# generating function of Laguerre polynomials

 $\displaystyle L_{n}(z)\;:=\;e^{z}\frac{d^{n}}{dz^{n}}e^{-z}z^{n}\qquad(n\;=\;0% ,\,1,\,2,\,\ldots).$ (1)

The consequence

 $\displaystyle f^{(n)}(z)\;=\;\frac{n!}{2\pi i}\oint_{C}\frac{f(\zeta)}{(\zeta-% z)^{n+1}}\ d\zeta$ (2)

of http://planetmath.org/node/1150Cauchy integral formula  allows to write (1) as the complex integral

 $L_{n}(z)\;=\;\frac{n!}{2i\pi}\oint_{C}\frac{e^{z}e^{-\zeta}}{(\zeta\!-\!z)^{n+% 1}}\,d\zeta\;=\;\frac{n!}{2i\pi}\oint_{C}\frac{e^{z-\zeta}\,d\zeta}{(1\!-\!% \frac{z}{\zeta})^{n}(\zeta\!-\!z)},$

where $C$ is any contour around the point $z$ and the direction is anticlockwise.  The http://planetmath.org/node/11373substitution

 $\zeta\!-\!z\;:=\;\frac{zt}{1\!-\!t},\quad\zeta\;=\;\frac{z}{1\!-\!t},\quad t\;% =\;1\!-\!\frac{z}{\zeta}\quad d\zeta\;=\;\frac{z\,dt}{(1\!-\!t)^{2}}$

here yields

 $L_{n}(z)\;=\;\frac{n!}{2i\pi}\oint_{C^{\prime}}\frac{e^{-\frac{zt}{1-t}}z\,dt}% {(1\!-\!t)^{2}t^{n}\cdot\frac{zt}{1-t}}\;=\;\frac{n!}{2i\pi}\oint_{C^{\prime}}% \frac{e^{-\frac{zt}{1-t}}\,dt}{(1\!-\!t)t^{n+1}}$

where the contour $C^{\prime}$ goes round the origin.  Accordingly, by (2) we can infer that

 $L_{n}(z)\;=\;\left[\frac{d^{\,n}}{dt^{n}}\frac{e^{-\frac{zt}{1-t}}}{1\!-\!t}% \right]_{t=0},$

whence we have found the generating function

 $\frac{e^{-\frac{zt}{1-t}}}{1\!-\!t}\;=\;\sum_{n=0}^{\infty}\frac{L_{n}(z)}{n!}% t^{n}$

of the Laguerre polynomials.

 Title generating function of Laguerre polynomials Canonical name GeneratingFunctionOfLaguerrePolynomials Date of creation 2013-03-22 19:06:51 Last modified on 2013-03-22 19:06:51 Owner pahio (2872) Last modified by pahio (2872) Numerical id 8 Author pahio (2872) Entry type Derivation Classification msc 33B99 Classification msc 30B10 Classification msc 26C05 Classification msc 26A09 Classification msc 33E30 Related topic ExampleOfFindingTheGeneratingFunction Related topic GeneratingFunctionOfHermitePolynomials Related topic VariantOfCauchyIntegralFormula