(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of 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[ 220843, 8413]*) (*NotebookOutlinePosition[ 221769, 8444]*) (* CellTagsIndexPosition[ 221725, 8440]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[ "APPENDIX \n\nSOME MATHEMATICAL TOOLS FOR POPULATION BIOLOGISTS"], "Subtitle",\ Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData["Introduction"], "Subsection", Evaluatable->False, PageBreakAbove->False, AspectRatioFixed->True], Cell[TextData[{ " I assume a basic knowledge of elementary algebra and calculus. \ This appendix discusses some other mathematical tools which are frequently \ used in the development of ecological and evolutionary models, and shows how \ they can be implemented in ", StyleBox["Mathematica", FontSlant->"Italic"], ". Some worked examples are shown. You should do these examples by \ activating the ", StyleBox["Mathematica", FontSlant->"Italic"], " input, and make up more examples of your own for any of this material \ which is unfamiliar to you." }], "Text", Evaluatable->False, PageBreakAbove->False, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["A Taylor series expansions"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ " If ", StyleBox["f(x) ", FontSlant->"Italic"], "is a function of ", StyleBox["x", FontSlant->"Italic"], ", it is often useful to linearize it about a value ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic"], "which is of particular interest. We can write\n ", StyleBox["f(x) ~ f(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + (x \[LongDash] x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") \t\t", FontSlant->"Italic"], "[1]", StyleBox["\n", FontSlant->"Italic"], "where ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") ", FontSlant->"Italic"], "is the derivative of ", StyleBox["f(x) ", FontSlant->"Italic"], "evaluated at ", StyleBox["x = x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ". This approximation is valid for values of ", StyleBox["x ", FontSlant->"Italic"], "sufficiently close to ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ". It relies on the function being locally linear; the derivative is the \ slope of the curve at ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ", so that the righthand side of [1] is the best linear approximation to \ the curve.\nThis linear approximation is given in ", StyleBox["Mathematica", FontSlant->"Italic"], " by the function ", StyleBox["Series[f[x], {x,x", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[", 1}]", FontWeight->"Bold"], ", the '1'", StyleBox[" ", FontWeight->"Bold"], "indicating that the approximation includes only the first (linear) term. \ For example, we find the linear approximation to ln x near x = 2 and assess \ its accuracy: " }], "Text", CellMargins->{{3, 107}, {Inherited, Inherited}}, Evaluatable->False, CellLabelMargins->{{4, Inherited}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[TextData["Series[Log[x],{x,2.,1}]"], "Input", AspectRatioFixed->True], Cell[TextData["f[x_]=Normal[Series[Log[x],{x,2.,1}]]"], "Input", AspectRatioFixed->True], Cell[TextData["Table[{x,Log[x],f[x]},{x,1.5,2.5,.1}] //TableForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ "The last term in ", StyleBox["Series ", FontWeight->"Bold"], "shows that the approximation is accurate to order (x\[LongDash]2)", StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "; it can be deleted to give an ordinary expression by using ", StyleBox["Normal", FontWeight->"Bold"], ".\n The linear approximation [1] depends on the linearity of the \ function near ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ", and will break down as soon as ", StyleBox["x ", FontSlant->"Italic"], "moves far enough from ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic"], "for any curvature to become appreciable. One might hope that a better \ approximation could be obtained by taking into account higher powers of ", StyleBox["(x\[LongDash]x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["), ", FontSlant->"Italic"], "leading to a power series expansion of the form\n ", StyleBox["f(x) ~ a", FontSlant->"Italic"], StyleBox["0", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" + a", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x\[LongDash]x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + a", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" (x\[LongDash]x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + a", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" (x\[LongDash]x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["+ ...\n", FontSlant->"Italic"], "For the power series to have the same derivatives as ", StyleBox["f(x) ", FontSlant->"Italic"], "at ", StyleBox["x=x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " requires that ", StyleBox["a", FontSlant->"Italic"], StyleBox["r", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" = f ", FontSlant->"Italic"], StyleBox["(r)", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")/r", FontSlant->"Italic"], "!, where ", StyleBox["f ", FontSlant->"Italic"], StyleBox["(r)", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") ", FontSlant->"Italic"], "denotes the ", StyleBox["r", FontSlant->"Italic"], "th derivative of ", StyleBox["f(x)", FontSlant->"Italic"], ". Thus the power series is" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " ", StyleBox["f(x) ~ f(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + (x\[LongDash]x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + ", FontSlant->"Italic"], StyleBox["(x\[LongDash]x", FontSlant->"Italic", FontVariations->{"Underline"->True}], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic", FontVariations->{"Underline"->True}], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], StyleBox[" f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["''(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + ", FontSlant->"Italic"], StyleBox["(x\[LongDash]x", FontSlant->"Italic", FontVariations->{"Underline"->True}], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic", FontVariations->{"Underline"->True}], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'''(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ") + ...", StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "[2]", StyleBox["\n ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" 2! ", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["3!", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "This is known as a Taylor series expansion. By taking more terms into \ account, one expects to get a better and better approximation. The Taylor \ series up to terms of order ", StyleBox["(x\[LongDash]x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic"], "can be generated with ", StyleBox["Series[f[x],{x,x", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[",n}]", FontWeight->"Bold"], "." }], "Text", Evaluatable->False, CellLabelMargins->{{414, Inherited}, {Inherited, Inherited}}, AspectRatioFixed->True], Cell[TextData[ "Find the Taylor series expansions to ln x near x = 2 including (a) quadratic \ and (b) cubic terms and compare their accuracy with that of the linear \ approximation."], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " Three important Taylor series expansions about x", StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = 0 are those for the exponential, sine and cosine functions. The \ divisors are the appropriate factorials:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Series[Exp[x],{x,0,5}]"], "Input", AspectRatioFixed->True], Cell[TextData["Series[Sin[x],{x,0,5}]"], "Input", AspectRatioFixed->True], Cell[TextData["Series[Cos[x],{x,0,5}]"], "Input", AspectRatioFixed->True], Cell[TextData[{ " This technique can be extended to a function of several \ variables, ", StyleBox["f(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[")", FontWeight->"Bold"], ", where ", StyleBox["x\:02da", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["=\:02da(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[",\:02dax", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[")", FontSlant->"Italic"], ", expanded about the point ", StyleBox["x = x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontWeight->"Bold", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ". The best linear approximation is\n ", StyleBox["f(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[") ~ f(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + ", FontSlant->"Italic"], StyleBox["\[CapitalSigma] (", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" - x", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") ", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["f/", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ " ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "[3]", StyleBox["\n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "where ", StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["f/", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "is the partial derivative of ", StyleBox["f ", FontSlant->"Italic"], "with respect to ", StyleBox["x", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], "evaluated at ", StyleBox["x = x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ". For example, we find the linear approximation to ln x", StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "/x", StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " near x", StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 2, x", StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "= 1:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "f[x1_,x2_]=Normal[Series[Log[x1/x2],\n\t\t\t\t{x1,2.,1},{x2,1.,1}]]"], "Input", AspectRatioFixed->True], Cell[TextData["f[1.9,1.1]"], "Input", AspectRatioFixed->True], Cell[TextData["Log[1.9/1.1]"], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["B Complex numbers "], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ "This section is addressed to readers who are unfamiliar with the use of \ complex numbers, which arise frequently in stability analysis in determining \ whether oscillatory behavior is likely to occur. The quadratic equation\n\t\t\ \t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["ax", FontSlant->"Italic"], StyleBox["2 ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["+ bx + c = 0\n", FontSlant->"Italic"], "has two solutions\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["x = ", FontSlant->"Italic"], "[", StyleBox["\[LongDash]b + (b", FontSlant->"Italic"], StyleBox["2 ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash] ", FontSlant->"Italic"], "4", StyleBox["ac)", FontSlant->"Italic"], StyleBox[".5", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "]", StyleBox["/", FontSlant->"Italic"], "2", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["x = ", FontSlant->"Italic"], "[", StyleBox["\[LongDash]b \[LongDash] (b", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" \[LongDash] ", FontSlant->"Italic"], "4", StyleBox["ac)", FontSlant->"Italic"], StyleBox[".5", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "]", StyleBox["/", FontSlant->"Italic"], "2", StyleBox["a\n", FontSlant->"Italic"], "When ", StyleBox["b", FontSlant->"Italic"], StyleBox["2 ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["< ", FontSlant->"Italic"], "4", StyleBox["ac", FontSlant->"Italic"], ", this presents the problem of finding the square root of a negative \ number. A similar problem can occur in solving higher order polynomial \ equations, and in other contexts. It is convenient to resolve this problem by \ inventing an \"imaginary\" number ", StyleBox["i ", FontSlant->"Italic"], "whose square is \[LongDash]1, in the same vein as earlier mathematicians \ invented negative numbers. In ", StyleBox["Mathematica", FontSlant->"Italic"], " ", StyleBox["i ", FontSlant->"Italic"], " is writtten ", StyleBox["I ", FontWeight->"Bold"], "in accordance with the convention that inbuilt quantities begin with \ capitals. Thus" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Solve[x^2 - 2x + 5 ==0]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "A number like 1 \[LongDash] 2", StyleBox["i ", FontSlant->"Italic"], ", with a real part (1) and an imaginary part (\[LongDash]2) are called ", StyleBox["complex ", FontSlant->"Italic"], "numbers.\nComplex numbers provide a consistent system when manipulated by \ the ordinary laws of algebra, provided that ", StyleBox["i", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "is replaced by \[LongDash]1 whenever it occurs. For example,\n\t\t\t(1 + \ 2", StyleBox["i", FontSlant->"Italic"], ") + (2 \[LongDash] ", StyleBox["i", FontSlant->"Italic"], ") = 3 + (2\[LongDash]1)", StyleBox["i", FontSlant->"Italic"], " = 3 +", StyleBox[" i", FontSlant->"Italic"], "\n\t\t\t(1 + 2", StyleBox["i", FontSlant->"Italic"], ")(2 \[LongDash] ", StyleBox["i", FontSlant->"Italic"], ") = 2 \[LongDash]", StyleBox[" i", FontSlant->"Italic"], " + 4", StyleBox["i", FontSlant->"Italic"], " -2", StyleBox["i", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Superscript"}], "= 4 + 3", StyleBox["i", FontSlant->"Italic"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["z1 = 1 + 2 I;\nz2 = 2 - I;"], "Input", AspectRatioFixed->True], Cell[TextData["z1 + z2"], "Input", AspectRatioFixed->True], Cell[TextData["z1 z2"], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ratio", StyleBox[" z", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["z", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "/", StyleBox["z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " is the number such that ", StyleBox["z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" z", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["z", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ". In the above example, if", StyleBox[" z", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["x", FontSlant->"Italic"], " + ", StyleBox["iy", FontSlant->"Italic"], "\n\t\t\t(2 \[LongDash]", StyleBox[" i", FontSlant->"Italic"], ")(", StyleBox["x", FontSlant->"Italic"], " +", StyleBox[" iy", FontSlant->"Italic"], ") = (2", StyleBox["x", FontSlant->"Italic"], " + ", StyleBox["y", FontSlant->"Italic"], ") +(\[LongDash]", StyleBox["x", FontSlant->"Italic"], " + 2", StyleBox["y", FontSlant->"Italic"], ")", StyleBox["i", FontSlant->"Italic"], " = 1 + 2", StyleBox["i", FontSlant->"Italic"], "\n\t\t\t2", StyleBox["x", FontSlant->"Italic"], " + ", StyleBox["y", FontSlant->"Italic"], " = 1; \[LongDash]", StyleBox["x", FontSlant->"Italic"], " + 2", StyleBox["y", FontSlant->"Italic"], " = 2\n\t\t\t", StyleBox["x", FontSlant->"Italic"], " = 0, ", StyleBox["y", FontSlant->"Italic"], " = 1\n\t\t\t", StyleBox["z", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " =", StyleBox[" i", FontSlant->"Italic"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["z3=z1/z2"], "Input", AspectRatioFixed->True], Cell[TextData["z2 z3"], "Input", AspectRatioFixed->True], Cell[TextData[{ " A complex number has two parts, real and imaginary, which can be \ represented in a two-dimensional diagram (the Argand diagram) with the real \ part along the horizontal axis and the imaginary part along the vertical \ axis. (See Figure B1 in the book.) It is sometimes convenient to represent a \ complex number by its polar coordinates in the Argand diagram, that is to say \ its ", StyleBox["modulus ", FontSlant->"Italic"], "or ", StyleBox["absolute value ", FontSlant->"Italic"], " (its distance from the origin) and its ", StyleBox["argument ", FontSlant->"Italic"], "(the angle between the line joining the point to the origin and the real \ axis):" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["r1=Abs[z1] //N"], "Input", AspectRatioFixed->True], Cell[TextData["theta1=Arg[z1] //N"], "Input", AspectRatioFixed->True], Cell[TextData["r2=Abs[z2] //N"], "Input", AspectRatioFixed->True], Cell[TextData["theta2=Arg[z2] //N"], "Input", AspectRatioFixed->True], Cell[TextData[{ "From the diagram in the book, for any complex number ", StyleBox["z", FontSlant->"Italic"], " = ", StyleBox["x", FontSlant->"Italic"], " + ", StyleBox["iy", FontSlant->"Italic"], " with ", StyleBox["r", FontSlant->"Italic"], " = Abs(", StyleBox["z", FontSlant->"Italic"], "), ", StyleBox["\[Theta] ", FontFamily->"Symbol"], "= Arg(", StyleBox["z", FontSlant->"Italic"], "), then \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " cos", StyleBox[" ", FontSlant->"Italic"], StyleBox["\[Theta]", FontFamily->"Symbol"], "\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["y", FontSlant->"Italic"], " =", StyleBox[" r", FontSlant->"Italic"], " sin ", StyleBox["\[Theta]\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t", FontFamily->"Symbol"], StyleBox["\t", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["z ", FontSlant->"Italic"], "=", StyleBox[" x", FontSlant->"Italic"], " +", StyleBox[" iy", FontSlant->"Italic"], " =", StyleBox[" r", FontSlant->"Italic"], "( cos", StyleBox[" ", FontSlant->"Italic"], StyleBox["\[Theta] ", FontFamily->"Symbol"], "+ ", StyleBox["i ", FontSlant->"Italic"], "sin ", StyleBox["\[Theta]", FontFamily->"Symbol"], ")" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["r1 Cos[theta1]"], "Input", AspectRatioFixed->True], Cell[TextData["r1 Sin[theta1]"], "Input", AspectRatioFixed->True], Cell[TextData[{ " It is possible to define functions of a complex variable by \ analogy with the corresponding definition for a real variable. Of particular \ importance is the exponential function, which can be defined by the series \ expansion\n \texp(", StyleBox["z", FontSlant->"Italic"], ") = 1 + ", StyleBox["z", FontSlant->"Italic"], " + ", StyleBox["z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/2! + ", StyleBox["z", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/3! + ..." