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Notebook[{
Cell["Differential Equations Lab I", "Title"],
Cell[TextData[{
"This lab ",
StyleBox["explores some of the ODE solving capabilities of ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Mathematica",
Evaluatable->False,
AspectRatioFixed->True,
FontSlant->"Italic"],
StyleBox[
". The notebook is organized into several sections. Each section is just a \
group of ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Mathematica",
Evaluatable->False,
AspectRatioFixed->True,
FontSlant->"Italic"],
StyleBox[
" cells containing explanatory text like this and actual commands like the \
one below. To execute a command, click the mouse in the cell containing the \
command then do a Shift-Return, that is, hold down the shift key and hit the \
Return key. Try it on the cell below, which evaluates a complicated \
integral.",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text"],
Cell[BoxData[
\(Integrate[\((1 - 2*x^2)\)^\((\(-3\)/2)\), x]\)], "Input"],
Cell[TextData[{
"To begin the actual lab work you should open the first group of cells \
below by double-clicking on the bracket to the far right of the title ",
StyleBox["Finding Explicit Solutions with DSolve.",
FontWeight->"Bold"],
" The bracket will open up to reveal lots of interesting text and commands. \
Just follow through the section, reading, executing commands and (hopefully) \
thinking about the results. When you get done you can close up the section \
again by double clicking on the expanded bracket at the right edge of the \
cells and move on to the next section. Have fun !"
}], "Text"],
Cell[CellGroupData[{
Cell["Finding Explicit Solutions with DSolve", "Section"],
Cell[CellGroupData[{
Cell[TextData[{
StyleBox[
"To find the exact formula for the solution of simple one-variable ODE's, \
use the ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["DSolve", "Input",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[" ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[
"command. Of course, not all equations can be solved. When using this \
command, you have to be careful to use the double equal sign and to write all \
the y's as ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["y[t]", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[" in the equation.",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],
Cell["rules = DSolve[ y'[t] == (-1/2)*y[t] + Sin[t], y[t], t]", "Input",
AspectRatioFixed->True],
Cell[TextData[{
StyleBox[
"The answer is always a list of one or more subtitution rules. A list in ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Mathematica",
Evaluatable->False,
AspectRatioFixed->True,
FontSlant->"Italic"],
StyleBox[
" is always enclosed in set brackets like this { first thing, second thing \
,... }. In this case the list contains just one thing, a substitution rule. \
A substitution rule is indicated by an arrow, -> and it is also enclosed in \
set brackets. Hence the double set brackets in the output. The first (and \
in this case the only) element of the list of solution rules is called ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["rules[[1]]", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[".",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text"],
Cell["rules[[1]]", "Input"],
Cell[TextData[{
StyleBox[
"To extract the solution as a function for plotting or further analysis, \
you have to apply one of the substitution rules to the expression ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["y[t]",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[". You apply a rule in ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Mathematica",
Evaluatable->False,
AspectRatioFixed->True,
FontSlant->"Italic"],
StyleBox[" using the cryptic",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[" ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox["/.", "Input",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[" ", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[
" notation. Give the result another name, ygen, so you can use it \
later.",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],
Cell["ygen = y[t]/.rules[[1]]", "Input",
AspectRatioFixed->True],
Cell[TextData[{
StyleBox["Here ygen is the general solution with the arbitrary constant, ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["C[1]",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[
". Note that this is a linear ODE and so the general solution consists of \
a particular solution plus and exponential term.To pick out a particular \
solution, just apply another substitution rule, substituting a particular \
constant for ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["C[1]",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[".",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
ypart = ygen/.C[1]->0
Plot[ypart,{t,-10.0,10}];\
\>", "Input",
AspectRatioFixed->True],
Cell[TextData[{
StyleBox[
"To see lots of integral curves , you can plot a list of several solutions \
with different values of ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["C[1]",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[". One quick way to do this is to use the ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Table ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[
"command to generate the list of particular solutions. For some reason you \
need the ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Evaluate",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[" command inside the Plot command.",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
sollist = Table[ ygen/.C[1]->i/5,{i,-5,5}]
Plot[Evaluate[sollist],{t,-10,10}, PlotRange->{-2,2} ];\
\>", "Input",
AspectRatioFixed->True],
Cell[TextData[{
"You can use the mouse to make the plot bigger. Note that all solutions \
converge to the particular solution we plotted earlier. This is due to the \
exponential decay of the term involving ",
StyleBox["C[1]", "Input"],
". In a case like this you would refer to the particular solution as the ",
StyleBox["steady state",
FontWeight->"Bold"],
" solution. "
}], "Text",
Evaluatable->False,
AspectRatioFixed->True]
}, Open ]],
Cell[TextData[{
"Go back to the top of this section and try some other solvable ODE's. For \
example, \n\t\t",
StyleBox["y'[t] == y[t]^2", "Input"],
"\t or \t",
StyleBox["y'[t] == -t*y[t]", "Input"],
"\nJust replace the differential equation inside ",
StyleBox["DSolve",
FontWeight->"Bold"],
" with the new equation. You may also want to change the plotting ranges."
