Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SPECIAL FUNCTIONS

and has three regular singular points, atx=− 1 , 1 ,∞. By comparing it with

(18.1), we see that the Chebyshev equation is very similar in form to Legendre’s

equation. Despite this similarity, equation (18.54) does not occur very often

in physical problems, though its solutions are of considerable importance in

numerical analysis. The parameterνis a given real number, but in nearly all

practical applications it takes an integer value. From here on we thus assume

thatν=n,wherenis a non-negative integer. As was the case for Legendre’s

equation, in normal usage the variablexis the cosine of an angle, and so

− 1 ≤x≤1. Any solution of (18.54) is called aChebyshev function.

The pointx= 0 is an ordinary point of (18.54), and so we expect to find

two linearly independent solutions of the formy=

∑∞

m=0amx

m. One could find

the recurrence relations for the coefficientsamin a similar manner to that used

for Legendre’s equation in section 18.1 (see exercise 16.15). For Chebyshev’s

equation, however, it is easier and more illuminating to take a different approach.

In particular, we note that, on making the substitutionx=cosθ, and consequently

d/dx=(− 1 /sinθ)d/dθ, Chebyshev’s equation becomes (withν=n)

d^2 y

dθ^2

+n^2 y=0,

which is the simple harmonic equation with solutions cosnθand sinnθ.The

corresponding linearly independent solutions of Chebyshev’s equation are thus

given by

Tn(x)=cos(ncos−^1 x)andVn(x)=sin(ncos−^1 x). (18.55)

It is straightforward to show that theTn(x)arepolynomialsof ordern,whereas

theVn(x)arenotpolynomials

Find explicit forms for the series expansions ofTn(x)andVn(x).

Writingx=cosθ, it is convenient first to form the complex superposition

Tn(x)+iVn(x)=cosnθ+isinnθ

=(cosθ+isinθ)n

=

(

x+i

√

1 −x^2

)n

for|x|≤1.

Then, on expanding out the last expression using the binomial theorem, we obtain

Tn(x)=xn−nC 2 xn−^2 (1−x^2 )+nC 4 xn−^4 (1−x^2 )^2 −···, (18.56)

Vn(x)=

√

1 −x^2

[n

C 1 xn−^1 −nC 3 xn−^3 (1−x^2 )+nC 5 xn−^5 (1−x^2 )^2 −···

]

, (18.57)

wherenCr=n!/[r!(n−r)!] is a binomial coefficient. We thus see thatTn(x) is a polynomial
of ordern, butVn(x) is not a polynomial.

It is conventional to define the additional functions

Wn(x)=(1−x^2 )−^1 /^2 Tn+1(x)andUn(x)=(1−x^2 )−^1 /^2 Vn+1(x).

(18.58)