the nearest distance in geometric sequence

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the nearest distance in geometric sequence

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Source: 2024 China High School Olympics A P1

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#1 h Sep 8, 2024, 6:09 AM

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A positive integer $r$ is given, find the largest real number $C$ such that there exists a geometric sequence $\{ a_n \}_{n\ge 1}$ with common ratio $r$ satisfying
$\| a_n \| \ge C$ for all positive integers $n$ . Here, $\| x \|$ denotes the distance from the real number $x$ to the nearest integer.

This post has been edited 1 time. Last edited by David-Vieta, Sep 8, 2024, 11:53 PM

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Quantum-Phantom

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Quantum-Phantom

#2 h Sep 8, 2024, 3:39 PM • 2 Y

Y by David-Vieta, ukj

@David-Vieta

If the sequence contains integral terms, then $C\le0$ , so now assume that it doesn't.

Note that $\|a_n\|\le C$ is equivalent to $C\le\{a_n\}\le1-C$ , so $C\le\tfrac12$ . ( $\{x\}=x-[x]$ )

If $r$ is odd, them $C=\tfrac12$ is feasible for $a_1=\tfrac12$ (the decimal part of all terms is $\tfrac12$ )

If $r$ is even, let $r=2t$ and $a_1=\tfrac t{2t+1}$ . Induction on $n$ shows that the decimal part of each term of the sequence alternates between $\tfrac t{2t+1}$ and $\tfrac{t+1}{2t+1}$ , so $C=\tfrac t{2t+1}$ is feasible. For the sake of contradiction, assume that $\tfrac t{2t+1}<C\le\tfrac12$ . Note that $\{a_2\}=\{2ta_1\}=\{2t\{a_1\}\}$ , then $\{a_1\}\ne\tfrac12$ (otherwise $a_2$ would be an integer).

$\bullet$ If $C\le\{a_1\}<\tfrac12$ , then $\{a_1\}>\tfrac t{2t+1}$ . So $\tfrac t{2t+1}\cdot2t<2t\{a_1\}<\tfrac12\cdot2t$ , implying that $[2t\{a_1\}]=t-1$ . Hence, $\{a_2\}=2t\{a_1\}-(t-1)$ , so $C+t-1\le2t\{a_1\}\le t-C$ . Thus, $\tfrac{t-C}{2t}\ge C$ , and solving gives $C\le\tfrac t{2t+1}$ .

$\bullet$ If $\tfrac12<\{a_1\}\le1-C$ , then imitating the previous case, we finally get $\tfrac{C+t}{2t}\le1-C$ , thereby $C\le\tfrac t{2t+1}$ .

Both cases yield a contradiction, so the assumption is false, and we have proven that $C\le\tfrac t{2t+1}=\tfrac r{2r+2}$ .

In conclusion, the maximum value of $C$ is $\tfrac12$ when $r$ is odd and $\tfrac r{2r+2}$ when $r$ is even.

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ukj

#3 h Sep 9, 2024, 1:02 PM

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Quantum-Phantom wrote:

@David-Vieta

If the sequence contains integral terms, then $C\le0$ , so now assume that it doesn't.

Note that $\|a_n\|\le C$ is equivalent to $C\le\{a_n\}\le1-C$ , so $C\le\tfrac12$ . ( $\{x\}=x-[x]$ )

If $r$ is odd, them $C=\tfrac12$ is feasible for $a_1=\tfrac12$ (the decimal part of all terms is $\tfrac12$ )

If $r$ is even, let $r=2t$ and $a_1=\tfrac t{2t+1}$ . Induction on $n$ shows that the decimal part of each term of the sequence alternates between $\tfrac t{2t+1}$ and $\tfrac{t+1}{2t+1}$ , so $C=\tfrac t{2t+1}$ is feasible. For the sake of contradiction, assume that $\tfrac t{2t+1}<C\le\tfrac12$ . Note that $\{a_2\}=\{2ta_1\}=\{2t\{a_1\}\}$ , then $\{a_1\}\ne\tfrac12$ (otherwise $a_2$ would be an integer).

$\bullet$ If $C\le\{a_1\}<\tfrac12$ , then $\{a_1\}>\tfrac t{2t+1}$ . So $\tfrac t{2t+1}\cdot2t<2t\{a_1\}<\tfrac12\cdot2t$ , implying that $[2t\{a_1\}]=t-1$ . Hence, $\{a_2\}=2t\{a_1\}-(t-1)$ , so $C+t-1\le2t\{a_1\}\le t-C$ . Thus, $\tfrac{t-C}{2t}\ge C$ , and solving gives $C\le\tfrac t{2t+1}$ .

$\bullet$ If $\tfrac12<\{a_1\}\le1-C$ , then imitating the previous case, we finally get $\tfrac{C+t}{2t}\le1-C$ , thereby $C\le\tfrac t{2t+1}$ .

Both cases yield a contradiction, so the assumption is false, and we have proven that $C\le\tfrac t{2t+1}=\tfrac r{2r+2}$ .

In conclusion, the maximum value of $C$ is $\tfrac12$ when $r$ is odd and $\tfrac r{2r+2}$ when $r$ is even.

Can you explain why you chose this a1? Are there other a1s that can also meet the conditions?

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Quantum-Phantom

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Quantum-Phantom

#4 h Sep 9, 2024, 1:56 PM

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Yes, other possibilities include $k+\tfrac t{2t+1}$ or $k+\tfrac{t+1}{2t+1}$ where $k$ is a natural number. ( $r=2t$ )

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#5 h Sep 10, 2024, 3:46 AM

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First note that if $r$ is odd then we can easly achieve $C=\frac{1}{2}$ (which is maximal by the definition of $\| \cdot \|$ ), with $a_1=\frac{1}{2}$ .
So now suppose that $r=2k$ , in this case we claim that $C=\frac{k}{2k+1}$ is the most we can achieve.
First to show we can actually reach it we let $a_1=\frac{k}{2k+1}$ , because then the integer part switches from $\frac{k}{2k+1}$ and $\frac{k+1}{2k+1}$ (when even it's $k+1$ on numerator and when odd it's $k$ on numerator).
Now to prove that it's indeed maximal suppose FTSOC that we can achieve some $\frac{k}{2k+1}<C \le \frac{1}{2}$ , then first notice that we have that $\{a_{n+1} \}=\{2k \cdot a_n \}=\{ 2k \cdot \{a_n \} \}$ so if we had that $C \le \{a_n \} \le \frac{1}{2}$ then we have that $\frac{1}{2} \ge \{ a_n \}>\frac{k}{2k+1}$ so $k \ge 2k \cdot \{ a_n \}>\frac{2k^2}{2k+1}$ notice if for some $i$ we had $\{a_i \}=\frac{1}{2}$ then $\{a_{i+1} \}=0$ contradicting that $C>\frac{k}{2k+1}$ , so we have $k>2k \cdot \{a_n \}>\frac{2k^2}{2k+1}$ and notice that $2k^2>(2k+1)(k-1)=2k^2-k-1$ so we get that $\lfloor 2k\cdot \{ a_n \} \rfloor=k-1$ which implies that $\{a_{n+1} \}=\{2k \cdot \{a_n \} \}=2k \cdot \{a_n \}-(k-1)>\frac{k+1}{2k+1}$ but we have that then $C \le \| a_{n+1} \|<\frac{k}{2k+1}$ , contradiction!.
A similar process applies when $\{a_n \} >\frac{1}{2}$ , you will also get a contradiction on $C$ , thus we are done :cool:

:cool:

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Nguyen

#6 h Sep 20, 2024, 5:13 AM

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It's too easy isn't it? Just consider everything in base $r$ .

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#7 h Sep 30, 2024, 2:28 PM

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hardest 1 of all time(many
CHN student think so).analyse $a_1$ and $a_2$ to prove even case.shall notice the answer can be very close to 1/2 and this construction produces oscillations rather than remaining constant.i did in 30min in the examination:)

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#9 h Mar 30, 2025, 3:38 AM

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It's not that easy since many of my classmates weren't able to guess the correct answer.
One of the smartest students in our class guessed it wrongly.
When I saw it, I chose to pass it first and moved to the geometry.(P2)
And I spent about twenty minutes to guess the correct answer.
With the right answer was that $C=\frac{r}{2r+2}$

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