# Animation of a geometric transformation

This code:

t = .25;
RegionPlot[TransformedRegion[Rectangle[{-1, -1}, {1, 1}], {Indexed[#1, {1}]
(1 + t ( Indexed[#1, {2}]^2 - 1)) + 2 t,Indexed[#1, {2}] (1 + t ( Indexed[#1,
{1}]^2 - 1)) + 2 t} &], PlotRange -> {{-0.5, 1.5}, {-0.5, 1.5}}]


gives me this:

Now, when I animate this transformation:

Animate[RegionPlot[TransformedRegion[Rectangle[{-1, -1}, {1, 1}],
{Indexed[#1, {1}] (1 + t ( Indexed[#1, {2}]^2 - 1)) + 2 t,
Indexed[#1, {2}] (1 + t ( Indexed[#1, {1}]^2 - 1)) + 2 t} &],
PlotRange -> {{-0.5, 1.5}, {-0.5, 1.5}}], {t, .25, .5}]


This is what I get:

I tried setting the "AnimationRunning" option for "Animate" to "False". When I do this, I get the transformation for the first value of "t", but the same thing happens when I try to get the figure by dynamically changing "t".

Any help is highly appreciated. Thanks in advance.

• fyi, the screen shot you show above (first one), is not what I get when I run your code under V10.1 !Mathematica graphics Commented Apr 7, 2015 at 18:22
• @Nasser That's strange. Could you please tell me what you get? I ran it again and nothing seemed to be incorrect!
– Ali
Commented Apr 7, 2015 at 19:07
• PlotRange->{{-0.5,1.5},{-0.5,1.5}} Commented Apr 7, 2015 at 19:10
• @TimothyWofford Did you mean to change PlotRange->All to PlotRange->{{-0.5. . .? I did that but it didn't work. Did it work for you?
– Ali
Commented Apr 7, 2015 at 19:37
• Could you please tell me what you get? the screen shot I pasted in the comment shows what I get. Clicking on it shows the output. I just copied and pasted your input, and the Plot I get, as you can see, is not the same as what you have. Windows 7, 64 bit, V 10.1 Commented Apr 7, 2015 at 20:42

The issue at least on my machine (Linux, M10.1) is that the plots take a long time to generate and so creating a smooth animation with on-the-fly generated plots is, well, impossible. You could generate the plots beforehand, though. Here's some code to show you the progress of plot generation as well.

Assuming that you have defined a function f[t] that generates the plot:

Clear[t];
Monitor[
(* using rationals as suggested by @Taiki *)
(plots = Table[f[t], {t, 25/100, 50/100, 1/100}]),
ProgressIndicator[t, {25/100, 50/100}]] // AbsoluteTiming


On my machine, it took over 2 minutes to generate them all!

Then you have to use the ListAnimate function:

ListAnimate[plots, DefaultDuration -> 2.5]


I guess there is some floating-point-related issue here...

This works:

f[t_] := RegionPlot[
TransformedRegion[
Rectangle[{-1, -1}, {1, 1}],
{
Indexed[#1, {1}] (1 + t (Indexed[#1, {2}]^2 - 1)) + 2 t,
Indexed[#1, {2}] (1 + t (Indexed[#1, {1}]^2 - 1)) + 2 t
} &
],
PlotRange -> {{-1 + 2 t, 1 + 2 t}, {-1 + 2 t, 1 + 2 t}}
]
Animate[f[t], {t, 25/100, 50/100, 1/100}]


The animation should look similar to this:

# Issue #1

The following doesn't work:

Animate[f[t], {t, 0.25, 0.50, 0.01}]


There are actually other strange issues as well.

# Issue #2

The first frame that

Animate[f[t], {t, 25/100, 50/100, 1/100}]


shows is

# Issue #3

The following produces a list of plots I expect (the frames of the animation above):

Table[f[t], {t, 25/100, 50/100, 1/100}]


But if I substitute the definition of f[t] for f[t], i.e.

Table[RegionPlot[...], {t, 25/100, 50/100, 1/100}]


the plots are incorrect (apart from the first one).

I'm using Mathematica 10.0.2 for Mac by the way.

• Thank you for the answer. I think Mathematica has many bugs in its region transformation functions. I had seen some others before. Hope they fix them soon!
– Ali
Commented Apr 8, 2015 at 20:00