Mathematica running very slowly with just 4 integrals. Memoization? - wolfram-mathematica

While playing with Fourier Integrals, I've noticed my computations taking an excessively long time to calculate. I just have two sets of a Piecewise function, two functions and two integrals. I wouldn't think Mathematica would need to take so long to compute these. What, in particular, is slowly my computations down so much? How can I improve computation time? Will using some combination of memoization help ( :=, for which I don't quite understand)?
Decreasing the Limit on the integrals helps slightly, but not as much as I'd think.
My code is the following:
MM = 50;
qf = 10;
(*First Set*)
f[t_] = Piecewise[{{1, t <= 0.5}, {0, t > 0.5}}];
A[w_] = qf Sin[w/2]/(Pi w);
B[w_] = qf (1 - Cos[w/2])/(Pi w);
f[x_] = Integrate[A[w] Cos[w x], {x, 0, MM}];
g[x_] = Integrate[B[w] Sin[w x], {x, 0, MM}];
Plot[f[x], {x, 0, 1}]
Plot[g[x], {x, 0, 1}]
Plot[f[x] + g[x], {x, 0, 1}]
(*Second Set*)
ff[t_] = Piecewise[{{1, t <= 0.5}, {0, t > 0.5}}];
AA[w_] = qf (Sin[w] - Sin[w/2])/(Pi w);
BB[w_] = -qf (Cos[w] - Cos[w/2])/(Pi w);
ff[x_] = Integrate[AA[w] Cos[w x], {x, 0, MM}];
gg[x_] = Integrate[BB[w] Sin[w x], {x, 0, MM}];
Plot[ff[x], {x, 0, 1}]
Plot[gg[x], {x, 0, 1}]
Plot[ff[x] + gg[x], {x, 0, 1}]

You have several problems with your code as shown:
You haven't defined w so e.g. f[x_] isn't fully evaluated
you define f[t_] then f[x_] which overwrites the first definition
You're misusing Piecewise. It should be something like pw = Piecewise[{{Sin[x]/x, x < 0}, {1, x == 0}}, -x^2/100 + 1]

Related

Ploting implicit variable of Nonanalytic Equation

I have a non-analytical equation. I could solve for different values of parameters but my program is not working at all. At the end i want to plot y vs x
f[x_] := y + Sqrt[3 + x*y - x^20 - y^4]
Table[f[x], {x, 0.1, 0.5, 0.1}]
NSolve[f[x] == 0, y]
f[x_] := y + Sqrt[3 + x*y - x^20 - y^4]
sol = Solve[f[x] == 0, y];
x0 = Table[i, {i, 0.1, 0.5, 0.1}];
subs = N[sol /. x -> #] & /# x0
This creates results from which we can see that the first and second solutions produce complex numbers. Plotting the two real solutions first.
y3 = subs[[All, 3, 1, 2]];
y4 = subs[[All, 4, 1, 2]];
ListLinePlot[{Transpose[{x0, y3}], Transpose[{x0, y4}]}]
Alternatively the plot can be produced from the solutions with
Plot[{sol[[3, 1, 2]], sol[[4, 1, 2]]}, {x, 0.1, 0.5}]
The complex solutions can be plotted like so:
ParametricPlot[{{Re[sol[[1, 1, 2]]], Im[sol[[1, 1, 2]]]},
{Re[sol[[2, 1, 2]]], Im[sol[[2, 1, 2]]]}}, {x, 0, Pi/2}]

How does this deviation come out after I use Evaluate and Plot in Mathematica?

