0
$\begingroup$

Find all functions $f, g$ that satisfy the functional equation $ f(x)-f(y)= (x -y)g(\sqrt{xy}) \quad \forall\ x,y>0. $

  • 0
    I do it and i have this equation: (x^2-1)g(x)-(y^2-1)g(y)=(x^2-y^2)g(xy).If F(x)=(x^2-1)g(x) then we have the equation: ((X^2)*(y^2)-1)(F(x)-F(y))=(x^2-y^2)F(xy). I can't find any solution.Can you help me?2012-11-09

2 Answers 2

1

Hint: First fix $y=1$, this shows what $f$ has to be, in terms of $g$. Then fix another $y$ as well.

  • 0
    I do it and i have this equation: (x^2-1)g(x)-(y^2-1)g(y)=(x^2-y^2)g(xy).If F(x)=(x^2-1)g(x) then we have the equation: ((X^2)*(y^2)-1)(F(x)-F(y))=(x^2-y^2)F(xy). I can't find any solution.Can you help me2012-11-09
0

Since the functional equation is $\;f(x)-f(y)= (x -y)g(\sqrt{xy})\;\;\forall x,y>0\;$ we can write it as $\;g(x) = (f(x t)-f(x/t))/(x t-x/t).\;$ Assuming that $f(x):=\sum_{n=0}^\infty c_nx^n\;$ then the equation is $g(x)=c_1+c_2x(1/t+t)+c_3x^2(1/t^2+1+t^2)+c_4x^3(1/t^3+1/t+t+t^3)+\cdots.$ The coefficients of $g(x)$ must be constants, so $\;c_n=0\;$ for all $n>1.$ Now $f(x)=c_0+c_1x.$