📂Functions

Rodrigues' Formula for Hermite Polynomial

Official

The explicit form of the Hermite polynomial is as follows.

Physicist’s Hermite Polynomial

$$H_{n} = (-1)^{n} e^{x^2} {{d^{n}} \over {dx^{n}}} e^{-x^2} \tag{1}$$

Probabilist’s Hermite Polynomial

$$H_{e_{n}} = (-1)^{n} e^{{x^2} \over {2}} {{d^{n}} \over {dx^{n}}} e^{- {{x^2} \over {2}}}$$

Derivation

The solution to the differential equation below

$$y_{n}^{\prime \prime} - x^{2}y_{n} = -(2n+1)y_{n} \tag{2}$$

is called the Hermite function, and is as follows.

$$y_{n} = e^{\frac{x^{2}}{2}}D^{n}e^{-x^{2}}$$

Here, $D = \dfrac{d}{dx}$ is the differential operator. The Hermite function and the Hermite polynomial satisfy the following equation.

$$y_{n} = (-1^{n})e^{-\frac{x^{2}}{2}}H_{n}(x)$$

Calculating the derivative of $y_{n}$ yields the following.

$$\begin{align*} y_{n}^{\prime} &= (-1^{n})(-xe^{-\frac{x^{2}}{2}})H_{n}(x) + (-1^{n})e^{-\frac{x^{2}}{2}}H_{n}^{\prime}(x) \\ y_{n}^{\prime \prime} &= (-1^{n})\left[ (-e^{-\frac{x^{2}}{2}})H_{n}(x) + x^{2}e^{-\frac{x^{2}}{2}}H_{n}(x) + (-xe^{-\frac{x^{2}}{2}})H_{n}^{\prime}(x)\right] \\ &\quad +(-1^{n})\left[ (-xe^{-\frac{x^{2}}{2}})H_{n}^{\prime}(x) + (-1^{n})e^{-\frac{x^{2}}{2}}H_{n}^{\prime \prime}(x) \right] \end{align*}$$

Substituting this result into $(2)$ gives the following.

$$\left[ e^{-\frac{x^{2}}{2}}H_{n}^{\prime \prime}(x) - 2xe^{-\frac{x^{2}}{2}}H_{n}^{\prime}(x) + (x^{2}-1)e^{-\frac{x^{2}}{2}}H_{n}(x) \right] - x^{2}e^{-\frac{x^{2}}{2}}H_{n}(x) + (2n + 1)e^{-\frac{x^{2}}{2}}H_{n}(x) = 0$$

In summary, we obtain the following equation.

$$H_{n}^{\prime \prime}(x) - 2xH_{n}^{\prime}(x) + 2nH_{n}(x) = 0$$

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