Culegere de probleme de Analiz˘a numeric˘a

= 1
5
h 12 h3f ′′
(µi,y(µi)) = 5
12 h2y ′′′ (µi,y(µi))
167
Pentru m = 3 avem
y(xi+1) ≈ y(xi)+h f(xi,y(xi))+ 1 5
∇f(xi,y(xi))+
2 12 ∇2
f(xi,y(xi)) =
= y(xi)+h{f(xi,yi)+ 1
2 [f(xi,y(xi))−f(xi−1,y(xi−1))]+
+ 5
12 [f(xi,y(xi))−2f(xi−1,y(xi−1))+f(xi−2,y(xi−2))]} =
= y(xi)+ 4
12 [23f(xi,yi)−16f(xi−1,y(xi−1))+5f(xi−2,yi−2)]
y0 = α, y1 = α1, y2 = α2
yi+1 = yi + h
12 [23f(xi,yi)−16f(xi−1,yi−1)+5f(xi−2,yi−2)]
h4f (3) (µi,y(µi))(−1) 3
1
0
f (3) (µi,y(µi)) = y (4) (µi)
τi+1 = y(xi+1)−y(xi)
4
−s
3
ds = 3h4
8 f(3) (µi,y(µi))
− 1
12 [23f(xi,y(xi))−hf(xi−1,y(xi−1))+
+5f(xi−2,y(xi−2))] = 1
4 3h
4 8 f(3) (µi,y(µi))
Pentru m = 4 obt¸inem
y(xi+1) = y(xi)+h
yi+1 = yi +h
= 3h3
8 y(4) (µi)
f(xi,yi)+ 1
2 ∇f(xi,y(xi))+
+ 5
12 ∇2f(xi,y(xi))+ 3
8 ∇3
f(xi,y(xi)) +
+h 5 f (4) (µi,y(µi))(−1) 4
1
0
−s
ds
4
f(xi,yi)+ 1
2 [f(xi,yi)−f(xi−1,yi−1)]+
+ 5
12 [f(xi,yi)−2f(xi−1,yi−1)+f(xi−2,yi−2)]+
+ 3
8 [f(xi,yi)−3f(xi−1,yi−1)+3f(xi−2,yi−2)−f(xi−3,yi−3)] =