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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[ 620961, 15442]*) (*NotebookOutlinePosition[ 621693, 15467]*) (* CellTagsIndexPosition[ 621649, 15463]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Half-String Coordinate Modes Derivation", "Subtitle"], "\nBen Sauerwine\n\nVersion 6: June 22, 2004\nThe newest version of this \ derivation can be downloaded from http://www.snakebyte.biz/string/day1.html\n\ \n", StyleBox["References Used", "Subsection"], "\n\nHalf-String Oscillator Approach to String Theory\nJ. Bordes, Chan \ Hong-Mo, Lukas Nellen, Tsou Sheung Tsun\n\nEric W. Weisstein. \"Fourier \ Series.\" \nFrom ", StyleBox["MathWorld", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}], "--A Wolfram Web Resource. ", StyleBox["http://mathworld.wolfram.com/FourierSeries.html", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}], " \n\nOperator Formulation of Interacting String Field Theory (I)\nDavid J. \ Gross, Antal Jevicki\n\n", StyleBox["Definitions and Boundaries", "Subsection"], "\n\nGiven a String defined from 0 to \[Pi] by the formula\n\n[1a]\nx(\ \[Sigma]) =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x\_0\)\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] x\_n\)]], "cos(n \[Sigma])\n[1b]\n", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " = ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "\[ImaginaryI] (", Cell[BoxData[ \(TraditionalForm\`a\_0\)]], "- ", Cell[BoxData[ FormBox[ SuperscriptBox[ FormBox[\(a\_0\), "TraditionalForm"], "+"], TraditionalForm]]], ")\n", Cell[BoxData[ \(TraditionalForm\`p\_0\)]], " = (", Cell[BoxData[ \(TraditionalForm\`a\_\(\(0\)\(\ \)\) + \ \(a\_0\^+\)\)]], ")\n[1c]\n", Cell[BoxData[ \(TraditionalForm\`x\_n\)]], "= ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ \(TraditionalForm\`\[ImaginaryI]\ \@\(2\/n\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`a\_n\)]], "- ", Cell[BoxData[ FormBox[ SuperscriptBox[ FormBox[\(a\_n\), "TraditionalForm"], "+"], TraditionalForm]]], ")\n", Cell[BoxData[ \(TraditionalForm\`p\_n\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\@\(n\/2\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`a\_\(\(n\)\(\ \)\) + \ \(a\_n\^+\)\)]], ")\n\n\nI wish to discuss it in terms of its left and right side, which I \ will call ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], " and", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\[Chi]\_R\)\)\)]], ", respectively. The left and right sides will be discussed in terms of \ displacement from a central reference point, and so I choose the central \ reference point to be ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " define\n\n[2a]\n", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_L\)(\[Sigma])\)]], " = x(\[Sigma]) - x(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ")\n[2b]\n", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_R\)(\[Sigma])\)]], " = x(\[Pi] - \[Sigma]) - x(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ")\n\nOver the interval \[Sigma] \[Element] [0, ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], "]\n\nNote that in these definitions, the meaning of \[Sigma] is distance \ inwards from the outer limit of the string and that I choose to include the \ border point, ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ", in both substrings.\n\nBy examination, I see from [1] that x'(0) = x'(\ \[Pi]) = 0. Therefore, from definitions [2a] and [2b], ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "'(0) = ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_R\)]], "'(0) = 0. Also, since the two half-strings share the common midpoint, ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ") = ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_R\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ") = 0 (this is obvious from the definitions).\n\n", StyleBox["Fourier Series\n\n", "Subsection"], "Keeping in mind that the goal is to find an expression for ", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_L\)(\[Sigma])\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_R\)(\[Sigma])\)]], " in terms of Fourier series and then connecting these expressions to that \ of the original x(\[Sigma]), I want solutions like\n\n[3a]\n", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_L\)(\[Sigma])\)]], "= ", Cell[BoxData[ \(TraditionalForm\`A\_\(L\ 0\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] A\_\(L\ n\)\)]], "Cos(n \[Sigma]) + ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] B\_\(L\ n\)\)]], "Sin(n \[Sigma])\n[3b]\n", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_R\)(\[Sigma])\)]], "= ", Cell[BoxData[ \(TraditionalForm\`A\_\(R\ 0\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] A\_\(R\ n\)\)]], "Cos(n \[Sigma]) + ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] B\_\(R\ n\)\)]], "Sin(n \[Sigma])\n\nHowever, noting the boundary conditions, I see that \ x'(0) = x'(\[Pi]) = 0 and so the B coefficients are all zero. I now have\n\n\ [4a]\n", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_L\)(\[Sigma])\)]], "= ", Cell[BoxData[ \(TraditionalForm\`A\_\(L\ 0\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] A\_\(L\ n\)\)]], "Cos(n \[Sigma]) \n[4b]\n", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_R\)]], "(\[Sigma]) = ", Cell[BoxData[ \(TraditionalForm\`A\_\(R\ 0\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] A\_\(R\ n\)\)]], "Cos(n \[Sigma]) \n\nFurther, the periodicity of the Cosine function plus \ the constraint that ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ") = ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_R\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ") = 0 implies that these half-wave functions must be odd about ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[Pi]\/2\) . \ \ Therefore\)\(,\)\)\)]], "the even indices A are all zero and these half-waves can be rewritten\n\n\ [5a]\n", Cell[BoxData[ \(TraditionalForm\`\(\[Chi]\_L\)(\[Sigma])\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity]\( 1\/2\) A\_\(L\ n\)\)]], "(1 - ", Cell[BoxData[ \(TraditionalForm\`\((\(-1\))\)\^n\)]], ")Cos(n \[Sigma]) \n[5b]\n", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_R\)]], "(\[Sigma]) = ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity]\( 1\/2\) A\_\(R\ n\)\)]], "(1 - ", Cell[BoxData[ \(TraditionalForm\`\((\(-1\))\)\^n\)]], ")Cos(n \[Sigma]) \n\nWith the new terms being added to eliminate even \ values of n without having to redefine the indices on A or modify the \ Cosines.\nBy the standard procedure for finding the coefficients of a Fourier \ series, then, (Let A subsequently be denoted by \[Chi])\n\n[6a]\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`1\/\[Pi]\)]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(-\[Pi]\)\%\[Pi]\(\( \[Chi]\_L\)(\ \[Sigma])\)\ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], "\nBut letting even-indices equal zero per [5]\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\(2 \[Pi]\)\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(-\[Pi]\)\%\[Pi]\(\( \[Chi]\_L\)(\ \[Sigma])\)\ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], "\nHowever, this requires ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\[Chi]\_L\)(\[Sigma])\)\(\ \)\)\)]], "to be extended. I will choose an odd extension about ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " and an even extension about 0, so that\n", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "(\[Sigma]) = ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "(\[Pi] - \[Sigma]) and ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "(-\[Sigma]) = ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], "(\[Sigma])\nAnd now\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\(2 \[Pi]\)\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(-\[Pi]\)\%0\(\( \[Chi]\_L\)(\[Sigma])\ \)\ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Pi]\(\( \[Chi]\_L\)(\[Sigma])\)\ \ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\(2 \[Pi]\)\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], "-", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(-\[Pi]\)\(\(\[Chi]\_L\)(\[Sigma])\)\ \ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Pi]\(\( \[Chi]\_L\)(\[Sigma])\)\ \ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\(2 \[Pi]\)\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Pi]\(\( \[Chi]\_L\)(\(-\[Sigma]\))\ \)\ \(Cos(\(-n\)\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Pi]\(\( \[Chi]\_L\)(\[Sigma])\)\ \ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\(2 \[Pi]\)\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], "2", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Pi]\(\( \[Chi]\_L\)(\[Sigma])\)\ \ \(Cos(n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\(\(\[Chi]\_L\)(\[Sigma])\ \)\ \(Cos( n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\ + \ \[Integral]\_\(\ \[Pi]\/2\)\%\[Pi]\(\( \[Chi]\_L\)(\[Sigma])\)\ \(Cos( n\ \[Sigma])\)\ \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\(\(\[Chi]\_L\)(\[Sigma])\ \)\ \(Cos( n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\ - \ \[Integral]\_0\%\ \(\[Pi]\/2\)\(\(\[Chi]\_L\)(\[Pi]\ - \[Sigma]\ )\)\ \(Cos( n\ \((\[Pi]\ - \[Sigma])\))\)\ \ \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(1\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", "("}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\(\(\[Chi]\_L\)(\[Sigma])\ \)\ \(Cos( n\ \[Sigma])\)\ \[DifferentialD]\[Sigma]\ - \ \[Integral]\_0\%\ \(\[Pi]\/2\)\(\(\[Chi]\_L\)(\[Sigma]\ )\)\ \(Cos( n\ \((\[Pi]\ - \[Sigma])\))\)\ \ \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " "}], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\(\(\[Chi]\_L\)(\[Sigma])\ \)\ \(Cos(n\ \[Sigma])\)\ \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)\)]], "\n[6b]\nAnd following similar arguments,\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(R\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " "}], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\(\(\[Chi]\_R\)(\[Sigma])\ \)\ \(Cos(n\ \[Sigma])\)\ \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)\)]], "\n\nSubstituting [2] into [6], \n\n[7a]\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " "}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ RowBox[{"(", RowBox[{\(x(\[Sigma])\), "-", RowBox[{"x", "(", FormBox[\(\[Pi]\/2\), "TraditionalForm"], ")"}]}], ")"}], " ", \(Cos(n\ \[Sigma])\), " ", \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}]}], TraditionalForm]]], "\n[7b]\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(R\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " "}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ RowBox[{"(", RowBox[{\(x(\[Pi] - \[Sigma])\), "-", RowBox[{"x", "(", FormBox[\(\[Pi]\/2\), "TraditionalForm"], ")"}]}], ")"}], " ", \(Cos(n\ \[Sigma])\), " ", \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}]}], TraditionalForm]]], "\n\nAnd [1a] into [7],\n\n[8a]\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " "}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ RowBox[{"(", " ", RowBox[{\(x\_0\), "+", RowBox[{ FormBox[\(\@2\), "TraditionalForm"], FormBox[\(\[Sum]\+\(m = 1\)\%\[Infinity] x\_m\), "TraditionalForm"], \(Cos(m\ \[Sigma])\)}], "-", RowBox[{"(", " ", RowBox[{\(x\_0\), "+", RowBox[{ FormBox[\(\@2\), "TraditionalForm"], FormBox[\(\[Sum]\+\(m = 1\)\%\[Infinity] x\_m\), "TraditionalForm"], \(Cos(m\ \[Pi]\/2)\)}]}], ")"}]}], " ", ")"}], \(Cos(n\ \[Sigma])\), " ", \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}]}], TraditionalForm]]], "\nRemember: n is odd\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{ FractionBox[ RowBox[{"2", " ", FormBox["", "TraditionalForm"]}], "\[Pi]"], RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " ", \(\[Sum]\+\(m = 1\)\%\[Infinity]\)}], TraditionalForm]]], " (", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ FormBox[\(x\_m\), "TraditionalForm"], \(Cos(m\ \[Sigma])\), \(Cos( n\ \[Sigma])\)}]}], "-", RowBox[{ FormBox[\(x\_m\), "TraditionalForm"], \(Cos(m\ \[Pi]\/2)\), \(Cos( n\ \[Sigma])\), \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}]}], TraditionalForm]]], ")\n\n(the ", Cell[BoxData[ FormBox[ FormBox[\(\@2\), "TraditionalForm"], TraditionalForm]]], " was incorporated into the definition of \[Chi])\n\nNow, looking at the \ integral, \n", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ FormBox[\(x\_m\), "TraditionalForm"], \(Cos(m\ \[Sigma])\), \(Cos( n\ \[Sigma])\)}]}], TraditionalForm]]], " \[DifferentialD]\[Sigma] \n\nis ", Cell[BoxData[ \(\(n\ \[Pi] + Sin \((n\ \[Pi])\)\)\/\(4\ n\)\)]], "if m = n, 0 if m is odd and m \[NotEqual] n, but is in general\n\n", Cell[BoxData[ \(\(m\ Cos \((\(n\ \[Pi]\)\/2)\)\ Sin \((\(m\ \[Pi]\)\/2)\) - n\ Cos \ \((\(m\ \[Pi]\)\/2)\)\ Sin \((\(n\ \[Pi]\)\/2)\)\)\/\(m\^2 - n\^2\)\)]], "\nAnd the integral \n", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\)]], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"-", FormBox[\(x\_m\), "TraditionalForm"]}], \(Cos(m\ \[Pi]\/2)\), \(Cos( n\ \[Sigma])\), \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}], TraditionalForm]]], "\nis zero if m is odd and ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(\[Pi]\/2\)\)]], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"-", SuperscriptBox[ RowBox[{ FormBox[\(x\_m\), "TraditionalForm"], "(", \(-1\), ")"}], \(m\/2\)]}], " ", \(Cos(n\ \[Sigma])\), \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}], TraditionalForm]]], " if m is even, or\n", Cell[BoxData[ \(\(-\(\(\((1\ + \ \((\(-1\))\)\^m)\) \((\(-1\))\)\^\(m\/2\)\ Sin \ \((\(n\ \[Pi]\)\/2)\)\ x\_m\)\/\(2 n\)\)\)\)]], "\nSo now \n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], \((\ \ \ \[Sum]\+\(m = 1\)\%\[Infinity] x\_m\)}], TraditionalForm]]], Cell[BoxData[ \(\(m\ Cos \((\(n\ \[Pi]\)\/2)\)\ Sin \((\(m\ \[Pi]\)\/2)\) - n\ Cos \ \((\(m\ \[Pi]\)\/2)\)\ Sin \((\(n\ \[Pi]\)\/2)\)\)\/\(m\^2 - n\^2\)\)]], " ", Cell[BoxData[ \(\(-\(\[Sum]\+\(m = 1\)\%\[Infinity]\(\((1\ + \ \((\(-1\))\)\^m)\) \ \((\(-1\))\)\^\(m\/2\)\ Sin \((\(n\ \[Pi]\)\/2)\)\ x\_m\)\/\(2 n\)\)\)\)]], ")\n\n\nSimplifying the sums individually,\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity] \ x\_m\)\)\)]], Cell[BoxData[ \(\(m\ Cos \((\(n\ \[Pi]\)\/2)\)\ Sin \((\(m\ \[Pi]\)\/2)\) - n\ Cos \ \((\(m\ \[Pi]\)\/2)\)\ Sin \((\(n\ \[Pi]\)\/2)\)\)\/\(m\^2 - n\^2\)\)]], "\n\nBecomes\n", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_m\)]], Cell[BoxData[ \(\(\(-n\)\ Cos \((\(m\ \[Pi]\)\/2)\)\ \((\(-1\))\)\^\(\(n - 1\)\/2\)\)\ \/\(m\^2 - n\^2\)\)]], "\n\nSince n is odd, except when m = n, when it becomes ", Cell[BoxData[ \(\(\((n\ \[Pi] + Sin \((n\ \[Pi])\))\)\ x\_n\)\/\(4\ n\)\)]], "\n\nFor ", Cell[BoxData[ \(\(\(\ \)\(\[Pi]\ x\_n\)\)\/\(\(4\)\(\ \)\)\)]], "+", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(\[Sum]\+\(m = 1, \ m\ \[NotEqual] \ n\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_m\)]], Cell[BoxData[ \(\(\(-n\)\ \((1\ + \ \((\(-1\))\)\^m)\)\ \((\(-1\))\)\^\(\(\(m\)\(\ \ \)\)\/2\)\ \((\(-1\))\)\^\(\(n - 1\)\/2\)\)\/\(2 \((m\^2 - n\^2)\)\)\)]], "\n\nOr ", Cell[BoxData[ \(\(\(\ \)\(\[Pi]\ x\_n\)\)\/\(\(4\)\(\ \)\)\)]], "+", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\(\(-n\)\ \ \((\(-1\))\)\^m\ \((\(-1\))\)\^\(\(n - 1\)\/2\)\)\/\(4 m\ \^2 - n\^2\)\)]], "since odd m yield 0 except when n = m\n\nAnd simplifying the other sum, \n\ \n", Cell[BoxData[ \(\[Sum]\+\(m = 1\)\%\[Infinity]\(\((1\ + \ \((\(-1\))\)\^m)\) \((\(-1\ \))\)\^\(m\/2\)\ Sin \((\(n\ \[Pi]\)\/2)\)\ x\_m\)\/\(2 n\)\)]], "\nyields\n", Cell[BoxData[ \(\[Sum]\+\(m = 1\)\%\[Infinity]\(\((1\ + \ \((\(-1\))\)\^m)\) \((\(-1\ \))\)\^\(m\/2\)\ \((\(-1\))\)\^\(\^\(\(n\ - \ 1\)\/2\)\)\ x\_m\)\/\(2 \ n\)\)]], "\nor\n", Cell[BoxData[ \(\[Sum]\+\(m = 1\)\%\[Infinity]\(\((\(-1\))\)\^m\ \((\(-1\))\)\^\(\^\(\ \(n\ - \ 1\)\/2\)\)\ x\_\(2 m\)\)\/n\)]], "\n\nRecombining, \n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}]}], TraditionalForm]]], "(", Cell[BoxData[ \(\(\(\ \)\(\[Pi]\ x\_n\)\)\/\(\(4\)\(\ \)\)\)]], "+", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\(\(-n\)\ \ \((\(-1\))\)\^m\ \((\(-1\))\)\^\(\(n - 1\)\/2\)\)\/\(4 m\ \^2 - n\^2\)\)]], "-", Cell[BoxData[ \(\[Sum]\+\(m = 1\)\%\[Infinity]\(\((\(-1\))\)\^m\ \((\(-1\))\)\^\(\^\(\ \(n\ - \ 1\)\/2\)\)\ x\_\(2 m\)\)\/n\)]], ")\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{\(\(\(2\)\(\ \)\)\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}]}], TraditionalForm]]], "(", Cell[BoxData[ \(\(\(\ \)\(\[Pi]\ x\_n\)\)\/\(\(4\)\(\ \)\)\)]], "-", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\((\(-1\))\)\^\(\(m\)\(\ \)\)\ \((\(-1\))\)\^\(\(n - 1\)\/2\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`n\/\(4 m\^2 - n\^2\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`1\/n\)]], ")))\n\nOr,\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "= 0 \nif n is even\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(x\_n\), "TraditionalForm"]}], TraditionalForm]]], "-", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\(4\)\(\ \)\)\/\[Pi]\ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\((\(-1\))\)\^\(\(n\ \ + \ 2 m\ - \ 1\)\/2\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`n\/\(4 m\^2 - n\^2\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`1\/n\)]], ")\nif n is odd\n\n\n[8b]\nBy a similar argument, noting that Cos(\[Pi] - \ \[Sigma]) = - Cos(\[Sigma]), \n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(R\ n\)\)\(\ \)\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{\(2\/\[Pi]\), RowBox[{"(", RowBox[{"1", "-", FormBox[\(\((\(-1\))\)\^n\), "TraditionalForm"]}], ")"}], " "}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{\(\[Integral]\_0\%\(\[Pi]\/2\)\), RowBox[{ RowBox[{"(", " ", RowBox[{\(x\_0\), "+", RowBox[{ FormBox[\(\@2\), "TraditionalForm"], FormBox[\(\[Sum]\+\(m = 1\)\%\[Infinity] x\_m\), "TraditionalForm"], \(Cos( m\ \((\[Pi]\ - \[Sigma])\))\)}], "-", RowBox[{"(", " ", RowBox[{\(x\_0\), "+", RowBox[{ FormBox[\(\@2\), "TraditionalForm"], FormBox[\(\[Sum]\+\(m = 1\)\%\[Infinity] x\_m\), "TraditionalForm"], \(Cos(m\ \[Pi]\/2)\)}]}], ")"}]}], ")"}], " ", \(Cos(n\ \[Sigma])\), " ", \(\(\[DifferentialD]\[Sigma]\)\(\ \)\)}]}], TraditionalForm]]], "\n\nSimplifies to \n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(R\ n\)\)\(\ \)\)\)]], "= 0 \nif n is even\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(R\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(-\ x\_n\), "TraditionalForm"]}], TraditionalForm]]], "-", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\(4\)\(\ \)\)\/\[Pi]\ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\((\(-1\))\)\^\(\(n\ + \ 2 m\ - 1\)\/2\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`n\/\(4 m\^2 - n\^2\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`1\/n\)]], ")\nif n is odd\n\n\n[9]\n\nLet ", Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ m\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\((\(-1\))\)\^\(\(n\ + \ m\ + \ \ 1\)\/2\)\/\[Pi]\)]], "(", Cell[BoxData[ \(TraditionalForm\`1\/\(n\ + \ m\)\)]], "- ", Cell[BoxData[ \(TraditionalForm\`1\/\(n\ - \ m\)\)]], ")\n\nSo, for odd n,\n\n", Cell[BoxData[ \(TraditionalForm\`B\_\(2 n\ - \ 1, \ 2 m\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\((\(-1\))\)\^\(\(2 n\ \ - \ 1\ + \ 2 m\ + \ \ 1\)\/2\)\/\[Pi]\)]], "(", Cell[BoxData[ \(TraditionalForm\`1\/\(2 n\ - \ 1\ + \ 2 