Demonstratie - metoda lui Jacobi
Contents
Consideram sistemul
Initializare
A = [2,1;5,7]; b = [4;19]; xn = [2;1];
Pregatirea matricelor metodei
M = diag(diag(A)) N = M -A
M = 2 0 0 7 N = 0 -1 -5 0
Prima iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 1.500000000000000 1.285714285714286 ea = 0.500000000000000
A doua iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 1.357142857142857 1.642857142857143 ea = 0.357142857142857
Dupa 25 de iteratii
for k=1:23 xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); end xn ea er =ea/norm(xn,inf)
xn = 1.000002153146738 1.999996924076088 ea = 2.153146738237410e-06 er = 1.076575024850136e-06