I ran into this problem when I try to solve a partial differential equation. Here is my code:
dd = NDSolve[{D[tes[t, x], t] ==D[tes[t, x], x, x] + Exp[-1/(tes[t, x])],
tes[t, 0] == 1, tes[t, -1] == 1, tes[0, x] == 1}, {tes[t, x]}, {t, 0, 5}, {x, -1, 0}]
f[t_, x_] = tes[t, x] /. dd
kkk = FunctionInterpolation[Integrate[Exp[-1.1/( Evaluate[f[t, x]])], {x, -1, 0}], {t, 0, 0.05}]
kkg[t_] = Integrate[Exp[-1.1/( Evaluate[f[t, x]])], {x, -1, 0}]
Plot[Evaluate[kkk[t]] - Evaluate[kkg[t]], {t, 0, 0.05}]
N[kkg[0.01] - kkk[0.01], 1]
It's strange that the deviation showed in the graph reaches up to more than 5*10^-7 around t=0.01, while it's only -3.88578*10^-16 when calculated by N[kkg[0.01] - kkk[0.01], 1], I wonder how this error comes out.
By the way, I feel it strange that the output of N[kkg[0.01] - kkk[0.01], 1] has so many decimal places, I've set the precision as 1, right?
Using Mathematica 7 the plot I get does not show a peak at 0.01:
Plot[kkk[t] - kkg[t], {t, 0, 0.05}, GridLines -> Automatic]
There is a peak at about 0.00754:
kkk[0.00754] - kkg[0.00754] // N
{6.50604*10^-7}
Regarding N, it does not change the precision of machine precision numbers as it does for exact or arbitrary precision ones:
N[{1.23456789, Pi, 1.23456789`50}, 2]
Precision /# %
{1.23457, 3.1, 1.2}
{MachinePrecision, 2., 2.}
Look at SetPrecision if you want to force (fake) a precision, and NumberForm if you want to print a number in a specific format.

Solution of non-linear differential equation

I don't use Mathematica in general and I need it to compare with an other program. I want to solve a system of three differential and non linear equations. For this I use Dsolve. Everything goes wrong when I put the nonlinear term (exponential).
Here is my code:
equa = {x'[t] == z[t] - Exp[y[t]],
y'[t] == z[t] - y[t],
z'[t] == x[t] + y[t] - z[t],
x[0] == 0,
y[0] == 0,
z[0] == 0};
slt = DSolve[equa, {x, y, z}, t]
Plot[{x[t] /. slt}, {t, 0, 10}]
And the errors are like this :
DSolve::dsvar: 0.1 cannot be used as a variable.
ReplaceAll::reps:{Dsolve[<<1>>]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing
Does someone know why the exponential term makes troubles ?
Thanks
You may try
s = NDSolve[equa, {x, y, z}, {t, 0, 10}];
Plot[Evaluate[({x[t], y[t], z[t]} /. s)], {t, 0, 1}]

An efficient data structure or method to manage plotting data that grow with time

