TeachingSpecial Functions – Orthogonal polynomials – Laguerre polynomials
The Laguerre polynomials are orthogonal on the interval \((0,\infty)\) with respect to the gamma distribution \(w(x)=e^{-x}x^{\alpha}\). They can be defined by means of their Rodrigues formula:
\[L_n^{(\alpha)}(x)=\frac{1}{n!}\,\frac{1}{w(x)}\,D^n\left[w(x)\,x^n\right] =\frac{1}{n!}\,e^x\,x^{-\alpha}\,D^n\left[e^{-x}x^{n+\alpha}\right],\quad n=0,1,2,\ldots.\tag1\]Using Leibniz' rule we have
\begin{align*} D^n\left[e^{-x}x^{n+\alpha}\right]&=\sum_{k=0}^n\binom{n}{k}D^ke^{-x}D^{n-k}x^{n+\alpha} =\sum_{k=0}^n\binom{n}{k}(-1)^ke^{-x}(n+\alpha)(n+\alpha-1)\cdots(\alpha+k+1)x^{\alpha+k}\\[2.5mm] &=e^{-x}\,x^{\alpha}\,\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{\Gamma(n+\alpha+1)}{\Gamma(k+\alpha+1)}\,x^k. \end{align*}Hence we have
\[L_n^{(\alpha)}(x)=\sum_{k=0}^n(-1)^k\binom{n+\alpha}{n-k}\frac{x^k}{k!},\quad n=0,1,2,\ldots,\tag2\]where
\[\binom{n+\alpha}{n-k}=\frac{\Gamma(n+\alpha+1)}{(n-k)!\,\Gamma(k+\alpha+1)} =\frac{(k+\alpha+1)_{n-k}}{(n-k)!},\quad k=0,1,2,\ldots,n.\]This proves that \(L_n^{(\alpha)}(x)\) is a polynomial of degree \(n\). Since
\[(-1)^k\binom{n+\alpha}{n-k}=\frac{(-1)^k}{(n-k)!}\,\frac{(\alpha+1)_n}{(\alpha+1)_k} =\frac{(\alpha+1)_n}{n!}\,\frac{(-n)_k}{(\alpha+1)_k},\quad k=0,1,2,\ldots,n\]we also have
\[L_n^{(\alpha)}(x)=\frac{(\alpha+1)_n}{n!}\sum_{k=0}^n\frac{(-n)_k}{(\alpha+1)_k}\,\frac{x^k}{k!} =\binom{n+\alpha}{n}\,{}_1F_1\left(\genfrac{}{}{0pt}{}{-n}{\alpha+1}\,;\,x\right),\quad n=0,1,2,\ldots.\tag3\]Note that we have
\[L_n^{(\alpha)}(0)=\binom{n+\alpha}{n}=\frac{(\alpha+1)_n}{n!},\quad n=0,1,2,\ldots\]and that the leading coefficient of the polynomial \(L_n^{(\alpha)}(x)\) equals
\[k_n=\frac{(-1)^n}{n!},\quad n=0,1,2,\ldots.\]Further we have
\begin{align*} \frac{d}{dx}L_n^{(\alpha)}(x)&=\frac{d}{dx}\sum_{k=0}^n(-1)^k\binom{n+\alpha}{n-k}\frac{x^k}{k!} =\sum_{k=1}^n(-1)^k\binom{n+\alpha}{n-k}\frac{x^{k-1}}{(k-1)!}\\[2.5mm] &=\sum_{k=0}^{n-1}(-1)^{k-1}\binom{n+\alpha}{n-k-1}\frac{x^k}{k!}=-L_{n-1}^{(\alpha+1)}(x),\quad n=1,2,3,\ldots.\tag4 \end{align*}Now we can prove the orthogonality relation
\[\int_0^{\infty}e^{-x}x^{\alpha}L_m^{(\alpha)}(x)L_n^{(\alpha)}(x)\,dx =\frac{\Gamma(n+\alpha+1)}{n!}\,\delta_{mn},\quad\alpha>-1\tag5\]for \(m,n\in\{0,1,2,\ldots\}\). First of all, the integral converges if the moments
\[\mu_n=\int_0^{\infty}e^{-x}x^{n+\alpha}\,dx\]exists for all \(n\in\{0,1,2,\ldots\}\). This leads to the condition \(\alpha>-1\). Note that \(\mu_n=\Gamma(n+\alpha+1)\).
