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Chebyshev Polynomials SCIENTIA Series A: Mathematical Sciences, Vol. 27 (2016), 31{45 Universidad T´ecnicaFederico Santa Mar´ıa Valpara´ıso,Chile ISSN 0716-8446 c Universidad T´ecnicaFederico Santa Mar´ıa2016 The integrals in Gradshteyn and Ryzhik. Part 29: Chebyshev polynomials Victor H. Moll and Christophe Vignat Abstract. The table of Gradshteyn and Ryzhik contains many integrals that involve Chebyshev polynomials. Some examples are discussed. 1. Introduction The Chebyshev polynomial of the first kind Tn(x) is defined by the relation (1.1) cos nθ = Tn(cos θ): The elementary recurrence (1.2) cos(n + 1)θ = 2 cos θ cos nθ − cos(n − 1)θ yields the three-term recurrence for orthogonal polynomials (1.3) Tn+1(x) = 2xTn(x) − Tn−1(x) and, with initial conditions T0(x) = 1 and T1(x) = x, shows that Tn(x) is indeed a polynomial in x. The polynomial Tn(x) is of degree n and its leading coefficient is 2n−1. These elementary facts follow directly from (1.3). The Chebyshev polynomial of the second kind Un(x) is defined by the relation sin(n + 1)θ (1.4) = U (cos θ): sin θ n This polynomial satisfies the recurrence (1.5) Un+1(x) = 2xUn(x) − Un−1(x); (the same recurrence as (1.3)), this time with initial conditions U0(x) = 1 and U1(x) = 2x. Some basic properties of Chebyshev polynomials are collected next. The first result gives the classical generating function for these polynomials. 2000 Mathematics Subject Classification. Primary 33. Key words and phrases. Integrals, Chebyshev polynomials. 31 32 V. MOLL AND C. VIGNAT Proposition 1.1. The generating function for the Chebyshev polynomials is given by 1 X 1 − xt (1.6) T (x)tn = n 1 − 2xt + t2 n=0 and 1 X 1 (1.7) U (x)tn = : n 1 − 2xt + t2 n=0 n Proof. Multiply the recurrence (1.3) by t and sum over n > 1. Binet's formula for Chebyshev polynomials follows directly from their generating functions (1.6) and (1.7). Corollary 1.2. The Chebyshev polynomial Tn(x) is given by 1 h p p i (1.8) T (x) = (x + x2 − 1)n + (x − x2 − 1)n : n 2 Similarly, the polynomial Un(x) is given by h i 1 p 2 n+1 p 2 n+1 (1.9) Un(x) = p (x + x − 1) − (x − x − 1) : 2 x2 − 1 Proof. Expand the right-hand side of (1.6) and (1.7) in partial fractions and expand the resulting terms. A useful expression for the Chebyshev polynomials is their Rodrigues formulas n (−1)n2nn!p d (1.10) T (x) = 1 − x2 (1 − x2)n−1=2 n (2n)! dx and n n n (−1) (n + 1)!2 1 d 2 n+1=2 (1.11) Un(x) = p (1 − x ) : (2n + 1)! 1 − x2 dx These will be used in some simplifications in the rest of the paper. 2. Some elementary examples The classical table of integrals [2] contains a small collection of integrals with Tn(x) or Un(x) in the integrand. The goal of this note is to provide self-contained proofs of these entries. The most elementary entry is 7:343:1 that is equivalent to the orthogonality of the family fcos nθg on the interval [0; 2π]. Indeed, define Z 1 f(x)g(x) (2.1) hf; gi = p dx; 2 −1 1 − x then the first example simply gives hTn;Tmi = 0 if n 6= m. CHEBYSHEV POLYNOMIALS 33 Example 2.1. Entry 7:343:1: 8 0 if m 6= n; Z 1 T (x)T (x) <> (2.2) np m = π if m = n 6= 0; 1 − x2 2 −1 :>π if m = n = 0: The next couple of examples computes integrals involving powers of Chebyshev polynomials. Example 2.2. The evaluation Z 1 (−1)n−1 − 1 (2.3) Tn(x) dx = ; for n > 2; −1 (n − 1)(n + 1) is not included in [2]. To confirm this formula, let x = cos θ and use the identity 1 (2.4) cos nθ sin θ = [sin(n + 1)θ − sin(n − 1)θ] 2 to produce Z 1 1 Z π (2.5) Tn(x) dx = [sin(n + 1)θ − sin(n − 1)θ] dθ: −1 2 0 The result follows by