fzero runs out of iterations

Hi,

Depending on where I start fzero searching, it either will, or will not find the zero that I know is there. If I start it closer, it finds it. Is there a way to just increase the number of tries it takes before giving up? It should be a well-behaved 4th order polynomial with 2 zeros. I'm willing to try something else more robust if needed. Haven't looked into fsolve yet....

Thanks!

Dave

Answer 1

[Roger Stafford]

You should try giving ‘fzero’ an x0 (estimate) value in the form of a two-element vector such that the function is of opposite signs at the two points of x0. That should be a guarantee of a fast convergence.

[David Pesetsky]

Problem is I can't be sure the 2 zeros will be inside a given range, and it will only look inside a range I believe. What I'm doing is doing fzero twice, once with a single negative start, and once with a single positive start. I'm hoping it knows which direction it knows to fire off in to look. I'm getting Nan's when i don't believe I should be, based on the nature of the polynomial.

[Roger Stafford]

If your estimate has two elements such that the function is positive on one side and negative on the other even if they are relatively far apart, you are almost guaranteed rapid convergence to a solution (assuming the function is continuous.) You should try it. Mathworks states: "Tip - Calling fzero with an interval (x0 with two elements) is often faster than calling it with a scalar x0."

[David Pesetsky]

But it will return only 1 zero within that range right? So since I'm looking for 2 zeros, I do the range method twice. But the zero(s) could be in positive or negative X-territory, so i don't have a generic range that will always work. If I could tell it to start very high, and walk to lower values, I know it would always work. And not run out of iterations :)

[Torsten]

If it's a polynomial, why don't you use "roots" to find its zeros ?

Best wishes

Torsten.

Answer 2

[John D'Errico]

It is claimed to be a polynomial. Use roots!

By the way, fzero will be far more robust than fsolve, because you should be supplying a bracket that bounds your root. fsolve is targeted at multidimensional problems.

But roots is better than any of these alternatives, if it is truly a polynomial.

[David Pesetsky]

Yes, I was trying to avoid roots, because the polynomial is created by matrix multiplication, and I haven't worked out what the coeeficients would be. See here, where at the end, Matt suggests that: https://www.mathworks.com/matlabcentral/answers/288602-fzero-or-solve-a-file-full-of-relations

[John D'Errico]

You want a more robust solution. Do the work and compute the coefficients. Roots will be seriously faster.

Or, use symbolic tools if you just want a solution and are not worried about it taking a bit more time. solve or vpasolve will give the solutions with no problems. Then you will not have any problems in computing the coefficients, since MATLAB will do all the work. It still will very possibly be faster than an iterative scheme that may or may not happily converge.

As for fighting for a symbolic license, use symbolic tools ONCE, upfront. I fail to see the problem here.

[David Pesetsky]

Once, upfront? You mean make an executable out of the .m to bake the license in somehow?