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Exp[z1]//N"], "Input", AspectRatioFixed->True], Cell[TextData[ "Use the following function to evaluate the series expansion for exp(z1) for \ different values of n."], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["approxexp[z_,n_]:=Sum[z^i/i!,{i,0,n}]//N"], "Input", AspectRatioFixed->True], Cell[TextData[{ "If ", StyleBox["z", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " and", StyleBox[" z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " are complex numbers, it can be verified from the series expansion \ definition that\n exp(", StyleBox["z", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " +", StyleBox[" z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ") = exp(", StyleBox["z", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ") exp(", StyleBox["z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ")\na well known property for real variables. For example," }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Exp[z1+z2] //N"], "Input", AspectRatioFixed->True], Cell[TextData["Exp[z1] Exp[z2] //N "], "Input", AspectRatioFixed->True], Cell[TextData[{ "In particular, if ", StyleBox["z", FontSlant->"Italic"], " = ", StyleBox["x ", FontSlant->"Italic"], "+ ", StyleBox["iy", FontSlant->"Italic"], "\n exp(", StyleBox["z", FontSlant->"Italic"], ") = exp(", StyleBox["x", FontSlant->"Italic"], " +", StyleBox[" iy", FontSlant->"Italic"], ") = exp(", StyleBox["x", FontSlant->"Italic"], ") exp(", StyleBox["iy", FontSlant->"Italic"], ")\nThe exponential of an imaginary number is\n exp ", StyleBox["iy", FontSlant->"Italic"], " = 1 + ", StyleBox["iy", FontSlant->"Italic"], " + ", StyleBox["i", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["y", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/2! + ", StyleBox["i", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["y", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/3! + ...\n = (1 \[LongDash] ", StyleBox["y", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/2! + ", StyleBox["y", FontSlant->"Italic"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/4! ...) + ", StyleBox["i", FontSlant->"Italic"], "(", StyleBox["y", FontSlant->"Italic"], " \[LongDash] ", StyleBox["y", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/3! + ", StyleBox["y", FontSlant->"Italic"], StyleBox["5", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/5! ...)\n = cos y + i sin y\n Hence \n \t\t", StyleBox["z", FontSlant->"Italic"], " =", StyleBox[" r", FontSlant->"Italic"], "(cos ", StyleBox["\[Theta] ", FontFamily->"Symbol"], "+", StyleBox[" i", FontSlant->"Italic"], " sin ", StyleBox["\[Theta]", FontFamily->"Symbol"], ") = ", StyleBox["r", FontSlant->"Italic"], " exp ", StyleBox["i", FontSlant->"Italic"], StyleBox["\[Theta]", FontFamily->"Symbol"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["z1"], "Input", AspectRatioFixed->True], Cell[TextData["r1 Exp[I theta1] "], "Input", AspectRatioFixed->True], Cell[TextData[{ "It follows that\n ", StyleBox["z", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " ", StyleBox["z", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["r", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " exp(", StyleBox["i ", FontSlant->"Italic"], StyleBox["\[Theta]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ") ", StyleBox["r", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " exp", StyleBox["(i ", FontSlant->"Italic"], StyleBox["\[Theta]", FontFamily->"Symbol"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ") = ", StyleBox["r", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " ", StyleBox["r", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " exp", StyleBox[" i", FontSlant->"Italic"], "(", StyleBox["\[Theta]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["\[Theta]", FontFamily->"Symbol"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ")\nThus the rule for multiplying complex numbers is simple in polar \ coordinates: multiply the absolute values and add the arguments." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Abs[z1 z2]"], "Input", AspectRatioFixed->True], Cell[TextData["r1 r2"], "Input", AspectRatioFixed->True], Cell[TextData["Arg[z1 z2] //N"], "Input", AspectRatioFixed->True], Cell[TextData["theta1 + theta2"], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["C Elementary vector operations"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ "\t A vector is an ordered list of numbers. For example, ", StyleBox["agefem ", FontWeight->"Bold"], "below is the vector giving the numbers of breeding female Great Tits of \ ages 1 through 8 shown in Table 4.2, and ", StyleBox["agemale ", FontWeight->"Bold"], "is the corresponding vector for males: " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "agefem={92,37,22,12,6,3,0,0};\nagemale={83,67,31,31,16,7,3,2};"], "Input", AspectRatioFixed->True], Cell[TextData[{ " Geometrically, a vector of length ", StyleBox["n", FontSlant->"Italic"], " (i.e. with ", StyleBox["n", FontSlant->"Italic"], " elements) can be thought of as a point in ", StyleBox["n", FontSlant->"Italic"], "-dimensional space; it may also be thought of as a directed line from the \ origin to the point, the line having a ", StyleBox["magnitude ", FontSlant->"Italic"], "(the distance from the origin) and a ", StyleBox["direction", FontSlant->"Italic"], ". These two ways of representing the vector {1.5, 1} in two dimensions are \ shown in Figure C1 in the book." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "\t To demonstrate elementary arithmetic operations on vectors, I \ define two symbolic vectors:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["v1={a1,a2,a3};\nv2={b1,b2,b3};"], "Input", AspectRatioFixed->True], Cell[TextData[ "Addition of two vectors of the same length is defined by adding their \ corresponding elements:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["v1+v2"], "Input", AspectRatioFixed->True], Cell[TextData[ "For example, the age distribution of Great Tits of both sexes is"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["agefem + agemale"], "Input", AspectRatioFixed->True], Cell[TextData[ "Multiplication by a scalar (an ordinary number) is defined by multiplying \ each element by that number:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["2 v1"], "Input", AspectRatioFixed->True], Cell[TextData[{ " The inner product of two vectors of the same length is a scalar \ obtained by multiplying corresponding elements and then summing these \ products. It is often called the dot product, and is represented in ", StyleBox["Mathematica", FontSlant->"Italic"], " by a dot (", StyleBox[".", FontWeight->"Bold"], "):" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["v1.v2"], "Input", AspectRatioFixed->True], Cell[TextData[ "Geometrically, the inner product of a vector with itself is the square of \ its magnitude:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["v1.v1"], "Input", AspectRatioFixed->True], Cell[TextData[{ "If the magnitudes of ", StyleBox["v", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "and ", StyleBox["v", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "are", StyleBox[" r", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " and", StyleBox[" r", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " and the angle between them is ", StyleBox["\[Theta]", FontFamily->"Symbol"], ", it can be shown that\n ", StyleBox["v", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".v", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "=", StyleBox[" r", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" r", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " cos ", StyleBox["\[Theta]\n", FontFamily->"Symbol"], "This can be used to find cos ", StyleBox["\[Theta] ", FontFamily->"Symbol"], "and hence", StyleBox[" \[Theta]", FontFamily->"Symbol"], ", as we now do for the age distributions of female and male Great Tits:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "agefem.agemale/Sqrt[agefem.agefem*agemale.agemale]//N"], "Input", AspectRatioFixed->True], Cell[TextData["ArcCos[%]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "(This is in radians; if you want the answer in degrees divide it by ", StyleBox["Degree ", FontWeight->"Bold"], "and take the numerical value.) If two vectors have the same direction, so \ that one is a multiple of the other, ", StyleBox["\[Theta] ", FontFamily->"Symbol"], "= 0 and cos\:02da", StyleBox["\[Theta] ", FontFamily->"Symbol"], "= 1; if the two vectors are at right angles (orthogonal), ", StyleBox["\[Theta] ", FontFamily->"Symbol"], "= ", StyleBox["\[Pi]/2", FontFamily->"Symbol"], " and cos\:02da", StyleBox["\[Theta] ", FontFamily->"Symbol"], "= 0.\n\t Another use of the inner product is illustrated by the \ following calculation of the mean age of the female Great Tits:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["agefem.Range[8]/Sum[agefem[[i]],{i,8}] //N"], "Input", AspectRatioFixed->True], Cell[TextData[{ " The outer product of two vectors ", StyleBox["v", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " and ", StyleBox["v", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "of lengths ", StyleBox["m ", FontSlant->"Italic"], "and ", StyleBox["n ", FontSlant->"Italic"], "is a matrix of order ", StyleBox["m ", FontSlant->"Italic"], "x ", StyleBox["n ", FontSlant->"Italic"], "whose ", StyleBox["ij", FontSlant->"Italic"], "th element is the product of the ", StyleBox["i", FontSlant->"Italic"], "th element of ", StyleBox["v", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "and the ", StyleBox["j", FontSlant->"Italic"], "th element of ", StyleBox["v", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ". It will be written ", StyleBox["v", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "x ", StyleBox["v", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "and is obtained in ", StyleBox["Mathematica", FontSlant->"Italic"], " as follows:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Outer[Times,v1,v2]"], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["D Elementary matrix operations"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ " Matrix algebra was invented by Cayley in the last century to \ provide a compact way of writing simultaneous linear equations. A matrix is a \ rectangular array of numbers, which is represented in ", StyleBox["Mathematica", FontSlant->"Italic"], " as a list of lists. For example, let us make a list of the two lists ", StyleBox["agefem ", FontWeight->"Bold"], "and ", StyleBox["agemale ", FontWeight->"Bold"], "defined above:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["age={agefem, agemale}"], "Input", AspectRatioFixed->True], Cell[TextData["This can be displayed in conventional matrix format:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["MatrixForm[age]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus ", StyleBox["age ", FontWeight->"Bold"], "is a 2 x 8 matrix (with 2 rows and 8 columns), the rows representing \ females and males and the columns representing age classes.\n We \ now set up some matrices to use as examples in discussing their properties. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "A1={{1,3,4},{9,-1,6}};\nA2={{3,-1,2},{10,3,4}};\nAa={{a11,a12},{a21,a22}};\n\ Ab={{b11,b12},{b21,b22}};"], "Input", AspectRatioFixed->True], Cell[TextData[ " Addition of two matrices of the same order is defined by adding \ their corresponding elements. "], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "Aa //MatrixForm\nAb //MatrixForm\nAa+Ab //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[ "A1 //MatrixForm\nA2 //MatrixForm\nA1+A2 //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ " The transpose of a matrix ", StyleBox["A", FontWeight->"Bold"], ", usually denoted ", StyleBox["A", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ", has the rows and columns transposed:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A1"], "Input", AspectRatioFixed->True], Cell[TextData["Transpose[A1]"], "Input", AspectRatioFixed->True], Cell[TextData["% //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[ " Multiplication of a matrix by a scalar is defined by multiplying \ each element by the scalar. "], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["1.5 A1 //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ " Matrix multiplication is an extension of the vector dot product. \ Suppose ", StyleBox["A ", FontWeight->"Bold"], "and ", StyleBox["B ", FontWeight->"Bold"], "are matrices of orders ", StyleBox["m", FontSlant->"Italic"], " x ", StyleBox["n", FontSlant->"Italic"], " and ", StyleBox["n", FontSlant->"Italic"], " x ", StyleBox["p", FontSlant->"Italic"], " respectively. Their product ", StyleBox["A.B ", FontWeight->"Bold"], "is defined as the matrix with dimensions ", StyleBox["m", FontSlant->"Italic"], " x ", StyleBox["p", FontSlant->"Italic"], ", whose ", StyleBox["ij", FontSlant->"Italic"], "th element is the dot product of the ", StyleBox["i", FontSlant->"Italic"], "th row of ", StyleBox["A ", FontWeight->"Bold"], "and the ", StyleBox["j", FontSlant->"Italic"], "th column of ", StyleBox["B", FontWeight->"Bold"], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Aa.Ab //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData["Transpose[A1].A2 //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ " Notice that the matrix product ", StyleBox["A.B ", FontWeight->"Bold"], "is only defined when the number of columns of ", StyleBox["A ", FontWeight->"Bold"], "is equal to the numbers of rows of ", StyleBox["B", FontWeight->"Bold"], ". If ", StyleBox["A ", FontWeight->"Bold"], "and ", StyleBox["B ", FontWeight->"Bold"], "are square matrices with the same dimensions, both ", StyleBox["A.B ", FontWeight->"Bold"], "and ", StyleBox["B.A ", FontWeight->"Bold"], "are defined, but they are not usually the same; that is to say, matrix \ multiplication is not commutative:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["{{1,3},{2,-1}}.{{4,6},{-2,8}}"], "Input", AspectRatioFixed->True], Cell[TextData["{{4,6},{-2,8}}.{{1,3},{2,-1}}"], "Input", AspectRatioFixed->True], Cell[TextData["\n"], "Input", AspectRatioFixed->True], Cell[TextData[{ "[NOTE: A vector in ", StyleBox["Mathematica", FontSlant->"Italic"], " is simply a list, and there is no distinction between row and column \ vectors. You can use \"dot\" for either left or right multiplication of \ vectors by matrices and it carries out whatever operation is possible. If ", StyleBox["A ", FontWeight->"Bold"], "is a vector of order ", StyleBox["m ", FontSlant->"Italic"], "x ", StyleBox["n ", FontSlant->"Italic"], "and ", StyleBox["v ", FontWeight->"Bold"], "is a vector of length ", StyleBox["n ", FontSlant->"Italic"], ", ", StyleBox["A.v ", FontWeight->"Bold"], "is a vector of length ", StyleBox["m ", FontSlant->"Italic"], "obtainedby multiplying the rows of ", StyleBox["A ", FontWeight->"Bold"], "into ", StyleBox["v", FontWeight->"Bold"], "; if ", StyleBox["v ", FontWeight->"Bold"], "is of length ", StyleBox["m", FontSlant->"Italic"], ", ", StyleBox["v.A ", FontWeight->"Bold"], "is a vector of length ", StyleBox["n ", FontSlant->"Italic"], "obtained by multiplying ", StyleBox["v ", FontWeight->"Bold"], "into the columns of ", StyleBox["A", FontWeight->"Bold"], ": " }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData["A1.{x1,x2,x3}"], "Input", AspectRatioFixed->True], Cell[TextData["{x1,x2}.A1"], "Input", AspectRatioFixed->True], Cell[TextData[{ "In matrix algebra it is usual to distinguish between ", StyleBox["row vectors ", FontSlant->"Italic"], "and ", StyleBox["column vectors", FontSlant->"Italic"], ". A row vector is really a matrix with one row, and likewise a column \ vector is a matrix with one column, and if they are to be used in ", StyleBox["Mathematica", FontSlant->"Italic"], " they must be defined in this way:" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData[ "v1={1,2,3};\nv2={4,5,6};\nrowv={{1,2,3}};\ncolv={{4},{5},{6}};"], "Input", AspectRatioFixed->True], Cell[TextData["Note that"], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData["colv.rowv"], "Input", AspectRatioFixed->True], Cell[TextData["is the same as the outer product"], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData["Outer[Times,v2,v1]"], "Input", AspectRatioFixed->True], Cell[TextData["However"], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData["rowv.colv"], "Input", AspectRatioFixed->True], Cell[TextData["is a matrix with one row and one column whereas"], "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData["v1.v2"], "Input", AspectRatioFixed->True], Cell["\<\ is a scalar.] \ \>", "Text", Evaluatable->False, AspectRatioFixed->True, FontSize->11], Cell[TextData[ " The motivation for defining matrix multiplication like this is \ that it leads to useful applications. Consider the problem of solving 3 \ linear equations in 3 unknowns:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t3", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 4", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " \[LongDash] ", StyleBox["x", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 4\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t4", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 12", StyleBox["x", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 12\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t9", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " \[LongDash] 3", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["x", FontSlant->"Italic"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 6" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "This can be written in matrix terminology more compactly as"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[" \t\t\t\t\t\t\t\t\t\t\t\t\t\t A.x = c", "Text", Evaluatable->False, AspectRatioFixed->True, FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], Cell[TextData[{ "where ", StyleBox["A", FontWeight->"Bold"], " is the matrix of known coefficients,", StyleBox[" x", FontWeight->"Bold"], " is the vector of unknowns and ", StyleBox["c", FontWeight->"Bold"], " is the vector of known constants on the right hand side:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A = {{3,4,-1},{4,1,12},{9,-3,1}}; \nc={4,12,6};"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Note the value of ", StyleBox["A.x", FontWeight->"Bold"], ":" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A.