bisection method code help

function root=bisection(func,x1,xu)
xr=x1;es=.0001;
while(1)
    xrold=xr;
    xr=(x1+xu)/2;
    if xr~=0
        ea=abs((xr-xrold)/xr)*100;
    else
        ea=100;
    end
    if func(x1)*func(xr)<0
        xu=xr;
    elseif func(x1)*func(xr)>0
        x1=xr;
    else
        ea=0;
    end
    if ea<=es,break,end
end
root=xr;
end

[Steven Lord]

What is your question or concern about the code you've posted and the text of (what appears to be) a homework problem?

[Angelina Encinias]

What do I need to change to the code to fit the problem?

Answer 1

[Sam Chak]

Hi @Angelina Encinias,

I did not change anything. The bisection code works for a simple test below to find :

fcn   = @(x) sin(x) - 0.5;  % if the equation is f(x) = g(x), then enter f(x) - g(x)
xL    = 0;                  % lower bound
xU    = pi/2;               % upper bound
x     = bisection(fcn, xL, xU);
x_deg = x*180/pi            % solution in degree

x_deg = 30.0000

function root=bisection(func,x1,xu)
xr=x1;es=.0001;
while(1)
    xrold=xr;
    xr=(x1+xu)/2;
    if xr~=0
        ea=abs((xr-xrold)/xr)*100;
    else
        ea=100;
    end
    if func(x1)*func(xr)<0
        xu=xr;
    elseif func(x1)*func(xr)>0
        x1=xr;
    else
        ea=0;
    end
    if ea<=es,break,end
end
root=xr;
end

[Angelina Encinias]

Okay. Is this for my question above? I dont see anything about angles for this question.