m\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`1\/\(2 n\ - \ 1\ - \ 2 m\)\)]], ")\n\nLet ", Cell[BoxData[ \(TraditionalForm\`B\_\(2 n\ - \ 1, \ 2 m\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\((\(-1\))\)\^\(n\ + \ m\)\/\[Pi]\)]], "(", Cell[BoxData[ \(TraditionalForm\`1\/\(2 n\ - \ 1\ + \ 2 m\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`1\/\(2 n\ - \ 1\ - \ 2 m\)\)]], ")\n\n[10a]\n\nAnd continuing for odd n,\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(x\_n\), "TraditionalForm"]}], TraditionalForm]]], "-", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\(4\)\(\ \)\)\/\[Pi]\ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\((\(-1\))\)\^\(\(n\ + \ 2 m\ - 1\)\/2\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`n\/\(4 m\^2 - n\^2\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`1\/n\)]], ")\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(x\_n\), "TraditionalForm"]}], TraditionalForm]]], "-", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\(4\)\(\ \)\)\/\[Pi]\ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\((\(-1\))\)\^\(\(n\ + \ 2 m\ - 1\)\/2\)\)]], Cell[BoxData[ \(TraditionalForm\`\(4 m\^2\)\/\(n\ \((4 m\^2\ - \ n\^2)\)\)\)]], "\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(x\_n\), "TraditionalForm"]}], TraditionalForm]]], "+", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\(2\)\(\ \)\)\/\[Pi]\ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(\((\(-1\))\)\^\(\(n\ + \ 2 m\ + 1\)\/2\)\)]], Cell[BoxData[ \(TraditionalForm\`\(2 m\)\/n\)]], Cell[BoxData[ \(TraditionalForm\`\(2 \((2 m)\)\)\/\(\((2 m)\)\^2\ - \ n\^2\)\)]], "\n\nNote that ", Cell[BoxData[ \(TraditionalForm\`\(2 \((2 m)\)\)\/\(\((2 m)\)\^2\ - \ n\^2\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`1\/\(n\ + \ 2 m\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`1\/\(n\ - \ 2 m\)\)]], "\n\n", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(\(L\)\(\ \)\(n\)\(\ \)\)\)]], "=", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(x\_n\), "TraditionalForm"]}], TraditionalForm]]], "+", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(2\ \ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(TraditionalForm\`\(2 m\)\/\(\(n\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ 2 m\)\)]], " (n odd)\n\n[10b]\n\nSimilarly,\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(R\ n\)\)\(\ \)\)\)]], "= -", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(x\_n\), "TraditionalForm"]}], TraditionalForm]]], "+", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(2\ \ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(TraditionalForm\`\(2 m\)\/\(\(n\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ 2 m\)\)]], " (n odd)\n\nThe final goal is to invert this in order to find the value of \ x in terms of ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_L\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_R\)]], ". Conveniently,\n\n[11]\n\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(\[Chi]\_\(L\ n\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ n\)\), "TraditionalForm"]}], TraditionalForm]]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(4\ \[Sum]\+\(m = \ 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(TraditionalForm\`\(2 m\)\/n\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ 2 m\)\)]], "(n odd)\n\n", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{" ", RowBox[{"(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ n\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ n\)\), "TraditionalForm"]}], ")"}]}], \(\(4\)\(\ \)\)], TraditionalForm]]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], Cell[BoxData[ \(TraditionalForm\`\(2 m\)\/n\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ 2 m\)\)]], "(n odd)\n\nwill help me find even-indexed values of x.\n\nInterestingly, \ this is analogous to a matrix operation with\n\n[\[Chi]] = [B] [x]\n\nWhere [\ \[Chi]] is a column vector with the n-th item equal to ", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{" ", RowBox[{"(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ n\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ n\)\), "TraditionalForm"]}], ")"}]}], "4"], TraditionalForm]]], "\n[x] is a column vector with the k-th item equal to ", Cell[BoxData[ \(TraditionalForm\`x\_\(2 k\)\)]], "\nand [B] being defined with n increasing down columns and k increasing \ across rows as ", Cell[BoxData[ \(TraditionalForm\`\(2\ k\)\/n\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ 2 k\)\)]], "\n\nClearly, knowledge of ", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(\(-1\)\(\ \)\)\)]], "would allow a solution for x in terms of \[Chi]. \n\nGuess: ", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(\(-1\)\(\ \)\)\)]], "is similar to ", Cell[BoxData[ \(TraditionalForm\`\(\([B]\)\^+\)\)]], "\n\n[12a]\nAs a side note, \n\n", Cell[BoxData[ \(TraditionalForm\`B\_\(a, \ b\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(-b\)\/a\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(b, \ a\)\)]], "\n\nVerify: Across the diagonal (n = m) the result is constant\nVerify by \ evaluating ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\(2\ n\)\/k\), FormBox[\(B\_\(2 k, \ n\)\), "TraditionalForm"]}], TraditionalForm]]], " (n is odd)\n[12b]\n\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ n\)\)]], "\n-", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`B\_\(n, \ 2 k\)\)]], " (using [12a])\n-", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), SuperscriptBox[ RowBox[{"(", RowBox[{\(\((\(-1\))\)\^\(\(n\ + \ 2 k\ + \ \ 1\)\/2\)\/\[Pi]\), RowBox[{"(", RowBox[{ FormBox[\(1\/\(n\ + \ 2 k\)\), "TraditionalForm"], "-", FormBox[\(1\/\(n\ - 2 k\)\), "TraditionalForm"]}], ")"}]}], ")"}], "2"]}], TraditionalForm]]], "\n-", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{\(1\/\[Pi]\^2\), SuperscriptBox[ RowBox[{"(", RowBox[{ FormBox[\(1\/\(n\ + \ 2 k\)\), "TraditionalForm"], "-", FormBox[\(1\/\(n\ - 2 k\)\), "TraditionalForm"]}], ")"}], "2"]}]}], TraditionalForm]]], "\n-", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%\[Infinity]\( 4\/\[Pi]\^2\) \((\((2 k)\)\^2\/\((\((2\ k)\)\^2 - \ n\^2)\)\^2)\)\)]], "\n-", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(TraditionalForm\`lim\_\(\[Phi] \[Rule] 1\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%\[Infinity]\((\((2 \ k)\)\^2\/\((\(\((2\ k)\)\^2\) \[Phi] - n\^2)\)\^2)\)\)]], "\n-", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(TraditionalForm\`lim\_\(\[Phi] \[Rule] 1\)\)]], "[-", Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\[Phi]\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%\[Infinity]\((1\/\(\(\((2\ \ k)\)\^2\) \[Phi] - n\^2\))\)\)]], "]\n-", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(TraditionalForm\`lim\_\(\[Phi] \[Rule] 1\)\)]], "[-", Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\[Phi]\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`1\/\[Phi]\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%\[Infinity]\((1\/\(\((2\ k)\)\^2 \ - \((n\/\@\[Phi])\)\^2\))\)\)]], "]\n\nBy contour integration,\n-", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(TraditionalForm\`lim\_\(\[Phi] \[Rule] 1\)\)]], "[", Cell[BoxData[ \(TraditionalForm\`\(-\[PartialD]\_\[Phi]\ \(1\/\[Phi]\)\) \(2\ \[Phi] \ - n\ \[Pi]\ \@\[Phi]\ \(Cot(\(n\ \[Pi]\)\/\(2\ \@\[Phi]\))\)\)\/\(4\ \ n\^2\)\)]], "]\n\n", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(TraditionalForm\`lim\_\(\[Phi] \[Rule] 1\)\)]], "[", Cell[BoxData[ \(\(\[Pi]\ Csc \((\(n\ \[Pi]\)\/\(2\ \@\[Phi]\))\)\^2\ \((\(-n\)\ \[Pi] \ + \@\[Phi]\ Sin \((\(n\ \[Pi]\)\/\@\[Phi])\))\)\)\/\(16\ n\ \[Phi]\^2\)\)]], "]\n\n", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(\(\[Pi]\ Csc \((\(n\ \[Pi]\)\/2)\)\^2\ \((\(-n\)\ \[Pi] + Sin \((n\ \ \[Pi])\))\)\)\/\(16\ n\)\)]], "\n\nSince n is odd, \n", Cell[BoxData[ \(TraditionalForm\`4\/\[Pi]\^2\)]], Cell[BoxData[ \(\(\[Pi]\ \ \((\(-n\)\ \[Pi])\)\)\/\(16\ n\)\)]], "\n\n-", Cell[BoxData[ \(TraditionalForm\`1\/4\)]], "\n\nIn other words, the inverse of ", Cell[BoxData[ FormBox[ RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}], TraditionalForm]]], " appears to be -4 ", Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ n\)\)]], ", henceforth ", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(-1\)\)]], "\n\nNow it is important to verify that the off-diagonal cells for the \ product of [B] and my tentative ", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(-1\)\)]], " is 0. \n\nVerify: Non-diagonal cells (n \[NotEqual] m) the result is \ zero. Constants multiplied across the sum will be ignored.\nVerify by \ evaluating ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ \ m\)\)]], " (n is odd)\n\nLet ", Cell[BoxData[ \(TraditionalForm\`\[Psi]\_c\)]], "(a) = ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(b = c\)\%\[Infinity]\(-1\)\/\(a\ + \ \ \ b\)\)]], "\n\nNote that the digamma function ", Cell[BoxData[ \(TraditionalForm\`\[Psi]\_c\)]], "(a) has the property that ", Cell[BoxData[ \(TraditionalForm\`\[Psi]\_c\)]], "(", Cell[BoxData[ \(TraditionalForm\`\(-n\)\/2\)]], ") = -", Cell[BoxData[ \(TraditionalForm\`\[Psi]\_c\)]], "(", Cell[BoxData[ \(TraditionalForm\`n\/2\)]], ") where n is an integer.