Convergence of a sequence + parametre

[Felafel]

Homework Statement

let $a_n$ be $a_{n+1}=\frac{1}{4-3a_n} \quad n≥1$
for which values of $a_1$ does the sequence converge? which is the limit?

The Attempt at a Solution

$0<a_1<\frac{4}{3}$ because if $a_1>\frac{4}{3}→a_2<0$ not possible.
Now let's assume $a_n$ converges to M.
I have:
$M=\frac{1}{4-3M}$
$3M^2-4M+1=0 → M_{1,2}= 1, \frac{1}{3}$
However the can't be two limits, but both could work. How do I decide which one's wrong?

 

[Michael Redei]

Why is $a_1>\frac43$ not possible? A quick test shows me that if $a_1=2$, your sequence converges on $\frac13$.

 

[Felafel]

really? for $a_1=2$ I had $a_2=-\frac{1}{2}$.
how did you get $\frac{1}{3}$?

 

[Michael Redei]

$$
\begin{eqnarray*}
2 & \to & \frac1{4-3\cdot2} = \frac1{-2} = -\frac12 \\
& \to & \frac1{4-3\cdot(-1/2)} = \frac1{11/2} = \frac2{11} \\
& \to & \frac1{4-3\cdot(2/11)} = \frac1{38/11} = \frac{11}{38} \\
& \to & \frac1{4-3\cdot(11/38)} = \frac1{119/38} = \frac{38}{119} \\
& \to & \ldots \to \frac13.
\end{eqnarray*}
$$
Did you stop work after $a_2$?

 

[Felafel]

yes, because i thought that all the elements of the sequence had to be positive.
the teacher didn't say this explicitly, but we normally work on all-positive sequences.
but maybe this is not the case.. does it still converge if there are negative numbers in it? I mean, it won't be monotonic then. Any hints on how to solve the exercise?

 

[pasmith]

Let $f(x) = 1/(4 - 3x)$, so that $a_{n+1} = f(a_n)$. Then you can show fairly quickly that
$$f((0,1)) \subset (0,1) \\<br /> f((-\infty,0]) = (0,1/4] \subset (0,1) \\<br /> f((1,4/3)) = (1,\infty) \\<br /> f((4/3,\infty)) = (-\infty,0)$$
and that $a_{n+1} > a_n$ if $a_n \in (1,4/3)$ so that it only takes a finite number of steps until $a_n > 4/3$. You can also show that for $x \in (0,1)$,
$$|f(x) - 1/3| < |x - 1/3|$$
which means that $|a_n - 1/3|$ is monotonic decreasing here, and so $a_n$ tends to a limit which must be 1/3.

Putting this together you can see that $a_n \to 1/3$ unless $a_1 = 1$, when $a_n = 1$ for all n, or $a_n = 4/3$ for some n, when there is a problem. But this will only happen if $a_1$ is in the sequence
$$b_1 = 4/3, b_{n+1} = \frac{4b_n - 1}{3b_n}.$$

 

[Felafel]

Let $f(x) = 1/(4 - 3x)$, so that $a_{n+1} = f(a_n)$.

thanks.
okay so far, but how do you exactly do the rest of the demonstration later on?

 

[Felafel]

I had it this way, could it be right? :
solving the equation above the two possible limits are 1, 1/3.
$a_1$ can't be 4/3, because the function doesn't exist in that point.
NOw, if
$a_1>\frac{4}{3} --> a_{n+1}<0$ it's impossible, because it wouldn't converge to any of the two limits.
$a_1<\frac{4}{3} --> a_{n+1}>0$ works.
I consider the following cases:
if $\frac{1}{3}<a_1<1$
the sequence is decreasing and converges to 1/3.
if $a_1<\frac{1}{3}$
the sequence is decreasing and converges to 1/3.
if $1<a_1<\frac{4}{3}$
the sequence is not monotone and so it doesn't converge.

if $a_1=1$ the sequence is constant.

 

[Michael Redei]

You seem to hate negative numbers. Don't worry: If $a_1>\frac43$, you do have $a_{n+1}<0$, but in the next step you get $0<a_{n+1}<\frac14$, and so the sequence should converge to $\frac13$, as you say yourself. (It's more complicated than that though, as you'll see in a moment.)

If $1<a_1<\frac43$, the sequence is not monotone (as you say), but that doesn't mean it can't converge. (If you think a converging seuqnce must be monotone, just look at the sequence $(-\frac12)^n$. It alternates between positive and negative values and so can't be monotone, but it does converge against 0.)

One case you must take care of is $a_1=\frac{13}{12}$, since this means $a_2=\frac43$ and so $a_3=\frac10$, which is impossible.

And you must beware of $a_1=\frac{40}{39}$ as well, since you get $a_2=\frac{13}{12}$, which leads you back to $a_4=\frac10$.

And you mustn't have $a_1=\frac{121}{120}$ either, since that will take you to $a_2=\frac{40}{13}$.

I fear this is a greater mess than any of us had expected. However, I can see a pattern in those "forbidden" numbers $\frac43, \frac{13}{12}, \frac{40}{30}, \frac{121}{120}, \ldots$ They are all fractions $\frac{x_n+1}{x_n}$ with $x_1=3$ and $x_{n+1}=3(x_n+1)$, or $x_n=\sum_{k=1}^n3^k$.