{x1,x2,x3} //ColumnForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ " It is not only more convenient to use this compact terminology, but also \ leads to new methodology such as solution of the equations by using the \ inverse matrix:\n\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\ \t\t\t\t\t\t\t\t\t\t", StyleBox["x = A", FontWeight->"Bold"], StyleBox["\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".c", FontWeight->"Bold"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Inverse[A].c //N"], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox[ "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\ \t\t\t\n", FontWeight->"Bold"], "This is the solution for ", StyleBox["x ", FontWeight->"Bold"], "as will be explained in the next section." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[ "E Determinants, linear independence and matrix inversion"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ " Suppose ", StyleBox["A ", FontWeight->"Bold"], "is a square matrix. Its ", StyleBox["determinant", FontSlant->"Italic"], " is a scalar which allows one to determine whether there are any linear \ relationships among its rows or columns. If such relationships exist the \ determinant is zero and the matrix is called ", StyleBox["singular", FontSlant->"Italic"], "; otherwise the determinant is non-zero and the matrix is called ", StyleBox["non-singular", FontSlant->"Italic"], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "A3={{1,2,4},{3,1,2},{7,4,8}};\nA4={{1.,2,3},{2.,1,2},{3.,4,5}};"], "Input", AspectRatioFixed->True], Cell[TextData["Det[A3]"], "Input", AspectRatioFixed->True], Cell[TextData["Det[A4]"], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["A", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "is singular because the third row is equal to twice the second row plus \ the third row. The rows (and columns) of ", StyleBox["A", FontWeight->"Bold"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "on the other hand are linearly independent; no such linear relationship \ exists between them.\n\t The definition of a determinant can be seen \ by calculating it for symbolic matrices of different dimensions. ", StyleBox["Array[a,{m,n}] ", FontWeight->"Bold"], "generates a matrix of dimensions {", StyleBox["m", FontSlant->"Italic"], ", ", StyleBox["n", FontSlant->"Italic"], "} with ", StyleBox["a", FontSlant->"Italic"], "[", StyleBox["i", FontSlant->"Italic"], ",", StyleBox[" j", FontSlant->"Italic"], "] in the ", StyleBox["ij", FontSlant->"Italic"], "th position." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Det[Array[a,{2,2}]]"], "Input", AspectRatioFixed->True], Cell[TextData["Det[Array[a,{3,3}]]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Each term contains one element from each row and from each column. The \ column subscripts in each term are a permutation of the integers {1, 2, ...,", StyleBox[" n", FontSlant->"Italic"], "} and the sum (with a + or \[LongDash] sign) is taken over all possible \ permutations, of which there are ", StyleBox["n", FontSlant->"Italic"], "!. A permutation is classified as odd or even depending on whether it can \ be obtained from the natural ordering by an odd or an even number of \ reciprocal interchanges (this can be determined with ", StyleBox["Signature", FontWeight->"Bold"], "), a + sign being given to even permutations and a \[LongDash] sign to odd \ ones.\n The following properties follow almost immediately from this \ definition. If all the elements of a row (or column) are multiplied by a \ constant, the determinant is multiplied by the same constant. If two rows (or \ columns) are interchanged the sign of the determinant changes (because even \ permutations become odd and ", StyleBox["vice versa", FontSlant->"Italic"], ") but its absolute value is unchanged. If two rows (or columns) are \ identical, the determinant is zero (because interchanging them leaves the \ matrix the same). Hence for any square matrix, adding one row (or column) to \ another leaves the determinant unchanged. It follows that if the rows are \ linearly dependent the determinant is zero. (Suppose that there is a linear \ dependence involving the first row, for example. Then by adding multiples of \ the other rows to it, it is possible, without changing the determinant, to \ make this row zero; a matrix with a zero row has zero determinant.) The \ converse is also true: if the rows are independent then the determinant is \ non-zero. Thus there is a one-one correspondence between the dependence or \ independence of the rows (and columns) and whether the determinant is zero or \ non-zero.\n If ", StyleBox["A ", FontWeight->"Bold"], "is a non-singular matrix with dimensions {", StyleBox["n", FontSlant->"Italic"], ", ", StyleBox["n", FontSlant->"Italic"], "} it has a unique inverse matrix ", StyleBox["A", FontWeight->"Bold"], StyleBox["\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSize->10], "such that ", StyleBox["A.A", FontWeight->"Bold"], StyleBox["\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = ", StyleBox["A", FontWeight->"Bold"], StyleBox["\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".A", FontWeight->"Bold"], " = ", StyleBox["I", FontWeight->"Bold"], StyleBox["n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ", where ", StyleBox["I", FontWeight->"Bold"], StyleBox["n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " is the identity matrix, a square matrix with dimensions {", StyleBox["n", FontSlant->"Italic"], ",", StyleBox[" n", FontSlant->"Italic"], "} with all its diagonal elements equal to 1 and all off-diagonal elements \ zero. Note how the identity matrix and the inverse are obtained in ", StyleBox["Mathematica", FontSlant->"Italic"], ": \t" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["IdentityMatrix[4] //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData["B=Inverse[A4] "], "Input", AspectRatioFixed->True], Cell[TextData["A4.B //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData["B.A4"], "Input", AspectRatioFixed->True], Cell[TextData[{ " A singular matrix cannot have an inverse. Suppose that ", StyleBox["A.B = I", FontWeight->"Bold"], StyleBox["n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ", write ", StyleBox["r", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " for the ", StyleBox["i", FontSlant->"Italic"], "th row of ", StyleBox["A ", FontWeight->"Bold"], "and ", StyleBox["c", FontWeight->"Bold"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " for the ", StyleBox["j", FontSlant->"Italic"], "th column of ", StyleBox["B", FontWeight->"Bold"], ", and suppose also that ", StyleBox["A ", FontWeight->"Bold"], "is singular with ", StyleBox["r", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["r", FontWeight->"Bold"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 2", StyleBox["r", FontWeight->"Bold"], StyleBox["5", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ". Then ", StyleBox["r", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".c", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 1 and (", StyleBox["r", FontWeight->"Bold"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "\:02da+ 2", StyleBox["r", FontWeight->"Bold"], StyleBox["5", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ")", StyleBox[".c", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " =\:02da", StyleBox["r", FontWeight->"Bold"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".", FontWeight->"Bold"], "c", StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 2", StyleBox["r", FontWeight->"Bold"], StyleBox["5", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".c", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " =\:02da0. This contradicts the assumption that ", StyleBox["r", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "\:02da=\:02da", StyleBox["r", FontWeight->"Bold"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "\:02da+\:02da2", StyleBox["r", FontWeight->"Bold"], StyleBox["5", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData["Solution of linear equations"], "Subsubsection", Evaluatable->False, PageBreakAbove->False, AspectRatioFixed->True], Cell[TextData[ " We return to the problem of solving n linear equations in n \ unknowns:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox[" A.x ", FontWeight->"Bold"], "= ", StyleBox["c", FontWeight->"Bold"] }], "Output", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "where ", StyleBox["A ", FontWeight->"Bold"], "is a matrix of known coefficients ", StyleBox["a", FontSlant->"Italic"], StyleBox["ij", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ", ", StyleBox["x ", FontWeight->"Bold"], "is the vector of unknowns and ", StyleBox["c ", FontWeight->"Bold"], "is a vector of known constants. \n Consider first the \ non-homogeneous equation with ", StyleBox["c ", FontWeight->"Bold"], StyleBox["=", FontWeight->"Bold", FontVariations->{"StrikeThrough"->True}], " / ", StyleBox["0", FontWeight->"Bold"], ". If ", StyleBox["A ", FontWeight->"Bold"], "is non-singular it has a unique inverse; pre-multiplying both sides by \ this inverse we find the unique solution" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox[" x ", FontWeight->"Bold"], "= ", StyleBox["A", FontWeight->"Bold"], StyleBox["-1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".c", FontWeight->"Bold"] }], "Output", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " If ", StyleBox["A ", FontWeight->"Bold"], "is singular, there are linear relationships among its rows. If these \ relationships do not hold among the ", StyleBox["c", FontSlant->"Italic"], "'s, the equations are inconsistent and have no solution; if the \ relationships among the rows of ", StyleBox["A ", FontWeight->"Bold"], "also happen to hold among the", StyleBox[" c", FontSlant->"Italic"], "'s, a range of solutions exists. For example, the equations" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 1\n2", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 2", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 3" }], "Output", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["have no solution whereas the equations"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 1\n2", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 2", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 2" }], "Output", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "are compatible with any values of ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " and", StyleBox[" x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " on the line", StyleBox[" x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 1 \[LongDash] ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ".\n Consider finally the homogeneous equation" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\t\t\t", StyleBox["A.x ", FontWeight->"Bold"], "= ", StyleBox["0", FontWeight->"Bold"] }], "Input", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "If ", StyleBox["A ", FontWeight->"Bold"], "is non-singular, premultiplication by its inverse shows that the only \ solution is the trivial solution ", StyleBox["x\:02da", FontWeight->"Bold"], "=\:02da", StyleBox["0", FontWeight->"Bold"], ". If ", StyleBox["A ", FontWeight->"Bold"], "is singular, then any linear relationship among the rows of ", StyleBox["A ", FontWeight->"Bold"], "automatically holds on the right-hand side so that a range of non-trivial \ solutions exists. For example, the equations" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 0\n2", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + 2", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = 0" }], "Output", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "are satisfied by any values on the line ", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " = \[LongDash]", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["F Eigenvalues and eigenvectors"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ " If ", StyleBox["A ", FontWeight->"Bold"], "is a square matrix", StyleBox[" ", FontWeight->"Bold"], "of dimensions ", StyleBox["n", FontSlant->"Italic"], " x ", StyleBox["n", FontSlant->"Italic"], " and ", StyleBox["x ", FontWeight->"Bold"], "is a vector of length ", StyleBox["n", FontSlant->"Italic"], ", then ", StyleBox["A.x ", FontWeight->"Bold"], "is also a vector of length ", StyleBox["n", FontSlant->"Italic"], ". Thus ", StyleBox["A ", FontWeight->"Bold"], "can be regarded as defining a linear transformation on points in ", StyleBox["n", FontSlant->"Italic"], "-dimensional space. Consider a simple example in two-dimensional space: " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A = {{1,2},{3,2}};"], "Input", AspectRatioFixed->True], Cell[TextData["A.{0,1}"], "Input", AspectRatioFixed->True], Cell[TextData["A.{1,1}"], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["A ", FontWeight->"Bold"], "has transformed the point {0,1} into the point {2,2} and the point {1,1} \ into {3,5}.\n A common application of this idea occurs in the set of \ linear difference equations:\n ", StyleBox["n", FontWeight->"Bold"], "(", StyleBox["t", FontSlant->"Italic"], "+1) = ", StyleBox["A.n", FontWeight->"Bold"], "(", StyleBox["t", FontSlant->"Italic"], ").\nIn this context ", StyleBox["n", FontWeight->"Bold"], "(", StyleBox["t", FontSlant->"Italic"], ") might be the vector of numbers of animals of different ages in year ", StyleBox["t", FontSlant->"Italic"], ", and ", StyleBox["A ", FontWeight->"Bold"], " defines how this year's numbers are transformed into next year's (the \ Leslie matrix, Chapter 3).\n We shall now consider an important tool \ in studying linear transformations. If ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "is a scalar and ", StyleBox["x ", FontWeight->"Bold"], "a non-zero vector such that\n ", StyleBox["A.x ", FontWeight->"Bold"], "= ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox[ "x \ ", FontWeight->"Bold"], "[1]", StyleBox["\n", FontWeight->"Bold"], "then ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "is called an ", StyleBox["eigenvalue ", FontSlant->"Italic"], "of ", StyleBox["A ", FontWeight->"Bold"], "and ", StyleBox["x ", FontWeight->"Bold"], "is the corresponding ", StyleBox["eigenvector", FontSlant->"Italic"], ". This equation can be written\n (", StyleBox["A ", FontWeight->"Bold"], "- ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["I", FontWeight->"Bold"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ")", StyleBox[".x ", FontWeight->"Bold"], "= ", StyleBox["0", FontWeight->"Bold"], ". \ [2]\nThis homogeneous equation can only have a \ non-zero solution if\n det(", StyleBox["A ", FontWeight->"Bold"], "- ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["I", FontWeight->"Bold"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ") = 0. \ [3]\nThe determinant defines a polynomial of \ degree ", StyleBox["n", FontSlant->"Italic"], " in ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "and the equation will in general have ", StyleBox["n", FontSlant->"Italic"], " solutions, though there may be fewer if there are repeated roots. In \ practice there are usually ", StyleBox["n", FontSlant->"Italic"], " distinct eigenvalues, and for simplicity we shall from now on assume that \ this is the case. These eigenvalues can be found thus:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Eigenvalues[A]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus there are two eigenvalues, \[LongDash]1 and 4, of the 2 x 2 matrix ", StyleBox["A", FontWeight->"Bold"], ".\n To each eigenvalue there corresponds an eigenvector. They can \ be found with ", StyleBox["Eigenvectors", FontWeight->"Bold"], ", but there is no guarantee that they will be listed in the same order as \ the eigenvalues, and it is better to use ", StyleBox["Eigensystem ", FontWeight->"Bold"], " which lists first the eigenvalues and then the corresponding \ eigenvectors. The following assignment sets ", StyleBox["vals ", FontWeight->"Bold"], "to the list of eigenvalues, and ", StyleBox["vecs ", FontWeight->"Bold"], "to the list of eigenvectors:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["{vals,vecs}=Eigensystem[A];"], "Input", AspectRatioFixed->True], Cell[TextData["vals"], "Input", AspectRatioFixed->True], Cell[TextData["vecs"], "Input", AspectRatioFixed->True], Cell[TextData[{ "The eigenvector {\[LongDash]1, 1} corresponds to the eigenvalue \ \[LongDash]1, and the eigenvector {2/3, 1) to the eigenvalue 4. In fact {2, \ 3} or {4, 6} or any multiple thereof is also an eigenvector corresponding to \ the eigenvalue 4; an eigenvector is arbitrary up to a multiplicative constant \ as is clear from the equation defining it. We now verify that 4 is an \ eigenvalue of ", StyleBox["A ", FontWeight->"Bold"], "with the associated eigenvector {2, 3} or any multiple thereof:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A.{2,3}"], "Input", AspectRatioFixed->True], Cell[TextData["A.{20,30}"], "Input", AspectRatioFixed->True], Cell[TextData[{ " An eigenvector can be thought of as a ", StyleBox["direction ", FontSlant->"Italic"], "in ", StyleBox["n", FontSlant->"Italic"], "-dimensional space which is conserved under the linear transformation ", StyleBox["A ", FontWeight->"Bold"], "while the ", StyleBox["magnitude ", FontSlant->"Italic"], "of the vector is multiplied by ", StyleBox["\[Lambda]", FontFamily->"Symbol"], ". In the set of difference equations\n ", StyleBox["n", FontWeight->"Bold"], "(", StyleBox["t", FontSlant->"Italic"], "+1) = ", StyleBox["A.n", FontWeight->"Bold"], "(", StyleBox["t", FontSlant->"Italic"], "),\nif the population vector ", StyleBox["n ", FontWeight->"Bold"], "lies in the direction of a particular eigenvector this year, it will lie \ in the same direction next year but will be multiplied in magnitude by ", StyleBox["\[Lambda]", FontFamily->"Symbol"], "; that is to say, all the elements of ", StyleBox["n ", FontWeight->"Bold"], "will be multiplied by the same constant ", StyleBox["\[Lambda]", FontFamily->"Symbol"], ". Linear difference equations will be considered in more detail in the \ section 6.