secant method matlab to solve x^2-5=0 in the interval [2 4]

Introduction : In this article we will write secant method matlab program to solve x^2-5 in the interval [2 4].

Question : Solve

\[x^2-5=0\]

in the interval [2, 4] by Secant method matlab.

Solution:

To solve

\[x^2-5=0\]

let \[f(x)=x^2-5\]

Secant method to find solution of \(f(x)=0\) is given below

\[x_{i}=x_{i-1}-f\left(x_{i-1}\right)\frac{\left(x_{i-1}-x_{i-2}\right)}{\left(f\left(x_{i-1}\right)-f\left(x_{i-2}\right)\right)} \]

for i =2,3,4,5,6,….n

provided two initial guesses \(x_0, x_1\) to the root are given.

%matlab program to solve x^2-5=0 with initial guesses x0=2, x1=4 with tolerance 10^(-8)
f=@(x) x^2-5 ;

x(1)=2;  
x(2)=4;
tolerance= 0.00000001; 
n=20; 

 
for i=3:n
   x(i) = x(i-1) - (f(x(i-1)))*((x(i-1) - x(i-2))/(f(x(i-1)) - f(x(i-2)))); %Secant formula 
   
    if abs((x(i)-x(i-1))/x(i))<tolerance
        root=x(i)
        
        break
    end
end
fplot(f,[2,4])
xlabel('x')
ylabel('f(x)')
title('graph of f(x)')

%output results of the program

root = 2.2361

Graph of f(x)