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# Corrected exercise: Hermite polynomials

Here is the statement of an exercise that will use the Hermite polynomials, of which we will demonstrate various properties. It is a family of classical polynomials. We will therefore put this exercise in the chapter on polynomials. This is a second year higher education exercise.

And here is the statement:

## A little culture

The first 6 Hermite polynomials are:

\begin{array}{l|l} n & H_n(x)\\ \hline 0 & 1\\ 1 & x\\ 2 & x^2 - 1\\ 3 & x^3-3x\\ 4 & x^4-6x^2+3\\ 5 & x^5 - 10x^3 +15x\\ 6 & x^6 - 15x^4 +45x^2-15 \end{array}

Note that the degree of Hn is n.

## Corrected

### Question 1

Let P be a polynomial of degree n and Q a polynomial of degree m. We have :

\lim_{x \to \pm \infty} x ^{n+m+2} e^{-\frac{x^2}{2}} = 0

It is therefore easy to deduce that:

x ^{n+m} e^{-\frac{x^2}{2}} =_{\pm \infty} o\left( \frac{1}{x^2} \right)

Yet

P(x) \sim_{\pm \infty} x^n ,Q(x) \sim_{\pm \infty} x^m,

So,

P(x)Q(x) e^{-\frac{x^2}{2}} =_{\pm \infty} o\left( \frac{1}{x^2} \right)

Urban artist

P,Q \in C^0(\R,\R)

So the integral is well defined.
Obviously, we have:

\angle P|Q \rangle = \angle Q|P \rangle

Moreover, taking P1 and P2 2 polynomials and λ a real number, we have:

\begin{array}{ll} \langle P_1+ \lambda P_2|Q \rangle & =\displaystyle \int_{-\infty}^{+\infty} (P_1+ \lambda P_2)(t)Q(t) e^ {- \frac{t^2}{2}}dt \\ & =\displaystyle \int_{-\infty}^{+\infty} P_1(t)Q(t) e^{- \frac{t^ 2}{2}}dt +\lambda \int_{-\infty}^{+\infty} P_2(t)Q(t) e^{- \frac{t^2}{2}}dt \\ & = \langle P_1 |Q \rangle +\lambda \langle P_2|Q \rangle \end{array}

Which proves the linearity on the left (and therefore on the right by symmetry)
We then have:

\langle P|P \rangle = \int_{-\infty}^{+\infty} P^2(t) e^{- \frac{t^2}{2}}dt > 0

which is indeed positive by positivity of the integral. Moreover,

\langle P|P \rangle= 0 \iff \int_{-\infty}^{+\infty} P^2(t) e^{- \frac{t^2}{2}}dt = 0

Which implies

P^2(t) e^{- \frac{x^2}{2}}

is zero almost everywhere because the quantity under the integral is positive. As'exponential is non-zero then P has infinitely many roots and is therefore zero. So

\angle .|. \rangle

is indeed a scalar product.

### Question 2 : Orthogonality

The second case is symmetric: we will assume to ask this question that n < m. We will therefore have:

\begin{array}{ll} \angle H_n | H_m \rangle &=\displaystyle \int_{-\infty}^{+\infty} H_n(x)(-1)^me^{\frac{t^2}{2}}\dfrac{d^m} {dx^m}\left(e^{-\frac{t^2}{2}}\right)e^{-\frac{t^2}{2}}dt\\ & =(-1) ^m\displaystyle \int_{-\infty}^{+\infty} H_n(x)\dfrac{d^m}{dx^m}\left(e^{-\frac{t^2}{2} }\right)dt \end{array}

We will then perform m integrations by parts: We integrate n times the right-hand side and we derive n times the left-hand side. Without writing all the calculations, note that a polynomial multiplied by

e^{-\frac{x^2}{2}}

Will give a zero limit in ±∞.
So we have :

\begin{array}{ll} &(-1)^{mn}\displaystyle \int_{-\infty}^{+\infty} H_n^{(n)}(t) \dfrac{d^{mn} }{dx^{mn}}\left(e^{-\frac{t^2}{2}}\right)dt\\ =&(-1)^{mn}n!\displaystyle \int_{- \infty}^{+\infty} \dfrac{d^{mn}}{dx^{mn}}\left(e^{-\frac{t^2}{2}}\right)dt \end{ array}

We have left to the reader the proof that

\forall t \in \mathbb{R},H_n^{(n)}(t) = n!

