# Hermite equation

 $\frac{d^{2}f}{dz^{2}}-2z\frac{df}{dz}+2nf\;=\;0,$

in which $n$ is a real , is called the Hermite equation.  Its general solution is  $f:=Af_{1}\!+\!Bf_{2}$  with $A$ and $B$ arbitrary and the functions $f_{1}$ and $f_{2}$ presented as

$f_{1}(z)\;:=\;z+\frac{2(1-n)}{3!}z^{3}+\frac{2^{2}(1-n)(3-n)}{5!}z^{5}+\frac{2% ^{3}(1-n)(3-n)(5-n)}{7!}z^{7}+\ldots\!,$

$f_{2}(z)\;:=\;1+\frac{2(-n)}{2!}z^{2}+\frac{2^{2}(-n)(2-n)}{4!}z^{4}+\frac{2^{% 3}(-n)(2-n)(4-n)}{6!}z^{6}+\ldots$

It’s easy to check that these power series satisfy the differential equation.  The coefficients $b_{\nu}$ in both series obey the recurrence

 $b_{\nu}\;=\;\frac{2(\nu\!-\!2\!-\!n)}{\nu(nu\!-\!1)}b_{\nu\!-\!2}.$

 $R\;=\;\lim_{\nu\to\infty}\left|\frac{b_{\nu-2}}{b_{\nu}}\right|\;=\;\lim_{\nu% \to\infty}\frac{\nu}{2}\!\cdot\!\frac{1\!-\!1/\nu}{1\!-\!(n\!+\!2)/\nu}\;=\;\infty.$
If the $n$ is a non-negative integer, then one of $f_{1}$ and $f_{2}$ is simply a polynomial function.  The polynomial solutions of the Hermite equation are usually normed so that the highest degree (http://planetmath.org/PolynomialRing) is $(2z)^{n}$ and called the Hermite polynomials.