Convergence of a sequence + parametre

Felafel
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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?
 
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Why is ##a_1>\frac43## not possible? A quick test shows me that if ##a_1=2##, your sequence converges on ##\frac13##.
 
really? for ##a_1=2## I had ##a_2=-\frac{1}{2}##.
how did you get ##\frac{1}{3}##?
 
$$
\begin{eqnarray*}
2 & \to & \frac1{4-3\cdot2} = \frac1{-2} = -\frac12 \\
& \to & \frac1{4-3\cdot(-1/2)} = \frac1{11/2} = \frac2{11} \\
& \to & \frac1{4-3\cdot(2/11)} = \frac1{38/11} = \frac{11}{38} \\
& \to & \frac1{4-3\cdot(11/38)} = \frac1{119/38} = \frac{38}{119} \\
& \to & \ldots \to \frac13.
\end{eqnarray*}
$$
Did you stop work after ##a_2##?
 
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?
 
Let f(x) = 1/(4 - 3x), so that a_{n+1} = f(a_n). Then you can show fairly quickly that
<br /> 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)<br />
and that a_{n+1} &gt; a_n if a_n \in (1,4/3) so that it only takes a finite number of steps until a_n &gt; 4/3. You can also show that for x \in (0,1),
<br /> |f(x) - 1/3| &lt; |x - 1/3|<br />
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
<br /> b_1 = 4/3, b_{n+1} = \frac{4b_n - 1}{3b_n}.<br />
 
pasmith said:
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?
 
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.
 
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##.
 
  • #10
Michael Redei said:
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?
 
  • #11
Felafel said:
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 &lt; a_2 = 1/(4 - 3a_1) \leq 1/4 so that a_2 \in I.

If a_1 &gt; 4/3 then a_2 &lt; 0 so from the above I know a_3 \in I.

The problem comes if 1 &lt; a_1 &lt; 4/3. Here x &lt; 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 &gt; 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
<br /> b_1 = 4/3,\quad b_{n+1} = \frac{4b_n-1}{3b_n}<br />
(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),
<br /> |f(x) - 1/3| &lt; |x - 1/3|<br />
which means that |a_n - 1/3| is monotonic decreasing here, and so a_n tends to a limit which must be 1/3.
 
  • #12
pasmith said:
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 &lt; a_2 = 1/(4 - 3a_1) \leq 1/4 so that a_2 \in I.

If a_1 &gt; 4/3 then a_2 &lt; 0 so from the above I know a_3 \in I.

The problem comes if 1 &lt; a_1 &lt; 4/3. Here x &lt; 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 &gt; 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
<br /> b_1 = 4/3,\quad b_{n+1} = \frac{4b_n-1}{3b_n}<br />
(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
 

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