(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 4.2' 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[ 48618, 1895]*) (*NotebookOutlinePosition[ 84626, 3126]*) (* CellTagsIndexPosition[ 84582, 3122]*) (*WindowFrame->Normal*) Notebook[{ Cell["Differential Equations", "Title", ImageRegion->{{0, 1}, {0, 1}}], Cell["R.G. Palmer", "Subsubtitle", TextAlignment->Center, ImageRegion->{{0, 1}, {0, 1}}], Cell[CellGroupData[{ Cell["Background and Copyright", "Subsection"], Cell[TextData[{ "This ", StyleBox["Mathematica", FontSlant->"Italic"], " Notebook was written by Richard G. Palmer (Physics Department, Duke \ University) for use in a course he taught. As of 1999, it has been made \ available for general non-profit use under the following copyright \ provision." }], "Text"], Cell[TextData[{ StyleBox["This Mathematica Notebook is Copyright Richard G. Palmer, 1996", FontWeight->"Bold"], ". It may be freely used by individuals, and by classes at academic \ institutions, provided:\n1. Credit is given to Richard Palmer as the original \ author; and\n2. It is not bought or sold or exchanged for profit, or \ incorporated into material that is bought or sold or exchanged for profit.\n\ Any other use requires the written permission of Richard Palmer, Dept. of \ Physics, Box 90305, Duke University, Durham, NC 27708, USA. ", "See ", StyleBox["http://www.phy.duke.edu/~palmer", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " for the email address.", "." }], "Text"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " Version: 4.\nDate: 7/21/00." }], "Text"], Cell[TextData[{ "See ", StyleBox["http://www.phy.duke.edu/~palmer/notebooks/", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " for other ", StyleBox["Mathematica", FontSlant->"Italic"], " noteboks by Richard Palmer." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Preface", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "This Notebook concerns solving ordinary differential equations ", StyleBox["analytically", FontSlant->"Italic"], " using Mathematica. A companion Notebook, ", StyleBox["NDE.nb", FontWeight->"Bold"], ", concerns solving differential equations (ODEs and PDEs) ", StyleBox["numerically", FontSlant->"Italic"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ There's also some further development of Mathematica skills, \ especially including pure functions. \ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["There are 5 problems embedded in the Notebook.", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "This Notebook is complete in itself (assuming you know Mathematica \ basics), but you might also want to look at sections 1.5.8, and 3.5.10 of the \ Mathematica book for more on ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["DSolve[]", "Section", ImageRegion->{{0, 1}, {0, 1}}], Cell[CellGroupData[{ Cell["Warning -- a common problem", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "In this section we'll use ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " over and over again as the ", StyleBox["unknown function", FontSlant->"Italic"], " that we're trying to find. It's crucial that Mathematica doesn't have a \ stored value for this (or for ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", or ", StyleBox["y'[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", etc). But it's very easy to mistakenly type e.g., the ", StyleBox["assignment:", FontSlant->"Italic"], "\n\t", StyleBox["y'[x] = x y[x]\n", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "instead of the ", StyleBox["equation:", FontSlant->"Italic"], "\n\t", StyleBox["y'[x] == x y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ",\nthus producing exactly such a problem." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "If you do this, or you're having other weird problems, try ", StyleBox["Remove[y]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Introduction to DSolve[]", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Mathematica has a built-in function for solving ODEs analytically, called \ ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". First let's look at its definition:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["?DSolve", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "1. The first form, ", StyleBox["DSolve[eqn, y[x], x] ", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "or", StyleBox[" DSolve[eqn, y, x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", is the form we'll normally use. Or specifying more than one equation, ", StyleBox[" DSolve[{eqn1, eqn2}, y[x], x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", allows you to give some boundary conditions, as we'll see." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "2. The second form shows that you can specify simultaneous differential \ equations. Specifying N more than one dependent variables and M\ \[GreaterEqual]N equations lets you solve a set of N simultaneous ODEs ", StyleBox["with the same independent variable", FontSlant->"Italic"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ 3. The third form shows that you specify more than one independent \ variable, which would let us solve a PDE.\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Examples", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "First we'll try a few examples to see what ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " can do. Here's an easy one:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["DSolve[y'[x] + y[x] == 1, y[x], x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Note the following:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "1. I specified the equation with ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " written out everywhere it appears, with derivatives specified by primes. \ It is ", StyleBox["crucial", FontSlant->"Italic"], " to use ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " instead of ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "; not doing so is the most common mistake in the use of ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox[".", FontSize->12] }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "2. The arbitrary constant in the solution is given as ", StyleBox["C[1]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". If there were two, they'd be ", StyleBox["C[1]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and ", StyleBox["C[2]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", etc. You can change the name ", StyleBox["C", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and the numbering scheme if necessary; that's the sort of thing to look \ up if you need it." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "3. The result is specified in the form of a ", StyleBox["rule", FontSlant->"Italic"], ", ", StyleBox["y[x] -> ", FontWeight->"Bold"], StyleBox["expr", FontSlant->"Italic"], ", inside a ", StyleBox["double", FontSlant->"Italic"], " list. If we had more than one dependent variable, like ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and ", StyleBox["z[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", then the inner list would have an entry for each. If there were more \ than one solution, there'd be more entries in the outer list. So for \ example, we might get" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["{{y[x]->expr1, z[x]->expr2},{y[x]->expr3, z[x]->expr4}}", "Print", CellMargins->{{23, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "if we had two solutions and two variables. ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", ", StyleBox["Solve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", ", StyleBox["NSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", and ", StyleBox["NDSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " all use this same form for their solutions." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "We need to get an ", StyleBox["expression", FontSlant->"Italic"], " for the answer, instead of the ", StyleBox["rule", FontSlant->"Italic"], " for ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". We can do that with a simple replacement:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["answer = y[x] /. %[[1]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Plea for understanding", "Subsubsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "If you don't fully understand the above expression, it's time for some \ review. ", StyleBox["Don't", FontSlant->"Italic"], " just go on regardless; you'll get rapidly lost. In brief, the %[[1]] \ took the first (and only) element of the previous expression, thus producing" }], "Text", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "{y[x] -> ", StyleBox["solution", FontSlant->"Italic"], "}" }], "Print", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell["Then we had", "Text", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "y[x] /. {y[x] -> ", StyleBox["solution", FontSlant->"Italic"], "}" }], "Print", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "which just produced ", StyleBox["solution", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], ". Finally ", StyleBox["answer", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " was set to this result. If you still feel hazy about the idiom" }], "Text", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["expression", FontSlant->"Italic"], " /. {", StyleBox["symbol", FontSlant->"Italic"], " -> ", StyleBox["replacement", FontSlant->"Italic"], "}" }], "Print", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ please read sections 1.4.2 and 2.4.1 of the Mathematica book before \ continuing.\ \>", "Text", CellMargins->{{55, Inherited}, {Inherited, Inherited}}, ImageRegion->{{0, 1}, {0, 1}}], Cell["Here's an exact (but not separable) equation:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["DSolve[x y'[x] + x + y[x] == 0, y[x], x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Here's another example from lectures. It isn't exact, but because the \ left hand side is linear in ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " it could be solved by finding an integrating factor depending only on ", StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " (I don't know whether Mathematica uses that approach though)." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["DSolve[x y'[x] + (1 + x)*y[x] == Exp[x], y[x], x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Here's a equation that's first order but ", StyleBox["second", FontSlant->"Italic"], " degree." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["DSolve[y'[x]^2 + x y'[x] - y[x] == 0, y[x], x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ And here's an example that Mathematica obviously knows about:\ \>", \ "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["DSolve[x^2 y''[x] + x y'[x] + x^2 y[x] == 0, y[x], x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Note that Mathematica uses ", StyleBox["BesselY[m,x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " for the N", StyleBox["m", FontVariations->{"CompatibilityType"->"Subscript"}], "(x) Bessel function. Physicists usually use N", StyleBox["m", FontVariations->{"CompatibilityType"->"Subscript"}], "(x) to reserve the letter Y for spherical harmonics, but Y", StyleBox["m", FontVariations->{"CompatibilityType"->"Subscript"}], "(x) is more common outside physics." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Here's a example that we actually got ", StyleBox["two", FontSlant->"Italic"], " solutions." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[BoxData[ \(DSolve[\(y'\)[x]^2\ + \ \(y'\)[x]\ - \ 2\ \ == \ 0, \ y[x], x]\)], "Input"], Cell["Let us extract the first and second solutions:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["y[x] /. %[[1]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[BoxData[ \(y[x] /. %%[\([2]\)]\)], "Input"], Cell[TextData[{ "Here's another exact equation with a boundary condition specified (note \ just how, especially that ", StyleBox["==", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", not ", StyleBox["=", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", must be used):" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ DSolve[{x y'[x]/y[x] + Log[y[x]] == 2, y[1] == 1}, y[x], x]\ \>", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ Note that the solution is complete, without any arbitrary constants \ remaining:\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[BoxData[ \(y[x] /. %[\([1]\)]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Problem 1", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["a.", FontWeight->"Bold"], " Use ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " to solve the general 2nd order homogeneous linear ODE with constant \ coefficients:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["ay'' + by' + cy = 0", "Print", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["Watch out, I'm trying to trick you with notation", FontSlant->"Italic"], ". Note that this can be solved