Question

script for Secant Method. Using Matlab or Octave.

The data:

y = x.^2 - 1.2

x_lower = 0.4

x_upper = 0.6
Es = 0.5*10^(2-n)

n = 5

Answer 1

Matlab Program-1:

a=input('Enter given function:','s');
f=inline(a)

x(1)=input('Enter the first point of guess interval: ');
x(2)=input('Enter the second point of guess interval: ');
n=input('Enter the allowed Error in calculation: ');
iteration=0;

for i=3:1000
x(i) = x(i-1) - (f(x(i-1)))*((x(i-1) - x(i-2))/(f(x(i-1)) - f(x(i-2))));
iteration=iteration+1;
if abs((x(i)-x(i-1))/x(i))*100<n
root=x(i)
iteration=iteration
break
end
end

Result :

Enter function:x.^2 -1.2

f =

Inline function:
f(x) = x.^2 -1.2

Enter the first point of guess interval: 0.4
Enter the second point of guess interval: .6
Enter the allowed Error in calculation: 0.5*10^-3 ( Es = 0.5*10^(2-n) where n = 5

root =

1.0954

iteration =

6

or

Matlab program- 2:

function secant_method()
f = @(x) x^2 - 1.2;
n=5;
eps = 0.5.*10^(2-n);
x0 =0.6; x1 = x0 - 0.001;
[solution,no_iterations] = secant(f, x0, x1, eps);
if no_iterations > 0 % Solution found
fprintf('Number of function calls: %d\n', 2 + no_iterations);
fprintf('A solution is: %f\n', solution)
else
fprintf('Abort execution.\n')
end
end

function [solution,no_iterations] = secant(f, x0, x1, eps)
f_x0 = f(x0);
f_x1 = f(x1);
iteration_counter = 0;
while abs(f_x1) > eps && iteration_counter < 100
try
denominator = (f_x1 - f_x0)/(x1 - x0);
x = x1 - (f_x1)/denominator;
catch
fprintf('Error! - denominator zero for x = \n', x1)
break
end
x0 = x1;
x1 = x;
f_x0 = f_x1;
f_x1 = f(x1);
iteration_counter = iteration_counter + 1;
end
% Here, either a solution is found, or too many iterations
if abs(f_x1) > eps
iteration_counter = -1;
end
solution = x1;
no_iterations = iteration_counter;
end

Result:

Number of function calls: 6
A solution is: 1.095563