(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 134731, 3819]*) (*NotebookOutlinePosition[ 135779, 3851]*) (* CellTagsIndexPosition[ 135735, 3847]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Cusp Catastrophe How-to", "Title"], Cell["by W. Garrett Mitchener", "Author"], Cell[TextData[{ "In this worksheet, I illustrate how to generate a bifurcation diagram for \ a two-parameter dynamical system in one dependent variable. The particular \ types of bifurcations that can happen form what's called a ", StyleBox["cusp catastrophe. ", FontSlant->"Italic"], StyleBox["T", FontVariations->{"CompatibilityType"->0}], "his worksheet shows several ways to generate the cusp picture that \ illustrates what happens." }], "Abstract"], Cell[CellGroupData[{ Cell["Setup", "Section"], Cell["Let's work with the dynamical system given by:", "Text"], Cell[BoxData[ \(\(x\& . \) \[Equal] h + r\ x - x\^3\)], "DisplayFormula"], Cell["\<\ First, here's how to define a function for the right hand \ side:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(f[x_, r_, h_] = h + r\ x - x^3\)], "Input", CellLabel->"In[1]:="], Cell[BoxData[ \(h + r\ x - x\^3\)], "Output", CellLabel->"Out[1]="] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["First method: Solve for fixed point collision directly", "Section"], Cell[TextData[{ "Fixed points happen when ", Cell[BoxData[ \(TraditionalForm\`\(x\& . \) \[Equal] 0\)]], ", and bifurcations happen when fixed points collide (and in other \ situations). But look what happens when we try to solve for the fixed \ points:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(fps = Solve[f[x, r, h] \[Equal] 0, x]\)], "Input", CellLabel->"In[2]:="], Cell[BoxData[ \({{x \[Rule] \(-\(\(\((2\/3)\)\^\(1/3\)\ r\)\/\((\(-9\)\ h + \@3\ \@\(27\ \ h\^2 - 4\ r\^3\))\)\^\(1/3\)\)\) - \((\(-9\)\ h + \@3\ \@\(27\ h\^2 - 4\ \ r\^3\))\)\^\(1/3\)\/\(2\^\(1/3\)\ 3\^\(2/3\)\)}, {x \[Rule] \(\((1 + \ \[ImaginaryI]\ \@3)\)\ r\)\/\(2\^\(2/3\)\ 3\^\(1/3\)\ \((\(-9\)\ h + \@3\ \ \@\(27\ h\^2 - 4\ r\^3\))\)\^\(1/3\)\) + \(\((1 - \[ImaginaryI]\ \@3)\)\ \ \((\(-9\)\ h + \@3\ \@\(27\ h\^2 - 4\ r\^3\))\)\^\(1/3\)\)\/\(2\ 2\^\(1/3\)\ \ 3\^\(2/3\)\)}, {x \[Rule] \(\((1 - \[ImaginaryI]\ \@3)\)\ r\)\/\(2\^\(2/3\)\ \ 3\^\(1/3\)\ \((\(-9\)\ h + \@3\ \@\(27\ h\^2 - 4\ r\^3\))\)\^\(1/3\)\) + \ \(\((1 + \[ImaginaryI]\ \@3)\)\ \((\(-9\)\ h + \@3\ \@\(27\ h\^2 - 4\ r\^3\))\ \)\^\(1/3\)\)\/\(2\ 2\^\(1/3\)\ 3\^\(2/3\)\)}}\)], "Output", CellLabel->"Out[2]="] }, Open ]], Cell[TextData[{ "Those expressions are pretty nasty, and you can't tell which roots are \ real or complex. Don't let the i's in those two lower expresssions fool you. \ The square roots may generate complex numbers depending on ", Cell[BoxData[ \(TraditionalForm\`r\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\)]], " that cancel out with the i's on top. For the moment, suppose that's not \ a problem. Now imagine trying to set two of these equal and solve for ", Cell[BoxData[ \(TraditionalForm\`r\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\((x /. fps[\([1]\)])\) \[Equal] \((x /. fps[\([2]\)])\), h]\)], "Input", CellLabel->"In[3]:="], Cell[BoxData[ \({{h \[Rule] \(-\(\(2\ r\^\(3/2\)\)\/\(3\ \@3\)\)\)}, {h \[Rule] \(2\ \ r\^\(3/2\)\)\/\(3\ \@3\)}}\)], "Output", CellLabel->"Out[3]="] }, Open ]], Cell[TextData[{ "We got lucky: That's the right answer, and the computer didn't crash. I \ don't recommend this method because if your dynamical system is more \ complicated than this example, the expressions for the roots will probably be \ so big and complicated that ", StyleBox["Mathematica", FontSlant->"Italic"], " can't deal so easily with them. It's worth a try in that case, but there \ are alternatives. I'll plot the solution below, after showing an alternative \ method of deriving it." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Second method: Combine fixed point solve with linear stability \ analysis failure\ \>", "Section"], Cell[TextData[{ "An alternative is to use the fact that ", StyleBox["Mathematica", FontSlant->"Italic"], " can handle systems of polynomial equations very easily. Here, we ", ButtonBox["Solve", ButtonStyle->"RefGuideLink"], " for the parameter values such that there is a fixed point at ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", and such that linear stability analysis fails there. " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(bifSol = Solve[{f[x, r, h] \[Equal] 0, D[f[x, r, h], x] \[Equal] 0}, {r, h}]\)], "Input", CellLabel->"In[4]:="], Cell[BoxData[ \({{h \[Rule] \(-2\)\ x\^3, r \[Rule] 3\ x\^2}}\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[TextData[{ "We want to plot ", Cell[BoxData[ \(TraditionalForm\`r\)]], " horizontally and ", Cell[BoxData[ \(TraditionalForm\`h\)]], " vertically to reproduce the figure on p. 71 of Strogatz. The nice thing \ about this result is that it gives ", Cell[BoxData[ \(TraditionalForm\`h\)]], " and ", Cell[BoxData[ \(TraditionalForm\`r\)]], " very simply in terms of the fixed point ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", so we can vary ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", and generate a plot of ", Cell[BoxData[ \(TraditionalForm\`h\)]], " vs. ", Cell[BoxData[ \(TraditionalForm\`r\)]], " parametrically. Here's how to set that up." }], "Text"], Cell[TextData[{ "Just so you'll know, the solution comes as a list of rule tables, because \ some systems of equations may have many solutions. So we use ", Cell[BoxData[ \(TraditionalForm\`bifSol[\([1]\)]\)]], " to fish out the single rule table we want:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(bifSol[\([1]\)]\)], "Input", CellLabel->"In[5]:="], Cell[BoxData[ \({h \[Rule] \(-2\)\ x\^3, r \[Rule] 3\ x\^2}\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[TextData[{ " Then we have to make an expression for ", ButtonBox["ParametricPlot", ButtonStyle->"RefGuideLink"], ". 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o`3ooon6ool0023oo`8000Koo`@00?ooohGoo`008Ooo0P00ooooS_oo000Rool00`00ooooo`3ooon< ool002;oo`03003ooooo0?ooohcoo`007?oo0P000ooo0P00ooooS_oo000Mool4003ooon@ool00?oo ok7oo`00oooo/Ooo003ooonaool00?oook7oo`00oooo/Ooo003ooonaool00?oook7oo`00oooo/Ooo 003ooonaool00?oook7oo`00oooo/Ooo0000\ \>"], ImageRangeCache->{{{109, 396}, {423.562, 136.562}} -> {-4.12177, -0.168114, \ 0.0155771, 0.0155771}}], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output", CellLabel->"Out[8]="] }, Open ]], Cell[TextData[{ "By the way, I just guessed that ", Cell[BoxData[ \(TraditionalForm\`x\)]], " should go from -5 to 5. Alternatively, you could solve for the fixed \ points explicitly for ", Cell[BoxData[ \(TraditionalForm\`r = 2\)]], " (the right-hand edge) and get an idea for the ", Cell[BoxData[ \(TraditionalForm\`x\)]], " range that way." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Third method: Discriminants", "Section"], Cell[TextData[{ "Every polynomial has a discriminant. (See ", StyleBox["http://mathworld.wolfram.com/PolynomialDiscriminant.html.", "Program"], ") The definition is kind of complicated: You start by numbering all the \ roots of a polynomial ", Cell[BoxData[ \(TraditionalForm\`r\_1, ... , r\_n\)]], ". Remember, every complex polynomial of degree ", Cell[BoxData[ \(TraditionalForm\`n\)]], " has ", Cell[BoxData[ \(TraditionalForm\`n\)]], " complex roots. Then the discriminant is defined to be the product of the \ squares of the differences between every possible pair of roots:" }], "Text"], Cell[BoxData[ \(D = \[Product]\+\(i = 1\)\%n\(\[Product]\+\(j = i + 1\)\%n\((r\_i - \ r\_j)\)\^2\)\)], "DisplayFormula"], Cell["\<\ So the discriminant is zero if two of the roots coincide. Luckily, \ there's a way to compute the discriminant just from the coefficients of the \ polynomial. The following incantation defines such a calculation. (Don't \ worry about how this works for now.)