symbolically, without giving numerical \ values for ", StyleBox["a", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", ", StyleBox["b", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", and ", StyleBox["c", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["b.", FontWeight->"Bold"], " Repeat for an ", StyleBox["inhomogeneous", FontSlant->"Italic"], " 2nd order linear ODEequation with a simple function like ", StyleBox["Cos[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " or ", StyleBox["Exp[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " or ", StyleBox["x^3", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " (or a product or sum of some of these) in the right hand side. This time \ you'd better use numerical values (e.g. 1, 3, 2) for ", StyleBox["a", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", ", StyleBox["b", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", and ", StyleBox["c", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", or it may take a very long time. To simplify your results, you can try \ ", StyleBox["Simplify[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["c.", FontWeight->"Bold"], " Extract the solution at the point where ", StyleBox["C[1]", FontFamily->"Courier", FontWeight->"Bold"], "equals 0 and ", StyleBox["C[2]", FontFamily->"Courier", FontWeight->"Bold"], "equals 1. Do ", StyleBox["not", FontSlant->"Italic"], " set ", StyleBox["C[1]", FontFamily->"Courier", FontWeight->"Bold"], " and ", StyleBox["C[2]", FontFamily->"Courier", FontWeight->"Bold"], " themselves to numbers, or that'll mess up future use of ", StyleBox["DSolve[]", FontFamily->"Courier", FontWeight->"Bold"], "; use a replacement instead." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Pure function solutions", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "There's actually another way to use ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", specifying just the name of the dependent variable (e.g. ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ") instead of a function (like ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "). This gives a solution as a ", StyleBox["pure function", FontSlant->"Italic"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "We studied pure functions in the last section of ", StyleBox["FFT", FontWeight->"Bold"], " notebook. For a review either look at that again or read section 2.2.5 \ of the Mathematica book. Or finish reading this section and see if you \ remember enough." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["For example", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["purefun = DSolve[y''[x] == - omega^2 y[x], y, x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Note the following:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "1. I still had to use ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " in specifying the equation; it was just the second argument of ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " that changed." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "2. The result is still in the form of a rule inside a double list, though \ the rule is for ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", not for ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox[".", FontSize->12] }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "3. The pure function is specified by the function is filled with ", StyleBox["#1", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " (meaning \"put the first argument here\"). Note also the ", StyleBox["&", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " at the end of the pure function; this is an essential part of the syntax, \ all too easy to forget. You also often need the parentheses to limit the \ scope of ", StyleBox["&", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "4. When we use a pure function with an argument, ", StyleBox["var", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], " is replaced by the actual argument. ", StyleBox["E.g.:", FontFamily->"Courier", FontSize->12, FontSlant->"Italic"] }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["y[t] /. purefun[[1]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["y[6.5] /. purefun[[1]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["5. There's an alternate syntax which would read: ", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ {{y -> Function[x, C[2] Cos[omega x] + C[1] Sin[omega x]]}}\ \>", \ "Output", CellEditDuplicate->False, Evaluatable->False, GeneratedCell->False, CellAutoOverwrite->False], Cell[TextData[{ "The syntax ", StyleBox["Function[x, ", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox["f(", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox[")", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " is equivalent to ", StyleBox["(", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox["f(", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["#1", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox[")", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["&)", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". Note that now there's dummy argument. Either syntax can be used to \ specify a pure function, but you need to recognize both." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "What's the advantage of using the pure function version of ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "? Obviously it makes it easier to specify an ODE with independent \ variable ", StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " (say) and then use the solution with ", StyleBox["t", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " (say), but that's not exactly a magnificent achievement (why not use ", StyleBox["t", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " in the first place?). The main advantage only comes when you want to use \ quantities related to ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " that don't actually contain an explicit ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". The simplest example is ", StyleBox["y'[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". Compare the following: " }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["1 + y'[x] /. y[x] -> x^3", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["1 + y'[x] /. y -> (#1^3 &)", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["1 + y'[x] /. y -> Function[x, x^3] (* same as previous *)", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "In the first case