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Convergence of a sequence + parametre

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data
    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?

    3. 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?
     
  2. jcsd
  3. Dec 14, 2012 #2
    Why is ##a_1>\frac43## not possible? A quick test shows me that if ##a_1=2##, your sequence converges on ##\frac13##.
     
  4. Dec 14, 2012 #3
    really? for ##a_1=2## I had ##a_2=-\frac{1}{2}##.
    how did you get ##\frac{1}{3}##?
     
  5. Dec 14, 2012 #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##?
     
  6. Dec 15, 2012 #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?
     
  7. Dec 15, 2012 #6

    pasmith

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    Homework Helper

    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]
     
  8. Dec 16, 2012 #7
    thanks.
    okay so far, but how do you exactly do the rest of the demonstration later on?
     
  9. Dec 16, 2012 #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.
     
  10. Dec 16, 2012 #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##.
     
  11. Dec 16, 2012 #10
    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?
     
  12. Dec 16, 2012 #11

    pasmith

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    Homework Helper

    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:

     
  13. Dec 17, 2012 #12
    awesome! thank you
     
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