📂Functions

Hermite Polynomials' Generating Function

Formulas

The generating function of Hermite Polynomials is as follows.

$$\Phi (x,t)=\sum \limits _{n=0}^{\infty} \frac{H_{n}(x)}{n!}t^{n}= e^{2xt-t^{2}}$$

Explanation

The generating function of Hermite Polynomials, simply put, is a polynomial that uses Hermite Polynomials as its coefficients.

$H_{n}(x)$ is a Hermite Polynomial, and can be obtained by multiplying Hermite function $y_{n}=e^{\frac{x^{2}}{2}}\frac{ \d ^{n} }{ \d x^{n} }e^{-x^{2}}$ with $(-1)^{n}e^{\frac{x^{2}}{2}}$ or by solving the Hermite Differential Equation.

$$H_{n}(x)=(-1)^{n}e^{x^{2}}\frac{ \d ^{n}}{ \d x^{n} }e^{-x^{2}}$$

Derivation

Let’s assume $f(x)=e^{-x^{2}}$. Then,

$$f^{(n)}(x)=\frac{ \d ^{n}}{ \d x^{n} }e^{-x^{2}}=(-1)^{n}e^{-x^{2}}H_{n}(x) \tag{1}$$

And by the Taylor series,

$$f(x)=\sum \limits _{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^{n}$$

Here, substituting $x-a=t$ and $a=y$, we get

$$f(y+t) = \sum \limits _{n=0}^{\infty} \frac{f^{(n)}(y)}{n!}t^{n}$$

Again, setting $y$ as $x$ and substituting $(1)$, we get

$$f(x+t)=\sum \limits _{n=0}^{\infty} \frac{f^{(n)}(y)}{n!}t^{n}=\sum \limits _{n=0}^{\infty}(-1)^{n}e^{-x^{2}}\frac{t^{n}}{n!}H_{n}(x)$$

Now, substituting $-t$ in place of $t$,

$$\begin{align*} f(x-t) &=\sum \limits _{n=0}^{\infty}e^{-x^{2}}\frac{t^{n}}{n!}H_{n}(x) \\ &= e^{-(x-t)^{2}}=e^{-x^{2}+2xt-t^{2}} \end{align*}$$

Summarizing,

$$e^{-x^{2}+2xt-t^{2}} = \sum \limits _{n=0}^{\infty}e^{-x^{2}}H_{n}(x)\frac{t^{n}}{n!}$$

Now, by multiplying both sides by $e^{x^{2}}$, we obtain the desired formula.

$$e^{2xt-t^{2}} = \sum \limits _{n=0}^{\infty}H_{n}(x)\frac{t^{n}}{n!}$$

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Theorem

Moreover, the generating function of Hermite Polynomials satisfies the following differential equation.

$$\frac{ \partial^{2} \Phi}{ \partial x^{2}}-2x\frac{ \partial \Phi}{ \partial x}+2t\frac{ \partial \Phi}{ \partial t }=0$$

Proof

By substituting $\Phi (x,t)=e^{2xt-t^{2}}$,

$$\begin{align*} &\frac{ \partial^{2} \Phi}{ \partial x^{2}}-2x\frac{ \partial \Phi}{ \partial x}+2t\frac{ \partial \Phi}{ \partial t } \\ &= 4t^{2}e^{2xt-t^{2}}-4xte^{2xt-t^{2}}+2t(2x-2t)e^{2xt-t^{2}} \\ &=0 \end{align*}$$

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