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Background

Consider the following iteration:

$$ \begin{aligned} x_{n+1} = \sin(x_n)\\ y_{n+1} = \cos(y_n)\\ z_{n+1} = \tan(z_n)\\ \end{aligned} $$

For any $(x_0, y_0, z_0)\in\mathbb{R}$, $x_n$ converges to $0$, $y_n$ converges to $0.739$, $z_n$ does not always converge.

And for almost all initial values, $y_n$ converges faster than $x_n$.

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Question

I would like to know if there is any way to measure the iterative convergence speed of two elementary functions.

These elementary functions are not always continuous, but are smooth on every piecewise interval.

Aster
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  • Asymptotically, $\ln x_{n+1} \approx \ln x_{n}-\frac16x_n^2$. Try and compare that to power laws for example, such as $n^{\epsilon-\frac12}$, with $\epsilon$ small, either negative or positive. – Axel Jan 12 '24 at 10:05
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    Do you mean $$ \begin{aligned} x_{n+1} = \sin(x_n)\ y_{n+1} = \cos(\color{red}y_n)\ z_{n+1} = \tan(\color{red}z_n)\ \end{aligned} $$ – NN2 Jan 12 '24 at 10:12
  • As for $z_n$, with Taylor expansion around $0$, you can see it is almost the same as $x_n$. You should compare it to the same power sequences. – Axel Jan 12 '24 at 10:21
  • $1/x_n^2$ increases by a roughly constant amount at each step – Empy2 Jan 12 '24 at 12:48
  • Oops, my bad about $z_n$. See my answer below. – Axel Jan 12 '24 at 16:15

1 Answers1

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Do Taylor expansion of you functions around their stable points to find the asymptotic behavior of the associated sequences and then compare the sequences to well chosen power sequences : $n^{\epsilon-\frac12}$ for $x_n$, with $\epsilon$ small, either negative or positive or even $0$.

Assume $\frac{k_1}{n^{\frac12}}<x_n<\frac{k_2}{n^{\frac12-\epsilon}}$ and show it stays true for $x_{n+1}$.

As for $|y_n - l|$, it should be compared to $e^{-\alpha n}$.

For $z_n$, the Taylor expansion around fixed point $0$ shows you that this fixed point is unstable: if you are close you are thrown away. You can check that it is also the case of the other fixed points of $\tan$. I am even willing to bet that for almost all starting values, $(z_n)$ is dense in $\mathbb{R}$.

I would further add that the set of starting points that make $(z_n)$ be undefined after a while is dense in $\mathbb{R}$. Also the set of starting points that make $(z_n)$ converge is dense in $\mathbb{R}$.

EDIT : Sorry for the numerous mistakes that I had to correct.

Axel
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  • Assume $\frac{k_1}{n^{\frac12}}<x_n<\frac{k_2}{n^{\frac12-\epsilon}}$ and show it stays true for $x_{n+1}$. – Axel Jan 12 '24 at 10:53