Article

Polynomial interpolation and sums of powers of integers

Taylor & Francis
International Journal of Mathematical Education In Science & Technology
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Abstract

In this note, we revisit the problem of polynomial interpolation and explicitly construct two polynomials in n of degree k + 1, Pk(n) and Qk(n), such that Pk(n) = Qk(n) = fk(n) for n = 1, 2,… , k, where fk(1), fk(2),… , fk(k) are k arbitrarily chosen (real or complex) values. Then, we focus on the case that fk(n) is given by the sum of powers of the first n positive integers Sk(n) = 1k + 2k + ⋅⋅⋅ + nk, and show that Sk(n) admits the polynomial representations Sk(n) = Pk(n) and Sk(n) = Qk(n) for all n = 1, 2,… , and k ≥ 1, where the first representation involves the Eulerian numbers, and the second one the Stirling numbers of the second kind. Finally, we consider yet another polynomial formula for Sk(n) alternative to the well-known formula of Bernoulli.

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... Example 1. As a simple example, we may use formula (2) to calculate the coefficients c p,ℓ for p = 5. As is readily verified, for this case the list of m-tuples satisfying the above conditions is: (0, 0, 0, 0) for ℓ = 0; (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) for ℓ = 1; (2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 0, 1, 0), (0, 1, 0, 1), and (1, 0, 0, 1) for ℓ = 2; (3, 0), (0, 3), (1, 0, 2), and (2, 0, 1) for ℓ = 3; and 4 for ℓ = 4. Hence, from (2) we get c 5,0 = 120, c 5,1 = 240, c 5,2 = 150, c 5,3 = 30, and c 5,4 = 1. ...
... Proposition 2. For p ≥ 1 and ℓ = 0, 1, . . . , p − 1, let c p,ℓ be the coefficient defined in (2). Then, ML conjecture ⇔ c p,ℓ = (p − ℓ)!S(p, p − ℓ). ...
... In Section 3, we prove that the ML conjecture is true. Specifically, by using the representation of the Stirling numbers of the second kind given in equation (12), we show that the coefficients defined in (2) are indeed identical to c p,ℓ = (p − ℓ)!S(p, p − ℓ) for all p ≥ 1 and ℓ = 0, 1, . . . , p − 1. ...
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