1995
DOI: 10.4064/am-23-3-319-323
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Recurrence relations with periodic coefficients and Chebyshev polynomials

Abstract: Abstract. We show that polynomials defined by recurrence relations with periodic coefficients may be represented with the help of Chebyshev polynomials of the second kind.Introduction. The class of orthogonal polynomials studied in this paper served as a starting point for several authors [1-4] when studying more general classes of orthogonal polynomials or continued fractions. The aim of this note is to show that, on the other hand, this class can be described with the help of the classical Chebyshev polynomi… Show more

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Cited by 10 publications
(19 citation statements)
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“…For this purpose we employ the results of Ref. [7]. Let p n (x) be a sequence of polynomials set by the recursion p n (x) = (x + b n−1 )p n−1 (x) − a n−1 p n−2 (x) (A2) supplied by the boundary conditions p −1 = 0 and p 0 = 1, and the periodic coefficients a n+ω = a n and b n+ω = b n .…”
Section: Appendix A: Three-term Periodic Recursionmentioning
confidence: 99%
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“…For this purpose we employ the results of Ref. [7]. Let p n (x) be a sequence of polynomials set by the recursion p n (x) = (x + b n−1 )p n−1 (x) − a n−1 p n−2 (x) (A2) supplied by the boundary conditions p −1 = 0 and p 0 = 1, and the periodic coefficients a n+ω = a n and b n+ω = b n .…”
Section: Appendix A: Three-term Periodic Recursionmentioning
confidence: 99%
“…The 8 phases (for N = even) are labelled by (σ 1 ,σ 2 ,σ 3 ) where In Sec. 2 we present finite SSH model for demonstrating the technique [7] of solving periodic three-term recurrence relation analytically in terms of Chebyshev polynomials. In Sec.…”
Section: Introductionmentioning
confidence: 99%
“…Let us remind some necessary results from [14]. We denote by {ϕ n (x)} ∞ n=0 the polynomial sequence defined by recurrence relations ϕ n (x) = (x + a n−1 )ϕ n−1 (x) − b n−1 ϕ n−2 (x), n ≥ 1, ϕ 0 (x) = 1, ϕ −1 (x) = 0,…”
Section: Introductionmentioning
confidence: 99%
“…It was proved in [14] that for any N ≥ 1 the polynomial ϕ N −1 (x) divides the polynomial ϕ 2N −1 (x) ϕ 2N −1 (x) = ϕ N −1 (t)P N (x),…”
Section: Introductionmentioning
confidence: 99%
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