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The problem states:

a sequence of real numbers $a_0, a_1, \dots $ is defined as follows. $a_0$ is an arbitrary real number and for $n \ge 0, a_{n +1} = \lfloor a_n \rfloor \{ a_n \}$ (where $\{x\} = x - \lfloor x \rfloor$).

Prove that for $a_n = a_{n+2}$ for $n$ large enough.

my attempt:

I notice that this is a decreasing sequence because $$\frac{a_{n+2}}{a_{n+1}} = \frac{\lfloor a_{n+1} \rfloor \{ a_{n+1} \}}{ a_{n+1}} < 1$$

since $\{ a_{n+1} \} \le 1$ (we can exclude the case where it's equal to one since then it becomes the null sequence that satisfies the thesis).

Moreover this sequence is bounded between $0$ and $a_0$ so by the monotone convergence theorem it must have a limit.

Taking the limit I obtain $$\ell = \lim \lfloor a_n \rfloor (\ell - \lim \lfloor a_n \rfloor) \iff \ell = - \lim \lfloor a_n \rfloor^2 / (1 - \lim \lfloor a_n \rfloor) $$

And so if I prove that there exists $n$ s.t. $ \lfloor a_n \rfloor < 1$ I am done. Because then the limit could only be $0$ and this implies that there exist infinite $a_n = a_{n+2} = 0$.

But I can't seem to prove it. Is my reasoning up to now ok? How could I proceed?

Edit: enter image description here Add original text of the problem.

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    $\begingroup$ Are you sure about the formulation of the problem? It seems that $a_n$ eventually becomes zero? ($a_0$ is positive I suppose?) $\endgroup$ – H. H. Rugh Jul 27 '16 at 14:42
  • $\begingroup$ Ok, in fact when $a_0<0$ it makes sense and is eventually two periodic. $\endgroup$ – H. H. Rugh Jul 27 '16 at 14:58
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Assuming that $a_0 \gt 0$, note that for $a_n \gt 1$, $a_{n+1} \lt \lfloor a_n \rfloor$, so $\lfloor a_n+1\rfloor \le \lfloor a_n\rfloor -1$ so the sequence is bounded above by $a_n, a_n-1, a_n-2\dots 1$ and must reach $0$ and be constant there.

Now assume $a_0 \lt 0$. We have $a_1=\lfloor a_0 \rfloor \{a_0\} \gt \lfloor a_0 \rfloor $ so the sequence cannot go below $\lfloor a_0 \rfloor $. If $\{a_n\} \lt 1-\frac 1{\lfloor a_n \rfloor }, \lfloor a_{n+1} \rfloor \gt \lfloor a_n \rfloor$ and we proceed toward zero. If $\{a_n\} = 1-\frac 1{\lfloor a_n \rfloor }, a_{n+1}= a_n$ and we are at a fixed point. In that case we will have $a_{n+2}=a_n$. If $\{a_n\} \gt 1-\frac 1{\lfloor a_n \rfloor },\{a_{n+1}\} \lt 1-\frac 1{\lfloor a_{n+1} \rfloor }$ and we will proceed toward zero next step. Once we get to $a_n \gt -1$ we have $a_{n+1}=(-1)(a_n+1)=-a_n-1 \gt -1, a_{n+2}=-a_{n+1}-1=-(-a_n-1)-1=a_n$ and we are done.

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  • $\begingroup$ For $a_0<0$, the situation is different. $\endgroup$ – xpaul Jul 27 '16 at 14:53
  • $\begingroup$ @xpaul: I included the restriction. Thanks. $\endgroup$ – Ross Millikan Jul 27 '16 at 14:55
  • $\begingroup$ I'm not sure I understand the argument.. $\endgroup$ – Monolite Jul 27 '16 at 14:57
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    $\begingroup$ For example, assume $a_0=300.999999$. Then $a_1=300 \cdot 0.999999 \lt 300,$ so $\lfloor a_1 \rfloor =299$. As the fractional part of $a_1 \lt 1$ the whole part of $a_2$ will be at most $298$. By $a_{300}$ (and in practice sooner) we will be at $0$. $\endgroup$ – Ross Millikan Jul 27 '16 at 15:05
  • $\begingroup$ Thanks! any idea for negative starting values? $\endgroup$ – Monolite Jul 27 '16 at 15:19
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graph of the iteration map All positive $a_0$ iterates to zero in finite time, see the graph of the iterating map. For negative $a_0$ there are fixed points at every $x=-m^2/(m+1)$, $m\geq 1$ (period 1 orbits). Either $a_0<-1$ eventually lands at one of those fixed points (which are unstable for $m\geq 2$), or it maps to the interval $[-1,0[$ after a finite number of iterations. The map is here $f_{-1}(x)=-(x+1)$. $-1$ maps to $0$ which is a fixed point, but every $x\in (-1,0)$ maps to $-(1+x)$ and then to $x$ (period 2 orbit). The exercise is very neat!

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  • $\begingroup$ Thanks! but how could I prove the negative case? $\endgroup$ – Monolite Jul 27 '16 at 15:40
  • $\begingroup$ Millikan's description is close but misses a few points. Notably that each fixed point $x_m=-m^2/(m+1)$ in $I_m=[-(m+1),-m)$, $m\geq 2$ is a repelling fixed point. If you start at $x_m$ you stay but otherwise you will wound around in an expanding spiral until you finally map out of $I_m$ and into another $I_\ell$ for some $\ell<m$. So either you will get stuck at a fixed point or eventually map into $I_1$ where you have the above mentioned 2-periodic orbits. It's a bit tricky to write down all the details but if you need it I could try to develop ? $\endgroup$ – H. H. Rugh Jul 27 '16 at 22:45
  • $\begingroup$ @Monolite Btw, what is the reference for the book containing the problem? $\endgroup$ – H. H. Rugh Jul 28 '16 at 7:34
  • $\begingroup$ International math olympiad 2006 SL. Never heard of repelling fixed points :). $\endgroup$ – Monolite Jul 28 '16 at 10:49
  • $\begingroup$ OK, thanks for the ref. An easy example: $f(x)=\lambda x$, some real $\lambda$. $f(0)=0$ is the only fixed point. If $|\lambda|<1$ and $x\neq 0$ then $f^n(x)=\lambda^n x$ goes to zero as $n\rightarrow +\infty$ and the fixed point is called attractive. If $|\lambda|>1$, $f^n(x)$ tends to zero as $n\rightarrow -\infty$ (thus backwards in time) and the fixed point is called repelling or repulsive. In general only local behavior matters, and is determined by the derivative $\lambda=f'(x)=-m$ at the fixed point. In our case $m \geq 2$ and the fixed point is repelling. Does it make sense? $\endgroup$ – H. H. Rugh Jul 28 '16 at 17:01

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