I'd like to ask if the following way I manage plotting result of simulation is efficient use of Mathematica and if there is a more 'functional' way to do it. (may be using Sow, Reap and such).
The problem is basic one. Suppose you want to simulate a physical process, say a pendulum, and want to plot the time-series of the solution (i.e. time vs. angle) as it runs (or any other type of result).
To be able to show the plot, one needs to keep the data points as it runs.
The following is a simple example, that plots the solution, but only the current point, and not the full time-series:
Manipulate[
sol = First#NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4, y'[0] == 0},
y, {t, time, time + 1}];
With[{angle = y /. sol},
(
ListPlot[{{time, angle[time]}}, AxesLabel -> {"time", "angle"},
PlotRange -> {{0, max}, {-Pi, Pi}}]
)
],
{{time, 0, "run"}, 0, max, Dynamic#delT, ControlType -> Trigger},
{{delT, 0.1, "delT"}, 0.1, 1, 0.1, Appearance -> "Labeled"},
TrackedSymbols :> {time},
Initialization :> (max = 10)
]
The above is not interesting, as one only sees a point moving, and not the full solution path.
The way currently I handle this, is allocate, using Table[], a buffer large enough to hold the largest possible time-series size that can be generated.
The issue is that the time-step can change, and the smaller it is, the more data will be generated.
But since I know the smallest possible time-step (which is 0.1 seconds in this example), and I know the total time to run (which is 10 seconds here), then I know how much to allocate.
I also need an 'index' to keep track of the buffer. Using this method, here is a way to do it:
Manipulate[
If[time == 0, index = 0];
sol = First#NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4,y'[0] == 0},
y, {t, time, time + 1}];
With[{angle = y /. sol},
(
index += 1;
buffer[[index]] = {time, angle[time]};
ListPlot[buffer[[1 ;; index]], Joined -> True, AxesLabel -> {"time", "angle"},
PlotRange -> {{0, 10}, {-Pi, Pi}}]
)
],
{{time, 0, "run"}, 0, 10, Dynamic#delT, AnimationRate -> 1, ControlType -> Trigger},
{{delT, 0.1, "delT"}, 0.1, 1, 0.1, Appearance -> "Labeled"},
{{buffer, Table[{0, 0}, {(max + 1)*10}]}, None},
{{index, 0}, None},
TrackedSymbols :> {time},
Initialization :> (max = 10)
]
For reference, when I do something like the above in Matlab, it has a nice facility for plotting, called 'hold on'. So that one can plot a point, then say 'hold on' which means that the next plot will not erase what is already on the plot, but will add it.
I did not find something like this in Mathematica, i.e. update a current plot on the fly.
I also did not want to use Append[] and AppendTo[] to build the buffer as it runs, as that will be slow and not efficient.
My question: Is there a more efficient, Mathematica way (which can be faster and more elegent) to do a typical task such as the above, other than what I am doing?
thanks,
UPDATE:
On the question on why not solving the ODE all at once.
Yes, it is possible, but it simplifies things alot to do it in pieces, also for performance reasons.
Here is an example with ode with initial conditions:
Manipulate[
If[time == 0, index = 0];
sol = First#
NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == y0,
y'[0] == yder0}, y, {t, time, time + 1}];
With[{angle = (y /. sol)[time]},
(
index += 1;
buffer[[index]] = {time, angle};
ListPlot[buffer[[1 ;; index]], Joined -> True,
AxesLabel -> {"time", "angle"},
PlotRange -> {{0, 10}, {-Pi, Pi}}])],
{{time, 0, "run"}, 0, 10, Dynamic#delT, AnimationRate -> 1,
ControlType -> Trigger}, {{delT, 0.1, "delT"}, 0.1, 1, 0.1,
Appearance -> "Labeled"},
{{y0, Pi/4, "y(0)"}, -Pi, Pi, Pi/100, Appearance -> "Labeled"},
{{yder0, 0, "y'(0)"}, -1, 1, .1, Appearance -> "Labeled"},
{{buffer, Table[{0, 0}, {(max + 1)*10}]}, None},
{{index, 0}, None},
TrackedSymbols :> {time},
Initialization :> (max = 10)
]
Now, in one were to solve the system once before, then they need to watch out if the IC changes. This can be done, but need extra logic and I have done this before many times, but it does complicate things a bit. I wrote a small note on this here.
