Proving the limiting behavior of functions containing iterated trigonometric functions.

You can compare the iteration to $x_{n+1}=x_n+ax_n^2$ or $x_{n+1}=x_n+ax_n^3$ where you get asymptotic behavior similar to the Bernoulli DE solution method, that is, consider $y_n=x_n^{-2}$ or some other suitable power. In your use case you would have to treat $x_n$ as function of $x_0$ and then insert the special $x_0$ into the asymptotic expression. See

  • Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$
  • Calculating $\lim_{n\to\infty}\sqrt{n}\sin(\sin...(\sin(x)..)$

One other method (which might also be used as refinement of the first one) is to find a conjugation map to transform the recursion into one with known behavior, see Schröder's equation, and as explored in

  • Asymptotic expansion of $v_n = 2^nu_n$ where $u_{n+1} = \dfrac{1}{2}\arctan(u_n)$
  • Asymptotic expansion of $u_{n + 1} = \frac12 \arctan(u_n)$

For the sine example you get for $x_{n+1}=\sin(x_n)=x_n-\frac16x_n^3+...$ that with $y_n=x_n^{-2}$ $$ y_{n+1}=\frac2{1-\cos(2x_n)} =\frac2{2x_n^2-\frac2{3}x_n^4+\frac4{45}x_n^6\pm...} =y_n+\frac13+\frac1{15}y^{-1}+O(y_n^{-2}) \\ \implies y_n=y_0+\frac n3+C+O(\log(3y_0+n)) $$ so that with $x_0=\frac1{\sqrt n}\implies y_0=n$ it follows that $$ \lim_{n\to\infty}\frac{y_n}{n}=\frac43 \implies \lim_{n\to\infty}\sqrt{n}x_n=\frac{\sqrt3}2 $$

In the case of the $\tanh$ iteration, the additive constant changes from $\frac13$ to $\frac23$, everything else remains largely the same, so that $\frac{y_n}n\to\frac53$.

Tags:

Trigonometry

Limits

Sequences And Series

Dynamical Systems

Iterated Function System

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