Functional Equation

by utkarshgupta, Jan 28, 2015, 4:31 PM

Problem :
Find all pairs $f,g : R \to R$ such that
\[f(x)-f(y)=(x-y)(g(x)+g(y))\]

Solution :

It is easy to see that
$f(x)=ax^2+2bx+c$ and $g(x)=ax+b$ satisfy the equation.
I will show that they are the only solutions.

Let $P(x,y$) be the assertion $f(x)-f(y)=(x-y)(g(x)+g(y))$

\[P(x,0) \implies f(x)-f(0)=x(g(x)+g(0))\]
\[\implies f(x)=x(g(x)+g(0))+f(0)\]

Replacing this value of $f(x),f(y)$ in $P(x)$, we get
\[x(g(x)+g(0))+f(0)-y(g(y)+g(0))-f(0)=(x-y)(g(x)+g(y))\]

Opening and cancelling...
\[xg(0)-yg(0)=xg(y)-yg(x)\]............ (I)
Introducing $h(x)=g(x)-g(0)$
(I) can be rewriiten as
\[xh(y)=yh(x)\]
\[xh(1)=h(x)\]
Setting $h(1)=a$
\[h(x)=ax\]
Setting $g(0)=b$ and using $h(x)=g(x)-g(0)$
We get, \[g(x)=ax+b\]

Now setting $f(0)=c$ and using $ f(x)=x(g(x)+g(0))+f(0)$,
\[f(x)=ax^2+2bx+c\]

Thus $\boxed{g(x)=ax+b}$ and $\boxed{f(x)=ax^2+2bx+c}$ are the only solutions.
This post has been edited 3 times. Last edited by utkarshgupta, Jan 30, 2015, 5:09 PM

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Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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