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  "body": "\n## Definition\n\nLet $\\kappa$ and $\\mu$ be complex numbers, let $z \\in \\C \\setminus (-\\infty, 0]$, and use principal branches of powers of $z$.\nLet $M(a, b, z)$ be [[Kummer's function|6AZ4ZBN1M8]] and $U(a, b, z)$ the [[Tricomi function|234F5DNAJH]].\nIf $2\\mu \\notin \\set{-1, -2, -3, \\dots}$, the *Whittaker function* $M_{\\kappa, \\mu}$ is\n$$M_{\\kappa, \\mu}(z) = e^{-z/2} z^{\\mu + \\frac12}\\, M\\left( \\mu - \\kappa + \\tfrac12,\\ 1 + 2\\mu,\\ z \\right).$$\nFor all $\\kappa, \\mu$, the *Whittaker function* $W_{\\kappa, \\mu}$ is\n$$W_{\\kappa, \\mu}(z) = e^{-z/2} z^{\\mu + \\frac12}\\, U\\left( \\mu - \\kappa + \\tfrac12,\\ 1 + 2\\mu,\\ z \\right).$$\n\n## Remarks\n\nBoth functions solve *Whittaker's equation*\n$$\\frac{d^2 w}{dz^2} + \\left( -\\frac14 + \\frac{\\kappa}{z} + \\frac{\\frac14 - \\mu^2}{z^2} \\right) w = 0.$$\nThis follows from Kummer's equation for $M$ and $U$ by the substitution $w = e^{-z/2} z^{b/2} y$ with $b = 1 + 2\\mu$.\nExample: for $\\kappa = 0$ and $\\mu = \\frac12$, one has $U(1, 2, z) = z^{-1}$, so $W_{0, 1/2}(z) = e^{-z/2}$.\nCheck: $w = e^{-z/2}$ gives $w'' = \\frac14 w$, and the equation becomes $\\left( \\frac14 - \\frac14 + 0 + 0 \\right) w = 0$.\nNon-example: for $\\kappa = 1$ and $\\mu = \\frac12$, the function $e^{-z/2}$ is not a solution, since the equation leaves the term $\\frac{1}{z} e^{-z/2} \\ne 0$.\nAs $z \\to \\infty$ with $\\abs{\\arg z} \u003c \\frac{3\\pi}{2} - \\delta$, one has $W_{\\kappa, \\mu}(z) \\sim e^{-z/2} z^{\\kappa}$.\nThe equation depends on $\\mu$ only through $\\mu^2$, and indeed $W_{\\kappa, -\\mu} = W_{\\kappa, \\mu}$.\n",
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