\n So far we have been considering ", StyleBox["right ", FontSlant->"Italic"], "eigenvalues and vectors. One can define in the same way ", StyleBox["left ", FontSlant->"Italic"], "eigenvalues and vectors satisfying\n ", StyleBox["x.A ", FontWeight->"Bold"], "= ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["x", FontWeight->"Bold"], ". \ [4]\nThe left eigenvalues satisfy the same \ determinantal equation as the right eigenvalues and are therefore the same, \ but the corresponding eigenvectors are different. ", StyleBox["Mathematica", FontSlant->"Italic"], " only calculates right eigenvectors, but the left eigenvectors can be \ found as the right eigenvectors of the transpose of the matrix since (", StyleBox["x.A", FontWeight->"Bold"], ")", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = ", StyleBox["A", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".x", FontWeight->"Bold"], ". Eigenvectors are arbitrary up to a multiplicative constant, and it is \ convenient to standardize each of the left eigenvectors so that its product \ with the corresponding right eigenvector is unity. As an example we find the \ standardized left eigenvectors of ", StyleBox["A", FontWeight->"Bold"], ":" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["{lvals, lvecs}=Eigensystem[Transpose[A]];"], "Input", AspectRatioFixed->True], Cell[TextData["lvals"], "Input", AspectRatioFixed->True], Cell[TextData[ "Do[lvecs[[i]]=lvecs[[i]]/(lvecs[[i]].vecs[[i]]),\n\t\t\t\t{i,2}]\t"], "Input",\ AspectRatioFixed->True], Cell[TextData["lvecs"], "Input", AspectRatioFixed->True], Cell[TextData[{ "A check should be made that the eigenvalues are listed in the same order \ in ", StyleBox["vals ", FontWeight->"Bold"], "and ", StyleBox["lvals", FontWeight->"Bold"], "; if not, the left eigenvectors should be rearranged to appear in the same \ order as the right eigenvectors. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " An important fact is that a left and a right eigenvector \ corresponding to different eigenvalues are orthogonal. Suppose that ", StyleBox["e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "is a right eigenvector corresponding to ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontFamily->"Symbol"], "and that ", StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold"], "is a left eigenvector corresponding to ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["2", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[". ", FontFamily->"Symbol"], "Then", StyleBox["\n\t\t\t", FontFamily->"Symbol"], StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".A.e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "= ", StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".", FontWeight->"Bold"], "(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ") = (", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ")", StyleBox[".e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n", FontWeight->"Bold"], "It follows that ", StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontWeight->"Bold"], "= 0 if ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontFamily->"Symbol"], "differs from", StyleBox[" ", FontFamily->"Symbol"], " ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[". ", FontFamily->"Symbol"], "We have standardized the left eigenvectors so that ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "= 1.If we define ", StyleBox["R ", FontWeight->"Bold"], "= {", StyleBox["e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ", ", StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ", ... , ", StyleBox["e", FontWeight->"Bold"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "} as the matrix of right eigenvectors and ", StyleBox["L\:02da", FontWeight->"Bold"], "=\:02da{", StyleBox["e", FontWeight->"Bold"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ", ", StyleBox["e", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ", ... , ", StyleBox["e", FontWeight->"Bold"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "}", StyleBox[" ", FontWeight->"Bold"], "as the matrix of standardized left eigenvectors, it follows that ", StyleBox["L ", FontWeight->"Bold"], "is the inverse of ", StyleBox["R", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ":\n ", StyleBox["L.R", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold"], "= ", StyleBox["I", FontWeight->"Bold"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ " \ [5]", FontVariations->{"CompatibilityType"->"Subscript"}], "\nSince a left inverse is also a right inverse\n ", StyleBox["R", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".L ", FontWeight->"Bold"], "= ", StyleBox["I", FontWeight->"Bold"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ " \ [6]", FontVariations->{"CompatibilityType"->"Subscript"}], "\nWe now verify this for the matrix ", StyleBox["A", FontWeight->"Bold"], ":" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["lvecs.Transpose[vecs]"], "Input", AspectRatioFixed->True], Cell[TextData["Transpose[vecs].lvecs"], "Input", AspectRatioFixed->True], Cell[TextData[{ " \tFinally, we consider the effect of a small perturbation of the \ elements of a matrix ", StyleBox["A ", FontWeight->"Bold"], "on an eigenvalue ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "with right and left", StyleBox[" ", FontFamily->"Symbol"], "eigenvectors ", StyleBox["e ", FontWeight->"Bold"], "and ", StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ". A perturbation from ", StyleBox["A ", FontWeight->"Bold"], "to ", StyleBox["A", FontWeight->"Bold"], "+d", StyleBox["A ", FontWeight->"Bold"], "will change ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "to ", StyleBox["\[Lambda]", FontFamily->"Symbol"], "+d", StyleBox["\[Lambda]", FontFamily->"Symbol"], " and ", StyleBox["e", FontWeight->"Bold"], " to ", StyleBox["e", FontWeight->"Bold"], "+d", StyleBox["e", FontWeight->"Bold"], " such that\n \t", StyleBox["A.e ", FontWeight->"Bold"], "= ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["e\n ", FontWeight->"Bold"], "(", StyleBox["A", FontWeight->"Bold"], "+d", StyleBox["A", FontWeight->"Bold"], ").(", StyleBox["e", FontWeight->"Bold"], "+d", StyleBox["e", FontWeight->"Bold"], ") = (", StyleBox["\[Lambda]", FontFamily->"Symbol"], "+d", StyleBox["\[Lambda])(", FontFamily->"Symbol"], StyleBox["e", FontWeight->"Bold"], "+d", StyleBox["e", FontWeight->"Bold"], ")\nIgnoring second order terms\n\t\t\td", StyleBox["A.e ", FontWeight->"Bold"], "+ ", StyleBox["A.", FontWeight->"Bold"], "d", StyleBox["e ", FontWeight->"Bold"], "= ", StyleBox["\[Lambda]", FontFamily->"Symbol"], "d", StyleBox["e ", FontWeight->"Bold"], "+", StyleBox["\t", FontWeight->"Bold"], "d", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["e\n", FontWeight->"Bold"], "Multiplying both sides by ", StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " and remembering that ", StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".A ", FontWeight->"Bold"], "= ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold"], "and that ", StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".e ", FontWeight->"Bold"], "= 1:\n ", StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".", FontWeight->"Bold"], "d", StyleBox["A.e", FontWeight->"Bold"], " = d", StyleBox["\[Lambda]", FontFamily->"Symbol"], " \ [7]", StyleBox["\n", FontFamily->"Symbol"], "If only one element in ", StyleBox["A ", FontWeight->"Bold"], "is perturbed, say ", StyleBox["a", FontSlant->"Italic"], StyleBox["ij ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " to ", StyleBox["a", FontSlant->"Italic"], StyleBox["ij ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "+ d", StyleBox["a", FontSlant->"Italic"], StyleBox["ij", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ", then\n d", StyleBox["\[Lambda] =", FontFamily->"Symbol"], StyleBox[" ", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["e", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " ", StyleBox["e", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " d", StyleBox["a", FontSlant->"Italic"], StyleBox["ij", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], " \ [8]\nwhere ", StyleBox["e", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " and", StyleBox[" e", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "are the ", StyleBox["i", FontSlant->"Italic"], "th and", StyleBox[" j", FontSlant->"Italic"], "th elements of ", StyleBox["e", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold"], "and ", StyleBox["e ", FontWeight->"Bold"], "respectively. This result is useful in sensitivity analysis discussed in \ Chapters 4 and 5.\n\t\t\t\t\t\t\t\t\t\t " }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["G Spectral decomposition of a matrix"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ StyleBox[" ", FontWeight->"Bold"], "It is often useful to represent a matrix in a particular way known as its \ ", StyleBox["spectral decomposition", FontSlant->"Italic"], ". We first observe that \n ", StyleBox["A.R", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold"], "= ", StyleBox["R", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".", FontWeight->"Bold"], StyleBox[ "\[CapitalLambda] \ ", FontFamily->"Symbol", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold"], "[1]", StyleBox["\n", FontFamily->"Symbol", FontWeight->"Bold"], "where ", StyleBox["R", FontWeight->"Bold"], " is the matrix of right eigenvectors defined in Section F and ", StyleBox["\[CapitalLambda] ", FontFamily->"Symbol", FontWeight->"Bold"], "is a diagonal matrix with the eigenvalues on the diagonal and zero \ everywhere else. This follows from the definitions of eigenvalues and \ eigenvectors together with the fact that post-multiplying a matrix by a \ diagonal matrix multiplies the", StyleBox[" i", FontSlant->"Italic"], "th column of that matrix by the ", StyleBox["i", FontSlant->"Italic"], "th diagonal element, so that the ", StyleBox["i", FontSlant->"Italic"], "th column of the right hand side is the", StyleBox[" i", FontSlant->"Italic"], "th eigenvalue multiplied by the ", StyleBox["i", FontSlant->"Italic"], "th eigenvector. Verifying this for ", StyleBox["A ", FontWeight->"Bold"], "defined above:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A.Transpose[vecs]"], "Input", AspectRatioFixed->True], Cell[TextData["Lambda = DiagonalMatrix[vals]"], "Input", AspectRatioFixed->True], Cell[TextData["Transpose[vecs].Lambda"], "Input", AspectRatioFixed->True], Cell[TextData[{ " Post-multiplying [1] by ", StyleBox["L ", FontWeight->"Bold"], "and using [F6] we find that\n ", StyleBox["A ", FontWeight->"Bold"], "= ", StyleBox["R", FontWeight->"Bold"], StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".", FontWeight->"Bold"], StyleBox["\[CapitalLambda].", FontFamily->"Symbol", FontWeight->"Bold"], StyleBox["L", FontWeight->"Bold"], " \ [2]\n This equation can be expressed as" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " ", StyleBox["A ", FontWeight->"Bold"], "= ", StyleBox[" \[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox[ "i \ ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "[3]", StyleBox[ " \ ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" \n", FontVariations->{"CompatibilityType"->"Subscript"}], "where ", StyleBox[ " \ ", FontVariations->{"CompatibilityType"->"Subscript"}], " \n ", StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], "= ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " x ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "," }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "the outer product of the right and left eigenvectors. This is called the ", StyleBox["spectral decomposition ", FontSlant->"Italic"], " of ", StyleBox["A", FontWeight->"Bold"], ". To verify that [3] follows from [2] we set up two arbitrary 3 x 3 \ matrices ", StyleBox["R ", FontWeight->"Bold"], "and ", StyleBox["L ", FontWeight->"Bold"], "and a diagonal matrix with 1 in the leading position and zero everywhere \ else which will isolate the coefficient of ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ":" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["R={{r11,r12,r13},{r21,r22,r23},{r31,r32,r33}}"], "Input", AspectRatioFixed->True], Cell[TextData["L={{l11,l12,l13},{l21,l22,l23},{l31,l32,l33}}"], "Input", AspectRatioFixed->True], Cell[TextData["Diag=DiagonalMatrix[{1,0,0}]"], "Input", AspectRatioFixed->True], Cell[TextData["Transpose[R].Diag.L //MatrixForm "], "Input", AspectRatioFixed->True], Cell[TextData["Outer[Times,R[[1]],L[[1]]] //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ " The significance of the spectral decomposition [3] is that \n \ ", StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".T", FontWeight->"Bold"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], "= ", StyleBox["0", FontWeight->"Bold"], ", ", StyleBox["i", FontSlant->"Italic"], StyleBox[" ", FontSize->10], StyleBox["/=", FontSize->10, FontVariations->{"StrikeThrough"->True}], StyleBox[" ", FontSize->10], StyleBox[" j", FontSize->10, FontSlant->"Italic"], " \ [4]\n(since ", StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".T", FontWeight->"Bold"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "= ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " x ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ". ", StyleBox["e", FontWeight->"Bold"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " x ", StyleBox["e", FontWeight->"Bold"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ", and ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ". ", StyleBox["e", FontWeight->"Bold"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "= 1 if ", StyleBox["i", FontSlant->"Italic"], " = ", StyleBox["j", FontSlant->"Italic"], " and 0 otherwise.) Thus the ", StyleBox["k", FontSlant->"Italic"], "th power of ", StyleBox["A ", FontWeight->"Bold"], "is\n ", StyleBox[" A", FontWeight->"Bold"], StyleBox["k", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " = (", StyleBox["\[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], ")", StyleBox["k", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " = ", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" \[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["k", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ " ", FontVariations->{"CompatibilityType"->"Subscript"}], " [5]", StyleBox["\n", FontVariations->{"CompatibilityType"->"Subscript"}], "This is useful for determining the behavior of large matrix powers. If ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " dominates the other eigenvalues in the sense that it is larger than any \ of them in absolute value, then [5] will be dominated by this eigenvalue when \ ", StyleBox["k ", FontSlant->"Italic"], "is large so that\n ", StyleBox["A", FontWeight->"Bold"], StyleBox["k", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold"], "~ ", StyleBox[" \[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["k", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["1 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ " \ ", FontVariations->{"CompatibilityType"->"Subscript"}], "[6]\n " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Do[T[i]=Outer[Times,vecs[[i]],lvecs[[i]]],{i,2}]"], "Input", AspectRatioFixed->True], Cell[TextData["T[1]"], "Input", AspectRatioFixed->True], Cell[TextData["T[1].T[1]"], "Input", AspectRatioFixed->True], Cell[TextData["T[1].T[2]"], "Input", AspectRatioFixed->True], Cell[TextData["MatrixPower[A,10]"], "Input", AspectRatioFixed->True], Cell[TextData["vals[[1]]^10 T[1] + vals[[2]]^10 T[2]"], "Input", AspectRatioFixed->True], Cell[TextData["vals[[2]]^10 T[2] //N"], "Input", AspectRatioFixed->True], Cell[TextData[{ " This idea can be extended to other functions of matrices. The \ exponential of a matrix, exp(", StyleBox["A", FontWeight->"Bold"], "), can be defined by the power series appropriate for the exponential of a \ real variable (see Section A): \n exp(", StyleBox["A", FontWeight->"Bold"], ") = ", StyleBox["I ", FontWeight->"Bold"], "+ ", StyleBox["A ", FontWeight->"Bold"], "+ ", StyleBox["A", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/2! + ", StyleBox["A", FontWeight->"Bold"], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/3! + ", StyleBox["A", FontWeight->"Bold"], StyleBox["4", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/4! + ... [7]\nUsing \ [5], we find that \n exp(", StyleBox["A", FontWeight->"Bold"], ")", StyleBox[" ", FontWeight->"Bold"], "= ", StyleBox[" \[CapitalSigma] (1 + \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "+ ", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/2! + ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/3! + ...)", StyleBox["T", FontWeight->"Bold"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "= ", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], " ", StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ")", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], " [8] \nWe \ shall be interested in exp(", StyleBox["A", FontWeight->"Bold"], StyleBox["t", FontSlant->"Italic"], ") where ", StyleBox["t ", FontSlant->"Italic"], "is time:\n exp(", StyleBox["A", FontWeight->"Bold"], StyleBox["t", FontSlant->"Italic"], ") = ", StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ")", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "~ ", StyleBox[" ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ")", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["1 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "[9]" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["MatrixExp[3.5 A] //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[ "Exp[3.5 vals[[1]]] T[1] + Exp[3.5 vals[[2]]] T[2] //\n\t\t\t\t\t\t\t\t\t\t\ MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData["Exp[3.5 vals[[2]]] T[2] \t//MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[{ " Another application is to find the square root of a matrix ", StyleBox["A", FontWeight->"Bold"], ", defined as a matrix ", StyleBox["B ", FontWeight->"Bold"], "such that ", StyleBox["B", FontWeight->"Bold"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSize->10, FontWeight->"Bold"], "= ", StyleBox["A", FontWeight->"Bold"], ". This can be calculated as\n ", StyleBox["B ", FontWeight->"Bold"], "= ", StyleBox["\[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["0.5", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " ", StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}] }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[ "H Linear differential equations and local stability analysis in \ continuous time"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData["Linear differential equations"], "Subsubsection", Evaluatable->False, PageBreakAbove->False, AspectRatioFixed->True], Cell[TextData[{ " The linear differential equation\n d", StyleBox["x/", FontSlant->"Italic"], "d", StyleBox[ "t = a x \ ", FontSlant->"Italic"], "[1]", StyleBox["\n", FontSlant->"Italic"], "expresses the model that the rate of change of ", StyleBox["x ", FontSlant->"Italic"], "is a constant multiple ", StyleBox["a ", FontSlant->"Italic"], "of its current value. Starting with the value ", StyleBox["x(", FontSlant->"Italic"], "0", StyleBox[") ", FontSlant->"Italic"], "at time zero, the solution is\n\t\t\t", StyleBox["x(t)= x(", FontSlant->"Italic"], "0", StyleBox[") ", FontSlant->"Italic"], "exp ", StyleBox["(at) ", FontSlant->"Italic"], "[2]" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Verify that this is the solution in two ways, (a) by using ", StyleBox["DSolve", FontWeight->"Bold"], ", (b) by showing that it satisfies both the initial condition and the \ differential equation." }], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " Consider now the set of ", StyleBox["n ", FontSlant->"Italic"], "linear differential equations in ", StyleBox["n ", FontSlant->"Italic"], "variables:\n d", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], "d", StyleBox["t = a", FontSlant->"Italic"], StyleBox["11", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" + a", FontSlant->"Italic"], StyleBox["12", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" x", FontSlant->"Italic"], StyleBox["2 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["+ ... + a", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n ", FontSlant->"Italic"], "d", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], "d", StyleBox["t = a", FontSlant->"Italic"], StyleBox["21", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["+ a", FontSlant->"Italic"], StyleBox["22", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["+ ... + a", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], StyleBox[ " \n . \ ", FontSlant->"Italic"], "[3]", StyleBox["\n .