Italian WinterCamps test07 Problem5

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26 posts

#1 h Jan 29, 2007, 8:40 AM • 12 Y

Y by Davi-8191, peace09, FaThEr-SqUiRrEl, donotoven, jhu08, megarnie, Mahmood.sy, son7, Adventure10, Mango247, buddyram, and 1 other user

A sequence of real numbers $a_{0},\ a_{1},\ a_{2},\dots$ is defined by the formula
$a_{i + 1} = \left\lfloor a_{i}\right\rfloor\cdot \left\langle a_{i}\right\rangle\qquad\text{for}\quad i\geq 0;$ here $a_0$ is an arbitrary real number, $\lfloor a_i\rfloor$ denotes the greatest integer not exceeding $a_i$ , and $\left\langle a_i\right\rangle=a_i-\lfloor a_i\rfloor$ . Prove that $a_i=a_{i+2}$ for $i$ sufficiently large.

Proposed by Harmel Nestra, Estionia

This post has been edited 2 times. Last edited by djmathman, Jun 27, 2015, 12:02 AM
Reason: changed wording and formatting to match that of english version of ISL2006

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N.T.TUAN

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N.T.TUAN

#2 h Jan 29, 2007, 9:07 AM • 5 Y

Y by peace09, FaThEr-SqUiRrEl, jhu08, son7, Adventure10

If $a_{0}\geq 0$ then $a_{n}\geq 0\forall n$ . By $xy\leq (\frac{x+y}{2})^{2}$ we have $a_{n+1}\leq \frac{a_{n}^{2}}{4}$ , therefore exists $n_{0}$ such that $0\leq a_{n}<1\forall n>n_{0}$ .

Now, for $n>n_{0}$ we have $a_{n+1}=0$ , and we have done!

If $a_{0}<0$ , Who can post ?

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pardesi

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pardesi

#3 h Mar 1, 2007, 2:13 PM • 5 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, son7, Mango247

say $a_{0}<0$
then since $0\leq a_{0}-[a_{0}]<1$
wehave if $a_{0}$ is negative other numbers are either $0$ or are negative
so let's then conside the series of nonegative real no.s

$-a_{0},-a_{1},-a_{2},\dots$
rest can be proved by N.T.TUAN's method

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Rust

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Rust

#4 h Mar 1, 2007, 4:39 PM • 5 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, son7, Mango247

N.T.TUAN wrote:

we have $a_{n+1}\leq \frac{a_{n}^{2}}{4}$ , therefore exists $n_{0}$ such that $0\leq a_{n}<1\forall n>n_{0}$ .

It is not correct.
We have $a_{n+1}=[a_{n}]\{a_{n}\}$ , therefore if $0\le a_{n}\le 1$ , then $a_{n+k}=0 \ \forall k\in N$ . If $a_{n}>1$ , then $a_{n+1}<[a_{n}]$ , therefore $a_{[a_{0}]+k}=0 \ \forall k\in N$ .
If $a_{n}<0$ , then $a_{n+1}\le 0$ . If $-1<a_{0}<0$ , then $a_{1}=-\{a_{0}\}=-a_{0}-1, a_{2}=-a_{1}-1=a_{0}$ , therefore $a_{n+2}=a_{n}$ .

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N.T.TUAN

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N.T.TUAN

#5 h Mar 2, 2007, 2:16 AM • 6 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, Adventure10, son7, Mango247

Sorry, because I think it is $a_{n+1}\leq \frac{a_{n}}{4}$

I am wrong!

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newcomer

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newcomer

#6 h Mar 8, 2007, 7:49 AM • 5 Y

Y by peace09, FaThEr-SqUiRrEl, son7, Adventure10, Mango247

if $[a_{0}]\geq 0$ ,then
if $a_{0}\geq 1$ we know that $a_{n+1}<[a_{n}]$
thus $[a_{n+1}]<[a_{n}]$
so there must be an $n$ St $[a_{n}]=0$
so $a_{n+1}=a_{n+2}=.......=0$
if $[a_{0}]<0$
we can easily prove that:
if $[a_{m}]=s<-1$ ,then there must exsist a $k$ ,St $[a_{k}]<[a_{m}]$
thus we can suppose $[a_{m}]=-1$
then $a_{n}+a_{n+1}=-1$ for any $n\geq m$

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edriv

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edriv

#7 h Mar 8, 2007, 6:18 PM • 5 Y

Y by peace09, FaThEr-SqUiRrEl, son7, Adventure10, Mango247

Quote:

a_{n+1}=\left[a_{n}\right]\cdot \left\{a_{n}\right\}

Prove it! I remember that we define: $[x]$ as the maximum integer n such that $n \le x$ , i.e.: [4] = 4, [4.5] = 4, [-3] = -3, [-3.5] = -4...

