(*********************************************************************** 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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That is, \ include\ncomments where appropriate, and use the Format menus to distinguish \ text from headings\nfrom actual computation. It might work best just to do \ all the work and then go back and insert comments, format things right, etc. \ Also, remember that when the instructions don't explicitly require the use of \ ", StyleBox["Mathematica,", FontSlant->"Italic"], " you're free to use pencil and paper to do some of the work--whatever gets \ the job done most effectively." }], "Text"], Cell[CellGroupData[{ Cell["Problem 2.", "Subsection"], Cell[CellGroupData[{ Cell["a. (2 points)", "Subsubsection"], Cell[BoxData[ \(soln[x_]\ = \ First[y[x] /. DSolve[{\(y'\)[x]\ + \ y[x]/x == 2, \ y[1] == c}, y[x], x]]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["b. (2 points)", "Subsubsection"], Cell[BoxData[ \(\(Plot[ Evaluate[Table[soln[x], {c, .8, 1.2, .1}]], {x, .75, 1.25}];\)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["c. (2 points)", "Subsubsection"], Cell[BoxData[ \(Table[soln[10^n], {n, \(-2\), 1}, {c, .8, 1.2, .1}]\)], "Input"], Cell[BoxData[ \(% // TableForm\)], "Input"], Cell[BoxData[{ \(TextForm\`The\ columns\ correspond\ to\ c\ = \ .8, \ .9, \ 1, \ 1.1, \ 1.2\ going\ from\ left\), \(TextForm \`to\ right . \ \ The\ rows\ correspond\ to\ x\ = \ \(10\^n = \ .01\), \ .1, \ 1, \ 10\ going\), \(TextForm\`from\ top\ to\ \(bottom . \)\)}], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (4 points)", "Subsubsection"], Cell[BoxData[ \(\(Plot[ Evaluate[Table[soln[x], {c, .8, 1.2, .1}]], {x, 0, 2.5}];\)\)], "Input"], Cell["\<\ All solutions are asymptotic to the same line (looks like y=x) as x goes to infinity--regardless of initial condition. Solutions with initial value greater than one tend to infinity as x goes to zero. Those with initial value less than one tend to negative infinity. The solution with initial value one tends to zero as x goes to zero.\ \>", "Text"], Cell[BoxData[ \(\(\n\n\n\n\n\n\n\)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 4.", "Subsection"], Cell[CellGroupData[{ Cell["a. (2 points) ", "Subsubsection"], Cell[BoxData[{ \(Clear[y]\n\), "\n", \(DSolve[\(y'\)[ x]\ == \ \((x\ - \ Exp[\(-x\)])\)/\((y[x] + Exp[y[x]])\), \n\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y[x], x]\)}], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["b. (2 points) ", "Subsubsection"], Cell["\<\ I'll cut and paste the left hand side and give it a convenient name (it's important to remove the [x]'s from the y[x]'s):\ \>", "Text"], Cell[BoxData[ \(lhs[x_, y_]\ = \ Simplify[\ 1\/2\ E\^\(-x\)\ \((\(-2\) + 2\ E\^\(x + y\) - E\^x\ x\^2 + E\^x\ y\^2)\)]\)], "Input"], Cell[BoxData[ \(\(ContourPlot[lhs[x, y], {x, \(-1\), 3}, {y, \(-2\), 2}, ContourShading -> False];\)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["c. (3 points) ", "Subsubsection"], Cell[BoxData[ \(\(plot1\ = ContourPlot[lhs[x, y], {x, \(-1\), 3}, {y, \(-2\), 2}, Contours\ -> \ {lhs[1.5, .5]}, \ ContourShading -> False];\)\)], "Input"], Cell["\<\ The top curve is the graph of the solution since it (evidently) passes through the point (1.5,.5). \t\t\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (3 points) ", "Subsubsection"], Cell["\<\ I'll use the previous picture to take a guess at the correct value of y and then use FindRoot to improve the guess: \ \>", "Text"], Cell[BoxData[{ \(FindRoot[lhs[0, y]\ == \ lhs[1.5, .5], {y, .5}]\), "\n", \(FindRoot[lhs[1, y]\ == \ lhs[1.5, .5], {y, .5}]\), "\n", \(FindRoot[lhs[1.8, y]\ == \ lhs[1.5, .5], {y, .5}]\), "\n", \(FindRoot[lhs[2.1, y]\ == \ lhs[1.5, .5], {y, .5}]\)}], "Input"], Cell["\<\ That is: \t y(0) = .318386... y(1) = .235633... \t y(1.8) = .682187... y(2.1) = .866212...