Hermite equation

The linear differential equation

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

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+\cdots\!,$

     $ 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+\cdots$

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

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

Thus we have the radii of convergence

$\displaystyle R = \lim_{\nu\to\infty}\left\vert\frac{b_{\nu-2}}{b_\nu}\right\ve...
...o\infty}\frac{\nu}{2}\!\cdot\!\frac{1\!-\!1/\nu}{1\!-\!(n\!+\!2)/\nu} = \infty.$

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.



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