And {x} = x - [x]

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newcomer

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newcomer

#8 h Mar 9, 2007, 12:53 PM • 7 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, Adventure10, megarnie, michaelwenquan, Mango247

i am sorry,but i don't know which part do you want me to prove

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edriv

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edriv

#9 h Mar 9, 2007, 1:58 PM • 4 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, Mango247

newcomer wrote:

we can easily prove that:
if $[a_{m}]=s<-1$ ,then there must exsist a $k$ ,St $[a_{k}]<[a_{m}]$

I cannot understand why is this so easy... during the test I wasn't able to solve this step.

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cefer

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cefer

#10 h Mar 9, 2007, 5:38 PM • 10 Y

Y by tenplusten, pablock, HolyMath, Rustem-E303, peace09, FaThEr-SqUiRrEl, AMMKMMRIR, guptaamitu1, Adventure10, and 1 other user

Note that if $a_{l} \in \mathbb{Z}$ , then $a_{i}=0$ for all $i > l$ and the claim obviously holds. Suppose $a_0\not \in \mathbb{Z}$ and consider the following two cases.

Case1: If $a_{0}<0$ , then $a_0 \in (-k, -k + 1)$ for some $k \in \mathbb{N}$ . We'll use induction on $k$ .

$(1)$ If $k=1$ , then $a_{0}=-1+\alpha$ for some $0<\alpha<1$ . Then $a_{1}=-\alpha$ and $a_{2}=\alpha-1=a_{0}$ . This means $a_{i}$ is periodic with period $2$ .
$(2)$ Suppose the claim holds up to some $k> 1$ .
$(3)$ Let $a_{0}\in (-k,-k+1)$ and write $a_{0}=-k+\alpha_{0}$ for some $\alpha_{0}\in (0,1)$ . Here we have two cases.

$(a)$ If $\alpha_{0}\leq \frac{k-1}{k}$ , then $a_1 = -k\alpha_0 \geq -k+1$ which means $a_{1}$ is either integer or it is in one of the intervals $(-1,0), (-2,-1), \ldots, (-k+1,-k+2)$ . So by induction hypothesis and above discussion we are done.
$(b)$ Now, suppose $\frac{k-1}{k}< \alpha_{0}$ . It is obvious that $a_{j}\geq-k$ for all $j$ . If one of these numbers $\geq -k+1$ , then by induction hypothesis we are done. So suppose that $a_{j}\in (-k,-k+1)$ for all $j$ and let $a_j = -k +\alpha_j$ . Then from recursion we get
$\alpha_{j}=k-k\alpha_{j-1}\, .$ This gives
$\alpha_{2l}=k-k^{2}+k^{3}-\cdots-k^{2l}+k^{2l}\alpha_{0}<1$ and
$\alpha_{2l+1}=k-k^{2}+k^{3}-\cdots+k^{2l+1}-k^{2l+1}\alpha_{0}<1\, .$ From the first inequality
$\begin{align*} \alpha_{0}<1- \frac{1}{k} + \cdots +\frac{1}{k^{2l}}\, . \end{align*}$ By letting $l \to \infty$ , we obtain $\alpha_0 \leq \frac{k}{k+1}$ . Similarly, from the second inequality
$\alpha_{0}>1-\frac{1}{k}+\frac{1}{k^{2}}-\cdots-\frac{1}{k^{2l+1}}\, .$ Again by letting $l\to \infty$ , we get $\alpha_{0} \geq \frac{k}{k+1}$ . Therefore $\alpha_0 =\frac{k}{k+1}$ which means $a_{0}=\frac{-k^{2}}{k+1}$ . Plugging this into recursion we see that $a_{i}=\frac{-k^{2}}{k+1}$ for all $i \geq 0$ . This completes induction.

