Special functions · Orthogonal polynomials
Legendre polynomials live on the finite interval \([-1, 1]\). What if the natural domain of a problem is the half-line \([0, \infty)\), as for the distance of an electron from a nucleus, or the time since a radioactive sample was prepared? Then the right family is the Laguerre polynomials, orthogonal on \([0, \infty)\) with the weight \(e^{-x}\). This post follows the same path as the one on Legendre polynomials (generating function, Rodrigues' formula, recurrence, orthogonality, differential equation), so the two can be compared side by side.
(In the book the lower limit of the sum is cut off; it is \(n = 0\). The series converges for \(|t| < 1\), since the generating function is singular at \(t = 1\).)
Expanding the left side in powers of \(t\) (or using Rodrigues' formula below) gives
\[ L_0 = 1,\quad L_1 = 1 - x,\quad L_2 = x^2 - 4x + 2,\quad L_3 = -x^3 + 9x^2 - 18x + 6,\quad L_4 = x^4 - 16x^3 + 72x^2 - 96x + 24. \]Notice: \( L_n(0) = n! \) (put \(x = 0\): the left side is \( \frac{1}{1 - t} = \sum t^n \)), and the leading coefficient is \((-1)^n\).
Compare with Legendre's \( P_n = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2 - 1)^n \): the polynomial \((x^2 - 1)^n\), which vanishes at the endpoints \(\pm1\), is replaced by \( x^ne^{-x} \), which vanishes at the endpoint \(0\) (to order \(n\)) and decays at \(\infty\); the factor \(e^x\) undoes the weight.
By Leibniz's rule, \( \frac{d^n}{dx^n}(x^ne^{-x}) = \sum_k\binom nk\frac{n!}{k!}x^k(-1)^ke^{-x} \), which gives the explicit form
\[ L_n(x) = \sum_{k=0}^n\binom nk\frac{n!}{k!}(-x)^k. \]Derivation. Let \( G(t, x) = \frac{1}{1 - t}e^{-xt/(1-t)} \). Differentiating with respect to \(t\) gives
\[ (1 - t)^2\frac{\partial G}{\partial t} = (1 - t - x)\,G. \]Substituting \( G = \sum\frac{L_n}{n!}t^n \) and comparing coefficients of \(t^n\) (then multiplying by \(n!\)) gives exactly the recurrence. The factor \(n^2\) comes from the \(\frac{1}{n!}\) in the generating function.
On an infinite interval, polynomials are not integrable by themselves, so a weight is needed; here \(e^{-x}\) makes every integral \( \int_0^\infty e^{-x}x^k\,dx = k! = \Gamma(k + 1) \) finite.
Why it holds (as for Legendre): for \(m < n\), write \( e^{-x}L_n = \frac{d^n}{dx^n}(x^ne^{-x}) \) and integrate by parts \(n\) times against \(L_m\). The boundary terms vanish (at \(0\) because of the factor \(x^n\), at \(\infty\) because of \(e^{-x}\)), and \( L_m^{(n)} = 0 \). For \(m = n\), only the leading term \((-1)^nx^n\) of \(L_n\) survives \(n\) differentiations, giving \( (-1)^n\cdot n!\cdot(-1)^n\int_0^\infty x^ne^{-x}dx = n!\cdot n! \).
So any reasonable function on \([0, \infty)\) (with \( \int_0^\infty e^{-x}f^2\,dx < \infty \)) can be expanded as \( f = \sum c_nL_n \) with \( c_n = \frac{1}{(n!)^2}\int_0^\infty e^{-x}f\,L_n\,dx \).
Why: let \( u = x^ne^{-x} \). Then \( xu' = (n - x)u \). Differentiating this \(n + 1\) times (Leibniz) gives \( xw'' + (1 + x)w' + (n + 1)w = 0 \) for \( w = u^{(n)} = e^{-x}L_n \); substituting \( w = e^{-x}y \) leads to Laguerre's equation for \(y = L_n\). Alternatively, check it directly for small \(n\): for \(L_1 = 1 - x\): \( 0 + (1 - x)(-1) + (1 - x) = 0 \). ✓
The equation has a regular singular point at \(x = 0\). The second solution has a logarithmic singularity there, so \(L_n\), the only polynomial solution, is the "physical" one, which is what "particular solution" means here (as for Legendre's equation).
They inherit everything, with \(x^s\) added to the weight:
| Legendre \(P_n\) | Laguerre \(L_n\) (book) | |
|---|---|---|
| Interval, weight | \([-1, 1]\), \(1\) | \([0, \infty)\), \(e^{-x}\) |
| Generating function | \( (1 - 2xt + t^2)^{-1/2} = \sum P_nt^n \) | \( \frac{1}{1 - t}e^{-\frac{xt}{1-t}} = \sum\frac{L_n}{n!}t^n \) |
| Rodrigues | \( \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2 - 1)^n \) | \( e^x\frac{d^n}{dx^n}(x^ne^{-x}) \) |
| Recurrence | \( (n + 1)P_{n+1} - (2n + 1)xP_n + nP_{n-1} = 0 \) | \( L_{n+1} + (x - 2n - 1)L_n + n^2L_{n-1} = 0 \) |
| Norm | \( \frac{2}{2n + 1} \) | \( (n!)^2 \) |
| Equation | \( (x^2 - 1)y'' + 2xy' - n(n + 1)y = 0 \) | \( xy'' + (1 - x)y' + ny = 0 \) |
| Physics | angular part (spherical symmetry) | radial part (hydrogen atom) |
Try each one before opening the solution. More solved problems from this topic are in Series.
\( (x^3e^{-x})''' = (6 - 18x + 9x^2 - x^3)e^{-x} \), so \( L_3 = -x^3 + 9x^2 - 18x + 6 \).
\( y = x^2 - 4x + 2 \), \( y' = 2x - 4 \), \( y'' = 2 \): \( 2x + (1 - x)(2x - 4) + 2(x^2 - 4x + 2) = 2x + 2x - 4 - 2x^2 + 4x + 2x^2 - 8x + 4 = 0 \). ✓
\( x = 2 \pm\sqrt2 \approx 0.586 \) and \( 3.414 \). (These are the nodes of 2-point Gauss–Laguerre quadrature for \( \int_0^\infty e^{-x}f(x)\,dx \).)
\( x = L_0 - L_1 \) (since \(L_1 = 1 - x\)). Check with the coefficient formula: \( c_1 = \frac{1}{1}\int_0^\infty e^{-x}x(1 - x)dx = 1 - 2 = -1 \). ✓
\( (3!)^2 = 36 \).
\( x^{-s}e^x\big(x^{1+s}e^{-x}\big)' = x^{-s}e^x\big((1 + s)x^s - x^{1+s}\big)e^{-x} = s + 1 - x \).
Laguerre polynomials are the half-line counterpart of Legendre polynomials. The same machinery (a generating function, a Rodrigues formula, a three-term recurrence, orthogonality proved by integrating by parts, a second-order differential equation) works with the weight \(e^{-x}\) on \([0, \infty)\), and with \(x^se^{-x}\) for the generalized family. Their best-known appearance is in quantum mechanics, where they shape the radial wavefunctions of the hydrogen atom. When comparing formulas from different books, watch the normalization (\(L_n(0) = n!\) or \(L_n(0) = 1\)).
Related posts: Legendre polynomials, Bernoulli polynomials (generating functions), and the Gamma function (\(\int_0^\infty x^ke^{-x}dx = k!\)).