bisection method code help

Question

Angelina Encinias 2022-8-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1791180-bisection-method-code-help

评论： Angelina Encinias 2022-8-31

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

2 个评论
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Steven Lord 2022-8-30

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 2022-8-31

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

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Answer 1

Sam Chak 2022-8-31

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1791180-bisection-method-code-help#answer_1038805

在 MATLAB Online 中打开

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