Bisection method (figuring out theta)

clc; clear;
%% Initialize
w = 30; h = 30; c = 3.5;   % In inches
 
%% Bisection 
    
a= 49.04; b = 50.5;
f = @(theta) w*sind(theta) - h*cosd(theta) - c;
err = 0.477; ii = 0;
ferr=0.176;
if f(a)*f(b)>0
    disp('Change initial values')
elseif abs(f(a))<ferr
    display('Your initial guess is correct')
elseif abs(f(b))<ferr
    display('Your second guess is correct')
    
else
    xmid= (a + b)/2;
    err_bisect = abs(f(xmid));
    while err_bisect > err  
        if (f(a))*f(xmid)<0
            b = xmid;
        else
            a = xmid;
        end
        xmid = (a + b)/2;
        err_bisect =abs(f(xmid));
        ii = ii + 1;
    end
end
fprintf("After %d iterations, final value using Bisection method is %f and error is %e \n", ii, xmid, err_bisect)

how do i get the code to calculate the iteration needed to come up with the value of theta??

Answer 1

[Geoff Hayes]

Ahmed - theta is your xmid and I think that you are calculating it correctly (see https://en.wikipedia.org/wiki/Bisection_method for details). One issue may be with your tolerance err

err = 0.477;

Why is this such a large value? If I leave your code as is, then the condition for

while err_bisect > err 

is never true so the bisection method doesn't do anything. Try reducing err to somethig much smaller (perhaps 0000477) and then the code will iterate 12 times and give you an answer of 49.732039 which f(49.732039) ~= 0. Same with

ferr=0.176;

Why such a large number? Shouldn't this be close to zero too?