Examples F

Section 4.12 Examples F

Laguerre’s Function.

Example 4.12.1.

Prove that

\begin{equation*} L_{n}(0)=n! \end{equation*}

Solution.

We have

\begin{equation*} \sum\limits_{n=0}^{\infty}\frac{L_{n}(x)}{n!}t^{n}=\frac{1}{1-t}e^{-xt/(1-t)} \end{equation*}

putting \(x=0\text{,}\) we get -

\begin{equation*} \sum\limits_{n=0}^{\infty}\frac{L_{n}(0)}{n!}t^{n}=\frac{1}{1-t}=(1-t)^{-1} = 1+t+t^{2}+\cdots+t^{n}+\cdots = \sum\limits_{n=0}^{\infty}t^{n} \end{equation*}

\begin{equation*} \therefore L_{n}(0) = n! \end{equation*}

Example 4.12.2.

Prove that

\begin{equation*} xL''_{n}(x) +(1-x)L'_{n}(x)+nL_{n}(x)=0 \end{equation*}

and hence deduce that \(L'_{n}(0) =-n(n!)\text{.}\)

Solution.

Since \(L_{n}(x)\) satisfies the Laguerre’s equation, \(xy''+(1-x)y'+ny =0\) [for \(\lambda =n\)].

\begin{equation*} \therefore, \quad xL''_{n}(x) +(1-x)L'_{n}(x)+nL_{n}(x)=0 \end{equation*}

putting \(x=0\text{,}\) we have \(L'_{n}(0)=-n L_{n}(0)\)

\begin{equation*} \therefore L'_{n}(0) =-n(n!). \end{equation*}

Example 4.12.3.

Evaluate

\begin{equation*} \int\limits_{0}^{\infty}e^{-x}[L_{n}(x)]^{2}\,dx. \end{equation*}

Solution.

We may have

\begin{equation*} \int\limits_{0}^{\infty}e^{-x}\sum\limits_{m,n =0}^{\infty}L_{m}(x)L_{n}(x)\frac{s^{m}t^{n}}{m!n!}\,dx \end{equation*}

\begin{equation*} =(1-s)^{-1}(1-t)^{-1}\int\limits_{0}^{\infty}exp\left[-x\frac{(1-st)}{(1-s)(1-t)}\right]\,dx \end{equation*}

\begin{equation*} =(1-st)^{-1}=\sum\limits_{n}(st)^{n} \hspace{1cm} \left[\because \int\limits_{0}^{\infty}e^{-\alpha x}\,dx=\frac{1}{\alpha}\right] \end{equation*}

put \(m=n\) and equate the coefficients of \((st)^{n}\) on both sides, we get -

\begin{equation*} \int\limits_{0}^{\infty}e^{-x}[L_{n}(x)]^{2}\,dx =(n!)^{2}. \end{equation*}

Alternative: In case of \(r=n\) in eqn. (4.11.8), we have

\begin{equation*} L_{n}(x) =(-1)^{n}x^{n} \end{equation*}

\begin{equation*} \therefore \int\limits_{0}^{\infty}e^{-x}[L_{n}(x)]^{2}\,dx =(-1)^{n}\int\limits_{0}^{\infty}e^{-x}x^{n}L_{n}(x)\,dx \end{equation*}

\begin{equation*} =(-1)^{n}\int\limits_{0}^{\infty}e^{-x}x^{n}e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x})\,dx \end{equation*}

\begin{equation*} =(-1)^{n}\int\limits_{0}^{\infty}x^{n}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x})\,dx \end{equation*}

Integrating by parts \(n\) times,

\begin{equation*} =(-1)^{n}n!\int\limits_{0}^{\infty}\frac{d^{n-m}}{dx^{n-m}}(x^{n}e^{-x})\,dx = (-1)^{n}n!\int\limits_{0}^{\infty}x^{n}e^{-x}\,dx=(n!)^{2}. \end{equation*}

Example 4.12.4.

Show that

\begin{equation*} L_{2n}(x) = n!\sum\limits_{m=0}^{n}\frac{2^{n-m}(-1)^{m}}{m!(n-m)!}L_{n-m}(x), \hspace{1cm} n \gt m. \end{equation*}

Solution.

We have

\begin{equation*} L_{n}(x)=\frac{1}{n!}e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x}) \end{equation*}

Therefore

\begin{equation*} L_{o}(x)=1, \hspace{1cm} L_{1}(x)=\frac{1}{1!}(1-x) =1-x. \end{equation*}

Replacing \(x\) by \(2x\text{,}\) we get -

\begin{equation*} L_{1}(2x)=1-2x =2(1-x)-1=2L_{1}(x)-2\frac{1}{1!}L_{o}(x) \end{equation*}

\begin{equation*} = 1!\sum\limits_{m=0}^{1}\frac{2^{1-m}(-1)^{m}}{m!(1-m)!}L_{1-m}(x) \end{equation*}

Also,

\begin{equation*} L_{2}(2x)=\frac{1}{2!}(2-8x+4x^{2}) = 1-4x+2x^{2} \end{equation*}

\begin{equation*} =4(1-2x+\frac{x^{2}}{2})-4(1-x)+1 \end{equation*}

\begin{equation*} =2^{2}L_{2}(x)-2\frac{2!}{1!1!}L_{1}(x)+\frac{2!}{0!2!}L_{o}(x) \end{equation*}

\begin{equation*} =2!\sum\limits_{m=0}^{2}\frac{2^{2-m}(-1)^{m}}{m!(2-m)!}L_{2-m}(x) \end{equation*}

similarly,

\begin{equation*} L_{3}(2x)=3!\sum\limits_{m=0}^{3}\frac{2^{3-m}(-1)^{m}}{m!(3-m)!}L_{3-m}(x) \end{equation*}

and in general,

\begin{equation*} L_{n}(2x)=3!\sum\limits_{m=0}^{n}\frac{2^{n-m}(-1)^{m}}{m!(n-m)!}L_{n-m}(x) \end{equation*}

Example 4.12.5.

Show that if

\begin{equation*} \int\limits_{0}^{\infty}e^{-x}x^{m}L_{n}(x)\,dx =0 \quad \text{if} \quad n \gt m. \end{equation*}

Solution.

\begin{equation*} \int\limits_{0}^{\infty}e^{-x}x^{m}L_{n}(x)\,dx= \int\limits_{0}^{\infty}x^{m}e^{-x}e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x})\,dx \end{equation*}

Rodrigue’s formula for \(L_{n}(x)\) is

\begin{equation*} e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x}) =\int\limits_{0}^{\infty}x^{m}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x})\,dx \end{equation*}

\begin{equation*} =\left.x^{m}\frac{d^{n-1}}{dx^{n-1}}(x^{n}e^{-x})\right\vert_{0}^{\infty}-\int\limits_{0}^{\infty}mx^{m-1}\frac{d^{n-1}}{dx^{n-1}}(x^{n}e^{-x})\,dx \end{equation*}

\begin{equation*} =0+(-1)m\int\limits_{0}^{\infty}x^{m-1}\frac{d^{n-1}}{dx^{n-1}}(x^{n}e^{-x})\,dx \end{equation*}

\begin{equation*} =(-1)^{2}m(m-1)\int\limits_{0}^{\infty}x^{m-2}\frac{d^{n-2}}{dx^{n-2}}(x^{n}e^{-x})\,dx \end{equation*}

\begin{equation*} =(-1)^{m}m!\int\limits_{0}^{\infty}\frac{d^{n-m}}{dx^{n-m}}(x^{n}e^{-x})\,dx =0. \quad if \quad n \gt m. \end{equation*}

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