A Quick Functional Equation | India IMO TST 2010

[josecarlosferreira]

At 2:26 we have:
1-x^2 = f(x) f(-x)

(1+x)(1-x) = f(x)f(-x)
Look 1+x is the same thing as (1-x), just the x is negative, so
f(x) = 1+x
or f(x) = 1-x

[rex_yourbud]

I used differential calculus to generate a LDE ( linear differential equation ) which generated me f(x) :))

[philippenachtergal]

Ok, I'm bad at those but let's try.
If (x, y) = (0, y) we have f(0+y) + 0 = f(0) * f(y) --> f(y) = f(0)* f(y)
As this must be true for all y, we have f(0) = 1
if we pose x=y we get that f(2y) + y² = f(y)² ---> f(2y) = f(y)² - y² = (f(y)-y) (f(y) + y) (and likewise of course for f(2x))
if we pose y=2x we get f(3x) + 2x² = f(x) * f(2x) but that doesn't really help as it doesn't simplify when injecting what we know about f(2x)
if we pose y=-x we get f(0) - x² = f(x) f(-x) --> f(x) * f(-x) = 1 - x²
hum if I pose y = -2x, i get f(-x) - 2x² = f(x)f(-2x) --> (1-x²) / f(x) - 2x²= f(-x) - 2x² = f(x) f(-2x) = f(x) ( f(-x)² - (-x)²) = f(x) ( f(-x)² - x²)
And I can expand and expand and expand but I can't find a way to simplify.

It should be possible to solve this with differential equations since f(x+1) = f(1) f(x) -x and it looks like a quadratic equation since the difference is of the order x but I don't remember how to solve those... Lets pose k = f(1)
Hum, a polynomial of the form f(x) = -x² /2 + b x + c has a derivate -2/2 x + b , looks good !
If my guess is a solution, f(0) = 1 gives that c=1

f(1) = -1/2 + b + 1 = b + 1/2
f(2) = -1 + 2b + 1 = 2b
So we need f(2) = 2b = f(1)² - 1 so 2b - 1/4 + b² + b - 1 so b² - b -0.75 = 0 so b= 3/2 or b = -1/2
Now let's try if either actually works or if my guess was wrong...
Hum, it seems it doesn't work.
Stuck.

[bait]

Interesting in first case didnt just sub f(x)=0 into first eqn to get 0=xy for all pairs. Nice the soln is 1+/-x

[pedrojose]

That one I could solve. But it is very difficult, for me, to solve this kind of problem. Often I do not solve. Can anyone help me telling me where to study to improve for this kind of problem?

[jameyatesmauriat]

No enough explanation, just copy paste