\ \>", "Text"], Cell[BoxData[ \(Discriminant[p_?PolynomialQ, x_] := With[{n = Exponent[p, x]}, Cancel[\((\((\(-1\))\)^\((n \((n - 1)\)/2)\) Resultant[p, D[p, x], x])\)/ Coefficient[p, x, n]^\((2 n - 1)\)]]\)], "Input", CellLabel->"In[9]:="], Cell[TextData[{ "Here's the familiar discriminant for a quadratic polynomial, up to a \ factor of ", Cell[BoxData[ \(TraditionalForm\`a\^2\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Discriminant[a\ x^2\ + \ b\ x\ + \ c, \ x]\)], "Input", CellLabel->"In[10]:="], Cell[BoxData[ \(\(b\^2 - 4\ a\ c\)\/a\^2\)], "Output", CellLabel->"Out[10]="] }, Open ]], Cell[TextData[{ "So if our function ", Cell[BoxData[ \(TraditionalForm\`f[x, r, h]\)]], " undergoes a bifurcation at parameter values ", Cell[BoxData[ \(TraditionalForm\`r\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\)]], ", two fixed points collide, which means the discriminant must be zero:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(discF = Discriminant[f[x, r, h], x]\)], "Input", CellLabel->"In[11]:="], Cell[BoxData[ \(\(-27\)\ h\^2 + 4\ r\^3\)], "Output", CellLabel->"Out[11]="] }, Open ]], Cell[TextData[{ "The equation ", Cell[BoxData[ \(TraditionalForm\`\(-27\)\ h\^2 + 4\ r\^3 \[Equal] 0\)]], " defines a curve in the ", Cell[BoxData[ FormBox[Cell[TextData[Cell[BoxData[ FormBox[Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\((r, h)\)\)]]]], TraditionalForm]]]]], TraditionalForm]]], "implicitly, and we want to get a picture of it. 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00<00?ooool00_oo00<00?ooool01?oo00<00?ooool0ooooUooo000"], ImageRangeCache->{{{109, 396}, {670.438, 383.438}} -> {-3.98531, 3.58008, \ 0.0153165, 0.0153165}}], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] ContourGraphics \[SkeletonIndicator]\), False, Editable->False]], "Output", CellLabel->"Out[13]="] }, Open ]], Cell["\<\ An alternative is to solve for one variable in terms of the other. \ This can get a little tricky:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(hSol = Solve[discF \[Equal] 0, h]\)], "Input", CellLabel->"In[14]:="], Cell[BoxData[ \({{h \[Rule] \(-\(\(2\ r\^\(3/2\)\)\/\(3\ \@3\)\)\)}, {h \[Rule] \(2\ \ r\^\(3/2\)\)\/\(3\ \@3\)}}\)], "Output", CellLabel->"Out[14]="] }, Open ]], Cell[TextData[{ "Now we need both solutions. This expression gives us a list of solutions \ for ", Cell[BoxData[ \(TraditionalForm\`h\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(h /. hSol\)], "Input", CellLabel->"In[15]:="], Cell[BoxData[ \({\(-\(\(2\ r\^\(3/2\)\)\/\(3\ \@3\)\)\), \(2\ r\^\(3/2\)\)\/\(3\ \@3\)}\ \)], "Output", CellLabel->"Out[15]="] }, Open ]], Cell[TextData[{ "The ", ButtonBox["/.", ButtonStyle->"RefGuideLink"], " command can work with a list of rule tables, and produces a list of what \ happens when you apply the different rule tables to the expression on the \ left. Just so you see what it's doing:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Foo[bar, baz] /. {{Foo \[Rule] f, bar \[Rule] x, baz \[Rule] y}, \[IndentingNewLine]{Foo \[Rule] g, bar \[Rule] a, baz \[Rule] b}}\)], "Input", CellLabel->"In[16]:="], Cell[BoxData[ \({f[x, y], g[a, b]}\)], "Output", CellLabel->"Out[16]="] }, Open ]], Cell[TextData[{ "Now for our plot. 