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