Generating function, Recurrence relation and Rodrigues formula of Laguerre's Differential Equation

Generating function of Laguerre's Differential equation:

The generating function of Laguerre polynomial is

$$\frac{1}{1-t}e^{\frac{-xt}{1-t}}=\sum_{x=0}^\infty t^n t_n(x)$$

L.H.S

$$\frac{1}{1-t}e^{\frac{-xt}{1-t}}=\frac{1}{1-t}\biggl[1+\frac{\biggl(\frac{-xt}{1-t}\biggr)}{1!}+\frac{\biggl(\frac{-xt}{1-t}\biggr)^2}{2!}+\dotsm\biggr]$$

$$or,\;\;\;\;\;\;\frac{1}{1-t}\sum_{j=0}^\infty\frac{\biggl(\frac{-xt}{1-t}\biggr)^j}{j!}=\sum_{j=1}^\infty\frac{(-1)^jx^jt^j}{(1-t)^{j+1}J!}$$

Now take,

$$\frac{1}{(1-t)^{1+j}}=(1-t)^{-(1+j)}$$

$$=1+\frac{(1+J)t}{1!}+\frac{(1+J)(2+J)t^2}{2!}+ \frac{(1+J)(2+J)(3+J)t^3}{3!}+\dotsm+\frac{(1+J)(2+J)\dotsm(J+s)+s}{s!}+\dotsm$$

$$=\sum_{s=0}^\infty\frac{(J+1)(J+2)\dotsm(J+s)t^s}{s!}\times\frac{J(J-1)(J-2)\dotsm 3.2.1}{J(J-1)(J-2)\dotsm 3.2.1}$$

$$=\sum_{s=0}^\infty \frac{(J+s)!t^s}{s!J!}$$

L.H.S

$$\sum_{j=0}\infty\frac{(-1)^J x^J t^J}{J!}\times \sum_{s+0}^\infty\frac{(J+S)!t^s}{s!J!}$$

$$\sum_{j=0}^\infty \sum_{s=0}^\infty \frac{(-1)^J x^J t^{J+S}}{(J!)^2S!}(J+S)!$$

Put J+S=n

$$=\sum_{n=0}^\infty \biggl[\sum_{j=0}^n \frac{(-1)^J x^J n!}{(J!)^2 (n-J)!}\biggr]t^n$$

$$=\sum_{n=0}^\infty \biggl[\sum_{j=0}^n \frac{(-1)^J x^J n!}{(J!)^2 (n-J)!}\biggr]t^n$$

$$=\sum_{n=0}^\infty L_n(x) t^n$$

$$R.H.S$$

Recurrence relation of Lagueree polynomial:

We know, the generating function of lagueree polynomial is

$$\frac{1}{1-t}e^{\frac{-xt}{1-t}}=\sum_{n=0}^\infty t^n L_n(x)\dotsm(1)$$

$$or\;\; e^{\frac{-xt}{1-t}}=\sum_{n=0}^\infty t^n L_n(x)-\sum_{n=0}^\infty t^{n+1}L_n(x)$$

Differentiating both sides with respect to 't', we get,

$$or,\;\; e^{\frac{-xt}{1-t}} (-x)\biggl[\frac{(1-t)\cdot (1-t) \cdot (-1)}{(1-t)^2}\biggr]=\sum_{n=0}^\infty nt^{n-1}L_n(x)-\sum_{n=0}^\infty (n+1)t^nL_n(x)$$

$$or,\;\; \frac{1}{1-t} \cdot e^{\frac{-xt}{1-t}}\cdot (-x)\biggl[\frac{1}{1-t}\biggr]=\sum_{n=0}^\infty nt^{n-1}L_n(x)-\sum_{n=0}^\infty(n+1)t^nL_n(x)$$

$$or,\;\;(-x) \sum_{n=0}^\infty t^n L_n(x)=\sum_{n=0}^\infty n\cdot t^{n-1} L_n(x)- \sum_{n=0}^\infty ntL_n(x)-\sum_{n=0}^\infty (n+1) t^n L_n(x) +\sum_{n=0}^\infty (n+1) t^{n+1}L_n(x)$$

From equation(1)

