Culegere de probleme de Analiz˘a numeric˘a

9.1. Derivare numerică 123
(Rmf)(x) = hm+1 q [m+1]
f ′ (x) ≈ (Lmf) ′ (x) = 1
h
(m+1)! f(m+1) (ξ) ξ ∈ (a,b)
m (−1) m−i [m+1] d q
f(xi)
i!(m−i)! dq q −i
i=0
(Rmf) ′ (x) = hm+1
(m+1)! f(m+1) (ξ) d
dq qm+1 + hm+1 d
q[m+1]
(m+1)! dq f(m+1) (ξ)
(Rmf) ′ (xi) = (−1) m−ihmi!(m−i)! (m+1)! f(m+1) (ξi)
m = 2 (3 puncte)
(L2f)(x) = 1
2 f(x0)(q −1)(q −2)−f(x1)q(q −2)+ 1
2 f(x2)q(q−1)
(L2f) ′ (x) = 1
1
h 2 f(x0)(2q −3)−(2q−1)f(x1)+ 1
2 f(x2)(2q−1)
f ′ (x0) = 1
1
[−3f(x0)+4f(x1)−f(x2)]+
2h 3 h2f ′′′ (ξ0)
f ′ (x1) = 1 1
[−f(x0)+f(x2)]−
2h 6 h2f ′′′ (ξ1)
f ′ (x2) = 1
1
[f(x0)−4f(x1)+3f(x2)]+
2h 3 h2f ′′′ (ξ2)
m = 3 4 puncte
(L3f) ′ (x) = 1
−
h
1
6 f(x0)[(q −1)(q −2)(q−3)] ′ +
+ 1
2 f(x1)[q(q −2)(q−3)] ′ − 1
2 f(x2)[q(q −1)(q−3)] ′ +
+ 1
6 f(x2)[q(q −1)(q−2) ′
]
f ′ (x0) = 1
h3
[−11f(x0)+18f(x1)−9f(x2)+2f(x3)]−
64 4 f(4) (ξ0)
f ′ (x1) = 1
h3
[−2f(x0)−3f(x1)+6f(x2)−f(x3)]+
6h 12 f(4) (ξ1)
f ′ (x2) = 1
h3
[f(x0)−6f(x1)+3f(x2)+2f(x3)]−
6h 12 f(4) (ξ2)
f ′ (x3) = 1
h3
[−2f(x0)+9f(x1)−18f(x2)+11f(x3)]+
6h 4 f(4) (ξ3)
m = 4 (5 puncte)
f ′ (x0) = 1
h4
[−25f(x0)+48f(x1)−36f(x2)+16f(x3)−3f(x4)]+
12h 5 f(5) (ξ0)
f ′ (x1) = 1
h4
[−3f(x0)−10f(x1)+18f(x2)−6f(x3)+f(x4)]−
12h 20 f(5) (ξ1)
f ′ (x2) = 1
h4
[f(x0)−8f(x1)+8f(x3)−f(x4)]+
12h 30 f(5) (ξ2)