Why does $\cos (\pi\cos (\pi \cos (\log (20+\pi)))) \approx -1$

$$\cos (\pi\cos (\pi \cos (\log (20+\pi)))) \approx -1$$

to a high degree of accuracy. Why is this true? Is this pure coincidence or is there some mathematical background?

• How much calculus do you know? Beyond the initial coincidence $\log (20 + \pi) \approx \pi$ the rest is more or less the Banach fixed point theorem (en.wikipedia.org/wiki/Banach_fixed-point_theorem) applied to $\cos \pi x$ near $x = -1$, but the easiest way to show this requires computing the derivative of $\cos \pi x$ at $x = -1$... – Qiaochu Yuan Jun 7 '12 at 20:43

It is a well known coincidence that

$$e^{\pi}-\pi \approx 20$$

Using this, we find

$$e^{\pi}-\pi \approx 20 \implies \pi\approx \log ( 20+\pi)$$

then

$$-1 =\cos (\pi) \approx \cos(\log ( 20+\pi))$$

$\cos (-\pi)=-1$, so a closer approximation of $-1$ can be found with

$$-1 =\cos(\pi\cos (\pi)) \approx \cos(\pi\cos(\log ( 20+\pi)))$$ and again

$$-1 =\cos(\pi \cos(\pi\cos (\pi))) \approx \cos(\pi\cos(\pi\cos(\log ( 20+\pi))))$$

In fact, if $x_0 \approx -1$ and $x_n=\cos (\pi x_{n-1})$ then $$\lim_{n \to \infty}x_n=-1$$