Question

Looking for Muller Method MatLab coding for these following equations:

f(x) = ?? - 2 f(x) = 23 – 3 f(0) = x+ - 0.06 f(0) = 3 + 3.22 - 1 f(1) = x3 – 22 – 24.0 – 32

This is the code I currently have but is not working

fun = @(x) x.^2-2;
x1 = 0;
x2 = 1;
tol = 0.0001;
kmax = 100;
function [x, y] = Muller(fun, x1, x2, tol, kmax)
x(1) = x1;
x(2) = x2;
x(3) = (x(2)+x(1))/2;
y(1) = feval(fun, x1);
y(2) = feval(fun, x2);
y(3) = feval(fun, x(3));
c(1) = (y(2)-y(1))/(x(2)-x(1));
for k = 3 : kmax
c(k-1) = (y(k)-y(k-1))/(x(k)-x(k-1));
d(k-2) = (c(k-1)-c(k-2))/(x(k)-x(k-2));
s = c(k-1)+(x(k)-x(k-1))*d(k-2);
x(k+1) = x(k)-2*y(k)/(s+sign(s)*sqrt(s^2-4*y(k)*d(k-2)));
y(k+1) = feval(fun, x(k+1));
if abs(x(k+1)-x(k)) < tol
disp('Muller method has converged'); break;
end
iter = k;
end
if iter >= kmax
disp('zero not found to desired tolerance');
end
end

Answer 1

clear all close all #1st function f=@(x) x.^2-2; xl=0;x2=1;tol=10^-3; kmax=1000 root=muller(f,x1,x2, tol, kmax); $2nd function f=@(x) x. *3-3; x1=0; x2=1;tol=10^-3; kmax=1000; root muller(f,x1, x2, tol, kmax); 3rd function f=@(x) x.4-0.06; xl=0;x2=1;tol=10^-3; kmax=1000; root=muller(f,x1,x2, tol, kmax); 84th function f=@(x) x. *3+3. *x.^2-1; x1=0; x2=1;tol=10^-3; kmax=1000; root=muller(f,x1,x2, tol, kmax); %5th function f=@(x) x. 3-x. 2-24. *x-32; xl=0;x2=1;tol=10^-3; kmax=1000; root=muller(f,x1,x2, tol, kmax); function for Muller method function root=muller (f,x1,x2, tol, kmax) terror value for initiation of while loop error=10; displaying the function fprintf('The function is f(x)=') disp(f) Loop for Muller's method x0=x1-0.5; cnt=0; while error>=tol function value corresponds to initial guess f0=f(x0); fl=f(xl); f2=f(x2); cnt-cnt+1; all other terms for Muller method h0=xl-x0; hl=x2-x1; del0=( f(x1)-f(x0) ) /h0; dell=(f(x2)-f(xl))/h1; a= (dell-de10)/(hl+h0); b=a*h1+dell;

c=f(x2); x31=x2+((-2*c)/(b+sqrt(b*b-4*a*c))); x32=x2+((-2*c)/(b-sqrt(b*b-4*a*c))); checking root closer to zero if abs(f(x31)) >=abs(f(x32)) x3=x32; else x3=x31; end percentage error after each iterations error=(abs ((x3-x2)/x3))*100; changing x values after each iteration x0=x1; x1=x2 ; x2=x3; if cnt>=kmax break end end root=x3; fprintf('\tThe root of the function is $f.\n',x3) end The function is f(x)= @(x)x. 2-2 The root of the function is 1.414214. The function is f(x)= @(x). *3-3 The root of the function is 1.442250. The function is f(x)= @(x)x. 4-0.06 The root of the function is -0.494923. The function is f(x)= (x) x. *3+3. *x. 2-1 The root of the function is -0.652 704. The function is f(x) = @(x)x. *3-x. 2-24. *x-32 ction is 70-532702-24. **- The root of the function is -1.619684. Published with MATLAB® R2018a

clear all
close all

%1st function
f=@(x) x.^2-2;
x1=0;x2=1;tol=10^-3; kmax=1000;
root=muller(f,x1,x2,tol,kmax);

%2nd function
f=@(x) x.^3-3;
x1=0;x2=1;tol=10^-3; kmax=1000;
root=muller(f,x1,x2,tol,kmax);

%3rd function
f=@(x) x.^4-0.06;
x1=0;x2=1;tol=10^-3; kmax=1000;
root=muller(f,x1,x2,tol,kmax);

%4th function
f=@(x) x.^3+3.*x.^2-1;
x1=0;x2=1;tol=10^-3; kmax=1000;
root=muller(f,x1,x2,tol,kmax);

%5th function
f=@(x) x.^3-x.^2-24.*x-32;
x1=0;x2=1;tol=10^-3; kmax=1000;
root=muller(f,x1,x2,tol,kmax);

%function for Muller method
function root=muller(f,x1,x2,tol,kmax)
    %error value for initiation of while loop
    error=10;
    %displaying the function
    fprintf('The function is f(x)=')
    disp(f)   
    %Loop for Muller's method
    x0=x1-0.5; cnt=0;
    while error>=tol
        %function value corresponds to initial guess
        f0=f(x0);
        f1=f(x1);
        f2=f(x2);
        cnt=cnt+1;
        %all other terms for Muller method
        h0=x1-x0;
        h1=x2-x1;

        del0=(f(x1)-f(x0))/h0;
        del1=(f(x2)-f(x1))/h1;

        a=(del1-del0)/(h1+h0);
        b=a*h1+del1;
        c=f(x2);

        x31=x2+((-2*c)/(b+sqrt(b*b-4*a*c)));
        x32=x2+((-2*c)/(b-sqrt(b*b-4*a*c)));

        %checking root closer to zero
        if abs(f(x31))>=abs(f(x32))
            x3=x32;
        else
            x3=x31;
        end
        %percentage error after each iterations
        error=(abs((x3-x2)/x3))*100;
        %changing x values after each iteration
        x0=x1; x1=x2;x2=x3;
        if cnt>=kmax
            break
        end
    end
    root=x3;
    fprintf('\tThe root of the function is %f.\n',x3)
end