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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[ 29373, 808]*) (*NotebookOutlinePosition[ 62640, 1915]*) (* CellTagsIndexPosition[ 62596, 1911]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Solutions to Exam 1", "Title", TextAlignment->Center, TextJustification->0], Cell["Physics 604", "Subtitle", TextAlignment->Center, TextJustification->0], Cell["Oct. 23, 2003", "Subsubtitle", TextAlignment->Center, TextJustification->0], Cell["\<\ This CLOSED BOOK exam. Three problems are to be completed in 75 minutes and \ carry equal weight. You must explain your reasoning to receive full \ credit.\ \>", "Text"], Cell[CellGroupData[{ Cell["Problem 1", "ExerciseTitle"], Cell[TextData[{ "Suppose that ", Cell[BoxData[ \(TraditionalForm\`f\_2[z]\)]], " is analytic at ", Cell[BoxData[ \(TraditionalForm\`z\)]], " and that ", Cell[BoxData[ \(TraditionalForm\`f\_1[w]\)]], " is analytic at ", Cell[BoxData[ \(TraditionalForm\`w = f\_2[z]\)]], ". Show that ", Cell[BoxData[ \(TraditionalForm\`f[z] = f\_1[\ f\_2[z]\ ]\)]], " is analytic at ", Cell[BoxData[ \(TraditionalForm\`z\)]], "." }], "ExerciseText"], Cell[CellGroupData[{ Cell["solution", "Subsubsection"], Cell["Let", "Text"], Cell[BoxData[{ FormBox[ FormBox[\(z = x + \[ImaginaryI]\ y\), "TraditionalForm"], TextForm], "\[IndentingNewLine]", FormBox[\(w = \(f\_2[z] = u[x, y] + \[ImaginaryI]\ v[x, y]\)\), TextForm], "\[IndentingNewLine]", FormBox[\(f\_1[w]\ = \ p[u, v]\ + \ \[ImaginaryI]\ q[u, v]\), TextForm]}], "DisplayFormula"], Cell["such that", "Text"], Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(f[z] = \ \(f\_1[w]\ = \ f\_1[\ f\_2[z]\ ]\)\), "TraditionalForm"], " ", "=", " ", \(p[u[x, y], v[x, y]]\ + \ \[ImaginaryI]\ q[u[x, y], v[x, y]]\)}]}], TextForm]], "DisplayFormula"], Cell[TextData[{ "Analyticity of ", Cell[BoxData[ \(TraditionalForm\`f\_2[z]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`f\_1[w]\)]], " requires that continuous first partial derivatives of the component \ functions exist at ", Cell[BoxData[ \(TraditionalForm\`z\)]], " for ", Cell[BoxData[ \(TraditionalForm\`\((u, v)\)\)]], " and at ", Cell[BoxData[ \(TraditionalForm\`w\)]], " for ", Cell[BoxData[ \(TraditionalForm\`\((p, q)\)\)]], " and satisfy the Cauchy-Riemann equations:" }], "Text"], Cell[BoxData[{ \(TextForm\`\[PartialD]u\/\[PartialD]x = \[PartialD]v\/\[PartialD]y\ , \ \ \ \ \ \ \[PartialD]u\/\[PartialD]y = \(-\(\[PartialD]v\/\[PartialD]x\)\)\), "\ \[IndentingNewLine]", \(TextForm\`\[PartialD]p\/\[PartialD]u = \[PartialD]q\/\[PartialD]v\ , \ \ \ \ \ \ \[PartialD]p\/\[PartialD]v = \(-\(\[PartialD]q\/\[PartialD]u\)\)\)}], \ "DisplayFormula"], Cell["\<\ Use of the chain rule for these real-valued functions of real variables\ \>", "Text"], Cell[BoxData[{ \(TextForm\`\[PartialD]p\/\[PartialD]x = \(\(\[PartialD]p\/\[PartialD]u\) \ \[PartialD]u\/\[PartialD]x\ + \ \(\[PartialD]p\/\[PartialD]v\) \ \[PartialD]v\/\[PartialD]x\ \ = \ \(\(\[PartialD]q\/\[PartialD]v\) \ \[PartialD]v\/\[PartialD]y\ + \ \(\[PartialD]q\/\[PartialD]u\) \ \[PartialD]u\/\[PartialD]y\ = \ \[PartialD]q\/\[PartialD]y\)\)\), "\ \[IndentingNewLine]", \(TextForm\`\[PartialD]p\/\[PartialD]y = \(\(\[PartialD]p\/\[PartialD]u\) \ \[PartialD]u\/\[PartialD]y\ + \ \(\[PartialD]p\/\[PartialD]v\) \ \[PartialD]v\/\[PartialD]y\ \ = \ \(\(-\(\[PartialD]q\/\[PartialD]v\)\) \ \[PartialD]v\/\[PartialD]x\ - \ \(\[PartialD]q\/\[PartialD]u\) \ \[PartialD]u\/\[PartialD]x\ = \ \(-\(\[PartialD]q\/\[PartialD]x\)\)\)\)\)}], \ "DisplayFormula"], Cell[TextData[{ "then results in the Cauchy-Riemann equations for ", Cell[BoxData[ \(TraditionalForm\`f[z]\)]], " and demonstrates the analyticity of composition of analytic functions." }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 2", "ExerciseTitle"], Cell["Use contour integration to evaluate", "ExerciseText"], Cell[BoxData[ \(TextForm\`\[Integral]\_0\%\[Pi]\( Sin[\[Theta]]\^\(2 n\)\) \[DifferentialD]\[Theta]\)], "DisplayFormula"], Cell[TextData[{ "for nonnegative integer ", Cell[BoxData[ \(TraditionalForm\`n\)]], "." }], "ExerciseText"], Cell[CellGroupData[{ Cell["solution", "Subsubsection"], Cell["Using symmetry and", "Text"], Cell[BoxData[ \(z = \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\ , \ \ \ \[DifferentialD]\[Theta] = \[DifferentialD]z\/\(\[ImaginaryI]\ z\), \ \ Sin[\ \[Theta]] = \(z - z\^\(-1\)\)\/\(2 \[ImaginaryI]\)\)], "DisplayFormula", FontFamily->"Times New Roman"], Cell["we obtain", "Text"], Cell[BoxData[ \(I\ = \ \(\[Integral]\_0\%\[Pi]\( Sin[\[Theta]]\^\(2 n\)\) \[DifferentialD]\[Theta]\ \ = \ \(\(1\/2\) \(\ \[Integral]\_0\%\(2 \[Pi]\)\(Sin[\[Theta]]\^\(2 n\)\) \[DifferentialD]\[Theta]\)\ = \ \(\(1\/\(2 \ \[ImaginaryI]\)\) \(\[ContourIntegral]\_C\(\((\(z - z\^\(-1\)\)\/\(2 \ \[ImaginaryI]\))\)\^\(2 n\)\) \[DifferentialD]z\/z\)\ = \ \ \(\((-)\)\^n\/\(\(2\^\(2 n + 1\)\) \[ImaginaryI]\)\) \ \(\[ContourIntegral]\_C\(\((z\^2 - 1)\)\^\(2 n\)\/z\^\(2 n + 1\)\) \ \[DifferentialD]z\)\)\)\)\)], "DisplayFormula", FontFamily->"Times New Roman"], Cell[TextData[{ "where ", StyleBox["C", FontSlant->"Italic"], " is the unit circle. The integrand has a pole at the origin of order ", Cell[BoxData[ \(TraditionalForm\`2 n + 1\)]], ". Using the binomial theorem" }], "Text"], Cell[BoxData[ RowBox[{\(\((z\^2 - 1)\)\^\(2 n\)\), " ", "=", " ", RowBox[{\(\[Sum]\+\(m = 0\)\%\(2 n\)\), RowBox[{ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(2 n\)}, {"m"} }], "\[NegativeThinSpace]", ")"}], \(\((-)\)\^m\), \(\((z\^2)\)\^\(2 n - m\)\)}]}]}]], "DisplayFormula", FontFamily->"Times New Roman"], Cell[TextData[{ "we identify the residue as the coefficient of ", Cell[BoxData[ \(TraditionalForm\`z\^\(2 n\)\)]], ", which is the term with ", Cell[BoxData[ \(TraditionalForm\`m = n\)]], ", and obtain" }], "Text"], Cell[BoxData[ RowBox[{"I", " ", "=", " ", RowBox[{ RowBox[{ "2", "\[Pi]", " ", "\[ImaginaryI]", " ", \(\((-)\)\^n\/\(\(2\^\(2 n + 1\)\) \[ImaginaryI]\)\), RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(2 n\)}, {"n"} }], "\[NegativeThinSpace]", ")"}], \(\((-)\)\^n\)}], " ", "=", " ", \(\[Pi]\/4\^n\ \(\((2 n)\)!\)\/\((\(n!