Case 2: If $a_{0}>0$ , then $a_0 \in (k, k+1)$ for some $k\geq 0$ . We'll again use induction on $k$ .
$(1)$ If $k=0$ , then $0<a_{0}<1 \Longrightarrow a_{i}=0$ for all $i \geq 0$ .
$(2)$ Suppose the claim holds up to $k$ .
$(3)$ Let $a_{0}\in (k,k+1)$ and write $a_{0}=k+\alpha$ for some $\alpha \in (0,1)$ . Then $0 < a_{1}=k\alpha <k$ which means $a_{1}$ is in one of the intervals $(0,1), (1,2), \ldots, (k-1,k)$ . So by induction hypothesis the claim also holds for $k$ .

This post has been edited 3 times. Last edited by cefer, Apr 30, 2017, 10:04 PM
Reason: minor changes

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barasawala

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barasawala

#11 h Apr 12, 2007, 7:53 AM • 4 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, Mango247

cefer wrote:

$a_{2l}=k-k^{2}+k^{3}-...-k^{2l}+k^{2l}\alpha_{0}<1$
$a_{2l+1}=k-k^{2}+k^{3}-...+k^{2l+1}-k^{2l+1}\alpha_{0}<1$
$(l \rightarrow \infty)$

I think $a_{2l}$ should be changed to $\alpha_{2l}$ and $a_{2l+1}$ to $\alpha_{2l+1}$

The rest is OK

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cefer

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cefer

#12 h Apr 12, 2007, 9:44 AM • 3 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10

barasawala wrote:

cefer wrote:

$a_{2l}=k-k^{2}+k^{3}-...-k^{2l}+k^{2l}\alpha_{0}<1$
$a_{2l+1}=k-k^{2}+k^{3}-...+k^{2l+1}-k^{2l+1}\alpha_{0}<1$
$(l \rightarrow \infty)$

I think $a_{2l}$ should be changed to $\alpha_{2l}$ and $a_{2l+1}$ to $\alpha_{2l+1}$

The rest is OK

You are right,i made typo mistake

.

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Hawk Tiger

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Hawk Tiger

#13 h Apr 12, 2007, 2:12 PM • 4 Y

Y by peace09, FaThEr-SqUiRrEl, Adventure10, Mango247

Well,THis is the first problem of IMO shortlist 2006

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jb05

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jb05

#14 h Jul 1, 2007, 3:06 PM • 6 Y

Y by mortezamoradi, peace09, FaThEr-SqUiRrEl, Adventure10, Mango247, and 1 other user

Here is a shorter proof:

If $a_{n}>1$ , we have $0\le\lfloor a_{n+1}\rfloor<\lfloor a_{n}\rfloor$ , as $a_{n+1}$ is $\lfloor a_{n}\rfloor$ times a nonnegative less than 1. And if $0\le a_{n}<1$ , then $a_{n+1}$ and all terms thereafter are 0. So if $a_{0}\ge 0$ , the sequence of floors is strictly decreasing, so it eventually becomes constant at 0 and we are done.

Suppose $a_{n}<0$ . Again, $0\ge\lfloor a_{n+1}\rfloor\ge\lfloor a_{n}\rfloor$ , for the same reason. Hence if $a_{0}<0$ , the sequence of floors is nondecreasing, so either it eventually reaches 0 and we are done or it eventually becomes constant at $m$ . Suppose that for all $n\ge j$ , $\lfloor a_{n}\rfloor=m$ . Also suppose that $a_{j}=m+x$ , for an integer $m$ and $x\in [0,1)$ . Then we prove by induction that for all $i\ge 0$ , $a_{j+i}=a+bm^{i}$ , where $a=\frac{m^{2}}{m-1}$ and $b=\frac{xm-x-m}{m-1}$ . It is obvious for $i=0$ . Suppose it is true for a certain $i$ . Then $a_{j+i+1}=\lfloor a_{j+i}\rfloor\cdot \{a_{j+i}\}=m(a_{j+i}-m)=m(a+bm^{i}-m)$ $=(am-m^{2})+bm^{i+1}=a+bm^{i+1}$ , as it may easily be verified that $am-m^{2}=a$ . Hence $a_{j+i}=a+bm^{i}$ for all $i\ge 0$ . Also, $b\neq 0$ since if it did, $xm-x-m=0$ and $(x-1)(m-1)=0$ , but neither $x$ nor $m$ may equal 1. Now suppose that $m<-1$ . Then the absolute value of $a_{j+i}$ becomes arbitrarily large, which is a contradiction to the fact that $\lfloor a_{j+i}\rfloor=m$ . Hence $m=-1$ . But then we are done, since $a_{j+i+2}=a+b(-1)^{i+2}=a+b(-1)^{i}=a_{j+i}$ .