\n ", FontSlant->"Italic"], "d", StyleBox["x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], "d", StyleBox["t = a", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["+ a", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" + ... + a", FontSlant->"Italic"], StyleBox["nn", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n", FontSlant->"Italic"], "which can be expressed in matrix notation as\n d", StyleBox["x", FontWeight->"Bold"], "/d", StyleBox["t= ", FontSlant->"Italic"], StyleBox["A", FontWeight->"Bold"], StyleBox[ ".x \ ", FontWeight->"Bold", FontSlant->"Italic"], "[4]", StyleBox["\n", FontSlant->"Italic"], "where differentiation of a vector means differentiating each of its \ elements. The solution subject to the initial values ", StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(", FontSlant->"Italic"], "0", StyleBox[") ", FontSlant->"Italic"], "is \n ", StyleBox[" ", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(t) = ", FontSlant->"Italic"], "exp(", StyleBox["A", FontWeight->"Bold"], StyleBox["t)", FontSlant->"Italic"], StyleBox[".", FontWeight->"Bold"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(", FontSlant->"Italic"], "0", StyleBox[ ") \ ", FontSlant->"Italic"], "[5]", StyleBox["\n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "This satisfies the initial conditions and it also satisfies the \ differential equations since d\:02daexp(", StyleBox["A", FontWeight->"Bold"], StyleBox["t", FontSlant->"Italic"], ")/d", StyleBox["t", FontSlant->"Italic"], "\:02da= ", StyleBox["A ", FontWeight->"Bold"], "exp(", StyleBox["A", FontWeight->"Bold"], StyleBox["t", FontSlant->"Italic"], "). Using the spectral decomposition in Equation[G9] and writing" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " ", StyleBox["c", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" = ", FontSlant->"Italic"], StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(", FontSlant->"Italic"], "0", StyleBox[")", FontSlant->"Italic"], " \ [6]" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["we find that"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " ", StyleBox["x", FontWeight->"Bold"], "(", StyleBox["t", FontSlant->"Italic"], ") = ", StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], StyleBox["c", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ") ", StyleBox["e", FontWeight->"Bold"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], " \ [7]" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "(since ", StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ")", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".x", FontWeight->"Bold"], "(0)", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ")(", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10, FontSlant->"Italic"], "x ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") ", StyleBox[".x", FontWeight->"Bold"], "(0) = ", StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ")", StyleBox[" ", FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " x (", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " ", StyleBox[".x", FontWeight->"Bold"], "(0)) \n\t =", StyleBox[" ", FontSlant->"Italic"], StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], StyleBox["c", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontFamily->"Symbol"], "exp(", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic"], ")", StyleBox[" ", FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "). " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Example: Solve\n d", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], "d", StyleBox["t = x", FontSlant->"Italic"], StyleBox["1 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["+ ", FontSlant->"Italic"], "2", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n ", FontSlant->"Italic"], "d", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], "d", StyleBox["t = ", FontSlant->"Italic"], "3", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" + ", FontSlant->"Italic"], "2", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n", FontSlant->"Italic"], "with ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], "0", StyleBox[") = x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], "0", StyleBox[") = ", FontSlant->"Italic"], "1", StyleBox[".", FontSlant->"Italic"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["x1[0]={1,1};\nA1={{1,2},{3,2}};"], "Input", AspectRatioFixed->True], Cell[TextData["{vals1,vecs1}=Eigensystem[A1]"], "Input", AspectRatioFixed->True], Cell[TextData["{lvals1,lvecs1}=Eigensystem[Transpose[A1]]"], "Input", AspectRatioFixed->True], Cell[TextData[ "Do[lvecs1[[i]]=lvecs1[[i]]/(vecs1[[i]].lvecs1[[i]]),\n\t\t\t\t\t{i,2}]"], "Input", AspectRatioFixed->True], Cell[TextData["c1=lvecs1.x1[0]"], "Input", AspectRatioFixed->True], Cell[TextData[ "x1[t]=Sum[c1[[i]] Exp[vals1[[i]] t] vecs1[[i]],{i,2}]\n"], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[ "Local stability analysis of nonlinear differential equations"], "Subsubsection", Evaluatable->False, PageBreakAbove->False, AspectRatioFixed->True], Cell[TextData[{ " Consider the differential equation\n d", StyleBox["x/", FontSlant->"Italic"], "d", StyleBox["t = f(x)", FontSlant->"Italic"], ".\nAt equilibrium there is no change so that ", StyleBox["f", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ")", StyleBox[" = ", FontSlant->"Italic"], "0; a value satisfying this condition is called an equilibrium (or \ stationary) value. Such a value is ", StyleBox["locally stable ", FontSlant->"Italic"], "if the system tends to return to it after a small perturbation; otherwise \ it is ", StyleBox["locally unstable", FontSlant->"Italic"], ". Local stability can be established by linearizing about the equilibrium \ value in a Taylor series expansion.\n Suppose that ", StyleBox["x", FontSlant->"Italic"], StyleBox["* ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "is an equilibrium value so that ", StyleBox["f(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") ", FontSlant->"Italic"], "= 0, and consider what happens when ", StyleBox["x ", FontSlant->"Italic"], "is near this value, so that ", StyleBox["x = x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + p", FontSlant->"Italic"], ", where ", StyleBox["p ", FontSlant->"Italic"], " is a small perturbation. Expanding ", StyleBox["f(x) ", FontSlant->"Italic"], "in a Taylor series expansion about ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ",\n ", StyleBox["f(x) ~ f(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") p.\n", FontSlant->"Italic"], "Also\n d", StyleBox["x/", FontSlant->"Italic"], "d", StyleBox["t = ", FontSlant->"Italic"], "d", StyleBox["(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + p)/", FontSlant->"Italic"], "d", StyleBox["t ", FontSlant->"Italic"], "= d", StyleBox["p/", FontSlant->"Italic"], "d", StyleBox["t\n", FontSlant->"Italic"], "since ", StyleBox["x", FontSlant->"Italic"], StyleBox["* ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "is a constant. Thus to a linear approximation\n d", StyleBox["p/", FontSlant->"Italic"], "d", StyleBox["t = f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") p\n", FontSlant->"Italic"], "whose solution is\n ", StyleBox["p", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], ") =", StyleBox[" p", FontSlant->"Italic"], "(0)", StyleBox[" ", FontSlant->"Italic"], "exp", StyleBox["(f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") t).", FontSlant->"Italic"], "\nThe behavior of the system depends on the sign of the derivative", StyleBox[" f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic"], ". If it is negative ", StyleBox["p", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], ")", StyleBox[" ", FontSlant->"Italic"], "tends to zero so that the system returns to ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic"], "after a small perturbation; the equilibrium is locally stable. If it is \ positive, ", StyleBox["p(t", FontSlant->"Italic"], ") increases exponentially and the system moves away from ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ", at least initially, after a small perturbation; the equilibrium is \ locally unstable.\n Consider for example the logistic model\n \ ", StyleBox[" ", FontSlant->"Italic"], "d", StyleBox["n/", FontSlant->"Italic"], "d", StyleBox["t = r(", FontSlant->"Italic"], "1", StyleBox[" \[LongDash] n/K)n\n", FontSlant->"Italic"], "We find the equilibria and evaluate the derivative at each of them:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["f[n_]:=r (1 - n/k) n"], "Input", AspectRatioFixed->True], Cell[TextData["equil=Solve[f[n]==0,n]"], "Input", AspectRatioFixed->True], Cell[TextData["D[f[n],n] /.equil"], "Input", AspectRatioFixed->True], Cell[TextData[{ "The equilibria are at ", StyleBox["n", FontSlant->"Italic"], " =", StyleBox[" K ", FontSlant->"Italic"], "and ", StyleBox["n ", FontSlant->"Italic"], "= 0; the first is stable and the second unstable.\n This analysis \ can be extended to a set of ", StyleBox["n ", FontSlant->"Italic"], "differential equations in ", StyleBox["n ", FontSlant->"Italic"], "variables:\n d", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "/d", StyleBox["t = f", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[")\n ", FontSlant->"Italic"], " d", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "/d", StyleBox["t = f", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[")\n .\n .\n ", FontSlant->"Italic"], "d", StyleBox["x", FontSlant->"Italic"], StyleBox["n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "/d", StyleBox["t = f", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[")\n", FontSlant->"Italic"], "Write ", StyleBox["x ", FontWeight->"Bold", FontSlant->"Italic"], "for the vector of variables. An equilibrium satisfies\n ", StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold"], StyleBox[") = 0, i ", FontSlant->"Italic"], "= 1, 2, ...,", StyleBox[" n.\n", FontSlant->"Italic"], "Suppose that ", StyleBox["x", FontWeight->"Bold"], StyleBox["* ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "is an equilibrium value and consider a small perturbation from it so that \ ", StyleBox["x\:02da", FontWeight->"Bold"], "=\:02da", StyleBox["x", FontWeight->"Bold"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\:02da", FontWeight->"Bold"], "+ ", StyleBox["p", FontWeight->"Bold"], ", where ", StyleBox["p ", FontWeight->"Bold"], "is a vector of perturbations. Then\n ", StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[") ~ ", FontSlant->"Italic"], StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")/", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[") p", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n", FontSlant->"Italic"], "Hence the set of differential equations governing the perturbations is \ approximately\n d", StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["/", FontSlant->"Italic"], "d", StyleBox["t = ", FontSlant->"Italic"], StyleBox["Jp\n", FontWeight->"Bold", FontSlant->"Italic"], "where ", StyleBox["J ", FontWeight->"Bold"], "is the Jacobian matrix whose ", StyleBox["ij", FontSlant->"Italic"], "th element is ", StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")/", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ". The asymptotic behavior of the solution to this equation is determined \ by the eigenvalue of ", StyleBox["J ", FontWeight->"Bold"], "with the largest real part, say ", StyleBox["\[Lambda]", FontFamily->"Symbol"], "\:02da=\:02da", StyleBox["a + bi", FontSlant->"Italic"], ". If ", StyleBox["a ", FontSlant->"Italic"], "is negative (that is to say, if all the eigenvalues have negative real \ parts), the perturbations will diminish and the equilibrium is locally \ stable; otherwise it is locally unstable. If ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "is real (", StyleBox["b ", FontSlant->"Italic"], "= 0), the perturbations will converge to (or diverge from ) 0 \ exponentially; if ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "is complex, the perturbations will tend to oscillate about the equilibrium \ (with decreasing or increasing amplitude depending on the sign of ", StyleBox["a", FontSlant->"Italic"], ").\n Consider for example the Lotka-Volterra prey-predator model:\ \n d", StyleBox["n", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "/d", StyleBox["t = (r", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\[LongDash] ", FontSlant->"Italic"], StyleBox["\[Alpha]", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["n", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[") n", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n ", FontSlant->"Italic"], "d", StyleBox["n", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "/d", StyleBox["t = (\[LongDash]r", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["+ ", FontSlant->"Italic"], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\[Alpha]", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["n", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[") n", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "The equilibrium with both species present is at (", StyleBox["n", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[", n", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[") = (r", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], StyleBox["\[Alpha]", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], ", ", StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["r", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["/", FontSlant->"Italic"], StyleBox["\[Alpha]", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[")", FontSlant->"Italic"], ". We find the eigenvalues of the Jacobian at this point:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "f[1][n1_,n2_]:=(r1-a1 n2) n1\nf[2][n1_,n2_]:=(-r2+a2 n1) n2"], "Input", AspectRatioFixed->True], Cell[TextData[ "Jac=Table[D[f[i][n[1],n[2]],n[j]],{i,2},{j,2}] /.\n\t\t\t\t\t\ {n[1]->r2/a2,n[2]->r1/a1}"], "Input", AspectRatioFixed->True], Cell[TextData["Eigenvalues[Jac]"], "Input", AspectRatioFixed->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[ "J Linear difference equations and local stability analysis in discrete \ time"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData["Linear difference equations"], "Subsubsection", Evaluatable->False, PageBreakAbove->False, AspectRatioFixed->True], Cell[TextData[{ " The linear difference equation\n ", StyleBox["x(t+", FontSlant->"Italic"], "1", StyleBox[ ") = a x(t) \ ", FontSlant->"Italic"], "[1]", StyleBox["\n", FontSlant->"Italic"], "expresses the model that the value of ", StyleBox["x ", FontSlant->"Italic"], " next time is a constant multiple ", StyleBox["a ", FontSlant->"Italic"], "of its value this time. Starting with the value ", StyleBox["x", FontSlant->"Italic"], "(0)", StyleBox[" ", FontSlant->"Italic"], "at time zero, we find\n ", StyleBox["x", FontSlant->"Italic"], "(1) ", StyleBox["= a x", FontSlant->"Italic"], "(0)", StyleBox["\n x", FontSlant->"Italic"], "(2)", StyleBox[" = a x", FontSlant->"Italic"], "(1)", StyleBox[" = a", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " ", StyleBox["x", FontSlant->"Italic"], "(0)", StyleBox[" \n", FontSlant->"Italic"], "and so on. The solution for any time ", StyleBox["t ", FontSlant->"Italic"], "is clearly\n", StyleBox[" x(t) = a", FontSlant->"Italic"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" x", FontSlant->"Italic"], "(0) ", StyleBox[ " \ ", FontSlant->"Italic"], "[2]", StyleBox["\n", FontSlant->"Italic"], "We can use ", StyleBox["NestList ", FontWeight->"Bold"], "to demonstrate this:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["f[x_]:=a x\nNestList[f,x[0],10]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Solve the linear difference equation with a = ", StyleBox[".5, 1.5, -.5, \[LongDash]1.5", FontSlant->"Plain"], " with n", StyleBox["(0)", FontSlant->"Plain"], " ", StyleBox["= 100", FontSlant->"Plain"], " in each case for 10 generations." }], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " Consider now the set of ", StyleBox["n ", FontSlant->"Italic"], "linear difference equations in ", StyleBox["n ", FontSlant->"Italic"], "variables:\n ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t", FontSlant->"Italic"], "+1", StyleBox[") = a", FontSlant->"Italic"], StyleBox["11", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + a", FontSlant->"Italic"], StyleBox["12 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + ... a", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t)\n x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t", FontSlant->"Italic"], "+1", StyleBox[") = a", FontSlant->"Italic"], StyleBox["21", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + a", FontSlant->"Italic"], StyleBox["22 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + ... a", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) ", FontSlant->"Italic"], "[3]", StyleBox["\n .