For any real α define f_{α}(x) = [αx]. Let n be a positive integer. Show that there exists an α such that for 1 ≤ k ≤ n, f_{α}^{k}(n^{2}) = n^{2} - k = f_{αk}(n^{2}), where f_{α}^{k} denotes the k-fold composition of f_{α}.

**Solution**

Answer: α = e^{-1/n2}.

*Fairly easy.*

We require α(n^{2} - k) < n^{2} - k, so α < 1. We also require α(n^{2} - k) ≥ n^{2} - (k + 1), so α ≥ 1 - 1/(n^{2} - k). This suggests trying a value near 1 - 1/n^{2}. However, we also need [α^{k}n^{2}] = n^{2} - k, so we would like an α such that α^{k} is easy to work with. This suggests trying e^{-1/n2}. Certainly 1 > e^{-1/n2} > 1 - 1/n^{2}, so it follows immediately that [α(n^{2} - k)] = n^{2} - (k+1) and hence that f_{α}^{k}(n^{2}) = n^{2} - 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 + x^{2}/2. Hence, in particular: 1 - k/n^{2} < α^{k} < 1 - k/n^{2} + k^{2}/(2n^{4}). Hence α^{k} lies between n^{2} - k and n^{2} - k + k^{2}/(2n^{2}), so [α^{k}n^{2}] = n^{2} - k, as required.

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998