Recursive Function - Bisection

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Tony Rankin
Tony Rankin 2021-3-1
编辑： Rik 2021-6-1
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.
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John D'Errico
John D'Errico 2021-3-21
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
Rena Berman 2021-5-6
(Answers Dev) Restored edit

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回答（1 个）

David Hill
David Hill 2021-3-1
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);
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David Hill
David Hill 2021-3-1
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
Steven Lord 2021-3-1
Ran in:
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.

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