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I was reading this question Limit of $(\sin\circ\sin\circ\cdots\circ\sin)(x)$ and nowhere did it talk about how to prove the existence of the limit.

In general, if you're given a function $f(x)$, how do you check for the existence of $\lim\limits_{n\to \infty} \underbrace{(f\circ f \circ \cdots \circ f)}_{n\text{ times}}(x)$?

When $f(x) = \sin(x)$, if the limit $c$ exists , then $f(c)$ must equal $c$ so $c = 0$ right? But isn't it also possible that the limit doesn't exist?

Or is it the case that when you have a monotonically decreasing function that's bounded below by some value, then $\lim\limits_{n\to\infty} \underbrace{(f \circ f \circ \cdots \circ f)}_{n\text{ times}}(x)$ is defined and necessarily a value that satisfies the equation $f(x) = x$?

Oh, and as a side note, is there a better way of denoting $f(x)$ composed with itself $n$ times?

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Herman Chau
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  • It's often denoted by $f^{(n)}$. – Hans Lundmark Sep 10 '11 at 13:51
  • "is there a better way of denoting $f(x)$ composed with itself $n$ times?" - generically, $f^{(n)} (x)$, but that would be quite confusing if applied to trigonometrics... and it might be confused with derivative notation. I suppose one should be able to infer from context. – J. M. ain't a mathematician Sep 10 '11 at 13:54
  • In Ross Millikan's answer to the question you're referring to, he does explain how to prove that the limit exists (in the case $f=\sin$). – Hans Lundmark Sep 10 '11 at 13:55
  • Yeah, I thought $f^{(n)}$ usually stands for the $n$th derivative or, in trigonometry, raising the function to the $n$th power. – Herman Chau Sep 10 '11 at 13:55
  • @HansLundmark I might not be understanding Ross Millikan's answer correctly, but he seems to be saying that you need to first prove that the limit exists and then the limit must be where $\sin(x) = x$. – Herman Chau Sep 10 '11 at 13:57
  • @Herman: Yes, and in between he explains that the sequence of numbers will be decreasing and bounded from below (which implies that it has a limit; this is a very well-known criterion, so I guess he didn't even bother to state that conclusion explicitly). – Hans Lundmark Sep 10 '11 at 14:10
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    May I opt for the notation $f^{\circ n}$ that I have seen on some occasions. – Lord_Farin Apr 22 '13 at 09:52

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As mentioned in the comments, Ross's answer in the linked question does prove that the limit exists. But with respect to your last comment, it might not be apparent. So I explain.

In general, when you have $\lim\limits_{n\to \infty} \underbrace{(f\circ f \circ \cdots \circ f)}_{n\text{ times}}(x)$, a standard way of showing this exists is to actually consider the sequence $f(x), f(f(x)), f^{(3)} (x), ...$ and so on. Then for any fixed x, this is just a sequence of real numbers. And then you do whatever you would do for that sequence of real numbers.

There is the possibility that there will be radically different behavior for different starting x. If that's the case, then you might have to be witty. Perhaps do something different, or perhaps break it up into cases and still consider the sequence above. It's a bit hard to say (giving general methods in mathematics is always very hard).

Implicitly, Ross did this. He also noted that when you iterate a few times, the sequence will monotonically decrease. And it's bounded. So it must converge.

davidlowryduda
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  • I see now. I was originally confused because I didn't know that if a sequence is monotonic and bounded, it must converge. And thanks for the answer to the more general case. But what if the function oscillates, like if $f(x) = \frac{1}{1-x}$ ($x \neq 1, 2$). Are there any techniques to evaluate the limit? – Herman Chau Sep 10 '11 at 14:37
  • @Herman: I'm pretty sure you chose poorly - as that function's iterative sequence does not converge (Why do I think this? If it converges, it will almost always converge to a fixed point - and it has none). But suppose it was $f(x) = \frac{1}{2-x}$, whose iterative sequence does converge. Yes, there are methods. I would recommend taking a course on analysis first, so that you know more methods about the convergence of sequences. One can prove that for this function, there is at least one interval in which the sequence is also monotone and bounded, and that there is a fixed point. But there... – davidlowryduda Sep 10 '11 at 14:45
  • Calculate $f(f(x))$, $f(f(f(x)))$ and you will get your answer. – André Nicolas Sep 10 '11 at 14:45
  • ... are many other nethods that you may have to use. Some functions are simply pathological. – davidlowryduda Sep 10 '11 at 14:45
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For positive $x$, $0 > \sin(x) > x$, so the iterated sequence is decreasing and bounded. Such a sequence always has a limit. Similarly for negative $x$ (except that the sequence is increasing and bounded.

hmakholm left over Monica
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