Question

Show that the functions f (x, y) = ln x - ln y and g(x, y) =(x2+2y2)/2xy are functionally dependent.

Answer 1

Answer on Question #81428, Math/Real Analysis

Show that the functions $f(x,y)=\ln x-\ln y$ and $g(x,y)=(x^{2}+2y^{2})/2xy$ are functionally dependent.

The answer: By definition, two functions $f(x,y)$ and $g(x,y)$ defined in a domain $D$ are functionally dependent of there exists a continuously differentiable function $\phi(u,v)$ such that :

- $\phi(f(x,y),g(x,y))\equiv 0$

- $\Delta\phi\neq 0$ at each point $(u,v)=(f(x,y),g(x,y))$ for $(x,y)$ in $D$.

Then following the theorem: ‘If $F$ is differentiable transformation defined as

\[ F:\begin{cases}u=f(x,y)\\

v=g(x,y)\end{cases} \]

and $f$ and $g$ are linearly dependent, then the Jacobian of $F$ is identically zero‘

So in our case $f(x,y)=\ln x-\ln y$ and $g(x,y)=(x^{2}+2y^{2})/2xy$, i.e. $x,y>0$. To get the Jacobian one needs to calculate the partial derivatives: $f_{x}$, $f_{y}$, $g_{x}$, $g_{y}$:

$f_{x}$ $=$ $\frac{1}{x}\ .,\quad f_{y}=-\frac{1}{y}$

$g_{x}$ $=$ $\frac{2x*2xy-2y*(x^{2}+2y^{2})}{4x^{2}y^{2}}=\frac{x^{2}-2y^{2}}{2x^{2}y}$

$g_{y}$ $=$ $\frac{4y*2xy-2x(x^{2}+2y^{2})}{4x^{2}y^{2}}=\frac{-x^{2}+2y^{2}}{2xy^{2}}$

then

$\det J=det\left[\frac{d(f,g)}{d(x,y)}\right]=f_{x}g_{y}-f_{y}g_{x}=\frac{-x^{2}+2y^{2}}{2x^{2}y^{2}}+\frac{x^{2}-2y^{2}}{2x^{2}y^{2}}=0$

Therefore the functions $f(x,y)=\ln x-\ln y$ and $g(x,y)=(x^{2}+2y^{2})/2xy$ are functionally dependent.

As $f(x,y)=\ln x-\ln y=\ln\frac{x}{y}$ then $\frac{x}{y}=e^{f(x,y)}$. Therfore

$g(x,y)=\frac{(x^{2}+2y^{2})}{2xy}=\frac{1}{2}\frac{x}{y}+\frac{y}{x}=\frac{1}{2}e^{f(x,y)}+\frac{1}{e^{f(x,y)}}$

that confirms a functional dependence of $f$ and $g$ functions.