Culegere de probleme de Analiz˘a numeric˘a

6.1. Interpolare polinomială 81
Cazuri particulare
1)rk = 0, k = 0,n Lagrange
2)n = 0, r0 = n Taylor
3)r0 = ··· = rn = 1 formula lui Hermite cu noduri duble
(H2m+1f)(x) =
f = H2m+1f +R2m+1f
m
hk0(x)f(xk)+
k=0
hx0(x) = uk(x)
uk(xk)
4) Dacă m = 1, x0 = a, x1 = b
m
hk1(x)f ′ (xk)
k=0
1−(x−xk) u′ k (xk)
uk(xk)
hk1(x) = (x−xk) uk(x)
uk(xk)
r0 = m, r1 = n
n+1 m
x−b (x−a)
(Hm+n+1f)(x) =
a−b
i=0
i
m−i
ν
n+ν x−a
f
i! ν b−a
ν=0
(i) (a)+
m+1 n
x−a (x−b)
+
b−a
j=0
j
n−j
µ
m+µ x−b
f
j! µ a−b
µ=0
(j) (b)
unde
xk ∈ [a,b], k = 0,m, xi = xk (i = j)
rk ∈ N, Ik ⊆ {0,1,...,rk}, k = 0,m
f : [a,b] → R ∃f (j) (xk) k = 0,m, j ∈ In
n = |I0|+···+|Im|−1
(Bnf)(x) =
m
bkj(x)f (j) (xk)
k=0 j∈Ik
f = Bnf +Rnf (formula de interpolare a lui Birkhoff)
Dacă f ∈ C n+1 [a,b] atunci
(Rnf) =
b
ϕn(x;s) = 1
(x−s)
n!
n + −
a
ϕn(x,s)f (n+1) (s)ds
m
k=0 j∈Ik
bkj(x)[(xk −s) n + ](j)