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Since $\cos (2x) = 2 \cos^2 (x) - 1$, I wonder about the iteration $(x,y) \mapsto (2x\mod 2\pi, 2y^2-1)$. Will it converge to the cosine graph? I tried it in mathematica and got quite a few points, if I cheat and give it an initial value.

x = {1, 0.540302306};
step[{a_, b_}] := {Mod[2 a, 2 Pi], 2 b^2 - 1}
Table[(x = step[x]; x), {k, 1, 30}];
ListPlot[%, AspectRatio -> Automatic]

Here's a variation based on the sum of angles formula gives nice convergenc to the graph.

x = {1, 0.540302306};
step[{a_, b_}] := (c = Mod[a + 1, 2 Pi]; {c , 
b Cos[1] - If[c < Pi, 1, -1] Sin[1] Sqrt[1 - b^2]})
Table[(x = step[x]; x), {k, 1, 500}];
ListPlot[%, AspectRatio -> Automatic]

These two algorithms have difference tolerances for error. In the first algorithm, if I start with a wrong initial value like {1, 0.4} I get a chaotic "cloud", but in the second case, it still converges to the cosine graph.

I guess I would like to know, does my original iteration have any other fixed points besides the cosine graph?

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2 Answers 2

up vote 4 down vote accepted

Because the $x$ and $y$ coordinates do not interact in your iteration, it does not make sense to expect any relation between them in the limit.

For example, the line $\{y=1\}$ is also invariant under your map.

You could ask the following: What are the real-valued continuous functions $h$, defined on the circle, such that the graph of $h$ is invariant under the map you define?

Then you are asking about semiconjugacies between angle doubling and the Chebyshev polynomial $T(x) = 2x^2-1$. That is, your function $h$ must satisfy $h(2x) = T(h(x))$.

Note that the function $h(x) = T^{\circ n}(\cos(x))$, where $T^{\circ n}$ denotes the $n$-th iterate, provides additional examples, since $$ h(2x) = T^{\circ n}(\cos(2x)) = T^{\circ (n+1)}(\cos(x))=T(h(x)).$$

It seems possible that these (together with the constant map $h(x)=1$) are the only examples, but I have not checked this.

In any case, you cannot expect convergence to these curves unless you start exactly on them.

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I gave the first over 400 iterations to settle down starting from (1, 0.4) and still got a chaotic cloud. Even starting with (1,cos(1)), Excel had a chaotic cloud after 400. I don't think this is surprising as you have two separate iterations that don't talk to each other, and the first coordinate one is about the most unstable iteration known.

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