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MysticTerminator

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MysticTerminator

#15 h Jul 2, 2007, 3:43 PM • 7 Y

Y by mortezamoradi, peace09, FaThEr-SqUiRrEl, Adventure10, megarnie, and 2 other users

you seem to reach a contradiction in the case of the seed value a_0 = -3.2 when there is no contradiction there

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Darvas

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Darvas

#46 h Jul 21, 2022, 3:54 PM

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We consider two cases.

Case 1: There is a $i$ such that $a_i\in\mathbb{Z}.$ Then $a_{k+1}=0$ for all $k\geq i$ and we have $a_{j+2}=a_j$ for all $j>i.$

Case 2: $a_i\not\in\mathbb{Z},\quad\forall i.$ Then $0<\{a_i\}<1$ and $[a_i]\not=0$ for all $i.$

+) If $a_0>0$ then $a_n>0$ for all $n$ , therefore $[a_n]\geq 1$ for all $n.$ From hypothesis we have $1\leq [a_{n+1}] <[a_n],\quad\forall n\geq 1,$ contradiction.

+) If $a_0<0$ then $a_n<0$ for all $n$ and $[a_n]\leq -1$ for all $n.$ Therefore $[a_{n+1}]\geq [a_n]$ for all $n,$ so there is an integer $m\leq -1$ such that $[a_n]=m$ from some where. We can assume that $[a_n]=m\leq -2$ for all $n\geq 0,$ then $\{a_{i+1}\}=m\{a_i\}-m,\quad\forall i\geq 0.$ If $\{a_0\}=\frac{m}{m-1}$ then $\{a_i\}=\frac{m}{m-1}$ for all $i$ and $a_i=\frac{m^2}{m-1}$ for all i, we are done. Else, $\{a_{n+1}\}=m^{n+1}\{a_0\}-\frac{m(m^{n+1}-1)}{m-1},\quad\forall n,$ and we have $\{a_{n}\}>1$ for some $n,$ contradiction.

This post has been edited 1 time. Last edited by Darvas, Jul 21, 2022, 3:54 PM
Reason: typo

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HamstPan38825

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HamstPan38825

#47 h Feb 11, 2023, 6:32 PM

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For $a_0 > 0$ the sequence $\{\lfloor a_i \rfloor\}$ strictly decreases, so there is nothing to prove.

When $a_0 < 0$ , it suffices to show that $\{\lfloor a_i \rfloor\}$ will increase after a finite amount of time or $a_i$ remains fixed. Let $n = \lfloor a_0 \rfloor$ . Consider the disjoint set of intervals $I_k = [x_k, y_k]$ determined recursively by $I_1 = \left[n-\frac{n-1}n, n\right]$ , $I_2 = \left[n-1, n-1 + \frac{(n-1)^2}{n^2}\right]$ , and $I_{2n+1} = \left[x_{2n-1} - \frac{(n-1)^2}{n^2} (y_{2n-1}-x_{2n-1}), x_{2n-1}\right] \text{ and } I_{2n} = \left[x_{2n-2}, x_{2n-2} + \frac{(n-1)^2}{n^2}(y_{2n-2}-x_{2n-2})\right].$ Every real number in $[n-1, n]$ belongs in one of these intervals except for $n - \frac n{n+1}$ , which is a fixed point. On the other hand, for $a_0 \in I_n$ , we will have $\left \lfloor a_n \right \rfloor > n$ , as needed.

Thus, eventually the sequence will reach a point where $\lfloor a_i \rfloor = 0$ for all $i > N$ . It is clear that the sequence alternates now.

