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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

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The last part is making use of Rdrigues' formula: $ H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}(e^{-x^2}).$ –  JamesT Feb 14 '13 at 17:49

1 Answer 1

up vote 2 down vote accepted

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).

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This is very helpful, thank you :) –  JamesT Feb 14 '13 at 18:20

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