Whittaker function

Remarks

Both functions solve Whittaker's equation $$\frac{d^2 w}{dz^2} + \left( -\frac14 + \frac{\kappa}{z} + \frac{\frac14 - \mu^2}{z^2} \right) w = 0.$$ This follows from Kummer's equation for $M$ and $U$ by the substitution $w = e^{-z/2} z^{b/2} y$ with $b = 1 + 2\mu$. Example: for $\kappa = 0$ and $\mu = \frac12$, one has $U(1, 2, z) = z^{-1}$, so $W_{0, 1/2}(z) = e^{-z/2}$. Check: $w = e^{-z/2}$ gives $w'' = \frac14 w$, and the equation becomes $\left( \frac14 - \frac14 + 0 + 0 \right) w = 0$. Non-example: for $\kappa = 1$ and $\mu = \frac12$, the function $e^{-z/2}$ is not a solution, since the equation leaves the term $\frac{1}{z} e^{-z/2} \ne 0$. As $z \to \infty$ with $\abs{\arg z} < \frac{3\pi}{2} - \delta$, one has $W_{\kappa, \mu}(z) \sim e^{-z/2} z^{\kappa}$. The equation depends on $\mu$ only through $\mu^2$, and indeed $W_{\kappa, -\mu} = W_{\kappa, \mu}$.