Re: FindRoot

*To*: mathgroup at smc.vnet.net*Subject*: [mg8902] Re: FindRoot*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Thu, 2 Oct 1997 22:56:54 -0400*Organization*: University of Western Australia*Sender*: owner-wri-mathgroup at wolfram.com

David Djajaputra wrote: > I need to plot the root of a transcendental equation > > sin(x)=x/a, for a from 2 to 3. > > I can find the root for each a using FindRoot. But how do I make a > continuous plot using FindRoot? Using ListPlot with PlotJoined->True is one way to do what you want, i.e., In[1]:= ListPlot[Table[{a,x/.FindRoot[Sin[x]==x/a,{x,2}]},{a,2,3,0.05}], PlotJoined->True]; For another approach, see the Mathematica Journal 3(3):28. I have attached a Mathematica 3.0 Notebook below relevant for your example. Cheers, Paul ____________________________________________________________________ Paul Abbott Phone: +61-8-9380-2734 Department of Physics Fax: +61-8-9380-1014 The University of Western Australia Nedlands WA 6907 mailto:paul at physics.uwa.edu.au AUSTRALIA http://www.pd.uwa.edu.au/~paul God IS a weakly left-handed dice player ____________________________________________________________________ Notebook[{ Cell[CellGroupData[{ Cell["Root Finding with a Parameter", "Section", CellTags->"root finding"], Cell[TextData[{ "Suppose that we wish to find the roots of the transcendental equation ", Cell[BoxData[ \(TraditionalForm\`Sin[x] = \ x/a\)]], " where ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is a parameter. For a fixed value of ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", this can be easily accomplished using the ", Cell[BoxData[ FormBox[ StyleBox["FindRoot", "Input"], TraditionalForm]]], " command. For example, if ", Cell[BoxData[ \(TraditionalForm\`a = 2\)]], " then to find the root near ", Cell[BoxData[ \(TraditionalForm\`x = 2\)]], " we use" }], "Text", CellTags->{"FindRoot", "transcendental equation"}], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`start = {2, x /. FindRoot(sin(x) == x\/2, {x, 2})}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{2, 1.89549426720871316`}\)], "Output"] }, Open ]], Cell[TextData[{ "An efficient technique for finding the values of a root as a function of \ the parameter is to use the functional iteration operation ", Cell[BoxData[ FormBox[ StyleBox["NestList", "Input"], TraditionalForm]]], ". First, define a function whose first argument is the parameter:" }], "Text", CellTags->{"functional iteration", "NestList"}], Cell[BoxData[ \(TraditionalForm\`f(a_, x_) := sin(x) - x\/a\)], "Input"], Cell[TextData[{ "Then step along the ", Cell[BoxData[ \(TraditionalForm\`a\)]], "-axis by ", Cell[BoxData[ \(TraditionalForm\`h\)]], " and find the root of the transcendental equation using the previous root \ as an initial guess:" }], "Text", CellTags->"iterative guessing"], Cell[BoxData[ \(TraditionalForm \`FRStep(f_, {a_, x_}, h_: 0.05) := Module({b}, {a + h, b /. FindRoot(f(a + h, b) == 0, {b, x})})\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`FRStep(f, start)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{2.04999999999999982`, 1.92367954032495482`}\)], "Output"] }, Open ]], Cell[TextData[{ "That is, at ", Cell[BoxData[ \(TraditionalForm\`a = 2.05\)]], ", ", StyleBox["x", FontSlant->"Italic"], " is approximately 1.92368. ", Cell[BoxData[ FormBox[ StyleBox["NestList", "Input"], TraditionalForm]]], " can then be used to create a list of the roots in the form ", Cell[BoxData[ \(TraditionalForm\`{a, x}\)]], ":" }], "Text"], Cell[BoxData[ \(TraditionalForm\`\(NestList(FRStep(f, #)&, start, 20); \)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(ListPlot(%, PlotJoined \[Rule] True); \)\)], "Input"] }, Open ]] } ]