Culegere de probleme de Analiz˘a numeric˘a

6.5. Interpolare rat¸ională 93
Tratând problema ca pe o PIB cu m = 2, I0 = {0}, I1 = {1}, I2 = {2}
obt¸inem
b00(x) = 1 b11(x) = x b22(x) = x2
2 −hx
(R3f)(x) =
2h
0
ϕ2(x;s)f ′′′ (s)ds
ϕ2(x;s) = 1
2! {(x−s)2 −b00(x)(0−s) 2 + −b11(x)[(h−s) 2 + ]′ −b22(x)[(2h−s) 2 + ]′′ }
unde
= 1
= 1
2
2 [(x−s)2 + −2x(h−s)+ −(x 2 −2hx)(2h−s) 0 + ]
⎧
⎪⎨
⎪⎩
s 2 x ≥ s s < h
s 2 +2x(h−x) x ≥ s s > h
x(2s−x) x < s s < h
−x(x−2h) x < s s > h
ϕ2(x;s) ≥ 0
Putem aplica corolarul la teorema lui Peano
E(x) = x3
6
∃ξ ∈ [0,2h] a.î. (R3f)(x) = E(x)f ′′′ (ξ),
1
−
2 h2b11(x)−24b22(x) = x3
6 − h2x 2 −2h
2 x
2 −hx
= x3
6 − h2 x
2 −hx2 +2h 2 x = x3
6 −hx2 + 3h2
2 x
6.5 Interpolare rat¸ională
Problema 6.5.1 Să se determine o aproximare Padé de grad 5 cu n = 2, n = 3
pentruf(x) = e x .
Solut¸ie.
r(x) = pn(x)
qm(x) , p ∈ Pn, q ∈ Pm
f (k) (0)−r (k) (0) = 0, k = 0,N, N = n+m = 5
f(x)−r(x) = f(x)− p(x)
q(x)
= f(x)q(x)−p(x)
q(x)
=