Bisection method code matlab

Question

Maria Galle 2020-10-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/610846-bisection-method-code-matlab

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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Answer 1

Steven Lord 2020-10-12

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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)