Now collecting the coefficient of \(t^n\) on both side. We get

$$or,\;\; (-x)L_n(x)=(n+1) L_{n+1}(x)-nL_n(x)-(n+1)L_n(x)+nL_{n-1}(x)$$

$$or\;\; e^{\frac{-xt}{1-t}}= (1-t) \sum_{n=0}^\infty t^nL_n(x)$$

$$or\;\; e^{\frac{-xt}{1-t}\times \frac{-t}{1-t}}=(1-t) \sum_{n=0}^\infty t^nL_n'(x)$$

$$or\;\; \sum_{n=0}^\infty t^{n+1} L_n(x)= \sum_{n=0}^\infty t^n L_n'(x)- \sum_{n=0}^\infty t^{n+1} L_n'(x)$$

Now, collecting the coefficient of \(t^n\) on both sides, we get,

$$-L_{n-1}(x)= L_n'(x)- L_{n-1}'(x)$$

$$L_{n-1}(x)= L_{n-1}'(x)- L_n'(x)$$

This is the 2nd recurrence relation of Laguree polynomial.

Rodrigues Formula:

We know, the generating function of Laguree polynomial is,

$$\frac{1}{1-t}e^{\frac{-xt}{1-t}}=\sum_{n=0}^\infty t^nL_n(x)$$

$$or, \sum_n=0^\infty t^n L_n(x)=\frac{1}{1-t} e^{\frac{-xt}{1-t}}$$

Differentiating both sides with respect to 't' for 'n' times and take limit t\(\to\ 0\)

$$\lim_{t\to0} \frac{d^n}{dt^n}\sum_{n=0}^\infty t^nL_n(x)=\lim_{t\to0}\frac{d^n}{dt^n}\biggl[\frac{1}{1-t} e^{\frac{-xt}{1-t}}\biggr]$$

$$or, \; 0+n!L_n(x)+0=\lim_{t\to0}\frac{d^n}{dt^n}\biggl[\frac{1}{1-t}\cdot e^{\frac{-xt}{1-t}}\biggr]$$

$$or,\; n!L_n(x)=\lim_{t\to0} \frac{d^n}{dt^n}\biggl[\frac{1}{1-t}e^{\frac{x(1-t-1}{1-t}}\biggr]$$

$$=e^x\lim_{t\to0}\frac{d^n}{dt^n}\biggl[\frac{1}{1-t} e^{frac{-x}{1-t}}\biggr]$$

$$=e^x\lim_{t\to0}\frac{d^n}{dt^n}\biggl[\frac{1}{1-t}\sum_{j=0}^\infty \frac{(\frac{-x}{1-t})^j}{j!}\biggr]$$

$$=\frac{e^x}{n!} \lim_{t\to0} \frac{d^n}{dt^n} \biggl[\sum_{j=0}^\infty \frac{(-1)^j x^j}{j!(1-t)^{j+1}}\biggr]$$

$$or,\;L_n(x)=\frac{e^x}{n!}\lim_t\to0 \sum_{j=0}^\infty \frac{(-1)^j x^j}{j!}\times\frac{(j+1)(j+2)\dotsm (j+n)}{(1-t)^{j+n-1}}$$

$$\frac{e^x}{n!}\sum_{j=0}^\infty \frac{(-1)^j x^j(j+1)(j+2)\dotsm(j+n)}{j!}\times \frac{j(j-1)(j-2)\dotsm 3.2.1}{j(j-1)(j-2)\dotsm 3.2.1)}$$

$$=\frac{e^x}{n!}\sum_{j=0}^\infty \frac{(-1)^j x^j (j+n)!}{j!j!}\dotsm(1)$$

Let us take,

$$\frac{d^n}{dx^n}(x^ne^{-x})$$

$$=\frac{d^n}{dx^n}\biggl(x^n \sum_{j=0}^\infty \frac{(-x)^j}{j!}\biggr)$$

$$=\frac{d^n}{dx^n}\biggl(\sum_{J=0}^\infty \frac{(-1)^j x^{n+1}}{j!}\biggr)$$

$$=\sum_{j=0}^\infty \frac{(-1)^j}{j!}\times (n+j) (n+j-1)\dotsm (j+1)x^j$$

$$=\sum_{j=0}^\infty \frac{(-1)^j}{j!} x^j (n+j) (n+j-1)\dotsm j+1\times \frac{(j-1)\dotsm 3.2.1}{j(j-1)\dotsm 3.2.1}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j x^j (j+n)!}{j! j!}\dotsm(2)$$

Now from equation (1) and (2) we get,

$$L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}(x^n e^{-x})$$

Reference:

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5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.