Editing Hermite polynomials (section)

===Recurrence relation===
The sequence of probabilist's Hermite polynomials also satisfies the [[recurrence relation]]
<math display="block">\operatorname{He}_{n+1}(x) = x \operatorname{He}_n(x) - \operatorname{He}_n'(x).</math>
Individual coefficients are related by the following recursion formula:
<math display="block">a_{n+1,k} = \begin{cases}
 - (k+1) a_{n,k+1} &  k = 0, \\
 a_{n,k-1} - (k+1) a_{n,k+1} & k > 0,
\end{cases}</math> 
and {{math|1=''a''<sub>0,0</sub> = 1}}, {{math|1=''a''<sub>1,0</sub> = 0}}, {{math|1=''a''<sub>1,1</sub> = 1}}.

For the physicist's polynomials, assuming 
<math display="block">H_n(x) = \sum^n_{k=0} a_{n,k} x^k,</math>
we have
<math display="block">H_{n+1}(x) = 2xH_n(x) - H_n'(x).</math>
Individual coefficients are related by the following recursion formula:
<math display="block">a_{n+1,k} = \begin{cases}
 - a_{n,k+1} & k = 0, \\ 
 2 a_{n,k-1} - (k+1)a_{n,k+1} & k > 0,
\end{cases}</math>
and {{math|1=''a''<sub>0,0</sub> = 1}}, {{math|1=''a''<sub>1,0</sub> = 0}}, {{math|1=''a''<sub>1,1</sub> = 2}}.

The Hermite polynomials constitute an [[Appell sequence]], i.e., they are a polynomial sequence satisfying the identity
<math display="block">\begin{align}
 \operatorname{He}_n'(x) &= n\operatorname{He}_{n-1}(x), \\
 H_n'(x) &= 2nH_{n-1}(x).
\end{align}</math>

An integral recurrence that is deduced and demonstrated in <ref>Hurtado Benavides, Miguel Ángel. (2020). De las sumas de potencias a las sucesiones de Appell y su caracterización a través de funcionales. [Tesis de maestría]. Universidad Sergio Arboleda.</ref> is as follows:
<math display="block">\operatorname{He}_{n+1}(x) = (n+1)\int_0^x \operatorname{He}_n(t)dt - He'_n(0),</math>

<math display="block">H_{n+1}(x) = 2(n+1)\int_0^x H_n(t)dt - H'_n(0).</math>

Equivalently, by [[Taylor series|Taylor-expanding]],
<math display="block">\begin{align}
 \operatorname{He}_n(x+y) &= \sum_{k=0}^n \binom{n}{k}x^{n-k} \operatorname{He}_{k}(y)
  &&= 2^{-\frac n 2} \sum_{k=0}^n \binom{n}{k} \operatorname{He}_{n-k}\left(x\sqrt 2\right) \operatorname{He}_k\left(y\sqrt 2\right), \\
 H_n(x+y) &= \sum_{k=0}^n \binom{n}{k}H_{k}(x) (2y)^{n-k}
  &&= 2^{-\frac n 2}\cdot\sum_{k=0}^n \binom{n}{k} H_{n-k}\left(x\sqrt 2\right) H_k\left(y\sqrt 2\right).
\end{align}</math>
These [[umbral calculus|umbral]] identities are self-evident and [[#Generalizations|included]] in the [[#Differential-operator representation|differential operator representation]] detailed below, 
<math display="block">\begin{align}
 \operatorname{He}_n(x) &= e^{-\frac{D^2}{2}} x^n, \\
 H_n(x) &= 2^n e^{-\frac{D^2}{4}} x^n.
\end{align}</math>

In consequence, for the {{mvar|m}}th derivatives the following relations hold:
<math display="block">\begin{align}
 \operatorname{He}_n^{(m)}(x) &= \frac{n!}{(n-m)!} \operatorname{He}_{n-m}(x)
  &&= m! \binom{n}{m} \operatorname{He}_{n-m}(x), \\
 H_n^{(m)}(x) &= 2^m \frac{n!}{(n-m)!} H_{n-m}(x)
  &&= 2^m m! \binom{n}{m} H_{n-m}(x).
\end{align}</math>

It follows that the Hermite polynomials also satisfy the [[recurrence relation]]
<math display="block">\begin{align}
 \operatorname{He}_{n+1}(x) &= x\operatorname{He}_n(x) - n\operatorname{He}_{n-1}(x), \\
 H_{n+1}(x) &= 2xH_n(x) - 2nH_{n-1}(x).
\end{align}</math>

These last relations, together with the initial polynomials {{math|''H''<sub>0</sub>(''x'')}} and {{math|''H''<sub>1</sub>(''x'')}}, can be used in practice to compute the polynomials quickly.

[[Turán's inequalities]] are
<math display="block">\mathit{H}_n(x)^2 - \mathit{H}_{n-1}(x) \mathit{H}_{n+1}(x) = (n-1)! \sum_{i=0}^{n-1} \frac{2^{n-i}}{i!}\mathit{H}_i(x)^2 > 0.</math>

Moreover, the following [[multiplication theorem]] holds:
<math display="block">\begin{align}
 H_n(\gamma x) &= \sum_{i=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \gamma^{n-2i}(\gamma^2 - 1)^i \binom{n}{2i} \frac{(2i)!}{i!} H_{n-2i}(x), \\
 \operatorname{He}_n(\gamma x) &= \sum_{i=0}^{\left\lfloor \tfrac{n}{2} \right\rfloor} \gamma^{n-2i}(\gamma^2 - 1)^i \binom{n}{2i} \frac{(2i)!}{i!}2^{-i} \operatorname{He}_{n-2i}(x).
\end{align}</math>