Can one run fminsearch for a function defined by cases?

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First, I have a function defined by cases like
if x<1 f(x)=x
if x>1 f(x)=sqrt(x)
Second, suppose I want to calculate the maximium of f for 0<x<2 (this problem itself is trivial, but as an example).
Third, can one run fminsearch for such problem?

Accepted Answer

Matt J
Matt J on 28 Jul 2020
Edited: Matt J on 28 Jul 2020
Because you are optimizing over a known bounded interval fminbnd is more appropriate:
x_optimal = fminbnd(@fun,0,2)
function out=fun(x)
if x<=1 out=x; else out=sqrt(x); end
out=-out; %because we're maximizing
end
  2 Comments
alpedhuez
alpedhuez on 28 Jul 2020
Edited: alpedhuez on 28 Jul 2020
Thank you. Then what should one do when one maximizes over a finite set of possible x's? Like 0,0.01,0.02,...,1?

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More Answers (1)

John D'Errico
John D'Errico on 28 Jul 2020
Edited: John D'Errico on 28 Jul 2020
Can you? Yes.
Should you? This may sometimes be a poor idea, although it depends on the specific segment functions. Are the segments continuous across the break? Are they differentiable across the break?
Fminsearch does not require a several times continuously differentiable objective. It is even reasonably robust in that respect. But if your function is discontinuous across a break, then expect that it can easily fail. The nastier is your function, the more chance of failure.
What does failure mean here? It does not mean the algorithm will fail due to an error message. It just means fminsearch may not consistently find the optimal solution. It could get more easily stuck in a sub-optimal solution. Note there is NO assurance of convergence to the globally optimal solution for fminsearch (or virtually any optimizer) even on a more well-behaved problem, just that if you make things messy it may fail more easily.
If your problem is a 1-dimensional one, then it is advisable to best break down the problem into sub-intervals, optimizing over each interval, and to then use fminbnd for each interval. This will yield the best possile result. Using fminbnd is good here, because it employs a bracket to search over. You then choose the overall best choice of all solutions returned. In the case of a bracket that extends to -inf or +inf, just choose a reasonably large number, but not excessively large, as that can cause numerical problems.
  1 Comment
alpedhuez
alpedhuez on 28 Jul 2020
Yes but the real problem is such that one cannot divide into subintervals of the real line.

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