Some expressions that allow us to calculate the Hermite polynomials are: $$\label{ec-34} H_v(\xi)=(2\xi)^vv(v-1)(2\xi)^{v- 2}+\frac{v(v-1)(v-2)(v-3)}{2}(2\xi)^{v-4}+......$$ The Equation (\ref{ec-34}) can be written more compactly as: $$\label{ec-35} H_v(\xi)=\sum_{k=0}^{v}(- 1)^k\frac{v!}{k!(v-2k)!}(2\xi)^{v-2k}$$

Now we will obtain the first and second derivatives of H from the equation (\ref{ec-34}). $$\label{ec-36} H'_v=2v(2\xi)^{v-1}-2v(v-1)(v-2)(2\xi)^{v-3 }+\frac{2v(v-1)(v-2)(v-3)(v-4)}{2}(2\xi)^{v-5}+......$$ From where we can see that: $$\label{ec-37} H'_v=2vH_{v-1}$$ Deriving the equation (\ref{ec-37}) we obtain the second derivative of H. $$\label{ec-38} H''_v=\frac{dH'_v}{d\xi}=2vH'_{v-1}=4v(v-1 )H_{v-2}$$ Substituting these derivatives into the Hermite equation (\ref{ec-20}) and taking into account that $\left(\frac{\alpha}{\beta}-1 \right)=2v$ yields: $$\label{ec-39} 4v(v-1)H_{v-2}-2\xi 2vH_{v-1} + 2vH_{v}=0$$ Dividing the equation (\ref{ec-39}) by $4v$: $$\label{ec-40} \xi H_{v-1}=(v-1)H_{ v-2}+\frac{1}{2}H_{v}$$ If in this last equation we change $v$ to $v+1$ we get: $$\label{ec- 41} \xi H_{v}=vH_{v-1}+\frac{1}{2}H_{v+1}$$ Another formula that may be useful is the Rodrigues equation. $$\label{ec-42} H_v =(-1)^{v}e^{\xi^2}\frac{d^ve^{-\xi^2}}{d\xi^ v}$$