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I'm struggling with this question, which says "given the generating funcion $g(x,z)=e^{-z^2 + 2xz} = \sum_{n=0}^{\infty}H_n(x) \frac{z^n}{n!}$ prove that the Hermite polynomials satisfy the Hermite equation".

So far I tried using the Rodrigues' expresion (which I already derived from the generating function, and it's the most explicit formula I could find for $H_n$) and tried to plug it into the Hermite equation, but I wasn't succesful. Is this the way to work it out? Any ideas?

Thanks a lot!

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  • $\begingroup$ It would help if you also wrote down the said equation they have to satisfy. $\endgroup$ – Clement C. Jan 22 '16 at 16:00
  • $\begingroup$ $\frac{d^2H_n(x)}{dx^2} - 2x\frac{dH_n(x)}{dx}+2nH_n(x) =0$ $\endgroup$ – Bouvet Island Jan 22 '16 at 16:06
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$$\cdot e^{-(x-z)^2}=(\sum_{n=0}^{\infty} \frac{z^n}{n !}\frac{d^n}{d z^n}e^{-(x-z)^2})|_{z=0}$$ Make the substitution $y=x-z$ $$\frac{d^n}{d z^n}e^{-(z-x)^2})|_{z=0}=(-1)^n\frac{d^n}{d y^n}e^{-y^2})|_{y=x}= e^{-x^2} H_n(x) \dots (\text{Rodrigue's Formula})$$

$$\rightarrow \cdot e^{-(x-z)^2}=e^{-x^2} (\sum_{n=0}^{\infty} \frac{z^n}{n !}H_n (x))$$ Multiply both sides by $e^{x^2}$ to get $g(x,z)$

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  • $\begingroup$ the generating function is already given. So I don't have to find it. The problem says "given the generating function, prove that the Hermite polynomials satisfy the Hermite equation". $\endgroup$ – Bouvet Island Jan 22 '16 at 16:34
  • $\begingroup$ Every function has a unique taylor series. Hence if $g = \sum H_n(x) \frac{z^n}{n!}$ true and $g = \sum a_n \frac{z^n}{n!}$ then $a_n = H_n$ $\endgroup$ – Shaswata Jan 22 '16 at 16:48
  • $\begingroup$ So it is sufficient to plug the generating function into the Hermite equation? $\endgroup$ – Bouvet Island Jan 22 '16 at 16:56
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$$g(x,z) = e^{-z^2+2zx} = \sum_{n=0}^{\infty}H_n(x)\frac{z^n}{n!}$$ $$\implies \frac{\partial g}{\partial x}=2ze^{-z^2+2zx}=\sum_{n=0}^{\infty}H_n(x)\frac{2z^{n+1}}{n!}=\sum_{n=0}^{\infty}H_n'(x)\frac{z^n}{n!}$$ Now equate coefficients of equal powers of $z$ in the last equality of the above expression. Doing that, one has $$H_n'(x)=2nH_{n-1}(x)$$
Similarly $$ \frac{\partial g}{\partial z}=(2x-2z)e^{-z^2+2zx}=\sum_{n=0}^{\infty}H_n(x)\frac{(2x-2z)z^{n}}{n!}=\sum_{n=0}^{\infty}H_n(x)\frac{z^{n-1}}{(n-1)!}$$Equating coefficients as previously, one has $$H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$$ Using these recursions, its easy to verify that $H_n(x)$ satisfies the DE $$H_n''(x)-2xH_n'(x) + 2nH_n(x) = 0.$$

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