Which is indeed 0 if n

\angle L_n | L_m \rangle = 0

### Question 3

Going back to the previous calculations, if n = m, we have:

\angle L_n | L_n \rangle = (-1)^{nn}n!\displaystyle \int_{-\infty}^{+\infty} \dfrac{d^{nn}}{dx^{nn}}\left(e^ {-\frac{t^2}{2}}\right)dt

Which simplifies to:

\angle L_n | L_n \rangle = n!\displaystyle \int_{-\infty}^{+\infty} e^{-\frac{t^2}{2}}dt

To calculate this integral, see our article on the Gaussian constant. We then obtain:

\angle L_n | L_n \rangle = n! \sqrt{\pi}

### Question 4: Recurrence relation

We have :

\dfrac{d^{n+1}}{dx^{n+1}} e^{-\frac{x^2}{2}} = \dfrac{d^{n}}{dx^{n }}(-xe^{-\frac{x^2}{2}} )

Thanks to the Leibniz formula, we get that:

 \dfrac{d^{n}}{dx^{n}}(-xe^{-\frac{x^2}{2}} )= -x\dfrac{d^{n}}{dx^{ n}}(e^{-\frac{x^2}{2}} ) -n \dfrac{d^{n-1}}{dx^{n-1}}( e^{-\frac{ x^2}{2}} )

We multiply this by

e^{\frac{x^2}{2}} (-1)^n

To get :

(-1)^ne^{\frac{x^2}{2}} \dfrac{d^{n+1}}{dx^{n+1}} e^{-\frac{x^2} {2}} =-(-1)^nxe^{-\frac{x^2}{2}} \dfrac{d^{n}}{dx^{n}}(e^{-\frac{ x^2}{2}} ) -(-1)^nn e^{-\frac{x^2}{2}} \dfrac{d^{n-1}}{dx^{n-1} }( e^{-\frac{x^2}{2}} )

We can therefore write:

-H_{n+1}(x) = - xH_n(x)+nH_{n-1}(x)

From where

H_{n+1}(x) - xH_n(x)+nH_{n-1}(x) = 0

This gives the desired recurrence relation!

### Question 5: Differential equation

#### Lemma

Let's show the following formula:

H'_n= nH_{n-1}

For this, we derive the following equality:

H_n(x) = (-1)^ne^{\frac{x^2}{2}}\dfrac{d^{n}}{dx^{n}}( e^{-\frac{x^ 2}{2}} )

To get :

\begin{array}{ll} H'_n(x) &= (-1)^nxe^{\frac{x^2}{2}}\dfrac{d^{n}}{dx^{n} }( e^{-\frac{x^2}{2}} )- (-1)^{n+1}e^{\frac{x^2}{2}}\dfrac{d^{n +1}}{dx^{n+1}}( e^{-\frac{x^2}{2}} )\\ &= xH_n(x) -H_{n+1}(x)\\ \end{array}

From where

H_{n+1} = xH_n -H'_n

Comparing with the result of the previous question, which is

H_{n+1} =xH_n -nH_{n-1}

We get well:

H'_n= nH_{n-1}

We then drift

H_{n+1} =xH_n -nH_{n-1}

To get

H'_{n+1} =xH'_n -nH'_{n-1}

Using the lemma, we get

(n+1)H_n = xH'n-n H''_n

Putting everything on the same side, we get:

(n+1)H_n - xH'n+ n H''_n = 0

Which concludes this exercise on Hermite polynomials

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