The linear differential equation $$\frac{d^2f}{dz^2}-2z\frac{df}{dz}+2nf \;=\; 0,$$ in which $n$ is a real constant, is called the Hermite equation. Its general solution is $f := Af_1\!+\!Bf_2$ with $A$ and $B$ arbitrary constants 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 formula $$b_\nu \;=\; \frac{2(\nu\!-\!2\!-\!n)}{\nu(nu\!-\!1)}b_{\nu\!-\!2}.$$ Thus we have the radii of convergence $$R \;=\; \lim_{\nu\to\infty}\left|\frac{b_{\nu-2}}{b_\nu}\right| \;=\; \lim_{\nu\to\infty}\frac{\nu}{2}\!\cdot\!\frac{1\!-\!1/\nu}{1\!-\!(n\!+\!2)/\nu} \;=\; \infty.$$ Therefore the series converge in the whole complex plane and define entire functions.

If the constant $n$ 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 term is $(2z)^n$ and called the Hermite polynomials.