Convergence of a sequence + parametre

In summary, the sequence converges to 1 if a_1=1, if 1<a_1<4/3 it does not converge, and for all other values except for 13/12, 40/39, 121/120, etc. the limit is 1/3. To determine if the sequence converges, we must find an interval I containing 1/3 such that 1 is not an element of I and f(I) is a subset of I. The interval (0,1) satisfies this condition. If a_1 equals 1, then a_n equals 1 for all values of n, resulting in a limit of 1. If a_1 is less than or
  • #1
Felafel
171
0

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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  • #2
Why is ##a_1>\frac43## not possible? A quick test shows me that if ##a_1=2##, your sequence converges on ##\frac13##.
 
  • #3
really? for ##a_1=2## I had ##a_2=-\frac{1}{2}##.
how did you get ##\frac{1}{3}##?
 
  • #4
$$
\begin{eqnarray*}
2 & \to & \frac1{4-3\cdot2} = \frac1{-2} = -\frac12 \\
& \to & \frac1{4-3\cdot(-1/2)} = \frac1{11/2} = \frac2{11} \\
& \to & \frac1{4-3\cdot(2/11)} = \frac1{38/11} = \frac{11}{38} \\
& \to & \frac1{4-3\cdot(11/38)} = \frac1{119/38} = \frac{38}{119} \\
& \to & \ldots \to \frac13.
\end{eqnarray*}
$$
Did you stop work after ##a_2##?
 
  • #5
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?
 
  • #6
Let [itex]f(x) = 1/(4 - 3x)[/itex], so that [itex]a_{n+1} = f(a_n)[/itex]. Then you can show fairly quickly that
[tex]
f((0,1)) \subset (0,1) \\
f((-\infty,0]) = (0,1/4] \subset (0,1) \\
f((1,4/3)) = (1,\infty) \\
f((4/3,\infty)) = (-\infty,0)
[/tex]
and that [itex]a_{n+1} > a_n[/itex] if [itex]a_n \in (1,4/3)[/itex] so that it only takes a finite number of steps until [itex]a_n > 4/3[/itex]. You can also show that for [itex]x \in (0,1)[/itex],
[tex]
|f(x) - 1/3| < |x - 1/3|
[/tex]
which means that [itex]|a_n - 1/3|[/itex] is monotonic decreasing here, and so [itex]a_n[/itex] tends to a limit which must be 1/3.

Putting this together you can see that [itex]a_n \to 1/3[/itex] unless [itex]a_1 = 1[/itex], when [itex]a_n = 1[/itex] for all n, or [itex]a_n = 4/3[/itex] for some n, when there is a problem. But this will only happen if [itex]a_1[/itex] is in the sequence
[tex]
b_1 = 4/3, b_{n+1} = \frac{4b_n - 1}{3b_n}.
[/tex]
 
  • #7
pasmith said:
Let [itex]f(x) = 1/(4 - 3x)[/itex], so that [itex]a_{n+1} = f(a_n)[/itex].
thanks.
okay so far, but how do you exactly do the rest of the demonstration later on?
 
  • #8
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.
 
  • #9
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 [itex]I \ni 1/3[/itex] such that [itex]1 \notin I[/itex] and [itex]f(I) \subset I[/itex] (ie, if [itex]x \in I[/itex] then [itex]f(x) \in I[/itex] also). The interval (0,1) will do nicely.

Because then if, given [itex]a_1 \in \mathbb{R}[/itex], there exists a finite [itex]N[/itex] such that [itex]a_N \in I[/itex], then [itex]a_n \in I[/itex] for all [itex]n \geq N[/itex]. In other words the sequence is trapped there, so to determine whether it converges I only have to look at what happens if [itex]a_1 \in I[/itex] (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 [itex]a_1[/itex] lead to sequences that get to that interval in a finite number of steps.

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

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

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

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

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

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

So now I can check what happens if [itex]a_1 \in I[/itex]. In general, it's possible that for different choices of [itex]a_1[/itex] 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 [itex]I[/itex] 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 [itex]x \in (0,1)[/itex],
[tex]
|f(x) - 1/3| < |x - 1/3|
[/tex]
which means that [itex]|a_n - 1/3|[/itex] is monotonic decreasing here, and so [itex]a_n[/itex] tends to a limit which must be 1/3.
 
  • #12
pasmith said:
The point is to find an interval [itex]I \ni 1/3[/itex] such that [itex]1 \notin I[/itex] and [itex]f(I) \subset I[/itex] (ie, if [itex]x \in I[/itex] then [itex]f(x) \in I[/itex] also). The interval (0,1) will do nicely.

Because then if, given [itex]a_1 \in \mathbb{R}[/itex], there exists a finite [itex]N[/itex] such that [itex]a_N \in I[/itex], then [itex]a_n \in I[/itex] for all [itex]n \geq N[/itex]. In other words the sequence is trapped there, so to determine whether it converges I only have to look at what happens if [itex]a_1 \in I[/itex] (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 [itex]a_1[/itex] lead to sequences that get to that interval in a finite number of steps.

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

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

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

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

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

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

So now I can check what happens if [itex]a_1 \in I[/itex]. In general, it's possible that for different choices of [itex]a_1[/itex] 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 [itex]I[/itex] 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
 

1. What is the definition of convergence of a sequence?

The convergence of a sequence refers to the behavior of a sequence of numbers as it approaches a fixed value, known as the limit. A sequence is said to converge if the terms of the sequence become closer and closer to the limit as the sequence progresses.

2. How is the convergence of a sequence determined?

The convergence of a sequence can be determined by checking if the terms of the sequence become arbitrarily close to the limit as the sequence progresses. This can be done by calculating the difference between each term and the limit and seeing if it approaches zero as the sequence progresses.

3. What role does the parameter play in the convergence of a sequence?

The parameter in the convergence of a sequence refers to a variable that can affect the behavior of the sequence. This parameter can change the limit, the rate of convergence, or the convergence itself. It is important to take into consideration the parameter when analyzing the convergence of a sequence.

4. Can a sequence converge to more than one limit?

No, a sequence can only converge to one limit. If a sequence has multiple limits, it is considered to be divergent. However, it is possible for a sequence to have a subsequence that converges to a different limit, but the original sequence must still converge to that one limit.

5. How does the rate of convergence of a sequence affect its behavior?

The rate of convergence of a sequence determines how quickly the terms of the sequence approach the limit. A sequence with a faster rate of convergence will have its terms become closer to the limit at a faster pace, while a sequence with a slower rate of convergence will take longer to approach the limit. Generally, a faster rate of convergence is preferred as it indicates a more efficient and stable behavior of the sequence.

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