Comments on: Exact chaos

By: Caos previsível | Bruno Kim

Caos previsível | Bruno Kim — Thu, 11 Apr 2013 02:40:52 +0000

[...] Do blog de John D. Cook: [...]

By: John

John — Tue, 26 Feb 2013 23:21:37 +0000

Here's a related paper: Method of constructing exactly solvable chaos.

By: Andrew Au

Andrew Au — Tue, 26 Feb 2013 23:18:57 +0000

If you let T_n = sin^2(U_n), translate the recurrence into the recurrence of U_n, the rest will be obvious.

Finding a good substitution for solving non-linear recurrence is hard – if I were only given the problem, I wouldn’t be able to even guess to substitute sine squared.

By: Dave Tate

Dave Tate — Tue, 26 Feb 2013 20:22:49 +0000

I’m actually more interested in how you derive the closed form sine expression from the recursion. Not obvious to me at all.

Also, might the relative ‘goodness’ of these depend on how sine and arcsine are implemented in a particular programming language? Neither one is ‘exact’…

By: Ran

Ran — Tue, 26 Feb 2013 13:39:13 +0000

John: I’m not sure I agree. Let’s say you use your high precision implementation but start from two different initial conditions that differ from one another by what would be one bit in the floating point representation, then you would still see divergence even though your calculation is “exact”.
These are two separate issues.
For chaotic systems the divergence does not rely on how fast you lose precision (in other words, the amount of noise) any two different initial conditions diverge even with infinite precision (without any noise).

All I’m saying is that the divergence may not be due to the fact that you lose too much precision…

By: John

John — Tue, 26 Feb 2013 13:10:25 +0000

Ran: Each iteration looses about a bit of precision in either method, so that’s why things fall apart around n=50. But the sensitivity of the system makes the loss of precision show up more dramatically.

By: Ran

Ran — Tue, 26 Feb 2013 09:38:19 +0000

Great demonstration of sensitivity to initial conditions!

Do you think the divergnce around n=50 is due to the properties of the floating point precision (The time when first differences appear) or due to the Lyapunov exponent of the system?

By: John

John — Mon, 25 Feb 2013 23:52:31 +0000

Anonymous: Good point. I recreated the last two figures to use the same vertical scale.

By: Anonymous

Anonymous — Mon, 25 Feb 2013 20:36:49 +0000

Hello John.

YMMV, but I prefer to fix the axis to the same limits when showing two graphs for comparison. Your last two graphs differ in scale in their x axis. Obviously, I can do that given that you provide the code. But I think it will help the readers to better understand the issue if you update the graphs.

Again, it is just my opinion. You blog is still awesome

By: John

John — Mon, 25 Feb 2013 17:56:02 +0000

Thomas: I updated the post to answer your question. Apparently the sine formula is better overall, but not much better and not uniformly better.

By: Thomas

Thomas — Mon, 25 Feb 2013 17:11:24 +0000

John: Yes, I forgot about contractions et al. I get your point now, I was just skimming the post and then thought to myself, that it’s just another floating point error. But in fact, it’s more than that.

Now, I wonder which version is closer to the “real” value. It probably is the closed form, but when you get to large n, then 2^n squeezes out a lot of precision. On the other hand, the other version accumulates a lot of error from the iteration. Interesting …

By: John

John — Mon, 25 Feb 2013 16:58:44 +0000

Thomas: It depends on the function being iterated. If the function is a contraction, then |f – g| would get smaller over time because both f and g would become more accurate.

Part of what’s going on here is that the two functions have different kinds of numerical error. The function f loses precision from simple arithmetic. The sine formula loses precision due to shifting. When you multiply the argument of sine by 2^n, you’re making noisy bits significant.

By: Thomas

Thomas — Mon, 25 Feb 2013 16:51:36 +0000

John: OK, so when you have a non-chaotic iteration, then h(n) = |f(n) – g(n)| should increase monotonically instead of just being a chaotic function itself for large enough n. (just some wild guess).

By: John

John — Mon, 25 Feb 2013 16:38:39 +0000

Thomas: I think that’s part of it. The limitations of floating point explain why things fall apart at around n = 50, approximately the number of bits of precision in a floating point number. But there are other functions you could iterate thousands of times with no floating point problems because their iterates are not chaotic.

By: Thomas

Thomas — Mon, 25 Feb 2013 16:25:24 +0000

In the end, this boils down to http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html