Talk:Dyadic transformation

Notation

Hi, what means : $\mathbb {R} ^{\infty },$ -- (talk) 18:14, 2 September 2009 (UTC)

Merge bit shift map into this article

Okay, I'll merge the bit shift map article into the dyadic transformation article. Duoduoduo (talk) 19:05, 27 May 2010 (UTC)

Done. Duoduoduo (talk) 17:21, 28 May 2010 (UTC)

Please can you clarify (within the article) your sentence "Equivalently, if we are able to view each iterate of as it actually occurs (as opposed to our simulated values), then we obtain an information flow of one bit per iteration, compared to what we would have known from our simulation." at the end of the section Rate of information loss and sensitive dependence on initial conditions. It is not clear what "as it actually occurs" means, and where the "information flow" is flowing from and to. Gandalf61 (talk) 10:30, 29 May 2010 (UTC)

Correspondence to logistic map

Not sure what the comment below means, but there is certainly an error in the section about r=4 case of logistic map being bit-shift under sine square change of variables. The r=4 case of the logistic map then looks like the tent map, not the bit-shift map. It is just as chaotic, but the manners in which these two transformations map (0.5,1) to (0,1) are reversed. I don't think these are topologically conjugate. (Disclaimer: I am actually a topologist, but not a dynamicist). -- Denis 192.16.204.74 (talk) 17:15, 2 June 2017 (UTC)

Bit-shift: 0.75->0.5 r=4: 0.75->0.75 -32ieww 10:04PM 11/19/2015 NY time 32ieww (talk) 03:04, 20 November 2015 (UTC)

Image color map

What does the colormap in the first image encode? Does the brightness correspond to the number of appearances or the darkness? Is it even accumulative? The caption should be improved. — Preceding unsigned comment added by Roxanabujack (talk • contribs) 00:04, 30 November 2017 (UTC)

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Periodicity via bit shifts rationality of periodic orbits

In the derivation to prove that periodic and preperiodic orbits are orbits of rational numbers, the derivation should read

$y\sum _{j=0}^{\infty }2^{-jm}={\frac {y}{1-2^{-m}}}$

instead of

$y\sum _{j=0}^{\infty }2^{-jm}={\frac {y}{1-2^{m}}}$

as it is now. — Preceding unsigned comment added by 54.37.18.136 (talk) 16:40, 5 May 2021 (UTC)