Form of Hermite Equation/Hermite Polynomials
The Hermite Equation is another class of second-order linear ODE similar to those before, with the form of
\begin{align}
y^{\prime\prime} -2xy’ + 2\nu y = 0 \tag{1}
\end{align}
where \( \nu \) is a given constant, usually as a non-negative integer. It can be shown to have an essential singularity at \(x = \infty\). Just as before, let’s derive its series solution about the ordinary point \(x = 0\). Substituting (1)-(3) in this tutorial into (1) leads to
\begin{align}
\sum_{n=2}^{\infty} n(n-1) a_nx^{n-2} -2x \sum_{n=1}^{\infty} n a_nx^{n-1} + 2\nu \sum_{n=0}^{\infty} a_n x^n &= 0 \\
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2}x^{n} -2 \sum_{n=1}^{\infty} n a_nx^n + 2\nu \sum_{n=0}^{\infty} a_n x^n &= 0 \tag{2}
\end{align}
by adjusting the index in the first summation by \(2\). Comparing the coefficients, the recurrence relation is then
\begin{align}
(n+2)(n+1) a_{n+2} -2 na_n +2 \nu a_n &= 0 \\
a_{n+2} &= -\frac{2(\nu-n)a_n}{(n+2)(n+1)} \tag{3}
\end{align}
This is also a two-term recurrence relationship with a stride of \( 2 \). We often just care about the case where \( \nu \) is a non-negative integer so that (3) terminates when \( n = \nu\ \), resulting in a polynomial. The general form of such a polynomial is then given by
\begin{align}
y &= x^n – \frac{n(n-1)}{2(2)}x^{n-2} + \frac{n(n-1)(n-2)(n-3)}{2^2(2)(4)}x^{n-4} + \cdots \\
&= \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} \frac{n!}{(n-2m)!}\frac{(-1)^m}{2^m (2^mm!)} x^{n-2m} \\
&= \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^mn!}{m!(n-2m)!} (2x)^{n-2m} \tag{4}
\end{align}
This is unsurprisingly called the Hermite polynomial of order \(n\), denoted by \(H_n(x)\). The Hermite polynomials are normalized in a similar way to the Chebyshev ones, by demanding that the leading coefficient be \( 2^n \). The first few Hermite polynomials are listed below:
\begin{align}
\begin{aligned}
P_0(x) &= 1 & & P_1(x) = 2x \\
P_2(x) &= 4x^2 -2 & & P_3(x) = 8x^3 -12x \\
P_4(x) &= 16x^4 -48x^2 + 12 & & P_5(x) = 32x^5 -160x^3 +120x \\
&\vdots\end{aligned} \tag{5}
\end{align}
Like the previous two types of polynomials, they are either odd or even, indicated by \(n\).
Hermite Equation in a Sturm-Liouville Form
The Hermite Equation can be readily put into the Sturm-Liouville Form by multiplying with the weighting factor \( e^{-x^2} \):
\begin{align}
e^{-x^2}y^{\prime\prime} -2x e^{-x^2}y’ + 2\nu e^{-x^2}y &= 0 \\
(e^{-x^2}y’)’ + 2\nu e^{-x^2}y &= 0 \tag{6}
\end{align}
where \( p(x) = \rho(x) = e^{-x^2} \), the eigenvalues are \( \lambda=2\nu \). The properties and results for Sturm-Liouville operators (this tutorial) then apply to the Hermite Equation/polynomials. However, this time, due to the form of \( p(x) \), the natural interval will be the entire \(x\)-axis \([-\infty,\infty]\), and this implies that the eigenfunctions cannot grow faster than \(e^{x^2}\) at large values, which the finite Hermite polynomials obviously satisfy.
