The generating function encodes all the Hermite polynomials in one formula. It is a function of $x$ and a dummy variable $t$ of the the form:
$e^{2xt-t^2}=\sum^\infty_{n=0}\frac{H_n(x)}{n!}t^n. $
We begin by considering
$f(t)=e^{-(x-t)^2}=e^{-x^2}e^{2xt-t^2}.$
The Taylor series for this function is
$f(t)=\sum^\infty_{n=0}\frac{f^{(n)}(0)}{n!}t^n.$
Here by using a substitution, $x-t=u$, we have
$f^{(n)}(0)=\bigg[ \frac{d^n}{dt^n}e^{-(x-t)^2} \bigg]_{t=0}=(-1)^n\bigg[ \frac{d^n}{dt^n}e^{-u^2} \bigg]_{u=x}=(-1)^n\frac{d^n}{dx^n}(e^{-x^2})=e^{-x^2}H_n(x).$
I am struggling to see how the substitution $x-t=u$ is implemented here and where the $(-1)^n$ arises from? Any help would be much appreciated.
Thank you
Best Answer
If $u=x-t$ then $t=x-u$ and $dt=d\left(x-u\right)=dx-du=-du$ ($x$ here is like constant).
Also $d^{n}t=\left(-1\right)^{n}d^{n}u$ because each differentiation changes sign to opposite.
$f^{\left(n\right)}\left(t\right)=\left[\frac{d^{n}}{dt^{n}}e^{-\left(x-t\right)^{2}}\right]_{t=0}=\left[\frac{d^{n}}{\left(-1\right)^{n}d^{n}u}e^{-u^{2}}\right]_{x-u=0}=\left[\left(-1\right)^{n}\frac{d^{n}}{d^{n}u}e^{-u^{2}}\right]_{u=x}=\\=\left(-1\right)^{n}\frac{d^{n}}{d^{n}x}e^{-x^{2}}=e^{-x^{2}}\cdot\left(-1\right)^{n}e^{x^{2}}\frac{d^{n}}{d^{n}x}e^{-x^{2}}=e^{-x^{2}}H_{n}\left(x\right)$
Last part I took from JamesT (see comment above).