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can someone please tell me why findroot performs poorly with jacobian ? in this example res1 is slower than res.

Clear[eqs, vars, function, jx, f, f1, res, res1]
eqs = {x^x + y^y - 20., x + y - 4.};
vars = {x, y};
function = Cases[Compile[##] &[vars, eqs], x_Function :> x] // First;

ClearAll[J];
J[HoldPattern[Function[vars : {__Symbol}, body_]]] := 
  Block[vars, Function @@ {vars, D[body, {vars}]}];
jx = N@J[function];
With[{code = jx[x, y]}, 
  cDf = Compile[{{x, _Real}, {y, _Real}}, code, 
    RuntimeAttributes -> {Listable}, Parallelization -> True, 
    RuntimeOptions -> "Speed"]];
f[x0_?NumericQ, y0_?NumericQ] := 
  FindRoot[{eqs[[1]] == 0, eqs[[2]] == 0}, {x, x0}, {y, y0}, 
   MaxIterations -> 5000, 
   Jacobian -> {"FiniteDifference", DifferenceOrder -> 1}, 
   AccuracyGoal -> 10];
f1[x0_?NumericQ, y0_?NumericQ] := 
  FindRoot[{eqs[[1]] == 0, eqs[[2]] == 0}, {x, x0}, {y, y0}, 
   MaxIterations -> 5000, Jacobian :> cDf[x0, y0], AccuracyGoal -> 10];
res = f[3., 2.] // RepeatedTiming
res1 = f1[3., 2.] // RepeatedTiming
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    $\begingroup$ I gues, the amount of numbercrunching within cDf is not enough to ammortized the calling overhead. This problem is just too small. $\endgroup$
    – Henrik Schumacher
    Commented Jun 18, 2018 at 12:34
  • 1
    $\begingroup$ You get some improvement by updating the Jacobian at each step, instead of using the constant Jac. at the starting point {x0, y0} (i.e., use Jacobian :> cDf[x, y]). $\endgroup$
    – Michael E2
    Commented Jun 18, 2018 at 17:32
  • $\begingroup$ @ Michael E2 it actually makes a difference and you are right i dont even know why i was evaluating Jacobian at initial points. now i have another question. if function and Jacobian being updated at each loop then what is the best way to tackle this problem? you can also post the answer and i will accept it . $\endgroup$
    – user49047
    Commented Jun 18, 2018 at 18:09

1 Answer 1

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Turning a comment into an answer (and correcting some typos).

I guess, the amount of number crunching within cDf is not enough to ammortize the calling overhead. This problem is just too small.

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