Bisection method code matlab

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Maria Galle
Maria Galle 2020-10-11
回答： Steven Lord 2020-10-12
I'm trying to write a code using the bisection method to find the root but I'm not sure I wrote it correctly.
t=4;
c_d=0.25; %kg/m
g=9.81;
v=36;
maxit=15;
k = 1;
n_bi(k) = 1; % iteration
x_l_bi(k) = 50; % initial guess lower bound
x_u_bi(k) = 200; % initial guess upper bound
f_xl_bi(k) = sqrt(g*x_l_bi(k)/c_d).*tan(tan(sqrt(g*c_d./x_l_bi(k))*t)) - v;
f_xu_bi(k) = sqrt(g*x_u_bi(k)/c_d).*tan(tan(sqrt(g*c_d./x_u_bi(k))*t))- v;
x_r_bi(k) = (x_l_bi(k) + x_u_bi(k))/2;
product_f_bi(k) = f_xl_bi(k)*x_r_bi(k);
ea_bi(k) = 1*100; % initial error setting in %
while ea_bi(k) > 0.005 & n_bi < maxit
if product_f_bi(k) < 0
x_l_bi(k+1) = x_r_bi(k);
x_u_bi(k+1) = x_u_bi(k);
else
x_l_bi(k+1) = x_l_bi(k);
x_u_bi(k+1) = x_r_bi(k);
end
n_bi(k+1) = k+1;
f_xl_bi(k+1) = sqrt(g*x_l_bi(k+1)/c_d).*tan(tan(sqrt(g*c_d./x_l_bi(k+1))*t))- v;
f_xu_bi(k+1) = sqrt(g*x_u_bi(k+1)/c_d).*tan(tan(sqrt(g*c_d./x_u_bi(k+1))*t))- v;
x_r_bi(k+1) = (x_l_bi(k+1) + x_u_bi(k+1))/2;
product_f_bi(k+1) = f_xl_bi(k+1)*x_r_bi(k+1);
ea_bi(k+1) = abs((x_r_bi(k+1) - x_r_bi(k))/x_r_bi(k+1))*100;
k = k+1;
end
result_bi = [n_bi; x_l_bi; x_u_bi; x_r_bi; ea_bi];
disp([' ']);
disp(['Bisection Method:']);
fprintf('%10s %12s %12s %12s %10s\r\n','iteration','x_l', 'x_u', 'x_r', 'ea');
fprintf('%10.0f %12.4f %12.4f %12.4f %10.2f\n', result_bi);

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回答（1 个）

Steven Lord
Steven Lord 2020-10-12
Rather than hard-coding the function you want to solve, you might want to create a function handle to the function and use that function handle to evaluate the function. That way you could specify a function from an example in your textbook (I'm assuming this is a homework assignment) and compare the results of your code with the textbook.
n = 3;
f = @(x) x.^n;
% This returns the cubes of the numbers 0 through 4
y = f(0:4)
f = @(x) 2*x+3;
% Now this is two times the numbers 0 through 4 plus 3.
% Same exact evaluation code, different function
y = f(0:4)

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