Homogeneous function

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In mathematics, a function f,

${\displaystyle f:\quad \mathbb {R} ^{n}\rightarrow \mathbb {R} }$

is homogeneous of degree p, if

${\displaystyle f(\lambda \mathbf {x} )=\lambda ^{p}f(\mathbf {x} ),\quad \mathbf {x} \in \mathbb {R} ^{n},\quad \lambda \in \mathbb {R} ,\quad p\in \mathbb {N} ^{*}.}$

The degree of homogeneity p is a positive integral number.

Examples[edit]

${\displaystyle {\begin{aligned}f(x)=&\;ax^{m}\Longrightarrow f(\lambda x)=a(\lambda x)^{m}=\lambda ^{m}(ax^{m})=\lambda ^{m}f(x)\quad ({\hbox{degree}}\;m).\\f(x,y,z)=&\;x^{2}yz+3xy^{2}z-5xyz^{2}\Longrightarrow f(\lambda x,\lambda y,\lambda z)=\lambda ^{4}x^{2}yz+3\lambda ^{4}xy^{2}z-5\lambda ^{4}xyz^{2}=\lambda ^{4}f(x,y,z)\quad ({\hbox{degree}}\;4)\\\end{aligned}}}$

Euler's theorem[edit]

Let f be differentiable and homogeneous of order p, then

${\displaystyle \sum _{i=1}^{n}x_{i}{\frac {\partial f(x_{1},\dots ,x_{n})}{\partial x_{i}}}=pf(x_{1},\dots ,x_{n})}$

Proof[edit]

By the chain rule

${\displaystyle {\frac {df(\lambda x_{1},\ldots ,\lambda x_{n})}{d\lambda }}=\sum _{i=1}^{n}{\frac {\partial f(\lambda x_{1},\dots ,\lambda x_{n})}{\partial (\lambda x_{i})}}{\frac {d\lambda x_{i}}{d\lambda }}=\sum _{i=1}^{n}x_{i}{\frac {\partial f(\lambda x_{1},\dots ,\lambda x_{n})}{\partial (\lambda x_{i})}}.\qquad \qquad \qquad (1)}$

From the homogeneity,

${\displaystyle {\frac {df(\lambda x_{1},\ldots ,\lambda x_{n})}{d\lambda }}={\frac {d\lambda ^{p}}{d\lambda }}f(x_{1},\dots ,x_{n})=p\lambda ^{p-1}f(x_{1},\dots ,x_{n}).\qquad \qquad \qquad \qquad \qquad \qquad (2)}$

Compare Eqs (1) and (2) for λ = 1 and the result to be proved follows.