3. Nelinearni sistemi

Jacobijeva iteracija in testni primer
Če vzamemo začetni približek x (0) = (0.4, 0.1, −0.4), potem iz
dobimo zaporedje
x (r+1)
1
= 1 3
x (r+1)
2
= 1 9
x (r+1)
3
= − 1
20
(
cos(x (r)
1 x(r)
√
(x (r)
2 ) + 0.6 )
1 )2 + sin(x (r)
3
) + 1.1 − 0.1
(
)
e −x(r) 1 x(r) 2 + 9.1 ,
r
3
‖x (r) − x (r−1) ‖ ∞ ‖F (x (r) )‖ ∞
1 0.533066702220 0.003672183829 −0.503039471957 1.3 · 10 −1 3.7 · 10 −1
2 0.533332694686 0.005530371351 −0.504902219788 1.9 · 10 −3 1.3 · 10 −3
3 0.533331883381 0.005451521829 −0.504852740886 7.9 · 10 −5 4.2 · 10 −5
4 0.533331924436 0.005454006133 −0.504854837609 2.5 · 10 −6 1.8 · 10 −6
5 0.533331923152 0.005453901274 −0.504854771543 1.0 · 10 −7 5.6 · 10 −8
6 0.533331923206 0.005453904579 −0.504854774331 3.3 · 10 −9 2.4 · 10 −9
7 0.533331923204 0.005453904439 −0.504854774243 1.4 · 10 −10 7.5 · 10 −11
8 0.533331923204 0.005453904444 −0.504854774247 4.4 · 10 −12 3.2 · 10 −12
x (r)
1
x (r)
2
x (r)
Bor Plestenjak - Numerična analiza 2004