Second Paper on Chebyshev Polynomials in the Unipodal Algebra P. Reany February 2, 2020 Abstract This paper is a follow-up to the paper Chebyshev Polynomials, Fibonacci, and Lucas Numbers in the Unipodal Algebra. It contains more information that could have been in the first paper, but wasn't to keep that paper short. It also breaks new ground by generalizing Chebyshev polynomials, called by some the `bivariate Chebyshev polynomials', and then that generalization is used to find the solutions to the cubic equation. It's hard to avoid Chebyshev polynomials. They appear in just about every branch of mathematics, including geometry, combinatorics, number theory, differential equations, approximation theory, numerical analysis, and statistics. | A. Benjamin & D. Walton 1 Introduction In the last paper, we saw how the Chebyshev polynomials of both first and second kinds can be reduced to the complex and uniplex parts of the following equation1 Xn = b ; (1) with one simple necessary constraint: XX− = 1 ; (2) where X− is the unegation of the unipodal number X. By working out the results of this pair of equations, we are able to ferret out the two unipodal versions of the Chebyshev polynomials of first and second kinds: 1 n −n Tn(x) = 2 X + X ; (3a) u n+1 −(n+1) Un(x) = X − X ; (3b) 2X1 where X = X0 + X1u and X0 has been replaced by x. A useful alternative formulation for Tn(X0) is given by 1 n n Tn(x) = 2 [X+ + X−] ; (4) 1It's unlikely that the reader will be able to follow this paper without first having a good knowledge of the contents of the previous paper. 1 the proof of which is provided in the previous paper. A rather different proof can be found in Appendix 1. An obvious generalization of the Chebyshev polynomials, which is accomplished by the more general constraint placed on X than given in (2), is given as XX− = r ; (5) where r is any nonzero complex number. This leads to the so-called `bivariate' Chebyshev polyno- mials, which we'll encounter later on. Lemma 1 1 n p p X t 2 2 p T (x) = 1 e(x+ x −1)t + e(x− x −1)t = etx cosh (t x2 − 1) : (6) n n! 2 n=0 Proof: We begin with (4) and multiply through by tn: n 1 n n Tn(X0)t = 2 [(tX+) + (tX−) ] : (7) On dividing through by n! and summing from n = 0 to 1, we get 1 n 1 1 n 1 n X t X [tX+] + [tX−] T (X ) = 2 2 n 0 n! n! n=0 n=0 1 n 1 n X [tX+] X [tX−] = 1 + 1 2 n! 2 n! n=0 n=0 1 tX+ 1 tX− = 2 e + 2 e p p 2 2 q 1 t(X0+ X0 −1) 1 t(X0− X0 −1) tX0 2 = 2 e + 2 e = e cosh (t X0 − 1) : (8) Then, on replacing X0 by x in this last equation, we get (6). Lemma 2 The Chebyshev polynomials of the first kind in X0 can be written as b n c X2 n T (X ) = Xn−2`(X2 − 1)` : (9) n 0 2` 0 0 `=0;1;::: Before we begin, given that Tn(X0) is pure complex (has no unisor part), it's good to anticipate that the uniplex part of the expansion of (3a) is identically zero. Also, note that for any nonnegative integer k: ( 2 ; for k even; [1 + (−1)k]uk = (10) 0 ; for k odd: Note: We will be using the floor function b·c, which operates on real numbers and returns the largest integer less than or equal to its argument. 2 Proof: Now, we expand both Xn and X−n = (X−)n with the binomial formula. 1 n − n Tn(X0) = 2 [X + (X ) ] 1 n n = 2 [(X0 + X1u) + (X0 − X1u) ] n h X n i = 1 Xn−kXk[1 + (−1)k]uk 2 k 0 1 k=0 n X n = Xn−kXk k 0 1 k=0;2;::: b n c X2 n = Xn−2`X2` 2` 0 1 `=0;1;::: b n c X2 n = Xn−2`(X2 − 1)` : 2` 0 0 `=0;1;::: Corollary (n) Since Tn(x) = b0 = cosh nθ, then we have from (9) that b n c X2 n cosh nθ = (coshn−2` θ)(cosh2 θ − 1)` : (11) 2` `=0;1;::: On replacing θ by iθ in this last equation, we get b n c X2 n cos nθ = (cosn−2` θ)(cos2 θ − 1)` : (12) 2` `=0;1;::: Lemma 3 For the Chebyshev polynomials of the second kind in X0 alone, we have b n c X2 n U (X ) = Xn−2`+1(X2 − 1)`−1 : (13) n−1 0 2` − 1 0 0 `=0;1;::: Note that for any nonnegative integer k: ( 0 ; for k even; [1 − (−1)k]uk = (14) 2u ; for k odd: Note: We will be using the floor function b·c, which was explained in the last lemma. 