(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.1' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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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[ 20496, 567]*) (*NotebookOutlinePosition[ 21170, 590]*) (* CellTagsIndexPosition[ 21126, 586]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["The theorem of Stern and Stolz", "Title"], Cell["\<\ Documenting the computer part of the proof on page 220 and following of \ theorem 12.1. \ \>", "Subsubtitle"], Cell[BoxData[ \(p[{x_, y_}] := {x, y, 0}/\((x^2 + y^2 + 1)\) + {0, 0, 1} \((1 - 1/\((1 + x^2 + y^2)\))\)\)], "Input", InitializationCell->True], Cell[CellGroupData[{ Cell[BoxData[ \(p[{x, y}]\)], "Input"], Cell[BoxData[ \({x\/\(1 + x\^2 + y\^2\), y\/\(1 + x\^2 + y\^2\), 1 - 1\/\(1 + x\^2 + y\^2\)}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(p[Infinity] = {0, 0}; p[{0, 0}] = {0, 0}\)], "Input", InitializationCell->True], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(p[{0, 0}]\)], "Input"], Cell[BoxData[ \({0, 0}\)], "Output"] }, Open ]], Cell["\<\ p maps {x,y} to the sphere of radius 1/2 about (0,0,1/2), stereographic \ projection. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(p[{1, 0}]\)], "Input"], Cell[BoxData[ \({1\/2, 0, 1\/2}\)], "Output"] }, Open ]], Cell[BoxData[ \(rhosqrd[u_, v_] := \((p[u] - p[v])\) . \((p[u] - p[v])\)\)], "Input", InitializationCell->True], Cell["\<\ rhosqrd[u,v] gives the squared euclidean distance between the p-stereographic \ projections of u and v onto our spherical model of the complex plane with \ infinity. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(expn = rhosqrd[t {z1, z2} + {u1, u2}, t {z1, z2} + {v1, v2}]\)], "Input",\ InitializationCell->True], Cell[BoxData[ \(\((\(-\(1\/\(1 + \((u1 + t\ z1)\)\^2 + \((u2 + t\ z2)\)\^2\)\)\) + \ 1\/\(1 + \((v1 + t\ z1)\)\^2 + \((v2 + t\ z2)\)\^2\))\)\^2 + \((\(u1 + t\ \ z1\)\/\(1 + \((u1 + t\ z1)\)\^2 + \((u2 + t\ z2)\)\^2\) - \(v1 + t\ z1\)\/\(1 \ + \((v1 + t\ z1)\)\^2 + \((v2 + t\ z2)\)\^2\))\)\^2 + \((\(u2 + t\ z2\)\/\(1 \ + \((u1 + t\ z1)\)\^2 + \((u2 + t\ z2)\)\^2\) - \(v2 + t\ z2\)\/\(1 + \((v1 + \ t\ z1)\)\^2 + \((v2 + t\ z2)\)\^2\))\)\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(expnprime = \((D[expn, t])\) /. t \[Rule] 0; expn0 = expn /. t \[Rule] 0\)], "Input", InitializationCell->True], Cell[BoxData[ \(\((\(-\(1\/\(1 + u1\^2 + u2\^2\)\)\) + 1\/\(1 + v1\^2 + v2\^2\))\)\^2 + \ \((u1\/\(1 + u1\^2 + u2\^2\) - v1\/\(1 + v1\^2 + v2\^2\))\)\^2 + \((u2\/\(1 + \ u1\^2 + u2\^2\) - v2\/\(1 + v1\^2 + v2\^2\))\)\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(logprime = expnprime/expn0\)], "Input", InitializationCell->True], Cell[BoxData[ \(\((2\ \((\(-\(1\/\(1 + u1\^2 + u2\^2\)\)\) + 1\/\(1 + v1\^2 + v2\^2\))\)\ \((\(2\ u1\ z1 + 2\ u2\ \ z2\)\/\((1 + u1\^2 + u2\^2)\)\^2 - \(2\ v1\ z1 + 2\ v2\ z2\)\/\((1 + v1\^2 + \ v2\^2)\)\^2)\) + 2\ \((u1\/\(1 + u1\^2 + u2\^2\) - v1\/\(1 + v1\^2 + v2\^2\))\)\ \((z1\/\(1 + u1\^2 + u2\^2\) - z1\/\(1 + v1\^2 + v2\^2\) - \(u1\ \((2\ u1\ z1 + 2\ u2\ z2)\)\ \)\/\((1 + u1\^2 + u2\^2)\)\^2 + \(v1\ \((2\ v1\ z1 + 2\ v2\ z2)\)\)\/\((1 + \ v1\^2 + v2\^2)\)\^2)\) + 2\ \((u2\/\(1 + u1\^2 + u2\^2\) - v2\/\(1 + v1\^2 + v2\^2\))\)\ \((z2\/\(1 + u1\^2 + u2\^2\) - z2\/\(1 + v1\^2 + v2\^2\) - \(u2\ \((2\ u1\ z1 + 2\ u2\ z2)\)\ \)\/\((1 + u1\^2 + u2\^2)\)\^2 + \(v2\ \((2\ v1\ z1 + 2\ v2\ z2)\)\)\/\((1 + \ v1\^2 + v2\^2)\)\^2)\))\)/\((\((\(-\(1\/\(1 + u1\^2 + u2\^2\)\)\) + 1\/\(1 + \ v1\^2 + v2\^2\))\)\^2 + \((u1\/\(1 + u1\^2 + u2\^2\) - v1\/\(1 + v1\^2 + \ v2\^2\))\)\^2 + \((u2\/\(1 + u1\^2 + u2\^2\) - v2\/\(1 + v1\^2 + \ v2\^2\))\)\^2)\)\)], "Output"] }, Open ]], Cell["\<\ logprime gives the logarithmic derivative with respect to t of the distance \ between u+t z and v+t z at t=0. This expression is complicated as it comes \ out of a direct evaluation from the definition, but it simplifies \ drastically, to the expression immediately below, we claim. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(claimed = \(-2\) \(({z1, z2} . {u1, u2}/\((1 + u1^2 + u2^2)\) + {z1, z2} . {v1, v2}/\((1 + v1^2 + v2^2)\))\)\)], "Input"], Cell[BoxData[ \(\(-2\)\ \((\(u1\ z1 + u2\ z2\)\/\(1 + u1\^2 + u2\^2\) + \(v1\ z1 + v2\ \ z2\)\/\(1 + v1\^2 + v2\^2\))\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[logprime - claimed]\)], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], "'s simplification engine reports that the two sides are identical. " }], "Text"], Cell[TextData[{ "The rest of this is scratch paper, indicating, if anyone is interested, \ some of the preliminary calculations enroute to the packaged verification. \ The cell below is closed, but can be opened from the ", StyleBox["Mathematica", FontSlant->"Italic"], " menus: cell/cell properties/open\n" }], "Text"], Cell[TextData[{ "\n\n", Cell[BoxData[ \(\(?D\)\)], "Input"], ";\n", Cell[BoxData[ RowBox[{"\<\"D[f, x] gives the partial derivative of f with respect to \ x. D[f, {x, n}] gives the nth partial derivative of f with respect to x. 