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Problem

Evaluate $$\lim\limits_{x \to 0} \frac{x-\sin\sin\cdots\sin x}{x^3},$$where $\sin\sin\cdots\sin x$ denotes n-fold composite sine function.

Solution

Consider applying Taylor's Formula with 3-order at $x=0$. We may obtain $$f_n(x)=\sin\sin\cdots\sin x=x-\frac{n}{6}x^3+\mathcal{O}(x^3).\tag{*}$$ To prove this, we can apply the mathematical induction. Let $n=1,$ then $$f(x)=\sin x=x-\frac{1}{6}x^3+\mathcal{O}(x^3),$$ It's true and shows that $(*)$ holds for $n=1$. Assume that $(*)$ holds for $n=k$. Then $$\begin{align*}f_{k+1}(x)&=\sin(f_k(x))\\&=x-\frac{k}{6}x^3+\mathcal{O}(x^3)-\frac{1}{6}\left(x-\frac{k}{6}x^3+\mathcal{O}(x^3)\right)^3+\mathcal{O}(x^3)\\&=x-\frac{k+1}{6}x^3+\mathcal{O}(x^3)\end{align*}.$$ This shows that $(*)$ holds for $n=k+1$. As a result, $(*)$ holds for all $n=1,2,\cdots.$ Now,let's go back to deal with the problem. $$\lim\limits_{x \to 0} \frac{x-\sin\sin\cdots\sin x}{x^3}=\lim\limits_{x \to 0} \dfrac{x-\left(x-\dfrac{n}{6}x^3+\mathcal{O}(x^3)\right)}{x^3}=\frac{n}{6}.$$

Please correct me if I'm wrong. Hope to see other solutions. Thanks.

Jean-Claude Arbaut
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mengdie1982
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  • Your solution is correct, and I don't know a better way. – Jean-Claude Arbaut May 29 '18 at 04:16
  • Maybe one can deduce a formula using the L'Hopital rule. Note that

    $$\frac{d}{dx}\sin(\sin(x))=\cos(x) \cos(\sin(x))$$ $$\frac{d}{dx}\sin(\sin(\sin(x)))=\cos(x) \cos(\sin(x)) \cos(\sin(\sin(x))),$$ and so on... Then, as we know that $$\lim_{x\to0}\frac{1-\cos{x}}{x^2}=\frac{1}{2}$$ maybe we can deduce your result. I don't know if it's works correctly, it's only an idea

     – jnaf May 29 '18 at 04:26
  • @nikilospuntajes But you have to differentiate $3$ times, and that's not going to be pretty. – Jean-Claude Arbaut May 29 '18 at 04:34
  • Your proof of $()$ seems fine to me, but I’m curious about whence $()$ itself originated. Did you use a calculator of some sort and then set about proving the relation, or did you try to deduce it algebraically? – gen-ℤ ready to perish May 29 '18 at 05:27
  • Your solution is very correct. Digging my old notes, I found that $$f_n(x)=\sin\sin\cdots\sin x=x-\frac{n}{6}x^3+\frac{n(5n-4)}{120}x^5+\mathcal{O}(x^7)$$ which shows how is approached the limit. By the way, $\to +1$. – Claude Leibovici May 29 '18 at 05:30
  • This is separate from my other comment and relates to an approach to the problem: Since the limit is taken as $x\to0$, why not just invoke that $$f_1(x)\sim f_2(x)\sim f_3(x)\sim\cdots\sim\sin x$$ for $x$ near $0$? – gen-ℤ ready to perish May 29 '18 at 05:32
  • @ChaseRyanTaylor Answer to first comment: almost obvious once you write the Taylor exansion at order $3$. Answer to second comment: that would give you no information on $x-\sin(\dots)$: you would know it's $o(x)$, and that's not enough. – Jean-Claude Arbaut May 29 '18 at 06:12
  • @Jean-ClaudeArbaut “almost obvious” I don’t think proof by intimidation is a valid problem-solving technique. – gen-ℤ ready to perish May 29 '18 at 12:38
  • @ChaseRyanTaylor Compute by hand the case $\sin(\sin(x))$, and you should see why I write that. – Jean-Claude Arbaut May 29 '18 at 12:47
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    @ChaseRyanTaylor In case the above is still not obvious, there is the mechanical way. Let $f_n(x)=\sin\dots\sin x$ with $n$ sines. It's an odd function, so it's expansion at order $3$ is $u_nx+v_nx^3+O(x^5)$. Since $f_n(x)\sim x$, $u_n=1$. Then write the Taylor formula for $\sin(f_n(x))$ to get a relation between $v_{n+1}$ and $v_n$. You'll find $v_{n+1}=v_n-\frac16$. You can check this at order $5$, and the coefficient of $x^5$ in $f_n(x)$ is a polynomial in $n$ of degree $2$. If I got the recurrence on $k$ right, the coef of $x^{2k+1}$ is always a polynomial in $n$ of degree at most $k$. – Jean-Claude Arbaut May 29 '18 at 16:12

1 Answers1

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I would like to share a more tricky method with you.

Use the same notation, and assume $f_0(x) = x$, then we can prove that $$ \lim_{x \to 0} \frac{x - f(x)}{x^3} = \frac{1}{6} $$ and $$ \lim_{x \to 0} \frac{f_k(x)}{x} = 1, ~ \forall k \in \mathbb{N}. $$ Thus we have $$ \begin{aligned} \lim_{x \to 0} \frac{f_0(x) - f_k(x)}{x^3} &= \lim_{x \to 0}\sum_{i = 1}^k \frac{f_{i - 1}(x) - f_{i}(x)}{x^3}\\ &=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f_{i}(x)}{x^3}\\ &=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f(f_{i-1}(x))}{(f_{i-1}(x))^3}\cdot \frac{(f_{i-1}(x))^3}{x^3}\\ &=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f(f_{i-1}(x))}{(f_{i-1}(x))^3} \cdot \lim_{x \to 0}\frac{(f_{i-1}(x))^3}{x^3}\\ &= \sum_{i = 1}^k \frac{1}{6}\\ &= \frac{k}{6}. \end{aligned} $$

Xiangxiang Xu
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