(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. 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[ 16118, 647]*) (*NotebookOutlinePosition[ 16822, 672]*) (* CellTagsIndexPosition[ 16778, 668]*) (*WindowFrame->Normal*) Notebook[{ Cell["Session #5", "Subtitle"], Cell[CellGroupData[{ Cell["The Physical Pendulum Problem", "Subsection"], Cell["\<\ The simple pendulum satisfies the nonlinear differential equation\ \ \>", "Text"], Cell[BoxData[ \(\(\(\[Theta]'\)'\)[t]\ \ == \ \(-Sin[\[Theta][t]]\)\)], "Input"], Cell[TextData[ "For small amplitudes \[Alpha] we expand the Sin[] to get the SHO. For \ larger amplitudes we know that the solution must be periodic in time, and \ that the period is of the form"], "Text"], Cell[TextData[ "2 \[Pi] f[\[Alpha]]=2 \[Pi](1+c \[Alpha]+d \[Alpha]^2+...)"], "DisplayFormula"], Cell[TextData[ "Where c and d are dimensionless constants we would like to find.\nAs a first \ pass, lets solve the equation numerically for a range of amplitudes \[Alpha], \ and fit the resulting period to extract the constant c."], "Text"], Cell[BoxData[ \(\(eqn\ = \ \(\(theta'\)'\)[t]\ == \ \(-Sin[theta[t]]\);\)\)], "Input",\ InitializationCell->True], Cell[BoxData[ \(\(bc1\ = \(theta'\)[0] == 0;\)\)], "Input", InitializationCell->True], Cell["Define the amplitude alpha here", "Text"], Cell[BoxData[ \(\(\n\(alpha\ = \ .1; \)\)\)], "Input"], Cell[BoxData[ \(\(soln\ = \ NDSolve[{eqn, bc1, theta[0] == alpha}, theta[t], {t, 0, 8}]; \)\)], "Input"], Cell[BoxData[ \(f[t_]\ = \ soln[\([1, 1, 2]\)]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(Plot[f[t], {t, 0, 8}]\)], "Input"], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(period\ = \ 4*\(FindRoot[f[t], {t, 1.5}]\)[\([1, 2]\)]\)], "Input"], Cell[BoxData[ \(6.28705045740466594`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\((period - 2\ Pi)\)/alpha^2\)], "Input"], Cell[BoxData[ \(0.386515022507971615`\)], "Output"] }, Open ]], Cell[TextData[StyleBox[ "Now lets put this all together in one function", "Text"]], "Text"], Cell[BoxData[ \(g[amp_]\ := \ Module[{z, x}, ff[t_]\ := \((NDSolve[{\(\(z'\)'\)[tt] == \(-Sin[z[tt]]\), \(z'\)[0] == 0, z[0] == amp}, z[tt], {tt, 0, 8}]\ /. \ tt -> t)\)[ \([1, 1, 2]\)]; \((\(FindRoot[ff[x], {x, 1.5}]\)[\([1, 2]\)])\)*4] \)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \({g[.1], \ g[.2]}\)], "Input"], Cell[BoxData[ \({6.28705045740466594`, 6.29889887250289159`}\)], "Output"] }, Open ]], Cell[BoxData[ \(Plot[\((g[a] - 2 Pi)\)/a^2, {a, .1, .5}, PlotRange -> {.35, .45}]\)], "Input"], Cell[TextData[ "So we learn that the period is\n T = 2\[Pi] (1 + .39 \[Alpha]^2)\nCan \ we get this constant analytically?\nHelpful insight: there is an integral of \ the motion, i.e. the energy:"], "Text"], Cell[TextData[{ "E = (Kinetic) + (Potential) =", Cell[BoxData[ \(1\/2\)]], " [\[Theta]']^2 - Cos[\[Theta]] = - Cos[\[Alpha]]" }], "DisplayFormula"], Cell[TextData[{ "Therefore the speed when at some angle \[Theta][t] is ", Cell[BoxData[ \(\@\(2 \((Cos[\[Theta]] - Cos[\[Alpha]])\)\)\)]], ", and the\nperiod is given exactly by the integral" }], "Text"], Cell[TextData[{ "T = ", Cell[BoxData[ \(4\/\@2\)]], " ", Cell[BoxData[ \(\[Integral]\_0 \%\[Alpha]\(\((Cos[\[Theta]] - Cos[\[Alpha]])\)\^\(\(-1\)/2\)\) \[DifferentialD]\[Theta]\)]] }], "DisplayFormula"], Cell[BoxData[ \(int\ := \ 4/Sqrt[2*\((Cos[th] - Sqrt[1 - Sin[alpha]^2])\)]\)], "Input"], Cell["\<\ Since straightforward integration doesn't help, try