$A(\lambda \rho )$=$1-\frac{1}{2\lambda \rho}-\frac{1}{8(\lambda)^2(\rho)^2}$
I need reference for this equation, can any one help me ? I used this equation in my work in statistic , another information about this function
$A(\lambda \rho)$=$\frac{I_1(\lambda\rho)}{I_0(\lambda \rho)}$ =$1-\frac{1}{2\lambda \rho}-\frac{1}{8(\lambda)^2(\rho)^2}$ where $I_0$ is the modified bessel function of the first kind and $I_1$ is the second Where i can find the proof of the previous function .
$\frac{I_1(\lambda\rho)}{I_0(\lambda \rho)}$ =$1-\frac{1}{2\lambda \rho}-\frac{1}{8(\lambda)^2(\rho)^2}\quad$ is false.
For example $\quad\frac{I_1(0)}{I_0(0)} =0\quad$ while $\left(1-\frac{1}{2(0)}-\frac{1}{8(0)^2}\right)$ is infinite.
You can check it for other values of $\lambda\rho$. In all cases $\quad\frac{I_1(\lambda\rho)}{I_0(\lambda \rho)}\neq 1-\frac{1}{2\lambda \rho}-\frac{1}{8(\lambda)^2(\rho)^2}$.
This seems an asymptotic approximate : $$\frac{I_1(x)}{I_0(x)}\simeq 1-\frac{1}{2x}-\frac{1}{8x^2}+...\qquad x\text{ large.}$$
This is consistent with the approximate for large argument (from "An Atlas of Functions", §50:9:3, p.495, J.Spanier, K.B.Oldham, Springer-Verlag, 1987 ).
$$I_\nu(x)\simeq \frac{e^x}{\sqrt{2\pi x}}\left(1-\frac{1}{x} \right)^\mu \qquad; \qquad \mu=\frac{\nu^2}{2}-\frac18$$
It can be derived that : $$\frac{I_1(x)}{I_0(x)}\simeq \sqrt{1-\frac{1}{x}}+...\qquad x\text{ large.}$$