applying generating function

The generating function of a function sequence carries information common to the members of the sequence.  It may be utilised for deriving various properties, such as recurrence relations, orthogonality properties etc.  We take as example

$e^{2zt-t^2}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{H_n(z)}{n!}}{t}^n,$(1)

the http://planetmath.org/node/11980generating function of the of Hermite polynomials, and derive from it a recurrence relation and the orthonormality (http://planetmath.org/Orthonormal) formula.

1.  First we form the partial derivative with respect to $t$ of both of (1):

$$(2z-2t)e^{2zt-t^2}=\sum _{m=1}^{\mathrm{\infty }}\frac{H_m(z)}{(m-1)!}{t}^{m-1}$$

Here we substitute (1) to the left hand side and rewrite the right hand side, getting

$$\sum _{n=0}^{\mathrm{\infty }}\frac{2z{H}_n(z)}{n!}{t}^n-\sum _{n=1}^{\mathrm{\infty }}\frac{2H_{n-1}(z)}{(n-1)!}{t}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{H_{n+1}(z)}{n!}{t}^n,$$

where we can compare the coefficients of $t^n$:

$$\frac{2z{H}_n}{n!}-\frac{2H_{n-1}}{(n-1)!}=\frac{H_{n+1}}{n!}\mathit{\hspace{1em}\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots })$$

Thus we have gotten the recurrence relation

$H_{n+1}(z)=\mathrm{\hspace{0.33em}2}z{H}_n(z)-2n{H}_{n-1}(z)\mathit{\hspace{1em}\hspace{1em}}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$(2)

Differentiating (1) partially with respect to $z$ enables respectively to find a formula expressing the derivative$H_n^{\prime }(z)$ via the themselves.

2.  We copy the equation (1) twice in the forms

$$\sum _{n=0}^{\mathrm{\infty }}\frac{H_n(x)}{n!}{t}^n=e^{2xt-t^2},\sum _{n=0}^{\mathrm{\infty }}\frac{H_n(x)}{n!}{u}^n=e^{2xu-u^2},$$

multiply these with each other and by $e^{-x^2}$ and then integrate the obtained equation termwise over$ℝ$:

${\displaystyle \sum _{m=0}^{\mathrm{\infty }}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-x^2}{H}_m(x)H_n(x)𝑑x\right){\displaystyle \frac{t^m{u}^n}{m!n!}}=$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-x^2}{e}^{2xt-t^2}{e}^{2xu-u^2}𝑑x$

$\mathrm{\hspace{0.33em}}=$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{2x(t+u)-t^2-u^2-x^2}𝑑x$

$\mathrm{\hspace{0.33em}}=$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-[{(t+u)}^2-2(t+u)x+x^2]+2tu}𝑑x$

$\mathrm{\hspace{0.33em}}=$$e^{2tu}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-{[x-(t+u)]}^2}𝑑x$

$\mathrm{\hspace{0.33em}}=$$e^{2tu}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-y^2}𝑑y$

$\mathrm{\hspace{0.33em}}=$$e^{2tu}\sqrt{\pi }$

$\mathrm{\hspace{0.33em}}=$${\displaystyle \sum _{ȷ=0}^{\mathrm{\infty }}}\sqrt{\pi }{\displaystyle \frac{2^j{t}^j{u}^j}{j!}}$

$\mathrm{\hspace{0.33em}}=$${\displaystyle \sum _{m=0}^{\mathrm{\infty }}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\left({\displaystyle \frac{\sqrt{\pi }}{n!}}\cdot 2^n{\delta }_{mn}\right)t^m{u}^n$

Thus we can infer that

$$\frac{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}{H}_m(x)H_n(x)𝑑x}{m!n!}=\frac{\sqrt{\pi }}{n!}\cdot 2^n{\delta }_{mn},$$

which implies the orthonormality relation

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-x^2}{H}_m(x)H_n(x)𝑑x={\mathrm{\hspace{0.33em}2}}^{m}m!{\delta }_{mn}\sqrt{\pi }.$(3)

Cf. Hermite polynomials.