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Math Help - 2 more questions about sequences..

  1. #1
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    2 more questions about sequences..

    i have a sequence which is defined by the following:

    a(1) ∈ (0, 0.5)

    a(n+1) = a(n) - 0.5a(n)^(n+1)

    1: Is it possible to form the ''equation'' of the sequence a(n) with the above conditions? if yes, how?
    2: with n -> infinite, is it possible to find a limit? Aren't there more than one possibilities?

    sorry for the fuzz..
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  2. #2
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    Re: 2 more questions about sequences..

    Quote Originally Posted by nappysnake View Post
    i have a sequence which is defined by the following:

    a_1 \in (0, 0.5)

    a_{n+1} = a_n - 0.5a_n^{n+1}

    1: Is it possible to form the ''equation'' of the sequence (a_n) with the above conditions? if yes, how?
    2: with n \to\infty, is it possible to find a limit? Aren't there more than one possibilities?
    This is an interesting sequence!

    The answer to 1) has to be No: it is not possible (as far as I can see) to give an explicit formula for a_n. You just have to work with the recursive formula that gives you a_{n+1} in terms of a_n.

    For 2) also, it does not seem possible to find the limit explicitly. The best that can be done is show that the limit must exist, and that it can be narrowed down to lie in some interval. The value of the limit will depend on the first term of the sequence.

    From the equation a_{n+1} = a_n - \tfrac12a_n^{n+1}, you can show by induction that a_{n+1}<a_n and also that a_{n+1}>0. Thus (a_n) is a decreasing sequence of positive terms. There is a theorem which says that any decreasing sequence that is bounded below must converge to a limit. So we know that this sequence must converge to some limit \ell\geqslant0. The big question is whether this limit is actually equal to 0, or whether the sequence converges to a strictly positive limit. In fact, it is true that \ell>0.

    To see that, write a=a_1 for the first term of the sequence, so that 0<a<\tfrac12. Since the sequence decreases, it follows that a_n<a for all n. Therefore a_n-a_{n+1} = \tfrac12a_n^{n+1} < \tfrac12a^{n+1} for each n=1,2,3,... . It follows that

    \begin{aligned}a-a_n &= (a_1-a_2)+(a_2-a_3)+\ldots+(a_{n-1}-a_n) \\&< \tfrac12(a^2+a^3+\ldots+a^n) = \frac{a^2- a^{n+1}}{2(1-a)}\end{aligned}

    (sum of geometric series). If you now let n\to\infty on both sides, you find that a-\ell\leqslant \frac{a^2}{2(1-a)}. Therefore

    \ell\geqslant a-\frac{a^2}{2(1-a)} = \frac{a(2-3a)}{2(1-a)} > 0.

    Thus the limit of the sequence is strictly positive, and lies between \tfrac{a(2-3a)}{2(1-a)} and a. I don't see any way of narrowing it down more closely than that.
    Last edited by Opalg; November 28th 2011 at 11:38 PM.
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    Re: 2 more questions about sequences..

    could you please show me how you perform induction in the above sequence?? everything else is pretty much clearer now, thanks!
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  4. #4
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    Re: 2 more questions about sequences..

    Quote Originally Posted by nappysnake View Post
    could you please show me how you perform induction in the above sequence?? everything else is pretty much clearer now, thanks!
    It is a very simple induction, to prove the fact that 0<a_n<1/2. For n=1 that is true by the initial assumption on a_1. Since a_{n+1} = a_n - \tfrac12a_n^{n+1} = a_n(1-\tfrac12a_n^n), it follows from the inductive hypothesis 0<a_n<1/2 that 0<a_{n+1}<1/2 (because the factor 1-\tfrac12a_n^n certainly lies between 0 and 1). That completes the inductive step.

    That is the only inductive proof that is needed. Once you know that a_n>0 it immediately follows from the equation a_{n+1} = a_n - \tfrac12a_n^{n+1} that a_{n+1}<a_n, and therefore the sequence (a_n) is decreasing.
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