This post has been edited 1 time. Last edited by HamstPan38825, Feb 11, 2023, 6:35 PM

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peppapig_

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peppapig_

#49 h Mar 18, 2023, 12:24 PM • 1 Y

Y by mulberrykid

We claim that this sequence eventually goes to $0$ for positive real numbers $a_0$ .

Notice that if $a_0$ is positive and less than $1$ , then we will have that $a_1=0$ , proving our claim. Now we will prove that for all $a_k>1$ , then we will have that $\lfloor{}a_{k+1}\rfloor{}<\lfloor{}a_k\rfloor{}$ , which is clearly true, since $a_{k+1}=\lfloor{}a_k\rfloor{}\{a_k\}$ . Therefore the sequence must go to zero, meaning that eventually, $a_i=a_{i+2}=0$ .

What about negative $a_0$ ? Notice that for $a_0<0$ , we have that the fractional part of $a_k$ is constantly increasing, meaning that eventually, we will have that $a_{i+1}=1-a_i$ . Therefore we have that $a_{i+2}=a_i$ , and we are done.

FS found by bobthegod78
What about negative $a_0$ ? Let $a_0=-k+\frac{k}{k+1}+x$ for some $k$ . If $x\neq{}0$ , we have that $a_1=-k+\frac{k}{k+1}-xk$ , $a_2=-k+\frac{k}{k+1}+xk^2$ , $a_3=-k+\frac{k}{k+1}-xk^3$ , and so on and so forth until we are left with $a_i=-k+\frac{k}{k+1}-x*(-k)^i$ . It is clear that this will eventually go to an integer, leaving us with a sequence of $n$ , $-1-n$ , and so on and so forth, proving that $a_i=a_i+2$ .

If $x=0$ , it is easy to see that the sequence is constant anyways, and we are done.

This post has been edited 3 times. Last edited by peppapig_, Apr 30, 2023, 3:02 PM
Reason: fakesolve fix + detail edits

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bobthegod78

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bobthegod78

#50 h Apr 27, 2023, 10:45 AM

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For $a_0\geq 0,$ it is obvious that the sequence becomes all $0$ s eventually.
For negative $a_i,$ let $a_i = -n + \frac{n}{n+1} + \epsilon$ . If $\epsilon =0$ , the sequence is constant and we are done. Notice that $a_{i+1} = -n+\frac{n}{n+1} - \epsilon \cdot n$ , and so on, so eventually, the value of $\epsilon \cdot n^k$ exceeds $\frac 1{n+1}$ , and the value the floor is incremented by 1, unless $n=1$ . If $n=1$ , we see that it oscillates between $-x, x-1$ , so $a_{i+2}=a_i$ .

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john0512

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john0512

#51 h May 12, 2023, 1:16 AM

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When $a_0$ is nonnegative, then $\lfloor a_{i+1}\rfloor <\lfloor a_{i}\rfloor$ , so it will eventually go down to zero. If $a_0$ is an integer, then clearly it will just go to 0. For the rest of this solution, assume $a_0$ is negative and non-integer.

Define a equlibrium point as a real number that is equal to $-\frac{(n-1)^2}{n}$ for some positive integer $n$ . The equilibrium points have the property that if $a_i$ is an equilibrium point, then $a_{i+1}=a_i$ . Furthermore, there is exactly one equilibrium point between any two nonpositive integers, and that equilibrium point is the only fixed point between those two numbers.

If $a_0$ starts between -1 and 0, then it will alternate between $a_0$ and $-1-a_0$ so the statement holds. From now on, assume $a_0$ is less than -1. Then, consider the equilibrium point in $a_0$ 's integer interval. If $a_0$ is at the equilibrium point, it will just stay there forever so the problem statement holds. Otherwise, its distance away from the equilibrium point will multiply by $-\lfloor a_i \rfloor$ but go in the opposite direction. Since we are assuming less than -1, the distance to the equilibrium point will keep increasing, and it will eventually escape the current integer range to the right (it's not possible for it to escape to the left), into either a nonnegative range or a less negative range. Of course, if it goes into a nonnegative range, it will then eventually go to zero as shown earlier. Thus, for any range below -1, as long as it doesn't land into the range at an equilibrium point, it will escape further to the right. Thus, if we keep doing this, it will either get stuck at an equilibrium point, get between -1 and 0 and alternate, or eventually go to zero, so we are done.