numerical:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Integrate[int, {th, 0, alpha}]\)], "Input"], Cell[BoxData[ \(2\ \@2\ \(\[Integral]\_0\%alpha\( 1\/\@\(Cos[th] - \@\(1 - Sin[alpha]\^2\)\)\) \[DifferentialD]th\ \)\)], "Output"] }, Open ]], Cell[BoxData[ \(alpha\ = \ . \)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(alpha\ = \ .1; \ NIntegrate[int, {th, 0, alpha}]\)], "Input"], Cell[BoxData[ \(6.28711385516261`\)], "Output"] }, Open ]], Cell[BoxData[ \(\((% - 2 Pi\ )\)/.1^2\)], "Input"], Cell["\<\ So we can get the same \"data\". Can we get the constant directly \ by expanding the integrand?\ \>", "Text"], Cell[BoxData[ \(ord = \ 10; alpha =. ; \nintxx\ = \ Normal[Series[int, {th, 0, ord}]]; \nq\ = \ Integrate[intxx, {th, 0, alpha}]; \)], "Input"], Cell[BoxData[ \(Series[q, {alpha, 0, 3}]\)], "Input"], Cell[BoxData[ \(N[%]\)], "Input"], Cell["\<\ OK this may work, but (a) it is too slow, and (b) in the end we \ still get just a number. Instead lets try changing variables \[Theta]=\[Alpha]x and seeing if we have \ more luck taking derivatives.\ \>", "Text"], Cell[BoxData[ \(int2\ := \ alpha*\((4/Sqrt[2])\)/Sqrt[Cos[x*alpha] - Cos[alpha]]\)], "Input"], Cell[BoxData[ \(Integrate[int2, {x, 0, 1}]\)], "Input"], Cell[BoxData[ \(\(\ NIntegrate[int2 /. alpha -> .1, {x, 0, 1}]\)\)], "Input"], Cell["\<\ The NIntegrate routine fails us! Curiously, the regular Integrate \ routine does better:\ \>", "Text"], Cell[BoxData[ \(Integrate[int2 /. alpha -> .1, {x, 0, 1}]\)], "Input"], Cell["\<\ OK lets just press on and expand the integrand. Remember to clear \ alpha first.\ \>", "Text"], Cell[BoxData[ \(alpha\ = \ .; \nq = Simplify[Series[int2, {alpha, 0, 12}]]\)], "Input"], Cell["Integrate the expansion term by term", "Text"], Cell[BoxData[ \(answer = Integrate[Normal[q], \ {x, 0, 1}]\)], "Input"], Cell[BoxData[ \(\(\nSimplify[answer]\)\)], "Input"], Cell[BoxData[ \(N[Pi/8]\)], "Input"], Cell[TextData[ "So there we have it: the .39 we \"measured\" earlier is in fact \[Pi]/8.\n\ Lets check how well the series does at alpha=.1: perfectly to the digits we \ have"], "Text"], Cell[BoxData[ \(N[answer\ /. \ alpha -> \ .1]\)], "Input"], Cell["\<\ The last thing is to guess the pattern in the coeffiecients. After \ some experimentation, I find that rewriting as a series in sa = Sin[alpha/2] \ simplifies the pattern:\ \>", "Text"], Cell[BoxData[ \(new\ = \ \(answer/2\)/Pi\ /. \ alpha -> Series[2*ArcSin[sa], {sa, 0, 12}]\)], "Input"], Cell[BoxData[ \(coeffs\ = \ Table[Sqrt[4^n\ Coefficient[new, sa^\((2\ n)\)]], {n, 1, 6}]\)], "Input"], Cell["\<\ OK the coefficients seem to be fractions N/D with N = 1 1*3 1*3*5 1*3*5*7 1*3*5*7*9 \ 1*3*5*7*9*11 D = 1 1*2 1*2*3 1*2*3*4 1*2*3*4*5 \ 1*2*3*4*5*6\ \>", "Text"], Cell[BoxData[ \(coeff[n_]\ := \ \((\(\(\((2\ n\ - 1)\)!!\)/\(n!\)\)/2^\ n)\)^2\)], "Input"], Cell[BoxData[ \(Table[coeff[n], {n, 0, 7}]\)], "Input"], Cell["So here is our best guess for the exact answer:", "Text"], Cell[BoxData[ \(pseries[th_, nterm_: 10]\ := \ 2\ Pi\ Sum[\ \((\((\ Sin[th/2]/2)\)^n\ \ \(\((2 n - 1)\)!!\)/\(n!