The Help page of FindRoot says: "by default, FindRoot uses Newton's method (Newton-Raphson) to solve a nonlinear system". But I still have the following puzzles:

  1. How will the Jacobian be calculated in each iteration: symbolically or using finite difference approximation only?

  2. Is there any damping method of the step being used when the Jacobian at any step is invertible?


1 Answer 1


You can control how the Jacobian is calculated via the Jacobian option:

Grid[Module[{s = 0, e = 0}, 
     FindRoot[ArcTan[1000 Cos[x]], {x, 1}, StepMonitor :> s++, 
      EvaluationMonitor :> e++, Jacobian -> #, Method -> {"Newton"}],
     "Steps" -> s, 
     "Evaluations" -> e
    }] & /@ {"Symbolic", "FiniteDifference"}]


Note the Sparse suboption of this option described on the linked Documentation page (see also this tutorial page for additional examples of use), it works both with "Symbolic" and "FiniteDifference" methods. The DampingFactor option may be also of help.

You can also find useful this MathGroups answer by Oliver Ruebenkoenig (Wolfram Research).

  • $\begingroup$ Thank you very much! that is what I need! $\endgroup$ Nov 26, 2013 at 11:15
  • 3
    $\begingroup$ @LCFactorization There's some additional info here (3rd and 4th sections), mostly about how the Jacobian is updated. $\endgroup$
    – Szabolcs
    Nov 26, 2013 at 15:40

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