# Suppose $f_1(x)=\sin(1+\sin(x))$, $f_2(x)=\sin(1+\sin(2+\sin(x)))$ and so on. Does $\lim_{n\to\infty}f_n(x)$ eventually become a constant function?

I came across this question on Quora. I'm rephrasing the question in the title.

A sequence of functions $$\{f_n(x)\}$$ is recursively defined such that $$f_{n}(x)=f_{n-1}(n+\sin x)$$ and $$f_0(x)=\sin(x)$$. The question is:

Does $$\lim\limits_{n\to\infty}{f_n(x)}$$ eventually become a constant function?

I intuitively thought that the answer is NO. This was my line of reasoning:

Whatever be the value of $$n$$, if we take a derivative of $$f_n(x)$$, we get another function in terms of $$x$$. But the derivative of a constant function is supposed to be zero, which is contradiction.

However, to my surprise, when I took to Python terminal to code this sequence, I found that the limit converges somewhere around $$0.9941666781206763$$ irrespective of the value of $$x$$ (which indicates it's indeed a constant function).

>>> from sys import setrecursionlimit
>>> setrecursionlimit(10**5)
>>> f = lambda n, x: f(n-1, n+sin(x)) if n>0 else sin(x)
>>> from math import sin
>>> f(1500, 10)
0.9941666781206763
>>> for x in range(0, 100):
...  print(f(1800, x))
...


I believe $$n=1800$$ iterations is fairly high i.e, sufficient for computing $$n\to\infty$$. Can anyone explain what was wrong with my intuition and/or if there's any bug in my code? I am not sure if the latter part of my query is in the scope of this site.

## 2 Answers

It is perfectly possible for a sequence of functions to approach a constant (point-wise) without the derivative eventually being close to zero. Take for example $$f_n(x)=\sin (nx)/n$$. Clearly $$f_n(x)\to 0$$ for every $$x$$ but $$f_n'(x)=\cos nx$$ which takes all the values between $$-1$$ and $$1$$.

As for the reason why your sequence approaches a constant, think about the following: every time you apply the $$\sin(\cdot)$$ function, you are shrinking the range of values to which the next $$\sin$$ will be applied, consequently shrinking the range again (this is because $$|\sin'(x)|=|\cos x|\le 1$$) The reason why your function is approaching a constant is that the outermost sine function is effectively acting over a very small range of values, thus becoming a constant in the limit.

As the first answer points out, it is feasible that $$f_n$$ approaches a constant even if $$f_n'$$ does not approach zero. Honestly, though, I think that rather misses the point since, in this particular situation, it's not to hard to show that $$f_n' \to 0$$ for most $$x$$. Indeed, we can apply the recursive definition

$$f_{n}(x)=f_{n-1}(n+\sin(x))$$

to obtain

$$f_n'(x) = \frac{d}{dx} f_{n-1}(n+\sin(x)) = f_{n-1}'(n+\sin(x))\cos(x).$$

From there, we see that a uniform bound on $$f_{n-1}'$$ implies a better uniform bound $$f_n'$$. In particular, $$f_n'$$ converges to zero uniformly on any closed interval contained in the open interval $$0. That, combined with a standard result in real analysis implies that, if $$f_n(x)$$ converges for some $$x$$ in one of those closed intervals, then it must converge for all $$x$$ in those intervals. But it's easy to see that $$f_n'(\pi/2)=0$$ for all $$n$$, since $$f_n$$ is symmetric about $$x=\pi/2$$. Thus, we certainly have that $$f_n$$ converges to a constant function on $$\{x:0 < x < \pi\}.$$ Symmetry and periodicity then dictates that $$f_n$$ converges to that same constant on $$\mathbb R$$, with the possible exceptions of the integer multiples of $$\pi$$.

Numerical experimentation certainly suggests that $$f_n'(0) \to 0$$. I don't quite see the proof of that right now, though it's not too hard to show that zero is definitely a cluster point.

• You are absolutely right. I made the point because the OP seems to believe that the derivative not stabilizing to zero is a reason to conclude that the limit is not a constant. Apr 20, 2023 at 0:21