Question

How can I code this problem in MATLAB:

a) Find the approximations to within 10^-4 to all real zeros of the following polynomials using Newton's method.?

f(x)=x³ - 2*x²- 5.

b) Find approximations to within 10^-5 to all the zeros of each of the following polynomials by first
finding the real zeros using Newton’s method and then reducing to polynomials of lower degree to
determine any complex zeros.

f(x)=x⁴ + 5x³ - 9*x² - 85*x - 136.

Answer 1

Matlab code for finding roots using Newton methods clear all close all function for which root have to find fun-e (x) the function f(x)=\n' ) 2.x. 2-5; For xx-1inspace(-2,4); %initial guess x0 10 terror limit n 10^-4 root-newton_method(fun,x0, n); fprintf tRoot of the function using Newton method is %f.\n',root) function for which root have to find fun @(x) x.^4 + 5.*x.^3- 9.*x.^2 -85.*x -136; fprintf(\nFor the function f(x)\n' disp (fun) xx-linspace(-2,4); %initial guess x0 10 error limit n 10^-4 [rootll newton_method (fun,x0,n); fprintf\tlst Root of the function using Newton method is %f \n',rootl) %initial guess x0-10; terror limit n 10^-4 root2 newton_method(fun, x0,n); fprintf\t2 nd Root of the function us ing Newton method is %f. \n',root2) fprintf\tNow complex root occured for the equation is syms x p x.^4 5.*x.^3- 9.*x.^2 -85. *x -136; p2 x-root1)*(x-root 2); a, r quorem(p, p2); disp (vpa (q,2 ) ) fprintf So that the complex 0ts a fprintf(\t-2.5 +1.32287565553 229i \n\t-2.5 -1.32287565553229i\n\n ' ) tt %$ % 1

Matlab function for Newton Method root ] =newton_method(fun,x0,n) function syms x gl (x) diff ( fun, x) хх3x0; Loop for all intial guesses maxit-10000; for i 1st Derivative of this function initial guess] (xx-(fun(xx )./g1 (xx)) ) ; $Newton Raphs on Formula x2 d abs (fun/x ire X2)); Error сс; хX-X2; if cc<-n break end end root-xx; end 888888885 8388888 For the function f(x)= в (x)х. "3-2.*х. "2-5 Root of the function using Newton method is 2.69064 8 For the function f(x) = @ (х)х. ^4+5. *x. ^3-9. *x. ^2-85. *x-136 1st Root of the function using Newton method is 4.123 106 2nd Root of the function using Newton method is -4.123106 Now complex root occured for the equation is x*2 + 5.0*x + 8.0 So that the complex roots are -2.5 +1.322875655 5322 9i -2.5 -1.322875655 53229i Published with MATLAB®R2018a 2

%Matlab code for finding roots using Newton methods
clear all
close all

%function for which root have to find
fun=@(x) x.^3 - 2.*x.^2- 5;
fprintf('For the function f(x)=\n')
disp(fun)
xx=linspace(-2,4);
%initial guess
x0=10;
%error limit
n=10^-4;
[root]=newton_method(fun,x0,n);
fprintf('\tRoot of the function using Newton method is %f.\n',root)

%function for which root have to find
fun=@(x) x.^4 + 5.*x.^3- 9.*x.^2 -85.*x -136;
fprintf('\nFor the function f(x)=\n')
disp(fun)
xx=linspace(-2,4);
%initial guess
x0=10;
%error limit
n=10^-4;
[root1]=newton_method(fun,x0,n);
fprintf('\t1st Root of the function using Newton method is %f.\n',root1)

%initial guess
x0=-10;
%error limit
n=10^-4;
[root2]=newton_method(fun,x0,n);
fprintf('\t2nd Root of the function using Newton method is %f.\n',root2)

fprintf('\tNow complex root occured for the equation is ')
syms x
p = x.^4 + 5.*x.^3- 9.*x.^2 -85.*x -136;
p2 = (x-root1)*(x-root2);

[q, r] = quorem(p, p2);
disp(vpa(q,2))

fprintf('So that the complex roots are \n')
fprintf('\t-2.5 +1.32287565553229i \n\t-2.5 -1.32287565553229i\n\n')

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%Matlab function for Newton Method
function [root]=newton_method(fun,x0,n)
syms x
g1(x) =diff(fun,x);   %1st Derivative of this function
xx=x0;            %initial guess]
%Loop for all intial guesses
    maxit=10000;
    for i=1:maxit
        x2=double(xx-(fun(xx)./g1(xx))); %Newton Raphson Formula
        cc=abs(fun(x2));                 %Error
        err(i)=cc;
        xx=x2;
        if cc<=n
            break
        end
      
    end
    root=xx;
end

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