HMMT February

Algebra Difficulty 5.3 AIME, harder Find the answer

Let $\{a_{i}\}_{i \geq 0}$ be a sequence of real numbers defined by $a_{n+1}=a_{n}^{2}-\frac{1}{2^{2020 \cdot 2^{n}-1}}$ for $n \geq 0$ . Determine the largest value for $a_{0}$ such that $\{a_{i}\}_{i \geq 0}$ is bounded.

Your answer

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Solution

Let $a_{0}=\frac{1}{\sqrt{2}^{2020}}\left(t+\frac{1}{t}\right)$ , with $t \geq 1$ . (If $a_{0}<\frac{1}{\sqrt{2}^{2018}}$ then no real $t$ exists, but we ignore these values because $a_{0}$ is smaller.) Then, we can prove by induction that $a_{n}=\frac{1}{\sqrt{2}^{2020 \cdot 2^{n}}}\left(t^{2^{n}}+\frac{1}{t^{2^{n}}}\right)$ . For this to be bounded, it is easy to see that we just need $\frac{t^{2^{n}}}{\sqrt{2}^{2020 \cdot 2^{n}}}=\left(\frac{t}{\sqrt{2}^{2020}}\right)^{2^{n}}$ to be bounded, since the second term approaches 0. We see that this is is equivalent to $t \leq 2^{2020 / 2}$ , which means $a_{0} \leq \frac{1}{\sqrt{2}^{2020}}\left(\sqrt{2}^{2020}+\left(\frac{1}{\sqrt{2}}\right)^{2020}\right)=1+\frac{1}{2^{2020}}$ .

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