}], "Text"],
Cell[TextData[{
"Some other ODE's are too difficult to solve explicitly. Try the equation \
",
StyleBox["y'[t] == (t^2-y[t]^2)/(1+y[t]^2)", "Input"],
". The program just spits the problem back at you."
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["Finding Approximate Solutions with NDSolve", "Section"],
Cell[TextData[{
"For problems which cannot be solved explicitly you can still approximate \
the solutions numerically on any desired interval of time. For this you use \
the ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" function ",
StyleBox["NDSolve. ",
FontWeight->"Bold"],
"The following command brings up information about this command."
}], "Text"],
Cell["?NDSolve", "Input"],
Cell[TextData[{
"We will apply it to the ODE which ",
StyleBox["DSolve", "Input"],
" failed to solve in the last section."
}], "Text"],
Cell["\<\
rules = NDSolve[
\t\t\t{y'[t] == (t^2-y[t]^2)/(1+y[t]^2),y[0]==1},
\t\t\ty[t],{t,-5,5}
\t\t\t]\
\>", "Input"],
Cell[TextData[{
"As before, you get a list of substitution rules, this time not given as a \
formula but as a so-called ",
StyleBox["Interpolating Function",
FontWeight->"Bold"],
". This is some function (whose formulas are hidden from you) which \
approximates the solution over the given time interval. You plot it just as \
you did the particular solutions above."
}], "Text"],
Cell["\<\
ypart = y[t]/.rules[[1]];
Plot[ypart,{t,-5,5},PlotRange->All];\
\>", "Input"],
Cell[TextData[{
"To draw lots of curves, make a list of solutions with different initial \
conditions. This is a bit more complicated than just choosing different \
values for the arbitrary constant as we did above. We will start with an \
empty list and append solutions to it using the ",
StyleBox["Append",
FontWeight->"Bold"],
" command The solutions are obtained by repeating the process we just went \
through. We will make the machine loop through these steps automatically \
using a ",
StyleBox["Do",
FontWeight->"Bold"],
" loop which runs through the initial conditions ",
StyleBox["y[0]=-3,-2.5,-2,...,0,...,2.5,3", "Input"],
", that is, ",
StyleBox["y[0]=i/2", "Input"],
" for ",
StyleBox["i=-6 ", "Input"],
"to",
StyleBox[" 6", "Input"],
"."
}], "Text"],
Cell["\<\
sollist = {};
Do[rules = NDSolve[{y'[t]==(t^2-y[t]^2)/(1+y[t]^2),y[0]==i/2},y,{t,-5,5}];
AppendTo[sollist, y[t]/.rules[[1]]],
{i,-6,6}]
sollist\
\>", "Input"],
Cell["Finally we plot the solutions on the list.", "Text"],
Cell["Plot[ Evaluate[sollist], {t,-5,5}];", "Input"],
Cell[TextData[{
"Now see if you can make a picture of the integral curves of the equation ",
StyleBox["y'[t]==Sin[Pi*y[t]]", "Input"],
". Just go back two cells and make some changes. Can you explain the \
resulting picture by thinking about the phase line ?"