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0?ooohkoo`008?oo00<00?ooool0ooooS_oo0000\ \>"], ImageRangeCache->{{{109, 396}, {650.562, 473.625}} -> {-0.50392, 2.88198, \ 0.00386179, 0.00624851}}], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output", CellLabel->"Out[17]="] }, Open ]], Cell[TextData[{ "That isn't good. What sometimes happens is that ", StyleBox["Mathematica", FontSlant->"Italic"], " evaluates the expression you're plotting in a funny order, so for \ example, it sets ", Cell[BoxData[ \(TraditionalForm\`r = \(-1.66\)\)]], ", and looks at ", Cell[BoxData[ \(TraditionalForm\`h /. hSol\)]], " but doesn't evaluate the ", ButtonBox["/.", ButtonStyle->"RefGuideLink"], " substitution, so there's nowhere to plug in ", Cell[BoxData[ \(TraditionalForm\`r\)]], ", and it gets confused. What usually works in this case is to tell ", StyleBox["Mathematica", FontSlant->"Italic"], " not to delay evaluating the expression to be plotted. You do that by \ wrapping it in ", ButtonBox["Evaluate", ButtonStyle->"RefGuideLink"], ". This is a good trick to know. ", ButtonBox["Solve", ButtonStyle->"RefGuideLink"], ", ", ButtonBox["Plot", ButtonStyle->"RefGuideLink"], ", and other functions are documented as having this odd behavior: They \ \"evaluate their arguments in a non-standard way\" which is a clue that you \ might need to use ", ButtonBox["Evaluate", ButtonStyle->"RefGuideLink"], ". 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003ooooo0?ooohKoo`00:?oo00<00?ooool0ooooQ_oo003ooonaool00?oook7oo`00oooo/Ooo003o oonaool00?oook7oo`00oooo/Ooo003ooonaool00?oook7oo`00oooo/Ooo0000\ \>"], ImageRangeCache->{{{109, 396}, {443.75, 266.812}} -> {-1.07623, 2.48007, \ 0.00789455, 0.0139062}}], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output", CellLabel->"Out[18]="] }, Open ]], Cell[TextData[{ "By the way, you can ignore the complaints about not getting a real number. \ The plotting process complains because the expression generates complex \ numbers for ", Cell[BoxData[ \(TraditionalForm\`r < 0\)]], " and nothing bad happens; they just don't show up in the plot." }], "Text"], Cell[TextData[{ "We can also solve for ", Cell[BoxData[ \(TraditionalForm\`r\)]], " as a function of ", Cell[BoxData[ \(TraditionalForm\`h\)]], ", which happens to be easier in this case." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(rSol = Solve[discF \[Equal] 0, r]\)], "Input", CellLabel->"In[19]:="], Cell[BoxData[ RowBox[{\(General::"spell1"\), \(\(:\)\(\ \)\), "\<\"Possible spelling \ error: new symbol name \\\"\\!\\(rSol\\)\\\" is similar to existing symbol \\\ \"\\!\\(hSol\\)\\\". \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"General::spell1\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[19]:="], Cell[BoxData[ \({{r \[Rule] 3\ \((\(-\(1\/2\)\))\)\^\(2/3\)\ h\^\(2/3\)}, {r \[Rule] \(3\ \ h\^\(2/3\)\)\/2\^\(2/3\)}, {r \[Rule] \(-\(\(3\ \((\(-1\))\)\^\(1/3\)\ \ h\^\(2/3\)\)\/2\^\(2/3\)\)\)}}\)], "Output", CellLabel->"Out[19]="] }, Open ]], Cell[TextData[{ "Two of the solutions are complex for every ", Cell[BoxData[ \(TraditionalForm\`h\)]], ", so the only one we need to worry about is:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(rSolReal = \(3\ h\^\(2/3\)\)\/2\^\(2/3\)\)], "Input", CellLabel->"In[20]:="], Cell[BoxData[ \(\(3\ h\^\(2/3\)\)\/2\^\(2/3\)\)], "Output", 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Now it should be clear why we get a \ cusp: Functions of the form ", Cell[BoxData[ \(TraditionalForm\`\((x\^2)\)\^\(1/n\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`n > 2\)]], " have cusp-shaped graphs. The plot is a little weird:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Plot[rSolReal, {h, \(-2\), 2}]\)], "Input", CellLabel->"In[21]:="], Cell[BoxData[ RowBox[{\(Plot::"plnr"\), \(\(:\)\(\ \)\), "\<\"\\!\\(rSolReal\\) is not \ a machine-size real number at \\!\\(h\\) = \\!\\(-1.9999998333333333`\\). \\!