\n[12c]\n\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}]}], TraditionalForm]]], "(-4) ", Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ \ m\)\)]], " (n is odd)\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", RowBox[{"k", " ", FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ \ m\)\)]], "(constant factors ignored)\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`B\_\(m, \ \ 2 k\)\)]], "(using [12a], constant factors ignored)\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[ RowBox[{\(\((\(-1\))\)\^\(\(n\ + \ 2 k\ + \ 1\)\/2\)\/\[Pi]\), RowBox[{"(", RowBox[{ FormBox[\(1\/\(n\ + \ 2 k\)\), "TraditionalForm"], "-", FormBox[\(1\/\(n\ - \ 2 k\)\), "TraditionalForm"]}], ")"}], \(\((\(-1\))\)\^\(\(m\ + \ 2 k\ + \ 1\)\/2\)\/\[Pi]\ \), RowBox[{"(", RowBox[{ FormBox[\(1\/\(m\ + \ 2 k\)\), "TraditionalForm"], "-", FormBox[\(1\/\(m\ - \ 2 k\)\), "TraditionalForm"]}], ")"}]}], "TraditionalForm"]}], TraditionalForm]]], "\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[ RowBox[{ RowBox[{"(", RowBox[{ FormBox[\(1\/\(n\ + \ 2 k\)\), "TraditionalForm"], "-", FormBox[\(1\/\(n\ - \ 2 k\)\), "TraditionalForm"]}], ")"}], RowBox[{"(", RowBox[{ FormBox[\(1\/\(m\ + \ 2 k\)\), "TraditionalForm"], "-", FormBox[\(1\/\(m\ - \ 2 k\)\), "TraditionalForm"]}], ")"}]}], "TraditionalForm"]}], TraditionalForm]]], "(constant factors ignored)\n\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[ RowBox[{ RowBox[{ FormBox[\(1\/\(n\ + \ 2 k\)\), "TraditionalForm"], FormBox[\(1\/\(m\ + \ 2 k\)\), "TraditionalForm"]}], "-"}], "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[ RowBox[{ RowBox[{ FormBox[\(1\/\(n\ - \ 2 k\)\), "TraditionalForm"], FormBox[\(1\/\(m\ + \ 2 k\)\), "TraditionalForm"]}], "-"}], "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[ RowBox[{ RowBox[{ FormBox[\(1\/\(n\ + \ 2 k\)\), "TraditionalForm"], FormBox[\(1\/\(m\ - \ 2 k\)\), "TraditionalForm"]}], "+"}], "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[ RowBox[{ FormBox[\(1\/\(n\ - \ 2 k\)\), "TraditionalForm"], FormBox[\(1\/\(m\ - \ 2 k\)\), "TraditionalForm"]}], "TraditionalForm"]}], TraditionalForm]]], "\n", Cell[BoxData[ \(1\/\(m\ - \ n\)\)]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\ + \ 2 k\) - 1\/\(m\ + \ 2 k\))\) - 1\/\(m\ + \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\ - \ 2 k\) + 1\/\(m\ + \ 2 k\))\) - 1\/\(m\ + \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\ + \ 2 k\) + 1\/\(m\ - \ 2 k\))\) + 1\/\(m\ - \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\((1\/\(m\ - \ 2 k\) - 1\/\(n\ - \ 2 k\))\), "TraditionalForm"]}], TraditionalForm]]], "\n\n", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "(", Cell[BoxData[ \(1\/\(m\ - \ n\)\)]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\/2\ + \ k\) - 1\/\(m\/2\ + \ k\))\) - 1\/\(m\ + \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\/2\ - \ k\) + 1\/\(m\/2\ + \ k\))\) - 1\/\(m\ + \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\/2\ + \ k\) + 1\/\(m\/2\ - \ k\))\) + 1\/\(m\ - \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\((1\/\(m\/2\ - \ k\) - 1\/\(n\/2\ - \ k\))\), "TraditionalForm"]}], TraditionalForm]]], ")\n", Cell[BoxData[ \(1\/\(m\ - \ n\)\)]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\/2\ + \ k\) - 1\/\(m\/2\ + \ k\))\) - 1\/\(m\ + \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\/2\ - \ k\) + 1\/\(m\/2\ + \ k\))\) - 1\/\(m\ + \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\(\((1\/\(n\/2\ + \ k\) + 1\/\(m\/2\ - \ k\))\) + 1\/\(m\ - \ n\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), " ", FormBox[\((1\/\(m\/2\ - \ k\) - 1\/\(n\/2\ - \ k\))\), "TraditionalForm"]}], TraditionalForm]]], " (constant factors ignored)\n", Cell[BoxData[ \(1\/\(m\ - \ n\)\)]], Cell[BoxData[ \(TraditionalForm\`\((\(\[Psi]\_1\)(n\/2)\ - \ \(\[Psi]\_1\)( m\/2))\) - 1\/\(m\ + \ n\)\)]], Cell[BoxData[ FormBox[ RowBox[{\((\(-\ \(\(\[Psi]\_1\)(\(-n\)\/2)\)\)\ + \ \(\[Psi]\_1\)( m\/2))\), " ", FormBox[\(-\(1\/\(m\ + \ n\)\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\((\(\[Psi]\_1\)( n\/2)\ - \ \(\[Psi]\_1\)(\(-m\)\/2))\)\ + 1\/\(m\ - \ n\)\)]], Cell[BoxData[ \(TraditionalForm\`\((\ \(\[Psi]\_1\)(\(-n\)\/2)\ - \ \ \(\[Psi]\_1\)(\(-m\)\/2))\)\)]], "\n", Cell[BoxData[ \(1\/\(m\ - \ n\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\(\[Psi]\_1\)(n\/2)\ - \ \(\[Psi]\_1\)(m\/2)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(+\ \(\(\[Psi]\_1\)(\(-n\)\/2)\)\)\ - \ \ \(\[Psi]\_1\)(\(-m\)\/2)\)]], ") - ", Cell[BoxData[ \(TraditionalForm\`1\/\(m\ + \ n\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\(-\ \(\(\[Psi]\_1\)(\(-n\)\/2)\)\)\ + \ \ \(\[Psi]\_1\)(m\/2)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`\(\[Psi]\_1\)( n\/2)\ + \ \(\[Psi]\_1\)(\(-m\)\/2)\)]], ")\n", Cell[BoxData[ \(1\/\(m\ - \ n\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\(\[Psi]\_1\)(n\/2)\ - \ \(\[Psi]\_1\)(m\/2)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(-\ \(\(\[Psi]\_1\)(n\/2)\)\)\ + \ \(\[Psi]\_1\)( m\/2)\)]], ") - ", Cell[BoxData[ \(TraditionalForm\`1\/\(m\ + \ n\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\[Psi]\_1\)(n\/2)\ + \ \(\[Psi]\_1\)( m\/2)\)\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`\(\[Psi]\_1\)(n\/2)\ - \ \(\[Psi]\_1\)(m\/2)\)]], ")\n0\n\nAnd so it has been shown that the inverse of the infinite matrix \ given by ", Cell[BoxData[ FormBox[ RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}], TraditionalForm]]], "is ", Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ n\)\)]], ", but should be multiplied by -4 on inversion as inversion creates a \ factor of ", Cell[BoxData[ \(TraditionalForm\`\(-1\)\/4\)]], ". In other words, -4", Cell[BoxData[ FormBox[ RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`B\_\(2 k, \ n\)\)]], "is the infinite identity matrix.\n\n[\[Chi]] = [B] [x]\n", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(-1\)\)]], "[\[Chi]] = [x]\n\nFor \n\n", Cell[BoxData[ \(TraditionalForm\`x\_\(2 m\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity]\)]], Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{"-", RowBox[{\(B\_\(2 m, \ 2 n - 1\)\), "(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2\ n - 1\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ \ 2 n - 1\)\), "TraditionalForm"]}], ")"}]}]}], TraditionalForm]]], " =", Cell[BoxData[ \(TraditionalForm\`\ \ \)]], "(for all positive integers m)\n\nThe knowledge of the values of \ odd-indexed values of \[Chi] will help to find odd-indexed values of x, from \ which I can find using the relation\n\n[13]\n\n", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(L\ n\)\)\(\ \)\)\)]], "-", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(\[Chi]\_\(R\ n\)\), "TraditionalForm"]}], TraditionalForm]]], " =", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(2 x\_n\), "TraditionalForm"]}], TraditionalForm]]], "(n odd)\n\n", Cell[BoxData[ \(TraditionalForm\`x\_n\)]], " = ", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ FormBox[\(\(\[Chi]\_\(L\ n\)\)\(\ \)\), "TraditionalForm"], "-", FormBox[ RowBox[{" ", FormBox[\(\[Chi]\_\(R\ n\)\), "TraditionalForm"]}], "TraditionalForm"]}], "2"], TraditionalForm]]], " (n odd)\n\nFinally, I need knowledge of ", Cell[BoxData[ \(TraditionalForm\`x\_0, \ which\ can\ be\ found\ from\ [ 1 a], \ \(\(by\)\(\ \)\(substituting\)\(\ \)\(the\)\(\ \)\(other\)\ \(\ \)\(coefficients\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`x\_n\)]], " and knowing x(", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ").\n\n[14]\n\nx(\[Sigma]) =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x\_0\)\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] x\_n\)]], "cos(n \[Sigma])\n", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x(\[Pi]\/2)\)\)\)]], "- ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] x\_n\)]], "cos(n ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ")\n(in this case, n must be even)\n", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x(\[Pi]\/2)\)\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] x\_\(2 n\)\)]], Cell[BoxData[ \(TraditionalForm\`\((\(-1\))\)\^n\)]], "\n", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x(\[Pi]\/2)\)\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(n = 1\)\%\[Infinity]\), " ", RowBox[{\(\((\(-1\))\)\^n\), " ", RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{\(B\_\(2 n, \ 2 m - 1\)\), "(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2\ m - 1\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ \ 2 m - 1\)\), "TraditionalForm"]}], ")"}]}]}]}], TraditionalForm]]], "\n\n", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x(\[Pi]\/2)\)\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = 1\)\%\[Infinity]\)]], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(n = 1\)\%\[Infinity]\), " ", RowBox[{\(\((\(-1\))\)\^n\), " ", RowBox[{\(B\_\(2 n, \ 2 m - 1\)\), "(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2\ m - 1\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ \ 2 m - 1\)\), "TraditionalForm"]}], ")"}]}]}], TraditionalForm]]], "\n", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(x(\[Pi]\/2)\)\)\)]], "- ", Cell[BoxData[ \(TraditionalForm\`\@2\/\[Pi]\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = 1\)\%\[Infinity]\)]], Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{\(\((\(-1\))\)\^m\/\(2\ m\ - \ 1\)\), " ", RowBox[{"(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2\ m - 1\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ \ 2 m - 1\)\), "TraditionalForm"]}], ")"}]}]}], TraditionalForm]]], "\n\nThe next goal is to find the conjugate momemntum P\n\nLet ", Cell[BoxData[ \(TraditionalForm\`P\_\(L\ n\) = \ \(-\ \[ImaginaryI]\)\ \ \(\(\[PartialD]\_\(\[Chi]\_\(L\ n\)\)\)\(\ \)\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`P\_\(R\ n\) = \ \(-\ \[ImaginaryI]\)\ \ \[PartialD]\_\(\[Chi]\_\(R\ n\)\)\)]], " and P = - \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\(x(\[Pi]\/2)\)\)\(\ \)\)\)], "DisplayFormula"], "\n", Cell[BoxData[ \(TraditionalForm\`\(\(p\)\(\ \)\)\_n = \ \(-\ \[ImaginaryI]\)\ \ \[PartialD]\_\(x\_\(\(\ \)\(n\)\)\)\)]], "\n\n[15a]\n\nI wish to evaluate\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(L\ n\) = \ \(-\ \[ImaginaryI]\)\ \ \[PartialD]\_\(\[Chi]\_\(L\ n\)\)\)]], "(x) ", Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_x\)\(\ \)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm\`P\_\(L\ n\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(\(-\ \[ImaginaryI]\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = \ 0\)\%\[Infinity]\[PartialD]\_\(\[Chi]\_\(L\ n\)\)\([x\_m]\)\)]], Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\(x\_m\)\)]], "\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\ \[PartialD]\_\(\[Chi]\_\(L\ \ n\)\)\)]], Cell[BoxData[ FormBox[ RowBox[{"[", " ", RowBox[{\(x(\[Pi]\/2)\), "-", RowBox[{ FormBox[\(\@2\/\[Pi]\), "TraditionalForm"], RowBox[{ FormBox[\(\[Sum]\+\(m = 1\)\%\[Infinity]\), "TraditionalForm"], FormBox[ RowBox[{" ", RowBox[{\(\((\(-1\))\)\^m\/\(2\ m\ - \ 1\)\), " ", RowBox[{"(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2\ m - 1\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ \ 2 m - 1\)\), "TraditionalForm"]}], ")"}]}], "]"}], "TraditionalForm"]}]}]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\(x\_0\)\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\ \[PartialD]\_\(\[Chi]\_\(L\ \ n\)\)\)]], " ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{ RowBox[{"[", RowBox[{\(\[Sum]\+\(q = 1\)\%\[Infinity]\), RowBox[{"-", RowBox[{\(B\_\(2 m, \ 2 q - 1\)\), "(", RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2 q - 1\)\)\(\ \)\), "TraditionalForm"], "+", " ", FormBox[\(\[Chi]\_\(R\ \ 2 q - 1\)\), "TraditionalForm"]}], ")"}]}]}], "]"}], \(\[PartialD]\_\(x\_\(2 m\)\)\)}]}], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\ \[PartialD]\_\(\[Chi]\_\(L\ \ n\)\)\)]], " [", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), FractionBox[ RowBox[{ FormBox[\(\(\[Chi]\_\(L\ 2 m\ - \ 1\)\)\(\ \)\), "TraditionalForm"], "-", FormBox[ RowBox[{" ", FormBox[\(\[Chi]\_\(R\ 2 m\ - \ 1\)\), "TraditionalForm"]}], "TraditionalForm"]}], "2"]}], TraditionalForm]]], "]", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\(x\_\(2 m - 1\)\)\)]], "\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " = 0 (n even, no dependence on ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(L\ n\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(\(-\ \[ImaginaryI]\)\(\ \)\)\)]], Cell[BoxData[ FormBox[ RowBox[{"[", " ", RowBox[{ RowBox[{"-", FormBox[\(\@2\/\[Pi]\), "TraditionalForm"]}], FormBox[\(\(\ \)\(\((\(-1\))\)\^\(\(n + 1\)\/2\)\/n\)\(\ \ \)\(]\)\), "TraditionalForm"]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\(x\_0\)\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`\(\(+\ \[ImaginaryI]\)\(\ \)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = 1\)\%\[Infinity]\([B\_\(2 m, \ n\)]\) \ \[PartialD]\_\(x\_\(2 m\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(\(-\ \[ImaginaryI]\)\(\ \)\)\)]], " [", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = 1\)\%\[Infinity] 1\/2\)]], "]", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\(x\_n\)\)]], "(n odd)\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\ \)]], Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ RowBox[{"-", FormBox[\(\@2\/\[Pi]\), "TraditionalForm"]}], FormBox[ RowBox[{" ", RowBox[{\(\(\((\(-1\))\)\^\(\(n + 1\)\/2\)\/n\) p\_\(\_0\)\), "-", FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{\([B\_\(2 m, \ n\)]\), \(p\_\(\_\(2 m\)\)\), " ", FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", FormBox[\(1\/2\), "TraditionalForm"], FormBox[\(p\_\(\_n\)\), "TraditionalForm"]}]}], "TraditionalForm"], " "}]}], "TraditionalForm"]}]}], TraditionalForm]]], " (n odd)\n\nAnd similarly,\n\n[15b]\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(R\ n\)\)\)]], " = 0 (n even, no dependence on ", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(R\ n\)\)]], ")\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(R\ n\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\ \)]], Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ RowBox[{"-", FormBox[\(\@2\/\[Pi]\), "TraditionalForm"]}], FormBox[\(\(\ \)\(\((\(-1\))\)\^\(\(n + 1\)\/2\)\/n\)\(\ \)\), "TraditionalForm"]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`\(\(p\_\(\_0\)\)\(\ \)\)\)]], " -", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = 1\)\%\[Infinity]\([B\_\(2 m, \ n\)]\) p\_\(\_\(2 m\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(\(-\)\(\ \)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ \(TraditionalForm\`p\_\(\_n\)\)]], "(n odd)\n\n[15c]\n\nAlso note\n\nP = - \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\(x(\[Pi]\/2)\)\)\(\ \)\)\)], "DisplayFormula"], "[", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(m = 0\)\%\[Infinity] x\_m\)]], Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\(x\_m\)\)]], "]\n\nP = - \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`\(\(\[PartialD]\_\(x(\[Pi]\/2)\)\)\(\ \)\)\)], "DisplayFormula"], "[", Cell[BoxData[ \(TraditionalForm\`x(\[Pi]\/2)\)]], "]", Cell[BoxData[ \(TraditionalForm\`\ \)]], Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\(x\_0\)\)]], "\n\nP =", Cell[BoxData[ \(TraditionalForm\`\ \)]], Cell[BoxData[ \(TraditionalForm\`p\_\(\_0\)\)]], "\n\nNow I need the full-string momenta as a function of these half-string \ momenta.\n\n[16]\n\nNote that \n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(R\ n\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`p\_\(\_n\)\)]], " (n odd)\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " + ", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(R\ n\)\)\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ RowBox[{ RowBox[{"-", FormBox[\(\(2 \@ 2\)\/\[Pi]\), "TraditionalForm"]}], FormBox[\(\(\ \)\(\((\(-1\))\)\^\(\(n + 1\)\/2\)\/n\)\(\ \)\), "TraditionalForm"], FormBox[\(\(p\_\(\_0\)\)\(\ \)\), "TraditionalForm"]}], "-", FormBox[\(2 \(\[Sum]\+\(m = 1\)\%\[Infinity]\( B\_\(2 m, \ n\)\) p\_\(\_\(2 m\)\)\)\), "TraditionalForm"], " "}]}], TraditionalForm]]], " (n odd)\n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " + ", Cell[BoxData[ FormBox[ RowBox[{\(P\_\(\(\ \)\(R\ n\)\)\), " ", "+", " ", RowBox[{ FormBox[\(\(2 \@ 2\)\/\[Pi]\), "TraditionalForm"], FormBox[\(\(\ \)\(\((\(-1\))\)\^\(\(n + 1\)\/2\)\/n\)\(\ \)\), "TraditionalForm"], "P"}]}], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(\(-2\) \(\[Sum]\+\(m = 1\)\%\[Infinity]\( B\_\(2 m, \ n\)\) p\_\(\_\(2 m\)\)\)\), "TraditionalForm"], " "}], TraditionalForm]]], " (n odd)\n\n-", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ FormBox[\(P\_\(\(\ \)\(L\ n\)\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(P\_\(\(\ \)\(R\ n\)\)\), " ", "+", RowBox[{ FormBox[\(\(2 \@ 2\)\/\[Pi]\), "TraditionalForm"], FormBox[\(\(\ \)\(\(\((\(-1\))\)\^\(\(n + 1\)\/2\)\/n\) \(P\ \)\(\ \)\)\), "TraditionalForm"]}]}], "TraditionalForm"]}], "2"], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(\[Sum]\+\(m = 1\)\%\[Infinity]\( B\_\(2 m, \ n\)\) p\_\(\_\(2 m\)\)\), "TraditionalForm"], " "}], TraditionalForm]]], " (n odd)\n\nis analogous to a matrix operation similar to the one \ encountered earlier.\n\n[P] = [B] [p]\n\nwhere [P] is a column vector with \ entry k defined by -", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ FormBox[\(P\_\(\(\ \)\(L\ 2 k\ - \ 1\)\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(P\_\(\(\ \)\(R\ 2 k\ - \ 1\)\)\), " ", "+", RowBox[{ FormBox[\(\(2 \@ 2\)\/\[Pi]\), "TraditionalForm"], FormBox[\(\(\ \)\(\(\((\(-1\))\)\^k\/\(2 k\ - \ 1\)\) \(P\)\(\ \)\)\), "TraditionalForm"]}]}], "TraditionalForm"]}], "2"], TraditionalForm]]], "\n[B] is an infinite matrix with j defined across rows and k defined down \ columns as ", Cell[BoxData[ \(TraditionalForm\`B\_\(2 j, \ 2 k\ - \ 1\)\)]], "\nand [p] is a column vector with entry k defined by ", Cell[BoxData[ \(TraditionalForm\`p\_\(\_\(2 k\)\)\)]], "\n\nknowledge of ", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(-1\), \ then, \ would\ allow\ me\ to\ invert\ this\ \(\(relation\)\(.\)\)\)]], "\n\nFrom my previous inversion, which showed that ", Cell[BoxData[ FormBox[ RowBox[{\(\(2\ k\)\/n\), FormBox[\(B\_\(n, \ 2 k\)\), "TraditionalForm"]}], TraditionalForm]]], "is the inverse of -4 ", Cell[BoxData[ \(TraditionalForm\`\(\(B\_\(2 k, \ n\)\)\(,\)\)\)]], "I could expect that -4", Cell[BoxData[ FormBox[ RowBox[{\(\(2\ j\)\/\(2 k\ - \ 1\)\), FormBox[\(B\_\(2 k\ - \ 1, \ 2 j\)\), "TraditionalForm"]}], TraditionalForm]]], " is the inverse of [B] so that\n\n", Cell[BoxData[ \(TraditionalForm\`\([B]\)\^\(-1\)\)]], " [P] = [p] \n\nfor\n\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{"-", RowBox[{"[", FractionBox[ RowBox[{ FormBox[\(P\_\(\(\ \)\(L\ 2 k\ - \ 1\)\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(P\_\(\(\ \)\(R\ 2 k\ - \ 1\)\)\), " ", "+", RowBox[{ FormBox[\(\(2 \@ 2\)\/\[Pi]\), "TraditionalForm"], FormBox[\(\(\ \)\(\(\((\(-1\))\)\^k\/\(2 k\ - \ 1\)\) \(P\)\(\ \)\)\), "TraditionalForm"]}]}], "TraditionalForm"]}], "2"]}]}]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(\(2 j\)\/\(2 k - \ 1\)\), RowBox[{"(", RowBox[{"-", FormBox[\(\(\(4\)\()\)\)\ B\_\(2 k\ - \ 1, \ 2 j\)\), "TraditionalForm"]}]}]}], TraditionalForm]]], " ]= ", Cell[BoxData[ \(TraditionalForm\`p\_\(2\ j\)\)]], " (j \[GreaterEqual] 1)\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{"-", RowBox[{"[", RowBox[{"(", RowBox[{ FormBox[\(P\_\(\(\ \)\(L\ 2 k\ - \ 1\)\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(P\_\(\(\ \)\(R\ 2 k\ - \ 1\)\)\), " ", "+", RowBox[{ FormBox[\(\(2 \@ 2\)\/\[Pi]\), "TraditionalForm"], FormBox[\(\(\ \)\(\(\((\(-1\))\)\^k\/\(2 k\ - \ 1\)\) \(P\)\(\ \)\)\), "TraditionalForm"]}]}], "TraditionalForm"]}], ")"}]}]}]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{\(j\/\(2 k - \ 1\)\), FormBox[\(\(\ \)\(\((\(-4\))\)\ B\_\(2 k\ - \ 1, \ 2 j\)\)\), "TraditionalForm"]}], TraditionalForm]]], " ]= ", Cell[BoxData[ \(TraditionalForm\`p\_\(2\ j\)\)]], " (j \[GreaterEqual] 1)\n\nOr evaluating the sum for the P term,\n\n", Cell[BoxData[ \(TraditionalForm\`p\_\(2\ j\)\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), RowBox[{"[", RowBox[{"2", " ", RowBox[{"(", RowBox[{ FormBox[\(P\_\(\(\ \)\(L\ 2 k\ - \ 1\)\)\), "TraditionalForm"], " ", "+", " ", FormBox[\(P\_\(\(\ \)\(R\ 2 k\ - \ 1\)\)\), "TraditionalForm"]}]}]}]}]}], TraditionalForm]]], ")", Cell[BoxData[ FormBox[ RowBox[{\(\(2 j\)\/\(2 k - \ 1\)\), FormBox[\(B\_\(2 k\ - \ 1, \ 2 j\)\), "TraditionalForm"]}], TraditionalForm]]], " ] + ", Cell[BoxData[ \(TraditionalForm\`\@2\)]], Cell[BoxData[ \(TraditionalForm\`\(\((\(-1\))\)\(\ \)\)\^j\)]], " P (j \[GreaterEqual] 1)\n\nand \n\n", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(L\ n\)\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`P\_\(\(\ \)\(R\ n\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`p\_\(\_n\)\)]], " (n odd)\n\nNext, I wish to show that the usual commutators hold:\n\n[", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(Q\ n\)\)\(,\)\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S\ m\)\)]], "] = i ", Cell[BoxData[ \(TraditionalForm\`\[Delta]\_\(Q\ S\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Delta]\_\(n\ m\)\)]], "\n\n[17a]\n\n[", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(Q\ n\)\)\(,\)\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S\ m\)\)]], "]\n", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ n\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S\ m\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`P\_\(S\ m\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ n\)\)]], "\n", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ n\)\)]], Cell[BoxData[ \(TraditionalForm\`\((\(-\ \[ImaginaryI]\)\ \[PartialD]\_\(\[Chi]\_\(S\ \ m\)\))\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`\((\(-\ \[ImaginaryI]\)\ \[PartialD]\_\(\[Chi]\_\(S\ \ m\)\))\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ n\)\)]], "\n0 + ", Cell[BoxData[ \(TraditionalForm\`\[ImaginaryI]\ \[PartialD]\_\(\[Chi]\_\(S\ \ n\)\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ m\)\)]], "\nWhich, of course, yields \[ImaginaryI] if S=Q and n=m, and 0 otherwise. \ \n\nand \n\n[17b]\n\nEvaluate to verify that \n\n[", Cell[BoxData[ \(TraditionalForm\`\(\(x(\[Pi]\/2)\)\(,\)\)\)]], Cell[BoxData[ \(TraditionalForm\`P\)]], "] =\[ImaginaryI]\n\n[", Cell[BoxData[ \(TraditionalForm\`\(\(x(\[Pi]\/2)\)\(,\)\)\)]], Cell[BoxData[ \(TraditionalForm\`P\)]], "]\n", Cell[BoxData[ \(TraditionalForm\`x(\[Pi]\/2)\)]], Cell[BoxData[ \(TraditionalForm\`P\)]], " - ", Cell[BoxData[ \(TraditionalForm\`P\)]], Cell[BoxData[ \(TraditionalForm\`x(\[Pi]\/2)\)]], "\n", Cell[BoxData[ \(TraditionalForm\`x(\[Pi]\/2)\)]], "(", Cell[BoxData[ FormBox[ RowBox[{\(-\[ImaginaryI]\), FormBox[\(\[PartialD]\_\(x(\[Pi]\/2)\)\), "TraditionalForm"]}], TraditionalForm]]], ") - ", Cell[BoxData[ FormBox[ RowBox[{\(-\[ImaginaryI]\), FormBox[\(\[PartialD]\_\(x(\[Pi]\/2)\)\), "TraditionalForm"]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`x(\[Pi]\/2)\)]], "\n0 + \[ImaginaryI]\n\[ImaginaryI]\n\nDefine the half-string creation and \ annihilation operators in the usual way\n\n[18a]\n", Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")\n\nand\n\n[18b]\n\n", Cell[BoxData[ \(TraditionalForm\`\(b\_\(Q\ n\)\^+\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\[ImaginaryI]\)\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")\n\nand\n\n[18c]\n\n", Cell[BoxData[ \(TraditionalForm\`\[Beta]\_\(Q\ n\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`n\^\(1\/2\)\)]], Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ \(n\ + \ 1\)\/2\)\)]], "\n\nand\n\n[18d]\n\n", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{ FormBox[\(\[Beta]\_\(Q\ n\)\), "TraditionalForm"], " "}], "+"], TraditionalForm]]], "= ", Cell[BoxData[ \(TraditionalForm\`\[Beta]\_\(Q\ - n\)\)]], "\n\nVerify that the commutator \n\n[", Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ n\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b\_\(S\ m\)\)]], "] = ", Cell[BoxData[ \(TraditionalForm\`\[Delta]\_\(Q\ S\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Delta]\_\(n\ m\)\)]], "\n\n[", Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ n\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(b\_\(S\ m\)\^+\)\)]], "]\n", Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ n\)\)]], Cell[BoxData[ \(TraditionalForm\`\(b\_\(S\ m\)\^+\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`\(b\_\(S\ m\)\^+\)\)]], Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ n\)\)]], "\n\n", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\[ImaginaryI]\)\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 m\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(S\ \ 2 m\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ m\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S2\ m\ - \ 1\)\)]], ") - ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\[ImaginaryI]\)\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 m\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(S\ \ 2 m\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ m\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S\ 2 m\ - \ 1\)\)]], ")", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 m\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "((", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")", Cell[BoxData[ \(TraditionalForm\`\ \)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(S\ \ 2 m\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ m\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S\ 2 m\ - \ 1\)\)]], ") - (", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(S\ \ 2 m\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ m\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(S\ 2 m\ - \ 1\)\)]], ")(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], "))\n\nThen, if Q \[NotEqual] S or n \[NotEqual] m,\n\n", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 m\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "((", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(S\ \ 2 m\ - \ 1\)\)]], ") - (", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(S\ \ 2 m\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "))\n0\n\nBut if Q = S and n = m,\n\n", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ FormBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}], TraditionalForm]]], "((", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ") + \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")) - (", Cell[BoxData[ FormBox[ RowBox[{\(\[Chi]\_\(Q\ \ 2 n\ - \ 1\)\), " ", RowBox[{"(", RowBox[{ FormBox[\(\[Chi]\_\(Q\ 2 n\ - \ 1\)\), "TraditionalForm"], "+", RowBox[{"\[ImaginaryI]", FormBox[\(2\/\(2\ n\ - \ 1\)\), "TraditionalForm"], FormBox[\(P\_\(Q\ 2 n\ - \ 1\)\), "TraditionalForm"]}]}], ")"}]}], TraditionalForm]]], "- \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ FormBox[ RowBox[{\(P\_\(\(Q\ 2 n\)\(\ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "(", RowBox[{ FormBox[\(\[Chi]\_\(Q\ 2 n\ - \ 1\)\), "TraditionalForm"], "+", RowBox[{"\[ImaginaryI]", FormBox[\(2\/\(2\ n\ - \ 1\)\), "TraditionalForm"], FormBox[\(P\_\(Q\ 2 n\ - \ 1\)\), "TraditionalForm"]}]}], ")"}], TraditionalForm]]], ")\n", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}], "("}], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], ") - (", Cell[BoxData[ \(TraditionalForm\`\(\(\[Chi]\_\(Q\ \ 2 n\ - \ 1\)\^2\)\(\ \)\)\)]], "- ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], "))\n", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], Cell[BoxData[ FormBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`4\/\(2\ n\ - \ 1\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`1\)]], "\n\nThe next goal is to rewrite the half-string creation and annihilation \ operators, b, from [18] in terms of the full-string creation and annihilation \ operators, a, from [1b]\n\n[19a]\n\n", Cell[BoxData[ \(TraditionalForm\`b\_\(Q\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(Q\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(Q\ 2 n\ - \ 1\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`b\_\(L\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ \(TraditionalForm\`\[Chi]\_\(L\ 2 n\ - \ 1\)\)]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ \(TraditionalForm\`P\_\(L\ 2 n\ - \ 1\)\)]], ")\n", Cell[BoxData[ \(TraditionalForm\`b\_\(L\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(x\_\(2 n\ - \ 1\)\), "TraditionalForm"], "+", RowBox[{ FormBox[\(\(\ \)\(2\ \ \[Sum]\+\(m = 1\)\%\[Infinity]\)\), "TraditionalForm"], FormBox[\(x\_\(2 m\)\), "TraditionalForm"], FormBox[\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "TraditionalForm"], FormBox[\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), "TraditionalForm"]}]}]}], TraditionalForm]]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{" ", RowBox[{ RowBox[{"-", FormBox[\(\@2\/\[Pi]\), "TraditionalForm"]}], FormBox[ RowBox[{" ", RowBox[{\(\(\((\(-1\))\)\^\(\(2 n\ - \ 1\ + 1\)\/2\)\/\ \(2 n\ \ - \ 1\)\) p\_\(\_0\)\), "-", FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{\([B\_\(2 m, \ 2 n\ \ - \ 1\)]\), \ \(p\_\(\_\(2 m\)\)\), " ", FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", FormBox[\(1\/2\), "TraditionalForm"], FormBox[\(p\_\(\_\(2 n\ \ - \ 1\)\)\), "TraditionalForm"]}]}], "TraditionalForm"], " "}]}], "TraditionalForm"]}]}], "TraditionalForm"], ")"}], TraditionalForm]]], ") \n", Cell[BoxData[ \(TraditionalForm\`b\_\(L\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/\@2\)]], Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"(", FormBox[\(\(2 n\ - \ 1\)\/2\), "TraditionalForm"], ")"}]], TraditionalForm]]], "(", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ FormBox[\(x\_\(2 n\ - \ 1\)\), "TraditionalForm"], "+", RowBox[{ FormBox[\(\(\ \)\(2\ \ \[Sum]\+\(m = 1\)\%\[Infinity]\)\), "TraditionalForm"], FormBox[\(x\_\(2 m\)\), "TraditionalForm"], FormBox[\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "TraditionalForm"], FormBox[\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), "TraditionalForm"]}]}]}], TraditionalForm]]], "+ \[ImaginaryI] ", Cell[BoxData[ \(TraditionalForm\`2\/\(2\ n\ - \ 1\)\)]], Cell[BoxData[ FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{" ", RowBox[{ RowBox[{"-", FormBox[\(\@2\/\[Pi]\), "TraditionalForm"]}], FormBox[ RowBox[{" ", RowBox[{\(\(\((\(-1\))\)\^n\/\(2 n\ \ - \ 1\)\) p\_\(\_0\)\), "-", FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{\([B\_\(2 m, \ 2 n\ \ - \ 1\)]\), \ \(p\_\(\_\(2 m\)\)\), " ", FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", FormBox[\(1\/2\), "TraditionalForm"], FormBox[\(p\_\(\_\(2 n\ \ - \ 1\)\)\), "TraditionalForm"]}]}], "TraditionalForm"], " "}]}], "TraditionalForm"]}]}], "TraditionalForm"], ")"}], TraditionalForm]]], ") \n", Cell[BoxData[ \(TraditionalForm\`b\_\(L\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(-\ \[ImaginaryI]\)\/2\)]], Cell[BoxData[ FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(x\_\(2 n\ - \ 1\)\), "TraditionalForm"], FormBox[ RowBox[{" ", RowBox[{ FormBox[\(-\ \[ImaginaryI]\), "TraditionalForm"], FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], "TraditionalForm"], " ", \(\[Sum]\+\(m = 1\)\%\[Infinity]\)}]}], "TraditionalForm"], FormBox[\(x\_\(2 m\)\), "TraditionalForm"], FormBox[\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "TraditionalForm"], FormBox[\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), "TraditionalForm"]}], TraditionalForm]]], "+ ", Cell[BoxData[ FormBox[ FractionBox["1", SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]]], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{" ", RowBox[{ RowBox[{"-", FormBox[\(\@2\/\[Pi]\), "TraditionalForm"]}], FormBox[ RowBox[{" ", RowBox[{\(\(\((\(-1\))\)\^n\/\(2 n\ \ - \ 1\)\) p\_\(\_0\)\), "-", FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{\([B\_\(2 m, \ 2 n\ \ - \ 1\)]\), \ \(p\_\(\_\(2 m\)\)\), " ", FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", FormBox[\(1\/2\), "TraditionalForm"], FormBox[\(p\_\(\_\(2 n\ \ - \ 1\)\)\), "TraditionalForm"]}]}], "TraditionalForm"], " "}]}], "TraditionalForm"]}]}], "TraditionalForm"], ")"}], TraditionalForm]]], ") \n", Cell[BoxData[ \(TraditionalForm\`b\_\(L\ n\)\)]], "= ", Cell[BoxData[ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], TraditionalForm]]], " -", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\[ImaginaryI]\)\)\/2\)]], Cell[BoxData[ FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(x\_\(2 n\ - \ 1\)\), "TraditionalForm"], FormBox[ RowBox[{" ", RowBox[{ FormBox[\(-\ \[ImaginaryI]\), "TraditionalForm"], FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], "TraditionalForm"], " ", \(\[Sum]\+\(m = 1\)\%\[Infinity]\)}]}], "TraditionalForm"], FormBox[\(x\_\(2 m\)\), "TraditionalForm"], FormBox[\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "TraditionalForm"], FormBox[ RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), FormBox["", "TraditionalForm"], FormBox[ RowBox[{ RowBox[{"-", FractionBox["1", SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]]]}], FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{\([B\_\(2 m, \ 2 n\ \ - \ 1\)]\), \ \(p\_\(\_\(2 m\)\)\), " ", FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", FormBox[\(1\/2\), "TraditionalForm"], FractionBox["1", SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]]], FormBox[\(p\_\(\_\(2 n\ \ - \ 1\)\)\), "TraditionalForm"]}]}], "TraditionalForm"], " "}], "TraditionalForm"]}], "TraditionalForm"]}], TraditionalForm]]], "\n\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{\(b\_\(L\ n\)\), "=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], "-", RowBox[{ FormBox[\(\(\(\ \)\(\[ImaginaryI]\)\)\/2\), "TraditionalForm"], FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], "TraditionalForm"], \(1\/2\), FormBox[\(\[ImaginaryI]\ \@\(2\/\(2 n - 1\)\)\), "TraditionalForm"], RowBox[{"(", RowBox[{ FormBox[\(a\_\(2 n - 1\)\), "TraditionalForm"], "-", FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 n - 1\)\), "TraditionalForm"], "+"], "TraditionalForm"]}], ")"}]}]}]}], "TraditionalForm"], FormBox[ RowBox[{" ", RowBox[{ FormBox[\(-\ \[ImaginaryI]\), "TraditionalForm"], FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], "TraditionalForm"], " ", \(\[Sum]\+\(m = 1\)\%\[Infinity](\)}]}], "TraditionalForm"], FormBox[ RowBox[{ FormBox[\(1\/2\), "TraditionalForm"], FormBox[\(\[ImaginaryI]\ \@\(2\/\(2 m\)\)\), "TraditionalForm"], RowBox[{"(", RowBox[{ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "-", FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], "TraditionalForm"]}], ")"}]}], "TraditionalForm"], FormBox[\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "TraditionalForm"], FormBox[ RowBox[{\(\(B\_\(2 n\ \ - \ 1, \ 2 m\)\)\()\)\), FormBox["", "TraditionalForm"], FormBox[ RowBox[{ RowBox[{"-", FractionBox["1", SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]]]}], FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{ RowBox[{"(", RowBox[{\([B\_\(2 m, \ 2 n\ \ - \ 1\)]\), \(\@\(\ \(2 m\)\/2\)\), RowBox[{"(", FormBox[\(a\_\(\(2\) \(m\)\(\ \)\) + \ \ \(a\_\(2 m\)\^+\)\), "TraditionalForm"], ")"}]}], " ", ")"}], FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", FormBox[\(1\/2\), "TraditionalForm"], FractionBox["1", SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]]], FormBox[ RowBox[{\(\@\(\(2 n\ - \ 1\)\/2\)\), RowBox[{"(", FormBox[\(a\_\(\(2 n\)\(\ \)\(-\)\(\ \)\(1\)\(\ \ \)\) + \ \(a\_\(2 n\ - \ 1\)\^+\)\), "TraditionalForm"], ")"}]}], "TraditionalForm"]}]}], "TraditionalForm"], " "}], "TraditionalForm"]}], "TraditionalForm"]}], TraditionalForm]]], "\n\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{\(b\_\(L\ n\)\), "=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], "+", RowBox[{ FormBox[\(\(\ \)\(\@2\/4\)\), "TraditionalForm"], RowBox[{"(", RowBox[{ FormBox[\(a\_\(2 n - 1\)\), "TraditionalForm"], "-", FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 n - 1\)\), "TraditionalForm"], "+"], "TraditionalForm"]}], ")"}]}]}]}], "TraditionalForm"], FormBox[ RowBox[{" ", RowBox[{ FormBox[\(-\ \[ImaginaryI]\), "TraditionalForm"], FormBox[ SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]], "TraditionalForm"], " ", \(\[Sum]\+\(m = 1\)\%\[Infinity](\)}]}], "TraditionalForm"], FormBox[ RowBox[{ FormBox[\(1\/2\), "TraditionalForm"], FormBox[\(\[ImaginaryI]\ \@\(2\/\(2 m\)\)\), "TraditionalForm"], RowBox[{"(", RowBox[{ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "-", FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], "TraditionalForm"]}], ")"}]}], "TraditionalForm"], FormBox[\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\), "TraditionalForm"], FormBox[ RowBox[{\(\(B\_\(2 n\ \ - \ 1, \ 2 m\)\)\()\)\), FormBox["", "TraditionalForm"], FormBox[ RowBox[{ RowBox[{"-", FractionBox["1", SqrtBox[ FormBox[\(2 n\ - \ 1\), "TraditionalForm"]]]}], FormBox[ RowBox[{\(\[Sum]\+\(m = 1\)\%\[Infinity]\), RowBox[{ RowBox[{"(", RowBox[{\([B\_\(2 m, \ 2 n\ \ - \ 1\)]\), \(\@\(\ \(2 m\)\/2\)\), RowBox[{"(", FormBox[\(a\_\(\(2\) \(m\)\(\ \)\) + \ \ \(a\_\(2 m\)\^+\)\), "TraditionalForm"], ")"}]}], " ", ")"}], FormBox[\(\(+\)\(\ \)\), "TraditionalForm"], " ", \(\@2\/4\), FormBox[ RowBox[{"(", FormBox[\(a\_\(\(2 n\)\(\ \)\(-\)\(\ \)\(1\)\(\ \ \)\) + \ \(a\_\(2 n\ - \ 1\)\^+\)\), "TraditionalForm"], ")"}], "TraditionalForm"]}]}], "TraditionalForm"], " "}], "TraditionalForm"]}], "TraditionalForm"]}], TraditionalForm]]], "\n\n", Cell[BoxData[ FormBox[ RowBox[{\(b\_\(L\ n\)\), "=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], "+", RowBox[{ FormBox[\(\(\ \)\(1\/\@2\)\), "TraditionalForm"], FormBox[ RowBox[{\(a\_\(2 n - 1\)\), FormBox[\(\(\ \ \)\(\(+\ \(1\/\@2\)\)\ \[Sum]\+\(m = 1\)\%\ \[Infinity]\)\), "TraditionalForm"], FormBox[ RowBox[{"(", RowBox[{ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "-", FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], "TraditionalForm"]}], ")"}], "TraditionalForm"], FormBox[\(\@\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\), "TraditionalForm"], FormBox[ RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), FormBox["", "TraditionalForm"], FormBox[ RowBox[{"-", FormBox[ RowBox[{" ", RowBox[{\(B\_\(2 m, \ 2 n\ \ - \ 1\)\), \(\@\(\(2 m\)\/\(2 n\ - \ 1\)\)\), RowBox[{"(", FormBox[\(a\_\(\(2\) \(m\)\(\ \)\) + \ \(a\ \_\(2 m\)\^+\)\), "TraditionalForm"], ")"}], " "}]}], "TraditionalForm"]}], "TraditionalForm"]}], "TraditionalForm"]}], "TraditionalForm"]}]}]}], TraditionalForm]]], "\n\n", Cell[BoxData[ FormBox[ RowBox[{\(b\_\(L\ n\)\), "=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], "+", RowBox[{ FormBox[\(\(\ \)\(1\/\@2\)\), "TraditionalForm"], FormBox[ RowBox[{\(a\_\(2 n - 1\)\), FormBox[\(\(\ \ \)\(\(+\ \(1\/\@2\)\)\ \(\[Sum]\+\(m = \ 1\)\%\[Infinity]\@\(\( 2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\)\ \)\), "TraditionalForm"], FormBox[ RowBox[{"(", " ", RowBox[{"(", RowBox[{ FormBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "TraditionalForm"], "-", " ", \(a\_\(2 m\)\^+\)}], ")"}]}], "TraditionalForm"], FormBox[ RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), FormBox["", "TraditionalForm"], FormBox[ RowBox[{"+", FormBox[ RowBox[{ RowBox[{" ", RowBox[{\(B\_\(2 m, \ 2 n\ \ - \ 