Orthogonality
Exactly as before, the orthogonality between Hermite polynomials follows from the Sturm-Liouville theory, with a weighting of \( \rho(x) = e^{-x^2} \):
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} H_k(x) H_l(x) dx &= 0 &\text{if \( k \neq l \)} \tag{7}
\end{align}
The direct way to show this is still straightforward, and is given below for the sake of completeness. Plugging in \(H_l\) that satisfies (6) with \( \nu = l \) and multiplying by \(H_k\), we get
\begin{align}
H_k(x)(e^{-x^2}H_l(x)’)’ + 2l e^{-x^2}H_k(x)H_l(x) &= 0 \tag{8}
\end{align}
Integrating by parts:
\begin{align}
\int_{-\infty}^{\infty} H_k(x)(e^{-x^2}H_l(x)’)’ dx + \int_{-\infty}^{\infty} 2l e^{-x^2}H_k(x)H_l(x) dx &= 0 \\
-\int_{-\infty}^{\infty} H_k'(x)e^{-x^2}H_l'(x) dx + \int_{-\infty}^{\infty} 2l e^{-x^2}H_k(x)H_l(x) dx &= 0 \tag{9}
\end{align}
Repeating with \( k,l\) interchanged and subtracting each other and we are done:
\begin{align}
\begin{aligned} (-\int_{-\infty}^{\infty} H_k'(x)e^{-x^2}H_l'(x) dx + \int_{-\infty}^{\infty} 2l e^{-x^2}H_k(x)H_l(x) dx) \\
-(-\int_{-\infty}^{\infty} H_k'(x)e^{-x^2}H_l'(x) dx + \int_{-\infty}^{\infty} 2k e^{-x^2}H_k(x)H_l(x) dx) \end{aligned}
&= 0 \\
2(l-k) \int_{-\infty}^{\infty} e^{-x^2}H_k(x)H_l(x) dx &= 0 \tag{10}
\end{align}
Rodrigues’ Formula
The Rodrigues’ formula for the Hermite polynomials is
\begin{align}
H_n(x) = (-1)^ne^{x^2} \frac{d^n}{dx^n}(e^{-x^2}) \tag{11}
\end{align}
The steps to establish this are very similar to those before. Let \( u = e^{-x^2} \), differentiate once to get
\begin{align}
\frac{d}{dx}(e^{-x^2}) &= -2xe^{-x^2} \\
u’ + 2xu &= 0 \tag{12}
\end{align}
Differentiating \(n+1\) times more using Leibniz Product Rule:
\begin{align}
u^{(n+2)} + 2xu^{(n+1)} + 2(n+1)u^{(n)} &= 0 \tag{13}
\end{align}
Let \( v = (-1)^ne^{x^2} u^{(n)} \), then
\begin{align}
\begin{aligned}
u^{(n+1)} &= (-1)^n \frac{d}{dx}(ve^{-x^2}) = (-1)^n (v’e^{-x^2} -2xve^{-x^2}) \\
u^{(n+2)} &= (-1)^n (v^{\prime\prime} e^{-x^2} – 4xv’e^{-x^2} +4x^2ve^{-x^2} – 2ve^{-x^2}) \\
\end{aligned} \tag{14}
\end{align}
Substituting these into (13) then yields
\begin{align}
\begin{aligned} (v^{\prime\prime} e^{-x^2} – 4xv’e^{-x^2} +4x^2ve^{-x^2} – 2ve^{-x^2}) \\
+ 2x (v’e^{-x^2} -2xve^{-x^2}) + 2(n+1)ve^{-x^2} \end{aligned} &= 0 \\
v^{\prime\prime} e^{-x^2} – (4x-2x)v’e^{-x^2} + (4x^2-2-4x^2+2(n+1))ve^{-x^2} &= 0 \\
v^{\prime\prime} -2xv’ + 2nv &= 0 \tag{15}
\end{align}
which is consistent with the original Hermite equation. It also matches the normalization condition where the leading coefficient is \( 2^n \) as the leading term of \( \frac{d^n}{dx^n}(e^{-x^2}) \) is clearly \( (-2x)^n \), omitting the \( e^{-x^2} \) common factor.
