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Math Help - recurrence equation question

  1. #1
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    recurrence equation question

    Hi,

    I was hoping that someone would kindly help me with the last part of this question. I've found the general solution, however I have no idea how to approach the last part 'show that the limiting value..."

    Many thanks!
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  2. #2
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    Re: recurrence equation question

    \text{Have: }

    p_t = (1+r)p_{t-1} + k (q^D(p_{t-1}) - q^S(p_{t-1})), \ q^D(p) = 3 - p, \ q^S(p) = p \ p_0 = 1.

    \text{Thus } p_t = (1+r)p_{t-1} + k ( \ (3 - p_{t-1}) - (p_{t-1}) \) = (1+r)p_{t-1} + k(3 - 2p_{t-1})

    = (1+r)p_{t-1} + 3k - 2kp_{t-1} = 3k + (1+r - 2k)p_{t-1}.

    \text{So your formula is } p_t = 3k + (1+r - 2k)p_{t-1}.

    \text{Since that's not homogeneous, do the following trick to "divvy up" 3k equally to both terms:}

    \text{If }(p_t+x) = (1+r - 2k)(p_{t-1}+x),

    \text{then } p_t = [-x + (1+r - 2k)x] + (1+r - 2k)p_{t-1},

    \text{so solve  } -x + (1+r - 2k)x = 3k \text{ to get } x = \frac{3k}{r-2k} \text{ for } r \ne 2k.

    \text{Let } a_t = p_t + x. \text{ Then } a_t = (1+r - 2k)a_{t-1}.

    \text{That can be easily and explicitly solved for } a_t,

    \text{ and so thus for } p_t = a_t - x.

    \text{Also note that the } r = 2k \text{ case is } p_t = 3k + p_{t-1},

    \text{which is both also explicitly solveable - and very divergent.}

    \text{Maybe try it again from here?}
    Last edited by johnsomeone; October 19th 2012 at 05:21 PM.
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  3. #3
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: recurrence equation question

    p(t)=\frac{3 k}{2 k-r}-\frac{(k+r) (-2 k+r+1)^t}{2 k-r}

    Convergence occurs for:

    |-2 k+r+1|\leq 1

    Simplify to obtain:

    r/2 < k\leq 1+r/2

    then for k=1+r/2:

    p(t)=\frac{3 (2+r)}{4}-\frac{1}{4} (-1)^t (2+3 r) = \left(2+\frac{3 r}{2}\right)

    and,

    \lim_{k\to \frac{r}{2}} \, p(t)\to  1+\frac{3 r t}{2}
    Last edited by MaxJasper; October 19th 2012 at 06:17 PM.
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  4. #4
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    Re: recurrence equation question

    \text{Assuming that solution, convergence fails at }-2k+r+1 = -1 \text

    \text{when }k + r \ne 0.

    p_t \text{ oscillates between } 1 \text{ and } \frac{4k+r}{2k - r} \text{ when } k = 1 + \frac{r}{2}.
    Last edited by johnsomeone; October 19th 2012 at 06:33 PM.
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