Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 13: Line 13:
 
Now,  
 
Now,  
 
<cmath>\begin{align*}
 
<cmath>\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 \\
+
u_{k+1} &= 2\cdot(\frac{1}{2}-\frac{1}{2^{n_k})-2\cdot(\frac{1}{2}-\frac{1}{2^{n_k}})^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}}-2\cdot(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}) \\
 
&= 1-\frac{1}{2^{n_k - 1}}-\frac{1}{2}+\frac{1}{2^{n_k-1}}-\frac{1}{2^{2 \cdot n_k-1}} \\
 
&= 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}} \\
 
&= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}} \\
−
\end{align*}
+
\end{align*}</cmath>
−
</cmath>
 
  
 
This means that <math>n_{k+1}=2 \cdot n_k-1</math>.  
 
This means that <math>n_{k+1}=2 \cdot n_k-1</math>.  

Revision as of 17:38, 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(\frac{1}{2}-\frac{1}{2^{n_k})-2\cdot(\frac{1}{2}-\frac{1}{2^{n_k}})^2 \\
&= 1-\frac{1}{2^{n_k - 1}}-2\cdot(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}) \\
&= 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 $n_{k+1}=2 \cdot n_k-1$.

We see that $n_k$ seems to always be $1$ above a power of $2$. We can prove this using induction.

Claim: $n_k = 2^k+1$

Base case: $n_0=2^0+1$

Induction: $n_{k+1}=2 \cdot 2^k+2-1=2^{k+1}+1$

It follows that $n_{10}=2^{10}+1>1000$, and $n_9=2^9+1<1000$. Therefore, the least value of $k$ would be $\boxed{\textbf{(A) }10}$.

-ConcaveTriangle

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_18&oldid=166233"