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.
$$ e^{2xt-t^{2}} = \sum \limits _{n=0}^{\infty}H_{n}(x)\frac{t^{n}}{n!} $$
■
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 $\Phi (x,t)=e^{2xt-t^{2}}$,
$$ \begin{align*} &\frac{ \partial^{2} \Phi}{ \partial x^{2}}-2x\frac{ \partial \Phi}{ \partial x}+2t\frac{ \partial \Phi}{ \partial t } \\ &= 4t^{2}e^{2xt-t^{2}}-4xte^{2xt-t^{2}}+2t(2x-2t)e^{2xt-t^{2}} \\ &=0 \end{align*} $$
■