MATHEMATICS OF COMPUTATION Volume 75, Number 255, July 2006, Pages 1217–1231 S 0025-5718(06)01859-X Article electronically published on May 1, 2006 INTEGRAL FORMULAS FOR CHEBYSHEV POLYNOMIALS AND THE ERROR TERM OF INTERPOLATORY QUADRATURE FORMULAE FOR ANALYTIC FUNCTIONS SOTIRIOS E. NOTARIS Abstract. 1 ∓ | | We evaluate explicitly the integrals −1 πn(t)/(r t)dt, r =1, with the πn being any one of the four Chebyshev polynomials of degree n. These integrals are subsequently used in order to obtain error bounds for in- terpolatory quadrature formulae with Chebyshev abscissae, when the function to be integrated is analytic in a domain containing [−1, 1] in its interior. 1. Introduction The usefulness of Chebyshev polynomials in numerical analysis is undisputed. This is largely due to their nice properties, many of which follow from the trigono- metric representations these polynomials satisfy on the interval [−1, 1]. An immedi- ate consequence of these representations is that the zeros of Chebyshev polynomials can be given by explicit formulas, and this makes them attractive choices for nodes of interpolatory quadrature formulae. Notable examples of the latter are the well- known Fej´er and Clenshaw–Curtis rules. In the following, we enlarge the list of properties of Chebyshev polynomials by showing that the integrals 1 πn(t) | | (1.1) ∓ dt, r =1, −1 r t with the πn being any one of the four Chebyshev polynomials of degree n,canbe computed explicitly. This finds immediate application in estimating the error of certain interpolatory formulae with Chebyshev abscissae, by means of Hilbert space or contour integration methods, when the function to be integrated is analytic in a domain containing [−1, 1] in its interior. The quality of these estimates, and a comparison with already existing ones, is illustrated with a numerical example. Received by the editor May 13, 2004 and, in revised form, October 3, 2004. 2000 Mathematics Subject Classification. Primary 33C45, 65D32. Key words and phrases. Integral formulas, Chebyshev polynomials, interpolatory quadrature formulae, error bounds. This work was supported in part by a grant from the Research Committee of the University of Athens, Greece, and in part by a “Pythagoras” O.P. Education grant to the University of Athens from the Ministry of National Education, Greece, and the European Union. c 2006 American Mathematical Society Reverts to public domain 28 years from publication 1217 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1218 S. E. NOTARIS 2. Integral formulas for Chebyshev polynomials The nth degree Chebyshev polynomials Tn,Un,Vn, and Wn of the first, second, third, and fourth kind, respectively, are characterized by the well-known represen- tations sin (n +1)θ (2.1) T (cos θ)=cosnθ, U (cos θ)= , n n sin θ cos (n +1/2)θ sin (n +1/2)θ (2.2) V (cos θ)= ,W(cos θ)= . n cos (θ/2) n sin (θ/2) We first give some simple relations among these polynomials, which will be useful in the subsequent development. Lemma 2.1. Let U−1(t)=0.Then 1 (2.3) T (t)= {U (t) − U − (t)},n=1, 2,..., n 2 n n 2 (2.4) Vn(t)=Un(t) − Un−1(t),n=0, 1, 2,..., (2.5) Wn(t)=Un(t)+Un−1(t),n=0, 1, 2,.... Proof. Formulas (2.3) and (2.4) follow from the corresponding formulas for the difference of sines sin (n +1)θ − sin (n − 1)θ =2cosnθ sin θ, sin (n +1)θ − sin nθ =2cos(n +1/2)θ sin (θ/2), and (2.1)–(2.2). Also, replacing θ by π + θ in the second equation in (2.1) and (2.2) gives n (2.6) Un(t)=(−1) Un(−t),n=0, 1, 2,..., n (2.7) Wn(t)=(−1) Vn(−t),n=0, 1, 2,..., which, together with (2.4), implies (2.5). We can now present the announced explicit forms for the integrals (1.1). The pattern was discovered by