Normalization Factor
With Rodrigues’ formula, we are then able to compute the normalization factor:
\begin{align}
\int_{-\infty}^{\infty} H_n(x)H_n(x) e^{-x^2} dx = 2^n n! \sqrt{\pi} \tag{16}
\end{align}
Substitute (11) leads to
\begin{align}
\int_{-\infty}^{\infty} H_n(x)H_n(x) e^{-x^2} dx &= (-1)^n \int_{-\infty}^{\infty} H_n(x)\frac{d^n}{dx^n}(e^{-x^2}) dx \\
&= \int_{-\infty}^{\infty} \frac{d^nH_n(x)}{dx^n}e^{-x^2} dx \tag{17}
\end{align}
where we have integrated by parts \(n\) times and all boundary terms vanish. As the leading term of \( H_n(x) \) is \( 2^nx^n \), \( \frac{d^nH_n(x)}{dx^n} = 2^n n! \), then by the famous result of the Gaussian integral (see (24) below), we recover (16).
Generating Function
The generating function of Hermite polynomials is easy to obtain:
\begin{align}
e^{2xt-t^2} = \sum_{n=0}^{\infty} \frac{H_n(x)}{n!}t^n \tag{18}
\end{align}
To show this, rewrite (18) and differentiate it with respect to \(t\) for \(k\) times:
\begin{align}
e^{2xt-t^2} &= e^{x^2}e^{-(x-t)^2} \\
\frac{\partial^k}{\partial t^k} \sum_{n=0}^{\infty} \frac{H_n(x)}{n!}t^n &= \frac{\partial^k}{\partial t^k}(e^{x^2}e^{-(x-t)^2}) \\
\sum_{n=k}^{\infty} \frac{H_n(x)}{(n-k)!}t^{n-k} &= e^{x^2}\frac{\partial^k}{\partial t^k}e^{-(x-t)^2} = (-1)^ke^{x^2}\frac{\partial^k}{\partial x^k}e^{-(x-t)^2} \tag{19}
\end{align}
Shifting the index by \(-k\), it becomes
\begin{align}
\sum_{n=0}^{\infty} \frac{H_{n+k}(x)}{n!}t^{n} &= (-1)^ke^{x^2}\frac{\partial^k}{\partial x^k}e^{-(x-t)^2} \tag{20}
\end{align}
Setting \( t = 0 \) results in
\begin{align}
H_k(x) &= (-1)^ke^{x^2}\frac{\partial^k}{\partial x^k}e^{-x^2} \tag{21}
\end{align}
which is consistent with (11).
Recurrence Relations
Two useful recurrence relations are given in Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence:
\begin{align}
\begin{aligned}
H_{n+1}(x) &= 2xH_n(x) -2nH_{n-1}(x) \\
H_n'(x) &= 2n H_{n-1}(x)
\end{aligned} \tag{22}
\end{align}
We will briefly show the first one. Differentiating (18) against \(t\) gives
\begin{align}
\frac{\partial}{\partial t}e^{2xt-t^2} &= \frac{\partial}{\partial t} \sum_{n=0}^{\infty} \frac{H_n(x)}{n!}t^n \\
(2x-2t)e^{2xt-t^2} &= \sum_{n=1}^{\infty} \frac{H_n(x)}{(n-1)!}t^{n-1} \\
2x\sum_{n=0}^{\infty} \frac{H_n(x)}{n!}t^n -2\sum_{n=0}^{\infty} \frac{H_n(x)}{n!}t^{n+1} &= \sum_{n=1}^{\infty} \frac{H_n(x)}{(n-1)!}t^{n-1} \\
2x\sum_{n=0}^{\infty} \frac{H_n(x)}{n!}t^n -2\sum_{n=0}^{\infty} \frac{H_{n-1}(x)}{(n-1)!}t^{n} &= \sum_{n=0}^{\infty} \frac{H_{n+1}(x)}{n!}t^{n} \tag{23}
\end{align}
Equating the coefficients of \(t^n\) then gives the desired formula.
