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I remember years ago coming across some seemingly non-trivial (ie. non-fixed point related) limits describing to the behavior of infinitely iterated trigonometric functions, but I can't for the life of me remember how to construct the proof.

Can someone point me in the right direction?

Specifically, I want to prove the following limits:

$$ \lim _{\left|n\right|\to \infty }\sqrt{\frac{4n}{3}}\left(\sin ^{\left\{n\right\}}\left(\frac{1}{\sqrt{n}}\right)\right) = 1 $$ $$\textbf{and}$$ $$ \lim _{\left|n\right|\to \infty }\sqrt{\frac{5n}{3}}\left(\tanh ^{\left\{n\right\}}\left(\frac{1}{\sqrt{n}}\right)\right) = 1 $$




I.e. that is to say:

$$ \sin \left(\sin \left(\sin \left(\sin \left(\sin \left(\frac{1}{\sqrt{5}}\right)\right)\right)\right)\right) \cdot \sqrt{\frac{4\cdot 5}{3}} \approx 1 $$
$$ \tanh \left(\tanh \left(\tanh \left(\tanh \left(\tanh \left(\tanh \left(\frac{1}{\sqrt{6}}\right)\right)\right)\right)\right)\right)\cdot \sqrt{\frac{5\cdot 6}{3}}\approx 1 $$
$$ \operatorname{arcsinh}\left(\operatorname{arcsinh}\left(\operatorname{arcsinh}\left(\frac{1}{\sqrt{3}}\right)\right)\right)\cdot \sqrt{\frac{4\cdot 3}{3}}\approx 1 $$

... and so on, noting the absolute value in the limits.

Note on Notation:

It seems people use a variety of different notations for expressing function iteration, but I went with this one since it felt most natural: $$ f^{\left\{0\right\}}\left(x\right)=x $$ $$ f^{\left\{1\right\}}\left(x\right)=f(x) $$ $$ ... $$ $$ f^{\left\{k\right\}}\left(x\right)=f\left(f^{\left\{k-1\right\}}\left(x\right)\right)\text{ } \forall k\in \mathbb{Z} $$

This has been bugging me for a while, but I can't seem to make any substantive progress (despite several hours of unsuccessful attempts to reconstruct the proof from old notes), so I will be forever grateful if you guys can give me some guidance!

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  • $\begingroup$ Interesting and un-intuitive to me. Your first one off-the-cuff suggests to use $\sin x \approx x$ for very small $x$, and so the iterated sine would give $\approx \sqrt{1/n}$ and the entire expression would then yield $\sqrt{4/3}$. Would be happy to see advice from more seasoned people. $\endgroup$ Commented Sep 23, 2020 at 13:17
  • $\begingroup$ Another curious limit from my notebook around this period that might help, seems to be a simplification: $$\lim_{n\to \infty }\frac{\sin ^{\left\{n^2\right\}}\left(\frac{1}{n}\right)}{\sinh ^{\left\{n^2\right\}}\left(\frac{1}{n}\right)}=\frac{1}{\sqrt{2}}$$ $\endgroup$ Commented Sep 23, 2020 at 15:38
  • $\begingroup$ just checked the $\sin()/\sinh()$ formula via taylorexpansion. Got the same result,seems to be correct. $\endgroup$ Commented Sep 26, 2020 at 17:21

2 Answers 2

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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

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

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 $\frac{y_n}n\to\frac53$.

Generalizing to $x_0=\frac{x}{\sqrt{n}}$ one finds $\frac{y_n}{n}=\frac1{x^2}+\frac1{3}+O(\frac{\log(n)}{n})$, so that $$ \lim_{n\to\infty}\sqrt{n}\sin^{\circ n}\left(\frac{x}{\sqrt n}\right)=\frac{x}{\sqrt{1+3x^2}}. $$

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  • $\begingroup$ Is there a simpler way to understand geometrically why the ratio of sin and hyperbolic sin iterated in this way yields 1/sqrt(2)? $\endgroup$ Commented Apr 4, 2022 at 17:38
  • $\begingroup$ With $f(x)=\frac{\sin(x)}{\sinh(x)}$ you get a different class of functions, as this has a constant term. I'm not sure where you get the value from, the fixed point in $[0,1]$ is at $0.8054$, which is not $\sqrt{0.5}$. But you get $\frac{\sin^2x}{\sinh x}=x-\frac12x^3+O(x^5)$, which would indeed converge to $\frac1{\sqrt{1+2·\frac12}}$. $\endgroup$ Commented Apr 4, 2022 at 18:57
  • $\begingroup$ Thank Lutz, also for anyone curious, there are some helpful visual diagrams in jstor.org/stable/10.4169/math.mag.87.5.338 "Iteration of Sine and Related Power Series" which also goes into proving the arctan identity which clarified the tanh^{-x} case where -x is a negative integer. $\endgroup$ Commented Apr 5, 2022 at 1:35
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Disclaimer: This isn't really an answer, but something that I tried.

I use $\sin_n$ to denote the sine function iterated $n$ times. I formulate the problem as: Show that $$ \sin_n (\frac{1}{\sqrt n}) \to \frac{\sqrt 3}{2} \frac{1}{\sqrt n}$$ I saw this post about Taylor approximation for iterated sine: here which says that $$ \sin_n(x) = x - \frac{n}{6}x^3 - \left(\frac{n}{30} - \frac{n^2}{24} \right)x^5 + \epsilon$$ I plug in $x = 1 / \sqrt n$ and get $$ \sin_n(\frac{1}{\sqrt n}) = \frac{1}{\sqrt n} \left( \frac{5}{6} - \left( \frac{\frac{4}{n} - 5}{120} \right) \right) + \epsilon$$ So as $n \to \infty$, the term inside the big brackets goes to $$ \frac{5}{6} + \frac{5}{120} = 5 \left(\frac{1}{3!} + \frac{1}{5!} \right)$$ I just make a wild guess that if more terms of the Taylor expansion are used, you going to get a pattern $$5 \left(\frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} \dotsm\right)$$ and Wolfram says this is $5(\sinh(1)-1) = 0.8760...$. Compare to $\sqrt 3 / 2 = 0.8660...$ and it seems pretty close...

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  • $\begingroup$ Interesting, thanks for the help! I'm gonna try the same method with tanh and see if this is the way to go! $\endgroup$ Commented Sep 23, 2020 at 15:39
  • $\begingroup$ I get for the $\tanh()$-version $0.774596669241 \approx \sqrt{3/5}$ . Here I use the method of Carleman-matrices for functions which have a taylorseries expansion and which exploit the concept of simple matrix-powers for function-iterates. It's a simple method and I can put example in an answer-box if still needed. $\endgroup$ Commented Sep 26, 2020 at 9:39

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