redsoxfan325 said: As I was messing around in Maple (yes I'm a nerd), I noticed something.



It appears as if the infinite chain \(\displaystyle \sin(\cos(\sin(\cos(\sin(\cos(.....(\sin(\cos(x)))))))))\) converges for all x.



It appears to be converging to a number very close to \(\displaystyle \ln(2)\), but I don't think it actually does, unless it's really, really slowly.



The number (calculated for 7 sine-cosine pairs) is approximately \(\displaystyle 0.69481969\).



Also, if you let the first term be cosine instead of sine, this chain gets close to \(\displaystyle 0.768169156736795977462\).



Does anyone know anything about this? Click to expand...

This kind of thing is called fixed point iteration . Starting with \(\displaystyle x_0\) being whatever, you let \(\displaystyle x_1=\sin(\cos x_0)\) and more generally \(\displaystyle x_{n+1}=\sin(\cos x_n)\) for every \(\displaystyle n\geq 0\).There are various assumptions that ensure that the sequence converges (for instance, here the function \(\displaystyle f(x)=\sin\cos x\) is contracting: \(\displaystyle |f(x)-f(y)|\leq k|x-y|\) where \(\displaystyle k\in(0,1)\), hence Banach's fixed point theorem applies). Then, if \(\displaystyle x_n\to\ell\), necessarily we have \(\displaystyle \ell=\sin\cos\ell\) (taking the limit in \(\displaystyle x_{n+1}=\sin\cos x_n\)).So that your constants are just the fixed points of \(\displaystyle \sin\cos\) and \(\displaystyle \cos\sin\): the numbers \(\displaystyle c_1,c_2\) such that \(\displaystyle \sin\cos c_1=c_1\) and \(\displaystyle \cos\sin c_2=c_2\) (they are unique). I don't think there exits a simple explicit formula for them.You probably can find more about this if you look for "Picard iteration", or "Banach fixed point theorem" with Google.