computing, analytically, enough of these integrals for the first few values of n. We examine separately the cases |r| > 1and|r| < 1. Proposition 2.2. Let r ∈ R with |r| > 1. (i) If πn = Tn,then [(n+1)/2] 1 πn(t) r +1 πn−2k+1(r) (2.8) dt = πn(r)ln − 4 , − r − t r − 1 2k − 1 1 k=1 while, if πn = Un,πn = Vn, or πn = Wn, then 1 [(n+1)/2] πn(t) r +1 πn−2k+1(r) (2.9) dt = πn(r)ln − 4 . − r − t r − 1 2k − 1 1 k=1 (ii) If πn = Tn,then [(n+1)/2] 1 πn(t) r +1 πn−2k+1(−r) (2.10) dt = πn(−r)ln +4 , − r + t r − 1 2k − 1 1 k=1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGRAL FORMULAS FOR CHEBYSHEV POLYNOMIALS 1219 while, if πn = Un,πn = Vn, or πn = Wn, then 1 [(n+1)/2] πn(t) r +1 πn−2k+1(−r) (2.11) dt = πn(−r)ln +4 . − r + t r − 1 2k − 1 1 k=1 By [·] we denote the integer part of a real number, while the notation means that the last term in the sum must be halved when n is odd. Proof. (i)Wefirstprove(2.9)withπn = Un. To this end, we apply induction on n.Forn = 1, an easy computation, using the second equation in (2.1), gives 1 U1(t) r +1 − − dt = U1(r)ln − 4. −1 r t r 1 Assume now that the formula is true for the indices n − 1andn, and we want to prove it for the index n + 1. It is well known that the Chebyshev polynomials of the second kind satisfy the three-term recurrence relation Um+1(t)=2tUm(t) − Um−1(t),m=0, 1, 2,..., (2.12) U−1(t)=0,U0(t)=1. Applying (2.12) with m = n, dividing by r−t and taking the integral on both sides, we have 1 1 1 Un+1(t) tUn(t) − Un−1(t) (2.13) − dt =2 − dt − dt. −1 r t −1 r t −1 r t Furthermore, adding and subtracting the term rUn(t) in the numerator of the first integral on the right side of (2.13), we get 1 1 1 1 Un+1(t) − Un(t) − Un−1(t) (2.14) − dt = 2 Un(t)dt +2r − dt − dt. −1 r t −1 −1 r t −1 r t We now consider two cases for n.Whenn is even, (2.14), by virtue of the induction hypothesis and ⎧ ⎨ 2 1 if m is even, Um(t)dt = ⎩ m +1 −1 0ifm is odd, yields 1 Un+1(t) r +1 dt = {2rU (r) − U − (r)} ln r − t n n 1 r − 1 −1 1 − 4 {2rUn−1(r) − Un−2(r)} + {2rUn−3(r) − Un−4(r)} 3 1 1 + ···+ {2rU (r) − U (r)} + , n − 1 1 0 n +1 which, in view of (2.12), gives 1 Un+1(t) r +1 dt = Un+1(r)ln − r − t r − 1 (2.15) 1 1 1 1 − 4 U (r)+ U − (r)+···+ U (r)+ . n 3 n 2 n − 1 2 n +1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1220 S. E. NOTARIS Also, for n odd, we obtain in a like manner, 1 U (t) r +1 n+1 dt = U (r)ln r − t n+1 r − 1 −1 1 1 1 − 4 U (r)+ U − (r)+···+ U (r)+ U (r) , n 3 n 2 n − 2 3 n 1 which, together with (2.15), combines to 1 [((n+1)+1) /2] Un+1(t) r +1 Un+1−2k+1(r) dt = Un+1(r)ln − 4 , − r − t r − 1 2k − 1 1 k=1 proving the induction claim, and concluding the induction. Also, from (2.3), 1 1 1 Tn(t) 1 Un(t) − Un−2(t) − dt = − dt − dt , −1 r t 2 −1 r t −1 r t and inserting (2.9) with πn = Un and πn−2 = Un−2, we find, taking into account that U−1(t)=0, 1 T (t) n dt r − t −1 1 r +1 = {U (r) − U − (r)} ln 2 n n 2 r − 1 1 1 1 − 4 {U − (r) − U − (r)} + {U − (r) − U − (r)} 2 n 1 n 3 3 2 n 3 n 5 1 1 { − } 1 1 { − } − U3(r) U1(r) + − U1(r) U−1(r) if n is even + ···+ n 3 2 n 1 2 , 1 1 { − } 1 1 n−2 2 U2(r) U0(r) + 2 n if n is odd which, by (2.3), shows (2.8). The proof of (2.9) with πn = Vn or πn = Wn is similar to that of (2.8), if we use (2.4) and (2.5) in place of (2.3). (ii) Replacing θ by π + θ in (2.1) and (2.2) gives n Tn(t)=(−1) Tn(−t),n=0, 1, 2,..., n Vn(t)=(−1) Wn(−t),n=0, 1, 2,..., which, inserted, together with (2.6) and (2.7), into (2.8) and (2.9), and changing variables from −t to t in the integrals involved there, yield (2.10) and (2.11). We now turn to the integrals (1.1) with |r| < 1, which, in view of the singularity at r or −r, will be computed in the Cauchy principal value sense.