\ \>", "Text"], Cell[BoxData[ \(\(Show[plot1, Graphics[{PointSize[0.03], Point[{0, .318386}], Point[{1, .235633}], \n\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Point[{1.8, .682187}], \ Point[{2.1, .866212}]}]];\)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 10.", "Subsection"], Cell[CellGroupData[{ Cell["a. (4 points)", "Subsubsection"], Cell["\<\ Before I plot the vector field, I'd like to check the graph of the right side of the ODE as a function of y. This will give me some idea of what vertical range to use. Since the equation is autonomous, the horizontal range is unimportant.\ \>", "Text"], Cell[BoxData[ \(\(Plot[3 Sin[y] + y - 2, {y, \(-10\), 10}];\)\)], "Input"], Cell["\<\ looks like I should expect equilibrium solutions at about y=1/2, y=4 and y=5. Therefore, I'll choose [-1,6] as a range for y.\ \>", "Text"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(\(PlotVectorField[{1, 3 Sin[y]\ + \ y\ - \ 2}, {t, 0, 3}, {y, \(-1\), 6}, Axes -> \ True, \n\t ScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15];\)\)], "Input"], Cell["To find the equilibrium solutions:", "Text"], Cell[BoxData[{ \(FindRoot[3 Sin[y] + y - 2 == 0, {y, .5}]\), "\n", \(FindRoot[3 Sin[y] + y - 2 == 0, {y, 4}]\), "\n", \(FindRoot[3 Sin[y] + y - 2 == 0, {y, 5}]\n\)}], "Input"], Cell["\<\ From this and the two plots above, it is apparent that the ODE has two unstable equilibrium solutions: y= 4.9295 and y=.517049 and one stable equilibrium solution: y= 3.77452.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b. (4 points)", "Subsubsection"], Cell["\<\ This ODE is not autonomous, so I cannot proceed as I did in the \ last example. I'll have to get along with just the direction field. Note that the right side vanishes (i.e. the direction field is horizontal) when y=0 and y=x.\ \>", "Text"], Cell[BoxData[ \(\(PlotVectorField[{1, y^2\ - x*y}, {x, \(-4\), 4}, {y, \(-4\), 4}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15];\)\)], "Input"], Cell["\<\ Looks like y=0 is a constant solution of the ODE--an observation that is easily verified by plugging into the ODE. All solutions y[x] starting below y= 0 appear to converge to y=0 as x goes to infinity. Solutions y[x] starting above y= 0 appear to converge to either infinity or 0. Solutions exhibiting these two types of behavior appear to be separated by a curve asymptotic to the line y=x as x goes to infinity and y=0 as x goes to negative infinity. At any rate it is clear from the right side of the differential equation that the dividing curve must lie at or above the line y=x, because the right side is negative when y < x (and y > 0) so that solutions lying between y=0 and y=x will slope downward as x becomes large (and positive). \t \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (4 points)", "Subsubsection"], Cell[BoxData[ \(DSolve[{\(y'\)[x] == \ y[x]^2\ - \ x\ *y[x], y[0]\ == \ c}, y[x], x]\)], "Input"], Cell["What the *&^&* is this Erf thing?", "Text"], Cell[BoxData[ \(\(Plot[Erf[x/Sqrt[2]], {x, \(-10\), 10}];\)\)], "Input"], Cell[BoxData[{ FormBox[\(Doesn' t\ really\ matter, \ I\ guess . \ \ The\ point\ is\ that\ Erf[x/\@2]\), TextForm], "\n", FormBox[ RowBox[{ FormBox[\(increases\ from\ - 1\ to\ 1\ as\ x\ goes\ from\ negative\ infinity\ to\), "TextForm"], " "}], TextForm], "\n", FormBox[ RowBox[{ FormBox[\(infinity . \ Therefore, \ if\ c = y[0]\ is\ less\ than\ \@\(2/\[Pi]\)\ then\ it' s\ clear\), "TextForm"], " "}], TextForm], "\n", FormBox[\(that\ y[x]\ is\ defined\ for\ all\ x\ \((that\ is, \ the\ denominator\ never\nvanishes)\) and\ that\ y[x]\ converges\ to\ zero\ as\ x\ goes\ to\ infinity\), TextForm], "\n", FormBox[\(\((because\ the\ numerator\ of\ y[ x]\ definitely\ goes\ to\ zero\ while\ the\n denominator\ doesn' t)\) . \ \ If\ y[0]\ > \@\(2/\[Pi]\), \ then\ at\ some\ finite\), TextForm], "\n", FormBox[\(x\ the\ denominator\ vanishes\ and\ the\ solution\ goes\ to\ \ infinity\ along\), TextForm], "\n", FormBox[\(a\ vertical\ asymptote\ \((this\ is\ different\ than\ what\ I\ \ predicted\ from\ the\ \ndirection\ field)\) . \ \ It' s\ harder\ to\ tell\ just\ what\ happens\ if\), TextForm], "\n", FormBox[\(y[0]\ = \ \@\(2/\[Pi]\), \ so\ I\ plot\ that\ particular\ \(\(solution\)\(:\)\)\), TextForm]}], "Text"], Cell[BoxData[ \(\(Plot[\(-\(\(2\ \@\(2/\[Pi]\)\ E\^\(-\(x\^2\/2\)\)\)\/\(\(-2\) + \ \(\@\(2/\[Pi]\)\) \@\(2\ \[Pi]\)\ Erf[x\/\@2]\)\)\), {x, 0, 7}];\)\)], "Input"], Cell["\<\ This looks pretty close to a straight line, so I compare with the \ line y=x that looked important in the direction field:\ \>", "Text"], Cell[BoxData[ \(\(Plot[{\(-\(\(2\ \@\(2/\[Pi]\)\ E\^\(-\(x\^2\/2\)\)\)\/\(\(-2\) + \(\@\ \(2/\[Pi]\)\) \@\(2\ \[Pi]\)\ Erf[x\/\@2]\)\)\), x}, {x, 0, 7}];\)\)], "Input"], Cell["\<\ Sure enough, the \"dividing\" solution is asymptotic to the line \ y=x.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 12.", "Subsection"], Cell[CellGroupData[{ Cell["a. (3 points)", "Subsubsection"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(PlotVectorField[{1, \(-t\)*x^2}, {t, \(-3\), 3}, {x, \(-3\), 3}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 19]\)], "Input"], Cell["\<\ Looks like x = 0 is a constant solution.Moreover, it looks like solutions x[t] that start above x=0 will converge to zero as t goes to either positive or negative infinity,whereas solutions x[t] that start below x=0 will converge to negative infinity.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b. (3 points) ", "Subsubsection"], Cell["\<\ I'm going to get the solution to the ODE in a slightly different form than is asked for in the book.\ \>", "Text"], Cell[BoxData[ \(soln\ = \ First[x[t] /. DSolve[{\(x'\)[t] == \(-t\)*x[t]^2, x[0]\ == \ c}, x[t], t]]\)], "Input"], Cell[BoxData[ \(\(Plot[ Evaluate[Table[soln, {c, \(-3\), 3, 1}]], {t, \(-5\), 5}];\)\)], "Input"], Cell["\<\ So in fact, my predictions from the direction field weren't quite right. It's true that any solution satisfying x[0] >0 will converge to 0 as t goes to positive or negative infinity. However, any negative solution appears to have to vertical asymptotes and to go to negative infinity along both of them--i.e. solutions that start \ below x-axis go to negative infinity at some *finite* positive and negative values of t.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (4 points)", "Subsubsection"], Cell["\<\ Five types of solution curves: 1. x[t] < 0, defined for all t in some finite interval symmetric about the y-axis, and x[t] tends to negative infinity at each end of the interval. 