Recursive Function - Bisection

I have written code to compute the flow of a fluid in a pipe using the bisection method. This uses a while loop and works. However, I have been provided with code for a recursive function – which although perhaps not as efficient in this case – I would like to test.

I am having problems calling the function though. I have named a .m file with MyBisection and I have the correlation equation, the upper and lower bounds of the interval, and also the tolerance I want to use for the problem.

[Stephen23]

Original question by Tony Rankin retrieved from Google Cache:

https://web.archive.org/web/20210306054256/https://webcache.googleusercontent.com/search?q=cache%3AF-2VCOSQn2EJ%3Ahttps%3A%2F%2Fwww.mathworks.com%2Fmatlabcentral%2Fanswers%2F759516-recursive-function-bisection

http://web.archive.org/web/20210306054705/https://www.mathworks.com/matlabcentral/answers/759516-recursive-function-bisection

Recursive Function - Bisection

I have written code to compute the flow of a fluid in a pipe using the bisection method. This uses a while loop and works. However, I have been provided with code for a recursive function – which although perhaps not as efficient in this case – I would like to test.

I am having problems calling the function though. I have named a .m file with MyBisection and I have the correlation equation, the upper and lower bounds of the interval, and also the tolerance I want to use for the problem.

[John D'Errico]

It is rude to delete your question when someone has spent a great deal of time to answer your problem. Others can benefit from the solution. But when you remove the question, you remove all context for the answer.

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[David Hill]

Your problem is with your function.

E=1;
D=5;
Re=1;
f=@(x)-1./sqrt(x)-2.*log10(((E/D)/3.7)+2.51./(Re*sqrt(x)));%is this the equation? E, D, and Re will have to be provided before the function call
R=MyBiscection(f,0,1,1e-4);

[Steven Lord]

Typo: you have one too many c's.

MyBiscection % should be
MyBisection

[Steven Lord]

I suspect you've modified Bisection_Trial significantly from the MyBisection function you posted originally. Can you post your current version of Bisection_Trial so we can see the whole function not just isolated pieces?

[David Hill]

I believe your problem is S=0 and log10(0)=-inf and a cannot equal 0 either.This works just fine.

p = 0.9*1000;
mu = 8*0.001;
E = .001;%change to something small other than zero
D = 0.1016;
S = E/D;
Q = (2000*42*3.785*10^-3)/(24*60*60);
A = (pi*(D.^2))/4;
U = Q/A;
Re = (p*D*U)/mu;
f = @(x) -1./sqrt(x)-2.*log10((S)/3.7)+2.51./(Re*sqrt(x));
R = Bisection_Trial(f,.01,1,1e-4);%change a to something small .01
function R=Bisection_Trial(f,a,b,tol)
if sign(f(a)) == sign(f(b))
    error('Scalars a and b do not bound a root.')
end
m = (a+b)/2;                        
if abs(f(m)) < tol
    R = m;                          
elseif sign(f(a)) == sign(f(m))
    R = MyBisection(f,m,b,tol);     
elseif sign(f(b)) == sign(f(m))
    R = MyBisection(f,a,m,tol);     
end
end

[Steven Lord]

First, please don't put "clear all" in your script files. When you need to clear the variables, execute it at the prompt in the MATLAB Command Window.

Second, your code does not handle the case where the absolute value of f(m) is not less than tol and the sign of f(m) is not the same as either the sign of f(a) or the sign of f(b).

sp1 = sign(1) % +1

sp1 = 1

sm1 = sign(-1) % -1

sm1 = -1

sc = sign(3+4i) % complex

sc = 0.6000 + 0.8000i

To detect this I might put some code in your function to display the values of a, m, and b along with f(m) before entering your if / elseif section, something along the lines of this untested code:

fm = f(m);
result = table(a, m, b, fm, 'VariableNames', ["a", "m", "b", "f(m)"]);
disp(result)

Or maybe pass this between the recursive calls and augment it in each one so you can see the progress.