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VicKmath7

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VicKmath7

#52 h May 12, 2023, 2:11 PM

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Solution

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Isometric

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Isometric

#53 h Jun 21, 2023, 6:59 PM

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I hope this is not a fakesolve

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huashiliao2020

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huashiliao2020

#54 h Jul 31, 2023, 8:39 PM

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It's easy to see that for $a_0\ge 0$ , the problem is trivial since the sequence is strictly decreasing. For $a_0<-1$ ( $-1\le a_0<0$ implies it iterates from a_i=c<0 to -1*(c-(-1))=-1-c and oscillates with period 2), if we let -b_i=floor[a_i]<-1, we want to show $a_{i+1}=-b_i(a_i+b_i)\ge a_i\iff a_i\ge \frac{b_i^2}{-b_i-1}=-b_i+\frac{b_i}{b_i+1}.$ Because of this experimenting, we're motivated to set $a_i =-b_i+\frac{b_i}{b_i+1}+c_i.$ Indeed, $a_{i+1}=-b_i+\frac{b_i}{b_i+1}-b_ic_i$ and so on. Since b_i>1, for some even amount of iterations eventually $b_i^kc_i>\frac{1}{b_i+1}$ , whence -b_i increases by 1 and so b_i decreases by 1 monotonically until b_i becomes 0, so all cases have been considered, and we are done! $\blacksquare$

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lelouchvigeo

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lelouchvigeo

#55 h Jan 2, 2024, 9:25 AM

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For a_0>0 Just observe that [a_i+1]<[a_i].Therefore the sequence is decreasing.We will find an [a_i]=0

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shendrew7

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shendrew7

#56 h Apr 30, 2024, 4:26 AM

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The sequence is strictly decreasing for $a_1 > 0$ and will eventually remain constant at 0. Hence we proceed by considering negative $a_1$ .

Consider the fixed points $a_i$ such that $a_i = a_{i+1}$ , which can be determined to be 0 and $-k\tfrac{1}{k+2}$ for $k \in \mathbb{Z}_{\ge 0}$ .

Once we hit/start with this fixed point, it remains constant, which finishes.
If we have $a_i$ greater than the corresponding fixed point, $a_{i+1}$ is less than the fixed point.
If we have $a_{i+1}$ less than the fixed point, either $\lfloor a_{i+1} \rfloor$ strictly increases, or $\lfloor a_i \rfloor < a_{i+2} < a_i$ where $\lfloor a_i \rfloor \leq -2$ , from which we eventually must increase $\lfloor a_{i+k} \rfloor$ .

In the end state, we either land on a fixed point (including 0), or $\lfloor a_i \rfloor = -1$ , from which we always have $a_{i+2} = a_i$ . $\blacksquare$

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ezpotd

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ezpotd

#57 h Aug 27, 2024, 10:55 PM

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Notice that if $a_0$ is nonnnegative, we can easily show $a_i$ is eventually $0$ , if $a_i$ is a noninteger the floor clearly decreases, so eventually the floor must be $0$ and we win, or if $a_i$ is an integer we floor and get $0$ anyways. Continue to ignore all integer values since those instantly finish.

So we take $a_i$ as negative, if $a_i = c$ is between $-1$ and $0$ , we see that the next term is just $-1 - c$ , and applying again give $c$ , so the sequence alternates. Thus we show $a_i$ eventually goes to something in $(-1, 0]$ . Take $a_i = x + (-k)$ , where $x,k$ are positive, and $k > 2$ . Then assume for contradictory purposes that the floor of $a_i$ never changes. If $x = \frac{k}{k + 1}$ , the sequence is constant and we can ignore this case. Otherwise, if $x > \frac{k}{k + 1}$ , $a_{i + 1}$ has floor $k$ and fractional part less than $\frac{k}{k + 1}$ , so take $a_j = -k + x$ where $x < \frac{k}{k + 1}$ , we then calculate the decrease in $x$ from $a_j$ to $a_{j + 2}$ . Clearly $a_{j + 1} = -kx$ , then since we assumed the floor never changes we have $a_{j + 2} = -k(-kx + k) = k^2(x - 1)$ , the decrease is then $-k + x + k^2 - k^2x = -k^2x + x + k^2 - k = (1 - k^2)x + k^2 - k$ , which is positive for $x < \frac{k}{k + 1}$ , moreover this decrease is decreasing in $x$ , so as we compute $a_{j + 2n}$ with the fixed floor being assumed, $x$ will continue to decrease at an ever faster rate, and eventually we will have negative $x$ , impossible. Thus the floor must decrease eventually unless the sequence is constant, done.

