Recurrence relation with periodic function

Question

$$x_{n+1} = x_n + \sin x_n$$ $$x_{n+1} = \sin \left(\frac {\pi} {2} x_n\right)$$

How to solve these? Or, at least, what can be said about their behavior and limits?

Answer 1

In general, nonlinear recurrences of the form $x_{n+1} = f(x_n)$ can't be solved in closed form. However, you can get information about possible limits by looking for fixed points, i.e. solving $x = f(x)$. To check stability of such a fixed point, you want to look at the value of $f'$ at the fixed point.

Answer 2

In general there is no simple answer for the general term of the sequence $(x_{n})$ defined by $x_{n+1} = f(x_{n})$. But the limiting behavior turns out to be relatively easy.

Assume that $I \subset \Bbb{R}$ is a closed interval and $f : I \to I$ is continuous and strictly increasing. Then for a sequence $(x_{n})$ defined by $x_{0} \in I$ and $x_{n+1} = f(x_{n})$, the following trichotomy holds:

1. If $x_{0} < f(x_{0})$, then $(x_{n})$ is strictly monotone-increasing and $$\lim_{n\to\infty} x_{n} = \inf \{ x \in I : x > x_{0} \text{ and } f(x) = x \}.$$ That is, $(x_{n})$ converges to the minimal fixed point of $f$ greater than $x_{0}$.
2. If $x_{0} > f(x_{0})$, then $(x_{n})$ is strictly monotone-decreasing and $$\lim_{n\to\infty} x_{n} = \sup \{ x \in I : x < x_{0} \text{ and } f(x) = x \}.$$ That is, $(x_{n})$ converges to the maximal fixed point of $f$ smaller than $x_{0}$.
3. If $x_{0} = f(x_{0})$, then $(x_{n})$ is a constant sequence.

Here, we adopt the convention that $\inf \varnothing = \infty$ and $\sup \varnothing = -\infty$. In other words, we extend the given configuration to the extended real line $\bar{\Bbb{R}}$ whenever $I$ is unbounded.

This result can be easily understood pictorially by considering the following sequence of points on the plane:

\begin{align*} & A_{0}=(x_{0}, x_{0}) \quad \longrightarrow \quad B_{0}=(x_{0}, f(x_{0}))=(x_{0},x_{1}) \\ \longrightarrow \quad & A_{1}=(x_{1}, x_{1}) \quad \longrightarrow \quad B_{1}=(x_{1}, f(x_{1}))=(x_{1},x_{2}) \\ \longrightarrow \quad & A_{2}=(x_{2}, x_{2}) \quad \longrightarrow \quad B_{2}=(x_{2}, f(x_{2}))=(x_{2},x_{3}) \\ \cdots \quad & \end{align*}

The following graph shows the case $f(x) = \sin(\pi x/2)$ with $x_{0} = 0.2$.

Here, the chain of blue arrows represent the chain of points $A_{0}, B_{0}, A_{1}, B_{1}, \cdots$.

Answer 3

We first proceed with the assumption that a limit exists, to get conditions about what must happen, to help with the conclusion. The case where $x_1 = k\pi $ is easily dealt with, so we'd ignore it.

Hint: If a limit exists, what do we know about $$\lim_{n\rightarrow \infty} \sin x_n $$

$$\lim_{n\rightarrow \infty} \sin x_n = 0$$

Hint: If a limit exists, what do we know about $$\lim_{n\rightarrow \infty} x_n $$

The limit is $k\pi$ for some integer $k$.

Now, let's make some guesses about which value it converges to, by observing the graph of $\sin \theta$. In particular, it cannot converge to 0 unless we already started with $0$.

If it does not start off at an even multiple of $\pi$, it must converge to a number of the form $(2k+1)\pi$ for some integer $k$.

Now prove that the limit converges to $(2k+1)\pi$, where $2k\pi < x_1 < 2(k+1)\pi$.