orthogonality of Legendre polynomials


We start from the first order differential equationMathworldPlanetmath

(1-x2)dudx+2nxu= 0, (1)

where one can separate the variables (http://planetmath.org/SeparationOfVariables) and then get the general solution

u=C(1-x2)n. (2)

Differentiating n+1 times the equation (1) it takes the form

(1-x2)dn+2udxn+2-2xdn+1udxn+1+n(n+1)dnudxn= 0

or

(1-x2)d2ydx2-2xdydx+n(n+1)y= 0 (3)

where

y=dnudxn=Cdndxn(1-x2)n.

Especially, the particular solution

y=Pn(x):=12nn!dndxn(1-x2)n, (4)

which which is the Legendre polynomialDlmfDlmfMathworldPlanetmath of degree n, has been seen to satisfy the Legendre’s differential equation (3).

The equality (4) is Rodrigues formula (http://planetmath.org/RodriguesFormula).  We use it to find the leading coefficient of Pn(x) and to show the orthogonality (http://planetmath.org/OrthogonalPolynomials) of the Legendre polynomials P0(x),P1(x),P2(x),

0.1 The coefficient of xn

By the binomial theorem,

Pn(x) =12nn!dndxnj=0n(nj)x2(n-j)(-1)j
=12nn!j=0n(nj)(2n-2j)(2n-2j-1)(2n-2j-n+1)xn-2j(-1)j.

From the term with  j=0  we get as the coefficient of xn the following:

12nn!(n0)(2n)(2n-1)(2n-2)(2n-n+1)(-1)0=12nn!(2n)!(2n-n)!=(2n)!2n(n!)2 (5)

0.2 Orthogonality

Let  fm(x):=a0+a1x++amxm  be any polynomial of degree  m<n.  Integrating by parts (http://planetmath.org/IntegrationByParts) m times we obtain

-11fm(x)Pn(x)𝑑x =12nn!-11fm(x)dndxn(x2-1)n𝑑x
=12nn!/-11fm(x)dn-1dxn-1(x2-1)n-12nn!-11fm(x)dn-1dxn-1(x2-1)n𝑑x
  
=(-1)mamm!2nn!-11dn-mdxn-m(x2-1)n𝑑x
=(-1)mamm!2nn!/-11fm(x)dn-m-1dxn-m-1(x2-1)n= 0,

since  x=±1  are zeros of the derivatives dn-kdxn-k(x2-1)n.

If, on the other hand,  m=n,  the calculation gives firstly

-11fn(x)Pn(x)𝑑x= 2(-1)nan2n01(x2-1)n𝑑x= 2(-1)nan2nIn, (6)

where the integralDlmfPlanetmath In is gotten from

In=/01x(x2-1)n-2n01x2(x2-1)n-1𝑑x=-2n01[(x2-1)n+(x2-1)n-1]𝑑x=-2nIn-2nIn-1,

Thus we infer the recurrence relation

In=-2n2n+1In-1.

Using this and  I0=1  one easily arrives at

In=(-1)n246(2n)357(2n+1)=(-1)n[246(2n)]2(2n+1)!=(-1)n22n(n!)2(2n+1)!. (7)

If fn(x) also is a Legendre polynomial Pn(x), we can in (6) by (5) put

an=(2n)!2n(n!)2

and taking into account (7), too, (6) reads

-11[Pn(x)]2𝑑x=(-1)n2n-1(2n)!2n(n!)2(-1)n22n(n!)2(2n+1)!=22n+1.

Our results imply the orthonormality (http://planetmath.org/Orthonormal) condition

-11Pm(x)Pn(x)𝑑x=22n+1δmn, (8)

where δmn is the Kronecker delta.

References

  • 1 K. Kurki-Suonio: Matemaattiset apuneuvot.  Limes r.y., Helsinki (1966).
Title orthogonality of Legendre polynomials
Canonical name OrthogonalityOfLegendrePolynomials
Date of creation 2013-03-22 18:55:30
Last modified on 2013-03-22 18:55:30
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 14
Author pahio (2872)
Entry type Derivation
Classification msc 33C45
Related topic OrthogonalPolynomials
Related topic OrthogonalityOfChebyshevPolynomials
Related topic SubstitutionNotation