3 Proof: As before, we expand both Xn and X−n with the binomial formula. u n − n Un−1(x) = [X − (X ) ] 2X1 u n n = [(X0 + X1u) − (X0 − X1u) ] 2X1 n u h X n n−k k k k i = X0 X1 [1 − (−1) u ] 2X1 k k=0 n X n = Xn−kXk−1 k 0 1 k=1;3;::: b n c X2 n = Xn−2`+1X2`−2 2` − 1 0 1 `=1;2::: b n c X2 n = Xn−2`+1(X2 − 1)`−1 : 2` − 1 0 0 `=1;2;::: 2 Introducing the unicoid of complexity r. For the rest of the lemmas in this paper, we will be using a generalization of the unipodes X available to us. So far, we have restricted our unipodes to the unicoid of complexity unity, by the constraint: XX− = 1 ; (15) which enforces the relation between X0 and X1: 2 2 X0 − X1 = 1 : (16) But now we must generalize: Definitions: The set of all unipodes X that satisfy the equation XX− = r, for r a nonzero complex number, is said to be the unicoid of complexity r, and such an X is called a unipode of complexity r. Also, every polynomial constructed from such an X and its unegate X− is called a polynomial of complexity r. So, now, our unipodes X will lie on the unicoid of complexity r, and X satisfies the constraint XX− = r where r 6= 0 ; (17) which enforces the relation between X0 and X1: 2 2 X0 − X1 = r : (18) We also retain the constraint on X with b by Xn = b : (19) Taken together, these last two equations imply that bb− = rn : (20) 4 For completeness, I need to comment on the relation between Equations (17) and (19). Given a nonzero complex number r, there are n distinct choices for the value of XX− that all produce − 2 the same value of bb . Let αn be any primitive nth root of unity in the complex numbers. Then n − αn = 1. Thus we could choose a different constraint on XX , namely, − j XX = αnr ; 0 ≤ j ≤ n − 1 ; (21) and produce the same value of bb−: − n − n − n j n n j n n bb = X (X ) = (XX ) = (αnr) = (αn) r = r : (22) Now, there are n2 roots to X in Eq. (19) over the unipodal numbers. For each value of n, Eq. (21) culls out n of those roots over the complex numbers. (More about this later.) Anyway, for the sake of simplicity, we will choose j = 0 in (21) to get (17).3 Note: Because of (17), unless r = 1, it's not true that X− = X−1. Instead, X−1 = r−1X− : (23) Lemma 4 Recurrence relation for Chebyshev polynomials of complexity r of the first kind. Given that X and b are consistent with the last few equations, the appropriate form for the recurrence relation for Chebyshev polynomials of complexity r of the first kind is Tn+1;r(X0) = 2X0Tn;r(X0) − rTn−1;r(X0) : (24) Note # 1: In the literature ([15], p.3), the previous relation is known as the `bivariate Chebyshev polynomials Tn(x; s) of the first kind', and defined by the recurrence relation Tn(x; s) = 2xTn−1(x; s) + sTn−2(x; s) ; (25) with T0(x; s) = 1 and T1(x; s) = x. Of course, (24) and (25) are essentially the same by letting r = −s. However, at the moment, I can't seem to bring myself to regard r in (24) as anything more than a parameter. n Note # 2: As we have done before, put simply, we define Tn;r(X0) as the complex part of X , and from that deduce the recurrence relation for Tn;r. Proof: As before, (n) Tn;r(X0) ≡ b0 ; where (26a) (n) 1 − b0 = b0 = 2 (b + b ) : (26b) n Once again, we see that Tn;r(X0) is the complex (or scalar) part of X . Thus, 1 n − n Tn;r(X0) = 2 [X + (X ) ] ; (27) 2A primitive nth root of unity is any complex number α such that the smallest positive integer power of α to be unity is n. 3 j Since we are free to choose the j in αn, we can think of this choice as discrete gauge freedom. 5 − which expresses Tn;r(X0) as a polynomial in X (and in X ). Again, a useful alternative is 1 n n Tn;r(x) = 2 [X+ + X−] ; (28) where q q 2 2 X+ = X0 − X0 + r and X− = X0 − X0 − r : (29) We have an easy corollary to the previous lemma, which requires the following fact.