\))\)^2, {n, 0, nterm}]\)], "Input"], Cell[BoxData[ \(pseries[x, 6]\)], "Input"], Cell["Check it out for alpha=2.0", "Text"], Cell[BoxData[ \(g[2.0]\)], "Input"], Cell[BoxData[ \(pseries[2] // N\)], "Input"], Cell[BoxData[ \(pseries[2, 20]\ // N\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Yet More Graphics", "Subsection"], Cell[CellGroupData[{ Cell["3D Graphics", "Subsubsection"], Cell[BoxData[ \(v\ = \ 1/Sqrt[\((x - 1)\)^2\ + \ y^2 + z^2]\ - \ 1/Sqrt[\((x + 1)\)^2 + y^2 + z^2]\)], "Input"], Cell[BoxData[ \(Plot3D[v /. z -> 1, {x, \(-3\), 3}, {y, \(-2\), 2}]\)], "Input"], Cell["Change the point of view:", "Text"], Cell[BoxData[ \(Plot3D[v /. z -> 1, {x, \(-3\), 3}, {y, \(-2\), 2}, ViewPoint -> {\(-1.271\), \ \(-1.665\), \ 2.657}]\)], "Input"], Cell["Contour plot", "Text"], Cell[BoxData[ \(Clear["\"]\)], "Input"], Cell[BoxData[ \(f\ = \ \((x^2 + y^2 - 3)\)^2 + \((\ \((x - 1)\)^2 + \((y + 1)\)^2\ - 3)\)^3\)], "Input"], Cell[BoxData[ \(q = Solve[{D[f, x] == 0, D[f, y] == 0}, {x, y}] // N\)], "Input"], Cell[BoxData[ \(Plot3D[f, {x, \(-1\), 2}, {y, \(-2\), 1}]\)], "Input"], Cell[BoxData[ \(DensityPlot[f, {x, \(-1\), 2}, {y, \(-2\), 1}]\)], "Input"], Cell[BoxData[ \(cplot\ = \ ContourPlot[f, {x, \(-1\), 2}, {y, \(-2\), 1}]\)], "Input"], Cell[BoxData[ \(lplot\ = \ ListPlot[{x, y} /. q, PlotStyle -> {PointSize[.03], RGBColor[1, 0, 0]}] \)], "Input"], Cell[BoxData[ \(Show[cplot, lplot]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Vector Fields", "Subsubsection"], Cell[BoxData[ \(\(?*Vector*\)\)], "Input"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(\(?*Vector*\)\)], "Input"], Cell[BoxData[ \(\(\(?PlotVectorField\)\n\)\)], "Input"], Cell[BoxData[ \(vec\ = \ {\ D[v, x], D[v, y]}\)], "Input"], Cell[BoxData[ \(PlotVectorField[vec /. z -> 1, {x, \(-3\), 3}, {y, \(-2\), 2}, \n\t ScaleFactor -> Automatic]\)], "Input"], Cell[BoxData[ \(PlotVectorField[vec /. z -> 1, {x, \(-3\), 3}, {y, \(-2\), 2}, \n\t ScaleFactor -> 1]\)], "Input"], Cell["Or vectors in 3D", "Text"], Cell[BoxData[ \(vec3\ = \ {D[v, x], D[v, y], D[v, z]}\)], "Input"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(PlotVectorField3D[vec3, {x, \(-3\), 3}, {y, \(-2\), 2}, {z, \(-2\), 2}] \)], "Input"], Cell[BoxData[ \(PlotGradientField3D[v, {x, \(-3\), 3}, {y, \(-2\), 2}, {z, \(-2\), 2}] \)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Surfaceof Revolution", "Subsubsection"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(f = x*\((1 - x)\)\)], "Input"], Cell[BoxData[ \(Plot[f, {x, 0, 1}]\)], "Input"], Cell[BoxData[ \(SurfaceOfRevolution[f, {x, 0, 1}]\)], "Input"], Cell[BoxData[ \(SurfaceOfRevolution[{Sin[x], x}, {x, 0, 4 Pi}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["3D Parametric Plot", "Subsubsection"], Cell[BoxData[ \(\(?Spher*\)\)], "Input"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(\(\(?Spher*\)\n\)\)], "Input"], Cell[BoxData[ \(\(plot[l_, m_]\ := \ SphericalPlot3D[\n\t\t \((Abs[SphericalHarmonicY[l, m, theta, phi]] // Evaluate)\), \n \t\t{theta, 0, Pi}, {phi, 0, 2\ Pi}, Boxed -> False, \n\t\t Axes -> False, DisplayFunction -> Identity]; \)\)], "Input"], Cell[BoxData[ \(\(pt1\ = \ Table[plot[el, m], {el, 0, 2}, {m, \(-el\), el}]; \)\)], "Input"], Cell[BoxData[ \(Show[GraphicsArray[pt1]]\)], "Input"] }, Closed]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Saving ", StyleBox["Mathematica", FontSlant->"Italic"], " Expressions" }], "Subsection"], Cell[CellGroupData[{ Cell[TextData[{ "Plain ", StyleBox["Mathematica", FontSlant->"Italic"], " expressions" }], "Subsubsection"], Cell["Send an expression out to a text file:", "Text"], Cell[BoxData[ \(Expand[\((1 + x)\)^10]\ >> \ outfile\)], "Input"], Cell["Append to the same file", "Text"], Cell[BoxData[ \(Factor[\((1 - x^2)\)]\ >>> \ outfile\)], "Input"], Cell["Take a look at the file from within Mathematica", "Text"], Cell[BoxData[ \(\(! 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