First four Lagueree polynomial, Orthogonality of Lagueree polynomial

First four Lagueree polynomial:

We know, the Rodrigue's formula for Lagueree polyomial is,

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

For first four polynomial put n=0,1,2,3

Then,

For n=0, \(L_0(x)=\frac{e^x}{0!} \frac{d^0}{dx^0} (x^0 e^{-x})\)

$$ =e^x(e^{-x})$$

$$=1$$

Put n=1,

$$L_1(x)=\frac{e^x}{1!} \frac{d}{dx} (x^1 e^{-x})$$

$$=e^x{1.e^{-x} + xe^{-x} (-1)}$$

$$= e^x[e^{-x}- xe^{-x}(-1)]$$

$$=e^x[e^{-x} - xe^{-x}]$$

$$=e^x\cdot e^{-x}[1-x]$$

$$=[1-x]$$

Put n=2

$$l_2(x)= \frac{e^x}{n!} \frac{d^2}{dx^2}[x^2 e^{-x}]$$

$$=\frac{e^x}{2!} \frac{d}{dx}[ 2xe^{-x}+ x^2(-1) e^{-x}]$$

$$=\frac{e^x}{2!} \frac{d}{dx}[2xe^{-x} - x^2 e^{-x}]$$

$$=\frac{e^x}{2!} [ 2e^{-x} + (-1) e^{-x} 2x- ( 2x.e^{-x} + (-1)e^{-x} x^2)]$$

$$=\frac{e^x}{2!} [ 2e^{-x} -- 2xe^{-x}- 2xe^{-x} + e^{-x} x^2]$$

$$=\frac{e^x}{2!} 2e^{-x} (1-x-x+x^2)$$

Orthogonality of Lagueree polynomial:

The form of Laguree differential equation as,

$$x\frac{d^2y}{dx^2}+ (1-x) \frac{dy}{dx}+ny=0$$

$$or,\;\; \frac{d^2y}{dx^2}+\frac{(1-x)}{x}\frac{dy}{dx}+\frac nx y=0$$

Since, \(y=L_n(x)\)

$$or, \frac{d^2L_n(x)}{dx^2} + \frac{(1-x)}{x}\cdot \frac{dL_n(x)}{dx}+\frac{n}{x} L_n(x)=0$$

Now, multiplying both sides by \(xe^{-x}\) on above equation as I.F we get,

$$ \frac{d}{dx}\biggl(xe^{-x} d\frac{dL_n(x)}{dx}\biggr) + ne^{-x} L_n(x)=0\dotsm(a)$$

For \(m^th\) order, we can write

$$ \frac{d}{dx} \biggl( xe^{-x} \frac{dL_m(x)}{d(x)}+ me^{-x} L_m(x)=0\dotsm(b)$$

Multiplying equation (a) by \(L_n(x)\) and subtracting we get,

$$or\;\;, L_m(x)\frac{d}{dx} \biggl(xe^{-x} d\frac{d_n(x)}{dx}\biggr)- l_n(x) \frac{d}{dx}\biggl( xe^{-x} \frac{dL_m(x)}{dx}\biggr)+e^{-x}(n-m) L_n(x) L_m(x)=0$$

$$or\;\; \frac{d}{dx} \biggl[ L_m(x) xe^{-x} \frac{dL_n(x)}{dx}- L_n(x) xe^{-x} \frac{dL_m(x)}{dx}\biggr]+ (n-m) e^{-x} L_n(x) L_m(x)=0$$

Integrating both side with respect to x from 0 to \(\infty\), we get

$$or\;\; \int_0^\infty \frac{d}{dx}\biggl[L_m(x) e^{-x} \frac{dL_n(x)}{dx}-L_n(x)\cdot xe^{-x} \frac{dL_m(x)}{dx}\biggr] dx+ (n-m) \int_0^\infty e^{-x} L_n(x) L_m(x) dx=0$$

$$or\;\;, \biggl| L_m(x)\cdot xe^{-x} \frac{dL_n(x)}{dx}- L_n(x)\cdot xe^{-x} \frac{dL_m(x)}{dx}\biggr|_0^\infty+ (n-m)\int_0^\infty e^{-x} L_n(x) L_m(x) dx=0$$

$$or\;\;, (n-m)\int_0^\infty e^{-x} L_n(x) L_m(x) dx=0\dot(d)$$

Case I:

For n\(\ne\)m

\(\int_0^\infty e^{-x} L_n(x) L_m(x) dx=0\dotsm(e)\)

Case II:

for n=m

\(\int_0^\infty e^{-x} L_n(x) L_m(x) dx\ne 0\)

We know the generating function of Laguree polynomial is:

$$\int_0^\infty t^n L_n(x)= \frac{1}{1-t} e^{\frac{-xt}{1-t}}$$

Squaring on both side, we get

$$\sum_0^\infty t^{2n} L_n(x) L_n(x) = \frac{1}{(1-t)^2} e^{\frac{-2xt}{1-t}}$$

When multiplying with \(e^{-x}\) and integrating from 0 to \(\infty\) cross term would be zero due to equation (d).

Multiplying both sides by \(e^{-x}\) , we get

$$or\;, \sum_{n=0}^\infty t^{2n} e^{-x} L_n(x) L_n(x) =\frac{1}{(1-t)^2} e^{-2x}\cdot e^{-2xt}{1-t}$$

$$or\;\; \sum_{n=0}^\infty t^{2n} e^{-x} L_n(x) L_n(x) = \frac{1}{(1-t)^2} e^{-x(\frac{1+t}{1-t})}$$

$$or\;\; \sum_{n=0}^\infty t^{2n} e^{-x} L_n(x) L_n(x) = \frac{1}{(1-t)^2} e^{-x(\frac{1+t}{1-t})}$$

Integrating both sides, we get,

$$\sum_{n=0}^\infty t^{2n} \int_0^\infty e^{-x} L_n(x) L_n(x) dx= \frac{1}{(1-t)^2} \int_0^\infty e^{{-x}(\frac{1+t}{1-t}) }dx$$

$$= \frac{1}{(1-t)^2}\frac{e^{-x(\frac{1+t}{1-t})}}{-(\frac{1+t}{1-t})}\biggr|_0^\infty$$

$$=\frac{1}({1-t)^2}\biggl[\frac{0-1}{-(\frac{1+t}{1-t})}\biggr]$$

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

$$=\frac{1}{t^2}$$

$$=(1-t^2)^{-1}$$

$$=1+\frac{1t^2}{1!}+\frac{1.2 (t^2)^2}{2!}+\frac{1.2.3}{3!}(t^2)^3+\dotsm+ (t^2)^n$$

$$=1+t^2+t^4+\dot t^{2n}$$

Collecting the coefficient of \(t^{2n}\) on both sides, we get,

$$\int_0^\infty e^{-x} L_n(x) L_n(x) dx=1\dot(1)$$

Now, combining (1) and (f), we get

$$\int_0^\infty e^{-x} L_n(x) L_m(x) dx= \delta _{mn}$$

Where \(\delta_{mn}\)= 1 for n=m

\(\delta_{mn}\)= 0 for n\(\ne\) m

this is the orthogonality relation for lagurre polynomial.

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.