Question

script for Secant Method. Using 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

Execute the following MATLAB code:

syms x
y=inline(x^2-1.2); %Define function y

x(1)=0.4; %x_lower
x(2)=0.6; %x_upper

n=5;
Es=0.5*10^(2-n); %Allowed error
iter=0; %Initiate the value of iteration with zero

for i=3:1000
x(i) = x(i-1) - (y(x(i-1)))*((x(i-1) - x(i-2))/(y(x(i-1)) - y(x(i-2))));
iter=iter+1;
if abs((x(i)-x(i-1))/x(i))*100<Es
root=x(i); %root in fraction
vpa(root,4) %root into decimal from fraction
iter=iter %number of iterations
break
end
end

Obtain the following output:

ans =

1.095

iter =

6

Conclusion:

The value of "x" is 1.095 using secant method.