Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 16B (Numerical Analysis)

Repeated differentiation of the Gaussian function gives $D^n{e}^{-x^2}=P_n(x)e^{-x^2}$, where $P_0=1$ and $P_{n+1}=P_n^{\prime }-2x{P}_n$. Induction shows $P_n$ has degree of a polynomial $n$ and leading coefficient $(-2{)}^n$: the differentiated term has lower degree, while multiplication by $-2x$ raises it by one. Thus the Hermite polynomial $H_n=(-1{)}^n{P}_n$ has degree $n$ and leading coefficient $2^n$.

For $m<n$, apply integration by parts $n$ times to the Rodrigues expression:All boundary terms vanish because each derivative of $e^{-x^2}$ is a polynomial times that exponential, and such products tend to zero at either infinity. The same result for $m>n$ follows by symmetry of the inner product. The diagonal value is positive; indeed $H_n^{(n)}=2^{n}n!$ givesThis proves the requested orthogonality as well as its normalization.

Differentiating the Rodrigues representation directly gives the identityWe prove simultaneously by induction that $H_n^{\prime }=2n{H}_{n-1}$ for $n\ge 1$. It holds for $H_0=1$, $H_1=2x$. Suppose it holds at $n$. Differentiating the preceding identity and using the same identity at $n-1$ givesThe induction is complete. Substituting the derivative identity into the first formula proves the Hermite polynomial three-term recurrence relationThe initial step $H_1=2x{H}_0$ supplies the recurrence at $n=0$ without needing to define $H_{-1}$.