Re: Using FindRoot

*To*: mathgroup at smc.vnet.net*Subject*: [mg83446] Re: [mg83429] Using FindRoot*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Wed, 21 Nov 2007 02:38:40 -0500 (EST)*Reply-to*: hanlonr at cox.net

eqn = .15*Pi*(R^2) == (2*R^2*ArcCos[45.9/(2*R)]) - (.5*45.9* Sqrt[4*R^2 - 45.9^2]); Solve[4*R^2 - 45.9^2 == 0, R] {{R -> -22.95}, {R -> 22.95}} The root is complex for |R| < 22.95 Plot[Evaluate[eqn[[1]] - eqn[[2]]], {R, 20, 35}] Use a closer starting value where R is real FindRoot[eqn, {R, 25}] {R -> 30.87736860938117} Bob Hanlon ---- tdude <stiletto at bigfoot.com> 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. >