Corrigé du test Équations fonctionnelles

f
x=y= 0 f(0) = 0
x=−y6= 0 0 = f(0) = 2x(f(x) + f(−x)) f(−x) =
−f(x)f
y−y
x, y f(x2−y2) = (x−y)(f(x)+f(y)) f(x2−y2) = (x+y)(f(x)−f(y))
xf(y) = yf(x)y= 1
f(x) = xf(1) f
x7→ x2
f k
x7→ kf(x)
f
x=y= 0 f(0) = 0 y
x= 0 f(y) + f(−y) = 2f(x)f
y=nx f((n+ 1)x) = 2f(x)+2f(nx)−f((n−1)x)
x n
f(2x) = 4f(x)f(3x) =
9f(x)
x∈Rn>1k∈ {0,1, . . . , n}
f(kx) = k2f(x)
f((n+1)x) = 2f(x)+2f(nx)−f((n−1)x) = (2+2n2−(n−1)2)f(x) = (n+1)2f(x)
n+ 1 f
x n
f(nx) = n2f(x)
f(1)
f(1) = a
f(n) = an2n
p, q q 6= 0
ap2=f(p) = fµq×p
q¶=q2fµp q¶

f(x) = ax2x
Q R f
x(xn)x
f x f(x) = limn→+∞f(xn) = ax2
x7→ ax2a
f
f a, b ∈N
f(a) = f(b) 3a=f(f(f(a))) + f(f(a)) + f(a) = f(f(f(b))) + f(f(b)) + f(b) =
3b a =b
n= 0 f(0) f(f(0)) f(f(f(0))) N
f(f(f(0))) + f(f(0)) + f(0) = 0 f(0) = 0
n= 1 f(1) f(f(1)) f(f(f(1))) N?
f(0) = 0 f f(f(f(1))) + f(f(1)) + f(1) = 3
f(1) = 1
n>1
p∈ {0, . . . , n −1}f(p) = p f f(n)
f(f(n)) f(f(f(n))) n f(f(f(n)))+ f(f(n))+
f(n)>3n
f(n) = n