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DEGREE N RELATIVES of the GOLDEN RATIO and RESULTANTS of the CORRESPONDING POLYNOMIALS

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DEGREE N RELATIVES of the GOLDEN RATIO and RESULTANTS of the CORRESPONDING POLYNOMIALS DEGREE n RELATIVES OF THE GOLDEN RATIO AND RESULTANTS OF THE CORRESPONDING POLYNOMIALS DAVID E. RUSH Abstract. Computations are given of the resultants Res(sm,sn) of pairs of Selmer polyno- n mials sn = sn(X) = X − X − 1. It is shown that for each fixed m ∈ N the sequence of k integers Res(sm,sm+k) = Res(sm(X), X − 1) satisfies a simple linear recursion which can be described in terms of higher Lucas sequences. 1. Introduction The object of this article is to give some computations of the resultants Res(sm,sn) of pairs n of Selmer polynomials sn = sn(X)= X X 1. In Section 2 the definition and properties of − − k resultants are reviewed and it is shown that Res(sm,sm k) = Res(sm(X), X 1). In Section + − 3, the Spectral Mapping Theorem is used to give formulas for Res(s2,s2+k) in terms of the Fibonacci sequence and the Lucas sequence. This shows in particular that Res(s2,s2+k) = k Res(s2(X), X 1), is a linear recursion sequence of degree 4 whose values are easily described in terms of the− Lucas sequence. In Section 4, we give formulas for the values of the simple k linear recurrence sequence Res(s3,s3+k) = Res(s3(X), X 1), k 1, which has order 6, in terms of the order 3 analogue of the Lucas sequence (namely− the Perrin≥ sequence). In Section 5, it is shown that if D is an integral domain and f D[X] is monic and separable of degree n, ∈ then there is an associated Lucas sequence Lf having f as its characteristic polynomial, and that Res(f(X), Xk 1), k 1, is a simple linear recurrence sequence, whose order can be 2n, − ≥ but its values are given in terms of the Lucas sequence Lf of order n. Finally, in Section 6, it 2 is noted that X X 1 is related to sn(X) by changing the number of mean proportionals b between a and a−+ b−from one to n 1. Some of the third order recursion sequences− having characteristic polynomial X3 X 1 have received quite a bit of attention in recent years. See for example [1, 5, 19, 21, 27,− 28,− 30, 31] and the references listed there. The polynomials Xn X 1, n 4 or some of their associated recursion sequences have also received some attention− rece−ntly [9,≥ 19, 20, 24, 26]. We were led n to consider the family of polynomials sn(X)= X X 1 n 2 and their discriminants in [24], by questions which arose in [23].{ To answer these− − motivating| ≥ } questions in [23] requires computing resultants of pairs of Selmer polynomials. Some results on resultants involving cyclotomic polynomials are given in [3, 4, 8] and [15]. However, the only results we are aware of where the resultants Res(fm,fn) have been computed for all pairs fm, fn in families fn(X) n 1 not involving cyclotomic polynomials are in [10]. { | ≥ } 2. The Resultants of Pairs of Selmer Polynomials Let R be an integral domain and let f, g R[X]. If ∈ NOVEMBER 2012 313 THE FIBONACCI QUARTERLY m m m−1 f = a X + a X + + am = a (X αi) and (2.1) 0 1 · · · 0 − Yi=1 n n n−1 g = b X + b X + + bn = b (X βj), (2.2) 0 1 · · · 0 − jY=1 then the resultant Res(f, g) of f and g is often defined as the determinant of the (n + m) by (n + m) Sylvester matrix a a am 0 0 0 1 · ··· · · · · 0 a a am 0 0 0 1 ··· · · · · 0 0 a0 a1 am 0 0 ·· · · · · . . 0 0 a0 a1 am 0 · · · · · · · 0 0 a0 a1 am · · ·· · · · · . b0 b1 bn−1 bn 0 0 · ··· · · · 0 b0 b1 bn−1 bn 0 0 ··· · · · 0 0 b0 b1 bn−1 bn 0 0 . .··. ·. · . .· . . . 