ISL 2011 A3

by utkarshgupta, Mar 3, 2015, 6:08 AM

And another FE :P
Something's wrong with me :P

Problem
]Determine all pairs $(f,g)$ of functions from the set of real numbers to itsels that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\] for all real numbers $x$ and $y$.

Solution
Let $P(x,y)$ be the assertion $g(f(x+y)) = f(x) + (2x + y)g(y)$.
$$P(x,0) \implies g(f(x))=f(x)+2xg(0)$$
$$P(0,x) \implies g(f(x))=f(0)+xg(x)$$
$$\implies xg(x)+f(0)=f(x)+2xg(0)$$
$$\implies f(x)=xg(x)+f(0)-2xg(0)$$

Replacing this in $P(x,y)$
$$g(f(x+y))=xg(x)+f(0)-2xg(0)+(2x+y)g(y)$$
Here setting $x=z$, $y=x+y-z$
$$g(f(x+y))=zg(z)+f(0)-2zg(0)+(x+y+z)g(x+y-z)$$

Comparing and setting $y=z=1$
$$xg(x)-2xg(0)+2xg(1)=2g(0)+(x+2)g(x)$$
$$\implies g(x)=Ax+B$$

Now the question is easy


Thus the solutions are $\boxed {f(x)=0=g(x)}$ or
$\boxed {f(x)=x^2+C}$ and $\boxed {g(x)=x}$

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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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  • I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it :D

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  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

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  • INTERSTING BLOG

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  • I have no plans for this blog right now....
    No time here people !
    But lets see....
    I may try some combinatorics :P

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