Newton - Verfahren f\374r nichtlineare Gleichungssysteme :restart : with(LinearAlgebra) :n:=3 :F:=(x1,x2,x3) -> evalf(Vector([ 3*x1 - cos(x2*x3) - 0.5 , x1^2 - 81*(x2+0.1)^2 +sin(x3) + 1.06 , exp(-x1*x2) + 20*x3 + (10*Pi-3)/3 ]) );

JSQuLi5H

J:=(x1,x2,x3) -> Matrix([ [ 3 , x3*sin(x2*x3) , x2*sin(x2*x3) ], [ 2*x1 , -162*(x2+0.1) , cos(x3) ], [ -x2*exp(-x1*x2) , -x1*exp(-x1*x2) , 20 ] ]) ; P:=< 0.1,0.1,-0.1 > ;

JSQuLi5HNiM+JSJQRy0lJ1JUQUJMRUc2JSIpKVtScSMtJSdNQVRSSVhHNiM3JTcjJCIiIiEiIkYtNyMkRjBGMCYlJ1ZlY3Rvckc2IyUnY29sdW1uRw==

Ein Newton - Schritt

eps:=10^(-8) : maxi:=1000: Pold:=P :

for j from 1 by 1 while (j <= 25 and maxi>=eps) do P:= P - MatrixInverse(J(P[1],P[2],P[3])) . F(P[1],P[2],P[3]) ; maxi:=0: printf("(%2.1d ) ", j ) ; printf("%15.8f\n", evalf(P , 8) ) ; for m from 1 by 1 while m <= n do maxi:= max( maxi , abs(P[m]-Pold[m]) ) ; end do : Pold:= P ; end do : Transpose( F(P[1],P[2],P[3]) );

( 1 ) 0.49986967 0.01946685 -0.52152047( 2 ) 0.50001424 0.00158859 -0.52355696( 3 ) 0.50000011 0.00001244 -0.52359845( 4 ) 0.50000000 0.00000000 -0.52359878( 5 ) 0.50000000 -0.00000000 -0.52359878NiMtJSdSVEFCTEVHNiUiKUdMRD8tJSdWRUNUT1JHNiM3JSQiIiFGLEYrRismJSdWZWN0b3JHNiMlJHJvd0c=