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

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

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