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

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This sequence tends to a limit; call it <math>L</math>. What is the least value of <math>k</math> such that <cmath>|u_k-L| \le \frac{1}{2^{1000}}?</cmath>
 
This sequence tends to a limit; call it <math>L</math>. What is the least value of <math>k</math> such that <cmath>|u_k-L| \le \frac{1}{2^{1000}}?</cmath>
  
−
<math>(\textbf{A})\: 10\qquad(\textbf{B}) \: 87\qquad(\textbf{C}) \: 123\qquad(\textbf{D}) \: 329\qquad(\textbf{E}) \: 401</math>
+
<math>\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401</math>
  
−
==Solution==
+
==Solution 1==
  
−
If we list out the first few values of k, we get the series <math>\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}</math>, which seem to always be a negative power of 2 away from <math>\frac{1}{2}</math>. We can test this out by setting <math>u_k</math> to <math>\frac{1}{2}-\frac{1}{2^{n_k}}</math>.  
+
Note that terms of the sequence <math>(u_k)</math> lie in the interval <math>\left(0,\frac12\right),</math> strictly increasing.
  
−
Now,  
+
Since the sequence <math>(u_k)</math> tends to the limit <math>L,</math> we set <math>u_{k+1}=u_k=L>0.</math>
 +
 
 +
The given equation becomes <cmath>L=2L-2L^2,</cmath> from which <math>L=\frac12.</math>
 +
 
 +
The given inequality becomes <cmath>\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},</cmath> and we only need to consider <math>\frac12-\frac{1}{2^{1000}} \leq u_k.</math>
 +
 
 +
We have
 +
<cmath>\begin{alignat*}{8}
 +
u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\
 +
u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\
 +
u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\
 +
u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\
 +
& \phantom{1111} \vdots
 +
\end{alignat*}</cmath>
 +
By induction, it can be proven that <cmath>u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac12-\frac{1}{2^{2^k+1}}.</cmath>
 +
We substitute this into the inequality, then solve for <math>k:</math>
 
<cmath>\begin{align*}
 
<cmath>\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 \\
+
\frac12-\frac{1}{2^{1000}} &\leq \frac12-\frac{1}{2^{2^k+1}} \\
−
&= 1-\frac{1}{2^{n_k - 1}}-2\cdot(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}) \\
+
-\frac{1}{2^{1000}} &\leq -\frac{1}{2^{2^k+1}} \\
 +
2^{1000} &\leq 2^{2^k+1} \\
 +
1000 &\leq 2^k+1.
 +
\end{align*}</cmath>
 +
Therefore, the least such value of <math>k</math> is <math>\boxed{\textbf{(A)}\: 10}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 2==
 +
 
 +
If we list out the first few values of <math>k</math>, we get the series <math>\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}</math>, which always seems to be a negative power of <math>2</math> away from <math>\frac{1}{2}</math>. We can test this out by setting <math>u_k=\frac{1}{2}-\frac{1}{2^{n_k}}</math>, where <math>n_0=2</math>.
 +
 
 +
Now, we get
 +
<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 \\
 +
&= 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}} \\
 
&= 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 this series approaches <math>\frac{1}{2}</math>, as the second term is decreasing. In addition, we find that <math>n_{k+1}=2 \cdot n_k-1</math>.
 +
 
 +
We claim that <math>n_k = 2^k+1</math>, which can be proven by induction:
 +
 
 +
<u><b>Base Case</b></u>
 +
 
 +
We have <math>n_0=2=2^0+1</math>.
 +
 
 +
<u><b>Induction Step</b></u>
 +
 
 +
Assuming that the claim is true, we have <math>n_{k+1}=2 \cdot (2^k+1)-1=2^{k+1}+1</math>.
 +
 
 +
It follows that <math>n_{10}=2^{10}+1>1000</math> and <math>n_9=2^9+1<1000</math>. Therefore, the least value of <math>k</math> would be <math>\boxed{\textbf{(A)}\: 10}</math>.
 +
 
 +
~ConcaveTriangle
 +
 
 +
==Solution 3==
 +
 
 +
We are given <math>u_{k+1} = 2u_k - 2{u_k}^2</math>. Multiply this equation by <math>2</math> and subtract <math>1</math> from both sides. The equations can then be written nicely as <math>2u_{k+1} - 1 = -(2u_k-1)^2</math>. Let <math>v_k = 2u_k - 1</math> so that <math>v_{k+1} = -(v_k)^2</math>.
 +
 
 +
Clearly, <math>v_0 = 2u_0 - 1 = -\frac{1}{2}</math>. Since the magnitude of <math>v_0</math> is less than <math>1</math> and because our recursive relation for <math>v_k</math> squares the previous term (and negates it), we see that as <math>k \rightarrow \infty, 2u_k - 1 = v_k \rightarrow 0</math>. This means <math>u_k \rightarrow \frac{1}{2}</math>, so <math>L = \frac{1}{2}</math>.
  
