Graph of Composite Sine and Cosine Function

Question

I was playing around with a graphing calculator and I noticed that if you repeatedly apply sines and cosines in this manner:

$$\sin(\cos(...{\sin (\cos{(\sin(x)}))))}$$

its graph flattens out. Why is this so?

Answer 1

There is an important idea in functional analysis called a contraction map. A contraction map is a function $f$ such that there is a positive number $a<1$ such that for any two distinct points $x$ and $y$, $$\left|\frac{f(x)-f(y)}{x-y}\right|<a.$$ If $f$ is differentiable, then you can think of this as the slope of $f$ is always less than $a$, which is in turn less than $1$. The importance of contraction maps comes from the fact that we re guaranteed the existence of a unique fixed point $x$ such that $f(x)=x$. Moreover, starting from any point $y$, the sequence $\{f(y),f(f(y)),f(f(f(y))),...\}$ converges to $x$. So if $\cos(\sin(x))$ is a contraction map, then it makes sense that repeatedly applying this function would pull all points close to the unique fixed point, making the graph almost constant. To see that $\cos(\sin(x))$ is a contraction map, we take a derivative: $$\frac{d}{dx}\cos(\sin(x))=-\sin(sin(x))\cos(x).$$ Though it is hardly a proof, a quick graph of the derivative should be enough to convince you that the absolute value of the derivative is less than $.5$. The mean value theorem then gives us that your function is a contraction map.

Answer 2

If you choose a function $f(x)$ like: $$ f(x)=\sin(\cos(x))) $$ Then the problem that you propose can be reduced to consider an iterative sequence of the form: $$ a_n = f(a_{n-1}) $$ And finally ask: $$ \lim_{n\to\infty} a_n = L? $$ You can solve this, without loss of generality for any function, through the following equation: $$ \dfrac{a_{n+1}}{a_n} = \ell $$

Where the value of $\ell$ can take 3 general cases:

1. $\ell > 1 \quad\Rightarrow\quad a_{n+1} > a_{n}$ (the "new" point $a_{n+1}$ "moves more positively" from $a_n$).
2. $\ell < 1 \quad\Rightarrow\quad a_{n+1} < a_{n}$ (the "new" point $a_{n+1}$ "moves more negatively" from $a_n$).
3. $\ell = 1 \quad\Rightarrow\quad a_{n+1} = a_{n}$ (the point "doesn't move anywhere").

Therefore, everything depends on the value that acquires $\ell$ and how is the behavior of each $a_{n+1}$ (behavior of $f(x)$ evaluated on $a_n$) from each $a_n$, respectively. So, you can calculate the "points that doesn't move anywhere" when $\ell$ satisfies the equation: $$ \ell = 1 \quad \Rightarrow \quad \dfrac{a_{n+1}}{a_n} = \dfrac{f(a_n)}{a_n} = 1 \quad \Rightarrow \quad f(a_n) = a_n $$

Note that this is what others users named as fixed point. In the case of the problem you proposed the fixed point is $x = 0.69482...$

However, it's important to remember that the fact that $f(x)$ has a fixed point, doesn't mean necessarily that the iterative sequence $a_n = f(a_{n-1})$ should converge to the fixed point from any $a_{0}\in\mathbb{R}$ (which is what you can see on your graphing calculator).

One way to ensure that the iterative sequence converge to the fixed point from any value $a_0\in\mathbb{R}$, it's to ensure that the sequence $a_1,a_2,a_3,...$ takes ever closer to the fixed point. Note that no matter if the sequence is approaching from a more positive or more negative value, or even if you are "jumping" around the fixed point (like your case); What matters is approaching from any $a_0\in\mathbb{R}$.

This concept extended to complex numbers is intuitively what is known as "a contraction map". (see the image below).

Intuitive idea of a Contraction Map of T(x) over a value x

At this point, you can resume Alex's response.