27. [15] Let $\left\{a_{i}\right\}_{i \geq 0}$ 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 $\left\{a_{i}\right\}_{i \geq 0}$ is bounded.
Answer: $\quad 1+\frac{1}{2^{2020}}$
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
$$
a_{0} \leq \frac{1}{\sqrt{2}^{2020}}\left(\sqrt{2}^{2020}+\left(\frac{1}{\sqrt{2}}\right)^{2020}\right)=1+\frac{1}{2^{2020}} .
$$