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Rodrigues' formula

Formula for the Legendre polynomials


Summary

Formula for the Legendre polynomials

In mathematics, Rodrigues' formula (formerly called the Ivory–Jacobi formula) generates the Legendre polynomials. It was independently introduced by , and . The name "Rodrigues formula" was introduced by Heine in 1878, after Hermite pointed out in 1865 that Rodrigues was the first to discover it. The term is also used to describe similar formulas for other orthogonal polynomials. describes the history of the Rodrigues formula in detail.

Statement

Let (P_n(x)){n=0}^\infty be a sequence of orthogonal polynomials on the interval [a, b] with respect to weight function w(x). That is, they have degrees deg(P_n) = n, satisfy the orthogonality condition \int_a^b P_m(x) P_n(x) w(x) , dx = K_n \delta{m,n} where K_n are nonzero constants depending on n, and \delta_{m,n} is the Kronecker delta. The interval [a, b] may be infinite in one or both ends.

w(x)=W(x)/B(x), \quad \frac{W'(x)}{W(x)} = \frac{A(x)}{B(x)}, where A(x) is a polynomial with degree at most 1 and B(x) is a polynomial with degree at most 2, and \lim_{x \to a} x^k W(x) = 0, \qquad \lim_{x \to b} x^k W(x) = 0. for any k = 0, 1, 2, \dots.

Then, if \frac{d^n}{dx^n} !\left[ B(x)^n w(x)\right] \neq 0 for all n = 0, 1, 2, \dots, then P_n(x) = \frac{c_n}{w(x)} \frac{d^n}{dx^n} !\left[ B(x)^n w(x)\right], for some constants c_n.

Let F_k := \frac 1w D_x^k(B^n w), then F_k = B^{n-k} p_k for all k \in 0:n for some polynomials p_k, such that deg(p_k) \leq k. Proven by induction on k: F_{k+1} = B^{n-k-1}(B p_k' + (n-k)B' p_k + (A-B')p_k)

Let Q_n := \frac 1w D_x^n(B^n w). We have shown that Q_n is a polynomial of degree \leq n. With integration by parts, we have for all n m, \int_a^b Q_m Q_n w dx = \int_a^b B^n w (D_x^n Q_m) dx = 0 since D_x^n Q_m=0. Thus, Q_0, Q_1, \dots make up an orthogonal polynomial series with respect to w. Thus, P_n = c_n Q_n for some constants c_n.

\lambda_n = -\frac{1}{2}n(n-1)B''-nA'

When n = 0, it is trivial. When n = 1, it simplifies to AP_1' = A'P_1, which is true since P_1 = \frac{c_1}{w}(Bw)' = c_1A. So assume n \geq 2. Define I_n(x) = \frac{d^n}{dx^n}(B^n(x) w(x)), then by direct computation and simplification, the equation to be proven is equivalent to

\frac{d^2}{dx^2} (B(x) I_n(x)) - \frac{d}{dx} (A(x) I_n(x)) + \lambda_n I_n(x) = 0

By Leibniz differentiation rule, we have

B(x) \frac{d^n}{dx^n} y = \frac{d^n}{dx^n} (B(x) y) - n \frac{d^{n-1}}{dx^{n-1}} (B'(x) y) + \frac{n(n-1)}{2} \frac{d^{n-2}}{dx^{n-2}} (B'' y)

A(x) \frac{d^n}{dx^n} y = \frac{d^n}{dx^n} (A(x) y) - n \frac{d^{n-1}}{dx^{n-1}} (A' y)

for arbitrary y. This allows us to move A(x), B(x) to the other side of the n-th derivative. Set y = B^n(x) w(x) , and define

J(x) = \frac{d^2}{dx^2} (B(x) y(x)) - n \frac{d}{dx} (B'(x) y(x)) + \frac{n(n-1)}{2} B'' y(x)

K(x) = -\frac{d}{dx} (A(x) y(x)) + n A' y(x)

L(x) = \lambda_n y(x)

Then the equation simplifies to \frac{d^n}{dx^n} (J+K + L) = 0

J(x) has three terms, call them in order J_1(x), J_2(x), J_3(x). K(x) has two terms, call them in order K_1(x), K_2(x).

J_3(x) + K_2(x) + L(x) = (\lambda_n + \frac{n(n-1)}{2} B'' + n A')y=0.

That J_1(x) + J_2(x) + K_1(x) = 0. follows from first writing J_1(x) as

J_1(x) = \frac{d^2}{dx^2} \left(B^n(x) \int \exp\left(\frac{A(x)}{B(x)}\right)dx \right)

and then taking the innermost first derivative to obtain

J_1(x) = \frac{d}{dx}\left[\bigg(nB'(x)B^{n-1}(x) + A(x)B^{n-1}(x)\bigg)\int \exp\left(\frac{A(x)}{B(x)}\right)dx\right]

and then rewriting this as

J_1(x) = \frac{d}{dx}\Big(nB'(x)B^{n}(x)w(x)+ A(x)B^{n}(x)w(x)\Big)

The first term is the negative of J_2(x) and the second term is the negative of K_1(x).

More abstractly, this can be viewed through Sturm–Liouville theory. Define an operator Lf := - \frac{1}{w} (Wf')', then the differential equation is equivalent to LP_n = \lambda_n P_n. Define the functional space X = L^2([a,b], w(x)dx) as the Hilbert space of functions over [a, b], such that \langle f, g\rangle := \int_a^b fgw. Then the operator L is self-adjoint on functions satisfying certain boundary conditions, allowing us to apply the spectral theorem.

Generating function

A simple argument using Cauchy's integral formula shows that the orthogonal polynomials obtained from the Rodrigues formula have a generating function of the form

G(x,u)=\sum_{n=0}^\infty u^nP_n(x)

The P_n(x) functions here may not have the standard normalizations. But we can write this equivalently as

G(x,u)=\sum_{n=0}^\infty \frac{u^n}{N_n}N_nP_n(x)

where the N_n are chosen according to the application so as to give the desired normalizations. The variable u may be replaced by a constant multiple of u so that

G(x,\alpha u)=\sum_{n=0}^\infty \frac{\alpha^n u^n}{N_n}N_nP_n(x)

This gives an alternate form of the generating function.

By Cauchy's integral formula, Rodrigues’ formula is equivalent toP_n(x)=\frac{n!}{2\pi i}\frac{c_n}{w(x)}\oint_C \frac{B^n(t) w(t)}{(t-x)^{n+1}},dtwhere the integral is along a counterclockwise closed loop around x. Let

u=\frac{t-x}{B(t)}

Then the complex path integral takes the form

P_n(x)=\frac{n!}{2\pi i}c_n\oint_C \frac{G(x,u)}{u^{n+1}},du

G(x,u)=\frac{w(t)\frac{dt}{du}}{w(x)B(t)}

where now the closed path C encircles the origin. In the equation for G(x,u), t is an implicit function of u. Expanding G(x,u) in the power series given earlier gives

\frac{1}{2\pi i}\oint_C \frac{G(x,u)}{u^{n+1}},du=\frac{1}{2\pi i}\oint_C \frac{\sum_{m=0}^\infty u^mP_m(x)}{u^{n+1}},du=P_n(x)

Only the m=n term has a nonzero residue, which is P_n(x). The n!,c_n coefficient was dropped since normalizations are conventions which can be inserted afterwards as discussed earlier.

By expressing t in terms of u in the general formula just given for G(x,u), explicit formulas for G(x,u) may be found. As a simple example, let B(x)=1 and A(x)=-x (Hermite polynomials) so that w(x)=\exp\left(-\frac{x^2}{2}\right), t=u+x, w(t)=\exp\left(-\frac{(u+x)^2}{2}\right) and so G(x,u)=\exp\left(-xu-\frac{u^2}{2}\right).

Examples

Family[a,b]wWABc_n
Legendre P_n[-1,+1]11-x^2-2x1-x^2\frac{(-1)^n}{2^n n!}
Chebyshev (of the first kind) T_n[-1,+1]1/\sqrt{1-x^2}\sqrt{1-x^2}-x1-x^2\frac{(-1)^n}{(2n-1)!!}
Chebyshev (of the second kind) U_n[-1,+1]\sqrt{1-x^2}(1-x^2)^{3/2}-3x1-x^2\frac{(-1)^n (n+1)}{(2n+1)!!}
Gegenbauer/ultraspherical C_n^{(\alpha)}(x)[-1,+1](1-x)^{\alpha-1/2} (1+x)^{\alpha-1/2}(1-x)^{\alpha+1/2} (1+x)^{\alpha+1/2}-(2\alpha + 1)x1-x^2\frac{(-1)^n (2\alpha)_n}{(\alpha+\frac{1}{2})_{n} 2^nn!}
Jacobi P_n^{(\alpha, \beta)}[-1,+1](1-x)^\alpha (1+x)^\beta(1-x)^{\alpha+1} (1+x)^{\beta +1}( \beta - \alpha ) - (\alpha+ \beta + 2) x1-x^2\frac{(-1)^n}{2^nn!}
associated Laguerre L^{(\alpha)}_n0, \infty)x^\alpha e^{-x}x^{\alpha+1} e^{-x}\alpha + 1 - xx\frac{1}{n!}
[physicist's Hermite H_n(-\infty, +\infty)e^{-x^2}e^{-x^2}-2x1(-1)^n

These formulae

Legendre

Source:

Rodrigues stated his formula for Legendre polynomials P_n:

P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} !\left[ (x^2 -1)^n \right]!. (1 - x^2) P_n''(x) - 2 x P_n'(x) + n (n + 1) P_n(x) = 0

For Legendre polynomials, the generating function is defined as

G(x,u)=\sum_{n=0}^\infty u^nP_n(x).

