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СЕМИНАРЫ |
Еженедельный исследовательский семинар по Математике в Университете АДА
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Allemands Edward Barbeau University of Toronto |
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Аннотация: A certain type problem in competitions presents the candidates with a recursion such as x_(n+1) = ((x_n)^2 + 1)/x_(n−1) with x_0 = x_1 = 1 and requires them to show that each entry is an integer. The trick is to find a linear recurrence that it satisfies, in this case, x_(n+1_ = 3x_n − x_(n−1). We now have both the sum and the product of x_(n+1) and x_(n−1), so we can recognize them as the solutions of the quadratic equation 0 = h(x, xn) = x^2 − 3(x_n)x + (x_n − 1) = (x^2 + (x_n)^2) − 3(x_n)x − 1. This suggests a method of defining a sequence that starts with a symmetric function h(x, y) that is quadratic in each variable and quadratic in each of them, along with a “seed” x_0, such that for each integer n, the entries x_(n+1) and x_(n−1) satisfy the quadratic equation h(x, xn) = 0. Thus 0 = h(x_n, x_(n+1)) = h(x_(n+1), x_n). Such sequences are called allemands. We look at some particular cases. While we have a few preliminary results, it is not clear that this is more than a novelty, and we leave as an open question as to whether a significant body of results can be laid on this foundation. The final frontier of more general h(x, y) has yet to be explored. Язык доклада: английский |