Hermite Polynomials: Generating Function, Recurrence Relations

Here we study Hermite polynomials along with its generating function and recurrence relations.

Note that e^ax = $\displaystyle \sum_{n=0}^\infty \dfrac{(ax)^n}{n!}$ …(I)

Generating Function of Hermite Polynomials

Theorem: The generating function for the Hermite Polynomials H_n(x) is given by $G(x, t) = e^{2tx-t^2}$. That is,

$e^{2tx-t^2} = \displaystyle\sum_{n=0}^{\infty} H_n(x) \dfrac{t^n}{n!}$

Proof:

One need to prove that one can get the Hermite polynomials H_n(x), for each n ∈ ℕ, out of this function G(x, t). We have

$G(x, t) = e^{2tx-t^2}$

Using the above Equation (I), we have:

$G(x, t) =e^{2tx} \cdot e^{-t^2}$ $= \displaystyle\sum_{s=0}^{\infty} \dfrac{(2tx)^s}{s!} \displaystyle\sum_{r=0}^{\infty} \dfrac{(-t^2)^r}{r!}$

⇒ $G(x, t) = \displaystyle\sum_{s=0}^{\infty} \displaystyle\sum_{r=0}^{\infty} (-1)^r \dfrac{(2x)^s}{r! s!} t^{s+2r}$.

Let n=s+2r.

Fixing r, we have s=n-2r.

Then the co-efficient of tⁿ in the above expression is given equal to

$(-1)^r \dfrac{(2x)^{n-2r}}{r! (n-2r)!}$.

Now, note that s≥0 ⇒ r≤ n/2 with r an integer. So the values of r lies between [0, n/2] or [0, (n-1)/2] depending upon n is even or odd.

Therefore, we obtain all the co-efficients of tⁿ as follows.

$\displaystyle\sum_{r=0}^{[n/2]} (-1)^r \dfrac{1}{r! (n-2r)!} (2x)^{n-2r}$ where [n/2] = $\begin{cases} n/2 & \text{if n is even} \\ (n-1)/2 & \text{if n is odd} \end{cases}$

$=\dfrac{1}{n!} \displaystyle\sum_{r=0}^{[n/2]} \dfrac{n!}{r! (n-2r)!} (2x)^{n-2r}$

$=\dfrac{1}{n!} H_n(x)$ (say).

Noting the coefficient of tⁿ in the expression $\displaystyle\sum_{n=0}^\infty H_n(x)\dfrac{t^n}{n!}$ is $\dfrac{1}{n!}H_n(x)$, it follows that $e^{2tx-t^2}$ is the generating function for Hermite polynomials.

Values at 0

Prove that $H_{2n}(0)=(-1)^n\dfrac{(2n)!}{n!}$ and $H_{2n+1}(0)=0$.

Answer:

As H_n(x) is the generating function for the Hermite polynomials, we have

$\displaystyle\sum_{n=0}^\infty \dfrac{t^n}{n!} H_n(x)=e^{2tx-t^2}$.

Put x=0 in the above. So

$\displaystyle\sum_{n=0}^\infty \dfrac{t^n}{n!} H_n(0)=e^{-t^2}$

⇒ $\displaystyle\sum_{n=0}^\infty \dfrac{t^n}{n!} H_n(0)=\displaystyle\sum_{n=0}^\infty \dfrac{(-t^2)^n}{n!}$, by Equation (I).

⇒ $\displaystyle\sum_{n=0}^\infty \dfrac{1}{n!} H_n(0) t^n = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{n!} t^{2n}$

Comparing the co-efficients of t²ⁿ and t²ⁿ⁺¹ on both sides of the above equation respectively, it follows that

$\dfrac{H_{2n}(0)}{(2n)!}=\dfrac{(-1)^n}{n!}$ and $H_{2n+1}(0)=0$.

⇒ $H_{2n}(0)=(-1)^n\dfrac{(2n)!}{n!}$ and $H_{2n+1}(0)=0$. (Proved)

Recurrence Relations

1. $H_n'(x)=2nH_{n-1}(x)$ for $n \geq 1$.
2. $H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$for $n \geq 1$.

Also Study: Charlier Polynomials

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