Try this beautiful problem from Singapore Mathematics Olympiad based on Functional Equation.
Let f and g be functions such that for all real numbers x and y,
\( g (f (x+y)) = f( x ) + (x+y) g (y)\).
Find the value of \( g(0) + g (1) + ........................+ g (2013) \)
Functional Equation
Funcion
Arbitrary Numbers
Answer: 0
Singapore Mathemaics Olympiad
Challenges and thrills
We can start this problem by considering y = -x.
Then \( g (f (0) ) = f (x) \) for all x. This \(f\) is is a constant function ; namely
\( f (x) = c \) for some c.
Try the rest of the sum ............................................................
For all value of x,y we have
\( (x+y) g(y) = g(f(x+y)) - f(x) = g(c) - c = 0 \)
Since x + y is arbitrary , we must have \( g (y) = 0 \) for all y .Hence
\( g (0) + g ( 1 ) + ................................+ g(2013) = 0 \) (Answer).
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