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Homogeneous function

 
Sci-Tech Dictionary: homogeneous function
(′hä·mə′jē·nē·əs ′fəŋk·shən)

(mathematics) A real function ƒ(x1,x2, …,xn) is homogeneous of degree r if ƒ(ax1,ax2, …,axn) = arƒ(x1,x2, …,xn) for every real number a.


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Wikipedia: Homogeneous function
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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.

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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

Linear functions

Any 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. Similarly, any 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 that the nth Fréchet derivative of a function f: X \rightarrow Y between two Banach spaces X and Y is homogeneous of degree n.

Homogeneous polynomials

Monomials in n real variables define homogeneous functions  f:\mathbb{R}^n \rarr \mathbb{R}. For example,

f(x,y,z) = x5y2z3

is homogeneous of degree 10 since

x)5y)2z)3 = α10x5y2z3.

A homogeneous polynomial is a polynomial made up of a sum of monomials of the same degree. For example,

x5 + 2x3y2 + 9xy4

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

Properties

 \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 α, 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= \left(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n} \right),

from which the stated result is obtained by setting α = 1.

For the proof of the converse, see [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 yi, 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 \frac{\partial}{\partial x_i}f(\mathbf{y})

and hence

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

References

  • Blatter, Christian (1979). "20. Mehrdimensionale Differentialrechnung, Aufgaben, 1." (in German). Analysis II (2nd ed.). Springer Verlag. pp. 188. ISBN 3-540-09484-9. 

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