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.

Title	applying generating function
Canonical name	ApplyingGeneratingFunction
Date of creation	2013-03-22 19:06:58
Last modified on	2013-03-22 19:06:58
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Example
Classification	msc 33B99
Classification	msc 33C45
Classification	msc 26C05
Classification	msc 26A42
Related topic	AreaUnderGaussianCurve