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["The Euler and Runge-Kutta Methods", "Section"],
Cell[TextData[{
StyleBox[
"Although the NDSolve command implements a complicated numerical method for \
you, it is easy and instructive to implement some of the simple methods \
discussed in class yourself. Below we define a function which carries out \
one step of Euler's method for any ODE. As input it takes the function ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["f[t,y] ", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["defining the ODE, the initial point ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["{t,y}", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[" and the step size ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["h", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[". As output it returns the point ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["{t+h,y+ h*f[t,y]}", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[".",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],
Cell["EulerStep[f_,{t_,y_},h_] := {t+h,y+h*f[t,y]}", "Input",
AspectRatioFixed->True],
Cell[TextData[{
"We can apply this method over and over again to step forward in time as \
far as we want. At each step we want to start at the point obtained from the \
previous step. The ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" function ",
StyleBox["NestList", "Input",
FontWeight->"Bold"],
" automates this in a simple way. It is designed to take a function and \
feed it's outputs back into it's inputs n times producing a list of all \
values obtained. Throughout the process we want to keep the ODE f and the \
stepsize h fixed. Only the point {t , y} is changing. The expression ",
StyleBox["EulerStep[f,#,h]&", "Input",
FontWeight->"Bold"],
" inside the ",
StyleBox["NestList", "Input",
FontWeight->"Bold"],
" command represents a function with one argument (denoted by the # symbol) \
into which we repeatedly feed back values. "
}], "Text"],
Cell["\<\
EulerList[f_,{t_,y_},h_,n_] :=
\tNestList[ EulerStep[f,#,h]&,{t,y},n]\
\>", "Input",
AspectRatioFixed->True],
Cell["\<\
We will try it out on the simple ODE y ' = y with initial condition \
y[0]=1. Of course we know the exact solution is the exponential function. \
We will try to approximate it over the time interval from 0 to 1 by taking 4 \
steps of size h = 0.25. The y value at the end of the list should be e = \
2.71828 ...\
\>", "Text"],
Cell["\<\
f[t_,y_] := y
h = 0.1;
n = 10;
elist = EulerList[f,{0,1},h,n]\
\>", "Input"],
Cell[TextData[{
"To improve the approximation you need to make h smaller and n larger. Go \
back to the last command and try taking ",
StyleBox["h=0.1", "Input"],
" and ",
StyleBox["n=10", "Input"],
" and also ",
StyleBox["h=0.01", "Input"],
" and ",
StyleBox["n=100", "Input"],
". The lists are rather long and to continue, it is better not to print \
them all out. You can simply print out the last element of the list using \
the ",
StyleBox["Last",
FontWeight->"Bold"],
" command. The command below shows the result of using 1000 steps. \
Experiment to see how many steps it takes to get an approximation for e \
which is correct in the first four decimal places."
}], "Text"],
Cell["\<\
h = 0.001;
n = 1000;
Last[ EulerList[f,{0,1},h,n ] ]\
\>", "Input"],
Cell[TextData[{
"Next we will compute an approximate solution to a more difficult problem \
and plot it. Plotting just involves applying the ",
StyleBox["Line ", "Input",
FontWeight->"Bold"],
"command to the list produced by ",
StyleBox["EulerList", "Input",
FontWeight->"Bold"],
" . We will try to compute one of the solutions to the ODE ",
StyleBox["y'[t]==(t^2-y[t]^2)/(1+y[t])^2 ", "Input"],
"that we got in the previous section of this notebook using ",
StyleBox["NDSolve",
FontWeight->"Bold"],
" . We will work on the time interval from -4 to 4."
