Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I read on Wikipedia that

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

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

share|improve this question
2  
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
add comment

1 Answer

up vote 11 down vote accepted

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 $$\lim_{n \to \infty}x_n=-1$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.