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1\/\(1 + \((3\ \ t + z)\)\^2 + \((4\ t + z)\)\^2\))\)\^2 + \((\(-\(\(3\ t + w\)\/\(1 + \((3\ t \ + w)\)\^2 + \((4\ t + w)\)\^2\)\)\) + \(3\ t + z\)\/\(1 + \((3\ t + z)\)\^2 + \ \((4\ t + z)\)\^2\))\)\^2 + \((\(-\(\(4\ t + w\)\/\(1 + \((3\ t + w)\)\^2 + \ \((4\ t + w)\)\^2\)\)\) + \(4\ t + z\)\/\(1 + \((3\ t + z)\)\^2 + \((4\ t + \ z)\)\^2\))\)\^2\)], "Output"], "\n", Cell[BoxData[ \(g[z_, w_, b_, t_] := D[f[z, w, b, t], t] /. t \[Rule] 0\)], "Input"], ";\n", Cell[BoxData[ \(g[z, w, b, t]\)], "Input"], ";\n", Cell[BoxData[ RowBox[{ RowBox[{\((\(-p[w]\) + p[z])\), ".", RowBox[{"(", RowBox[{ RowBox[{\(-b\), " ", RowBox[{ SuperscriptBox["p", "\[Prime]", MultilineFunction->None], "[", "w", "]"}]}], "+", RowBox[{"b", " ", RowBox[{ SuperscriptBox["p", "\[Prime]", MultilineFunction->None], "[", "z", "]"}]}]}], ")"}]}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(-b\), " ", RowBox[{ SuperscriptBox["p", "\[Prime]", MultilineFunction->None], "[", "w", "]"}]}], "+", RowBox[{"b", " ", RowBox[{ SuperscriptBox["p", "\[Prime]", MultilineFunction->None], "[", "z", "]"}]}]}], ")"}], ".", \((\(-p[w]\) + p[z])\)}]}]], "Output"], "\n", Cell[BoxData[ \(g[{z1, z2}, {w1, w2}, {b1, b2}, t]\)], "Input"], ";\n", Cell[BoxData[ \(2\ \((\(-\(\(2\ b1\ w1 + 2\ b2\ w2\)\/\((1 + w1\^2 + w2\^2)\)\^2\)\) + \(2\ b1\ \ z1 + 2\ b2\ z2\)\/\((1 + z1\^2 + z2\^2)\)\^2)\)\ \((1\/\(1 + w1\^2 + w2\^2\) \ - 1\/\(1 + z1\^2 + z2\^2\))\) + 2\ \((\(w1\ \((2\ b1\ w1 + 2\ b2\ w2)\)\)\/\((1 + w1\^2 + w2\^2)\)\^2 \ - b1\/\(1 + w1\^2 + w2\^2\) - \(z1\ \((2\ b1\ z1 + 2\ b2\ z2)\)\)\/\((1 + \ z1\^2 + z2\^2)\)\^2 + b1\/\(1 + z1\^2 + z2\^2\))\)\ \((\(-\(w1\/\(1 + w1\^2 + w2\^2\)\)\) + z1\/\(1 + z1\^2 + z2\^2\))\) + 2\ \((\(w2\ \((2\ b1\ w1 + 2\ b2\ w2)\)\)\/\((1 + w1\^2 + w2\^2)\)\^2 \ - b2\/\(1 + w1\^2 + w2\^2\) - \(z2\ \((2\ b1\ z1 + 2\ b2\ z2)\)\)\/\((1 + \ z1\^2 + z2\^2)\)\^2 + b2\/\(1 + z1\^2 + z2\^2\))\)\ \((\(-\(w2\/\(1 + w1\^2 + w2\^2\)\)\) + z2\/\(1 + z1\^2 + z2\^2\))\)\)], "Output"], "\n", Cell[BoxData[ \(dg = %\)], "Input"], ";\n", Cell[BoxData[ \(2\ \((\(-\(\(2\ b1\ w1 + 2\ b2\ w2\)\/\((1 + w1\^2 + w2\^2)\)\^2\)\) + \(2\ b1\ \ z1 + 2\ b2\ z2\)\/\((1 + z1\^2 + z2\^2)\)\^2)\)\ \((1\/\(1 + w1\^2 + w2\^2\) \ - 1\/\(1 + z1\^2 + z2\^2\))\) + 2\ \((\(w1\ \((2\ b1\ w1 + 2\ b2\ w2)\)\)\/\((1 + w1\^2 + w2\^2)\)\^2 \ - b1\/\(1 + w1\^2 + w2\^2\) - \(z1\ \((2\ b1\ z1 + 2\ b2\ z2)\)\)\/\((1 + \ z1\^2 + z2\^2)\)\^2 + b1\/\(1 + z1\^2 + z2\^2\))\)\ \((\(-\(w1\/\(1 + w1\^2 + w2\^2\)\)\) + z1\/\(1 + z1\^2 + z2\^2\))\) + 2\ \((\(w2\ \((2\ b1\ w1 + 2\ b2\ w2)\)\)\/\((1 + w1\^2 + w2\^2)\)\^2 \ - b2\/\(1 + w1\^2 + w2\^2\) - \(z2\ \((2\ b1\ z1 + 2\ b2\ z2)\)\)\/\((1 + \ z1\^2 + z2\^2)\)\^2 + b2\/\(1 + z1\^2 + z2\^2\))\)\ \((\(-\(w2\/\(1 + w1\^2 + w2\^2\)\)\) + z2\/\(1 + z1\^2 + z2\^2\))\)\)], "Output"], "\n", Cell[BoxData[ \(dg[\([\)\(1\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(2\ \((\(-\(\(2\ b1\ w1 + 2\ b2\ w2\)\/\((1 + w1\^2 + w2\^2)\)\^2\)\) + \(2\ b1\ z1 \ + 2\ b2\ z2\)\/\((1 + z1\^2 + z2\^2)\)\^2)\)\ \((1\/\(1 + w1\^2 + w2\^2\) - 1\/\(1 + z1\^2 + z2\^2\))\)\)], "Output"], "\n", Cell[BoxData[ \(dg[\([\)\(2\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(2\ \((\(w1\ \((2\ b1\ w1 + 2\ b2\ w2)\)\)\/\((1 + w1\^2 + w2\^2)\)\^2 \ - b1\/\(1 + w1\^2 + w2\^2\) - \(z1\ \((2\ b1\ z1 + 2\ b2\ z2)\)\)\/\((1 + \ z1\^2 + z2\^2)\)\^2 + b1\/\(1 + z1\^2 + z2\^2\))\)\ \((\(-\(w1\/\(1 + w1\^2 + w2\^2\)\)\) + z1\/\(1 + z1\^2 + z2\^2\))\)\)], "Output"], "\n", Cell[BoxData[ \(dg[\([\)\(3\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(2\ \((\(w2\ \((2\ b1\ w1 + 2\ b2\ w2)\)\)\/\((1 + w1\^2 + w2\^2)\)\^2 \ - b2\/\(1 + w1\^2 + w2\^2\) - \(z2\ \((2\ b1\ z1 + 2\ b2\ z2)\)\)\/\((1 + \ z1\^2 + z2\^2)\)\^2 + b2\/\(1 + z1\^2 + z2\^2\))\)\ \((\(-\(w2\/\(1 + w1\^2 + w2\^2\)\)\) + z2\/\(1 + z1\^2 + z2\^2\))\)\)], "Output"], "\n", Cell[BoxData[ \(Together[dg]\)], "Input"], ";\n", Cell[BoxData[ \(\(-\(\((2\ \((b1\ w1\^3 + b2\ w1\^2\ w2 + b1\ w1\ w2\^2 + b2\ w2\^3 - b1\ w1\^2\ z1 + b1\ w1\^4\ z1 - 2\ b2\ w1\ w2\ z1 + b1\ w2\^2\ z1 + 2\ b1\ w1\^2\ w2\^2\ z1 + b1\ w2\^4\ z1 - b1\ w1\ z1\^2 - b1\ w1\^3\ z1\^2 + b2\ w2\ z1\^2 + b2\ w1\^2\ w2\ z1\^2 - b1\ w1\ w2\^2\ z1\^2 + b2\ w2\^3\ z1\^2 + b1\ z1\^3 - b1\ w1\^2\ z1\^3 - 2\ b2\ w1\ w2\ z1\^3 + b1\ w2\^2\ z1\^3 + b1\ w1\ z1\^4 + b2\ w2\ z1\^4 + b2\ w1\^2\ z2 + b2\ w1\^4\ z2 - 2\ b1\ w1\ w2\ z2 - b2\ w2\^2\ z2 + 2\ b2\ w1\^2\ w2\^2\ z2 + b2\ w2\^4\ z2 - 2\ b2\ w1\ z1\ z2 - 2\ b2\ w1\^3\ z1\ z2 - 2\ b1\ w2\ z1\ z2 - 2\ b1\ w1\^2\ w2\ z1\ z2 - 2\ b2\ w1\ w2\^2\ z1\ z2 - 2\ b1\ w2\^3\ z1\ z2 + b2\ z1\^2\ z2 + b2\ w1\^2\ z1\^2\ z2 - 2\ b1\ w1\ w2\ z1\^2\ z2 - b2\ w2\^2\ z1\^2\ z2 + b1\ w1\ z2\^2 + b1\ w1\^3\ z2\^2 - b2\ w2\ z2\^2 - b2\ w1\^2\ w2\ z2\^2 + b1\ w1\ w2\^2\ z2\^2 - b2\ w2\^3\ z2\^2 + b1\ z1\ z2\^2 - b1\ w1\^2\ z1\ z2\^2 - 2\ b2\ w1\ w2\ z1\ z2\^2 + b1\ w2\^2\ z1\ z2\^2 + 2\ b1\ w1\ z1\^2\ z2\^2 + 2\ b2\ w2\ z1\^2\ z2\^2 + b2\ z2\^3 + b2\ w1\^2\ z2\^3 - 2\ b1\ w1\ w2\ z2\^3 - b2\ w2\^2\ z2\^3 + b1\ w1\ z2\^4 + b2\ w2\ z2\^4)\))\)/\((\((1 + w1\^2 + w2\^2)\)\^2\ \((1 + \ z1\^2 + z2\^2)\)\^2)\)\)\)\)], "Output"], "\n", Cell[BoxData[ \(dd = distsqd[p[{z1, z2}], p[{w1, w2}]]\)], "Input"], ";\n", Cell[BoxData[ \(\((1\/\(1 + w1\^2 + w2\^2\) - 1\/\(1 + z1\^2 + z2\^2\))\)\^2 + \((\(-\ \(w1\/\(1 + w1\^2 + w2\^2\)\)\) + z1\/\(1 + z1\^2 + z2\^2\))\)\^2 + \ \((\(-\(w2\/\(1 + w1\^2 + w2\^2\)\)\) + z2\/\(1 + z1\^2 + z2\^2\))\)\^2\)], "Output"], "\n", Cell[BoxData[ \(Together[dd]\)], "Input"], ";\n", Cell[BoxData[ \(\(w1\^2 + w2\^2 - 2\ w1\ z1 + z1\^2 - 2\ w2\ z2 + z2\^2\)\/\(\((1 + \ w1\^2 + w2\^2)\)\ \((1 + z1\^2 + z2\^2)\)\)\)], "Output"], "\n", Cell[BoxData[ \(Together[dg/dd]\)], "Input"], ";\n", Cell[BoxData[ \(\(-\(\((2\ \((b1\ w1 + b2\ w2 + b1\ z1 + b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b2\ z2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2)\))\)/\((\((1 + w1\^2 + w2\^2)\)\ \((1 + z1\^2 + z2\^2)\))\)\)\)\)], "Output"], "\n", Cell[BoxData[ \(wr = \((1 + w1^2 + w2^2)\); wz = 1 + z1^2 + z2^2; \)], "Input"], "\n", Cell[BoxData[ \(\(Simplify[ dist[p[{z1, z2}], p[{w1, w2}]] - \((\((w1 - z1)\)^2 + \((w2 - z2)\)^2)\)/\((wr\ zr)\)];\)\)], "Input"], "\n", Cell[BoxData[ \(\(-\(\(\((w1 - z1)\)\^2 + \((w2 - z2)\)\^2\)\/\(\((1 + w1\^2 + w2\^2)\)\ zr\)\)\) + dist[{z1\/\(1 + z1\^2 + z2\^2\), z2\/\(1 + z1\^2 + z2\^2\), 1 - 1\/\(1 + z1\^2 + z2\^2\)}, {w1\/\(1 + w1\^2 + w2\^2\), w2\/\(1 + w1\^2 + w2\^2\), 1 - 1\/\(1 + w1\^2 + w2\^2\)}]\)], "Output"], "\n", Cell[BoxData[ \(rg = Together[dg/dd]\)], "Input"], ";\n", Cell[BoxData[ \(\(-\(\((2\ \((b1\ w1 + b2\ w2 + b1\ z1 + b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b2\ z2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2)\))\)/\((\((1 + w1\^2 + w2\^2)\)\ \((1 + z1\^2 + z2\^2)\))\)\)\)\)], "Output"], "\n", Cell[BoxData[ \(rg[\([\)\(1\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(\(-2\)\)], "Output"], "\n", Cell[BoxData[ \(rg[\([\)\(2\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(1\/\(1 + w1\^2 + w2\^2\)\)], "Output"], "\n", Cell[BoxData[ \(rg[\([\)\(3\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(1\/\(1 + z1\^2 + z2\^2\)\)], "Output"], "\n", Cell[BoxData[ \(rg[\([\)\(4\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1 + b2\ w2 + b1\ z1 + b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b2\ z2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2\)], "Output"], "\n", Cell[BoxData[ \(rg4 = rg[\([\)\(4\)\(]\)]\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1 + b2\ w2 + b1\ z1 + b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b2\ z2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2\)], "Output"], "\n", Cell[BoxData[ \(Length[rg]\)], "Input"], ";\n", Cell[BoxData[ \(4\)], "Output"], "\n", Cell[BoxData[ \(rg4 - {b1, b2} . \(({w1, w2} + {z1, z2})\)\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1 + b2\ w2 + b1\ z1 + b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 - b1\ \((w1 + z1)\) + b2\ z2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2 - b2\ \((w2 + z2)\)\)], "Output"], "\n", Cell[BoxData[ \(Expand[%]\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2\)], "Output"], "\n", Cell[BoxData[ \(% - {b1, b2} . {z1, z2} \(({w1, w2} . {w1, w2})\)\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1\^2\ z1 + b1\ w2\^2\ z1 + b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b2\ w1\^2\ z2 + b2\ w2\^2\ z2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2 - \((w1\^2 + w2\^2)\)\ \((b1\ z1 + b2\ z2)\)\)], "Output"], "\n", Cell[BoxData[ \(Expand[%]\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2\)], "Output"], "\n", Cell[BoxData[ \(% - {b1, b2} . {w1, w2} \(({z1, z2} . {z1, z2})\)\)], "Input"], ";\n", Cell[BoxData[ \(b1\ w1\ z1\^2 + b2\ w2\ z1\^2 + b1\ w1\ z2\^2 + b2\ w2\ z2\^2 - \((b1\ w1 + b2\ w2)\)\ \((z1\^2 + z2\^2)\)\)], "Output"], "\n", Cell[BoxData[ \(Expand[%]\)], "Input"], ";\n", Cell[BoxData[ \(0\)], "Output"], "\nThe point of this is that the logarithmic derivative of the spherical \ distance between points is just -(b.w)/(1+|w|^2)-(b.z)/(1+|z|^2) and this is \ at worst -|b|. Thus, the change in the log of the spherical distance is at \ most Sum[|b_j|,{j,1,N}]. Hence, if the sum of the norms is convergent, the \ continued fractions starting with 0 and Infinity never come together, but \ remain separated by some fraction of their original distance. \n\n", Cell[BoxData[ \(f[{2, 1}, {3, 1}, {0, \(-1\)}, t]\)], "Input"], ";\n", Cell[BoxData[ \(\((2\/\(5 + \((1 - t)\)\^2\) - 3\/\(10 + \((1 - t)\)\^2\))\)\^2 + \((\ \(-\(1\/\(5 + \((1 - t)\)\^2\)\)\) + 1\/\(10 + \((1 - t)\)\^2\))\)\^2 + \ \((\(1 - t\)\/\(5 + \((1 - t)\)\^2\) - \(1 - t\)\/\(10 + \((1 - \ t)\)\^2\))\)\^2\)], "Output"], "\n", Cell[BoxData[ \(pdistsqd[{2, 1}, {3, 1}]\)], "Input"], ";\n", Cell[BoxData[ \(1\/66\)], "Output"], "\n", Cell[BoxData[ \(f[{2, 1}, {3, 1}, {0, \(-1\)}, t] - f[{2, 1}, {3, 1}, {0, \(-1\)}, 0]\)], "Input"], ";\n", Cell[BoxData[ \(\(-\(1\/66\)\) + \((2\/\(5 + \((1 - t)\)\^2\) - 3\/\(10 + \((1 - t)\)\ \^2\))\)\^2 + \((\(-\(1\/\(5 + \((1 - t)\)\^2\)\)\) + 1\/\(10 + \((1 - \ t)\)\^2\))\)\^2 + \((\(1 - t\)\/\(5 + \((1 - t)\)\^2\) - \(1 - t\)\/\(10 + \ \((1 - t)\)\^2\))\)\^2\)], "Output"] }], "Text", CellOpen->False] }, Open ]] }, FrontEndVersion->"5.1 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 685}}, AutoGeneratedPackage->None, WindowSize->{495, 527}, WindowMargins->{{218, Automatic}, {58, Automatic}} ] (******************************************************************* Cached data follows. 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