A Bound on the Distance from Approximation Vectors to the Plane

Transcription

1 A Bound on the Distance from Approximation Vectors to the Plane T. Cheslack-Postava, A. Diesl, M. Lepinski, A. Schuyler August 2, 999 Abstract In this paper, we will begin by reviewing triangle sequences. One interpretation of these sequences is as a sequence of approximation vectors approaching the plane x + αy + βz = 0. This paper establishes an upper bound on the distance from the n th approximation vector to the plane x + αy + βz = 0. Introduction This paper will start with an overview of triangle sequences as outlined in work by Garrity [2]. We begin with a geometrical interpretation of triangle sequences. We define an iteration T on the triangle: = (x, y) : x y > 0. This triangle is partitioned into an infinite set of disjoint subtriangles k = (x, y) : x ky 0 > x (k + )y, where k is any nonnegative integer. Define the map T : (x, 0) : 0 x by ( β T (α, β) = α, α kβ ), α where (α, β) k. The triangle sequence is recovered from this iteration by keeping track of the number of the triangle that the point is mapped into at each step. In other words, if T k (α, β) ak, then the point (α, β) will have the triangle sequence (a, a 2, a 3,... ). We will recursively define a sequence of vectors as follows: SetC 2 = (, 0, 0), C = (0,, 0), C 0 = (0, 0, ) and C n = C n 3 C n 2 a n C n

2 Let the components of C n be denoted by C n = (p n, q n, r n ). These vectors C n can be thought of as integer vectors approximating the plane x + αy + βz = 0. We thus refer to the C n vectors as approximation vectors. We define positive numbers d n in the following manner: d n = (, α, β) C n These numbers are an indication of close the approximation vectors are to the plane x+αy+βz = 0. (In fact, these numbers differ from the Euclidean distance to the plane by a constant factor). The rest of the paper concerns itself with bounding d n from above. Let X n = C n C n+ as defined in []. We shall denote the components of X n as (x n, y n, z n ). In the next section we establish the bound of: d n < x n+ 2 A Bound on d n Let N n be the matrix, x n y n z n x n x n y n y n z n z n x n 2 y n 2 z n 2 We set M n = (C n 2, C n, C n ). Recall from [2] that det(m n ) = C n 2 (C n C n ) =. Lemma M n = N n Proof: We know that x n y n z n p n 2 p n p n N n M n = x n x n y n y n z n z n q n 2 q n q n x n 2 y n 2 z n 2 r n 2 r n r n By performing the above multiplication, we get C n 2 X n C n X n C n X n N n M n = C n 2 X n C n 2 X n C n X n C n X n C n X n C n X n C n 2 X n 2 C n X n 2 C n X n 2 Since for all k, we know that X k = C k C k+, then C k X k = 0 C k X k = 0 C k X k+ = 2

3 Additionally, C n 2 X n = C n 2 (X n 3 + a n X n 2 + X n ) and C n 2 X n = C n 2 X n 3 + a n C n 2 X n 2 + C n 2 X n C n 2 X n = C n X n 2 = (C n 3 C n 2 a n C n ) X n 2 Therefore, C n X n 2 = C n 3 X n 2 C n 2 X n 2 + a n C n X n 2 C n X n 2 = Hence, N n = M n N n M n = Lemma 2 The following three equalities are true: Proof: We know that Therefore, since M n = N n, Therefore, = d n x n 2 + x n + d n x n d n x n α = d n y n 2 + y n + d n y n d n y n β = d n z n 2 + z n + d n z n d n z n (, α, β)m n = (, d n, d n ) (, α, β) = (, d n, d n )N n = d n x n 2 + x n + d n x n d n x n α = d n y n 2 + y n + d n y n d n y n β = d n z n 2 + z n + d n z n d n z n Lemma 3 α = y n + β n y n 2 + α n y n α n y n x n + β n x n 2 + α n x n α n x n β = z n + β n z n 2 + α n z n α n z n x n + β n x n 2 + α n x n α n x n 3

4 Proof: From Lemma 2, we know that = x n + d n x n 2 + d n x n d n x n By recalling that α n = d n and β n = d n, we can divide through by to get = x n + β n x n 2 + α n x n α n x n Or equivalently, = Also from Lemma 2, we have that x n + β n x n 2 + α n x n α n x n () α = y n + d n y n 2 + d n y n d n y n β = z n + d n z n 2 + d n z n d n z n Factoring out a from the right hand sides yields α = (y n + β n y n 2 + α n y n α n y n ) β = (z n + β n z n 2 + α n z n α n z n ) Substituting in for from Equation (), we get α = y n + β n y n 2 + α n y n α n y n x n + β n x n 2 + α n x n α n x n β = z n + β n z n 2 + α n z n α n z n x n + β n x n 2 + α n x n α n x n Theorem d n < x n+ Proof: From Lemma 2, we know that d n+ x n + d n x n + d n x n+ d n x n = Since d n+ > 0 and x n > 0, we can write the inequality: d n x n + d n x n+ d n x n < Since the d n terms are decreasing, we can substitute d n for d n and maintain the inequality: d n x n + d n x n+ d n x n < This implies that d n < x n+ 4

5 References [] T. Cheslack-Postava, A. Diesl, M. Lepinski, and A. Schuyler. Some results concerning uniqueness of triangle sequences [2] Thomas Garrity. On periodic sequences for algebraic numbers