Part 1: The Perrin Conjugate and the Laguerre Orthogonal Polynomial 3 2 3 2 I defined the conjugate of a cubic polynomial G(x) = x - Bx – Cx - D as G(x)c = x + Bx – Cx + D. By multiplying the polynomial with its’ conjugate one obtains the first polynomial of order 2; [1] P2(x,2) = x6 + (B2-2C) x4 + (C2-2BD) x2 - D2 3 3 The Perrin sequence is associated with solutions to G(x) = x –x – 1 = 0. Its’ Perrin conjugate, G(x)c = x –x + 1 multiplied by G(x) yields a degree six polynomial, [2] P2{x,2) = x6 -2 x4 + x2 – 1 Successive multiplication of P2(x,2) with its conjugates gives the recursive formula, [3] P2(x,n)* P2(x,n)c = P2(x,2n). In Part 1, I investigate another sequence which is related to the Perrin conjugate. Many well- known functions can be expanded as summations in powers of x. As an example, the exponential function ex is expanded as, 푥2 푥3 푥4 푥5 푥6 푥7 푥8 푥9 푥10 푥11 [4] F(x) = ex = 1 + 푥 + + + + + + + + + + + ⋯ .. 2 6 24 120 720 5040 40320 362880 3628800 39916800 or in terms of a summation, 푥푛 [5] F(x) = ex = ∑∞ 푛=0 푛! In this case the denominator of the expansion is a series of factorials in n. Many other exponential functions can be expressed as an expansion of factorials. For example, 푥3 푥5 푥7 푥9 푥11 푥13 [6] F(x) = Sinh[x] = 푥 + + + + + + + ⋯ .. 6 120 5040 362880 39916800 6227020800 For some functions, the denominator of the expansion does not need to be a common factorial. An x 3 example I will discuss further is the expansion of e *G(x)c, specifically when G(x)c = x –x + 1. 2 3 4 5 6 7 8 9 10 11 x 푥 2푥 7푥 7푥 23푥 17푥 47푥 31푥 79푥 7푥 [7] F(x) = e *G(x)c = 1 − + + + + + + + + + + ⋯ 2 3 8 15 144 420 5760 22680 403200 285120 In this case the denominator is not an obvious factorial and the numerator is an unknown series of coefficients. If these sequences are searched on OEIS then the numerator sequence 2,7,7,23,17,47.. is close to sequence A164314 described as the largest prime factor of (n-1)2-2. For example for n = 10, (10- 1)2-2 = 79 which is a prime and the largest factor of itself. Unfortunately, as the number of terms increases the numerator is not the largest prime factor for all terms (e.g. at n = 29, (29-1)2-2 = 782 = 23*17*2) but the factor is indicated as 1. The reason for these discrepancies will be discussed. A search of the denominator sequence 2,3,8,15,144,420,5760… on OEIS is described as the value of the x denominator of the Laguerre polynomial of degree n at x = 1. The expansion of e *G(x)c is then related indirectly to the Laguerre polynomial. The Laguerre Polynomial The Laguerre polynomial is one of several classical orthogonal polynomials found in mathematics. A standard source that I will be using is found in reference (1). The closed form definition of the generalized Laguerre polynomial of degree n is 푛 + 훼 푥푖 [8] 퐿훼 (푥) = ∑푛 (−1)푖 ∗ ( ) ∗ 푛 푖=0 푛 − 푖 푖! 푛 + 훼 Where ( ) is the binomial coefficient equal to (n+)! /((n-i)! (+i)!). 푛 − 푖 0 The Laguerre polynomial for =0 is defined as 퐿푛(푥) = 퐿푛(푥). The exponential generating function for 퐿푛(1) is 푥2 2푥3 5푥4 7푥5 37푥6 17푥7 887푥8 1405푥9 168919푥10 [9] ex/x-1/(1-x) = 1 − − − − − − + + + + ⋯ .. 2 3 8 15 144 420 5760 4536 403200 Comparing equation [9] with [7] notice that some denominators are different but are integer multiples. Although this disproves that the numerator of [7] is the largest prime factor of (n-1)2-2 it strengths the hypothesis that the denominator is a denominator of 퐿푛(1). The polynomial for a given n with =0 can be calculated from [8]. A cubic polynomial at n=3 is 1 2 [10] 퐿 (푥) = (6 − 18푥 + 9푥2 − 푥3) with 퐿 (1) = − 3 6 3 3 rd agreeing with the 3 coefficient term in [9]. Note that the coefficient for 퐿1(1) = 0 An exponential generating function can be used to find the expansion coefficients for the generalized Laguerre polynomial. 푐푥 훼−1 − 훼 [11] 퐿푛 (푐) = 푒 1−푥/(1 − 푥) Where c is an integer and is any rational number ≥0 or ≤0. 훼−1 Equation [7] can also be generalized to give the magnitude of the denominator of 퐿푛 (푐) 훼−1 푐푥 3 훼 [12] 퐷푒푛표푚푖푛푎푡표푟[퐿푛 (푐)] = 푓 ∗ 푒 ∗ (1 − 푥 − 푥 ) where f is a positive integer generally equal to 1 and ≥0 or ≤0. As with most orthogonal functions, any monomial term can be expressed as a sum of Laguerre polynomials. By adding these monomial terms together and multiplying by the associated coefficient any polynomial can be expressed as a sum of Laguerre polynomials. 