Services & Resources / Wolfram Forums / MathGroup Archive

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Using FindRoot

  • To: mathgroup at
  • Subject: [mg83458] Re: Using FindRoot
  • From: Jens-Peer Kuska <kuska at>
  • Date: Wed, 21 Nov 2007 02:45:01 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <fhu7qo$7dv$>
  • Reply-to: kuska at


FindRoot[0.15*\[Pi]*(R^2) == (2*R^2*ArcCos[45.9/(2*R)]) - (0.5*45.9*
      Sqrt[4*R^2 - 45.9^2]), {R, 40.1}, MaxIterations -> 100000]

may help ..


tdude wrote:
> I am trying to find the root of the following equation
> FindRoot[.15*[Pi]*(R^2)==(2*R^2*ArcCos[45.9/(2*R)])-(.5*45.9*Sqrt[4*R^2-45.9^2]),{R,.1}, MaxIterations->100000]
> When I try this with two other systems, both give me an answer of 30.8773686.
> However, with Mathematica, the answer I get appears to be a complex root, along with this message:
> FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function.  You may need more than MachinePrecision digits of working precision to meet these tolerances.
> {R -> 20.1137- 4.87774*10^-14 \[ImaginaryI]}
> Why would the answer be a complex root when using Mathematica, yet real when using the other two packages?
> Thanks Tony.

  • Prev by Date: Re: Fast way of checking for perfect squares?
  • Next by Date: Re: Extra spacing in GraphicsGrid
  • Previous by thread: Re: Using FindRoot
  • Next by thread: Re: Using FindRoot