2021 Fall AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 17:36, 24 November 2021

Problem

Set $u_0 = \frac{1}{4}$ , and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence $u_{k+1} = 2u_k - 2u_k^2.$

This sequence tends to a limit; call it $L$ . What is the least value of $k$ such that $|u_k-L| \le \frac{1}{2^{1000}}?$

$(\textbf{A})\: 10\qquad(\textbf{B}) \: 87\qquad(\textbf{C}) \: 123\qquad(\textbf{D}) \: 329\qquad(\textbf{E}) \: 401$

Solution

If we list out the first few values of k, we get the series $\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}$ , which seem to always be a negative power of 2 away from $\frac{1}{2}$ . We can test this out by setting $u_k$ to $\frac{1}{2}-\frac{1}{2^{n_k}}$ .

Now,

\begin{align*}
u_{k+1} &= 2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_k}\right)-2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_k}}\right)^2 \\
&= 1-\frac{1}{2^{n_k - 1}}-2\cdot\left(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}\right) \\
&= 1-\frac{1}{2^{n_k - 1}}-\frac{1}{2}+\frac{1}{2^{n_k-1}}-\frac{1}{2^{2 \cdot n_k-1}} \\
&= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}} \\
\end{align*} (Error compiling LaTeX. Unknown error_msg)

This means that $n_{k+1}=2 \cdot n_k-1$ .