The contour integral gives the Schläfli integral for Legendre polynomials:

P_n(x) = \frac{1}{2\pi i 2^n} \oint_C \frac{(t^2-1)^n}{(t-x)^{n+1}} dt

Summing up the integrand

G(x,u) = \frac{1}{\sqrt{1 - 2ux + u^2}} \frac{1}{2\pi i} \oint_C \left(\frac{1}{t - t_-} - \frac{1}{t - t_+}\right) dt

where t_\pm = \frac{1}{u} (1 \pm \sqrt{1 - 2ux + u^2}). For small u, we have t_- \approx x, t_+ \to \infty, which heuristically suggests that the integral should be the residue around t_-, thus giving

G(x,u) = \frac{1}{\sqrt{1 - 2ux + u^2}}

Hermite

Source:

Physicist's Hermite polynomials:

H_n(x)=(-1)^n e^{x^2} \frac{d^n}{dx^n} !\left[e^{-x^2}\right] = \left(2x-\frac{d}{dx} \right)^n\cdot 1.H_n'' - 2xH_n' + 2nH_n = 0

The generating function is defined as

G(x,u)=\sum_{n=0}^\infty \frac{H_n(x)}{n!}, u^n.

The contour integral gives

H_n(x)=(-1)^n e^{x^2}\frac{n!}{2\pi i}\oint_C \frac{e^{-t^2}}{(t-x)^{n+1}},dt.

\begin{aligned} G(x,u) &= \sum_{n=0}^\infty \frac{(-1)^n e^{x^2}}{n!}\frac{n!}{2\pi i}, u^n \oint_C \frac{e^{-t^2}}{(t-x)^{n+1}},dt \ &= e^{x^2}\frac{1}{2\pi i}\oint_C e^{-t^2}\left(\sum_{n=0}^\infty \frac{(-1)^n u^n}{(t-x)^{n+1}}\right)dt \ &= e^{x^2}\frac{1}{2\pi i}\oint_C e^{-t^2} \frac{1}{t-x+u}\ &= e^{x^2}, e^{-(x-u)^2} \ & = e^{2xu- u^2} \end{aligned}

Laguerre

Source:

For associated Laguerre polynomials

L_n^{(\alpha)}(x) = {x^{-\alpha} e^x \over n!}{d^n \over dx^n} \left(e^{-x} x^{n+\alpha}\right) = \frac{x^{-\alpha}}{n!}\left( \frac{d}{dx}-1\right)^nx^{n+\alpha}.

xL^{(\alpha)}_n(x)'' + (\alpha + 1 - x)L^{(\alpha)}_n(x)' + nL^{(\alpha)}_n(x) = 0~.

The generating function is defined as

G(x,u) := \sum_{n=0}^\infty u^n L^{(\alpha)}_n(x)

By the same method, we have G(x,u) = \frac{1}{(1-u)^{\alpha+1}} e^{-\frac{ux}{1-u}}.

Jacobi

Source:{{cite web

P_n^{(\alpha,\beta)}(x) = \frac{(-1)^n}{2^n n!} (1-x)^{-\alpha} (1+x)^{-\beta} \frac{d^n}{dx^n} \left{ (1-x)^\alpha (1+x)^\beta \left (1 - x^2 \right )^n \right}. \left (1-x^2 \right)P_n^{(\alpha,\beta)}{}'' + ( \beta-\alpha - (\alpha + \beta + 2)x )P_n^{(\alpha,\beta)}{}' + n(n+\alpha+\beta+1) P_n^{(\alpha,\beta)} = 0.

: \sum_{n=0}^\infty P_n^{(\alpha,\beta)}(x) u^n = 2^{\alpha + \beta} R^{-1} (1 - u + R)^{-\alpha} (1 + u + R)^{-\beta},

where R = \sqrt{1 - 2ux + u^2} , and the branch of square root is chosen so that R(x, 0) = 1.

References

References

  1. Shapiro 2016, p. 2.
  2. Shapiro 2016, p. 2.
  3. Schläfli, Ludwig. (1881). "Über die zwei Heineschen Kugelfunktionen mit beliebigem Parameter und ihre ausnahmslose Darstellung durch bestimmte Integrale". Springer Basel.
  4. Arfken and Weber 2005, p. 817.
  5. Arfken and Weber 2005, p. 837.
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