0 0 b b bn− bn 0 · · · 0 1 · · · 1 0 0 0 b b bn− bn · · · 0 1 · · · 1 We also have the following formulas [18, pp. 135-138]: m n n m Res(f, g)= a b (αi βj) (2.3) 0 0 − Yi=1 jY=1 m n = a0 g(αi) (2.4) Yi=1 n mn m = ( 1) b f(βj). (2.5) − 0 jY=1 Thus, if f is monic, we have m Res(f, g)= g(αi). (2.6) Yi=1 It follows that Res(f, g) R, Res(f, g)= fh1 + gh2 for some hi R[X] and that Res(f, g) = 0 if and only if f and g have∈ a common root in some field containing∈ R. The following lemma connects the computation of the resultants Res(sm,sm+k) to Mahler measure which has been much studied since it was defined in [19]. For some recent work on Mahler measure see for example [2, 7, 11, 14, 29], and the references listed there. k Lemma 2.1. Res(sm,sm k) = Res(sm, X 1) for m 2 and k 0. + − ≥ ≥ 314 VOLUME 50, NUMBER 4 DEGREE n RELATIVES OF THE GOLDEN RATIO Proof. From Equation (2.6), if α1,...,αm are the roots of sm, we get m m m m+k k m Res(sm,sm k)= sm k(αj)= (α αj 1) = (α α αj 1), + + j − − j j − − jY=1 jY=1 jY=1 and since αj is a root of sm, this is m m k k k (α (αj + 1) αj 1) = (α 1)(αj + 1) = Res(sm, (X 1)(X + 1)). j − − j − − jY=1 jY=1 But from the property Res(f,gh) = Res(f, g) Res(f, h), which is immediate from formula k (2.4), this is Res(sm, X 1)Res(sm, X + 1), and applying formula (2.5) to the second factor, k − m k 2m k we get Res(sm, X 1))( 1) sm( 1) = Res(sm, X 1))( 1) = Res(sm, X 1). − − − − − − We obtain the following result on the behavior of Res(sm,sm+1) which is sufficient to answer a motivating question which arises from [23]. Proposition 2.2. For any integer m 2, we have m+1 ≥ (a) Res(sm,sm ) = ( 1) , and +1 − (b) limk→∞ Res(sm,sm k) = . | + | ∞ m Proof. (a) From Lemma 2.1 and formula (2.5), Res(sm,sm+1) = Res(sm, X 1) = ( 1) sm(1) = ( 1)m+1. − − − (b) If α1,...,αm are the roots of sm, we get from Lemma 2.1 that Res(sm,sm+k) = k m k | | Res(sm, X 1) = j=1 αj 1 , and for one index j, say j = 1, we have αj > 1 [26]. Then | k − | | − | k α 1 goes to infinityQ with k. For the j with αj < 1, the factor α 1 goes to 1 as n | 1 − | | | | j − | goes to infinity. Thus it suffices to show that no αj has modulus 1. Suppose some αj = α has modulus 1. Then 1 = αm = α + 1 . But the only points α on the unit circle such that α + 1 | | | ±| 2πi is also on the unit circle are α = e 3 . But since sm is irreducible over Q [26, Theorem 1], ± 2πi if some αj has modulus 1, this would imply that sm is the minimal polynomial of α = e 3 , ± 2πi which is clearly not the case. (Alternately, one could just show directly that α = e 3 are not roots of sm.) 3. Formulas for Res(s2,sn) in Terms of Fibonacci and Lucas Sequences For small values of m and n it is easy to compute Res(sm,sn) directly from the definition in terms of the Sylvester matrix. However, this matrix increases with both m and n. So to compute Res(s2,sn) for all n it is useful to convert equation (2.6) into a form that involves only two-by-two matrices of integers. For this we can use the following result which is often called the Spectral Mapping Theorem. See [12] for example. If R is a commutative ring and M is a matrix with entries in R, we denote the characteristic polynomial of M by PM (X). Theorem 3.1. [18, Theorem 14, p. 404] If R is a commutative ring, g R[X], M is a m m∈ matrix with entries in R and PM (X)= i=1(X αi), then Pg(M)(X)= i=1(X g(αi)). In m − − particular, det(g(M)) = i=1 g(αi). Q Q Q NOVEMBER 2012 315 THE FIBONACCI QUARTERLY m To apply this to equation (2.6) requires choosing a matrix M with f = PM (X)= i=1(X m m−1 − αi). For f = X + am−1X + + a0 R[X], we take the companion matrix ofQf to be · · · ∈ 0 100 . 0 0 0 0 010 . 0 0 0 0 001 0 . 0 0 . . C(f)= . . 0 0 0 . 01 0 0 0 0 0 . 00 1 0 0 0 0 . 00 0 1 a a . am− am− − 0 − 1 − 2 − 1 Some other matrices that are often called “the companion matrix of f” are the transpose C(f)T of C(f), and the matrices obtained by reversing the rows and columns of C(f) or C(f)T . An easy induction, expanding on the first column, shows that the characteristic polynomial of C(f) is f. Thus by Theorem 3.1 and equation (2.6), we have the following companion matrix form of the resultant. (See [6, Theorem 1.3, p. 15], [12, Theorem 5.1].) m Res(f, g)= g(αi) = det(g(C(f)).
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