Demonstratie - metoda Gauss-Seidel
Contents
Consideram sistemul
Initializare
A = [2,1;5,7]; b = [4;19]; xn = [2;1];
Pregatirea matricelor metodei
M = tril(A) N = M -A
M = 2 0 5 7 N = 0 -1 0 0
Prima iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 1.500000000000000 1.642857142857143 ea = 0.642857142857143
A doua iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 1.178571428571429 1.872448979591837 ea = 0.321428571428571
Dupa 20 de iteratii
for k=1:19 xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); end xn ea er =ea/norm(xn,inf)
xn = 1.000000000569914 1.999999999592918 ea = 1.025845408619830e-09 er = 5.129227044143157e-10