Culegere de probleme de Analiz˘a numeric˘a

86 Interpolare
h02(x) = x(x−1)3 (x+1) 2
16
h10(x) = (1−x 2 ) 3
h20(x) = x(x+1) 3
1 5(x−1)
− +
8 16
(x+1)2
2
h21(x) = x(x+1) 3
1 3(x−1)
(x−1) −
8 16
h22(x) = x(x+1)3 (x−1) 2
16
Problema 6.3.2 Aceeas¸i problemă, pentru aceleas¸i noduri ca mai sus, dar duble.
Solut¸ie.
r0 = r1 = r2 = 1, m = 2, n = 5, x0 = −1, x1 = 0, x2 = 1
(H2m+1f)(x) =
u0(−1) = 4 u ′ 0
m
hk0(x)f(xk)+
k=0
hk0(x) = uk(x)
uk(xk)
m
hk1(x)f ′ (xk)
k=0
1−(x−xk) u′ k (xk)
uk(xk)
hn1(x) = (x−xn) uk(x)
uk(xk) u0(x) = x 2 (x−1) 2
h00(x) = x2 (x−1) 2
(x) = 2x(x−1)(x−1+x) = 2x(x−1)(2x−1)
1−
4
12
4 (x+1)
= x2 (x−1) 2
(−3x−2)
4
h01(x) = (x+1)x2 (x−1) 2
4
u1(x) = (x+1) 2 (x−1) 2
u ′ 1(x) = 2(x+1)(x−1) 2 +2(x+1) 2 (x−1) =
= 2(x+1)(x−1)(x−1+x+1) = 4x(x−1)(x+1)
h10(x) = (x+1)2 (x−1) 2
[1−x·0] = (x+1)
1
2 (x−1) 2
h11(x) = (x+1) 2 (x−1) 2 x
u2(x) = (x+1) 2 x 2 u2(1) = 4