Culegere de probleme de Analiz˘a numeric˘a

168 Rezolvarea numerică ecuat¸iilor diferent¸iale
= yi + h
24 [55f(xi,yi)−55f(xi−1,yi−1)+37f(xi−2,yi−2)−9f(xi−3,yi−3)]
h 5 f (4) (µi,y(µi))(−1) 4
1
−s
ds =
4
251
720 f(4) (µi,y(µi))
0
τi+1 = 251
720 f(4) y (5) (µi)
Observat¸ia 11.0.7 Am integrat polinomul lui Newton cu diferent¸e regresive cu
nodurile
(xi,y(xi)),(xi−1,y(xi−1)),...,(xi+1−n,y(xi+1−m))
pentrumpas¸i.
Pentru corectorul cumpas¸i vom folosi formula lui Newton cu diferent¸e regresive
(xi+1,f(xi+1)),(xi,f(xi)),...,(xi−m+1,f(xi−m+1))
Pm(x) =
1
dk =
0
yi+1 = yi +h
= yi +4
m
s+k −2
∇
k
k f(xi+1,y(xi+1))
k=0
yi+1 = yi +h
s+k −2
k
m
dk∇ k f(xi+1,y(xi+1))
k=0
ds = (−1) k
1
d0 = 1, d1 = − 1
2 , d2 = − 1
12
d3 = − 1
24 , d4 = − 19
720
s = x−xi
4
x = xi +sh−m ≤ s ≤ 0
xi+1 = xi +h−m+1 ≤ s ≤ 1
f(xi+1,yi+1− 1
2
m = 2
0
−s+1
k
ds
∇f(xi+1,yi+1)− 1
12 ∇2 f(xi+1,yi+1)
f(xi+1,yi+1)− 1
2 [f(xi+1,yi+1)−f(xi,yi)]−
=