Newton-Raphson fractal script problem

matlab doesn't give results to this script and keeps running it non stop (i waited for +10 min)

any thoughts ?

f = @(z) z.^3-1;
df = @(z) 3*z.^2;
e = 0.00001;
x=-3:0.001:3;
y=x*1i;
[r,k]=ndgrid(x,y);
z=r+k; % setting a complexe plane 

for i=1:6000
    for j=1:6000
        while f(z(i,j))<e %newton methode
        dorime(i,j)=z(i,j)-f(z(i,j))/df(z(i,j));

        end
    end
end
        %coloring in white,black,grey 
     if  abs(dorime(i,j)-(-1+sqrt(3))/2) <e
         fractale (i,j)=1;
     end
       if dorime (i,j)-1<e
         fractale (i,j)=1/2;
       end
         if  abs(dorime(i,j)-(-1-sqrt(3))/2) <e
         fractale (i,j)=0;
         end
% showing the fractal
        imshow(fractal);

Answer 1

[Matt J]

I would recommend doing the Newton's method iterations as below.

f = @(z) z.^3-1;

df = @(z) 3*z.^2;

e = 0.00001;

x=-3:0.001:3;

y=x*1i;

[r,k]=ndgrid(x,y);

z=r+k; % setting a complexe plane

maxiter=15;

iter=0;

while any( abs ( f(z) ) > e ,'all') && iter < maxiter %newton methode

z=z-f(z)./df(z);

iter=iter+1

end

dorime=z;

The rest of the code, I don't understand. You seem to think z.^3-1 will have the 3 real roots,

z = [ (-1-sqrt(3))/2) 1 (-1+-sqrt(3))/2)]

but it doesn't

roots([1 0 0 -1])

ans = 
  -0.5000 + 0.8660i
  -0.5000 - 0.8660i
   1.0000 + 0.0000i

[hamza kharbouch]

<< You seem to think z.^3-1 will have the 3 real roots,

>> oh my bad . a typo

thank u very much

[Matt J]

You're welcome, but if the answer works for you, please Accept-click it.