computing the elementary trigonometric integrals. The indefinite version of this entry appears in [4] as entry 1:14:2:1 in the form Z 1 T (x) T (x) (2.6) T (x) dx = n+1 − n−1 : n 2 n + 1 n − 1 To verify this evaluation, let x = cos θ and observe that Z Z (2.7) Tn(x) dx = − cos(nθ) sin θ dθ: The result now follows from (2.4). Example 2.3. Entry 7:341 states that Z 1 2 2 1 2(2n − 1) (2.8) Tn (x) dx = 1 − 2 = : −1 4n − 1 (2n − 1)(2n + 1) The evaluation starts with (1.8) to obtain Z 1 Z 1 Z 1 2 1 p 2 2n 1 p 2 2n (2.9) Tn (x) dx = (x + x − 1) dx + (x − x − 1) dx + 1: −1 4 −1 4 −1 The change of variables x = cos θ gives Z 1 p Z π (2.10) (x + x2 − 1)2n dx = e2inθ sin θ dθ: −1 0 1 iθ −iθ The last integral is evaluated by writing sin θ = 2i e − e . The second integral is evaluated in the same manner and the stated formula is obtained from here. A generalization of this result is given in Section 8. 34 V. MOLL AND C. VIGNAT Example 2.4. Entry 7:341:2 is Z 1 1 1 (2.11) Tm(x)Tn(x) dx = 2 + 2 if m + n is even −1 1 − (m − n) 1 − (m + n) and Z 1 (2.12) Tm(x)Tn(x) dx = 0 if m + n is odd: −1 The proof is based on the identity 1 (2.13) T (x)T (x) = [T (x) + T (x)] n m 2 n−m n+m coming from its trigonometric counterpart 1 (2.14) cos nθ cos mθ = [cos(n + m)θ + cos(n − m)θ] : 2 The result now follows from (2.6). Example 2.5. The integral Z n−3 1 n−1 (2.15) (1 − x2) 2 T (x) dx = − (1 − x2) 2 T (x) n n − 1 n−1 appears as entry 1:14:2:3 in [5]. It does not appear in [2]. The proof is elementary: the change of variables x = cos θ gives Z n−3 Z 2 n−2 (2.16) (1 − x ) 2 Tn(x) dx = − sin θ cos nθ dθ and the elementary identity 1 d (2.17) sinn−2 θ cos nθ = sinn−1 θ T (cos θ) : n − 1 dθ n−1 The companion entry 1:14:2:4 in [5] Z n+3 1 n+1 (2.18) (1 − x2)− 2 T (x) dx = (1 − x2)− 2 T (x) n n + 1 n+1 is established in a similar manner. 3. The evaluation of a Mellin transform The Mellin transform of a function f(x) is defined by Z 1 (3.1) M(f)(s) = xs−1f(x) dx: 0 p In examples concerning Chebyshev polynomials, with kernel 1= 1 − x2, it is natural to consider their restriction to [−1; 1]. Entry 7:346 states that Z 1 −1 s−1 dx π 1 + s + n 1 + s − n (3.2) x Tn(x)p = B ; ; for Re s > 0: 2 s 0 1 − x s2 2 2 CHEBYSHEV POLYNOMIALS 35 This entry gives the Mellin transform of the function ( p T (x)= 1 − x2 if 0 x 1 (3.3) f(x) = n 6 6 0 otherwise: The change of variables x = cos θ transforms (3.2) to Z π=2 −1 s−1 π 1 + s + n 1 + s − n (3.4) cos(nθ) cos θ dθ = s B ; : 0 s2 2 2 This entry will be established in a future publication. Only a special case is required here. Special Case. Assume s = m + 1 is a positive integer. Then (3.4) becomes Z π=2 −1 m π 2 + m + n 2 + m − n (3.5) cos(nθ) cos θ dθ = m+1 B ; : 0 (m + 1)2 2 2 The reduction of this integral requires a simple trigonometric formula. This appears as entry 1:320 in [2]. The proof of (3.5) is presented next. Lemma 3.1. For m 2 N, j m−1 k 2 ( 1 X m 0 if m ≡ 1 mod 2 (3.6) xm = T (x) + 2m−1 k m−2k 2−m m if m ≡ 0 mod 2: k=0 m=2 Proof. Let x = cos θ and start with (eiθ + e−iθ)m (3.7) cosm θ = 2m m 1 X m = ei(2k−m)θ: 2m k k=0 By symmetry, since this a real function, the real part yields m 1 X m (3.8) cosm θ = cos(m − 2k)θ: 2m k k=0 To obtain the stated formula, split the sum in half to obtain m b 2 c m 1 X m 1 X m (3.9) cosm θ = cos(m − 2k)θ + cos(m − 2k)θ: 2m k 2m k k=0 m k=b 2 c+1 In the case m odd, both sums have the same number of elements and the change of indices j = k − m=2 shows that they are equal. In the case m even, there is an extra term corresponding to the index m=2.
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