what should we do in matlab code for newtons raphson method in the case f'(xn)=0 ?

This code doesnot work if f'(xn)=0 in some iteration so what should be the general code for Newtons raphson method?

Answer 1

[Maneet Kaur Bagga]

Hi,

As per my understanding from the given code, following are the errors which can cause runtime errors:

1. When using the Newton Raphson method if the derviative "f'(xn) = 0" at any iteration, the algorithm will attempt to divide the expression by zero leading to a runtime error.
2. If the derivative is zero, the while loop would become infinite as the "err" variable will never decrease.
3. If the function value "f(x(i+1)) is exactly zero, it breaks the loop without reporting about convergence or the number of iterations taken.
4. The convergence check is based on the error between iterations. You can add a check for the function value approaching zero to indicate of the method is convergin to the root.

To overcome the above edge cases you can check if the derivative is zero before computing the next approximation. Also, you can add a small pertubation to the point if the derivative is very close to zero to avoid division by zero.

Please refer to the modified code below for better understanding:

format long
f = @(x) x^3 - 2*x - 5;    
f1 = @(x) 3*x^2 - 2;        
tol = 0.001;                
i = 1;                      
x(i) = 2;                   
err = 1;                    

while err >= tol
    % Calculate the derivative at the current point
    derivative = f1(x(i));
    
    % Check if the derivative is close to zero
    if abs(derivative) < 1e-10   % Adjust tolerance as needed
        disp('Derivative is zero or very close to zero. Switching method or adjusting guess.')
        % Option 1: Adjust the guess slightly (perturbation)
        x(i) = x(i) + 1e-6; % Small perturbation
        derivative = f1(x(i)); % Recalculate derivative after perturbation
    end
    
    % Newton-Raphson formula
    x(i+1) = x(i) - (f(x(i)) / derivative);
    err = abs(x(i+1) - x(i));
    
    % Check for exact root
    if f(x(i+1)) == 0
        break;
    end
    
    i = i + 1;
end

% Display the result
fprintf('Root found: %f\n', x(end));
fprintf('Function value at root: %f\n', f(x(end)));
fprintf('Number of iterations: %d\n', i);

I hope this helps!