Hermite equation

The linear differential equation

$$\frac{d^{2}f}{d{z}^2}-2z\frac{df}{dz}+2nf=\mathrm{\hspace{0.33em}0},$$

in which $n$ is a real , is called the Hermite equation.  Its general solution is  $f:=A{f}_1+B{f}_2$  with $A$ and $B$ arbitrary and the functions $f_1$ and $f_2$ presented as

$f_1(z):=z+\frac{2(1-n)}{3!}{z}^3+\frac{2^2(1-n)(3-n)}{5!}{z}^5+\frac{2^3(1-n)(3-n)(5-n)}{7!}{z}^7+\mathrm{\dots },$

$f_2(z):=\mathrm{\hspace{0.33em}1}+\frac{2(-n)}{2!}{z}^2+\frac{2^2(-n)(2-n)}{4!}{z}^4+\frac{2^3(-n)(2-n)(4-n)}{6!}{z}^6+\mathrm{\dots }$

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

$$b_{\nu }=\frac{2(\nu -2-n)}{\nu (nu-1)}{b}_{\nu -2}.$$

Thus we have the radii of convergence (http://planetmath.org/RadiusOfConvergence)

$$R=\underset{\nu \to \mathrm{\infty }}{lim}\left|\frac{b_{\nu -2}}{b_{\nu }}\right|=\underset{\nu \to \mathrm{\infty }}{lim}\frac{\nu }{2}\cdot \frac{1-1/\nu }{1-(n+2)/\nu }=\mathrm{\infty }.$$

Therefore the series converge in the whole complex plane and define entire functions.

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.