\))\)\^2\)}]}]], \ "DisplayFormula", FontFamily->"Times New Roman"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 3", "ExerciseTitle"], Cell["Evaluate the principal-value integral", "ExerciseText"], Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalP] \(\[Integral]\_0\%\[Infinity]\( \ x\^\(1/4\)\/\(x\^2 - x - 2\)\) \[DifferentialD]x\)\)], "DisplayFormula"], Cell["\<\ Sketch your contour and justify clearly any portions that are neglected.\ \>", "ExerciseText"], Cell[CellGroupData[{ Cell["solution", "Subsubsection"], Cell["Let", "Text"], Cell[BoxData[ \(f[z]\ = \ z\^\(1/4\)\/\(\((z + 1)\) \((z - 2)\)\)\ , \ \ z\^\(1/4\) = \ \(\ \[LeftBracketingBar]z\[RightBracketingBar]\^\(1/4\)\) Exp[\(\[ImaginaryI]\/4\) Arg[z]]\)], "DisplayFormula", FontFamily->"Times New Roman"], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`0 \[LessEqual] Arg[z] < 2 \[Pi]\)]], " is cut along the positive real axis. There is a simple pole at ", Cell[BoxData[ \(TraditionalForm\`z = \(-1\)\)]], " on the negative real axis and a pole at ", Cell[BoxData[ \(TraditionalForm\`z = 2\)]], " on the cut. Applying the residue theorem to the contour sketched below \ we find" }], "Text"], Cell[BoxData[ \(TextForm\`\[ContourIntegral]\_C\( z\^\(1/4\)\/\(\((z + 1)\) \((z - 2)\)\)\) \[DifferentialD]z\ = \ 2 \[Pi]\ \[ImaginaryI]\ \[ExponentialE]\^\(\[ImaginaryI]\ \ \[Pi]/4\)\/\(-3\)\)], "DisplayFormula"], Cell[TextData[{ "using the residue of the pole at ", Cell[BoxData[ \(TraditionalForm\`z = \(-1\)\)]], ". Note that" }], "Text"], Cell[BoxData[ \(TextForm\`z = R\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\ \ \ \[DoubleLongRightArrow]\ \ \[LeftBracketingBar]f[ z] \[DifferentialD]z\/\[DifferentialD]\[Theta]\ \[RightBracketingBar]\ \[TildeEqual] \ R\^\(\(-3\)/4\) \[Rule] 0\)], "DisplayFormula"], Cell[TextData[{ "is negligible on a great circle with ", Cell[BoxData[ \(TraditionalForm\`R \[Rule] \[Infinity]\)]], " and" }], "Text"], Cell[BoxData[ \(TextForm\`z = \[CurlyEpsilon]\ \[ExponentialE]\^\(\[ImaginaryI]\ \ \[Theta]\)\ \ \[DoubleLongRightArrow]\ \ \[LeftBracketingBar]f[ z] \[DifferentialD]z\/\[DifferentialD]\[Theta]\ \[RightBracketingBar]\ \[TildeTilde] \ \(1\/2\) \[CurlyEpsilon]\^\(5/4\) \ \[Rule] 0\)], "DisplayFormula"], Cell["\<\ is negligible on an infinitesimal circle around the origin. Therefore, we \ can divide the contours along the real axis into principal value and pole \ contributions\ \>", "Text"], Cell[BoxData[ FormBox[ RowBox[{\(\[ContourIntegral]\_C\( z\^\(1/4\)\/\(\((z + 1)\) \((z - 2)\)\)\) \[DifferentialD]z\), " ", "=", " ", RowBox[{ RowBox[{\((1 - \[ImaginaryI])\), " ", FormBox[\(\[ScriptCapitalP] \(\[Integral]\_0\%\[Infinity]\( \ x\^\(1/4\)\/\(x\^2 - x - 2\)\) \[DifferentialD]x\)\), "TraditionalForm"]}], " ", "-", " ", \(\[ImaginaryI]\ \[Pi]\ \(2\^\(1/4\)\/3\) \((1 + \ \[ImaginaryI])\)\)}]}], TextForm]], "DisplayFormula"], Cell[TextData[{ "where the phases on opposite sides differ by ", Cell[BoxData[ \(TraditionalForm\`2 \[Pi]\)]], ". 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below define styles useful for making hypertext ButtonBoxes. 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If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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