\n .\n x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t", FontSlant->"Italic"], "+1", StyleBox[") = a", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + a", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + ... a", FontSlant->"Italic"], StyleBox["nn", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t)\n", FontSlant->"Italic"], "which can be expressed in matrix notation as\n ", StyleBox["x", FontWeight->"Bold"], StyleBox["(t", FontSlant->"Italic"], "+1", StyleBox[") = ", FontSlant->"Italic"], StyleBox["A", FontWeight->"Bold"], StyleBox[".x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[ "(t) \ ", FontSlant->"Italic"], "[4]", StyleBox["\n", FontSlant->"Italic"], "where ", StyleBox["x", FontWeight->"Bold"], StyleBox["(t) ", FontSlant->"Italic"], "is the vector {", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t)", FontSlant->"Italic"], "} and ", StyleBox["A ", FontWeight->"Bold"], "is the matrix of coefficients ", StyleBox["a", FontSlant->"Italic"], StyleBox["ij", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ". We want to solve this set of equations starting from the initial value ", StyleBox["x", FontWeight->"Bold"], "(0)", StyleBox[" ", FontSlant->"Italic"], "at time zero. By analogy with the univariate case the solution is\n \ ", StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(t) = ", FontSlant->"Italic"], StyleBox["A", FontWeight->"Bold"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".x", FontWeight->"Bold", FontSlant->"Italic"], "(0)", StyleBox[ " \ ", FontSlant->"Italic"], "[5]\nUsing Equation [G5] and writing\n ", StyleBox["c", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" = ", FontSlant->"Italic"], StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[".x", FontWeight->"Bold", FontSlant->"Italic"], "(0) \ [6]\nwe find" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " ", StyleBox[" ", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(t) = ", FontSlant->"Italic"], StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox[" ", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["c", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["i ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " \ [7]", StyleBox[" ", FontSlant->"Italic"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "(since as before ", StyleBox["\[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["T", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[".x", FontWeight->"Bold"], "(0)", StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["\[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "(", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " x ", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") ", StyleBox[".x", FontWeight->"Bold"], "(0) = ", StyleBox["\[CapitalSigma] \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " x (", StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["*", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " ", StyleBox[".x", FontWeight->"Bold"], "(0)) =", StyleBox[" ", FontSlant->"Italic"], StyleBox["\[CapitalSigma] ", FontFamily->"Symbol"], StyleBox["c", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" \[Lambda]", FontFamily->"Symbol"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontWeight->"Bold", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["e", FontWeight->"Bold"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Subscript"}], "). ", StyleBox[" ", FontSlant->"Italic"], "\n We shall now consider as an example the model\n\t\t", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1)", StyleBox[" = x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) +", FontSlant->"Italic"], " 2 ", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t)\n\t\tx", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1)", StyleBox[" = ", FontSlant->"Italic"], "3", StyleBox[" x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) +", FontSlant->"Italic"], " 2", StyleBox[" x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ "(t) ", FontSlant->"Italic"], "[8]", StyleBox["\n", FontSlant->"Italic"], "with the initial values ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(0)", StyleBox[" = x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(0) = 1", StyleBox[". ", FontSlant->"Italic"], "These equations can be solved by iterating fron one generation to the next \ recursively without using any of the above formulae:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["x[t_]:=x[t]=A.x[t-1]\nx[0]={1,1};\nA={{1,2},{3,2}};"], "Input", AspectRatioFixed->True], Cell[TextData["Table[x[t],{t,0,10}] //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[" They can also be solved by using Eqn [5]:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "Table[MatrixPower[A,t].x[0],{t,0,10}] //MatrixForm"], "Input", AspectRatioFixed->True], Cell[TextData[" Finally, they can be solved using Eqn [7]:"], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["{vals,vecs}=Eigensystem[A]"], "Input", AspectRatioFixed->True], Cell[TextData["{lvals,lvecs}=Eigensystem[Transpose[A]]"], "Input", AspectRatioFixed->True], Cell[TextData[ "Do[lvecs[[i]]=lvecs[[i]]/(vecs[[i]].lvecs[[i]]),\n\t\t\t\t\t{i,2}]"], "Input",\ AspectRatioFixed->True], Cell[TextData["c=lvecs.x[0]"], "Input", AspectRatioFixed->True], Cell[TextData["lvecs"], "Input", AspectRatioFixed->True], Cell[TextData[ "xform[t_]:=N[Sum[c[[i]] vals[[i]]^t vecs[[i]],{i,2}]]"], "Input", AspectRatioFixed->True], Cell[TextData["MatrixForm[xform[t]]"], "Input", AspectRatioFixed->True], Cell[TextData[ "Verify that the formula given by xform generates the same solutions as found \ above"], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " The advantage of this approach is that it shows immediately what \ the asymptotic behavior of the system will be. In this example, the term in \ 4", StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Superscript"}], "will eventually dominate the term in (\[LongDash]1)", StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ", and the system will asymptotically lie in the direction of the dominant \ right eigenvector {2/3, 1}, increasing fourfold each year (since the dominant \ eigenvalue is 4). The constant ", StyleBox["c ", FontSlant->"Italic"], "multiplying the dominant term (in this case ", StyleBox["c", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "= 1.2) is the inner product of the dominant left eigenvector and the \ vector of initial values. Thus the elements of this eigenvector", StyleBox[" ", FontSlant->"Italic"], "reflect the relative importance of the different initial values (which \ might, for example, represent age classes) in determining ", StyleBox["c ", FontSlant->"Italic"], "which decides the absolute population sizes. In the present case this \ eigenvector is {0.6, 0.6}, which means that the two initial values are \ equally important in determining the ultimate population sizes.\n We \ now consider an example in which the eigenvalues are complex:\n\t ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t)+ ", FontSlant->"Italic"], "2", StyleBox["x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t)\n\tx", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1) = ", StyleBox["\[LongDash]", FontSlant->"Italic"], "2", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t) + x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[ "(t) \ ", FontSlant->"Italic"], "[9]\nwith initial values ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(0) =", StyleBox[" x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(0) = 1. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["A={{1,2},{-2,1}};\nx[0]={1,1};"], "Input", AspectRatioFixed->True], Cell[TextData["{vals,vecs}=Eigensystem[A] "], "Input", AspectRatioFixed->True], Cell[TextData["{lvals,lvecs}=Eigensystem[Transpose[A]] "], "Input", AspectRatioFixed->True], Cell[TextData[ "Do[lvecs[[i]]=lvecs[[i]]/(vecs[[i]].lvecs[[i]]),\n\t\t\t\t\t{i,2}]"], "Input",\ AspectRatioFixed->True], Cell[TextData["c=lvecs.x[0]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Since the eigenvalues are complex, it is simpler to express them in polar \ coordinates,", StyleBox[" r", FontSlant->"Italic"], " and ", StyleBox["\[Theta]", FontFamily->"Symbol", FontSlant->"Italic"], ", so that \n ", StyleBox["\[Lambda]", FontFamily->"Symbol"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " =", StyleBox[" r", FontSlant->"Italic"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "(cos", StyleBox[" ", FontSlant->"Italic"], StyleBox["\[Theta] ", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["t", FontSlant->"Italic"], " + ", StyleBox["i", FontSlant->"Italic"], " sin ", StyleBox["\[Theta] ", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["t", FontSlant->"Italic"], ")" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["r=Abs[vals] //N"], "Input", AspectRatioFixed->True], Cell[TextData["theta=Arg[vals] //N"], "Input", AspectRatioFixed->True], Cell[TextData[ "func[i_,t_]:=r[[i]]^t (Cos[theta[[i]] t] +\n\t\t\t\t\tI Sin[theta[[i]] t]) \ "], "Input", AspectRatioFixed->True], Cell[TextData["x[t_]:=Sum[c[[i]] func[i,t] vecs[[i]],{i,2}];"], "Input", AspectRatioFixed->True], Cell[TextData["x[t]=Chop[Simplify[x[t]]]"], "Input", AspectRatioFixed->True], Cell[TextData[{ "Verify that this solution agrees with that found by direct iteration of ", StyleBox["[4]", FontSlant->"Plain"], ". Plot the solution. Quantify the period of the oscillations. Note that \ imaginary numbers appear in the intermediate calculations but not in the \ final answer." }], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " This method can be used for sets of equations with any number of \ variables. The asymptotic behavior of the system depends on the eigenvalues \ of the matrix ", StyleBox["A ", FontWeight->"Bold"], "and can be", StyleBox[" ", FontWeight->"Bold"], "determined as follows. (We suppose that these eigenvalues are distinct.) \ Rank the eigenvalues by their ", StyleBox["absolute value", FontSlant->"Italic"], ". There are two common situations. (i) There may be a", StyleBox[" real", FontSlant->"Italic"], " eigenvalue", StyleBox[" \[Lambda]", FontFamily->"Symbol"], " greater in absolute value than any of the others. The system will \ increase or decrease exponentially according as |", StyleBox["\[Lambda]", FontFamily->"Symbol"], "| >1 or |", StyleBox["\[Lambda]", FontFamily->"Symbol"], "| <1; it will alternate in sign or not according as ", StyleBox["\[Lambda] ", FontFamily->"Symbol"], "is negative or positive. (ii) There may be a pair of complex conjugate \ eigenvalues greater in absolute value than any of the others. The system will \ oscillate with increasing or decreasing amplitude according as |", StyleBox["\[Lambda]", FontFamily->"Symbol"], "| >1 or |", StyleBox["\[Lambda]", FontFamily->"Symbol"], "| <1.\n" }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[ "Local stability analysis of nonlinear difference equations"], "Subsubsection",\ Evaluatable->False, PageBreakAbove->Automatic, AspectRatioFixed->True], Cell[TextData[{ " Consider the difference equation\n ", StyleBox["x", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" f(x(t)).\n ", FontSlant->"Italic"], "At equilibrium ", StyleBox["x", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "+1)", StyleBox[" = x(t)", FontSlant->"Italic"], ", so that ", StyleBox["f(x) = x", FontSlant->"Italic"], ". Suppose that ", StyleBox["x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic"], " is an equilibrium, and write ", StyleBox["x(t)\:02da=\:02dax", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + p(t). ", FontSlant->"Italic"], "Then\n ", StyleBox["x", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" f(x(t)) = f(x", FontSlant->"Italic"], StyleBox["* ", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["+ p(t)) ~ f(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") p(t) = x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") p(t)\n", FontSlant->"Italic"], "whence to a first order approximation\n ", StyleBox["p", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") p(t).\n", FontSlant->"Italic"], "The solution of this equation is\n ", StyleBox["p(t) = ", FontSlant->"Italic"], "[", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")", FontSlant->"Italic"], "]", StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["p", FontSlant->"Italic"], "(0)", StyleBox["\n", FontSlant->"Italic"], "which converges to zero or diverges from it (giving stability or \ instability) according as the absolute value of ", StyleBox["f", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'(x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") ", FontSlant->"Italic"], "is less than or greater than unity.\n Consider now the set of ", StyleBox["n ", FontSlant->"Italic"], "difference equations in ", StyleBox["n ", FontSlant->"Italic"], "unknowns\n ", StyleBox["x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" f", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t))\n x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" f", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t))\n .\n .\n x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1 )=", StyleBox[" f", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(x", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), x", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t), ..., x", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(t))\n", FontSlant->"Italic"], "An equilibrium satisfies \n ", StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold"], StyleBox[") = ", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", i ", FontSlant->"Italic"], "= 1, 2, ...", StyleBox[", n.\n", FontSlant->"Italic"], "Thus near an equilibrium\n ", StyleBox["x", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(t)) = f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + ", FontSlant->"Italic"], StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(t)) ~ f(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[") + ", FontSlant->"Italic"], StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")/", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[") p", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[". = x", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" + ", FontSlant->"Italic"], StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["f", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["(", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["*", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[")/", FontSlant->"Italic"], StyleBox["\[Delta]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[") p", FontSlant->"Italic"], StyleBox["j", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\n", FontSlant->"Italic"], "so that to a first approximation\n ", StyleBox["p", FontWeight->"Bold", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], "+1) =", StyleBox[" ", FontSlant->"Italic"], StyleBox["J", FontWeight->"Bold"], StyleBox[" p", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["(t)\n", FontSlant->"Italic"], "where ", StyleBox["J", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], "is the Jacobian matrix of derivatives. The ultimate behavior of the \ solution to this equation depends on the eigenvalue of ", StyleBox["J ", FontWeight->"Bold"], "with the largest absolute value. The solution converges to zero (giving \ local stability) or diverges from it (giving instability) according as this \ absolute value is less than or greater than unity. If this eigenvalue is \ real, the perturbations will decrease or increase geometrically (with \ alternating sign if it is negative). If this eigenvalue is complex with \ argument ", StyleBox["\[Theta] ", FontFamily->"Symbol"], "(which implies that it has a complex conjugate with the same absolute \ value but with argument \[LongDash]", StyleBox["\[Theta]", FontFamily->"Symbol"], "), the perturbations will oscillate with period 2", StyleBox["\[Pi]/\[Theta]", FontFamily->"Symbol"], " (being damped or explosive according as the absolute value is less than \ or greater than unity).\n" }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["K The Poisson distribution"], "Subsection", Evaluatable->False, PageBreakAbove->True, AspectRatioFixed->True], Cell[TextData[{ "The Poisson distribution was first derived by S.D.Poisson in 1837 and has \ many applications in biology. It is a discrete probability distribution with \ the formula\n\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") = e", StyleBox["-", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "/", StyleBox["x", FontSlant->"Italic"], "!, ", StyleBox["x ", FontSlant->"Italic"], "= 0, 1, 2, ...\n", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") represents the probability that some random event occurs exactly x \ times. It depends on the parameter ", StyleBox["\[Mu] ", FontFamily->"Symbol"], "which is the mean of the distribution, representing the average number of \ occurences of the event; a characteristic of the Poisson distribution is that \ its mean is also its variance. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Write a function to generate the Poisson probabilities from x = 0 through \ 50 given the parameter ", StyleBox["\[Mu]", FontFamily->"Symbol"], ". Implement it with ", StyleBox["\[Mu]", FontFamily->"Symbol"], " = 0.5, 1, 2, 5, 20. Display the results, and in each case find the sum of \ the probabilities and the mean and variance, and verify that they agree with \ their theoretical values." }], "Special1", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ " The Poisson distribution is a limiting case of the binomial \ distribution which gives the probability of getting ", StyleBox["x", FontSlant->"Italic"], " \"successes\" and ", StyleBox["n", FontSlant->"Italic"], " \[LongDash] ", StyleBox["x", FontSlant->"Italic"], " \"failures\" in ", StyleBox["n", FontSlant->"Italic"], " trials if each trial results in one of two outcomes with a constant \ probability ", StyleBox["p", FontSlant->"Italic"], " of success at each trial, which is independent of the outcomes of \ previous trials. For example, the number of boys in families of fixed size is \ to a good approximation binomial because the probability of a male birth is \ very nearly the same for all families and independent of birth order. The \ formula for this distribution is\n \t", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") = (", StyleBox["n", FontSlant->"Italic"], "!/", StyleBox["x", FontSlant->"Italic"], "!(", StyleBox["n", FontSlant->"Italic"], " \[LongDash] ", StyleBox["x", FontSlant->"Italic"], ")!) ", StyleBox["p", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "(1 \[LongDash]", StyleBox[" p", FontSlant->"Italic"], ")", StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ", ", StyleBox["x", FontSlant->"Italic"], " = 0, 1, 2, ... , ", StyleBox["n", FontSlant->"Italic"], StyleBox["\n", FontVariations->{"CompatibilityType"->"Superscript"}], "The reason is that the probability of getting ", StyleBox["x", FontSlant->"Italic"], " successes and ", StyleBox["n", FontSlant->"Italic"], " \[LongDash] ", StyleBox["x", FontSlant->"Italic"], " failures ", StyleBox["in a particular order", FontSlant->"Italic"], " is ", StyleBox["p", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "(1 \[LongDash] ", StyleBox["p", FontSlant->"Italic"], ")", StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Superscript"}], "and there are", StyleBox[" n", FontSlant->"Italic"], "!/", StyleBox["x", FontSlant->"Italic"], "!(", StyleBox["n ", FontSlant->"Italic"], "\[LongDash] ", StyleBox["x", FontSlant->"Italic"], ")! ways in which this can happen. \n\tThe Poisson distribution is the \ limiting case of this distribution when the probability ", StyleBox["p", FontSlant->"Italic"], " becomes very small but the number of trials", StyleBox[" n", FontSlant->"Italic"], " becomes very large. Write ", StyleBox["\[Mu] ", FontFamily->"Symbol"], "= ", StyleBox["np", FontSlant->"Italic"], " for the mean of the binomial distribution (this is clearly its mean since \ ", StyleBox["p", FontSlant->"Italic"], " is the probability of success in each trial, whence", StyleBox[" np", FontSlant->"Italic"], " is the Expected number of successes in ", StyleBox["n", FontSlant->"Italic"], " trials), and let ", StyleBox["n", FontSlant->"Italic"], " increase and ", StyleBox["p", FontSlant->"Italic"], " decrease with ", StyleBox["\[Mu] ", FontFamily->"Symbol"], "held constant. Then with ", StyleBox["\[Mu] ", FontFamily->"Symbol"], "constant and ", StyleBox["n", FontSlant->"Italic"], "->infinity,\n\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["n", FontSlant->"Italic"], "!/", StyleBox["x", FontSlant->"Italic"], "!(", StyleBox["n ", FontSlant->"Italic"], "\[LongDash]", StyleBox[" x", FontSlant->"Italic"], ")! -> ", StyleBox["n", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "/", StyleBox["x", FontSlant->"Italic"], "!\t\t\t\t\t\t\t\t\t\t", StyleBox["p", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " = ", StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "/", StyleBox["n", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "\t\t\t\t\t\t\t\t\t\t\t(1 \[LongDash] ", StyleBox["p", FontSlant->"Italic"], ")", StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " = (1 \[LongDash] ", StyleBox["\[Mu]", FontFamily->"Symbol"], "/", StyleBox["n", FontSlant->"Italic"], ")", StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], " -> e", StyleBox["-", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontVariations->{"CompatibilityType"->"Superscript"}], "\nHence\n\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") -> e", StyleBox["-", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], "/", StyleBox["x", FontSlant->"Italic"], "!,\nthe Poisson distribution with parameter ", StyleBox["\[Mu]", FontFamily->"Symbol"], ".\n\tA classic example of the Poisson distribution is the numbers of \ deaths from horse kicks in the Prussian army between 1875 and 1894. Data on \ 14 army corps were collected by von Bortkewiecz for each of these years, \ giving 280 observations in all on the number of deaths from this cause in a \ corps in a year. The data are shown below:\n\tnumber of deaths 0 \ 1 2 3 4 5+ Total\n\tFrequency 144 \ 91 32 11 2 0 280\nOne would expect them to follow a \ Poisson distribution since there is a very large number of men at risk each \ of whom has a very small chance of being killed by a horse in a year; \ otherwise the conditions for the binomial distribution hold. The Poisson \ distribution can be fitted to these data by equating the observed mean number \ of deaths with the theoretical mean:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "data={144,91,32,11,2};\nmean = Sum[(i-1)*data[[i]],{i,5}]/280 //N"], "Input",\ AspectRatioFixed->True], Cell[TextData["expected=Table[280 Exp[-mean] mean^x/x!,{x,0,10}] "], "Input", AspectRatioFixed->True], Cell[TextData[{ " The Poisson distribution can be derived by an independent \ argument as the distribution of the number of random events occurring in a \ fixed time. Suppose that in a short time interval d", StyleBox["t ", FontSlant->"Italic"], "there is a chance ", StyleBox["\[Mu]", FontFamily->"Symbol"], " d", StyleBox["t", FontSlant->"Italic"], " that an event occurs, which does not depend on the number of events which \ have already occurred. (This is what we mean by saying that events occur ", StyleBox["at random", FontSlant->"Italic"], ".) Write ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ",", StyleBox[" t", FontSlant->"Italic"], ") for the probability that exactly ", StyleBox["n", FontSlant->"Italic"], " events have occurred by time ", StyleBox["t", FontSlant->"Italic"], ", starting with zero events at time zero. Now ", StyleBox["n", FontSlant->"Italic"], " events can occur by time (", StyleBox["t", FontSlant->"Italic"], " + d", StyleBox["t", FontSlant->"Italic"], ") in two ways: (1) ", StyleBox["n", FontSlant->"Italic"], " events occur by time ", StyleBox["t ", FontSlant->"Italic"], "and none between ", StyleBox["t", FontSlant->"Italic"], " and ", StyleBox["t", FontSlant->"Italic"], " + d", StyleBox["t", FontSlant->"Italic"], ", (2) ", StyleBox["n", FontSlant->"Italic"], " \[LongDash] 1 events occur by time ", StyleBox["t ", FontSlant->"Italic"], "and another event occurs between ", StyleBox["t ", FontSlant->"Italic"], "and", StyleBox[" t", FontSlant->"Italic"], " + d", StyleBox["t", FontSlant->"Italic"], ". (The second possibility only applies when n >= 1.) Thus\n ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ", ", StyleBox["t ", FontSlant->"Italic"], "+ d", StyleBox["t", FontSlant->"Italic"], ") = ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ",", StyleBox[" t", FontSlant->"Italic"], ") (1 \[LongDash] ", StyleBox["\[Mu]", FontFamily->"Symbol"], " d", StyleBox["t", FontSlant->"Italic"], ") + ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], " \[LongDash] 1,", StyleBox[" t", FontSlant->"Italic"], ") ", StyleBox["\[Mu]", FontFamily->"Symbol"], " d", StyleBox["t", FontSlant->"Italic"], "\nwhence\n d", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ",", StyleBox[" t", FontSlant->"Italic"], ")/d", StyleBox["t", FontSlant->"Italic"], " = ", StyleBox["\[Mu]", FontFamily->"Symbol"], " (", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], " \[LongDash] 1, ", StyleBox["t", FontSlant->"Italic"], ") \[LongDash] ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ", ", StyleBox["t", FontSlant->"Italic"], ")) for", StyleBox[" n", FontSlant->"Italic"], ">=1\n d", StyleBox["P", FontSlant->"Italic"], "(0, ", StyleBox["t", FontSlant->"Italic"], ")/d", StyleBox["t", FontSlant->"Italic"], " = \[LongDash]", StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox[" ", FontFamily->"Symbol", FontSlant->"Italic"], StyleBox["P", FontSlant->"Italic"], "(0, ", StyleBox["t", FontSlant->"Italic"], "),\nwith initial conditions\n ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["n", FontSlant->"Italic"], ", 0) = 0 for n>=1\n \ ", StyleBox[" P", FontSlant->"Italic"], "(0, 0) = 1.\nIt is straightforward to verify that the solution to these \ differential equations is the Poisson distribution with parameter", StyleBox[" \[Mu]", FontFamily->"Symbol"], StyleBox["t", FontSlant->"Italic"], ":" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["p[n_,t_]:=Exp[-mu*t] (l t)^n/n!"], "Input", AspectRatioFixed->True], Cell[TextData["p[n,0]"], "Input", AspectRatioFixed->True], Cell[TextData["Limit[p[0,t],t->0]"], "Input", AspectRatioFixed->True], Cell[TextData["D[p[0,t],t] + mu*p[0,t]"], "Input", AspectRatioFixed->True], Cell[TextData["Simplify[D[p[n,t],t] - mu*(p[n-1,t]-p[n,t])]"], "Input", AspectRatioFixed->True], Cell[TextData["%/.n/n!->1/(n-1)!"], "Input", AspectRatioFixed->True], Cell[TextData[{ " The properties of the Poisson distribution are most easily \ investigated by calculating its probability generating function (pgf), ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], "). The pgf of a discrete random variable ", StyleBox["X", FontSlant->"Italic"], " with probability function\n Prob(", StyleBox["X", FontSlant->"Italic"], " = ", StyleBox["x", FontSlant->"Italic"], ") = ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ")\nis defined as\n\t", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ") = E(", StyleBox["s", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ") = ", StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" s", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSize->10, FontSlant->"Italic"], ").\nFor the Poisson distribution\n\t", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ") = ", StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " e", StyleBox["\[LongDash]", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "(", StyleBox["s", FontSlant->"Italic"], StyleBox["\[Mu]", FontFamily->"Symbol"], ")", StyleBox["x", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "/", StyleBox["x", FontSize->10], "! = e", StyleBox["\[LongDash]", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "e", StyleBox["s", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" =", FontFamily->"Symbol"], StyleBox[" ", FontFamily->"Symbol", FontVariations->{"CompatibilityType"->"Superscript"}], "e", StyleBox["(", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["s", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]1)", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\n ", FontFamily->"Symbol"], " If we differentiate", StyleBox[" ", FontFamily->"Symbol"], "the pgf with respect to ", StyleBox["s", FontSlant->"Italic"], " and then set ", StyleBox["s", FontSlant->"Italic"], " = 1, we obtain the expected value of the random variable (the mean of the \ distribution):\n ", StyleBox["G", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["'", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ")", StyleBox[" at ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["s", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["=1", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " ", StyleBox["xs", FontSlant->"Italic"], StyleBox["x", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["P", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") ", StyleBox["at ", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["s", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["=1", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", StyleBox["\[CapitalSigma]", FontFamily->"Symbol"], StyleBox["x", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " ", StyleBox["xP", FontSlant->"Italic"], "(", StyleBox["x", FontSlant->"Italic"], ") = E(", StyleBox["X", FontSlant->"Italic"], ").\nTaking the second derivative at ", StyleBox["s", FontSlant->"Italic"], " = 1 gives\n\t", StyleBox["G", FontSlant->"Italic"], StyleBox[" ", FontSize->6, FontSlant->"Italic"], StyleBox["''", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ")", StyleBox[" at", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" s", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["=1 ", FontVariations->{"CompatibilityType"->"Subscript"}], "= E[", StyleBox["X", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], " \[LongDash] 1)] = E(", StyleBox["X", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") \[LongDash] E(", StyleBox["X", FontSlant->"Italic"], ")\nfrom which the variance can be calculated as\n Var(X) = E(", StyleBox["X", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") - E", StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "(", StyleBox["X", FontSlant->"Italic"], ") = E[", StyleBox["X", FontSlant->"Italic"], "(", StyleBox["X", FontSlant->"Italic"], " \[LongDash] 1)] + E(", StyleBox["X", FontSlant->"Italic"], ") - E", StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "(", StyleBox["X", FontSlant->"Italic"], ").\nFor the Poisson distribution we can calculate: " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["G=Exp[(s-1)*mu];"], "Input", AspectRatioFixed->True], Cell[TextData["mean=D[G,s] /.s->1"], "Input", AspectRatioFixed->True], Cell[TextData["variance=D[G,{s,2}] + mean - mean^2 /.s->1"], "Input", AspectRatioFixed->True], Cell[TextData[{ "This verifies that the mean and the variance of the distribution are both \ equal to", StyleBox[" \[Lambda]", FontFamily->"Symbol"], ".\n\tThe ", StyleBox["r", FontSlant->"Italic"], "th derivative of the pgf evaluated at ", StyleBox["s", FontSlant->"Italic"], " = 0 is equal to ", StyleBox["r", FontSlant->"Italic"], "! ", StyleBox["P", FontSlant->"Italic"], "(", StyleBox["r", FontSlant->"Italic"], "). For example consider a variable which can take values 0, 1, ..., 5:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "G[s_]:=P[0]+s*P[1]+s^2*P[2]+s^3*P[3]+s^4*P[4]+s^5*P[5]"], "Input", AspectRatioFixed->True], Cell[TextData["Table[D[G[s],{s,r}]/r!,{r,0,5}] /.s->0"], "Input", AspectRatioFixed->True], Cell[TextData[{ " This enables the probability distribution to be calculated from \ the pgf, which is useful if the pgf has been obtained indirectly without \ direct knowledge of the probability distribution.\n Another useful \ property is that the pgf of the sum of two independent random variables is \ equal to the product of their pgf's. Suppose ", StyleBox["X", FontSlant->"Italic"], " and ", StyleBox["Y", FontSlant->"Italic"], " are independent random variables with pgf's ", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") and ", StyleBox["G", FontSlant->"Italic"], StyleBox["Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], "); the pgf of their sum, ", StyleBox["X", FontSlant->"Italic"], " + ", StyleBox["Y", FontSlant->"Italic"], ", is\n ", StyleBox["G", FontSlant->"Italic"], StyleBox["X+Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") = E(", StyleBox["s", FontSlant->"Italic"], StyleBox["X+Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ") = E(", StyleBox["s", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["s", FontSlant->"Italic"], StyleBox["Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ") (under independence) = ", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") ", StyleBox["G", FontSlant->"Italic"], StyleBox["Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ")\nIf ", StyleBox["X", FontSlant->"Italic"], " and ", StyleBox["Y ", FontSlant->"Italic"], "are independent Poisson variates with parameters ", StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " and ", StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["2 ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "respectively, the pgf of their sum is\n e", StyleBox["(s\[LongDash]1)", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["1 ", FontSize->10], "e", StyleBox["(s\[LongDash]1)", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["2 ", FontSize->10], "= e", StyleBox["(s\[LongDash]1)(", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["1", FontSize->10], StyleBox["+", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\[Mu]", FontFamily->"Symbol", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["2", FontSize->10], StyleBox[")\n", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "which is the pgf of a Poisson distribution with parameter ", StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "+ ", StyleBox["\[Mu]", FontFamily->"Symbol"], StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ". Hence the sum of two independent Poisson variates is itself a Poisson \ variate, with mean equal to the sum of the means.\n The sum of ", StyleBox["n", FontSlant->"Italic"], " identically and independently distributed random variables ", StyleBox["X", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], " with pgf ", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], "), say ", StyleBox["Y", FontSlant->"Italic"], " = ", StyleBox["X", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["X", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ... ", StyleBox["X", FontSlant->"Italic"], StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], ", has pgf ", StyleBox["G", FontSlant->"Italic"], StyleBox["Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") = (", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], "))", StyleBox["n", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], ". Suppose however that the number of ", StyleBox["X", FontSlant->"Italic"], StyleBox["i", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "'s is not fixed but is itself a random variable ", StyleBox["N", FontSlant->"Italic"], " with probability distribution ", StyleBox["P", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["n", FontSlant->"Italic"], ") and pgf ", StyleBox["G", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], "). The pgf of ", StyleBox["Y", FontSlant->"Italic"], " = ", StyleBox["X", FontSlant->"Italic"], StyleBox["1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ", StyleBox["X", FontSlant->"Italic"], StyleBox["2", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], " + ... ", StyleBox["X", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" ", FontSize->10], "is", StyleBox["\n ", FontSize->10], StyleBox["G", FontSlant->"Italic"], StyleBox["Y", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") = ", StyleBox["P", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(0) + ", StyleBox["P", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(1) ", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") + ", StyleBox["P", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(2) [", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ")]", StyleBox["2 ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "+ ", StyleBox["P", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(3) [", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ")]", StyleBox["3", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" ", FontVariations->{"CompatibilityType"->"Superscript"}], "+ ...