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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.

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ken
ken 2022-5-9
关闭： Dyuman Joshi 2023-12-14
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

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回答（1 个）

Nivedita
Nivedita 2023-12-14
Ran in:
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.

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