1\)\), "(", FormBox[\(\(-a\_\(\(2\) \(m\)\(\ \)\)\) - \ \ \(a\_\(2 m\)\^+\)\), "TraditionalForm"], ")"}], " ", ")"}], " "}], "TraditionalForm"]}], "TraditionalForm"]}], "TraditionalForm"]}], "TraditionalForm"]}]}]}], TraditionalForm]]], "\n", Cell[BoxData[ FormBox[ RowBox[{\(b\_\(L\ n\)\), "=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], "+", RowBox[{ FormBox[\(\(\ \)\(1\/\@2\)\), "TraditionalForm"], FormBox[ RowBox[{ RowBox[{\(a\_\(2 n - 1\)\), FormBox[\(\(\ \ \)\(\(+\ \(1\/\@2\)\)\ \(\[Sum]\+\(m = \ 1\)\%\[Infinity]\@\(\( 2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\)\ \)\), "TraditionalForm"], FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{ RowBox[{ "(", \(B\_\(2 n\ \ - \ 1, \ 2 m\)\ - \ B\_\(2 m, \ 2 n\ \ - \ 1\)\), " ", FormBox[")", "TraditionalForm"]}], \(a\_\(2 m\)\)}], "TraditionalForm"]}], "TraditionalForm"]}], "-", " ", RowBox[{"(", RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), " ", "+", " ", RowBox[{\(B\_\(2 m, \ 2 n\ \ - \ 1\)\), FormBox[ RowBox[{ RowBox[{")", SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], FormBox["", "TraditionalForm"]}], ")"}], "TraditionalForm"]}]}]}]}], "TraditionalForm"]}]}]}], TraditionalForm]]], "\n\nAnd similarly\n\n[19b]\n\n", Cell[BoxData[ FormBox[ RowBox[{\(b\_\(R\ n\)\), "=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], "-", RowBox[{ FormBox[\(\(\ \)\(1\/\@2\)\), "TraditionalForm"], FormBox[ RowBox[{ RowBox[{\(a\_\(2 n - 1\)\), FormBox[\(\(\ \ \)\(\(+\ \(1\/\@2\)\)\ \(\[Sum]\+\(m = \ 1\)\%\[Infinity]\@\(\( 2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\)\ \)\), "TraditionalForm"], FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{ RowBox[{ "(", \(B\_\(2 n\ \ - \ 1, \ 2 m\)\ - \ B\_\(2 m, \ 2 n\ \ - \ 1\)\), " ", FormBox[")", "TraditionalForm"]}], \(a\_\(2 m\)\)}], "TraditionalForm"]}], "TraditionalForm"]}], "-", " ", RowBox[{"(", RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), " ", "+", " ", RowBox[{\(B\_\(2 m, \ 2 n\ \ - \ 1\)\), FormBox[ RowBox[{ RowBox[{")", SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], FormBox["", "TraditionalForm"]}], ")"}], "TraditionalForm"]}]}]}]}], "TraditionalForm"]}]}]}], TraditionalForm]]], "\n\nwith ", Cell[BoxData[ FormBox[ SuperscriptBox[ FormBox[\(b\_\(Q\ n\)\), "TraditionalForm"], "+"], TraditionalForm]]], " obtainable by using the ", Cell[BoxData[ \(TraditionalForm\`\(\[Placeholder]\^+\)\)]], " operation on each a present.\n\nNow I would like to invert this \ relationship.\n\n[20a]\n\nClearly, \n\n", Cell[BoxData[ \(TraditionalForm\`b\_\(L\ n\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`b\_\(R\ n\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(2\/\@2\) a\_\(2 n - 1\)\)]], "\n", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\), "TraditionalForm"], " ", "-", " ", FormBox[\(b\_\(R\ n\)\), "TraditionalForm"]}], \(\@2\)], TraditionalForm]]], "= ", Cell[BoxData[ \(TraditionalForm\`a\_\(2 n - 1\)\)]], " \n\nfinds odd-indexed a. To find the corresponding ", Cell[BoxData[ FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 n - 1\)\), "TraditionalForm"], "+"], TraditionalForm]]], " operator, simply use the ", Cell[BoxData[ \(TraditionalForm\`\(\[Placeholder]\^+\)\)]], " operator on each side.\n\n", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\), "TraditionalForm"], " ", "+", " ", FormBox[\(\(b\_\(R\ n\)\)\(\ \)\), "TraditionalForm"]}], "2"], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{"=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], FormBox[ RowBox[{ RowBox[{ FormBox[\(\(\ \ \)\(\(+\ \(1\/\@2\)\)\ \(\[Sum]\+\(m = \ 1\)\%\[Infinity]\@\(\( 2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\)\ \)\), "TraditionalForm"], FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{ RowBox[{ "(", \(B\_\(2 n\ \ - \ 1, \ 2 m\)\ - \ B\_\(2 m, \ 2 n\ \ - \ 1\)\), " ", FormBox[")", "TraditionalForm"]}], \(a\_\(2 m\)\)}], "TraditionalForm"]}], "TraditionalForm"]}], "-", " ", RowBox[{"(", RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), " ", "+", " ", RowBox[{\(B\_\(2 m, \ 2 n\ \ - \ 1\)\), FormBox[ RowBox[{ RowBox[{")", SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], FormBox["", "TraditionalForm"]}], ")"}], "TraditionalForm"]}]}]}]}], "TraditionalForm"]}]}], TraditionalForm]]], "\n", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\^+\), "TraditionalForm"], " ", "+", " ", FormBox[\(\(b\_\(R\ n\)\^+\)\(\ \)\), "TraditionalForm"]}], "2"], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{"=", RowBox[{ FormBox[ RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], \(p\_\(\_0\)\)}], "TraditionalForm"], FormBox[ RowBox[{ RowBox[{ FormBox[\(\(\ \ \)\(\(+\ \(1\/\@2\)\)\ \(\[Sum]\+\(m = \ 1\)\%\[Infinity]\@\(\( 2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\)\ \)\), "TraditionalForm"], FormBox[ RowBox[{"(", " ", FormBox[ RowBox[{ RowBox[{ "(", \(B\_\(2 n\ \ - \ 1, \ 2 m\)\ - \ B\_\(2 m, \ 2 n\ \ - \ 1\)\), " ", FormBox[")", "TraditionalForm"]}], \(a\_\(2 m\)\^+\)}], "TraditionalForm"]}], "TraditionalForm"]}], "-", " ", RowBox[{"(", RowBox[{\(B\_\(2 n\ \ - \ 1, \ 2 m\)\), " ", "+", " ", RowBox[{\(B\_\(2 m, \ 2 n\ \ - \ 1\)\), FormBox[ RowBox[{ RowBox[{")", FormBox[\(a\_\(2 m\)\), "TraditionalForm"], FormBox["", "TraditionalForm"]}], ")"}], "TraditionalForm"]}]}]}]}], "TraditionalForm"]}]}], TraditionalForm]]], "\n\n[21a]\n\n", Cell[BoxData[ FormBox[ RowBox[{\(-\@2\), RowBox[{"(", RowBox[{ FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(b\_\(R\ n\)\), " ", "+", " ", FormBox[\(b\_\(L\ n\)\^+\), "TraditionalForm"], " ", "+", " ", FormBox[\(\(b\_\(R\ n\)\^+\)\(\ \)\), "TraditionalForm"]}], "TraditionalForm"]}], "4"], " ", "-", " ", RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], "P"}]}], ")"}]}], TraditionalForm]]], "=", Cell[BoxData[ FormBox[ FormBox[ RowBox[{ RowBox[{ FormBox[\(\(\ \ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\), "TraditionalForm"], FormBox[ FormBox[\(\(B\_\(2 m, \ 2 n\ \ - \ 1\)\)\((\)\(\(\@\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\) \ \((a\_\(2 m\)\)\)\), "TraditionalForm"], "TraditionalForm"]}], "+", FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(a\_\(2 m\)\^+\), "TraditionalForm"], FormBox["", "TraditionalForm"]}], ")"}], ")"}], "TraditionalForm"]}], "TraditionalForm"], TraditionalForm]]], "\n\n[21b]\n\n", Cell[BoxData[ FormBox[ RowBox[{\(\@2\), RowBox[{"(", FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(b\_\(R\ n\)\), " ", "-", " ", FormBox[\(b\_\(L\ n\)\^+\), "TraditionalForm"], " ", "-", " ", FormBox[\(\(b\_\(R\ n\)\^+\)\(\ \)\), "TraditionalForm"], " "}], "TraditionalForm"]}], "4"], ")"}]}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{"=", FormBox[ RowBox[{ RowBox[{ FormBox[\(\(\ \ \ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\), "TraditionalForm"], FormBox[ FormBox[\(B\_\(2 n\ \ - \ 1, \ 2 m\)\ \((\(\@\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\) \ \((a\_\(2 m\)\)\)\), "TraditionalForm"], "TraditionalForm"]}], "-", " ", FormBox[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], FormBox["", "TraditionalForm"]}], ")"}], ")"}], "TraditionalForm"]}], "TraditionalForm"]}], TraditionalForm]]], "\n\nI'll have to invert both 21a and 21b first, then from the results I'll \ be able to find ", Cell[BoxData[ \(TraditionalForm\`a\_\(2 m\)\)]], " and ", Cell[BoxData[ FormBox[ SuperscriptBox[ FormBox[\(a\_\(2 m\)\), "TraditionalForm"], "+"], TraditionalForm]]], ". As I've seen earlier, the inverse of the infinite matrix given by ", Cell[BoxData[ \(TraditionalForm\`B\_\(a, \ b\)\)]], "is ", Cell[BoxData[ FormBox[ RowBox[{\(a\/b\), FormBox[\(B\_\(b, \ a\)\), "TraditionalForm"]}], TraditionalForm]]], " and requires a factor of -4 to be multiplied after inversion\n\n[21c]: \ Inverse of [21a]\n\n", Cell[BoxData[ \(TraditionalForm\`B\_\(a, \ b\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(-b\)\/a\)]], Cell[BoxData[ \(TraditionalForm\`B\_\(b, \ a\)\)]], "\n\n[b] = [B][a]\n", Cell[BoxData[ \(TraditionalForm\`\(\([B]\)\^\(-1\)\)[b]\ = \ \([a]\)\)]], "\n\n", Cell[BoxData[ FormBox[ RowBox[{\(-\@2\), RowBox[{"(", RowBox[{ FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(b\_\(R\ n\)\), " ", "+", " ", FormBox[\(b\_\(L\ n\)\^+\), "TraditionalForm"], " ", "+", " ", FormBox[\(\(b\_\(R\ n\)\^+\)\(\ \)\), "TraditionalForm"]}], "TraditionalForm"]}], "4"], " ", "-", " ", RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], "P"}]}], ")"}]}], TraditionalForm]]], "=", Cell[BoxData[ FormBox[ FormBox[ RowBox[{ RowBox[{ FormBox[\(\(\ \ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\), "TraditionalForm"], FormBox[ FormBox[\(\(B\_\(2 m, \ 2 n\ \ - \ 1\)\)\((\)\(\(\@\(\(2 m\)\/\(\(2 n\)\(\ \ \)\(-\)\(\ \)\(1\)\(\ \)\)\)\) \ \((a\_\(2 m\)\)\)\), "TraditionalForm"], "TraditionalForm"]}], "+", FormBox[ RowBox[{ RowBox[{ RowBox[{ FormBox[\(a\_\(2 m\)\^+\), "TraditionalForm"], FormBox["", "TraditionalForm"]}], ")"}], ")"}], "TraditionalForm"]}], "TraditionalForm"], TraditionalForm]]], "\n\n", Cell[BoxData[ FormBox[ RowBox[{\(-\@2\), RowBox[{"(", RowBox[{ FractionBox[ RowBox[{ FormBox[\(b\_\(L\ n\)\), "TraditionalForm"], " ", "+", " ", FormBox[ RowBox[{\(b\_\(R\ n\)\), " ", "+", " ", FormBox[\(b\_\(L\ n\)\^+\), "TraditionalForm"], " ", "+", " ", FormBox[\(\(b\_\(R\ n\)\^+\)\(\ \)\), "TraditionalForm"]}], "TraditionalForm"]}], "4"], " ", "-", " ", RowBox[{ FractionBox[\(\(\@2\) \((\(-1\))\)\^\(n - 1\)\), RowBox[{"\[Pi]", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(2 n\ - \ 1\), "TraditionalForm"], ")"}], \(3\/2\)]}]], "P"}]}], ")"}]}], TraditionalForm]]], "=", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(\(\ \ \ \)\(\[Sum]\+\(m = 1\)\%\[Infinity]\)\), "TraditionalForm"], \(1\/\@\(2\ n\ - \ 1\)\), FormBox[ FormBox[\(\(B\_\(2 m, \