Question

Determinati f: N->N, cu proprietatea [f(n-1)]^2=2f(n-1)-1, irice ar fi n apartineN*

Answer 1

Fie f(n-1)=y ⇒ y**2 - 2y +1 = 0 ⇔ (y-1)**2 = 0 ⇒ I y-1 I = 0 ⇒ y=1
Deci functia cautata este functia constanta f(n)=1, ∀ n∈N*

Answer 2

[f(n-1)]^2-2f(n-1)+1=0
 [f(n-1)-1]²=0
f(n-1)=1 ∀n∈N
deci f(1-1)=f(0)=1
f(2-1)=f(1)=1
f(3-1)=f(2)=1
n∈N*
n-1∈N
deci f(n)=1  f(n) :N->{1} f(n)=1