Gram-Schmidt Orthogonalization
In a similar essence as before, we can also generate Hermite polynomials by Gram-Schmidt Orthogonalization, where the weighting in the inner products is now \( \rho = e^{-x^2} \) and the integration interval is \( [-\infty,\infty] \). As before, we first establish the appropriate reduction formula. It is a well-known result (Gaussian Integral) that
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \tag{24}
\end{align}
In general, for
\begin{align}
I_n = \int_{-\infty}^{\infty} x^n e^{-x^2} dx \tag{25}
\end{align}
\(I_{2m+1}\) is always \(0\) due to the odd symmetry, while
\begin{align}
I_{2m} &= \int_{-\infty}^{\infty} x^{2m} e^{-x^2} dx \\
&= -\frac{1}{2} \int_{-\infty}^{\infty} x^{2m-1} e^{-x^2} d(-x^2) \\
&= [-\frac{1}{2} x^{2m-1}e^{-x^2}]_{-\infty}^{\infty} + \frac{1}{2} \int_{-\infty}^{\infty} e^{-x^2} d(x^{2m-1}) \\
&= (0) + \frac{1}{2} \int_{-\infty}^{\infty} (2m-1) x^{2m-2} e^{-x^2} dx = \frac{2m-1}{2} I_{2m-2} \tag{26}
\end{align}
Then, orthogonalization for \( n=0,1 \) is trivial. Now recall the normalization property (16). For \(n = 2\),
\begin{align}
H_2(x) &= x^2 -\frac{\langle x^2, H_0 \rangle_{\rho}}{\lVert H_0 \rVert_{\rho}^2} H_0 -\frac{\langle x^2, H_1 \rangle_{\rho}}{\lVert H_1 \rVert_{\rho}^2} H_1 \\
&= x^2 -(\frac{1}{\sqrt{\pi}}\int_{-1}^{1} x^2e^{-x^2} dx) (1) -(\frac{1}{2\sqrt{\pi}}\int_{-1}^{1} (x^2)(2x)e^{-x^2} dx) (2x) \\
&= x^2 -\frac{1}{\sqrt{\pi}}\frac{\sqrt{\pi}}{2} -(0) \\
&= x^2 -\frac{1}{2} \tag{27}
\end{align}
Remember to normalize the result at each step. For \(n = 3\),
\begin{align}
H_3(x) &= x^3 -\frac{\langle x^3, H_0 \rangle_{\rho}}{\lVert H_0 \rVert_{\rho}^2} H_0 -\frac{\langle x^3, H_1 \rangle_{\rho}}{\lVert H_1 \rVert_{\rho}^2} H_1 -\frac{\langle x^3, H_2 \rangle_{\rho}}{\lVert H_2 \rVert_{\rho}^2} H_2 \\
&= x^3 -(\frac{1}{\sqrt{\pi}}\int_{-1}^{1} x^3e^{-x^2} dx) (1) -(\frac{1}{2\sqrt{\pi}}\int_{-1}^{1} x^3(2x)e^{-x^2} dx) (2x) \\
&\quad -(\frac{1}{(2^2)(2)\sqrt{\pi}}\int_{-1}^{1} x^3(4x^2-2)e^{-x^2} dx) (4x^2-2) \\
&= x^3 -(0) -\frac{1}{2\sqrt{\pi}}\frac{3(2)\sqrt{\pi}}{4}(2x) -(0) \\
&= x^3 -\frac{3}{2}x \tag{28}
\end{align}
and so on.
Exercise
The so-called probabilist’s Hermite polynomials are a variant of the original that satisfies the Sturm-Liouville form of
\begin{align}
(e^{-x^2/2}y’)’ + \nu e^{-x^2/2}y &= 0 \tag{29}
\end{align}
Show that it is related to the original by the change of variable \( x\to x/\sqrt{2} \). What is the corresponding Rodrigue’s formula for the probabilist’s version?
Answer
Replacing \( x \to x/\sqrt{2} \) in (6) produces (remember to take care of the differentials as well)
\begin{align}
\sqrt{2}(e^{-(x/\sqrt{2})^2}\sqrt{2}y’)’ + 2\nu e^{-(x/\sqrt{2})^2}y &= 0 \\
2(e^{-x^2/2}y’)’ + 2\nu e^{-x^2/2}y &= 0 \\
\end{align}
which matches (29). Then the Rodrigues’ formula is also given by the same change of variable for (11):
\begin{align}
(-1)^ne^{x^2/2} \frac{d^n}{dx^n}(e^{-x^2/2})
\end{align}
The factor arising from the differential is absorbed so that this always has a leading coefficient of \(1\).
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