For any real α define fα(x) = [αx]. Let n be a positive integer. Show that there exists an α such that for 1 ≤ k ≤ n, fαk(n2) = n2 - k = fαk(n2), where fαk denotes the k-fold composition of fα.
Answer: α = e-1/n2.
We require α(n2 - k) < n2 - k, so α < 1. We also require α(n2 - k) ≥ n2 - (k + 1), so α ≥ 1 - 1/(n2 - k). This suggests trying a value near 1 - 1/n2. However, we also need [αkn2] = n2 - k, so we would like an α such that αk is easy to work with. This suggests trying e-1/n2. Certainly 1 > e-1/n2 > 1 - 1/n2, so it follows immediately that [α(n2 - k)] = n2 - (k+1) and hence that fαk(n2) = n2 - k for 1 ≤ k ≤ n.
For 0 < x < 1, the Taylor series for e-x has decreasing terms of alternating sign, so 1 - x < e-x < 1 - x + x2/2. Hence, in particular: 1 - k/n2 < αk < 1 - k/n2 + k2/(2n4). Hence αk lies between n2 - k and n2 - k + k2/(2n2), so [αkn2] = n2 - k, as required.
© John Scholes
12 Dec 1998