Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1997
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1997

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

Search the Archive

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  ]]
}
]


  • Prev by Date: Re: prograMing: split a list
  • Next by Date: Re: a[[2]][[3]] = 1 error? why?
  • Previous by thread: Re: FindRoot
  • Next by thread: Fwd: FindRoot