Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 7.9 National Olympiad, round 2 Prove it Romania

Let $a$ be a real number in the open interval $(0, 1)$ , let $n$ be a positive integer and let $f_n: \mathbb{R} \to \mathbb{R}$ , $f_n(x) = x + x^2/n$ . Show that

$\frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2} < \underbrace{(f_n \circ \dots \circ f_n)}_{n} (a) < \frac{an + a^2}{(1-a)n + a}.$

Solution

Let $a_k = \underbrace{(f_n \circ \cdots \circ f_n)}_{k}(a)$ , $k \in \mathbb{N}$ , and notice that
$1/a_{k+1} = 1/a_k - 1/(a_k + n), \quad k \in \mathbb{N},$
to deduce that $1/a_n = 1/a - \sum_{k=0}^{n-1} 1/(a_k + n)$ , so
$1/a - n/(a+n) < 1/a_n < 1/a - n/(a_n + n), \quad (*)$
$a_n < \frac{an + a^2}{(1-a)n + a}$ .
Plugged into the rightmost expression in $(*)$ , this upper bound yields the required lower bound,
$a_n > \frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2}.$

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