.$ Note that $N = 2^{k+1} = 2\cdot 2^{k} = 2M.$ Thus we may split the given set of $N$ numbers into two groups of $M$ elements each. Let these be $x, y, z, \ldots$ and $u, v, w, \ldots.$ We have

$\begin{align} a(x, y, z, ..., u, v, w, \ldots ) &= (x + y + z + \ldots + u + v + w + \ldots )/N\\ &= [(x + y + z + \ldots )/M + (u + v + w + \ldots )/M]/2\\ &= (a(x, y, z, \ldots ) + a(u, v, w, \ldots ))/2\\ &= a(a(x, y, z, \ldots ), a(u, v, w, \ldots )). \end{align}$

Quite similarly $g(x, y, z, \ldots , u, v, w, \ldots ) = g(g(x, y, z, \ldots ),g(u, v, w, \ldots )).$ Since, by the inductive assumption,

$a(x, y, z, \ldots ) \ge g(x, y, z, \ldots )$ and $a(u, v, w, \ldots ) \ge g(u, v, w, \ldots )$

we finally have

$\begin{align} a(x, y, z, \ldots , u, v, w,\ldots ) &= a(a(x, y, z, \ldots ), a(u, v, w, \ldots ))\\ & \ge a(g(x, y, z, \ldots ), g(u, v, w, \ldots ))\\ & \ge g(g(x, y, z, \ldots ), g(u, v, w, \ldots ))\\ &= g(x, y, z, \ldots , u, v, w, \ldots ).\\ \end{align}$

Q.E.D.

### References

- G. H. Hardy, J. E. Littlewood, G. Polya,
*Inequalities*, Cambridge University Press (2^{nd}edition) 1988. - G. Polya,
*Mathematical Discovery*, John Wiley & Sons, 1981

|Contact| |Front page| |Contents| |Generalizations| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny