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arXiv:2106.14585v2 [math.CA] 29 Jun 2021 Factoring Variants of Chebyshev Polynomials with Minimal Polynomials of cos(2π d ) D.A. Wolfram College of Engineering & Computer Science The Australian National University, Canberra, ACT 0200 David.Wolfram@anu.edu.au Abstract We solve the problem of factoring polynomials Vn(x)±1 and Wn(x)±1 where Vn(x) and Wn(x) are Chebyshev polynomials of the third and fourth kinds. The method of proof is based on previous work by Wolfram [12] for factoring variants of Chebyshev polynomials of the first and second kinds, Tn(x) ± 1 and Un(x) ± 1. We also show that, in general, there are no factorizations of variants of Chebyshev polynomials of the fifth and sixth kinds, Xn(x) ± 1 and Yn(x) ± 1 using minimal polynomials of cos( 2π d ). 1 Background. Chebyshev polynomials of the third and fourth kinds were named by Gautschi [7] and are also called airfoil polynomials [6, 11]. They are used in areas such as solving differential equations [1], numerical integration [4, 6, 11], approxima- tions [11], interpolation [7] and combinatorics [5]. The significance of their applications in mathematics, engineering and numerical modeling provides a motivation for studying the properties of these polynomials. In previous work, Wolfram [12] solved an open factorization problem for Chebyshev polynomials of the second kind Un(x) ± 1, and gave a more direct proof of the result for Chebyshev polynomials of the first kind, Tn(x) ± 1 . We apply this method to solve the analogous factorization problems for Chebyshev polynomials of the third and fourth kinds. All of these factorizations can be expressed in terms of the minimal polynomials of cos( 2π d ). MSC: Primary 12E10, Secondary 12D05 1
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1.1 Chebyshev Polynomials of the Second Kind Chebyshev polynomials of the second kind can be defined by Un(x) = sin((n + 1)θ) sin(θ) (1) where x = cos θ and n ≥0. This is equation (1.4) of Mason and Handscomb [11]. It follows that Un(x)2 −1 = Un−1(x)Un+1(x) (2) where n ≥1, by applying the trigonometric identity sin2 A −sin2 B = sin(A + B) sin(A −B) with A = (n + 1)θ and B = θ. These polynomials satisfy the following recurrence, for example equations (1.6a)–(1.6b) of [11]: U0(x) = 1, U1(x) = 2x, Un(x) = 2xUn−1(x) −Un−2(x) (3) where n > 1. G¨urta¸s [8] showed that Un−1(x) = Y d|2n d>2 Ψd(x) (4) where n ≥1. The polynomials Ψd(x) are Ψd(x) = Y k∈Sd/2 2 x −cos 2π k d (5) where Sd/2 = {k | (k, d) = 1, 1 ≤k < d/2} and d > 2. They have degree φ(d)/2 where φ is Euler’s totient function [8]. The minimal polynomial in Q[x] of cos( 2π d ) is 2−φ(d) 2 Ψd(x) where d > 2 which follows from the proof of Theorem 1 of D. H. Lehmer [9]. Definition 1. The polynomial Ψ1(x) = 2(x −1) and Ψ2(x) = 2(x + 1). These are polynomials with roots cos(2π) and of cos(π), respectively. 1.2 Chebyshev Polynomials of the Third and Fourth Kinds Chebyshev polynomials of the third kind can be defined by Vn(x) = cos(n + 1 2)θ cos θ 2 (6) and of the fourth kind by Wn(x) = sin(n + 1 2)θ sin θ 2 (7) 2
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where x = cos θ and n ≥0. They can also be defined with respect to Chebyshev polynomials of the second kind, Un(x): Vn(x) = Un(x) −Un−1(x) (8) and Wn(x) = Un(x) + Un−1(x) (9) where n ≥1. These are equations (1.17)–(1.18) of Mason and Handscomb [11]. 2 Solution The method of solution follows that by Wolfram [12].The first step is to express Vn(x)2 −1 and Wn(x)2 −1 in terms of the minimal polynomials Ψd(x) where d ≥1. Lemma 1. Where n ≥1, Vn(x)2 −1 = Ψ1(x) Y d|2n d>2 Ψd(x) Y d|2n+2 d>2 Ψd(x) (10) Wn(x)2 −1 = Ψ2(x) Y d|2n d>2 Ψd(x) Y d|2n+2 d>2 Ψd(x). (11) Proof. From equation (8), we have Vn(x)2 −1 =(Un(x) −Un−1(x) + 1)(Un(x) −Un−1(x) −1) from equation (8) =Un(x)2 −1 −2Un(x)Un−1(x) + Un−1(x)2 =Un+1(x)Un−1(x) −2Un(x)Un−1(x) + Un−1(x)2 from equation (2) =Un−1(x)(Un+1(x) −2Un(x) + Un−1(x)) =Un−1(x)(2xUn(x) −2Un(x)) from equation (3) =Ψ1(x)Un−1(x)Un(x) from Definition 1 =Ψ1(x) Y d|2n d>2 Ψd(x) Y d|2n+2 d>2 Ψd(x) from equation (4). 3
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Similarly, we have Wn(x)2 −1 =(Un(x) + Un−1(x) + 1)(Un(x) + Un−1(x) −1) from equation (9) =Un(x)2 −1 + 2Un(x)Un−1(x) + Un−1(x)2 =Un+1(x)Un−1(x) + 2Un(x)Un−1(x) + Un−1(x)2 from equation (2) =Un−1(x)(Un+1(x) + 2Un(x) + Un−1(x)) =Un−1(x)(2xUn(x) + 2Un(x)) from equation (3) =Ψ2(x)Un−1(x)Un(x) from Definition 1 =Ψ2(x) Y d|2n d>2 Ψd(x) Y d|2n+2 d>2 Ψd(x) from equation (4) as required. The following theorem solves the factorization problem for Vn(x)2 −1. The second step of the method concerns defining the mapping that splits the 2n factors of Vn(x)2 −1 into the n factors of Vn(x) + 1 and the other n factors of Vn(x) −1. The factorizations are unique up to associativity and commutativity of multiplication. Theorem 1. If n ≥1, Vn(x) + 1 = Y d|2n d>2 2n/d odd Ψd(x) Y d|2n+2 d>2 (2n+2)/d odd Ψd(x) (12) and Vn(x) −1 = Ψ1(x) Y d|2n d>2 2n/d even Ψd(x) Y d|2n+2 d>2 (2n+2)/d even Ψd(x). (13) Proof. The polynomial Ψ1(x) = 2(x −1) is a factor of Vn(x)2 −1 from equa- tion (10), and Ψ1(cos(2π)) = 0. It follows from equation (6) that Vn(cos(2π)) = 1 and so Ψ1(x) is a factor of Vn(x) −1. If d | 2n and d > 2, let θ = 2πk d where (k, d) = 1, 1 ≤k < d 2 and a = 2n d . We have θ = πak n and Ψd(cos(θ)) = 0. From equation (6), Vn(cos(θ)) =cos((n + 1 2) πak n ) cos( θ 2) =cos(πak) cos( θ 2) −sin(πak) sin( θ 2) cos( θ 2) . The denominator cos( θ 2) ̸= 0 because θ 2 = πk d cannot equal π 2 when d > 2. The numbers ak and a have the same parity. This is immediate when a is even. If 4
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a is odd, it follows that d is even and k is odd because (k, d) = 1. We have cos(πak) = cos(πa), and Vn(cos(θ)) = cos(πa). Hence, if a is even, then Vn(cos(θ)) = 1 and Ψd(x) is a factor of Vn(x) −1. Similarly, if a is odd, then Vn(cos(θ)) = −1 and Ψd(x) is a factor of Vn(x) + 1. If d | 2n + 2 and d > 2, let b = 2n+2 d . We have θ = πbk n+1 where k is such that (k, d) = 1 and 1 ≤k < d 2. From equation (6), Vn(cos(θ)) =cos((n + 1 2)θ) cos( θ 2) =cos((n + 1)θ −θ 2) cos( θ 2) =cos(πbk) cos( θ 2) + sin(πbk) sin( θ 2) cos( θ 2) . Similarly to the previous case the denominator cos( θ 2) ̸= 0, the numbers bk and b have the same parity, and Vn(cos(θ)) = cos(πb). Ift follows that if b is odd then Ψd(x) is a factor of Vn(x) + 1 and if b is even then Ψd(x) is a factor of Vn(x) −1. From equations (3) and (8), Vn(x) has degree n. It follows that the right side of equation (10) of the factorization of Vn(x)2 −1 has degree 2n. It has 2n factors of the form 2(x −cos(θ)) from equation (5) and Definition 1, half of which are the factors of Vn(x)+1 and the other half are the factors of Vn(x)−1. The mapping defined above maps every such factor of Vn(x)2 −1 to either Vn(x)+ 1 or Vn(x)−1 depending on whether cos(θ) is either a root of Vn(x)+ 1 or Vn(x) −1, respectively. The right sides of equations (12) and (13) are the products of these mapped factors and so both have degree equal to n. From equations (3) and (8) and, the leading coefficients of Vn(x) ± 1 are 2n. The expansions of the factorizations on the right sides of equations (12) and (13) both have 2n as leading coefficients also, because each is a product of n factors of the form 2(x −cos(θ)), as required. 3 Examples with V The polynomial V12(x)2 −1 has 24 factors, and V12(x) + 1 and V12(x) −1 each are the products of half of these factors. The mapping in the proof of Theorem 1 gives V12(x) + 1 =Ψ8(x)Ψ24(x)Ψ26(x) V12(x) −1 =Ψ1(x)Ψ3(x)Ψ4(x)Ψ6(x)Ψ12(x)Ψ13(x) =(2(x −1))(2x + 1)(2x)(2x −1)(4x2 −3) (64x6 + 32x5 −80x4 −32x3 + 24x2 + 6x −1) 5
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The following theorem solves the factorization problem for Wn(x)2−1. These factorizations are also unique up to associativity and commutativity of multi- plication. Theorem 2. If n ≥1, Wn(x) + 1 = Y d|2n d>1 2n/d odd Ψd(x) Y d|2n+2 d>2 (2n+2)/d even Ψd(x) (14) and Wn(x) −1 = Y d|2n d>1 2n/d even Ψd(x) Y d|2n+2 d>2 (2n+2)/d odd Ψd(x). (15) Proof. The structure of the proof is similar to that of Theorem 1. If d | 2n and d > 1, let a = 2n d and k be such that (k, d) = 1 where 1 ≤k < d 2. From equation (7), Wn(cos(θ)) =sin((n + 1 2) πak n ) sin( θ 2) =cos(πak) sin( θ 2) + sin(πak) cos( θ 2) sin( θ 2) . The denominator sin( θ 2) ̸= 0 because θ 2 = πk d cannot equal π when d > 1. Similarly, we have that ak and a have the same parity, and Wn(cos(θ)) = cos(πa). Hence, if a is even, then Wn(cos(θ)) = 1 and Ψd(x) is a factor of Wn(x)−1. If a is odd, then Wn(cos(θ)) = −1 and Ψd(x) is a factor of Wn(x)+1. If d | 2n + 2 and d > 2, let b = 2n+2 d and k be such that (k, d) = 1 where 1 ≤k < d 2. We have θ = πbk n+1 and θ 2 = πk d . From equation (7), Wn(cos(θ)) =sin((n + 1 2)θ) sin( θ 2) =sin((n + 1)θ −θ 2) sin( θ 2) =−cos(πbk) sin( θ 2) + sin(πbk) cos( θ 2) sin( θ 2) . The denominator sin( θ 2) ̸= 0, and b and bk have the same parity, as above. Hence, if b is even, then Wn(cos(θ)) = −1 and Ψd(x) is a factor of Wn(x) + 1. If b is odd, then Wn(cos(θ)) = 1 and Ψd(x) is a factor of Wn(x) −1. It is straightforward to show that the degrees of the right sides of equa- tions (14) and (15) are both n, and the leading coefficients of both sides of these equations is 2n. 6