Now we use (1) to obtain
\[\int_0^{\infty}e^{-x}x^{\alpha}L_m^{(\alpha)}(x)L_n^{(\alpha)}(x)\,dx =\frac{1}{n!}\,\int_0^{\infty}L_m^{(\alpha)}(x)D^n\left[e^{-x}x^{n+\alpha}\right]\,dx.\]We apply integration by parts to obtain
\[\int_0^{\infty}L_m^{(\alpha)}(x)D^n\left[e^{-x}x^{n+\alpha}\right]\,dx =(-1)^n\int_0^{\infty}D^nL_m^{(\alpha)}(x)e^{-x}x^{n+\alpha}\,dx\]which equals zero for \(m < n\). For \(m=n\) we find
\[\int_0^{\infty}D^nL_n^{(\alpha)}(x)e^{-x}x^{n+\alpha}\,dx =k_n\,n!\,\int_0^{\infty}e^{-x}x^{n+\alpha}\,dx=(-1)^n\,\Gamma(n+\alpha+1).\]This proves the orthogonality relation (5).
The Laguerre polynomials can also be defined by their generating function
\[(1-t)^{-\alpha-1}\exp\left(-\frac{xt}{1-t}\right)=\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^n.\tag6\]The proof is straightforward. Start with (3) and change the order of summation to obtain
\begin{align*} \sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^n&=\sum_{n=0}^{\infty}\frac{(\alpha+1)_n}{n!}\,t^n \sum_{k=0}^n\frac{(-n)_k}{(\alpha+1)_k}\,\frac{x^k}{k!} =\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{(\alpha+1)_n}{(\alpha+1)_k}\,\frac{(-1)^kx^kt^n}{k!\,(n-k)!} =\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\frac{(\alpha+1)_{n+k}}{(\alpha+1)_k}\,\frac{(-1)^kx^kt^{n+k}}{k!\,n!}\\[2.5mm] &=\sum_{k=0}^{\infty}\frac{(-xt)^k}{k!}\sum_{n=0}^{\infty}\frac{(\alpha+k+1)_n}{n!}\,t^n =\sum_{k=0}^{\infty}\frac{(-xt)^k}{k!}\,(1-t)^{-\alpha-k-1} =(1-t)^{-\alpha-1}\sum_{k=0}^{\infty}\frac{1}{k!}\left(-\frac{xt}{1-t}\right)^k. \end{align*}This proves (6).
If we define
\[F(x,t):=(1-t)^{-\alpha-1}\exp\left(-\frac{xt}{1-t}\right),\]then we easily obtain
\[\frac{\partial}{\partial x}F(x,t)=-t(1-t)^{-\alpha-2}\exp\left(-\frac{xt}{1-t}\right) \quad\Longrightarrow\quad(1-t)\frac{\partial}{\partial x}F(x,t)+tF(x,t)=0\]and
\begin{align*} \frac{\partial}{\partial t}F(x,t)&=\left\{(\alpha+1)(1-t)^{-\alpha-2}+(1-t)^{-\alpha-1}\cdot\frac{-x(1-t)-xt}{(1-t)^2}\right\} \exp\left(-\frac{xt}{1-t}\right)\\[2.5mm] &=\left\{\alpha+1-\frac{x}{1-t}\right\}(1-t)^{-\alpha-2}\exp\left(-\frac{xt}{1-t}\right), \end{align*}which implies that
\[(1-t)^2\frac{\partial}{\partial t}F(x,t)+\left[x-(\alpha+1)(1-t)\right]F(x,t)=0.\]The first relation leads to
\[(1-t)\sum_{n=0}^{\infty}\frac{d}{dx}L_n^{(\alpha)}(x)t^n+t\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^n=0\]or equivalently
\[\sum_{n=0}^{\infty}\frac{d}{dx}L_n^{(\alpha)}(x)t^n-\sum_{n=0}^{\infty}\frac{d}{dx}L_n^{(\alpha)}(x)t^{n+1} +\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^{n+1}=0.\]Hence we have
\[\frac{d}{dx}L_{n+1}^{(\alpha)}(x)-\frac{d}{dx}L_n^{(\alpha)}(x)+L_n^{(\alpha)}(x)=0,\quad n=0,1,2,\ldots\tag7\]or equivalently, by using (4),