2. x[t] = 0 for all t 3. x[t] > 0, defined for all t and tending to zero as t goes to \[PlusMinus]\ \[Infinity]. 4. x[t] > 0 everywhere decreasing, and defined for t in an interval of the \ form [a,\[Infinity]) where a > 0; asymptote at t = a. 5. x[t] > 0 everywhere, increasing and defined for t in an interval of the \ form (-\[Infinity],-a] where a > 0; asymptote at t = -a.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (2 points)", "Subsubsection"], Cell["\<\ This `Show' command is a little tricky--took some playing around to \ get a good picture.\ \>", "Text"], Cell[BoxData[ \(\(\(\ \ \)\(plot1\ = \ PlotVectorField[{1, \(-t\)\ *x^2}, {t, \(-3\), 3}, {x, \(-4\), 4}, \ PlotPoints\ -> \ 17, \n\t\ \ \ ScaleFunction \[Rule] \((1 &)\), Axes -> \ True, Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1];\)\)\)], "Input"], Cell[BoxData[ \(\(\(\t\ \)\(plot2\ = \ Plot[Evaluate[Table[soln, {c, \(-3\), 3, 1}]], {t, \(-3\), 3}, \n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ PlotRange -> {\(-4\), 4}, AspectRatio -> 1];\)\)\)], "Input"], Cell[BoxData[ \(\(Show[plot1, plot2, PlotRange -> {\(-4\), 4}, Frame -> False, Ticks -> None];\)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 16.", "Subsection"], Cell[CellGroupData[{ Cell["a. (3 points)", "Subsubsection"], Cell[BoxData[ \(pop[t_, a_, b_, c_]\ = \ Expand[First[ p[t] /. DSolve[\n\t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\(p'\)[t]\ == \ a\ p[t] + b, \ p[0]\ == \ c}, \n\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p[ t], t]]]\)], "Input"], Cell["\<\ The parameter c is simply the population of Nevada in 1950--i.e. \ the initial value. The parameter a determines the contribution to population growth coming from births - deaths , and b is the contribution to growth coming from (people moving in) - (people moving out).\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b (3 points)", "Subsubsection"], Cell[BoxData[ \(Solve[{pop[0, a, b, c] == 285.3, pop[10, a, b, c] == 488.7, \n\t\t\ \ \ \ \ \ \ \ \ \ \ \ pop[20, a, b, c] == 800.5}, {a, b, c}]\)], "Input"], Cell["\<\ Only the first in this list of solutions gives real values for a,b, and c. I'll use that solution.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (4 points)", "Subsubsection"], Cell[BoxData[ \(population[t_]\ = \ pop[t, 0.042718, \ 4.1162, 285.3]\)], "Input"], Cell["\<\ Now Icheck the population in 1950 and 1990 predicted by this model:\ \>", "Text"], Cell[BoxData[ \(population[\(-10\)]\)], "Input"], Cell[BoxData[ \(population[30]\)], "Input"], Cell["\<\ These are at least close to the true values. To predict when the \ population reaches three million, I'll guess that it'll be fifteen years after 1990 (since the \ population nearly tripled in the twenty years between 1970 and 1990).\ \>", "Text"], Cell[BoxData[ \(FindRoot[population[t] == 3000, {t, 45}]\)], "Input"], Cell["\<\ So I'm predicting a population of three million by 2009 Given the real data for 1990 and 1950 this model seems to be projecting too much population growth. I'd probably guess that it'll be a few years after 2009 when the population really does reach three million. Based on the formula above, it looks like I'd have to decrease both a and b to get a more accurate population model.\ \>", "Text"], Cell[BoxData[ \(\(Plot[population[t], {t, \(-10\), 90}];\)\)], "Input"], Cell["\<\ If the model is right, Nevada will get pretty crowded in time!