Culegere de probleme de Analiz˘a numeric˘a

− 1
12 [f(xi+1,yi+1)−2f(xi,yi)+f(xi−1,yi−1)]
=
= yi + 4
12 [5f(xi+1,yi+1)+8f(xi,yi)−f(xi−1,yi−1)]
τi+1 = y(xi+1)−y(xi)
h
= h4
3!
f (3) (µi,y(µi))
(−1)
3!
3
yi+1 = yi +h
− 1
= yi +h
− 1
12 [5f(xi+1,yi+1)+8f(xi,yi)−f(xi−1,yi−1)] =
1
0
(−s+1)(−s)(−s−1)ds = − 1
24 h4 y (IV) (µi)
m = 4
f(xi+1,yi+1)− 1
2 ∇f(xi+1,yi+1)−
12 ∇2f(xi+1,yi+1)− 1
24 ∇3f(xi+1,yi+1)
f(xi+1,yi+1)− 1
2 [f(xi+1,yi+1)−f(xi,yi)]−
− 1
12 [f(xi+1,yi+1)−2f(xi,yi)+f(xi−1,yi−1)]−
− 1
24 [f(xi+1,yi+1)−3f(xi,yi)+3f(xi−1,yi−1)−f(xi−2,yi−2)]
= yi + h
24 [9f(xi+1,yi+1)+19f(xi,yi)−5f(xi−1,yi−1)+f(xi−2,yi−2)]
τi+1 = − 19
720 y(5) (µi)h 4
Problema 11.0.8 Deducet¸i următoarea formulă predictor-corector
y (0)
i+1 = yi−3 + 4h
3 [2f(xi,yi)−f(xi−1,yi−1)+2f(xi−2,yi−2)]
y (c)
τi+1 = 14
45 h4 y (5) (ξi), ξi ∈ (ti−1,ti+1) (Milne)
i+1 = yi−1 + h
3
[f(xi+1,y (p)
i+1 )+4f(xi,yi)+f(xi−1,yi−1)]
τi+1 = − h4
90 y(5) (ξi), ξi ∈ (ti−1,ti+1) (Simpson)
=
=
169