Template:Short description Template:For In mathematics, Rodrigues' formula (formerly called the Ivory–Jacobi formula) generates the Legendre polynomials. It was independently introduced by Template:Harvs, Template:Harvs and Template:Harvs. 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. Template:Harvtxt describes the history of the Rodrigues formula in detail.
StatementEdit
Let <math>(P_n(x))_{n=0}^\infty</math> be a sequence of orthogonal polynomials on the interval <math>[a, b]</math> with respect to weight function <math>w(x)</math>. That is, they have degrees <math>deg(P_n) = n</math>, satisfy the orthogonality condition <math display="block">\int_a^b P_m(x) P_n(x) w(x) \, dx = K_n \delta_{m,n}</math> where <math>K_n</math> are nonzero constants depending on <math>n</math>, and <math>\delta_{m,n}</math> is the Kronecker delta. The interval <math>[a, b]</math> may be infinite in one or both ends.
Template:Math proof{dx^{n-1}} (B'(x) y) + \frac{n(n-1)}{2} \frac{d^{n-2}}{dx^{n-2}} (B y)</math>
<math display="block">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)</math>
for arbitrary <math>y</math>. This allows us to move <math>A(x), B(x)</math> to the other side of the <math>n</math>-th derivative. Set <math>y = B^n(x) w(x) </math>, and define
<math display="block">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)</math>
<math display="block">K(x) = -\frac{d}{dx} (A(x) y(x)) + n A' y(x)</math>
<math display="block">L(x) = \lambda_n y(x)</math>
Then the equation simplifies to <math>\frac{d^n}{dx^n} (J+K + L) = 0</math>
<math>J(x)</math> has three terms, call them in order <math>J_1(x), J_2(x), J_3(x)</math>. <math>K(x)</math> has two terms, call them in order <math>K_1(x), K_2(x)</math>.
<math>J_3(x) + K_2(x) + L(x) = (\lambda_n + \frac{n(n-1)}{2} B + n A')y=0</math>.
That <math>J_1(x) + J_2(x) + K_1(x) = 0</math>. follows from first writing <math>J_1(x)</math> as
<math>J_1(x) = \frac{d^2}{dx^2} \left(B^n(x) \int \exp\left(\frac{A(x)}{B(x)}\right)dx \right)</math>
and then taking the innermost first derivative to obtain
<math>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]</math>
and then rewriting this as
<math>J_1(x) = \frac{d}{dx}\Big(nB'(x)B^{n}(x)w(x)+ A(x)B^{n}(x)w(x)\Big)</math>
The first term is the negative of <math>J_2(x)</math> and the second term is the negative of <math>K_1(x)</math>. }}
More abstractly, this can be viewed through Sturm–Liouville theory. Define an operator <math>Lf := - \frac{1}{w} (Wf')'</math>, then the differential equation is equivalent to <math>LP_n = \lambda_n P_n</math>. Define the functional space <math>X = L^2([a,b], w(x)dx)</math> as the Hilbert space of functions over <math>[a, b]</math>, such that <math>\langle f, g\rangle := \int_a^b fgw</math>. Then the operator <math>L</math> is self-adjoint on functions satisfying certain boundary conditions, allowing us to apply the spectral theorem.
Generating functionEdit
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
<math display="”block”">G(x,u)=\sum_{n=0}^\infty u^nP_n(x)</math>
The <math>P_n(x)</math> functions here may not have the standard normalizations. But we can write this equivalently as
<math display="”block”">G(x,u)=\sum_{n=0}^\infty \frac{u^n}{N_n}N_nP_n(x)</math>
where the <math>N_n</math> 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
<math display="”block”">G(x,\alpha u)=\sum_{n=0}^\infty \frac{\alpha^n u^n}{N_n}N_nP_n(x)</math>
This gives an alternate form of the generating function.
By Cauchy's integral formula, Rodrigues’ formula is equivalent to<math display="block">P_n(x)=\frac{n!}{2\pi i}\frac{c_n}{w(x)}\oint_C \frac{B^n(t) w(t)}{(t-x)^{n+1}}\,dt</math>where the integral is along a counterclockwise closed loop around <math>x</math>. Let
<math display=”block”>u=\frac{t-x}{B(t)}</math>
Then the complex path integral takes the form
<math display=”block”>P_n(x)=\frac{n!}{2\pi i}c_n\oint_C \frac{G(x,u)}{u^{n+1}}\,du</math>
<math display=”block”>G(x,u)=\frac{w(t)\frac{dt}{du}}{w(x)B(t)}</math>
where now the closed path C encircles the origin. In the equation for <math>G(x,u)</math>, <math>t</math> is an implicit function of <math>u</math>. Expanding <math>G(x,u)</math> in the power series given earlier gives
<math>\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)</math>
Only the <math>m=n</math> term has a nonzero residue, which is <math>P_n(x)</math>. The <math>n!\,c_n</math> 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 <math>G(x,u)</math>, explicit formulas for <math>G(x,u)</math> may be found. As a simple example, let <math>B(x)=1</math> and <math>A(x)=-x</math> (Hermite polynomials) so that <math>w(x)=\exp\left(-\frac{x^2}{2}\right)</math>, <math>t=u+x</math>, <math>w(t)=\exp\left(-\frac{(u+x)^2}{2}\right)</math> and so <math>G(x,u)=\exp\left(-xu-\frac{u^2}{2}\right)</math>.
