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Is it possible to evaluate limits involving sequences of function compositions?

For example, given the expression

$$g(x, n) = \sin(x)_1 \circ \sin(x)_2 \circ [...] \circ \sin(x)_n$$

is it possible to calculate the following limit?

$$\lim_{n \to +\infty} g(x, n)$$

(Intuitively, this limit would equal 0.)

haroba
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    I don't understand what you mean by the composition. Usually when you write $f \circ g (x)$ it takes only one argument, as in $f(g(x))$. Do you mean $(\sin(x))^n$? Then in the limit, does $x$ go to infinity? Or does $n$ go to infinity for a fixed $x$? – Ross Millikan Jun 02 '11 at 17:44
  • Are you asking about something like, say, the contraction mapping theorem? – Zhen Lin Jun 02 '11 at 17:47
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    This answer by David Speyer might be useful: http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n/3220#3220 – Aryabhata Jun 02 '11 at 19:35
  • Ross: I think the right way to interpret my composition is

    $$g(x, n) = \sin(x)1 \circ (\sin(x)_2 \circ ([...] \circ (\sin(x){n-1} \circ \sin(x)_n))$$

    I got curious about this question when I looked at graphs of sin(sin(sin(sin(x)))) and so on.

     – haroba Jun 02 '11 at 21:49

2 Answers2

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I will make a guess about what the question intended to ask. If I am told that my interpretation is not the intended one, this answer will be deleted.

Let $f(x)$ be a function, and in general define $f^{(n)}(x)$ by $f^{(1)}(x)=f(x)$ and $$f^{(n+1)}(x)= f(f^{(n)}(x)).$$

Let $g(x,n)=f^{(n)}(x)$. I interpret the question as asking whether one ever is interested in $$\lim_{n\to\infty}g(x,n).$$

A quick answer is yes, often, this is a very important kind of question, with many applications. You will find a guide to a possible exploration in the following Wikipedia article. The iteration of functions, and the possible limiting behaviour, is a frequent theme in many branches of mathematics, both pure and applied.

Your specific question: Indeed, if we let $f(x)=\sin x$, and interpret "sequence of function compositions" as I did, the limit, as you conjectured, is $0$ for all $x$. But the situation with $f(x)=\cos x$ is different. You can explore this by putting your calculator into radian mode, starting at some number, and pushing the cos button repeatedly.

André Nicolas
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  • Thank you! You interpreted my answer correctly. It seems I did not understand the topic well enough to ask the question properly. – haroba Jun 02 '11 at 21:47
  • @Aqwis: Note that many numerical procedures, including the Newton-Raphson method, and more generally fixed point iteration, use the idea that you described. – André Nicolas Jun 02 '11 at 21:58
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For your specific question (with the same interpretation as user6312): we have $|\sin(x)|\le |x|$ for every $x\in\mathbb{R}$. So if we fix $x\in\mathbb{R}$ then $|\sin^{(n)}(x)|$ is a decreasing sequence of non-negative real numbers, so it converges to some $s\ge0$. Since $|\sin|$ is continuous and even, we have $$ |\sin(s)|=|\sin(\lim\limits_{n\to \infty}|\sin^{(n)}(x)|)|=\lim\limits_{n\to \infty}|\sin(|\sin^{(n)}(x)|)|=\lim\limits_{n\to \infty}|\sin^{(n+1)}(x)|=s, $$ but if $t>0$ then $|\sin(t)|<t$. So $s=0$.

mac
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