let $\psi:[0,1]\times \{0,1\} \rightarrow [0,1]$ be defined by: $$ \psi(x,\beta) = \beta \cos x + (1-\beta) \sin x $$

define $B_n$ as the set of $2^n$ binary strings $b=b_0b_1\dots b_{n-1}$ where each $b_j$ is a $0$ or a $1$.

now, for a given $b \in B_n$ we define a set of $n$ functions: $\psi_{b,k}:[0,1] \rightarrow [0,1] \;(k=0,\dots n-1)$ as follows: $$ \psi_{b,0}(x) = \psi(x,b_0) $$ and for $k=1,\dots,n-1$ $$\psi_{b,k}(x) = \psi(\psi_{b,k-1}(x),b_k) $$ an example may clarify this definition. let $n=5$ and take $b=01101$ then $$ \psi_{b,0}(x) = \psi(x,0) = \sin x \\ \psi_{b,1} (x)= \psi(\sin x,1) = \cos \circ \sin x \\ \psi_{b,2} (x)= \psi(\cos \circ \sin x ,1) = \cos \circ \cos \circ \sin x \\ \psi_{b,3} (x)= \psi(\cos \circ \cos \circ \sin x ,0) = \sin \circ \cos \circ \cos \circ \sin x \\ \psi_{b,4} (x)= \psi(\sin \circ \cos \circ \cos \circ \sin x ,1) = \cos \circ\sin \circ \cos \circ \cos \circ \sin x $$ in this fashion, for any bit-string $b$ of length $n$, we may define $\psi_b:[0,1] \rightarrow [0,1]$ as the last of these functions $\psi_{b,n-1}$

QUESTION my recent question about the convergence of the iterated cosine was answered by user44197 by applying Brower's fixed point theorem. this problem was in fact the case $\psi_1$. can the same method be used to assure the iterative convergence of any $\psi_b$?

  • $\begingroup$ I have retagged your question as dynamical systems. From a quick look at your question, it appears you are studying the behaviour of a certain iterated function system, hence the retagging. $\endgroup$ – Willie Wong Jan 6 '14 at 12:23
  • $\begingroup$ @WillieWong thanks for the tip, and the edit. sorry didn't get back sooner - been offline a day or two $\endgroup$ – David Holden Jan 8 '14 at 22:22

Cool problem. Since $\sin$ and $\cos$ map $[0,1]$ to $[0,1]$, any finite composition of these two will still map $[0,1]$ to $[0,1]$. So we can establish the existence of a fixed point.

Secondly, the derivative of $\psi$ has to be product of $\sin$s and $\cos$s. So the derivative is less than 1 in magnitude. So the convergence would be geometric with a rate approximately the magnitude of the derivative at the fixed point.

One final comment: Longer the chain of composition, more $\sin$ and $\cos$ are multiplied to get the derivative, so the derivative will become smaller as the length of chain increases. So the convergence will be more rapid.

  • $\begingroup$ glad you like it! your insights are most helpful. I have one or two further questions on related themes. $\endgroup$ – David Holden Dec 30 '13 at 7:53
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    $\begingroup$ Ask away but I need to get to bed. It is 3AM here. I am sure there are lot of others who will provide a far better answer than me. $\endgroup$ – user44197 Dec 30 '13 at 7:56
  • $\begingroup$ thanks very much for your input, Natarajan. I've been up all night on this. there has been considerable progress in my understanding and terminology. $\endgroup$ – David Holden Dec 30 '13 at 8:28

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