TU Wien:Computernumerik VU (Schranz-Kirlinger)/Übungen SS16/Beispiel 26

Lösung

Matlab Code

function Bsp6()
x = -8;
newX = x;
eps = 0.5;
p = [1 -3 3-eps^2 eps^2-1];


for i=1:3
    for j=1:20
        der = polyder(p);
        newX = newX - (polyval(p,newX)/polyval(der,newX));
    end
    
    fprintf('Nst = %0.10f\n', newX);
    p = deconv(p,(x-newX));
    %newX = (newX+eps);
    newX = (newX + eps);
end


end