the replacement operator ", StyleBox["/.", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " did ", StyleBox["not", FontSlant->"Italic"], " regard ", StyleBox["y'[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " as an instance of ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", so no replacement was made. This is a consequence of the internal \ representation of ", StyleBox["y'[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", which you can read about in section 3.5.4 of the Mathematica book if you \ wish. In the second and third cases I defined the equivalent pure function \ and it worked." }], "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Problem 2", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell["Solve the differential equation", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["y'[x] + Cos[x] y[x] == Sin[2x]/2", "Print", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "with the boundary condition ", StyleBox["y[0] = 0", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", and numerically evaluate ", StyleBox["y''[Pi/2]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". ", StyleBox["Do the whole problem as one evaluation.", FontSlant->"Italic"] }], "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Implicit solutions", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "An interesting feature is that ", StyleBox["Mathematica", FontSlant->"Italic"], " can sometimes get implicit solutions when no explicit ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " solution can be found." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["DSolve[x^2 y'[x] + y[x] == x y[x] y'[x], y[x], x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "The error messages essentially say that it couldn't get an explicit \ solution, but then it gives us a solution in terms of ", StyleBox["Solve", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "'ing an implicit equation. In this particular case we could solve it for \ ", StyleBox["x[y]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " instead:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Solve[%[[1]] /. y[x]->y, x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Make sure you understand how that worked! Note that ", StyleBox["[[", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox["n", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["]]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " will take the ", StyleBox["n", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], "th argument of any function call:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["fun[arg1, arg2, arg3][[2]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Its most common use in taking the ", StyleBox["n", FontFamily->"Courier", FontSize->12, FontWeight->"Bold", FontSlant->"Italic"], "th element of a list is just a special case, since lists are represented \ internally as ", StyleBox["List[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " \"functions\":" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["{a, b, c} //FullForm", "Input", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Problem 3", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell["Solve:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["1 + y'[x]^2 + 2 y[x] y''[x] == 0", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["with the boundary conditions:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ y[0] == 1; y'[0] == 1;\ \>", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "and make a plot of ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " against ", StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " over the natural domain. Hints: (a) aim for ", StyleBox["x[y]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "; (b) ", StyleBox["x'[y]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " is relatively simple; (c) only one solution works; (d) ", StyleBox["Solve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " for ", StyleBox["C[1]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and ", StyleBox["C[2]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " yourself; (e) ", StyleBox["ParametricPlot[{x[y], y}, {y, y1, y2}]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " will make the plot; (f) you should figure out what ", StyleBox["y1", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and ", StyleBox["y2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " need to be." }], "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Laplace Transforms", "Section", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ Another analytic technique for solving ODEs is to use a Laplace \ transform to turn them into algebraic equations. That works for some PDEs \ too.\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["I'll do one example and you'll do another. ", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[CellGroupData[{ Cell["Example", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " ", StyleBox["4", FontSize->18], " has Laplace transforms function called ", StyleBox["LaplaceTransform", FontFamily->"Courier", FontWeight->"Bold"], ". ", StyleBox["Mathematica", FontSlant->"Italic"], " ", StyleBox["3", FontSize->18], " doesn't have Laplace transforms built into the kernel, but they are \ available in a standard package called ", StyleBox["Calculus`LaplaceTransform`", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". In ", StyleBox["Mathematica", FontSlant->"Italic"], " ", StyleBox["3", FontSize->18], ", you must first read in the package: ", StyleBox["Needs[\"Calculus`LaplaceTransform`\"]", FontFamily->"Courier", FontWeight->"Bold"] }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Let's check the definition of the Laplace transform:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["?LaplaceTransform", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ Now let's suppose we want to solve the following differential \ equation, in which (with hindsight) I'll name the left and right hand sides \ separately: \ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ lhs = y''[x] + 3 y'[x] + 2 y[x]; rhs = 4 DiracDelta[x] + Exp[x]; eqn = (lhs == rhs)\ \>", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "OK? (Notice that there were actually 3 evaluations in that cell and so \ the ", StyleBox["In[n]:=", FontSlant->"Italic"], " and ", StyleBox["Out[n]=", FontSlant->"Italic"], " numbers both advanced by 3). I put a delta function on the right hand \ side just because I could." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Now we want to:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "1. Take a Laplace transform of the whole equation (both sides), defining \ the Laplace transform of ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " to be ", StyleBox["g[s]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", say;" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "2. Solve the resulting ", StyleBox["algebraic", FontSlant->"Italic"], " equation for ", StyleBox["g[s]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "; and" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "3. Take an inverse Laplace transform to find ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox[".", FontSize->12] }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "The very first step is the trickiest; telling Mathematica that the \ Laplace transform of ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " is to be ", StyleBox["g[s]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ":" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["LaplaceTransform[y[x_], x_, s_, ___] ^:= g[s]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Note: The ^:= operator and the", " ", StyleBox["_ _ _", FontWeight->"Plain"], " pattern" }], "Subsubsection", CellMargins->{{17, Inherited}, {Inherited, Inherited}}], Cell[TextData[{ "The ", StyleBox["^:=", FontFamily->"Courier", FontWeight->"Bold"], " operator has basically the same function as ", StyleBox[":=", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", specifying that the left hand side should be replaced by the right hand \ side wherever it occurs. But it stores the rule with the first inner \ argument, ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", instead of with the outer function ", StyleBox["LaplaceTransform", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". This will ensure that the rule is applied immediately that ", StyleBox["y[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " appears in the above context, and is much more efficient than using ", StyleBox[":=", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". (It works with ", StyleBox[":=", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " too, but is much slower)." }], "Text", CellMargins->{{53, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Check what's known about ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], StyleBox[":", FontSize->12] }], "Text", CellMargins->{{53, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, ImageRegion->{{0, 1}, {0, 1}}], Cell["??y", "Input", CellMargins->{{53, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "The left hand side of the above expression is a ", StyleBox["pattern", FontSlant->"Italic"], " which would match, for example: " }], "Text", CellMargins->{{53, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ LaplaceTransform[y[x],x,s] LaplaceTransform[y[t],t,u] LaplaceTransform[y[t],t,u,Pi,SillyArgument]\ \>", "Print", CellMargins->{{53, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "The single underline (called ", StyleBox["Blank", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ") matches any expression (and the two ones labeled ", StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " must be the same), while the triple underline (called ", StyleBox["BlankNullSequence", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", or ", StyleBox["triple blank", FontSlant->"Italic"], ") matches any sequence of zero or more expressions. We need the triple \ blank because the standard definition of ", StyleBox["LaplaceTransform[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " contains an optional 4th argument." }], "Text", CellMargins->{{53, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, TextJustification->0, ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Now let's take the Laplace transform of each side of the equation. (We \ can't easily do it all together, because the first argument of ", StyleBox["LaplaceTransform[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " is supposed to be an expression, and it wouldn't know what to do with an \ equation. Experts should consider using ", StyleBox["Map[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " to do it altogether, but we'll play it simple and do the two sides \ separately.)" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["lhsLT = LaplaceTransform[lhs,x,s]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["rhsLT = LaplaceTransform[rhs,x,s]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Ignore the spelling error warning; they're sometimes useful, but usually \ just annoying. Note that the Laplace transform of the left hand side \ involves the initial conditions ", StyleBox["y[0]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and ", StyleBox["y'[0]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". That's one of the nice features of the Laplace transform approach -- \ the boundary conditions get built in. " }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Now we can solve the equation for ", StyleBox["g[s]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", turn it into an expression (rather than a list of rules), and take its \ inverse Laplace transform:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Solve[lhsLT == rhsLT, g[s]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["ansLT = g[s] /. %[[1]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["InverseLaplaceTransform[ansLT,s,x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ Pretty nice, huh? Of course we could evaluate this further if we \ were given explicit boundary conditions.\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Problem 4", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Solve the differential equation ", StyleBox["y'' + 9y = sin 3x", FontWeight->"Bold"], " with the boundary conditions ", StyleBox["y(0) = 1", FontWeight->"Bold"], " and ", StyleBox["y(", FontWeight->"Bold"], "\[Pi]", StyleBox["/2) = -1", FontWeight->"Bold"], ". Yes, this is a two-point boundary condition." }], "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Series Solutions", "Section", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Here we explore using the Frobenius method to generate a power series \ solution to a differential equation. This can be used even where there's no \ closed form solution (so ", StyleBox["DSolve[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " would necessarily fail). Fuch's theorem tells us that we will always get \ at least one power series solution around any point that's not too singular \ (no worse than a regular singular point)." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["I'll do one example and you'll do another. ", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[CellGroupData[{ Cell["Legendre equation", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Here's the Legendre equation for ", StyleBox["l=3", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ":" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["legendre3 = (1-x^2) y''[x] - 2 x y'[x] + 12 y[x] == 0", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ First, let's set up a general series expansion, to 7th order say:\ \ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["genericseries[x_] := Sum[a[i] x^i, {i,0,7}] + O[x]^8", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Here it is:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["genericseries[x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Note how:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "1. I made ", StyleBox["genericseries", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " a function of ", StyleBox["x", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", so that ", StyleBox["genericseries", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " itself can used as a direct replacement for ", StyleBox["y", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", and things like ", StyleBox["y'[x]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " will be interpreted correctly. Alternatively I could have defined ", StyleBox["genericseries", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " itself directly as a pure function." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "2. I used ", StyleBox["Sum", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " as a shortcut to generating the terms, using ", StyleBox["a[i]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " for the coefficients." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "3. I added on ", StyleBox["O[x]^8", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " at the end to make the result a proper series expansion in Mathematica's \ sense (a ", StyleBox["SeriesData", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " object, technically). We discussed this in the ", StyleBox["Series.nb", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " Notebook; see also in section 3.6.2 in the Mathematica book. To check it \ worked, we could look at the ", StyleBox["FullForm", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " of the result:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["% // FullForm", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["OK, now we can substitute:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["legendre3 /. y -> genericseries", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ How about that! We can see the equations that need to be satisfied \ to make each coefficient vanish. And Mathematica will solve them all at once \ for us:\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Solve[%]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ Ignore the warning. Let's look at our generic series with these \ substitutions:\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["genericseries[x] /. %[[1]]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Notice that there are only only two coefficients left, ", StyleBox["a[3]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " and ", StyleBox["a[6]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". Let's collect up their terms; ", StyleBox["Collect[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " is a useful function for such things:" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Collect[%,{a[3],a[6]}]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Note that there's also an ", StyleBox["O[x]^8", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " term, which got dropped because ", StyleBox["Collect[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " had to convert the series to the form of an ordinary expression (like \ applying ", StyleBox["Normal[]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ")." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "So look at what we got; the first term (multiplying by ", StyleBox["a[3]", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ") is essentially the l=3 Legendre polynomial P", StyleBox["3", FontSize->13, FontVariations->{"CompatibilityType"->"Subscript"}], "(x):" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["LegendreP[3,x]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "We'd need to set ", StyleBox["a[3] = 5/2 ", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "to get the standard normalization." }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "And the other part is the start of the infinite series expansion for the \ l=3 Legendre function of the second kind Q", StyleBox["3", FontSize->13, FontVariations->{"CompatibilityType"->"Subscript"}], "(x):" }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["Series[LegendreQ[3,x], {x,0,6}]", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Multiply it by 15/8 to check it's the same:", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["15/8 %", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell["Yes. It almost makes you believe in Legendre functions.", "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["\<\ OK course you could study the results a little harder to convince \ yourself that going to higher order in the original expansion would only add \ terms to the Q part, not the P part.\ \>", "Text", ImageRegion->{{0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Problem 5", "Subsection", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "Generate a power series solution of the ", StyleBox["m=2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " Bessel's equation " }], "Text", ImageRegion->{{0, 1}, {0, 1}}], Cell["bessel2 = x^2 y''[x] + x y'[x] + (x^2 - 4) y[x] == 0", "Input", ImageRegion->{{0, 1}, {0, 1}}], Cell[TextData[{ "by expanding around ", StyleBox["x = 0", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". The indicial equation tells you to use an ", StyleBox["x^k", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " prefactor with ", StyleBox["k = 2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], " (or ", StyleBox["-2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], "), so use an expansion with powers ", StyleBox["x^2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", ", StyleBox["x^3", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ", up to ", StyleBox["x^8", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], ". 