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4 Examples with W The polynomial W12(x)2 −1 has 24 factors, and W12(x)+1 and W12(x)−1 each are the products of half of these factors. The mapping in the proof of Theorem 2 gives W12(x) + 1 =Ψ8(x)Ψ24(x)Ψ13(x) W12(x) −1 =Ψ2(x)Ψ3(x)Ψ4(x)Ψ6(x)Ψ12(x)Ψ26(x) =(2(x + 1))(2x + 1)(2x)(2x −1)(4x2 −3) (64x6 −32x5 −80x4 + 32x3 + 24x2 −6x −1) When n is odd, Ψ2(x) is a factor of Wn(x) + 1: W11(x) + 1 =Ψ2(x)Ψ22(x)Ψ3(x)Ψ4(x)Ψ6(x)Ψ12(x). 5 Chebyshev Polynomials of the Fifth and Sixth Kinds Chebyshev polynomials of the fifth kind, Xn(x), and sixth kind, Yn(x), were de- fined by Masjed-Jamei [10]. Similarly to the other four kinds of Chebyshev poly- nomials, they are orthogonal polynomials with integer coefficients and Xn(x) and Yn(x) have degree n where n ≥0 [2, 3]. They have the form ⌊n 2 ⌋ X v=0 avxn−2v. Monic Chebyshev polynomials of the fifth and sixth kinds, ¯Xn(x) and ¯Yn(x), can be defined by the following recurrences which we simplify from [2]. G0,m(x) =1 G1,m(x) =x Gn,m(x) =xGn−1,m(x) + An−1,m Gn−2,m(x), n > 1 (16) where An,m =(2n + m −2)(−1)n + (2n −(m −2)) −nm −n2 (2n + m −1)(2n + m −3) , and (17) ¯Xn(x) =Gn,3(x) (18) ¯Yn(x) =Gn,5(x). (19) 7
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The first six Chebyshev polynomials of the fifth kind over Z are X0(x) =1 X1(x) =x X2(x) =4x2 −3 X3(x) =6x3 −5x X4(x) =16x4 −20x2 + 5 X5(x) =80x5 −112x3 + 35x X6(x) =64x6 −112x4 + 56x2 −7. The first six Chebyshev polynomials of the sixth kind over Z are Y0(x) =1 Y1(x) =x Y2(x) =2x2 −1 Y3(x) =8x3 −5x Y4(x) =16x4 −16x2 + 3 Y5(x) =24x5 −28x3 + 7x Y6(x) =16x6 −24x4 + 10x2 −1. The polynomials X5(x) ± 1 and Y5(x) ± 1 are irreducible over Z and none is a minimal polynomial of cos( 2π d ), of which there are only two of degree 5: Ψ11(x) =32x5 + 16x4 −32x3 −12x2 + 6x + 1 Ψ22(x) =32x5 −16x4 −32x3 + 12x2 + 6x −1. These counter examples for Xn(x) ± 1 and Yn(x) ± 1, show that, in general, they do not have factorizations using the minimal polynomials of cos( 2π d ). 6 Conclusion We have solved the problem of factoring variants of Chebyshev polynomials of the third and fourth kinds, Vn(x) ± 1 and Wn(x) ± 1, in terms of minimal polynomials for cos( 2π d ). This was done by applying the method of Wolfram [12] for factoring Tn(x) ± 1 and Un(x) ± 1 in a similar way. We have shown that there are no generalizations of this factorization to similar variants of Chebyshev polynomials of the fifth and sixth kinds, Xn(x) ± 1 and Yn(x) ± 1. Acknowledgment I am grateful to the College of Engineering & Computer Science at The Aus- tralian National University for research support. 8
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