Culegere de probleme de Analiz˘a numeric˘a

98 Interpolare
3
i=1
bix r i = 0, r = 0,m−1, m = 2
s ′′
i(x) = 6b1x+ +6b2
s ′′
i
x− 1
+6b3(x−1)+
2 +
(0) = s′′ (1) = 0
s ′′′
i (x) = 6(b1 +b2 +b3) = 0 ⇒ b1 +b2 +b3 = 0 (x ≥ 1)
i
s ′′
i (0) = 0
s ′′
i (1) = 6b1 +3b2 = 0
b2 = −2b1
b1 +b2 +b3 = 0 ⇒ b3 = b1
si(x) = a0 +a1x+b x 3 + −2
x− 1
3 2
s1(0) = a0 = 1
1
s1 = 1+
2
a1 1
+b· = 0
2 8
s1(1) = 1+a1 +b 1− 1
= 0
4
s1(x) = 1− 5
2 x+2
x 3
+ −2 x− 1
3 2
s2(0) = a0 = 0
1
s2 =
2
a1 b
+ = 1
2 8
s2(1) = a1 +b 1− 1
= 0
4
s2(x) = 3x−4 x 3 + −2
x− 1
3 2
s3
s3(0) = a0 = 0
1
=
2
a1 b
+ = 0
2 8
s3(1) = a1 + 3b
4
= 1
+(x−1) 3 +
+(x−1)
+
3 +
+(x−1)
+
3 +