\n = ", StyleBox["G", FontSlant->"Italic"], StyleBox["N", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["G", FontSlant->"Italic"], StyleBox["X", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], "))\nFor example, suppose an individual has a Poisson number of offspring \ with mean ", StyleBox["\[Mu]", FontFamily->"Symbol"], " and then dies, and that each of these offspring has a Poisson number of \ offspring with the same mean. Then the number of grandoffspring of the \ original individual has pgf ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ")), where ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ") is the pgf of the Poissson distributioin; this can be used to find the \ probability distribution of the number of grandoffspring. Suppose for example \ that ", StyleBox["\[Mu]", FontFamily->"Symbol"], " = 1; we compare the distribution of the number of offspring (Poisson with \ mean 1) and the number of grandoffspring:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["G[s_]:=Exp[(s-1)]"], "Input", AspectRatioFixed->True], Cell[TextData["Table[Exp[-1]/r!,{r,0,10}] //N"], "Input", AspectRatioFixed->True], Cell[TextData["Table[D[G[G[s]],{s,r}]/r!,{r,0,10}] /.s->0 //N"], "Input", AspectRatioFixed->True], Cell[TextData[{ "This process is called a Poisson branching process. If ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ") is the Poisson pgf, the pgf of the descendants of a single individual \ after t generations is\n\t", StyleBox["G", FontSlant->"Italic"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") = ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["G", FontSlant->"Italic"], "(...", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ")...)))\nwhere there are", StyleBox[" t G", FontSlant->"Italic"], "'s on the right hand side; this can be calculated from the recurrence \ relationship\n\t", StyleBox["G", FontSlant->"Italic"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ") = ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["\tG", FontSlant->"Italic"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox["\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["s", FontSlant->"Italic"], ")) \nIn particular if ", StyleBox["q", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], ") is the probability that the line has gone extinct by ", StyleBox["t ", FontSlant->"Italic"], "years, which is the probability that there are no descendants, then ", StyleBox["q", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], ") = ", StyleBox["G", FontSlant->"Italic"], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Subscript"}], "(0), which satisfies the recurrence\n\t", StyleBox["q(t", FontSlant->"Italic"], ") = ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["q", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], " \[LongDash] 1)).\nIf ", StyleBox["G", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ") is the pgf of a Poisson distribution with mean 1, this becomes\n\t", StyleBox["q", FontSlant->"Italic"], "(", StyleBox["t", FontSlant->"Italic"], ") = e", StyleBox["q", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["(", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["t", FontSize->10, FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\[LongDash]1)\[LongDash]1", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["\nwith the initial condition that ", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["q", FontSlant->"Italic", FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox["(0) = 0.", FontVariations->{"CompatibilityType"->"Superscript"}] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["q[0]=0.0;\nq[t_]:=q[t]=Exp[q[t-1]-1]"], "Input", AspectRatioFixed->True], Cell[TextData["Table[{t,q[t]},{t,10,100,10}] //TableForm"], "Input", AspectRatioFixed->True] }, Closed]] }, Open ]] }, FrontEndVersion->"NeXT 3.0", ScreenRectangle->{{0, 1053}, {0, 832}}, WindowToolbars->{"RulerBar", "EditBar"}, CellGrouping->Automatic, WindowSize->{520, 600}, WindowMargins->{{216, Automatic}, {Automatic, 82}}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, 128}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, CharacterEncoding->"MacintoshAutomaticEncoding" ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. ***********************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1731, 51, 144, 4, 133, "Subtitle"], Cell[CellGroupData[{ Cell[1900, 59, 117, 3, 43, "Subsection", PageBreakAbove->False], Cell[2020, 64, 663, 16, 110, "Text", PageBreakAbove->False] }, Closed]], Cell[CellGroupData[{ Cell[2720, 85, 132, 3, 27, "Subsection", PageBreakAbove->True], Cell[2855, 90, 3127, 110, 242, "Text"], Cell[5985, 202, 76, 1, 24, "Input"], Cell[6064, 205, 90, 1, 24, "Input"], Cell[6157, 208, 103, 1, 24, "Input"], Cell[6263, 211, 4375, 153, 202, "Text"], Cell[10641, 366, 3711, 128, 122, "Text"], Cell[14355, 496, 245, 5, 36, "Special1"], Cell[14603, 503, 346, 9, 52, "Text"], Cell[14952, 514, 75, 1, 24, "Input"], Cell[15030, 517, 75, 1, 24, "Input"], Cell[15108, 520, 75, 1, 24, "Input"], Cell[15186, 523, 4165, 156, 132, "Text"], Cell[19354, 681, 124, 3, 36, "Input"], Cell[19481, 686, 63, 1, 24, "Input"], Cell[19547, 689, 65, 1, 24, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[19649, 695, 123, 3, 27, "Subsection", PageBreakAbove->True], Cell[19775, 700, 2520, 89, 256, "Text"], Cell[22298, 791, 76, 1, 24, "Input"], Cell[22377, 794, 1474, 55, 126, "Text"], Cell[23854, 851, 79, 1, 36, "Input"], Cell[23936, 854, 60, 1, 24, "Input"], Cell[23999, 857, 59, 1, 24, "Input"], Cell[24061, 860, 2253, 104, 106, "Text"], Cell[26317, 966, 61, 1, 24, "Input"], Cell[26381, 969, 58, 1, 24, "Input"], Cell[26442, 972, 770, 19, 110, "Text"], Cell[27215, 993, 67, 1, 24, "Input"], Cell[27285, 996, 71, 1, 24, "Input"], Cell[27359, 999, 68, 1, 24, "Input"], Cell[27430, 1002, 72, 1, 24, "Input"], Cell[27505, 1005, 1451, 70, 167, "Text"], Cell[28959, 1077, 67, 1, 24, "Input"], Cell[29029, 1080, 67, 1, 24, "Input"], Cell[29099, 1083, 739, 25, 84, "Text"], Cell[29841, 1110, 63, 1, 24, "Input"], Cell[29907, 1113, 179, 4, 22, "Special1"], Cell[30089, 1119, 93, 1, 24, "Input"], Cell[30185, 1122, 1096, 41, 74, "Text"], Cell[31284, 1165, 68, 1, 24, "Input"], Cell[31355, 1168, 73, 1, 24, "Input"], Cell[31431, 1171, 2491, 108, 159, "Text"], Cell[33925, 1281, 55, 1, 24, "Input"], Cell[33983, 1284, 71, 1, 24, "Input"], Cell[34057, 1287, 1827, 72, 87, "Text"], Cell[35887, 1361, 63, 1, 24, "Input"], Cell[35953, 1364, 58, 1, 24, "Input"], Cell[36014, 1367, 68, 1, 24, "Input"], Cell[36085, 1370, 68, 1, 24, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[36190, 1376, 136, 3, 27, "Subsection", PageBreakAbove->True], Cell[36329, 1381, 406, 11, 62, "Text"], Cell[36738, 1394, 116, 2, 36, "Input"], Cell[36857, 1398, 696, 21, 78, "Text"], Cell[37556, 1421, 178, 4, 46, "Text"], Cell[37737, 1427, 83, 1, 36, "Input"], Cell[37823, 1430, 170, 4, 30, "Text"], Cell[37996, 1436, 58, 1, 24, "Input"], Cell[38057, 1439, 139, 3, 30, "Text"], Cell[38199, 1444, 69, 1, 24, "Input"], Cell[38271, 1447, 179, 4, 46, "Text"], Cell[38453, 1453, 57, 1, 24, "Input"], Cell[38513, 1456, 414, 12, 62, "Text"], Cell[38930, 1470, 58, 1, 24, "Input"], Cell[38991, 1473, 165, 4, 30, "Text"], Cell[39159, 1479, 58, 1, 24, "Input"], Cell[39220, 1482, 1744, 68, 117, "Text"], Cell[40967, 1552, 107, 2, 24, "Input"], Cell[41077, 1556, 62, 1, 24, "Input"], Cell[41142, 1559, 825, 24, 120, "Text"], Cell[41970, 1585, 96, 1, 24, "Input"], Cell[42069, 1588, 1686, 72, 80, "Text"], Cell[43758, 1662, 71, 1, 24, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[43866, 1668, 136, 3, 27, "Subsection", PageBreakAbove->True], Cell[44005, 1673, 540, 15, 78, "Text"], Cell[44548, 1690, 74, 1, 24, "Input"], Cell[44625, 1693, 126, 2, 30, "Text"], Cell[44754, 1697, 68, 1, 24, "Input"], Cell[44825, 1700, 359, 9, 62, "Text"], Cell[45187, 1711, 157, 3, 60, "Input"], Cell[45347, 1716, 182, 4, 46, "Text"], Cell[45532, 1722, 115, 2, 48, "Input"], Cell[45650, 1726, 115, 2, 48, "Input"], Cell[45768, 1730, 359, 13, 36, "Text"], Cell[46130, 1745, 55, 1, 24, "Input"], Cell[46188, 1748, 66, 1, 24, "Input"], Cell[46257, 1751, 68, 1, 24, "Input"], Cell[46328, 1754, 180, 4, 46, "Text"], Cell[46511, 1760, 72, 1, 24, "Input"], Cell[46586, 1763, 1028, 47, 78, "Text"], Cell[47617, 1812, 74, 1, 24, "Input"], Cell[47694, 1815, 84, 1, 24, "Input"], Cell[47781, 1818, 714, 26, 78, "Text"], Cell[48498, 1846, 82, 1, 24, "Input"], Cell[48583, 1849, 82, 1, 24, "Input"], Cell[48668, 1852, 55, 1, 36, "Input"], Cell[48726, 1855, 1321, 55, 99, "Text"], Cell[50050, 1912, 66, 1, 24, "Input"], Cell[50119, 1915, 63, 1, 24, "Input"], Cell[50185, 1918, 503, 15, 65, "Text"], Cell[50691, 1935, 116, 2, 60, "Input"], Cell[50810, 1939, 99, 3, 31, "Text"], Cell[50912, 1944, 62, 1, 24, "Input"], Cell[50977, 1947, 122, 3, 31, "Text"], Cell[51102, 1952, 71, 1, 24, "Input"], Cell[51176, 1955, 97, 3, 31, "Text"], Cell[51276, 1960, 62, 1, 24, "Input"], Cell[51341, 1963, 137, 3, 31, "Text"], Cell[51481, 1968, 58, 1, 24, "Input"], Cell[51542, 1971, 102, 6, 48, "Text"], Cell[51647, 1979, 256, 5, 46, "Text"], Cell[51906, 1986, 1485, 58, 128, "Text"], Cell[53394, 2046, 134, 3, 30, "Text"], Cell[53531, 2051, 254, 8, 46, "Text"], Cell[53788, 2061, 353, 13, 46, "Text"], Cell[54144, 2076, 102, 1, 36, "Input"], Cell[54249, 2079, 149, 7, 30, "Text"], Cell[54401, 2088, 79, 1, 24, "Input"], Cell[54483, 2091, 520, 14, 132, "Text"], Cell[55006, 2107, 70, 1, 24, "Input"], Cell[55079, 2110, 321, 11, 94, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[55437, 2126, 163, 4, 27, "Subsection", PageBreakAbove->True], Cell[55603, 2132, 602, 18, 78, "Text"], Cell[56208, 2152, 117, 2, 36, "Input"], Cell[56328, 2156, 60, 1, 24, "Input"], Cell[56391, 2159, 60, 1, 24, "Input"], Cell[56454, 2162, 1182, 43, 122, "Text"], Cell[57639, 2207, 72, 1, 24, "Input"], Cell[57714, 2210, 72, 1, 24, "Input"], Cell[57789, 2213, 3470, 90, 396, "Text"], Cell[61262, 2305, 84, 1, 24, "Input"], Cell[61349, 2308, 69, 1, 24, "Input"], Cell[61421, 2311, 71, 1, 24, "Input"], Cell[61495, 2314, 57, 1, 24, "Input"], Cell[61555, 2317, 3140, 127, 104, "Text"], Cell[CellGroupData[{ Cell[64720, 2448, 136, 3, 40, "Subsubsection", PageBreakAbove->False], Cell[64859, 2453, 157, 4, 30, "Text"], Cell[65019, 2459, 177, 8, 24, "Output"], Cell[65199, 2469, 907, 34, 84, "Text"], Cell[66109, 2505, 311, 13, 31, "Output"], Cell[66423, 2520, 579, 18, 78, "Text"], Cell[67005, 2540, 650, 27, 50, "Output"], Cell[67658, 2569, 112, 2, 30, "Text"], Cell[67773, 2573, 650, 27, 50, "Output"], Cell[68426, 2602, 759, 28, 52, "Text"], Cell[69188, 2632, 180, 9, 24, "Input"], Cell[69371, 2643, 624, 21, 78, "Text"], Cell[69998, 2666, 650, 27, 50, "Output"], Cell[70651, 2695, 413, 16, 36, "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[71113, 2717, 136, 3, 27, "Subsection", PageBreakAbove->True], Cell[71252, 2722, 814, 34, 62, "Text"], Cell[72069, 2758, 71, 1, 24, "Input"], Cell[72143, 2761, 60, 1, 24, "Input"], Cell[72206, 2764, 60, 1, 24, "Input"], Cell[72269, 2767, 3257, 116, 340, "Text"], Cell[75529, 2885, 67, 1, 24, "Input"], Cell[75599, 2888, 793, 22, 110, "Text"], Cell[76395, 2912, 80, 1, 24, "Input"], Cell[76478, 2915, 57, 1, 24, "Input"], Cell[76538, 2918, 57, 1, 24, "Input"], Cell[76598, 2921, 580, 12, 94, "Text"], Cell[77181, 2935, 60, 1, 24, "Input"], Cell[77244, 2938, 62, 1, 24, "Input"], Cell[77309, 2941, 2836, 87, 328, "Text"], Cell[80148, 3030, 94, 1, 24, "Input"], Cell[80245, 3033, 58, 1, 24, "Input"], Cell[80306, 3036, 122, 3, 36, "Input"], Cell[80431, 3041, 58, 1, 24, "Input"], Cell[80492, 3044, 374, 12, 62, "Text"], Cell[80869, 3058, 7127, 265, 286, "Text"], Cell[87999, 3325, 74, 1, 24, "Input"], Cell[88076, 3328, 74, 1, 24, "Input"], Cell[88153, 3331, 5805, 245, 310, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[93995, 3581, 142, 3, 27, "Subsection", PageBreakAbove->True], Cell[94140, 3586, 1903, 65, 172, "Text"], Cell[96046, 3653, 70, 1, 24, "Input"], Cell[96119, 3656, 82, 1, 24, "Input"], Cell[96204, 3659, 75, 1, 24, "Input"], Cell[96282, 3662, 703, 24, 71, "Text"], Cell[96988, 3688, 1858, 64, 85, "Text"], Cell[98849, 3754, 723, 25, 87, "Text"], Cell[99575, 3781, 98, 1, 24, "Input"], Cell[99676, 3784, 98, 1, 24, "Input"], Cell[99777, 3787, 81, 1, 24, "Input"], Cell[99861, 3790, 86, 1, 24, "Input"], Cell[99950, 3793, 92, 1, 24, "Input"], Cell[100045, 3796, 6497, 236, 239, "Text"], Cell[106545, 4034, 101, 1, 24, "Input"], Cell[106649, 4037, 57, 1, 24, "Input"], Cell[106709, 4040, 62, 1, 24, "Input"], Cell[106774, 4043, 62, 1, 24, "Input"], Cell[106839, 4046, 70, 1, 24, "Input"], Cell[106912, 4049, 90, 1, 24, "Input"], Cell[107005, 4052, 75, 1, 24, "Input"], Cell[107083, 4055, 4451, 166, 166, "Text"], Cell[111537, 4223, 83, 1, 24, "Input"], Cell[111623, 4226, 140, 3, 36, "Input"], Cell[111766, 4231, 91, 1, 24, "Input"], Cell[111860, 4234, 1055, 41, 77, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[112952, 4280, 195, 5, 42, "Subsection", PageBreakAbove->True], Cell[CellGroupData[{ Cell[113172, 4289, 137, 3, 40, "Subsubsection", PageBreakAbove->False], Cell[113312, 4294, 958, 36, 94, "Text"], Cell[114273, 4332, 289, 8, 36, "Special1"], Cell[114565, 4342, 6993, 261, 246, "Text"], Cell[121561, 4605, 882, 34, 36, "Text"], Cell[122446, 4641, 86, 2, 30, "Text"], Cell[122535, 4645, 953, 39, 39, "Text"], Cell[123491, 4686, 3241, 132, 64, "Text"], Cell[126735, 4820, 1737, 76, 96, "Text"], Cell[128475, 4898, 84, 1, 36, "Input"], Cell[128562, 4901, 82, 1, 24, "Input"], Cell[128647, 4904, 95, 1, 24, "Input"], Cell[128745, 4907, 127, 3, 36, "Input"], Cell[128875, 4912, 68, 1, 24, "Input"], Cell[128946, 4915, 109, 2, 36, "Input"] }, Closed]], Cell[CellGroupData[{ Cell[129092, 4922, 172, 5, 26, "Subsubsection", PageBreakAbove->False], Cell[129267, 4929, 5693, 220, 448, "Text"], Cell[134963, 5151, 73, 1, 24, "Input"], Cell[135039, 5154, 75, 1, 24, "Input"], Cell[135117, 5157, 71, 1, 24, "Input"], Cell[135191, 5160, 10932, 407, 528, "Text"], Cell[146126, 5569, 113, 2, 36, "Input"], Cell[146242, 5573, 143, 3, 36, "Input"], Cell[146388, 5578, 69, 1, 24, "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[146506, 5585, 184, 5, 27, "Subsection", PageBreakAbove->True], Cell[CellGroupData[{ Cell[146715, 5594, 135, 3, 40, "Subsubsection", PageBreakAbove->False], Cell[146853, 5599, 1908, 75, 170, "Text"], Cell[148764, 5676, 84, 1, 36, "Input"], Cell[148851, 5679, 346, 13, 36, "Special1"], Cell[149200, 5694, 7280, 267, 296, "Text"], Cell[156483, 5963, 1220, 45, 39, "Text"], Cell[157706, 6010, 5156, 196, 137, "Text"], Cell[162865, 6208, 104, 1, 48, "Input"], Cell[162972, 6211, 88, 1, 24, "Input"], Cell[163063, 6214, 125, 2, 30, "Text"], Cell[163191, 6218, 107, 2, 24, "Input"], Cell[163301, 6222, 127, 2, 30, "Text"], Cell[163431, 6226, 79, 1, 24, "Input"], Cell[163513, 6229, 92, 1, 24, "Input"], Cell[163608, 6232, 122, 3, 36, "Input"], Cell[163733, 6237, 65, 1, 24, "Input"], Cell[163801, 6240, 58, 1, 24, "Input"], Cell[163862, 6243, 107, 2, 24, "Input"], Cell[163972, 6247, 73, 1, 24, "Input"], Cell[164048, 6250, 163, 4, 22, "Special1"], Cell[164214, 6256, 3282, 104, 278, "Text"], Cell[167499, 6362, 83, 1, 36, "Input"], Cell[167585, 6365, 81, 1, 24, "Input"], Cell[167669, 6368, 94, 1, 24, "Input"], Cell[167766, 6371, 122, 3, 36, "Input"], Cell[167891, 6376, 65, 1, 24, "Input"], Cell[167959, 6379, 1113, 46, 81, "Text"], Cell[169075, 6427, 69, 1, 24, "Input"], Cell[169147, 6430, 73, 1, 24, "Input"], Cell[169223, 6433, 131, 3, 36, "Input"], Cell[169357, 6438, 98, 1, 24, "Input"], Cell[169458, 6441, 78, 1, 24, "Input"], Cell[169539, 6444, 367, 9, 50, "Special1"], Cell[169909, 6455, 1393, 40, 194, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[171339, 6500, 173, 5, 26, "Subsubsection", PageBreakAbove->Automatic], Cell[171515, 6507, 12398, 473, 628, "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[183962, 6986, 131, 3, 27, "Subsection", PageBreakAbove->True], Cell[184096, 6991, 1281, 41, 160, "Text"], Cell[185380, 7034, 490, 13, 69, "Special1"], Cell[185873, 7049, 7842, 276, 612, "Text"], Cell[193718, 7327, 122, 3, 36, "Input"], Cell[193843, 7332, 103, 1, 24, "Input"], Cell[193949, 7335, 4203, 187, 327, "Text"], Cell[198155, 7524, 84, 1, 24, "Input"], Cell[198242, 7527, 59, 1, 24, "Input"], Cell[198304, 7530, 71, 1, 24, "Input"], Cell[198378, 7533, 76, 1, 24, "Input"], Cell[198457, 7536, 97, 1, 24, "Input"], Cell[198557, 7539, 70, 1, 24, "Input"], Cell[198630, 7542, 6710, 265, 320, "Text"], Cell[205343, 7809, 69, 1, 24, "Input"], Cell[205415, 7812, 71, 1, 24, "Input"], Cell[205489, 7815, 95, 1, 24, "Input"], Cell[205587, 7818, 600, 23, 67, "Text"], Cell[206190, 7843, 108, 2, 24, "Input"], Cell[206301, 7847, 93, 1, 24, "Input"], Cell[206397, 7850, 10598, 409, 503, "Text"], Cell[216998, 8261, 70, 1, 24, "Input"], Cell[217071, 8264, 84, 1, 24, "Input"], Cell[217158, 8267, 102, 1, 24, "Input"], Cell[217263, 8270, 3362, 133, 238, "Text"], Cell[220628, 8405, 89, 1, 36, "Input"], Cell[220720, 8408, 95, 1, 24, "Input"] }, Closed]] }, Open ]] } ] *) (*********************************************************************** End of Mathematica Notebook file. ***********************************************************************)