- 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$f:\; V\; arr\; W\; qquadqquad$is a function between two

vector space s over a field $F\; qquadqquad$.We say that $f\; qquadqquad$ is "homogeneous of degree $k\; qquadqquad$" if :$f(alpha\; mathbf\{v\})\; =\; alpha^k\; f(mathbf\{v\})$for all nonzero $alpha\; isin\; F\; qquadqquad$ and $mathbf\{v\}\; isin\; V\; qquadqquad$.

**Examples***A

linear function $f:\; V\; arr\; W\; qquadqquad$ is homogeneous of degree 1, since by the definition of linearity:$f(alpha\; mathbf\{v\})=alpha\; f(mathbf\{v\})$for all $alpha\; isin\; F\; qquadqquad$ and $mathbf\{v\}\; isin\; V\; qquadqquad$.*A

multilinear function $f:\; V\_1\; imes\; ldots\; imes\; V\_n\; arr\; W\; qquadqquad$ 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 $alpha\; isin\; F\; qquadqquad$ and $mathbf\{v\}\_1\; isin\; V\_1,ldots,mathbf\{v\}\_n\; isin\; V\_n\; qquadqquad$.*It follows from the previous example that the $n$th

Fréchet derivative of a function $f:\; X\; ightarrow\; Y$ between two Banach spaces $X$ and $Y$ is homogeneous of degree $n$.*

Monomials in $n$ real variables define homogeneous functions $f:mathbb\{R\}^n\; arr\; mathbb\{R\}$. 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 homogeneousfunctions.**Elementary theorems***Euler's theorem: Suppose that the function $f:mathbb\{R\}^n\; arr\; mathbb\{R\}$ is

differentiable and homogeneous of degree $k$. Then:$mathbf\{x\}\; cdot\; abla\; f(mathbf\{x\})=\; kf(mathbf\{x\})\; qquadqquad$.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:$frac\{partial\}\{partial\; x\_1\}f(alphamathbf\{y\})frac\{mathrm\{d\{mathrm\{d\}alpha\}(alpha\; y\_1)+\; cdotsfrac\{partial\}\{partial\; x\_n\}f(alphamathbf\{y\})frac\{mathrm\{d\{mathrm\{d\}alpha\}(alpha\; y\_n)\; =\; k\; alpha\; ^\{k-1\}\; f(mathbf\{y\})$,so that:$y\_1frac\{partial\}\{partial\; x\_1\}f(alphamathbf\{y\})+\; cdotsy\_nfrac\{partial\}\{partial\; x\_n\}f(alphamathbf\{y\})\; =\; k\; alpha^\{k-1\}\; f(mathbf\{y\})$.The above equation can be written in thedel notation as:$mathbf\{y\}\; cdot\; abla\; f(alpha\; mathbf\{y\})\; =\; k\; alpha^\{k-1\}f(mathbf\{y\}),\; qquadqquad\; abla=(frac\{partial\}\{partial\; x\_1\},ldots,frac\{partial\}\{partial\; x\_n\})$,from which the stated result is obtained by setting $alpha=1$.*Suppose that $f:mathbb\{R\}^n\; arr\; mathbb\{R\}$ is

differentiable and homogeneous of degree $k$. Then its first-order partial derivatives $partial\; f/partial\; x\_i$ are homogeneous of degree $k-1\; qquadqquad$.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:$frac\{partial\}\{partial\; x\_i\}f(alphamathbf\{y\})frac\{mathrm\{d\{mathrm\{d\}y\_i\}(alpha\; y\_i)\; =\; alpha\; ^k\; frac\{partial\}\{partial\; x\_i\}f(mathbf\{y\})frac\{mathrm\{d\{mathrm\{d\}y\_i\}(y\_i)$,so that:$alphafrac\{partial\}\{partial\; x\_i\}f(alphamathbf\{y\})\; =\; alpha\; ^k\; frac\{partial\}\{partial\; x\_i\}f(mathbf\{y\})$and hence:$frac\{partial\}\{partial\; x\_i\}f(alphamathbf\{y\})\; =\; alpha\; ^\{k-1\}\; frac\{partial\}\{partial\; x\_i\}f(mathbf\{y\})$.**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 theseparable differential equation :$x\; frac\{mathrm\{d\}v\}\{mathrm\{d\}x\}=-frac\{J(1,v)\}\{I(1,v)\}-v$.**References***cite book | author=Blatter, Christian | title=Analysis II (2nd ed.) | publisher=Springer Verlag | year=1979 |language=German |isbn=3-540-09484-9 | pages=p. 188 | chapter=20. Mehrdimensionale Differentialrechnung, Aufgaben, 1.

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### Look at other dictionaries:

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