Re: Using FindRoot

*To*: mathgroup at smc.vnet.net*Subject*: [mg83465] Re: Using FindRoot*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Wed, 21 Nov 2007 02:48:44 -0500 (EST)*References*: <fhu7qo$7dv$1@smc.vnet.net>

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^2-4 > 5.9^2]),{R,.1}, MaxIterations->100000] You've already gotten two good answers. But instead of graphing in order to find a good initial value for R, one could just think about a geometric interpretation of the problem: The distance between the centers of two circles of radius R is 45.9. Find R such that the area of the lens-shaped region intersection of the circles is .15 times the area of one of the circles. >From this interpretation, it's obvious that R must be greater than 45.9/2; otherwise, the circles would not intersect at all. And BTW, in this problem, the initial value for R can actually be much greater than the root and FindRoot will still work fine. For example, In[6]:= FindRoot[.15*Pi*R^2 == 2*R^2*ArcCos[45.9/(2*R)] - .5*45.9*Sqrt[4*R^2 - 45.9^2], {R, 1000}] Out[6]= {R -> 30.8774} > 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? You might be interested to know that your equation actually does have two other complex roots: In[7]:= FindRoot[.15*Pi*R^2 == 2*R^2*ArcCos[45.9/(2*R)] - .5*45.9*Sqrt[4*R^2 - 45.9^2],{R, -10+14I}] Out[7]= {R -> -9.73876 + 13.5078 I} and its conjugate. But of course I doubt that those are what you wanted. David