Homogeneous function

In mathematics, a homogeneous function is a function with multiplicative scaling behaviour: if the argument is multiplied by a factor, then the result is multiplied by some power of this factor.

Formal definition
Suppose that $$ f: V \rarr W \qquad\qquad $$ is a function between two vector spaces over a field $$ F \qquad\qquad$$.

We say that $$ f \qquad\qquad$$ is homogeneous of degree $$ k \qquad\qquad$$ if
 * $$ f(\alpha \mathbf{v}) = \alpha^k f(\mathbf{v}) $$

for all nonzero $$ \alpha \isin F \qquad\qquad$$ and $$ \mathbf{v} \isin V \qquad\qquad$$.

Examples

 * A linear function $$ f: V \rarr W \qquad\qquad $$ is homogeneous of degree 1, since by the definition of linearity
 * $$f(\alpha \mathbf{v})=\alpha f(\mathbf{v})$$

for all $$ \alpha \isin F \qquad\qquad$$ and $$ \mathbf{v} \isin V \qquad\qquad$$.


 * A multilinear function $$ f: V_1 \times \ldots \times V_n \rarr W \qquad\qquad $$ is homogeneous of degree n, since by the definition of multilinearity
 * $$f(\alpha \mathbf{v}_1,\ldots,\alpha \mathbf{v}_n)=\alpha^n f(\mathbf{v}_1,\ldots, \mathbf{v}_n)$$

for all $$ \alpha \isin F \qquad\qquad$$ and $$ \mathbf{v}_1 \isin V_1,\ldots,\mathbf{v}_n \isin V_n \qquad\qquad$$.


 * It follows from the previous example that the $$n$$th Fréchet derivative of a function $$f: X \rightarrow Y$$ between two Banach spaces $$X$$ and $$Y$$ is homogeneous of degree $$n$$.


 * Monomials in $$n$$ real variables define homogeneous functions $$ f:\mathbb{R}^n \rarr \mathbb{R}$$. For example,
 * $$f(x,y,z)=x^5y^2z^3$$

is homogeneous of degree 10 since
 * $$(\alpha x)^5(\alpha y)^2(\alpha z)^3=\alpha^{10}x^5y^2z^3$$.


 * A homogeneous polynomial is a polynomial made up of a sum of monomials of the same degree. For example,
 * $$x^5 + 2 x^3 y^2 + 9 x y^4$$

is a homogeneous polynomial of degree 5. Homogeneous polynomials also define homogeneous functions.

Elementary theorems

 * Euler's theorem: Suppose that the function $$ f:\mathbb{R}^n \rarr \mathbb{R}$$ is differentiable and homogeneous of degree $$ k $$. Then
 * $$ \mathbf{x} \cdot \nabla f(\mathbf{x})= kf(\mathbf{x}) \qquad\qquad $$.

This result is proved as follows. Writing $$f=f(x_1,\ldots,x_n) $$ and differentiating the equation
 * $$f(\alpha \mathbf{y})=\alpha^k f(\mathbf{y})$$

with respect to $$\alpha$$, we find by the chain rule that
 * $$\frac{\partial}{\partial x_1}f(\alpha\mathbf{y})\frac{\mathrm{d}}{\mathrm{d}\alpha}(\alpha y_1)+ \cdots

\frac{\partial}{\partial x_n}f(\alpha\mathbf{y})\frac{\mathrm{d}}{\mathrm{d}\alpha}(\alpha y_n) = k \alpha ^{k-1} f(\mathbf{y})$$, so that
 * $$y_1\frac{\partial}{\partial x_1}f(\alpha\mathbf{y})+ \cdots

y_n\frac{\partial}{\partial x_n}f(\alpha\mathbf{y}) = k \alpha^{k-1} f(\mathbf{y})$$. The above equation can be written in the del notation as
 * $$ \mathbf{y} \cdot \nabla f(\alpha \mathbf{y}) = k \alpha^{k-1}f(\mathbf{y}), \qquad\qquad \nabla=(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n})$$,

from which the stated result is obtained by setting $$\alpha=1$$.


 * Suppose that $$ f:\mathbb{R}^n \rarr \mathbb{R}$$ is differentiable and homogeneous of degree $$ k $$. Then its first-order partial derivatives $$\partial f/\partial x_i$$ are homogeneous of degree $$ k-1 \qquad\qquad$$.

This result is proved in the same way as Euler's theorem. Writing $$f=f(x_1,\ldots,x_n) $$ and differentiating the equation
 * $$f(\alpha \mathbf{y})=\alpha^k f(\mathbf{y})$$

with respect to $$y_i$$, we find by the chain rule that
 * $$\frac{\partial}{\partial x_i}f(\alpha\mathbf{y})\frac{\mathrm{d}}{\mathrm{d}y_i}(\alpha y_i) = \alpha ^k \frac{\partial}{\partial x_i}f(\mathbf{y})\frac{\mathrm{d}}{\mathrm{d}y_i}(y_i)$$,

so that
 * $$\alpha\frac{\partial}{\partial x_i}f(\alpha\mathbf{y}) = \alpha ^k f(\mathbf{y})$$

and hence
 * $$\frac{\partial}{\partial x_i}f(\alpha\mathbf{y}) = \alpha ^{k-1} f(\mathbf{y})$$.

Application to ODEs
The substitution $$v=y/x$$ converts the ordinary differential equation
 * $$I(x, y)\frac{\mathrm{d}y}{\mathrm{d}x} + J(x,y) = 0,$$

where $$I$$ and $$J$$ are homogeneous functions of the same degree, into the separable differential equation
 * $$x \frac{\mathrm{d}v}{\mathrm{d}x}=-\frac{J(1,v)}{I(1,v)}-v$$.