Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

SPECIAL FUNCTIONS


Prove the expression (18.14) for the coefficients in the Legendre polynomial expansion
of a functionf(x).

If we multiply (18.13) byPk(x) and integrate fromx=−1tox= 1 then we obtain
∫ 1


− 1

Pk(x)f(x)dx=

∑∞


=0

a

∫ 1


− 1

Pk(x)P(x)dx

=ak

∫ 1


− 1

Pk(x)Pk(x)dx=

2 ak
2 k+1

,


where we have used the orthogonality property (18.12) and the normalisation property
(18.11).


Generating function

A useful device for manipulating and studying sequences of functions or quantities


labelled by an integer variable (here, the Legendre polynomialsP(x) labelled by


)isagenerating function. The generating function has perhaps its greatest utility


in the area of probability theory (see chapter 30). However, it is also a great


convenience in our present study.


The generating function for, say, a series of functionsfn(x)forn=0, 1 , 2 ,...is

a functionG(x, h) containing, as well asx, a dummy variablehsuch that


G(x, h)=

∑∞

n=0

fn(x)hn,

i.e.fn(x) is the coefficient ofhnin the expansion ofGin powers ofh. The utility


of the device lies in the fact that sometimes it is possible to find a closed form


forG(x, h).


For our study of Legendre polynomials let us consider the functionsPn(x)

defined by the equation


G(x, h)=(1− 2 xh+h^2 )−^1 /^2 =

∑∞

n=0

Pn(x)hn. (18.15)

As we show below, the functions so defined are identical to the Legendre poly-


nomials and the function (1− 2 xh+h^2 )−^1 /^2 is in fact the generating function for


them. In the process we will also deduce several useful relationships between the


various polynomials and their derivatives.


Show that the functionsPn(x)defined by (18.15) satisfy Legendre’s equation

In the followingdPn(x)/dxwill be denoted byPn′. Firstly, we differentiate the defining
equation (18.15) with respect toxand get


h(1− 2 xh+h^2 )−^3 /^2 =


Pn′hn. (18.16)

Also, we differentiate (18.15) with respect tohto yield


(x−h)(1− 2 xh+h^2 )−^3 /^2 =


nPnhn−^1. (18.17)
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