Help with simple bisection method function while loop

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Timothy Monroe
Timothy Monroe 2021-2-24
回答： Steven Lord 2021-2-25
I am writing a simple root finding bisection method function using a while loop and can't seem to get it to cycle through once i get first correction. Any help is much appreciated.
function [x] = bisection2(x1,x2,tol)
%UNTITLED2 Summary of this function goes here
f=@(x) cos((pi/2)*x)/(1-x^2);
x= (x1+x2)/2;
p=f(x);
i=0;
if p<=tol
fprintf('A Root at x = %f was found in 1 iterations with an error of \n' , x,i)
return
end
while p > tol
i= i+1;
x= (x1+x2)/2;
p=f(x);
if sign(p)==sign(f(x1))
x1=x;
fprintf('Iteration: %f x tested: %f Error Found: %f \n',i,x,p)
return
end
if sign(p)== sign(f(x2))
x2=x;
fprintf('Iteration: %f x tested: %f Error Found: %f \n',i,x,p)
return
else
disp('Root not Bounded!')
return
end
end
end

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回答（1 个）

Steven Lord
Steven Lord 2021-2-25
Ran in:
Let's say your function was:
f = @(x) -x.^2;
and your tolerance was 1e-6. Is the tolerance satisfied if I evaluate f at x = 1?
tol = 1e-6
tol = 1.0000e-06
y = f(1)
y = -1
isYSmallerThanTol = y < tol
isYSmallerThanTol = logical
1
So good enough, I've found my root, right?

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