ExamplesEdit
Family | <math>[a,b]</math> | <math>w</math> | <math>W</math> | <math>A</math> | <math>B</math> | <math>c_n</math> |
---|---|---|---|---|---|---|
Legendre <math>P_n</math> | <math>[-1,+1]</math> | <math>1</math> | <math>1-x^2</math> | <math>-2x</math> | <math>1-x^2</math> | <math>\frac{(-1)^n}{2^n n!}</math> |
Chebyshev (of the first kind) <math>T_n</math> | <math>[-1,+1]</math> | <math>1/\sqrt{1-x^2}</math> | <math>\sqrt{1-x^2}</math> | <math>-x</math> | <math>1-x^2</math> | <math>\frac{(-1)^n}{(2n-1)!!}</math> |
Chebyshev (of the second kind) <math>U_n</math> | <math>[-1,+1]</math> | <math>\sqrt{1-x^2}</math> | <math>(1-x^2)^{3/2}</math> | <math>-3x</math> | <math>1-x^2</math> | <math>\frac{(-1)^n (n+1)}{(2n+1)!!} </math> |
Gegenbauer/ultraspherical <math>C_n^{(\alpha)}(x)</math> | <math>[-1,+1]</math> | <math>(1-x)^{\alpha-1/2} (1+x)^{\alpha-1/2} </math> | <math>(1-x)^{\alpha+1/2} (1+x)^{\alpha+1/2} </math> | <math>-(2\alpha + 1)x </math> | <math>1-x^2</math> | <math>\frac{(-1)^n (2\alpha)_n}{(\alpha+\frac{1}{2})_{n} 2^nn!}</math> |
Jacobi <math>P_n^{(\alpha, \beta)}</math> | <math>[-1,+1]</math> | <math>(1-x)^\alpha (1+x)^\beta</math> | <math>(1-x)^{\alpha+1} (1+x)^{\beta +1}</math> | <math>( \beta - \alpha ) - (\alpha+ \beta + 2) x</math> | <math>1-x^2</math> | <math>\frac{(-1)^n}{2^nn!}</math> |
associated Laguerre <math>L^{(\alpha)}_n</math> | <math>[0, \infty)</math> | <math>x^\alpha e^{-x}</math> | <math>x^{\alpha+1} e^{-x}</math> | <math>\alpha + 1 - x</math> | <math>x</math> | <math>\frac{1}{n!}</math> |
physicist's Hermite <math>H_n</math> | <math>(-\infty, +\infty)</math> | <math>e^{-x^2}</math> | <math>e^{-x^2}</math> | <math>-2x</math> | <math>1</math> | <math>(-1)^n</math> |
Similar formulae hold for many other sequences of orthogonal functions arising from Sturm–Liouville equations, and these are also called the Rodrigues formula (or Rodrigues' type formula), especially when the resulting sequence is polynomial.
LegendreEdit
Rodrigues stated his formula for Legendre polynomials <math>P_n</math>: <math display="block">P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \!\left[ (x^2 -1)^n \right]\!.</math><math display="block">(1 - x^2) P_n(x) - 2 x P_n'(x) + n (n + 1) P_n(x) = 0</math>For Legendre polynomials, the generating function is defined as <math display="”block”">G(x,u)=\sum_{n=0}^\infty u^nP_n(x)</math>.
The contour integral gives the Schläfli integral<ref>Template:Citation</ref> for Legendre polynomials:<math display="block">P_n(x) = \frac{1}{2\pi i 2^n} \oint_C \frac{(t^2-1)^n}{(t-x)^{n+1}} dt</math> Summing up the integrand,<math display="block">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</math>where <math>t_\pm = \frac{1}{u} (1 \pm \sqrt{1 - 2ux + u^2})</math>. For small <math>u</math>, we have <math>t_- \approx x, t_+ \to \infty</math>, which heuristically suggests that the integral should be the residue around <math>t_-</math>, thus giving<math display="block">G(x,u) = \frac{1}{\sqrt{1 - 2ux + u^2}}</math>
HermiteEdit
Physicist's Hermite polynomials:<math display="block">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.</math><math display="block">H_n - 2xH_n' + 2nH_n = 0</math>
The generating function is defined as<math display="block">G(x,u)=\sum_{n=0}^\infty \frac{H_n(x)}{n!}\, u^n.</math>The contour integral gives<math display="block"> H_n(x)=(-1)^n e^{x^2}\frac{n!}{2\pi i}\oint_C \frac{e^{-t^2}}{(t-x)^{n+1}}\,dt. </math><math display="block"> \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} </math>
LaguerreEdit
For associated Laguerre polynomials,<math display="block">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}.</math><math display="block">xL^{(\alpha)}_n(x) + (\alpha + 1 - x)L^{(\alpha)}_n(x)' + nL^{(\alpha)}_n(x) = 0~.</math>
The generating function is defined as<math display="block">G(x,u) := \sum_{n=0}^\infty u^n L^{(\alpha)}_n(x)</math>By the same method, we have <math>G(x,u) = \frac{1}{(1-u)^{\alpha+1}} e^{-\frac{ux}{1-u}}</math>.
JacobiEdit
<math display="block"> 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\}.</math><math display="block"> \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.</math>
- <math> \sum_{n=0}^\infty P_n^{(\alpha,\beta)}(x) u^n = 2^{\alpha + \beta} R^{-1} (1 - u + R)^{-\alpha} (1 + u + R)^{-\beta}, </math>
where <math display="inline"> R = \sqrt{1 - 2ux + u^2} </math>, and the branch of square root is chosen so that <math>R(x, 0) = 1</math>.