Use the bisection method to approximate the first negative solution, the negative root that is closest to the origin. The accuracy must be of the order 10−4.

the given bisection code. and the given equation()

function [c, n, err] = bisection_method(f, a, b, tol, N)
    c = [];
    n = 0;
    err = inf;
    FA = f(a);
    FB = f(b);
    if(a > b)
        err = inf;
        c = [];
    elseif (FA*FB>= 0)
    else
        while ((abs(err) > abs(tol)) && (n <= N))
            n = n+1;
            c = (a + b) / 2;
            fmid = f(c);
            err = abs(fmid);
            if(fmid * f(a) > 0)
                a = c;       
            
            else
                b = c;
             
            end
        end
    end
end

[Torsten]

Anything wrong with the solution given here ?

https://de.mathworks.com/matlabcentral/answers/1714655-how-to-find-first-negative-solution-with-the-bisection-method?s_tid=srchtitle

Answer 1

[Nivedita]

Hi Ken,

I understand that you are trying to find the first negative solution using the bisection method, which is closest to the origin and has an accuracy of the order 10^(-4). We need to identify an interval [a, b] that brackets this root in a manner that "f(a)*f(b)<0". I have considered the interval [-1.5, -0.5] as the starting point for the bisection method.

Here is an updated version of the code to do the same:

%% Calling the function within the interval [-1.5, 0.5]
% Define the function f(x)
f = @(x) x + 1 - 2 * sin(pi * x);

% Define the initial interval [a, b]
a = -1.5;
b = -0.5;

% Define the tolerance
tol = 1e-4;

% Define the maximum number of iterations
N = 1000;

% Call the bisection_method function
[c, n, err] = bisection_method(f, a, b, tol, N);

% Display the result
if isnan(c)
    fprintf('The root was not found within the specified tolerance and iteration limit.\n');
else
    fprintf('The root is approximately at x = %.4f after %d iterations with an error of %.4f\n', c, n, err);
end

The root is approximately at x = -1.0001 after 14 iterations with an error of 0.0001

%% Function to find the first negative root using the bisection method
function [c, n, err] = bisection_method(f, a, b, tol, N)
    c = (a + b) / 2; % Initial midpoint
    n = 0;           % Iteration counter
    err = inf;       % Initial error
    FA = f(a);
    FB = f(b);
    
    % Check if the interval is valid
    if (FA * FB > 0)
        c = NaN; % Return NaN to indicate the root is not bracketed
        return;
    end
    
    % Bisection method loop
    while ((b - a) > tol) && (n < N)
        n = n + 1;
        c = (a + b) / 2;
        FC = f(c);
        
        if FC == 0
            % c is a root
            err = 0;
            break;
        elseif (FA * FC < 0)
            b = c;
            FB = FC;
        else
            a = c;
            FA = FC;
        end
        
        err = b - a; % Update the error estimate
    end
    
    if (n == N) && (err > tol)
        c = NaN; % Set c to NaN to indicate the root was not found within tolerance
    end
end

I hope this helps!

Regards,

Nivedita.