}], "Text"],
Cell["\<\
f[t_,y_] := (t^2-y^2)/(1+y^2)
n = 4;
h = 8.0/n;
t0 = -4;
y0 = -2;
elist = EulerList[f,{t0,y0},h,n];
Show[ Graphics[Line[elist]], Axes->True ];\
\>", "Input"],
Cell["\<\
You should go back to the last cell and try a sequence of larger n \
values so you can see the method converging to the actual solution.\
\>",
"Text"],
Cell[TextData[{
"For fun, you can make a movie showing the convergence of Euler's method by \
making a table of pictures and then using ",
StyleBox["Animate Selected Graphics",
FontWeight->"Bold"],
" command in the ",
StyleBox["Cell",
FontWeight->"Bold"],
" menu. First we produce a table of 20 pictures with n values ranging from \
4 to 80 and appropriately chosen h values."
}], "Text"],
Cell[BoxData[
\(Table[\n
Show[Graphics[Line[EulerList[f, {t0, y0}, 8.0/\((4*n)\), 4*n]]], \n\t
Axes -> True, \ PlotRange -> {{\(-4\), 4}, {\(-2\), 4}}],
\n{n, 1, 20}]\)], "Input"],
Cell[TextData[{
"To animate them you have to select them by clicking with the mouse on the \
blue bracket on the right of the cell. Once they are selected, go into the \
Cell menu and choose the item called ",
StyleBox["Animate Selected Graphics",
FontWeight->"Bold"],
". You should see a movie of your 20 plots showing how Euler's method \
converges. To stop the movie just click the mouse. \n\nIf your movie goes \
too fast or too slow you can fix it as follows. Go to the ",
StyleBox["Preferences",
FontWeight->"Bold"],
" item in the ",
StyleBox["Edit",
FontWeight->"Bold"],
" menu. Select",
StyleBox[" GraphicsOptions",
FontWeight->"Bold"],
" and then ",
StyleBox["Animation",
FontWeight->"Bold"],
". You will find an item called ",
StyleBox["AnimationDisplayTime",
FontWeight->"Bold"],
". Click on the old value, enter a new one and hit ",
StyleBox["Return",
FontWeight->"Bold"],
". Increasing it will slow the movie down."
}], "Text"],
Cell[TextData[{
"We will now write our own version of the Runge-Kutta method. The only \
difference is that the method for taking one step is more complicated. To \
carry it out we will use a ",
StyleBox["Module ",
FontWeight->"Bold"],
"command. This allows us to use temporary variable to hold the four \
necessary slope values."
}], "Text"],
Cell["\<\
RKStep[f_,{t_,y_},h_] := Module[{m1,m2,m3,m4},
\t\t\tm1 = f[t,y];
\t\t\tm2 = f[t+h/2,y+h/2*m1];
\t\t\tm3 = f[t+h/2,y+h/2*m2];
\t\t\tm4 = f[t+h,y+h*m3];
\t\t\t{t+h, y+(m1 + 2*m2 + 2*m3 + m4)*h/6.0 }
\t\t\t]
\t\t\t
RKList[f_,{t_,y_},h_,n_] :=
\tNestList[ RKStep[f,#,h]&,{t,y},n]\
\>", "Input"],
Cell[TextData[{
"This can be used in the same way as ",
StyleBox["EulerList",
FontWeight->"Bold"],
" to produce an approximate solution, The difference is that it does a \
better job for a given step size, h. Here is a comparison of the two methods \
for the ODE we have been looking at with Euler in red and Runge-Kutta in \
green. After you see the plot, try increasing n and watching them converge \
(the formula for h is adjusted automatically when you change n)."