\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 17.", "Subsection"], Cell[CellGroupData[{ Cell["a. (2 points)", "Subsubsection"], Cell["Mathematica is really overkill for this part...", "Text"], Cell[BoxData[ \(Solve[\((\[Alpha] - 1)\) x - \ x^3\ == \ 0, x]\)], "Input"], Cell[TextData[ "That is, we get three real roots when \[Alpha]>1, and only one real\nroot \ when \[Alpha]<1. When \[Alpha]=1, we also get one real root, but with\n\ multiplicity three."], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b. (2 points)", "Subsubsection"], Cell[BoxData[ \(\(Table[ PlotVectorField[\n\t\t{1, \((\[Alpha] - 1)\) x - \ x^3}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\t ScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15], {\[Alpha], \(-2\), 0}];\)\)], "Input"], Cell["\<\ The picture is more or less the same in all three cases: x=0 is a stable equilibrium solution.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (2 points)", "Subsubsection"], Cell[BoxData[ \(PlotVectorField[\n\t\t{1, \(-x^3\)}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15]\)], "Input"], Cell["x=0 is still a (barely) stable equilibrium.", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (3 points)", "Subsubsection"], Cell[BoxData[ \(PlotVectorField[\n\t\t{1, .5 x\ - \ x^3}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\t ScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 17]\)], "Input"], Cell[BoxData[{ \(TextForm\`Now\ x = 0\ is\ an\ unstable\ equilibrium, \ but\ there\ are\ two\ stable\ equilibrium\), \(TextForm\`solutions\ at\ x\ = \ \(\[PlusMinus]\@ .5\)\)}], "Text"], Cell[BoxData[ \(PlotVectorField[\n\t\t{1, x\ - \ x^3}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1 &)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 17]\)], "Input"], Cell["\<\ This is similar to the last case, except that the stable equilibria have migrated to x=\[PlusMinus]1.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["e. ( 3 points)", "Subsubsection"], Cell[TextData[ "When \[Alpha]<0, we had a single stable equilibrium solution at x=0.\nWhen \ \[Alpha] increased past 0, we suddenly had two stable equilibrium\nthat \ started just above and below x=0 and moved further from\nx=0 as \[Alpha] \ increased."], "Text"] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"4.0 for X", ScreenRectangle->{{0, 1280}, {0, 1024}}, WindowSize->{632, 833}, WindowMargins->{{164, Automatic}, {Automatic, 48}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "PostScriptOutputFile":>FrontEnd`FileName[{$RootDirectory, "home", "jeff", \ "teaching", "win00", "m230", "nb"}, "solved2.nb.ps", CharacterEncoding -> \ "ISO8859-1"], "Magnification"->1} ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. ***********************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1739, 51, 98, 4, 89, "Subsubtitle"], Cell[1840, 57, 738, 14, 122, "Text"], Cell[CellGroupData[{ Cell[2603, 75, 32, 0, 45, "Subsection"], Cell[CellGroupData[{ Cell[2660, 79, 38, 0, 42, "Subsubsection"], Cell[2701, 81, 153, 4, 27, "Input"] }, Open ]], Cell[CellGroupData[{ Cell[2891, 90, 38, 0, 42, "Subsubsection"], Cell[2932, 92, 125, 3, 27, "Input"] }, Open ]], Cell[CellGroupData[{ Cell[3094, 100, 38, 0, 42, "Subsubsection"], Cell[3135, 102, 86, 1, 27, "Input"], Cell[3224, 105, 47, 1, 27, "Input"], Cell[3274, 108, 309, 6, 64, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[3620, 119, 38, 0, 42, "Subsubsection"], Cell[3661, 121, 121, 3, 27, "Input"], Cell[3785, 126, 363, 7, 122, "Text"], Cell[4151, 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