Newton Raphson!but with multiple initial values :/

This works for one initial value. Now I want to do the exact thing but for 2 more initial values. Where they all are x0= [0.5,1.5,2.5]; and the outputs will be r = [r1 , r2 , r3]. Any help would be appreciated, thanks.

r1 = newtonr()
function[r]= newtonr
 i = 0; 
 x0=0.5; 
 n = 5;
 
 while i < 5
      f = x0^4 - 9*x0^3+29*x0^2-39*x0+18;
 df= 4*x0^3 - 27*x0^2 +58*x0 -39;
     xnew= x0 - f/df;
     i= i+1
     x0=xnew; 
     disp(x0); 
     if i== 5
         r= xnew;
         disp('root is')
         disp(x0)
         break 
     end 
     
 end 
end
     

Answer 1

[David Hill]

 i = 0; 
 x=[0.5,1.5,2.5]; 
 n = 5;
 f = @(x)x.^4 - 9*x.^3+29*x.^2-39*x+18;
 df= @(x)4*x.^3 - 27*x.^2 +58*x -39;%functions should not be in loop
 while i < 5
     x= x - f(x)./df(x);
     i= i+1
     disp(x); 
 end 
 disp('roots are: ');
 disp(x);

Answer 2

[Michael Soskind]

Hi Alex,

Edited: Please note, David Hill's answer is the one I recommend.

As a note:

To make calculations work on multiple values within an array, the operator should be preceded with a '.'

This is the case for multiplication, division, and raising to a power.

This is the main modification that David makes in his code, as well as implementing functions for the function and its derivative. This is in general better practice.