Also, I noticed that I can get much better speed by solving the system for smaller time segments as time marches on, than the whole thing at once. NDSolve call overhead is very small. But when the time duration to NDsolve for is large, problems can result when one ask for higher accuracy from NDSolve, as in options AccuracyGoal ->, PrecisionGoal ->, which I could not when time interval is very large.
Overall, the overhead of calling NDSolve for smaller segments seems to much less compare to the advantages it makes in simplifing the logic, and speed (may be more accurate, but I have not checked on this more). I know it seems a bit strange to keep calling NDSolve, but after trying both methods (all at once, but add logic to check for other control variables) vs. this method, I am now leaning towards this one.
UPDATE 2
I compared the following 4 methods for 2 test cases:
tangle[j][j] method (Belisarius)
AppendTo (suggested by Sjoerd)
Dynamic linked list (Leonid) (with and without SetAttributes[linkedList, HoldAllComplete])
preallocate buffer (Nasser)
The way I did this, is by running it over 2 cases, one for 10,000 points, and the second for 20,000 points. I did leave the Plot[[] command there, but do not display it on the screen, this is to eliminate any overhead of the actual rendering.
I used Timing[] around a Do loop which iterate over the core logic which called NDSolve and iterate over the time span using delT increments as above. No Manipulate was used.
I used Quit[] before each run.
For Leonid method, I changed the Column[] he had by the Do loop. I verified at the end, but plotting the data using his getData[] method, that the result is ok.
All the code I used is below. I made a table which shows the results for the 10,000 points and 20,000. Timing is per seconds:
result = Grid[{
{Text[Style["method", Bold]],
Text[Style["number of elements", Bold]], SpanFromLeft},
{"", 10000, 20000},
{"", SpanFromLeft},
{"buffer", 129, 571},
{"AppendTo", 128, 574},
{"tangle[j][j]", 612, 2459},
{"linkedList with SetAttribute", 25, 81},
{"linkedList w/o SetAttribute", 27, 90}}
]
Clearly, unless I did something wrong, but code is below for anyone to verify, Leonid method wins easily here. I was also surprised that AppendTo did just as well as the buffer method which pre-allocated data.
Here are the slightly modified code I used to generate the above results.
buffer method
delT = 0.01; max = 100; index = 0;
buffer = Table[{0, 0}, {(max + 1)*1/delT}];
Timing[
Do[
sol = First#
NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4,
y'[0] == 0}, y, {t, time, time + 1}];
With[{angle = y /. sol},
(index += 1;
buffer[[index]] = {time, angle[time]};
foo =
ListPlot[buffer[[1 ;; index]], Joined -> True,
AxesLabel -> {"time", "angle"},
PlotRange -> {{0, 10}, {-Pi, Pi}}]
)
], {time, 0, max, delT}
]
]
AppendTo method
Clear[y, t];
delT = 0.01; max = 200;
buffer = {{0, 0}}; (*just a hack to get ball rolling, would not do this in real code*)
Timing[
Do[
sol = First#
NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4,
y'[0] == 0}, y, {t, time, time + 1}];
With[{angle = y /. sol},
(AppendTo[buffer, {time, angle[time]}];
foo =
ListPlot[buffer, Joined -> True, AxesLabel -> {"time", "angle"},
PlotRange -> {{0, 10}, {-Pi, Pi}}]
)
], {time, 0, max, delT}
]
]
tangle[j][j] method
Clear[y, t];
delT = 0.01; max = 200;
Timing[
Do[
sol = First#
NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4,
y'[0] == 0}, y, {t, time, time + 1}];
tangle[time] = y /. sol;
foo = ListPlot[
Table[{j, tangle[j][j]}, {j, .1, max, delT}],
AxesLabel -> {"time", "angle"},
PlotRange -> {{0, max}, {-Pi, Pi}}
]
, {time, 0, max, delT}
]
]
dynamic linked list method
Timing[
max = 200;
ClearAll[linkedList, toLinkedList, fromLinkedList, addToList, pop,
emptyList];
SetAttributes[linkedList, HoldAllComplete];
toLinkedList[data_List] := Fold[linkedList, linkedList[], data];
fromLinkedList[ll_linkedList] :=
List ## Flatten[ll, Infinity, linkedList];
addToList[ll_, value_] := linkedList[ll, value];
pop[ll_] := Last#ll;
emptyList[] := linkedList[];
Clear[getData];
Module[{ll = emptyList[], time = 0, restart, plot, y},
getData[] := fromLinkedList[ll];
plot[] := Graphics[
{
Hue[0.67`, 0.6`, 0.6`],
Line[fromLinkedList[ll]]
},
AspectRatio -> 1/GoldenRatio,
Axes -> True,
AxesLabel -> {"time", "angle"},
PlotRange -> {{0, 10}, {-Pi, Pi}},
PlotRangeClipping -> True
];
DynamicModule[{sol, angle, llaux, delT = 0.01},
restart[] := (time = 0; llaux = emptyList[]);
llaux = ll;
sol :=
