ON SHANKS’ ALGORITHM FOR COMPUTING THE CONTINUED FRACTION OF logb a TERENCE JACKSON AND KEITH MATTHEWS
نویسنده
چکیده
We give a more practical variant of Shanks’ 1954 algorithm for computing the continued fraction of logb a, for integers a > b > 1, using the floor and ceiling functions and an integer parameter c > 1. The variant, when repeated for a few values of c = 10, enables one to guess if logb a is rational and to find approximately r partial quotients. 1. Shanks’ algorithm In his article [1], Shanks gave an algorithm for computing the partial quotients of logb a, where a > b are positive integers greater than 1. Construct two sequences a0 = a, a1 = b, a2, . . . and n0, n1, n2, . . ., where the ai are positive rationals and the ni are positive integers, by the following rule: If i ≥ 1 and ai−1 > ai > 1, then a ni−1 i ≤ ai−1 < a ni−1+1 i (1.1) ai+1 = ai−1/a ni−1 i . (1.2) Clearly (1.1) and (1.2) imply ai > ai+1 ≥ 1. Also (1.1) implies ai ≤ a 1/ni−1 i−1 for i ≥ 1 and hence by induction on i ≥ 0, ai+1 ≤ a 1/n0···ni 0 . (1.3) Also by induction on j ≥ 0, we get a2j = a r 0/a s 1, a2j+1 = a u 1/a v 0, (1.4) where r and u are positive integers and s and v are non–negative integers. Two possibilities arise: 1 2 TERENCE JACKSON AND KEITH MATTHEWS (i) ar+1 = 1 for some r ≥ 1. Then equations (1.4) imply a relation a q 0 = a p 1 for positive integers p and q and so loga1 a0 = p/q. (ii) ai+1 > 1 for all i. In this case the decreasing sequence {ai} tends to a ≥ 1. Also (1.3) implies a = 1, unless perhaps ni = 1 for all sufficiently large i; but then (1.2) becomes ai+1 = ai−1/ai and hence a = a/a = 1. If ai−1 > ai > 1, then from (1.1) we have
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