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equation solving - best way for finding approximate root of monotonic sawtooth like function


First I will generate such kind of monotonic sawtooth like function called func


n = 50;

block = Symmetrize[RandomReal[1., {n, n}], Symmetric[{1, 2}]];
diagF = SparseArray[Band[{1, 1}] -> Normal /@ #] &;
mat = diagF[{block, block, block, block}];
dim = Length@mat;
eigs = Transpose@Sort@Transpose@Eigensystem@mat;
eigsdensity = {eigs[[1]], Abs[eigs[[2]]]^2};
fermiCom =
Compile[{{e, _Real}, {u, _Real}, {t, _Real}},
Module[{tmp}, tmp = Exp[(u - e)/t]; tmp = 1./(1. + tmp);
tmp = 1. - tmp], RuntimeOptions -> "Speed"];

Clear[fermi];
fermi[e_?NumericQ, u_?NumericQ, t_?NumericQ] := fermiCom[e, u, t];
num = Length@Select[eigs[[1]], # < 0 &] + 1;
Clear[func];
func[u_] :=
Total[Sum[
eigsdensity[[2, i]] fermi[eigs[[1, i]], u, 0.0001], {i, 1, dim}]]

The plot of func is like this


Plot[func[u], {u, -3, 3}]


enter image description here


This is what I call monomotic sawtooth like function


Now I want to know at which u value, func[u] is closest to num ( num = Length@Select[eigs[[1]], # < 0 &] + 1)


Normally, FindRoot fastest for this kind of task. However it is not working for this problem


FindRoot[func[u] - num == 0, {u, 0}]

gives error



FindRoot::jsing: Encountered a singular Jacobian at the point {u} = {0.}. Try perturbing the initial point(s). >>




This is because num is a value guaranteed to be between two plateaus (the fermi function is very close to step function, while not exactly the same), so it can't find such a root.


Then, we can transform the problem into a minimization problem. That is to minimize Abs[func[u] - num]


HoweverFindMinimum is also not working.


FindMinimum[Abs[func[u] - num], u]


FindMinimum::fmgz: Encountered a gradient that is effectively zero. The result returned may not be a minimum; it may be a maximum or a saddle point. >>



The only working, I know at the moment is NMinimize



NMinimize[Abs[func[u] - num], u] // AbsoluteTiming
(*{0.934564, {1., {u -> 0.0116856}}}*)

But as you can see NMinimize is really expensive. So I wonder, since the func such a good monotonic property, is there a better way than direct NMinimize?



Answer



From the FindRoot documentation:



FindRoot[lhs == rhs, {x, x0, x1}] searches for a solution using $x_0$ and $x_1$ as the first two values of $x$, avoiding the use of derivatives. ... If you specify two starting values, FindRoot uses a variant of the secant method.



Indeed, that does the job pretty much instantly:



FindRoot[func[u] - num == 0, {u, -3, 3}]
(* {u -> 0.00249415} *)

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