}], "Text"],
Cell["\<\
n = 4;
h = 8.0/n;
elist = EulerList[f,{t0,y0},h,n];
rklist = RKList[f,{t0,y0},h,n];
Show[ Graphics[{RGBColor[1,0,0],Line[elist],
\t\tRGBColor[0,1,0],Line[rklist]}],Axes->True ];\
\>", "Input"],
Cell[TextData[{
StyleBox[
"Finally, we will compare the errors made by the two methods in \
approximating ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["e ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[
"by solving the differential equation y ' = y with y[0] = 1 for the time \
interval from 0 to 1. We will plot the differences between the values of \
y[1] produced by the two methods and the actual value of ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["e ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[
"for various values of h. To read off the value of y[1] we apply the ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Last",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[" method to get the pair {1 , y [1] } and then apply ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Last",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox[" again.",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
e = N[Exp[1]];
f[t_,y_] := y
errortable = Table[
\t\t{1.0/2^n,\t
\t\te - Last[Last[ EulerList[f,{0,1},1.0/2^n,2^n]]],
\t\te - Last[Last[ RKList[f,{0,1},1.0/2^n,2^n]]]},
\t\t{n,0,10}];
PrependTo[errortable,{\"Stepsize\",\"Euler\",\"Runge-Kutta\"}];
TableForm[errortable]\t\
\>", "Input",
AspectRatioFixed->True],
Cell["\<\
Recall that the three methods have errors of orders h and h^4 \
respectively. This is reflected in the table where dividing h by 2 causes \
the errors to be divided by factors near 2 and 16 respectively.\
\>", "Text",\
Evaluatable->False,
AspectRatioFixed->True]
}, Closed]],
Cell[CellGroupData[{
Cell["Picard Iteration", "Section"],
Cell[TextData[{
StyleBox[
"Next we will set up a function which does Picard iteration and create a \
movie illustrating the convergence of the successive approximations. Except \
for the ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Evaluate ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox["command, the ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Picard ",
Evaluatable->False,
AspectRatioFixed->True,
FontWeight->"Bold"],
StyleBox["function below is just a translation into ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["Mathematica",
Evaluatable->False,
AspectRatioFixed->True,
FontSlant->"Italic"],
StyleBox[" of Picard's formula. You feed it the current approximation ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["yn", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[", the ODE ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["f[t,y]", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[", and the initial conditions ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["t0,y0", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[". It returns the next approximation, ",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox["yn+1", "Input",
Evaluatable->False,
AspectRatioFixed->True],
StyleBox[
". This is an example of a function whose return value is a function, a \
possibility disallowed in most programming languages.",
Evaluatable->False,
AspectRatioFixed->True]
}], "Text"],
Cell[BoxData[
\(Picard[yn_, f_, t0_, y0_]\ := \
Function[t, \n\t\t\t\t
Evaluate[y0\ + \ Integrate[f[u, yn[u]], {u, t0, t}]]]\)], "Input"],
Cell[TextData[{
"Now we'll try it out on the ODE y' = y using the initial conditions x0=0 \
and y0 = 1. We will call the successive approximations ",
StyleBox["pic[0], pic[1]", "Input"],
" and so on. To evaluate them, you use expressions like ",
StyleBox["pic[0][t]", "Input"],
"."
}], "Text"],
Cell[BoxData[{
\(t0\ = \ 0; \ny0\ = \ 1; \n\(pic[0]\)[x_]\ := \ y0\ ; \n
f[t_, y_]\ := \ y; \npic[1]\ = \ Picard[pic[0], f, t0, y0]; \n
pic[2]\ = \ Picard[pic[1], f, t0, y0]; \n
pic[3]\ = \ Picard[pic[2], f, t0, y0]; \n\(pic[1]\)[t]\),
\(\(pic[2]\)[t]\),
\(\(pic[3]\)[t]\)}], "Input"],
Cell["\<\
Now we will generate several graphs of these approximations and \
animate them to make the movie.\
\>", "Text"],
Cell[BoxData[{
\(Do[\ pic[i]\ = \ Picard[pic[i - 1], f, 0, 1]; \n\t
Plot[\(pic[i]\)[t], {t, \(-5\), 5}, PlotRange -> {\(-5\), 5}], \n
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solutions don't exist for all time. For example, here is an attempt to \
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\>", "Text"],
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If you restrict the plot range when you display the results, no one \
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