First#NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4,
y'[0] == 0}, y, {t, time, time + 1}];
angle := y /. sol;
ll := With[{res =
If[llaux === emptyList[] || pop[llaux][[1]] != time,
addToList[llaux, {time, angle[time]}],
(*else*)llaux]
},
llaux = res
];
Do[
time += delT;
plot[]
, {i, 0, max, delT}
]
]
]
]
thanks for everyone help.
I don't know how to get what you want with Manipulate, but I seem to have managed getting something close with a custom Dynamic. The following code will: use linked lists to be reasonably efficient, stop / resume your plot with a button, and have the data collected so far available on demand at any given time:
ClearAll[linkedList, toLinkedList, fromLinkedList, addToList, pop, emptyList];
SetAttributes[linkedList, HoldAllComplete];
toLinkedList[data_List] := Fold[linkedList, linkedList[], data];
fromLinkedList[ll_linkedList] := List ## Flatten[ll, Infinity, linkedList];
addToList[ll_, value_] := linkedList[ll, value];
pop[ll_] := Last#ll;
emptyList[] := linkedList[];
Clear[getData];
Module[{ll = emptyList[], time = 0, restart, plot, y},
getData[] := fromLinkedList[ll];
plot[] :=
Graphics[{Hue[0.67`, 0.6`, 0.6`], Line[fromLinkedList[ll]]},
AspectRatio -> 1/GoldenRatio, Axes -> True,
AxesLabel -> {"time", "angle"}, PlotRange -> {{0, 10}, {-Pi, Pi}},
PlotRangeClipping -> True];
DynamicModule[{sol, angle, llaux, delT = 0.1},
restart[] := (time = 0; llaux = emptyList[]);
llaux = ll;
sol := First#
NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0, y[0] == Pi/4, y'[0] == 0},
y, {t, time, time + 1}];
angle := y /. sol;
ll := With[{res =
If[llaux === emptyList[] || pop[llaux][[1]] != time,
addToList[llaux, {time, angle[time]}],
(* else *)
llaux]},
llaux = res];
Column[{
Row[{Dynamic#delT, Slider[Dynamic[delT], {0.1, 1., 0.1}]}],
Dynamic[time, {None, Automatic, None}],
Row[{
Trigger[Dynamic[time], {0, 10, Dynamic#delT},
AppearanceElements -> { "PlayPauseButton"}],
Button[Style["Restart", Small], restart[]]
}],
Dynamic[plot[]]
}, Frame -> True]
]
]
Linked lists here replace your buffer and you don't need to pre-allocate and to know in advance how many data points you will have. The plot[] is a custom low-level plotting function, although we probably could just as well use ListPlot. You use the "Play" button to both stop and resume plotting, and you use the custom "Restart" button to reset the parameters.
You can call getData[] at any given time to get a list of data accumulated so far, like so:
In[218]:= getData[]
Out[218]= {{0,0.785398},{0.2,0.771383},{0.3,0.754062},{0.4,0.730105},{0.5,0.699755},
{0.6,0.663304},{0.7,0.621093},{0.8,0.573517},{0.9,0.521021},{1.,0.464099},
{1.1,0.403294},{1.2,0.339193},{1.3,0.272424}}
I just wonder why you want to solve the DE in pieces. It can be solved for the whole interval at once. There is also no need to place the NDSolve in the Manipulate then. It doesn't need to be solved time and again when the body of the Manipulateis triggered. Plot itself is sufficiently fast to plot the growing graph at each time step. The following code does what you want without the need for any storage.
sol = First#
NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]]==0,y[0] == Pi/4,y'[0] == 0}, y, {t, 0, 10}];
eps = 0.000001;
Manipulate[
With[{angle = y /. sol},
Plot[angle[t], {t, 0, time + eps},
AxesLabel -> {"time", "angle"},
PlotRange -> {{0, max}, {-Pi, Pi}}
]
],
{{time, 0, "run"}, 0, max,Dynamic#delT, ControlType -> Trigger},
{{delT, 0.1, "delT"}, 0.1, 1, 0.1, Appearance -> "Labeled"}, TrackedSymbols :> {time},
Initialization :> (max = 10)
]
BTW: AppendTo may be vilified as slow, but it is not that slow. On a typical list suitable for plotting it takes less than a milisecond, so it shouldn't slow plotting at all.
Not memory efficient at all, but its virtue is that it only needs a slight modification of your first code:
Clear[tangle];
Manipulate[
sol = First#NDSolve[{y''[t] + 0.1 y'[t] + Sin[y[t]] == 0,
y[0] == Pi/4,
y'[0] == 0},
y, {t, time, time + 1}];
(tangle[time] = y /. sol;
ListPlot[Table[{j, tangle[j][j]}, {j, .1, max, delT}],
AxesLabel -> {"time", "angle"},
PlotRange -> {{0, max}, {-Pi, Pi}}]),
{{time, 0, "run"}, 0, max, Dynamic#delT, ControlType -> Trigger},
{{delT, 0.1, "delT"}, 0.1, 1, 0.1, Appearance -> "Labeled"},
TrackedSymbols :> {time},
Initialization :> {(max = 10); i = 0}]