 

[Felafel]

If $1<a_1<\frac43$, the sequence is not monotone (as you say), but that doesn't mean it can't converge. (If you think a converging seuqnce must be monotone, just look at the sequence $(-\frac12)^n$. It alternates between positive and negative values and so can't be monotone, but it does converge against 0.)

yes but this work only for $(a)^n$ with $|a|<1$, right? this isn't the case.
so, in a nutshell, the answer would be that the sequence converges to 1 if $a_1=1$,
if $1<a_q<\frac{4}{3}$ it doesn't converge and for all the other values (except the forbidden ones) the limit is 1/3
is everything correct now?

 

[pasmith]

thanks.
okay so far, but how do you exactly do the rest of the demonstration later on?

The point is to find an interval $I \ni 1/3$ such that $1 \notin I$ and $f(I) \subset I$ (ie, if $x \in I$ then $f(x) \in I$ also). The interval (0,1) will do nicely.

Because then if, given $a_1 \in \mathbb{R}$, there exists a finite $N$ such that $a_N \in I$, then $a_n \in I$ for all $n \geq N$. In other words the sequence is trapped there, so to determine whether it converges I only have to look at what happens if $a_1 \in I$ (because if I don't start here or at a special case then I'm just inserting terms at the beginning of the sequence and that cannot affect convergence, which is about what happens eventually).

But first I have to check what other choices of $a_1$ lead to sequences that get to that interval in a finite number of steps.

Clearly if $a_1 = 1$ then $a_n = 1$ for all $n$, so that doesn't (and the limit will be 1).

If $a_1 \leq 0$ then $0 < a_2 = 1/(4 - 3a_1) \leq 1/4$ so that $a_2 \in I$.

If $a_1 > 4/3$ then $a_2 < 0$ so from the above I know $a_3 \in I$.

The problem comes if $1 < a_1 < 4/3$. Here $x < 1/(4 - 3x)$ so initially the sequence will be increasing. But that means that for some finite $n$, $a_n \geq 4/3$. If $a_n > 4/3$ then from the above I know that $a_{n+2} \in I$.

If $a_n = 4/3$ then the sequence blows up. To find the $a_1$ where the sequence hits 4/3, I started at 4/3 and worked backwards, which gave me the sequence
$$b_1 = 4/3,\quad b_{n+1} = \frac{4b_n-1}{3b_n}$$
(You can check that if $y = 1/(4 -3x)$ then $x = (4y-1)/(3y)$.)

So if $a_1 \neq 1$ and there is no $a_n$ such that $a_n = 4/3$ then the sequence is in $I$ after a finite number of steps.

So now I can check what happens if $a_1 \in I$. In general, it's possible that for different choices of $a_1$ the sequence might tend to a limit (the possible such limits in this case are 1/3 and 1, which I can't exclude because $I$ is open), or tend to a periodic cycle, or might be chaotic. But in this case I can exclude every possibility other than tending to 1/3:

You can also show that for $x \in (0,1)$,
$$|f(x) - 1/3| < |x - 1/3|$$
which means that $|a_n - 1/3|$ is monotonic decreasing here, and so $a_n$ tends to a limit which must be 1/3.

 

[Felafel]

The point is to find an interval $I \ni 1/3$ such that $1 \notin I$ and $f(I) \subset I$ (ie, if $x \in I$ then $f(x) \in I$ also). The interval (0,1) will do nicely.

Because then if, given $a_1 \in \mathbb{R}$, there exists a finite $N$ such that $a_N \in I$, then $a_n \in I$ for all $n \geq N$. In other words the sequence is trapped there, so to determine whether it converges I only have to look at what happens if $a_1 \in I$ (because if I don't start here or at a special case then I'm just inserting terms at the beginning of the sequence and that cannot affect convergence, which is about what happens eventually).

But first I have to check what other choices of $a_1$ lead to sequences that get to that interval in a finite number of steps.

Clearly if $a_1 = 1$ then $a_n = 1$ for all $n$, so that doesn't (and the limit will be 1).

If $a_1 \leq 0$ then $0 < a_2 = 1/(4 - 3a_1) \leq 1/4$ so that $a_2 \in I$.

If $a_1 > 4/3$ then $a_2 < 0$ so from the above I know $a_3 \in I$.

The problem comes if $1 < a_1 < 4/3$. Here $x < 1/(4 - 3x)$ so initially the sequence will be increasing. But that means that for some finite $n$, $a_n \geq 4/3$. If $a_n > 4/3$ then from the above I know that $a_{n+2} \in I$.

If $a_n = 4/3$ then the sequence blows up. To find the $a_1$ where the sequence hits 4/3, I started at 4/3 and worked backwards, which gave me the sequence
$$b_1 = 4/3,\quad b_{n+1} = \frac{4b_n-1}{3b_n}$$
(You can check that if $y = 1/(4 -3x)$ then $x = (4y-1)/(3y)$.)

So if $a_1 \neq 1$ and there is no $a_n$ such that $a_n = 4/3$ then the sequence is in $I$ after a finite number of steps.

So now I can check what happens if $a_1 \in I$. In general, it's possible that for different choices of $a_1$ the sequence might tend to a limit (the possible such limits in this case are 1/3 and 1, which I can't exclude because $I$ is open), or tend to a periodic cycle, or might be chaotic. But in this case I can exclude every possibility other than tending to 1/3:

awesome! thank you