# How fast does $f_n = \sin f_{n-1}$ approach zero? [duplicate]

The sequence $f(n) = \sin(\sin(\sin(......(1)......)))$ approaches zero like $\sqrt{3/n}$, as has been asked and answered here a few times.

So $f(n)$ would get below $1/n$ after $3n^2$ steps, but it seems to get there about $\ln n$ steps earlier: 0.1 after 295, 0.01 after 29992, 0.001 after 2999989 steps.

Is that numerical roundoff error or not?

• What is your measurement of the speed of convergence? Jun 20, 2013 at 14:52
• I think we should attempt to characterize the speed of convergence by some other simpler functions. For example, show that f(n) ~ g(n) as n tends to inf, where g(n) is easier to work with than sin(sin(sin... (1.. ) Jun 20, 2013 at 14:54
• this may help: math.stackexchange.com/questions/45283/…
– Alex
Jun 20, 2013 at 15:12
• This question has been asked before. 1 2 3. Jun 20, 2013 at 15:16
• Anyway, the answer is that $f(n) \sim \sqrt{3/n}$, and also the stronger result that $f(n) < \sqrt{3/n}$. Jun 20, 2013 at 15:17

Here is an incomplete answer which may still satisfy you: By a Taylor series approximation, when $x$ is small, $$\sin x = x - \frac {x^2} 2 + O(x^3)\\ \sin \sin x = \sin x - \frac { (\sin x)^2 } 2 + O((\sin x)^3) = x - x^2 + O(x^3) \\ \sin \sin \sin x = x - x^2 - \frac { (x - x^2)^2 } 2 + O(x^3) = x - \frac 3 2 x^2 + O(x^3)$$ and so on... One could show by induction that $$\underbrace {\sin \sin \ldots \sin }_{n}\ x=x-\frac n 2 x^2 + O(x^3)$$ The constant hidden in the $O$-notation depends on $n$ of course, so this cannot be used to give you precise answers, but you can get an order of magnitude.
For example, how many sines does one need to take to get from $x = \epsilon$ to $x = \frac \epsilon 2$ when $\epsilon$ is small? Let's say $\epsilon - \frac n 2 \epsilon^2 \approx \frac \epsilon 2$; then you need $n \approx \frac 1 \epsilon$ steps.