in mathematica, how to make initial condition as a variable in ndsolve?

i'd like to have something like this
w[w1_] :=
NDSolve[{y''[x] + y[x] == 2, y[0] == w1, y'[0] == 0}, y, {x, 0, 30}]
this seems like it works better but i think i'm missing smtn
w := NDSolve[{y''[x] + y[x] == 2, y[0] == w1, y'[0] == 0},
y, {x, 0, 30}]
w2 = Table[y[x] /. w, {w1, 0.0, 1.0, 0.5}]
because when i try to make a table, it doesn't work:
Table[Evaluate[y[x] /. w2], {x, 10, 30, 10}]
i get an error:
ReplaceAll::reps: {<<1>>[x]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>
ps: is there a better place to ask questions like that? mathematica doesn't have supported forums and only has mathGroup e-mail list. it would be nice if stackoverflow would have more specific mathematica tags like simplify, ndsolve, plot manipulation
There are a lot of ways to do that. One is:
w[w1_] := NDSolve[{y''[x] + y[x] == 2,
y'[0] == 0}, y[0] == w1,
y[x], {x, 0, 30}];
Table[Table[{w1,x,y[x] /. w[w1]}, {w1, 0., 1.0, 0.5}]/. x -> u, {u, 10, 30, 10}]
Output:
{{{0., 10, {3.67814}}, {0.5, 10, {3.25861}}, {1.,10, {2.83907}}},
{{0., 20, {1.18384}}, {0.5, 20, {1.38788}}, {1.,20, {1.59192}}},
{{0., 30, {1.6915}}, {0.5, 30, {1.76862}}, {1.,30, {1.84575}}}}
I see you already chose an answer, but I want to toss this solution for families of linear equations up. Specifically, this is to model a slight variation on Lotka-Volterra.
(*Put everything in a module to scope x and y correctly.*)
Module[{x, y},
(*Build a function to wrap NDSolve, and pass it
the initial conditions and range.*)
soln[iCond_, tRange_, scenario_] :=
NDSolve[{
x'[t] == -scenario[[1]] x[t] + scenario[[2]] x[t]*y[t],
y'[t] == (scenario[[3]] - scenario[[4]]*y[t]) -
scenario[[5]] x[t]*y[t],
x[0] == iCond[[1]],
y[0] == iCond[[2]]
},
{x[t], y[t]},
{t, tRange[[1]], tRange[[2]]}
];
(*Build a plot generator*)
GeneratePlot[{iCond_, tRange_, scen_,
window_}] :=
(*Find a way to catch errors and perturb iCond*)
ParametricPlot[
Evaluate[{x[t], y[t]} /. soln[iCond, tRange, scen]],
{t, tRange[[1]], tRange[[2]]},
PlotRange -> window,
PlotStyle -> Thin, LabelStyle -> Medium
];
(*Call the plot generator with different starting conditions*)
graph[scenario_, tRange_, window_, points_] :=
{plots = {};
istep = (window[[1, 2]] - window[[1, 1]])/(points[[1]]+1);
jstep = (window[[2, 2]] - window[[2, 1]])/(points[[2]]+1);
Do[Do[
AppendTo[plots, {{i, j}, tRange, scenario, window}]
, {j, window[[2, 1]] + jstep, window[[2, 2]] - jstep, jstep}
], {i, window[[1, 1]] + istep, window[[1, 2]] - istep, istep}];
Map[GeneratePlot, plots]
}
]
]
We can then use Animate (or table, but animate is awesome)
tRange = {0, 4};
window = {{0, 8}, {0, 6}};
points = {5, 5}
Animate[Show[graph[{3, 1, 8, 2, 0.5},
{0, t}, window, points]], {t, 0.01, 5},
AnimationRunning -> False]

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