Not really optimistic about getting a reply for a question tagged under "Bessel function" but here goes,
I have
$$J_{\frac{1}{2}} = (a_1 \cos(z) + a_2 \sin(z))Z^{-\frac{1}{2}} $$ and $$Y_{\frac{1}{2}} = (b_1\cos(z) + b_2\sin(z))Z^{-\frac{1}{2}} $$
$$J_{\frac{1}{2}} \sim \frac{z^{\frac{1}{2}}}{z^{\frac{1}{2}}(\frac{1}{2})!}$$ The asymptotic formula is:
Asymptotic formula:
$$J_m =\left\{\begin{matrix} 1 &m=0 \\ \frac{z^{m}}{z^{m}m!}&, m>0 \end{matrix}\right.$$
$$Y_m =2 \left\{\begin{matrix} \frac{2}{\pi} (\ln(z))&,m=0 \\ \frac{-2^{m}(m-1)!z^{m}}{\pi}&, m>0 \end{matrix}\right.$$
How do I see that $$J_\frac{1}{2} = (a_{1}(1-\frac{z^{2}}{z!}+ \cdots )+a_{2}(z - \frac{z^{3}}{3!}+\cdots ) )z^{\frac{-1}{2}}$$
This solution came up in a tut sheet I was working up but I'm not seeing how the taylor expansion came about.
You just have to notice that the Bessel differential equation
$$ x^2 y'' + x y'+ \left(x^2-\frac{1}{4}\right) y = 0, $$ for which $y=J_{\frac{1}{2}}$ is a solution, is mapped into a well-known homogeneous differential equation with constant coefficients by the substitution $y(x) = \frac{f(x)}{\sqrt{x}}$.