3.7 Laguerre Polynomials

Definition 3.7.1. Laguerre polynomials are defined by \[L_n^{\alpha }(x) = (-1)^n\,e^x\, x^{-\alpha }\, \frac {d^n}{dx^n}\left (x^{n+\alpha }\, e^{-x}\right )\hspace {0.3cm} \text {for}\hspace {0.2cm} n= 0,\, 1, \, 2,\, \cdots \] and \(\alpha > -1\).
We shall simply write \(L_n^{\alpha }\) by \(L_n\).

Theorem 3.7.2 (Recurrence Relations).

(a).: \(L_{n+1}(x) = (x - \alpha - n - 1)\, L_n(x) - x\, L_n'(x), \hspace {0.3cm} n=0, \, 1, \, 2, \, \cdots \cdots \)
(b).: \(L_{n+1}(x) = (x-\alpha -2n - 1)\, L_n(x) - n(\alpha +n)\, L_{n-1}(x)\).

Proof.

(a).: By definition, we have \begin {equation} (-1)^n\, x^{\alpha }\, e^{-x}\, L_n(x) = \frac {d^n}{dx^n}\left (x^{n+\alpha }\, e^{-\alpha }\right ) \end {equation} Replace \(n\) by \((n+1)\) in (4.7), get \begin {equation} (-1)^{n+1}\, x^{\alpha }\, e^{-x}\, L_{n+1}(x) = \frac {d^{n+1}}{dx^{n+1}}\left (x^{n+1+\alpha }\, e^{-x}\right ) \end {equation} But we can write \begin {equation} x^{n + 1 + \alpha }\, e^{-x} = x\left (x^{n+\alpha }\, e^{-x}\right ) \end {equation} Applying Leibniz formula for the \((n+1)\) of derivative of the product in the right hand side of (4.9), get \[\frac {d^{n+1}}{dx^{n+1}}\left (x^{n + \alpha + 1}\, e^{-x}\right ) = \frac {d^{n+1}}{dx^{n+1}}\left [x\left (x^{n + \alpha } \, e^{-x}\right )\right ] = \sum _{k =0}^{n+1} \binom {n+1}{k}\, x^{(k)}\, \frac {d^{n+1 -K}}{dx^{n+1-k}}\left (x^{n+k}\, e^{-x}\right )\] since \[\frac {d^n}{dx^n}[f(x)\,g(x)] = \sum _{k=0}^nf^{(k)}(x)\, \cdot \,g^{(n-k)}(x)\] \[\frac {d^{n+1}}{dx^{n+1}}\left (x^{n + \alpha + 1}\, e^{-x}\right ) = x\, \frac {d^{n+1}}{dx^{n+1}}\left (x^{n+k} \, e^{-x}\right ) + (n +1) \, \frac {d^n}{dx^n}\left (x^{n+\alpha } \, e^{-x}\right )\] Since \[\frac {d^k}{dx^k}(x) = 0\hspace {0.3cm} \text {for}\hspace {0.3cm} k \geq 2.\] \begin {equation} \frac {d^{n+1}}{dx^{n+1}}\left (x^{n + \alpha + 1}\, e^{-x}\right ) = x\, \frac {d}{dx}\left [\frac {d^n}{dx^n}\left (x^{n+\alpha }\, e^{-x}\right )\right ] + (n+1)\, \frac {d^n}{dx^n}\left (x^{n+\alpha }\, e^{-x}\right ) \end {equation} Using (4.7) and (4.8) in (4.10), get \begin {align*} (-1)^{n+1}\, x^{\alpha }\, e^{-x}\, L_{n+1}(x) & = x\, \frac {d}{dx}\left [(-1)^n x^{\alpha } e^{-x}L_n(x)\right ] + (-1)^n(n+1) x^{\alpha } e^{-x} L_n(x)\\ & = (-1)^nx^{\alpha }e^{-x}\left [(\alpha + n + 1 -x) L_n(x) + x L_n'(x)\right ]. \end {align*}
Multiplying through by \((-1)^{n+1}e^xx^{-\alpha }\), get \[L_{n+1}(x) = (x - \alpha -n -1)\, L_n(x) - xL_n'(x).\]
(b).: (Exercise)
Use the generating function \[g(x,t) = (1 - t)^{-\alpha -1}e^{-\frac {xt}{(1 - t)}} = \sum _{n=0}^{\infty } L_n(x)\, t^n\] for \(|t| <1\), to sow that \begin {equation} (1 -t^2)\frac {\partial g}{\partial t} + [x - (1-t)(1 +\alpha )]\, g(x,t) = 0 \end {equation} then substitute for \[g(x,t) = \sum _{n=0}^{\infty } L_n(x)\, t^n,\] in (4.11) to get the desired result after comparing coefficients.

□

Theorem 3.7.3 (Orthogonality). The weight function is \(w(x) = e^{-x}\, x^{\alpha }\) over \((0,\infty )\), so thats \[\int _0^{\infty }e^{-x}\, x^{\alpha }\, L_m(x)\, L_n(x)\, dx = \begin {cases} 0, & m\neq n\\ n!\, \Gamma (n + \alpha + 1), & n= m. \end {cases}\]

Proof. Exercise. □

Theorem 3.7.4 (Series Expansion). \[f(x) = \sum _{n=0}^{\infty } C_n\, L_n(x)\] where \[C_n = \frac {1}{n!\, \Gamma (\alpha + n + 1)}\int _0^{\infty }e^{-x}\, x^{\alpha }\, L_n(x)\, f(x)\, dx\, , \hspace {0.3cm} n=0, \, 1,\, 2, \, \cdots \cdots \]

Exercise 3.7.5.

(a).: Show that \(U = L_n(x)\) is a solution of the differential equation \[x\, U'' + (\alpha + 1 - x)\, U' + n\ U = 0.\]
(b).: If \(f(x) = e^{-\alpha x}\), \(\, \alpha >\frac {-1}{2}\), show that \[e^{-a x} = \sum _{n=0}^{\infty } C_n\, L_n(x)\] where \[C_n = \frac {a^n}{n!\, (a + 1)^{n + \alpha + 1}}.\]
(c).: Find the value of \(L_n''(0)\).
(d).: If \[g(x,t) = (1-t)^{-\alpha -t}\, e^{-\frac {xt}{(1-t)}} = \sum _{n=0}^{\infty } t^n\, L_n(x).\] Show that \[(1 - t)\, \frac {\partial g}{\partial t} + t\, g(x,t) = 0.\] Then use this to show that \[L_n'(x) - L_{n-1}'(x) + L_{n-1}(x) = 0, \hspace {0.3cm} n = 1, \, 2, \, \cdots \cdots \]

[next] [prev] [prev-tail] [front] [up]

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.