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{{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Subtitle", CounterAssignments->{{"Section", 0}, {"Equation", 0}, {"Figure", 0}, { "Subsubtitle", 0}}, FontFamily->"Helvetica", FontSize->24, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subtitle", "Presentation"], CellMargins->{{24, 10}, {20, 20}}, LineSpacing->{1, 0}, FontSize->36], Cell[StyleData["Subtitle", "Condensed"], CellMargins->{{8, 10}, {4, 4}}, FontSize->14], Cell[StyleData["Subtitle", "Printout"], CellMargins->{{2, 10}, {12, 8}}, FontSize->18] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubtitle"], CellMargins->{{12, Inherited}, {20, 15}}, CellGroupingRules->{"TitleGrouping", 20}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Subsubtitle", CounterAssignments->{{"Section", 0}, {"Equation", 0}, {"Figure", 0}}, FontFamily->"Helvetica", FontSize->14, FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subsubtitle", "Presentation"], CellMargins->{{24, 10}, {20, 20}}, LineSpacing->{1, 0}, FontSize->24], Cell[StyleData["Subsubtitle", "Condensed"], CellMargins->{{8, 10}, {8, 8}}, FontSize->12], Cell[StyleData["Subsubtitle", "Printout"], CellMargins->{{2, 10}, {12, 8}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Section"], CellDingbat->"\[FilledSquare]", CellMargins->{{25, Inherited}, {8, 24}}, CellGroupingRules->{"SectionGrouping", 30}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Section", CounterAssignments->{{"Subsection", 0}, {"Subsubsection", 0}}, FontFamily->"Helvetica", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Section", "Presentation"], CellMargins->{{40, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->24], Cell[StyleData["Section", "Condensed"], CellMargins->{{18, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Section", "Printout"], CellMargins->{{13, 0}, {7, 22}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 20}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, FontFamily->"Times", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->22], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Subsubsection", FontFamily->"Times", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], Cell[StyleData["Subsubsection", "Presentation"], CellMargins->{{34, 10}, {11, 26}}, LineSpacing->{1, 0}, FontSize->18], Cell[StyleData["Subsubsection", "Condensed"], CellMargins->{{17, Inherited}, {6, 12}}, FontSize->10], Cell[StyleData["Subsubsection", "Printout"], CellMargins->{{9, 0}, {7, 14}}, FontSize->11] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Body Text", "Section"], Cell[CellGroupData[{ Cell[StyleData["Text"], CellMargins->{{12, 10}, {7, 7}}, InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, Hyphenation->True, LineSpacing->{1, 3}, CounterIncrements->"Text"], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}, TextJustification->0.5, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellMargins->{{12, 10}, {6, 6}}, DefaultNewInlineCellStyle->"None", Hyphenation->True, LineSpacing->{1, 3}, LanguageCategory->"NaturalLanguage", CounterIncrements->"SmallText", FontFamily->"Helvetica", FontSize->9], Cell[StyleData["SmallText", "Presentation"], CellMargins->{{24, 10}, {8, 8}}, LineSpacing->{1, 5}, FontSize->12], Cell[StyleData["SmallText", "Condensed"], CellMargins->{{8, 10}, {5, 5}}, LineSpacing->{1, 2}, FontSize->9], Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, TextJustification->0.5, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Input/Output", "Section"], Cell["\<\ The cells in this section define styles used for input and output \ to the kernel. Be careful when modifying, renaming, or removing these \ styles, because the front end associates special meanings with these style \ names. Some attributes for these styles are actually set in FormatType Styles \ (in the last section of this stylesheet). \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Input"], CellMargins->{{45, 10}, {5, 7}}, Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, LinebreakAdjustments->{0.85, 2, 10, 0, 1}, CounterIncrements->"Input", FontWeight->"Bold"], Cell[StyleData["Input", "Presentation"], CellMargins->{{72, Inherited}, {8, 10}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Input", "Condensed"], CellMargins->{{40, 10}, {2, 3}}, FontSize->11], Cell[StyleData["Input", "Printout"], CellMargins->{{39, 0}, {4, 6}}, LinebreakAdjustments->{0.85, 2, 10, 1, 1}, FontSize->9] }, Closed]], Cell[StyleData["InputOnly"], Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, LinebreakAdjustments->{0.85, 2, 10, 0, 1}, CounterIncrements->"Input", StyleMenuListing->None, FontWeight->"Bold"], Cell[CellGroupData[{ Cell[StyleData["Output"], CellMargins->{{47, 10}, {7, 5}}, CellEditDuplicate->True, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, CounterIncrements->"Output"], Cell[StyleData["Output", "Presentation"], CellMargins->{{72, Inherited}, {10, 8}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Output", "Condensed"], CellMargins->{{41, Inherited}, {3, 2}}, FontSize->11], Cell[StyleData["Output", "Printout"], CellMargins->{{39, 0}, {6, 4}}, FontSize->9] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Message"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Message", StyleMenuListing->None, FontSize->11, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Message", "Presentation"], CellMargins->{{72, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Message", "Condensed"], CellMargins->{{41, Inherited}, {Inherited, Inherited}}, FontSize->11], Cell[StyleData["Message", "Printout"], CellMargins->{{39, Inherited}, {Inherited, Inherited}}, FontSize->7, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Print"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Print", StyleMenuListing->None], Cell[StyleData["Print", "Presentation"], CellMargins->{{72, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Print", "Condensed"], CellMargins->{{41, Inherited}, {Inherited, Inherited}}, FontSize->11], Cell[StyleData["Print", "Printout"], CellMargins->{{39, Inherited}, {Inherited, Inherited}}, FontSize->8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Graphics"], CellMargins->{{4, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"GraphicsGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, LanguageCategory->None, FormatType->InputForm, CounterIncrements->"Graphics", ImageMargins->{{43, Inherited}, {Inherited, 