Prove that a closed form expression is a polynomial

Question

Prove that

$ h_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} \qquad \qquad (1)$

is a polynomial of n-th degree (in fact it's the n-th Hermite polynomial).

Evaluating for n=1 with the chain rule

$ - e^{x^2} \left( e^{-x^2} (-2x) \right) = 2x $

and for n=2

$ e^{x^2} \left( e^{-x^2} (-2x)^2 + e^{-x^2} (-2) \right) = 4x^2 -2 $

and for n=3

$ -e^{x^2} \left( e^{-x^2} (-2x)^3 + e^{-x^2} (\text{lower terms}) \right) = 8x^3 + \text{lower terms} $

so for general n we have $ 2^n x^n + \text{lower terms} $.

It becomes clear that the proposition is true, but I am not sure how much is required for the formal proof and how to do it. There is a condition given here in the first answer for whether any function is a polynomial , but that does not help me. I can't find a clue on how to attack the general expression (1) to do a formal proof "in closed form". Could anyone give me a hint? Is this possible at all?

Do you know about induction? – Hyperbolic PDE friend Feb 16 '22 at 17:59 — Hyperbolic PDE friend, Feb 16 '22 at 17:59
Have you tried induction? – Jair Taylor Feb 16 '22 at 17:59 — Jair Taylor, Feb 16 '22 at 17:59
It may be related to Chebyshev polynomials. – markvs Feb 16 '22 at 18:03 — markvs, Feb 16 '22 at 18:03

score 2 · Accepted Answer · answered Feb 16 '22 at 18:18

If we prove that the $n$th derivative of $e^{-x^2}$ is in the form $e^{-x^2}\cdot P(x)$ where $P(x)$ is a polynomial, $h_n(x) = (-1)^nP(x)$ which is a polynomial.

We'll prove the claim by induction:

Let's look at the base case, $n=0$. The $0$th derivative of $e^{-x^2}$ is itself, which is in the form $e^{-x^2}P(x)$ where $P(x) = 1$.

Now, let's get to the inductive step:

If the claim holds for $n=k, \frac{d^k}{dx^k}e^{-x^2} = e^{-x^2}P(x) = f_k(x)$

Differentiating $f_k(x)$ once with respect to $x$ would yield (by the chain rule): \begin{equation} \frac{d}{dx}f_k(x) = \frac{d}{dx}e^{-x^2}P_k(x) = P_k(x)\frac{d}{dx}e^{-x^2} + e^{-x^2}\frac{d}{dx}P_k(x) = e^{-x^2}(-2x)P(x) + e^{-x^2}\frac{d}{dx}P_k(x) = e^{-x^2}(-2xP_k(x)+P_k'(x)) \end{equation} The sums(1), products(2) and derivatives(3) of polynomials are also polynomials.

The first term in parantheses is a polynomial because of (2), the second term is a polynomial because of (3), and the inside of the paranthesis is a polynomial because of (1).

Result: Because both the base case and the inductive step have been proved as true, by mathematical induction the statement holds true.

Prove that a closed form expression is a polynomial

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