\[\frac{d}{dx}L_n^{(\alpha)}(x)=L_n^{(\alpha)}(x)-L_n^{(\alpha+1)}(x),\quad n=0,1,2,\ldots.\]The second relation leads to
\[(1-t)^2\sum_{n=1}^{\infty}nL_n^{(\alpha)}(x)t^{n-1}+\left[x-(\alpha+1)(1-t)\right]\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^n=0\]or equivalently
\[\sum_{n=1}^{\infty}nL_n^{(\alpha)}(x)t^{n-1}-2\sum_{n=1}^{\infty}nL_n^{(\alpha)}(x)t^n +\sum_{n=1}^{\infty}nL_n^{(\alpha)}(x)t^{n+1}+x\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^n-(\alpha+1)\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^n +(\alpha+1)\sum_{n=0}^{\infty}L_n^{(\alpha)}(x)t^{n+1}=0.\]Equating the coefficients of equal powers of \(t\) we obtain the three-term recurrence relation
\[(n+1)L_{n+1}^{(\alpha)}(x)+(x-2n-\alpha-1)L_n^{(\alpha)}(x)+(n+\alpha)L_{n-1}^{(\alpha)}(x)=0,\quad n=1,2,3,\ldots.\tag8\]Note that this can also be written as
\[xL_n^{(\alpha)}(x)+(n+1)\left[L_{n+1}^{(\alpha)}(x)-L_n^{(\alpha)}(x)\right] -(n+\alpha)\left[L_n^{(\alpha)}(x)-L_{n-1}^{(\alpha)}(x)\right]=0,\quad n=1,2,3,\ldots.\]Now we differentiate and use (7) to obtain
\[x\frac{d}{dx}L_n^{(\alpha)}(x)+L_n^{(\alpha)}(x)-(n+1)L_n^{(\alpha)}(x)+(n+\alpha)L_{n-1}^{(\alpha)}(x)=0,\quad n=1,2,3,\ldots.\]This implies that
\[x\frac{d}{dx}L_n^{(\alpha)}(x)=nL_n^{(\alpha)}(x)-(n+\alpha)L_{n-1}^{(\alpha)}(x),\quad n=1,2,3,\ldots.\tag9\]Now we differentiate (9) and use (7) and (9) to find
\begin{align*} x\frac{d^2}{dx^2}L_n^{(\alpha)}(x)+\frac{d}{dx}L_n^{(\alpha)}(x) &=n\frac{d}{dx}L_n^{(\alpha)}(x)-(n+\alpha)\frac{d}{dx}L_{n-1}^{(\alpha)}(x) =(n+\alpha)\left[\frac{d}{dx}L_n^{(\alpha)}(x)-\frac{d}{dx}L_{n-1}^{(\alpha)}(x)\right]-\alpha\,\frac{d}{dx}L_n^{(\alpha)}(x)\\[2.5mm] &=-(n+\alpha)L_{n-1}^{(\alpha)}(x)-\alpha\,\frac{d}{dx}L_n^{(\alpha)}(x) =x\frac{d}{dx}L_n^{(\alpha)}(x)-nL_n^{(\alpha)}(x)-\alpha\,\frac{d}{dx}L_n^{(\alpha)}(x). \end{align*}This proves that the polynomial \(L_n^{(\alpha)}(x)\) satisfies the second-order linear differential equation
\[xy''(x)+(\alpha+1-x)y'(x)+ny(x)=0,\quad n\in\{0,1,2,\ldots\}.\]Finally, we use the generating function (6) to prove the addition formula
\[L_n^{(\alpha+\beta+1)}(x+y)=\sum_{k=0}^nL_k^{(\alpha)}(x)L_{n-k}^{\beta)}(y),\quad n=0,1,2,\ldots.\tag{10}\]The generating function (6) implies that
\begin{align*} \sum_{n=0}^{\infty}L_n^{(\alpha+\beta+1)}(x+y)t^n&=(1-t)^{-\alpha-\beta-2}\exp\left(-\frac{(x+y)t}{1-t}\right) =(1-t)^{-\alpha-1}\exp\left(-\frac{xt}{1-t}\right)\cdot(1-t)^{-\beta-1}\exp\left(-\frac{yt}{1-t}\right)\\[2.5mm] &=\sum_{k=0}^{\infty}L_k^{(\alpha)}(x)t^k\cdot\sum_{m=0}^{\infty}L_m^{(\beta)}(y)t^m =\sum_{n=0}^{\infty}\left(\sum_{k=0}^nL_k^{(\alpha)}(x)L_{n-k}^{(\beta)}(y)\right)t^n. \end{align*}This proves (10).
The Laguerre polynomials \(L_2^{(0)}(x)\), \(L_3^{(0)}(x)\) and \(L_4^{(0)}(x)\).
Last modified on May 22, 2021