0}}, StyleMenuListing->None, FontFamily->"Courier", FontSize->10], Cell[StyleData["Graphics", "Presentation"], ImageMargins->{{62, Inherited}, {Inherited, 0}}], Cell[StyleData["Graphics", "Condensed"], ImageMargins->{{38, Inherited}, {Inherited, 0}}, Magnification->0.6], Cell[StyleData["Graphics", "Printout"], ImageMargins->{{30, Inherited}, {Inherited, 0}}, Magnification->0.8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["CellLabel"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->9, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["CellLabel", "Presentation"], FontSize->12], Cell[StyleData["CellLabel", "Condensed"], FontSize->9], Cell[StyleData["CellLabel", "Printout"], FontFamily->"Courier", FontSize->8, FontSlant->"Italic", FontColor->GrayLevel[0]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Inline Formatting", "Section"], Cell["\<\ These styles are for modifying individual words or letters in a \ cell exclusive of the cell tag.\ \>", "Text"], Cell[StyleData["RM"], StyleMenuListing->None, FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["BF"], StyleMenuListing->None, FontWeight->"Bold"], Cell[StyleData["IT"], StyleMenuListing->None, FontSlant->"Italic"], Cell[StyleData["TR"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["TI"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["TB"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["TBI"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["MR"], StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["MO"], StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["MB"], StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["MBO"], StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["SR"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["SO"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SB"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["SBO"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[StyleData["SO10"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "Printout"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->7, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "EnhancedPrintout"], StyleMenuListing->None, FontFamily->"Futura", FontSize->7, FontWeight->"Plain", FontSlant->"Italic"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Formulas and Programming", "Section"], Cell[CellGroupData[{ Cell[StyleData["InlineFormula"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, LanguageCategory->"Formula", ScriptLevel->1, SingleLetterItalics->True], Cell[StyleData["InlineFormula", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["InlineFormula", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["InlineFormula", "Printout"], CellMargins->{{2, 0}, {6, 6}}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["DisplayFormula"], CellMargins->{{42, Inherited}, {Inherited, Inherited}}, CellHorizontalScrolling->True, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, LanguageCategory->"Formula", ScriptLevel->0, SingleLetterItalics->True, UnderoverscriptBoxOptions->{LimitsPositioning->True}], Cell[StyleData["DisplayFormula", "Presentation"], LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["DisplayFormula", "Condensed"], LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["DisplayFormula", "Printout"], FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Program"], CellFrame->{{0, 0}, {0.5, 0.5}}, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, Hyphenation->False, LanguageCategory->"Formula", ScriptLevel->1, FontFamily->"Courier"], Cell[StyleData["Program", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["Program", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["Program", "Printout"], CellMargins->{{2, 0}, {6, 6}}, FontSize->9] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Hyperlink Styles", "Section"], Cell["\<\ The cells below define styles useful for making hypertext \ ButtonBoxes. The \"Hyperlink\" style is for links within the same Notebook, \ or between Notebooks.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Hyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonNote->ButtonData}], Cell[StyleData["Hyperlink", "Presentation"], FontSize->16], Cell[StyleData["Hyperlink", "Condensed"], FontSize->11], Cell[StyleData["Hyperlink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell["\<\ The following styles are for linking automatically to the on-line \ help system.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["MainBookLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "MainBook", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["MainBookLink", "Presentation"], FontSize->16], Cell[StyleData["MainBookLink", "Condensed"], FontSize->11], Cell[StyleData["MainBookLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["AddOnsLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "AddOns", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["AddOnsLink", "Presentation"], FontSize->16], Cell[StyleData["AddOnsLink", "Condensed"], FontSize->11], Cell[StyleData["AddOnsLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RefGuideLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "RefGuide", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["RefGuideLink", "Presentation"], FontSize->16], Cell[StyleData["RefGuideLink", "Condensed"], FontSize->11], Cell[StyleData["RefGuideLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GettingStartedLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "GettingStarted", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["GettingStartedLink", "Presentation"], FontSize->16], Cell[StyleData["GettingStartedLink", "Condensed"], FontSize->11], Cell[StyleData["GettingStartedLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["OtherInformationLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "OtherInformation", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["OtherInformationLink", "Presentation"], FontSize->16], Cell[StyleData["OtherInformationLink", "Condensed"], FontSize->11], Cell[StyleData["OtherInformationLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Headers and Footers", "Section"], Cell[StyleData["Header"], CellMargins->{{0, 0}, {4, 1}}, DefaultNewInlineCellStyle->"None", LanguageCategory->"NaturalLanguage", 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