N = 100;
format long
[A,b] = twodlaplace(N);
disp('Konditionszahl von A');
condest(A),
x = A \ b;

DL = tril(A); f = DL \ b;
U = triu(A,1);
xk = zeros((N-1)^2,1); err = []; res = [];
for k = 1:1000,
   xk = -DL \ ( U*xk ) + f;
   err = [err;norm(x-xk)];
   res = [res;norm(A*xk - b)];
end
semilogy(err,'r-'),
hold on
semilogy(res,'b-'),
hold off
makelarge
err(end) / norm(x),