# 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

 $\displaystyle e^{2zt-t^{2}}\;=\;\sum_{n=0}^{\infty}\frac{H_{n}(z)}{n!}t^{n},$ (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):

 $(2z\!-\!2t)e^{2zt-t^{2}}\;=\;\sum_{m=1}^{\infty}\frac{H_{m}(z)}{(m\!-\!1)!}t^{% m-1}$

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

 $\sum_{n=0}^{\infty}\frac{2zH_{n}(z)}{n!}t^{n}-\sum_{n=1}^{\infty}\frac{2H_{n-1% }(z)}{(n\!-\!1)!}t^{n}\;=\;\sum_{n=0}^{\infty}\frac{H_{n+1}(z)}{n!}t^{n},$

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

 $\frac{2zH_{n}}{n!}-\frac{2H_{n-1}}{(n\!-\!1)!}\;=\;\frac{H_{n+1}}{n!}\qquad(n=% 1,\,2,\,\ldots)$

Thus we have gotten the recurrence relation

 $\displaystyle H_{n+1}(z)\;=\;2zH_{n}(z)-2nH_{n-1}(z)\qquad(n=1,\,2,\,\ldots).$ (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

 $\sum_{n=0}^{\infty}\frac{H_{n}(x)}{n!}t^{n}\;=\;e^{2xt-t^{2}},\quad\sum_{n=0}^% {\infty}\frac{H_{n}(x)}{n!}u^{n}\;=\;e^{2xu-u^{2}},$

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

 $\displaystyle\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\left(\int_{-\infty}^{% \infty}\!e^{-x^{2}}H_{m}(x)H_{n}(x)\,dx\right)\frac{t^{m}u^{n}}{m!n!}\;=$ $\displaystyle\int_{-\infty}^{\infty}\!e^{-x^{2}}e^{2xt-t^{2}}e^{2xu-u^{2}}\,dx$ $\displaystyle\;=$ $\displaystyle\int_{-\infty}^{\infty}\!e^{2x(t+u)-t^{2}-u^{2}-x^{2}}\,dx$ $\displaystyle\;=$ $\displaystyle\int_{-\infty}^{\infty}\!e^{-[(t+u)^{2}-2(t+u)x+x^{2}]+2tu}\,dx$ $\displaystyle\;=$ $\displaystyle e^{2tu}\!\int_{-\infty}^{\infty}\!e^{-[x-(t+u)]^{2}}\,dx$ $\displaystyle\;=$ $\displaystyle e^{2tu}\!\int_{-\infty}^{\infty}\!e^{-y^{2}}\,dy$ $\displaystyle\;=$ $\displaystyle e^{2tu}\sqrt{\pi}$ $\displaystyle\;=$ $\displaystyle\sum_{\j=0}^{\infty}\sqrt{\pi}\frac{2^{j}t^{j}u^{j}}{j!}$ $\displaystyle\;=$ $\displaystyle\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\left(\frac{\sqrt{\pi}}{n!}% \cdot 2^{n}\delta_{mn}\right)t^{m}u^{n}$

Thus we can infer that

 $\frac{\int_{-\infty}^{\infty}\!e^{-x^{2}}H_{m}(x)H_{n}(x)\,dx}{m!n!}\;=\;\frac% {\sqrt{\pi}}{n!}\cdot 2^{n}\delta_{mn},$

which implies the orthonormality relation

 $\displaystyle\int_{-\infty}^{\infty}\!e^{-x^{2}}H_{m}(x)H_{n}(x)\,dx\;=\;2^{m}% m!\,\delta_{mn}\sqrt{\pi}.$ (3)

Cf. Hermite polynomials.

Title applying generating function ApplyingGeneratingFunction 2013-03-22 19:06:58 2013-03-22 19:06:58 pahio (2872) pahio (2872) 11 pahio (2872) Example msc 33B99 msc 33C45 msc 26C05 msc 26A42 AreaUnderGaussianCurve