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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 18"
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==Solution==
 
==Solution==
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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>.
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Now,
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<cmath>\begin{align*}
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u_{k+1} &= 2\cdot(\frac{1}{2}-\frac{1}{2^{n_k}})-2\cdot(\frac{1}{2}-\frac{1}{2^{n_k}})^2 \\
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&= 1-\frac{1}{2^{n_k - 1}}-2\cdot(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}) \\
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&= 1-\frac{1}{2^{n_k - 1}}-\frac{1}{2}+\frac{1}{2^{n_k-1}}-\frac{1}{2^{2 \cdot n_k-1}} \\
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&= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}} \\
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\end{align*}
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</cmath>
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This means that <math>n_{k+1}=2 \cdot n_k-1</math>.
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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.
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Claim: <math>n_k = 2^k+1</math>
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Base case: <math>n_0=2^0+1</math>
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Induction: <math>n_{k+1}=2*2^k+2-1=2^{k+1}+1</math>
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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>.

Revision as of 17:19, 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*}

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: $n_{k+1}=2*2^k+2-1=2^{k+1}+1$

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

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