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Mathematical Methods for Physics and Engineering - Matematica.NET

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18.4 CHEBYSHEV FUNCTIONSin which the coefficients a n are given bya n = 2 π∫ 1−1f(x)U n (x)(1 − x 2 ) 1/2 dx.Generating functionsThe generating functions <strong>for</strong> the Chebyshev polynomials of the first <strong>and</strong> secondkinds are given, respectively, by1 − xh∞G I (x, h) =1 − 2xh + h 2 = ∑T n (x)h n , (18.63)n=0∞1G II (x, h) =1 − 2xh + h 2 = ∑U n (x)h n . (18.64)These prescriptions may be proved in a manner similar to that used in section18.1.2 <strong>for</strong> the generating function of the Legendre polynomials. For theChebyshev polynomials, however, the generating functions are of less practicaluse, since most of the useful results can be obtained more easily by takingadvantage of the trigonometric <strong>for</strong>ms (18.55), as illustrated below.Recurrence relationsThere exist many useful recurrence relationships <strong>for</strong> the Chebyshev polynomialsT n (x) <strong>and</strong>U n (x). They are most easily derived by setting x =cosθ <strong>and</strong> using(18.55) <strong>and</strong> (18.58) to writeT n (x) =T n (cos θ) =cosnθ, (18.65)sin(n +1)θU n (x) =U n (cos θ) = . (18.66)sin θOne may then use st<strong>and</strong>ard <strong>for</strong>mulae <strong>for</strong> the trigonometric functions to derivea wide variety of recurrence relations. Of particular use are the trigonometricidentitiesn=0cos(n ± 1)θ =cosnθ cos θ ∓ sin nθ sin θ, (18.67)sin(n ± 1)θ =sinnθ cos θ ± cos nθ sin θ. (18.68)◮Show that the Chebyshev polynomials satisfy the recurrence relationsT n+1 (x) − 2xT n (x)+T n−1 (x) =0, (18.69)U n+1 (x) − 2xU n (x)+U n−1 (x) =0. (18.70)Adding the result (18.67) with the plus sign to the corresponding result with a minus signgivescos(n +1)θ +cos(n − 1)θ =2cosnθ cos θ.601

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