# Proof using the orthogonality of the Hermite Polynomials

Given a suitable function $h(x)$ defined on $\mathbb R$ admits an orthogonal expansion in terms of the Hermite polynomials as

$h(x)=\sum_{m=0}^\infty d_mH_m(x)$.

By considering

$\int_{-\infty}^{\infty} h(x)H_n(x)exp(-x^2)dx$

and using orthogonality of the Hermite polynomials, prove that

$d_n= \frac{\int_{-\infty}^{\infty} h(x)H_n(x)exp(-x^2)dx}{\int_{-\infty}^{\infty} H_n^2(x)exp(-x^2)dx }$.

So to begin with I am aware for two different Hermite polynomials to be orthogonal we have the following relation:

$\langle H_n,H_m\rangle = \int_{-\infty}^{\infty}H_nH_{m}exp(-x^2)=0$

Now inputting our $h(x)$ in to the equation up for consideration we get:

$\int_{-\infty}^{\infty} \sum_{m=0}^\infty d_mH_m(x)H_n(x)exp(-x^2)dx$

But now I'm not really sure what can be done with this, any help would be appreciated.

## 1 Answer

Your orthogonality relations aren't quite right. They should say:

$$\int_{-\infty}^\infty H_n H_m \exp(-x^2) = 0 {\rm \ \ if \ }m \neq n.$$

Now look at your integral:

$$\int_{-\infty}^\infty \sum_{m=1}^\infty d_m H_n H_m \exp(-x^2)$$

Can you see that only the $m = n$ term contributes?

So

$$\int_{-\infty}^\infty \sum_{m=1}^\infty d_m H_n H_m \exp(-x^2) = d_n \int_{-\infty}^\infty H_n^2 \exp(-x^2)$$

Rearranging this, you get

$$d_n = \frac{\int_{-\infty}^\infty \sum_{m=1}^\infty d_m H_n H_m \exp(-x^2)}{\int_{-\infty}^\infty H_n^2 \exp(-x^2) }$$

• So the only time we get a value for this integral is when $m=n$ based on the relations you have mentioned. However I don't see what this means in terms of $d_m$ which becomes $d_n$ of course when $m=n$. In which case we get $\int_{-\infty}^{\infty} d_nH_n^2(x)exp(-x^2)dx$, can we then say this is equal to $1$? or is it equal to $d_n$ as we have the $d_n$ multiplier? – Evan Feb 28 '17 at 22:28
• Yes, that's the idea. I edited the answer above. – Kenny Wong Feb 28 '17 at 22:59