code on finding the roots using simple bisection method

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B 2015-5-7

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评论： Geoff Hayes 2015-5-7

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what's wrong in the following code:

function s=bisectionpart(s);

xl=5;

xu=100;

xr=xl;

while(1)

    xrold=xr;
xr=(xl+xu)/2;
error=abs((xr-xrold)/xr)*100;
if error<0.1
      break,
end;
test=fs(xl)*fs(xr);
if test<0;
      xu=xr;
else,
    xl=xr;
end
end

s=xr;

plot(t,s);

end

function f=fs(s);
s0=10;
vm=0.5;
ks=2;

t=0:50;

f=s0-(vm*t)+ks*log(s0/s)-s;

end

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Geoff Hayes 2015-5-7

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B - try formatting your above code that it is readable by others, or attach the code to your question using the paperclip button.

A couple of comments: why is s an input parameter to your bisectionpart function. How is it used, and what would it be when you call this function? Also, don't use error as the name of a variable as this is also the name of a built-in MATLAB function.

When I try to run the bisectionpart function, I get an error at line 17

 Error using  * 
 Inner matrix dimensions must agree.
 Error in bisectionpart (line 17)
 test=fs(xl)*fs(xr);

This is because the result returned from fs is an array of 51 elements. Is this expected? What is the purpose of fs i.e. what does it represent or model?

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