Examine the convergence of a sequence ${a_{n}}$: $a_{n+1}=a_{n}-\sin(a_{n}),0\le a_{1}<\pi$

Question

${a_{n}}$: $a_{n+1}=a_{n}-\sin(a_{n}),0\le a_{1}<\pi$

One way to do it is to show that the sequence is bounded and monotonous.

How to show that it is bounded?

If

$$-1\le \sin(a_{n})\le 1$$

then

$$0\le a_{n}<2\pi$$

Is it right that then

$$0\le a_{n+1}\le 2\pi$$

and the sequence is bounded (below and above)?

From

$$a_{n+1}-a_{n}=a_{n}-\sin(a_{n})-a_{n}=-\sin(a_{n}),a_{n}\in[0,2\pi]$$

the sequence is monotonous and non-increasing.

Can someone check this problem?

Answer 1

Your first implication does not hold. A counterexample would be $a_n=100$. Its sine is certainly between $-1$ and $1$, but $100$ itself is not between $0$ and $2\pi$.

I think it would be easier to say that if $a_n\in[0,\pi]$ then $a_{n+1}\in[0,\pi]$ because $0\le\sin x\le x$ when $x\in[0,\pi]$. Then, by induction, every $a_n$ is in this interval.

It is also true that if $a_n\in[0,2\pi)$ then $a_{n+1}\in[0,2\pi)$, but on one hand it is more work to argue that (because $\sin a_n$ can then be both positive or negative), and on the other hand it doesn't tell you enough that you can argue that the sequence is monotonic. For that you need to know that the $a_n$s stay within the range where the sine is nonnegative.

Answer 2

Hint: It suffices to show $0 \leq \sin a_n < a_n$ for $a_n$ in your range between $0$ and $\pi$.

Answer 3

First of all, the sine of any number is in $[-1,1]$, not only of numbers in $[0,2\pi)$. But that does not interest us.

To show this is monotonic, we will show that $\sin(a_n)\geq0$ for all $n$. For $a_1$, this is easy, since the sine is positive in $[0,\pi]$. $a_2$ is still in that interval since it is bounded by $a_1-0$ and $a_1-1$ which are both there. So $\sin(a_2)\geq0$. If $a_n\in[0,\pi]$ for all $n$, we are done. But it is easy to show $x-\sin x\geq0$ for $x\geq0$, since if you draw a unit circle and choose a point at angle $x$ from the positive $x$-axis, the $y$ of that point is $\sin x$ and the arc is $x$ in length since the radius is 1, and the arc is clearly longer than the segment. This proves our claim for $x\in[0,\frac{\pi}{2}]$, and after that $x\geq\frac{\pi}{2}>1\geq\sin x$. This means our sequence is bounded by 0, and therefore cannot ever escape $[0,\pi]$, thus the sines are all positive, and the sequence is monotonic nonincreasing.

While proving monotony, we found a lower bound, 0.

Lower bound plus nonincreasing monotony implies convergence, thus $a_n$ converges to some limit, probably 0.

Answer 4

This problem can be solved in a more generic manner.

Let $g(x) = x - \sin(x)$. The key observation is

• $g'(x) = 1 - \cos x \ge 0 $ for all $x$, so $g(x)$ is increasing over whole $\mathbb{R}$.
• $g'(x) = 0$ at and only at isolated points $2n\pi$ for $n \in \mathbb{Z}$, this implies $g(x)$ is strictly increasing.

In other words, $g$ is order preserving!

Start from any real number $b$, one can construct a sequence $( b_n )_{n\in\mathbb{Z}}$ by $$b_n = \begin{cases} b, & n = 1\\ g(b_{n-1}), & n > 1\end{cases}$$ There are three possibilities:

1. $b_2 = g(b_1) = b$ : we will have $b_n = b$ for all $n$.

2. $b_2 = g(b_1) > b$ : we will have $$b_2 > b_1 \implies b_3 = g(b_2) > g(b_1) = b_2 \implies b_4 > b_3 \implies \ldots$$ i.e. $b_n$ will be an increasing sequence.

3. $b_2 = g(b_1) < b$ : we will have $$b_2 < b_1 \implies b_3 = g(b_2) < g(b_1) = b_2 \implies b_4 < b_3 \implies \ldots$$ i.e. $b_n$ will be an decreasing sequence.

Furthermore, if $m, M$ are any pair of successive fixed points of $g(x)$, then for any $b \in (m, M)$, $$ m < b_1 < M \implies m = g(m) < b_2 = g_(b_2) < M = g(M)\\ \implies m < b_3 < M \implies m < b_4 < M \implies \ldots$$ i.e. the whole sequences $b_n$ will falls inside $(m,M)$. Being a bounded monotonic sequence, $b_n$ will converge to some numbers $b_\infty \in [m,M]$. Since $g(x)$ is continuous, $g(b_\infty) = b_\infty$. This implies $b_\infty \ne (m,M)$ and hence $b_\infty = m$ or $M$.

For those cases where $g(x)$

• has a largest fixed points $m'$ and $b$ falls between $(m',\infty)$
• or a smallest fixed points $M'$ and $b$ falls between $(-\infty,M')$,

the analysis will be similar. In general, we will have

$$b_{\infty} \stackrel{def}{=} \lim_{n\to\infty} b_n = \begin{cases} +\infty,& \text{ if } g(b) > b, b \in (m', \infty)\\ m',& \text{ if } g(b) < b, b \in (m', \infty)\\ M,& \text{ if } g(b) > b, b \in (m, M)\\ m,& \text{ if } g(b) < b, b \in (m, M)\\ M',& \text{ if } g(b) > b, b \in (-\infty,M')\\ -\infty,& \text{ if } g(b) < b, b \in (-\infty, M') \end{cases} $$

Apply this to our function $g(x) = x - \sin(x)$. Notice the fixed points of our $g(x)$ are located at $n\pi$ with $n \in \mathbb{Z}$. Over the two successive fixed points $0$ and $\pi$, $$g(x) = x - \sin(x) < x, \quad\text{ for } x \in (0,\pi)$$

By above argument, we find $\;\lim\limits_{n\to\infty} a_n = 0$.

As one can see, what we really need from $g(x)$ is the properties that it is continuous and strictly increasing. If you generate a sequence by functional iteration of a strictly increasing continuous function, the sequences converges to either one of its nearby fixed points or $\pm \infty$. Which fixed point to converge is controlled by the sign of $b_2 - b_1$.