3 1 As an example, 푥 = −6퐿3(푥) + 18퐿2(푥) − 18퐿1(푥) + 6퐿0(푥) , 푥 = −퐿1(푥) + 퐿0(푥) and 1 = 퐿0(푥) and the summation gives for the Perrin conjugate: 3 [13] 푥 − 푥 + 1 = −6퐿3(푥) + 18퐿2(푥) − 17퐿1(푥) + 6퐿0(푥) The value of the function on the LHS can then be verified for x=1 from associated Laguerre coefficients found in equation [9] or for any x = c from equation [11] The orthogonality of the Laguerre polynomial can be useful for integrating polynomials formed from the summation of Laguerre polynomials. For =0 Laguerre polynomials satisfy the condition, ∞ −푥 [14] ∫0 퐿푛(푥) ∗ 푒 ∗ 퐿푚(푥)푑푥 = 훿푚푛 where 훿푚푛 = 1 if n=m and 0 otherwise. −푥 Multiplication of both sides of [13] by 퐿푛(푥) ∗ 푒 and integrating in the limit 0 to infinity results in a simplification of the integration of the LHS due to orthogonality on the RHS. In general, ∞ −푥 ( ) [15] ∫0 퐿푛(푥) ∗ 푒 ∗ 퐺 푥 푑푥 = 푎푛 Where 푎푛 is the coefficient of 퐿푛(푥) for expansion of G(x) with Laguerre polynomials. The expansion coefficients 푎푛 for each monomial can conveniently be found either in tables such as in reference (1) [Table 22.10], or by application of a Groebner basis. In Mathematica, expand the series in unknown coefficients of decreasing n using the Laguerre command and then use the Groebner basis command to solve for the coefficients. An example for obtaining x3 is, [16] In = Expand[푎 ∗ LaguerreL[3, 푥] + 푏 ∗ LaguerreL[2, 푥] + 푐 ∗ LaguerreL[1, 푥] + 푑 ∗ LaguerreL[0, 푥]] 3푎푥2 푏푥2 푎푥3 Out = 푎 + 푏 + 푐 + 푑 − 3푎푥 − 2푏푥 − 푐푥 + + − 2 2 6 3푎 푏 In = GroebnerBasis[{ + , −3푎 − 2푏 − 푐, 푎 + 푏 + 푐 + 푑, 푎 + 6}, {푎, 푏, 푐, 푑}] 2 2 Out = {−6 + 푑, 18 + 푐, −18 + 푏, 6 + 푎} If the expansion is made with the generalized Laguerre polynomial then the integration becomes ∞ 훼 −푥 훼 ( ) [17] ∫0 퐿푛(푥) ∗ 푒 ∗ 푥 ∗ 퐺 푥 푑푥 = 푎푛 ∗ (n + + 1)/n! and the infinite range of integration on the LHS can be avoided by using the Gamma function of the RHS. Expansion of Classic Orthogonal Polynomials with Generalized Laguerre Polynomials Although a method such as [16] can be used for polynomials G(x) of any degree, as the degree increases the number of individual unknown coefficients to be solved increases as (n+1) (n+2)/2. For a polynomial of the 6th degree, 28 coefficients need to be found. The problem posed in this section is: Can an orthogonal polynomial be expanded in terms of the generalized Laguerre polynomials? The answer is yes for the following polynomials; Legendre, Hermite, and Chebyshev. We seek a solution such that given an orthogonal polynomial Xm(x), there exists an expansion; 푋 (푥) = ∑푚 퐴 ∗ 퐿훼푖 (0) ∗ 푥푖 [18] 푚 푖=0 푚,푖 푚푖 퐴 퐿훼푖 (0) where 푚,푖 is a coefficient dependent on m and i, and 푚푖 is the generalized Laguerre polynomial evaluated at x = 0 with coefficients mi and also dependent on i. I will use the notation of reference (1) where 푃푚(푥) is the Legendre polynomial, 퐻푚(푥) the Hermite polynomial, 푇푚(푥) the Chebyshev T polynomial, 푈푚(푥) the Chebyshev U polynomial, 푆푚(푥) the Chebyshev S polynomial, and 퐶푚(푥) the Chebyshev C polynomial. 푖 Using the orthogonality property, once these expressions are found then the nth term 푋푚푖푥 of 푃푚(푥), 퐻푚(푥), 푇푚(푥), 푈푚(푥), 푆푚(푥) or 퐶푚(푥) can be found by integration with the appropriate generalized Laguerre polynomial as in [19] and the Gamma function. ∞ 훼(푚,푖) −푧 훼(푚,푖) 훼(푚,푖) 푖 ∫ 퐿 (푧)∗푒 ∗푥 ∗푐(푚,푖)∗퐿 (푧)∗푥 푑푧 (훼(푚,푛)+푏(푚,푛)+1)∗푐(푚,푛)∗푥푖 [19] 푋 푥푖 = 0 푏(푚,푖) 푏(푚,푖) = 푚푖 (m,n)! 푏(푚,푛)!∗훼(푚,푛)! where (m,i), b(m,i) and c(m,i) are constant coefficients dependent on the degree m and the desired ith term of 푋푚(푥). In general, these integrals converge rapidly and integration to infinity is not required if performed numerically. 6 6 As an example, compute the coefficient for x (I = 6) in the degree 12 Chebyshev S polynomial, 푆12,6푥 . In equation [19] the 6th power of x is x (m-2n), n = 3 and the constants are (12,6) = n = 3, b (12,6) = m-2n = 6 and c (12,6) = (−1)푛 = -1. ∞ ∫ 퐿3(푧)∗푒−푧∗푥3∗(−1)∗퐿3(푧)∗푥6푑푧 (6+3+1)∗(−1)∗푥6 [20] 푆 푥6 = 0 6 6 = = −84푥6 12,6 3! 6!∗3! 6 The result agrees with Table 22.8 in reference (1) for matrix element A[S12(x), x ].