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ihatemath123

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ihatemath123

#58 h Sep 22, 2024, 5:36 AM • 1 Y

Y by OronSH

If $a_0$ is positive, then the sequence is nonincreasing and will eventually become just $0$ , so clearly $a_i = a_{i+2}$ .

If $a_0$ is negative, then we have $\lfloor a_i \rfloor \leq a_{i+1} \leq 0$ , so eventually $\lfloor a_i \rfloor$ is constant. Let this constant value be $C$ . Then, we have
$\left \langle a_{i+1} \right \rangle = C \left \langle a_i \right \rangle - C \implies \left( \left \langle a_{i+1} \right \rangle - \tfrac{C}{C-1} \right) = C \left( \left \langle a_i \right \rangle - \tfrac{C}{C-1} \right)$ for sufficiently large $i$ . If $C \leq -2$ , we have a contradiction since $\langle a_{i} \rangle - \frac{C}{C-1}$ is an unbounded geometric sequence, so $C = -1$ or $C = 0$ . The latter case is trivial, and the former case implies that $\langle a_i \rangle + \langle a_{i+1} \rangle = 1$ for sufficiently large $i$ , which also implies the problem statement.

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Maximilian113

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Maximilian113

#59 h Jan 2, 2025, 1:30 AM

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If $a_0 \geq 0,$ it is clear that the sequence is strictly decreasing and thus will eventually hit $0,$ so this case is finished.

Meanwhile, for $a_0 < 0,$ observe that $a_{i+1} =\lfloor a_i \rfloor \{ a_i \} \geq \lfloor a_i \rfloor$ so $\lfloor a_i \rfloor$ is decreasing (not strictly). If the sequence eventually reaches zero, it stays there so assume otherwise. Then the sequence is always negative meaning that its floor will eventually converge to a negative integer, say $-n$ for some positive integer $n.$

Then for sufficiently large $i,$ let $a_i = x_i-n$ for $0 \leq x_i < 1.$ If at some point $x_i=0,$ the sequence stays at zero so assume otherwise. Then the recurrence becomes $x_{i+1}-n = -nx_i \implies x_{i+1}=(1-x_i)n.$ Letting $r_i = \frac{x_i}{n^i}$ yields $r_{i+1} = \frac{1}{n^i} - r_i \implies r_{i+2}=r_i.$ Therefore $x_{i+2} = n^2 x_i$ for sufficiently large $i,$ but if $n \geq 2$ this is a contradiction as eventually $x_i > 1.$ Therefore, $n=1$ so $x_{i+2}=x_i,$ and thus $a_{i+2}=a_i.$ QED

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study1126

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study1126

#60 h Jan 2, 2025, 6:01 AM

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When $a_0\geq 0$ , we know that $\lfloor a_{i+1} \rfloor<\lfloor a_i \rfloor,$ therefore we eventually reach $0$ with sufficiently large $i$ .

If $a_0<0$ , then $\lfloor a_i \rfloor \leq a_{i+1} \leq 0,$ therefore eventually $\lfloor a_i \rfloor$ will reach a constant value, say $c$ . In this case, we see that ${a_{i+1}}=c{a_i}-c,$ which by algebraically manipulating gives ${a_{i+1}}-\frac{c}{c-1}=c ({a_i}-\frac{c}{c-1} ).$ We see that if $c\leq -2$ , then the sequence of decimal parts is unbounded, a contradiction. Thus, $c=-1$ or $c=0$ . If $c=0$ , then we will eventually reach a sequence of 0s. On the other hand, $c=-1$ gives an eventual sequence of period 2, which proves $a_{i+2}=a_i.$

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Marcus_Zhang

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Marcus_Zhang

#63 h Apr 14, 2025, 7:38 PM

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