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Math Help - limit of a sequence -> possibly squeeze theorem

  1. #1
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    limit of a sequence -> possibly squeeze theorem

    http://img96.imageshack.us/img96/8606/qqqqbj.jpg


    so what i'm thinking is that you let


    x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1


    then show that the first term can eventually go to the last term


    squeezing the term into the middle


    but the problem with this is i dont see where the 1/3 comes from


    any ideas?
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  2. #2
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    Re: limit of a sequence -> possibly squeeze theorem

    Quote Originally Posted by Angela11 View Post
    so what i'm thinking is that you let


    x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1
    What do you mean by "let"? One uses the word "let" during a definition: e.g., let w = 6. Why do you think these inequalities are true?

    Quote Originally Posted by Angela11 View Post
    any ideas?
    One way is to find an explicit formula for x_n. It is convenient to define a_n and b_n so that xₙ = (aₙx₁ + bₙx₂) / 2ⁿ⁻. Then it is possible to write recurrence relations and initial conditions on aₙ and bₙ, solve them and find the limits of aₙ / 2ⁿ⁻ and bₙ / 2ⁿ⁻.
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  3. #3
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    Re: limit of a sequence -> possibly squeeze theorem

    The way to do simple linear recursions is to try x_n = ar^n. This is very much analogous to ODEs, where you'd try y(x) = Ae^{f(x)}, seeing what f must be, and then using superposition to add together the different possibilities for f.

    Solve: x_{n+2} = \frac{x_{n+1} + x_n}{2}, n \in \mathbb{N}.

    Solution: Try x_n = ar^n for some constants a, r \in \mathbb{C}

    From x_{n+2} = \frac{x_{n+1} + x_n}{2} get  \ ar^{n+2} = \frac{ar^{n+1} + ar^n}{2}, so 2r^2 = r + 1.

    Solving that gives r = 1 or r = -1/2. Now just like with ODEs, put them together to get the general form of the solution:

    x_n = a_0(1)^n + a_1(\frac{-1}{2})^n = a_0 + \frac{a_1}{(-2)^n}.

    Note the required 2 unknown constants for a degree 2 linear recursion.

    When you plug that in to the recursion equation, it works. It's just algebra - I won't work it out here. Thus we have the generic solution to that recursion.

    Just like with ODEs, we use the initial conditions to get the specific solution:

    x_1 = a_0 + (\frac{a_1}{(-2)^1}) = a_0 - a_1/2, and x_2 = a_0 + (\frac{a_1}{(-2)^2}) = a_0 + a_1/4. Thus

    2x_1 = 2a_0 - a_1, and 4x_2 = 4a_0 + a_1, which add to give 2x_1 + 4x_2 = 6a_0, so a_0 = \frac{x_1 + 2x_2}{3}.

    To solve the problem, we don't even need to bother to find a_1, though it would be simple to do.

    \underset{n \rightarrow \infty}{\lim} x_n = \underset{n \rightarrow \infty}{\lim} \left(a_0 + \frac{a_1}{(-2)^n}\right) = a_0 = \frac{x_1 + 2x_2}{3}
    Last edited by johnsomeone; September 18th 2012 at 04:03 AM.
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