Displaying the answer of Newton's method for multiple roots?

Hello. I'm looking for a way to display my solutions for a Newton's method. There are 4 roots in my interval.

roots=[-0.0505 1.3232 1.3636 2.3333] %initial guesses corresponding to the 4 roots.

The trouble i'm having is that the script only shows the results for first root using -0.0505, while m reaches the value 4. I don't know why it cant display the other results for m=2,3,4. What could be the problem? Thanks!

m=1;

while m<=length(roots)
    x=roots(m);  % initial guess for the root
    tol=10e-8;
    fprintf('  k        xk            fx           dfx  \n');
    for k=1:50  % iteration number
        fx=sin(x^(2))+x^(2)-2*x-0.09;
        dfx=(2*(x*cos(x^2)+x-1));
        fprintf('%3d  %12.8f  %12.8f  %12.8f \n', k,x,fx,dfx);
        x=x-fx/dfx;
        m=m+1;
        if abs(fx/dfx)<tol
            return;
        end
    end
end
RUN
k        xk            fx           dfx  
1   -0.05050505    0.01611162   -2.20201987 
2   -0.04318831    0.00010707   -2.17275307 
3   -0.04313903    0.00000000   -2.17255596

Answer 1

[Sophie]

for k=1:50  
        fx=sin(x^(2))+x^(2)-2*x-0.09;
        dfx=(2*(x*cos(x^2)+x-1));
        fprintf('%3d  %12.8f  %12.8f  %12.8f \n', k,x,fx,dfx);
        x=x-fx/dfx;
        m=m+1;

You increase m on each iteration step.

[Sophie]

Maybe the best option is to seperate Newton and main code

function [k,x,fx,dfx]=Newton(initialguess)
x=initialguess;
k=0;fx=0;dfx=0;
    tol=10e-8;
    for k=1:50  % iteration number
        fx=sin(x^(2))+x^(2)-2*x-0.09;
        dfx=(2*(x*cos(x^2)+x-1));
        x=x-fx/dfx;

          if abs(fx/dfx)<tol
              return;
          end
    end
end

Main program

m=1;
  fprintf('  k        xk            fx           dfx  \n');
roots=[-0.0505 1.3232 1.3636 2.3333];
while m<=length(roots)
k=0;x=0;fx=0;dfx=0;
[k,x,fx,dfx]=Newton(roots(m));
      fprintf('%3d  %12.8f  %12.8f  %12.8f \n', k,x,fx,dfx);
      m=m+1;
end

There is a mistake in the code above, but the idea is clear, i think.

[Sophie]

I've found the mistake. Here Newton function returns only results from the last iteration. So that you have to make k,x,fx,dfx in the functions vectors.

[Roger L]

Thanks Sophie. It turns out that the loop wasn't working because of the "return;" command in the if loop. I changed that to a break command and it fixed the code.

Answer 2

[Sophie]

Obtained solution: Main code

m=1;
  fprintf('  k        xk            fx           dfx  \n');
roots=[-0.0505 1.3232 1.3636 2.3333];
while m<=length(roots)
k=[];x=[];fx=[];dfx=[];
[k,x,fx,dfx]=Newton(roots(m));
for i=1:length(k)
      fprintf('%3d  %12.8f  %12.8f  %12.8f \n', k(i),x(i),fx(i),dfx(i));
      end
      m=m+1;
 fprintf('\n');
end

Newton

function [kk,x,fx,dfx]=Newton(initialguess)
x=initialguess;
kk=[];fx=[];dfx=[];
    tol=10e-8;
    for k=1:50  % iteration number
    kk(end+1)=k;
        fx(end+1)=sin(x(end)^(2))+x(end)^(2)-2*x(end)-0.09;
        dfx(end+1)=(2*(x(end)*cos(x(end)^2)+x(end)-1));
        x(end+1)=x(end)-fx(end)/dfx(end);
        if abs(fx(end)/dfx(end))<tol
            return;
end
end
end