Why is my Muller Methods program fail?

i tried to cunstruct muller method program for polinomial x^3-x^2+2*x-2 using initial guest x0=1.3 x1=1.5 and x2=2. When i try using excell, i get approriate result, x=1. Unfortunately, when i use my mathlab program, the result is different nor the expected roots of the polinomial. i hope someone can help me and explain whats wrong with my program.

From the deepest part of my heart, i'm kindly express my highest gratitude for your cooperation.

x=[1.3 1.5 2];
p=[1 -1 2 -2];
y=polyval(p,x);

es=0.0001;
ea=10;
while ea>es
    h0=x(2)-x(1);
    h1=x(3)-x(2);
    delta0=(y(2)-y(1))/h0;
    delta1=(y(3)-y(1))/h1;
    a=(delta1-delta0)/h1+h0;
    b=a*h1+delta1;
    c=y(3);
    rad=sqrt(b^2-4*a*c);
    if abs(b+rad)>abs(b-rad)
        den=b+rad;
    else 
        den=b-rad;
    end
    x3=x(3)+(-2*c)/den;
    ea=abs((x3-x(3))/x3);
    x(1)=x(2);
    x(2)=x(3);
    x(3)=x3;
end
x3

Answer 1

[David Hill]

Not sure if you are doing divided differences correctly.

x=[1.3 1.5 2];%input x values
p=[1 -1 2 -2];%input polynomial
y=polyval(p,x);
es=1e-10;
ea=10;
c=3;
while ea>es
   w=div_diff(p,x(c-2:c-1))+div_diff(p,x([c-2,c]))-div_diff(p,x(c-1:c));
   Rad=sqrt(w^2-4*y(c)*div_diff(p,x(c-2:c)));
   Div=[w+Rad,w-Rad];
   [~,idx]=max(abs(Div));
   Div=Div(idx);
   x(c+1)=x(c)-2*y(c)/Div;
   y(c+1)=polyval(p,x(c+1));
   ea=abs((x(c+1)-x(c))/x(c));
   c=c+1;
end
Anser=x(end);

function y=div_diff(p,x)
  if length(x)==2
     y=diff(polyval(p,x))/diff(x);
  else
     y=div_diff(p,x(2:end))-div_diff(p,x(1:2))/(x(2)-x(1));
  end
end

[Fauzanul Ardhi]

ooh, i see. i think its also because i evaluate y before while loop, so its only give value for the initial guest. thanks a lot, sir.