It is known that Bessel functions for large arguments will behave as exp or cos/sin however I was wondering how does one arrive at those results. The motivation being that I would like to use these better approximations to evaluate them for large values analytically.
From wikipedia
From here http://upload.wikimedia.org/math/0/3/2/0327db30fc62796e50cced64f823128d.png
On
Primarily the asymptotic expansions are obtained by analyzing integral representations of the functions.
For reference, the leading-order term in the asymptotic expansion for $J_\nu(x)$ as $x \to \infty$ is derived in appendix A.1. of Stein and Shakarchi's Complex Analysis. The complete asymptotic expansions of the various Bessel functions are derived and discussed in great detail in chapter 7 of Watson's A Treatise on the Theory of Bessel Functions.
Answers part of the question. After looking into the textbook @Antonio Vargas suggested I found the equations I was after:
From: G. N. Watson, "A Treatise on the Theory of Bessel Functions", 2nd Edition, 1948.
$$ J_v (z) = \left(\frac{2}{πz} \right)^{1/2} \left[ \cos \left(z - \frac{\nu \pi}{2} - \frac{\pi}{4} \right) \sum_{m=0}^\infty \frac{(-1)^m (\nu, 2m)}{(2z)^{2m}} \\ -\sin \left(z - \frac{\nu \pi}{2} - \frac{\pi}{4} \right) \sum_{m=0}^\infty \frac{(-1)^m (\nu, 2m+1)}{(2z)^{2m+1}} \right] $$ $$ Y_v (z) = \left(\frac{2}{πz} \right)^{1/2} \left[ \sin \left(z - \frac{\nu \pi}{2} - \frac{\pi}{4} \right) \sum_{m=0}^\infty \frac{(-1)^m (\nu, 2m)}{(2z)^{2m}} \\ +\cos \left(z - \frac{\nu \pi}{2} - \frac{\pi}{4} \right) \sum_{m=0}^\infty \frac{(-1)^m (\nu, 2m+1)}{(2z)^{2m+1}} \right] $$ $$ K_v(z) = \left(\frac{\pi}{2z}\right)^{1/2} e^{-z} \sum_{m=0}^\infty \frac{(\nu,m)}{(2z)^m} $$ $$ I_v(z) = \frac{1}{(2\pi z)^{1/2}} e^{z} \sum_{m=0}^\infty \frac{(-1)^m(\nu,m)}{(2z)^m} + \frac{1}{(2\pi z)^{1/2}} e^{-z+(\nu+1/2)\pi i} \sum_{m=0}^\infty \frac{(\nu,m)}{(2z)^m} $$
Also:
$$ (\nu, m) = \prod_{i=1}^m \frac{4 \nu ^2 - (2i-1)^2}{4 i} $$