Culegere de probleme de Analiz˘a numeric˘a

9.5. Formule de cuadratură de tip Gauss 149
Problema 9.5.7 Deducet¸i o formulă de tip Cebâs¸ev pe [−1,1] cu w(x) = 1 s¸i cu
3 noduri.
Solut¸ie.
⎧
⎨
⎩
A = 2
3
t1 +t2 +t3 = 0
t2 1 +t2 2 +t2 3 = 1
t3 1 +t32 +t33 = 0
C1 = t1 +t2 +t3
C2 = t1t2 +t1t3 +t2t3
C3 = t1t2t3
C1 = 0
C2 = 1
2 [(t1 +t2 +t3) 2 −(t 2 1 +t22 +t23 )] = −1
2
C3 = 1
6 [(t1+t2+t3) 3 −3(t1+t2+t3)(t 2 1 +t2 2 +t2 3 )+2(t3 1 +t3 2 +t3 3
1
Deoarece
−1
2
3
t 3 −C1t 2 +C2t−C3 = 0
t 3 − 1
2 t = 0, t1
√
2
= −
2 , t2
√
2
= 0, t3 =
2
f(t)dt = 2
√ √
2 2
f − +f(0)+f +R3(f)
3 2 2
R3(f) =
1
−1
K3(t) = 1
(1−t)
6
4
−
4
2
3
3
(ti −t) 3 =
i=1
1
−1
K3(f)f (4) (t)dt
K3(t) = 1
6
3
i=1
(ti −t) 3 +
(x−t) 3 dx = (1−t)4
4
1
)] = (0−0+0) = 0
6
− (1+t)4
4