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Math Help - Proof of Convergence

  1. #1
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    Proof of Convergence

    I have a sequence of real numbers for which I wish to prove convergence.
    s_{n+1} = \sqrt[3]{s_{n}^{2} + s_{n-1} + 2} and 0 < s_{0} < s_{1} < 2

    So, if the sequence \{s_{n}\} converges to a limit s then
    s^3 - s^2 - s - 2 = 0 \; \Rightarrow \; \left( s - 2 \right)\left( s^2 + s + 1 \right) = 0 \; \Rightarrow \; s = 2
    Also, we have that t^3 - t^2 - t - 2 < 0 \; \Rightarrow \; t < s \forall t > 0

    But I have yet to prove convergence.
    \begin{align*}s_{n+1}^3 - s^3 &= -s^3 + s_{n}^{2} + s_{n-1} + 2 \\ &= -\left( s^3 - s^2 - s - 2 \right) + \left( s_{n}^{2} - s^2 \right) + \left( s_{n-1} - s \right) \\ &= \left( s_{n}^{2} - s^2 \right) + \left( s_{n-1} - s \right) \\ \end{align*}
    So if s_{n-1} and s_{n} are both less than s, then s_{n+1} < s and our initial conditions give us s_{n} < s \; \forall n \in \mathbb{N}.
    And thus the sequence \{s_{n}\} is bounded above.

    So now I need to prove that sequence \{s_{n}\} is increasing.
    \begin{align*} s_{n+1}^3 - s_{n}^3 &= -s_{n}^3 + s_{n}^{2} + s_{n-1} + 2 & \text{but} \\ s_{n} &< s \; \Rightarrow \; s_{n}^3 - s_{n}^{2} - s_{n} + 2 < 0 \\ \Rightarrow \; s_{n}^3 - s_{n}^{2} + 2 &< s_{n} & \text{so} \\ s_{n+1}^3 - s_{n}^3 &> s_{n-1} - s_{n} \\ \end{align*}
    But that doesn't show that s_{n+1}^3 - s_{n}^3 > 0 as I require.

    Can you help?
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  2. #2
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    Re: Proof of Convergence

    Have you learned that f:\mathbb{R} \to \mathbb{R} is continuous if and only if \{f(a_n)\} converges for all sequences \{a_n\} which converge in \mathbb{R}? Well, you know that f(x) = \sqrt[3]{x} is continuous. So, let r_n = s_n^3. If you can show that r_n converges, then since f is continuous, f(r_n) = s_n converges, as well.

    Anyway, to show convergence, you will want to end with something like:

    Given \varepsilon>0, let N be a positive integer such that N > \text{expression} (where the expression is some function of \varepsilon).

    Then show that for all n > N, |s_n - 2| < \varepsilon.
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  3. #3
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    Re: Proof of Convergence

    The text I'm looking at ("A First Course in Mathematical Analysis", Burkill) hasn't introduced the concept of a function at this point. So although your idea would work, I'm looking for something using much more basic ideas.

    I don't intend to go into \epsilon notation for this, since I have already shown that if the sequence converges, it must converge to 2. Given that, all I need is to show that it does converge. And the standard approach in the text is to show that the sequence is bounded (which I've done) and that it is monotonic. The combination of these results being sufficient for convergence. Another approach would be to show that \left| s_n - 2 \right| converges to zero, but that doesn't look very practical here, the analysis of s_{n+1}^3 - s^3 proving only that the sequence is bounded.
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  4. #4
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    Re: Proof of Convergence

    Suppose s_0 = 0.1, s_1 = 1.9. Then s_2 = \sqrt[3]{1.9^2 + 0.1 + 2} \approx 1.78736 < s_1, so the sequence is not necessarily monotonically increasing. You might need to prove that |s_n - 2| converges to zero.
    Last edited by SlipEternal; December 23rd 2013 at 08:58 PM.
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  5. #5
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    Re: Proof of Convergence

    You can, however, use the squeeze theorem. Let r_{n+1} = \sqrt[3]{r_n^2 + r_{n-1} + 2}, r_0 = r_1 = 0. Let t_{n+1} = \sqrt[3]{t_n^2 + t_{n-1} + 2}, t_0 = t_1 = 2. Obviously, r_n < s_n < t_n for all n. It is pretty easy to show that t_n = 2 for all n. So, if you can show that r_n converges to 2, then by the squeeze theorem, s_n also converges to 2. Now, you can show that r_n is bounded and monotonically increasing.
    Last edited by SlipEternal; December 23rd 2013 at 09:10 PM.
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  6. #6
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    Re: Proof of Convergence

    To show that r_n converges, you know that r_1 = 0, r_2 = \sqrt[3]{2} \approx 1.26, r_3 = \sqrt[3]{\sqrt[3]{4}+2} \approx 1.53, so r_1 < r_2 < r_3. Assume that for all positive integers greater than or equal to 3 and up to n, 2 > r_n > r_{n-1} > 1 and r_{n-1}>r_{n-2} > 0 (which is true for r_1,r_2,r_3). Then r_{n+1}^3 - r_n^3 = (r_n^2 + r_n + 2) - (r_{n-1}^2 + r_{n-2} + 2) = (r_n^2 - r_{n-1}^2) + (r_{n-1} - r_{n-2}) > 0, so r_{n+1} > r_n. Hence, by mathematical induction, r_n is monotone increasing. Then, since it is bounded, it converges. Finally, as you showed above, if it converges, it converges to 2. So, by the Squeeze Theorem, so too does s_n.
    Last edited by SlipEternal; December 23rd 2013 at 09:41 PM.
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  7. #7
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    Re: Proof of Convergence

    Haha, I hadn't realised that the initial values were sensitive like that. The original question has something like s_0 = 1, s_1 = 1.3

    One thing I have started to think about is that
    \begin{align*}<br />
s_{n+1}^3 - 2 &= s_{n}^2 + s_{n-1} + 2 - 2 \\<br />
&= s_{n}^2 + s_{n-1} \\<br />
\end{align*}
    So s_{n} > 0, s_{n-1} > 0 (which is easily seen from the definition of s_{n+1}) gives s_{n} \ge \sqrt[3]{2} \forall n > 0

    But then we also have
    \begin{align*}<br />
s_{n+1}^3 - 4 &= s_{n}^2 + s_{n-1} + 2 - 4 \\<br />
&= s_{n}^2 + s_{n-1} - 2 \\<br />
\end{align*}
    So s_{n} > \sqrt[3]{2}, s_{n-1} > \sqrt[3]{2} gives s_{n} \ge 4 \forall n > 2.

    And this is obviously something that can be repeated ad nauseam. I have yet to decide how useful it is though (if at all).
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  8. #8
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    Re: Proof of Convergence

    Ah! This is the trick I missed. Thanks a lot for your efforts.
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