Culegere de probleme de Analiz˘a numeric˘a

6.6. Interpolare spline 101
⎡ ⎤
3(3−3π)
b = ⎣ ⎦
− 9
2
3(2−π)
3
40 (a1 −a0)−3f ′ (0) = 3
1 ·
6
1
−3π = 3(3−3π)
2
−3π − 3 1(−1)
= 6−3π = 3(2−π)
2
Problema 6.6.4 Fie f : [a,b] → R, f ∈ C 1 [a,b], a < 0, b > 1. Să se
scrie o funct¸ie spline naturală de interpolare care verifică s(0) = f(0), s ′ (0) =
f ′ (0), s(1) = f(1), s ′ (1) = f ′ (1).
Solut¸ie. Funct¸ia căutată este de forma
s(x) = pm−1(x)+
n ri
i=1
j=0
cij(x−xi) 2m−1−j
+
s(x) = a0 +a1x+c10x 3 +c11x 2 +c20(x−1) 3 +c21(x−1) 2 +
Avem 6 necunoscute s¸i 4 condit¸ii
s ′ (x) = a1 +3c10x 2 + +2c11x+ +3c20(x−1) 2 + +2c21(x−1)+
s(0) = a0 = f(0)
s ′ (0) = a1 = f ′ (0)
s(1) = f(0)+f ′ (0)+c10 +c11 = f(1)
s ′ (1) = f ′ (0)+3c10 +2c11 = f ′ (1)
s ′′ (1) = 0
s ′′ (x) = 6c10x+ +2c11x 0 + +6c20(x−1)+ +2c21(x−1) 0 +
3c10 +c11 +c21 = 0
s ′′′ (x) = 6c10x 0 + +6c20(x−1) 0 +
s ′′′ (1) = c10 +c20 = 0 c20 = −c10
c10 +c11 = f(1)−f(0)−f ′ (0)
3c10 +2c11 = f ′ (1)−f ′ (0)
c10 = 2f(0)+f ′ (0)−2f(1)+f ′ (1)