Question

Write a program using the following root-finding methods: Mullers Method
Use your programs to find approximations to within 10^(-4) to all zeros of the following cubic polynomials.
Use |P_(n+1)-P_n| as a measure of the error in the iteration. Save all of the iterations. What are your conclusions?
(a) f(x) = x^3-5x^2 + 2x

(b) f(x) = x^3-2x^2-5

The program has to be used with MATLAB. I'm still learning how to use the program. I would love some help and tips on solving these methods. Thank you

Answer 1

INPUT endpoints a, b; tolerance TOL; the maximum number of iterations N₀ = log₂[(b-a)/TOL].
OUTPUT approximate solution p or message of failure.
Step 1 Set i = 1;
FA = f (a).
Step 2 While i ≤ N₀ do Steps 3–6.
Step 3 Set p = a + (b − a)/2; (Compute p_i)
FP = f ( p).
Step 4 If FP = 0 or (b − a)/2 < TOL then
OUTPUT (p); (Procedure completed successfully.)
STOP.
Step 5 Set i = i + 1.
Step 6 If FA · FP > 0 then set a = p; (Compute ai, bi.)
FA = FP
else set b = p. (FA is unchanged.)
Step 7 OUTPUT (‘Method failed after N0 iterations, N0 =’, N0);
(The procedure was unsuccessful.)
STOP

The actual code for the above question

% Find the root of f(x) using bisection method

f=@(x) x^3 -5*x^2+2*x; % If f(x) is different then we can change it accordingly.

a = 4; % Initially given the first point, It can change if we start with a different inial guess,

b = 5; % Initially given the second point, It can change if we start with different inial guess.

TOL = 10^(-4); % Given error tolerance.

N_o = log₂[(b-a)/ε]; % Calculation of number of iteration required from error bound equation.

if f(a)*f(b)>0

disp('Wrong choice of a and b; please change the value of and b such that f(a)*f(b)<0')

else

For i=1:N_o %total number of iteration is N_o.

c=(a+b)/2; %finding the mid point of a and b.

if f(c)*f(b)>0; %checking the sign of f(c) with respect to f(b)

b=c; %if sign of f(b) and f(c) are the same then change the value of endpoint b as c.

else a=c;%if sign of f(b) and f(c) are not the same then change the value of endpoint a as c.

end

end

fprintf('Root of given equation is %f',c)

% so we got the root of our function f(x).

This is a general code, which you can refer to for your reference.

Here key points -

f(x) = x^3-5x^2 + 2x

1. There is 3 root for the above cubic. It means we have to supply 3 initial pair of guess value, a & b.
2. The number of iteration is dependent on b-a.

f(x) = x^3-2x^2-5

1. It has only one real root. It means our bisection method doesn't solve for the imaginary root of the same and we have to provide only one pair of the initial guess.