\ \\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Plot::plnr\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[21]:="], Cell[BoxData[ RowBox[{\(Plot::"plnr"\), \(\(:\)\(\ \)\), "\<\"\\!\\(rSolReal\\) is not \ a machine-size real number at \\!\\(h\\) = \\!\\(-1.8377320337083367`\\). \\!\ \\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Plot::plnr\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[21]:="], Cell[BoxData[ RowBox[{\(Plot::"plnr"\), \(\(:\)\(\ \)\), "\<\"\\!\\(rSolReal\\) is not \ a machine-size real number at \\!\\(h\\) = 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That's because ", StyleBox["Mathematica", FontSlant->"Italic"], " is extremely careful with complex numbers, and knows that ", Cell[BoxData[ \(TraditionalForm\`x\^\(p/q\)\)]], " is generally not well defined for negative ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". The case of ", Cell[BoxData[ \(TraditionalForm\`p\/q \[Equal] 1\/2\)]], "is an obvious example. So we end up getting only half the plot we wanted. \ The parametric form is better because it doesn't have this problem." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Fourth method: Template polynomials", "Section"], Cell["\<\ Just so you'll know, \"template polynomial\" is a term I made up. \ As far as I know, there's no standard name for this technique.\ \>", "Text"], Cell[TextData[{ "We suppose that a bifurcation occurs at ", Cell[BoxData[ \(TraditionalForm\`\((r, h)\)\)]], " such that two fixed points coincide. That means there's a double root. \ So, to trace out such values of ", Cell[BoxData[ \(TraditionalForm\`r\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\)]], ", we set our ", Cell[BoxData[ \(TraditionalForm\`f\)]], " equal to a polynomial with the same degree, and same highest-power \ coefficient, but with overlapping roots. That's what I call the template:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(template = \(-\((x - p1)\)^2\) \((x - p2)\)\)], "Input", CellLabel->"In[22]:="], Cell[BoxData[ \(\(-\((\(-p1\) + x)\)\^2\)\ \((\(-p2\) + x)\)\)], "Output", CellLabel->"Out[22]="] }, Open ]], Cell[TextData[{ "Here, ", Cell[BoxData[ \(TraditionalForm\`p1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`p2\)]], " are the two roots, and by putting in ", Cell[BoxData[ \(TraditionalForm\`\((x - p1)\)\^2\)]], ", I've specified that ", Cell[BoxData[ \(TraditionalForm\`p1\)]], " is a double root. I put in the ", Cell[BoxData[ \(TraditionalForm\`-\)]], " sign because the highest-order term of ", Cell[BoxData[ \(TraditionalForm\`f\)]], " is ", Cell[BoxData[ \(TraditionalForm\`\(-x\^3\)\)]], " and the template has to match that. To make use of this template, what \ we need to do is equate the coefficients of our template to the coefficients \ of ", Cell[BoxData[ \(TraditionalForm\`f\)]], ". That gives a bunch of equations for ", Cell[BoxData[ \(TraditionalForm\`r\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\)]], ". We can get the coefficients by using the ", ButtonBox["CoefficientList", ButtonStyle->"RefGuideLink"], " function:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(CoefficientList[a\ x^2 + b\ x + c, x]\)], "Input", CellLabel->"In[23]:="], Cell[BoxData[ \({c, b, a}\)], "Output", CellLabel->"Out[23]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(CoefficientList[f[x, r, h], x]\)], "Input", CellLabel->"In[24]:="], Cell[BoxData[ \({h, r, 0, \(-1\)}\)], "Output", CellLabel->"Out[24]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(CoefficientList[template, x]\)], "Input", CellLabel->"In[25]:="], Cell[BoxData[ \({p1\^2\ p2, \(-p1\^2\) - 2\ p1\ p2, 2\ p1 + p2, \(-1\)}\)], "Output", CellLabel->"Out[25]="] }, Open ]], Cell[TextData[{ "We now need to turn these into a list of equations. The ", ButtonBox["Thread", ButtonStyle->"RefG