Mathematical Methods for Physics and Engineering - Matematica.NET

SPECIAL FUNCTIONSand has three regular singular points, at x = −1, 1, ∞. By comparing it with(18.1), we see that the Chebyshev equation is very similar in form to Legendre’sequation. Despite this similarity, equation (18.54) does not occur very oftenin physical problems, though its solutions are of considerable importance innumerical analysis. The parameter ν is a given real number, but in nearly allpractical applications it takes an integer value. From here on we thus assumethat ν = n, wheren is a non-negative integer. As was the case for Legendre’sequation, in normal usage the variable x is the cosine of an angle, and so−1 ≤ x ≤ 1. Any solution of (18.54) is called a Chebyshev function.The point x = 0 is an ordinary point of (18.54), and so we expect to findtwo linearly independent solutions of the form y = ∑ ∞m=0 a mx m . One could findthe recurrence relations for the coefficients a m in a similar manner to that usedfor Legendre’s equation in section 18.1 (see exercise 16.15). For Chebyshev’sequation, however, it is easier and more illuminating to take a different approach.In particular, we note that, on making the substitution x =cosθ, and consequentlyd/dx =(−1/ sin θ) d/dθ, Chebyshev’s equation becomes (with ν = n)d 2 ydθ 2 + n2 y =0,which is the simple harmonic equation with solutions cos nθ and sin nθ. Thecorresponding linearly independent solutions of Chebyshev’s equation are thusgiven byT n (x) =cos(n cos −1 x) and V n (x) =sin(n cos −1 x). (18.55)It is straightforward to show that the T n (x) arepolynomials of order n, whereasthe V n (x) arenot polynomials◮Find explicit forms for the series expansions of T n (x) and V n (x).Writing x =cosθ, it is convenient first to form the complex superpositionT n (x)+iV n (x) =cosnθ + i sin nθ=(cosθ + i sin θ) n(= x + i √ ) n1 − x 2 for |x| ≤1.Then, on expanding out the last expression using the binomial theorem, we obtainT n (x) =x n − n C 2 x n−2 (1 − x 2 )+ n C 4 x n−4 (1 − x 2 ) 2 − ··· , (18.56)V n (x) = √ 1 − x 2 [n C 1 x n−1 − n C 3 x n−3 (1 − x 2 )+ n C 5 x n−5 (1 − x 2 ) 2 − ···] , (18.57)where n C r = n!/[r!(n−r)!] is a binomial coefficient. We thus see that T n (x) is a polynomialof order n, but V n (x) is not a polynomial. ◭It is conventional to define the additional functionsW n (x) =(1− x 2 ) −1/2 T n+1 (x) and U n (x) =(1− x 2 ) −1/2 V n+1 (x).596(18.58)