Re: Using FindRoot

*To*: mathgroup at smc.vnet.net*Subject*: [mg83458] Re: Using FindRoot*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Wed, 21 Nov 2007 02:45:01 -0500 (EST)*Organization*: Uni Leipzig*References*: <fhu7qo$7dv$1@smc.vnet.net>*Reply-to*: kuska at informatik.uni-leipzig.de

Hi, 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 .. Regards Jens 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. >