Skip to main content

Compute $\sin(\sin(...\sin(x)))$

Problem

Find the limit $\sin(\sin(...\sin(x)))$ for any real number $x$.

Solution

The problem can be precisely formulated as follows: Find the limit $L$ such that $$ L \equiv\lim_{n\rightarrow\infty}{f_n(x)}, $$ where $x\in\mathbb{R}$, $n\in\mathbb{N}$, and $$\begin{align} f_1(x) &= \sin(x)\\ f_n(x) &= \sin(f_{n-1}(x)). \end{align}$$ We observe that, if the limit $L$ exists, $$\begin{align} \lim_{n\rightarrow\infty} f_{n-1}(x) &= L\\ \lim_{n\rightarrow\infty} f_n(x) &= L, \end{align}$$ and thus, $$ L = \sin{L}, $$ which we show to be $L=0$ in the following.

First, we show that the limit exists. We accept the Axiom of Continuity of the Real Line, which says that any bounded monotonic sequence converges. Thus, to show that $L$ exists, we need only show that (1) $f_n(x)$ is a bounded sequence of $n$, and that (2) $f_n(x)$ is a monotonic sequence of $n$.

  1. $f_n(x)$ is clearly bounded because $f_n(x) = \sin{y}$, where $y\equiv f_{n-1}(x) \in \mathbb{R}$ and from trigonometry, $-1 \leq \sin{y}\leq 1$ for $\forall y\in\mathbb{R}$. Consequently, $f_n(x)$ is also bounded; in particular, $-1 \leq f_n(x) \leq 1.$
  2. We now show that if $f_1(x)\in[0,1]$, then $f_n(x)\in[0,1]$ and if $f_1(x)\in[-1,0]$, then $f_n(x)\in[-1,0]$. If $f_1(x)\in[0,1]$, then $\sin(f_2(x))\in[0,1]$ because $\sin{x}$ is an odd function and because of the preceding section. Likewise, for all $k\in\mathbb{N}$, $f_k(x)\in[0,1]$ implies $f_{k+1}(x)\in[0,1]$, which, by mathematical induction, means that $\forall n\in\mathbb{N}$, $f_n(x)\in[0,1]$. The same proof can be employed to show the case for $f_1(x)\in[-1,0]$.
  3. In the interval $[0,\frac{\pi}{2})$, $f_n(x)$ is monotonic decreasing because $f_n(x) = \sin(f_{n-1}(x))$ and $\sin(f_{n-1}(x)) \leq f_{n-1}(x)$, which follows from the geometry of the definition of $\sin{x}$ as shown in the figure, which clearly shows that $\sin{\theta} \leq h $ because $h$ is the hypotenuse of the right triangle whose one leg is $\sin{\theta}$, and $h \leq \theta$ because $h$ measures a line segment while $\theta$ measures an arc both of which have the same endpoints. Thus, $ \sin{\theta} \leq \theta$.
  4. In the interval $(-\frac{\pi}{2},0]$, $f_n(x)$ is monotonic increasing because if we let $y \equiv -f_n(x) $, then $y \in [0,\frac{\pi}{2})$, and the preceding section applies so that we can conclude that $$\begin{align} \sin y &\leq y\\ \sin (-f_{n-1}(x)) &\leq -f_{n-1}(x)\\ -\sin (f_{n-1}(x)) &\leq -f_{n-1}(x)\\ \sin (f_{n-1}(x)) &\geq f_{n-1}(x). \end{align}$$
  5. The last two sections imply that $f_n(x)$ is a bounded and monotonic sequence in each of the intervals $(-\frac{\pi}{2},0]$ and $[0, \frac{\pi}{2})$. This means that the limit $L$ exists in those intervals. Furthermore, the monotonicity impies that $L = \sin L$ is a one-to-one function in those intervals, which means that $L = 0$ is the only solution to $L = \sin{L}$. ■

References

Labels:

Comments

Post a Comment

Popular posts from this blog

Find the limit of the sequence. $a_1=1, a_{n+1} = \frac{3a_n+4}{2a_n+3}$

We take the limits of sequences defined recursively by first showing the existence of the limits then, actually computing them. Existence is demonstrated by showing that the sequence is both (1) monotonic and (2) bounded. Problem Find the limit of the following sequence. $$\begin{align} $a_1&=1\\ a_{n+1} &= \frac{3a_n+4}{2a_n+3}$ \end{align}$$ Solution Monotonicity : We use mathematical induction to show that for all $n\in\mathbb{N}$, $a_{n+1} - a_n \geq 0$ and therefore, $\{a_n\}$ is monotonic increasing. For the base case at $n=1$, it is easy to see that $a_{n+1}-a_n = \frac{7}{5} - 1 = \frac{2}{5} >0.$ For the inductive case at $n=k$, we assume that $$\begin{align} a_{k+1} - a_{k} & = \frac{3a_k+4}{2a_k+3} - \frac{3a_{k-1}+4}{2a_{k-1}+3}\\ & = \frac{a_k - a_{k-1}}{(2a_k+3)(2a_{k-1}+3)} \geq 0. \end{align}$$ For $n=k...
Post a Comment
Read more

Prove: $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$, $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$

We prove identities involving inverse trigonometric functions. Sum formulas from trigonometry are exploited. $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$ Suppose that $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} \neq \frac{\pi}{4}. $$ Then, $$\begin{align} \tan{\left(\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}\right)} &\neq \tan{\frac{\pi}{4}}\\ \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} &\neq 1 \\ \frac{5}{6-5} &\neq 1 \\ 1 &\neq 1, \end{align}$$ which is a contradiction. Thus, $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}.□ $$ $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$ Suppose that $$ \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} \neq \arcsin{\frac{3}{5}}. $$ Then, $$\begin{align} \sin{\left(\arcsin{\frac{16}{65}}+\arcsin{\fra...
Post a Comment
Read more