−
This means that <math>n_{k+1}=2 \cdot n_k-1</math>.  
+
Isolating <math>u_k</math> in our relation <math>2u_k - 1 = v_k</math> gives us <math>u_k = \frac{v_k + 1}{2}</math>.
 +
Substituting into the inequality, we have <math>\left|\frac{v_k + 1}{2}-\frac{1}{2}\right| \le \frac{1}{2^{1000}}</math>. Rewriting this, we get <math>|v_k| \le \frac{1}{2^{999}}</math>.
  
−
We see that <math>n_k</math> seems to always be <math>1</math> above a power of <math>2</math>. We can prove this using induction.  
+
The sequence <math>\{v_k\}</math> is much easier to handle because of its simple recursive relation. Writing out a few terms shows that <math>|v_k| = \frac{1}{2^{2^k}}</math>. Now it just comes down to having <math>2^k > 999</math>, so <math>k = \boxed{\textbf{(A)}\: 10}</math>.
  
−
Claim: <math>n_k = 2^k+1</math>
+
~chezpotato
  
−
Base case: <math>n_0=2^0+1</math>
+
==Video Solution==
 +
https://youtu.be/ZfHql_vhNes
  
−
Induction: <math>n_{k+1}=2*2^k+2-1=2^{k+1}+1</math>
+
~MathProblemSolvingSkills.com
  
−
It follows that <math>n_{10{=2^{10}+1>1000</math>, and <math>n_9=2^9+1<1000</math>. Therefore, the least value of <math>k</math> would be <math>\boxed{\textbf{(A) }10}</math>.
+
==See Also==
 +
{{AMC12 box|year=2021 Fall|ab=B|num-a=19|num-b=17}}
 +
{{MAA Notice}}

Latest revision as of 09:36, 5 November 2022

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 1

Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right),$ strictly increasing.

Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$

The given equation becomes \[L=2L-2L^2,\] from which $L=\frac12.$

The given inequality becomes \[\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},\] and we only need to consider $\frac12-\frac{1}{2^{1000}} \leq u_k.$

We have \begin{alignat*}{8} u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \phantom{1111} \vdots \end{alignat*} By induction, it can be proven that \[u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac12-\frac{1}{2^{2^k+1}}.\] We substitute this into the inequality, then solve for $k:$ \begin{align*} \frac12-\frac{1}{2^{1000}} &\leq \frac12-\frac{1}{2^{2^k+1}} \\ -\frac{1}{2^{1000}} &\leq -\frac{1}{2^{2^k+1}} \\ 2^{1000} &\leq 2^{2^k+1} \\ 1000 &\leq 2^k+1. \end{align*} Therefore, the least such value of $k$ is $\boxed{\textbf{(A)}\: 10}.$

~MRENTHUSIASM

Solution 2

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 always seems to be a negative power of $2$ away from $\frac{1}{2}$. We can test this out by setting $u_k=\frac{1}{2}-\frac{1}{2^{n_k}}$, where $n_0=2$.

Now, we get \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*} This means that this series approaches $\frac{1}{2}$, as the second term is decreasing. In addition, we find that $n_{k+1}=2 \cdot n_k-1$.

We claim that $n_k = 2^k+1$, which can be proven by induction:

Base Case

We have $n_0=2=2^0+1$.

Induction Step

Assuming that the claim is true, we have $n_{k+1}=2 \cdot (2^k+1)-1=2^{k+1}+1$.

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

Solution 3

We are given $u_{k+1} = 2u_k - 2{u_k}^2$. Multiply this equation by $2$ and subtract $1$ from both sides. The equations can then be written nicely as $2u_{k+1} - 1 = -(2u_k-1)^2$. Let $v_k = 2u_k - 1$ so that $v_{k+1} = -(v_k)^2$.

Clearly, $v_0 = 2u_0 - 1 = -\frac{1}{2}$. Since the magnitude of $v_0$ is less than $1$ and because our recursive relation for $v_k$ squares the previous term (and negates it), we see that as $k \rightarrow \infty, 2u_k - 1 = v_k \rightarrow 0$. This means $u_k \rightarrow \frac{1}{2}$, so $L = \frac{1}{2}$.

Isolating $u_k$ in our relation $2u_k - 1 = v_k$ gives us $u_k = \frac{v_k + 1}{2}$. Substituting into the inequality, we have $\left|\frac{v_k + 1}{2}-\frac{1}{2}\right| \le \frac{1}{2^{1000}}$. Rewriting this, we get $|v_k| \le \frac{1}{2^{999}}$.

The sequence $\{v_k\}$ is much easier to handle because of its simple recursive relation. Writing out a few terms shows that $|v_k| = \frac{1}{2^{2^k}}$. Now it just comes down to having $2^k > 999$, so $k = \boxed{\textbf{(A)}\: 10}$.

~chezpotato

Video Solution

https://youtu.be/ZfHql_vhNes

~MathProblemSolvingSkills.com

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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