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Math Help - Recurrence relation

  1. #1
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    Recurrence relation

    Hello,

    I cannot solve the following problem:
    What can we say about the asymptotic behavior, depending on the initial conditions, of a recurrence relation that has a inhomogeneous part (log n)^(log n) and has roots 1,2,3 ?
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  2. #2
    Senior Member bkarpuz's Avatar
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    Hi aa2010,

    I am assuming that you are asking about the asymptotic nature of the difference equation
    (\mathcal{L}y)(n):=\Delta^{3}y(n)-3\Delta^{2}y(n)+2\Delta y(n)-6y(n)=\big(\ln(n)\big)^{\ln(n)} for n\geq n_{0}(\geq2)
    under the initial conditions
    \Delta^{2}y(n_{0})=a, \Delta y(n_{0})=b and y(n_{0})=c.
    Here, \Delta is the forward difference operator, i.e., \Delta y(n):=y(n+1)-y(n) and \Delta^{k}y(n):=\Delta^{k-1}\Delta y(n).
    For such problems, you need to compute the fundamental solutions, i.e.,
    the solutions of the following initial value problems
    \mathcal{Y}_{1}(n,m): (\mathcal{L}y)(n)=0 for n\geq m\geq n_{0} with \Delta^{2}y(m)=1, \Delta y(m)=0 and y(m)=0
    \mathcal{Y}_{2}(n,m): (\mathcal{L}y)(n)=0 for n\geq m\geq n_{0} with \Delta^{2}y(m)=0, \Delta y(m)=1 and y(m)=0
    \mathcal{Y}_{3}(n,m): (\mathcal{L}y)(n)=0 for n\geq m\geq n_{0} with \Delta^{2}y(m)=0, \Delta y(m)=0 and y(m)=1.
    Then, by the variation of constants formula, we can write the solution as follows
    y(n):=a\mathcal{Y}_{1}(n,n_{0})+b\mathcal{Y}_{2}(n  ,n_{0})+c\mathcal{Y}_{3}(n,n_{0})+\displaystyle\su  m_{k=n_{0}}^{n-1}\mathcal{Y}_{1}(n,k+1)\big(\ln(k)\big)^{\ln(k)} for n\geq n_{0}.
    Here, we have
    \mathcal{Y}_{1}(n,m)=\dfrac{1}{2}3^{n-m}-2^{n-m}+\dfrac{1}{2} for n\geq m\geq n_{0}
    \mathcal{Y}_{2}(n,m)=-\dfrac{1}{2}3^{n-m}+22^{n-m}-\dfrac{1}{3} for n\geq m\geq n_{0}
    and
    \mathcal{Y}_{3}(n,m)\equiv1 for n\geq m\geq n_{0}.
    Assume that n_{0}:=9, then \big(\ln(n)\big)^{\ln(n)}\geq n for n\geq9, which yields
    y(n)\geq a\mathcal{Y}_{1}(n,9)+b\mathcal{Y}_{2}(n,9)+c\math  cal{Y}_{3}(n,9)+\displaystyle\sum_{k=n_{0}}^{n-1}\mathcal{Y}_{1}(n,k+1)k
    ......... =a\mathcal{Y}_{1}(n,9)+b\mathcal{Y}_{2}(n,9)+c\mat  hcal{Y}_{3}(n,9)+\dfrac{19}{8}3^{n-9}-\dfrac{5}{2}2^{n-9}+\dfrac{1}{8}\big(2n^{2}+4n-137\big)
    for all n\geq9.
    Thus, if a/2-b/2+19/8>0 (the coefficients of the dominant term 3^{n-9}), then y(n)\to\infty as n\to\infty.
    Obtaining a complete result for your problem requires a deep inspection since the forcing term is not a nice one.
    Last edited by bkarpuz; March 30th 2011 at 12:14 AM.
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  3. #3
    Senior Member bkarpuz's Avatar
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    As I said, better inequalities will give you better results.
    Note that for n\geq 2, we have \big(\ln(n)\big)^{\ln(n)}\geq\dfrac{1}{n^{4}}3^{n+  1},
    then the contribution of the forcing term to the solution can not be less than
    \displaystyle\sum_{k=2}^{n-1}\mathcal{Y}_{1}(n,k+1)\frac{1}{k^{4}}3^{k+1} for all n\geq 2.
    As n\to\infty,
    \dfrac{1}{3^{n-2}}\displaystyle\sum_{k=2}^{n-1}\mathcal{Y}_{1}(n,k+1)\frac{1}{k^{4}}3^{k+1}\to\  dfrac{\pi^{4}}{10}-9\approx0.74,
    which implies that a/2-b/2+\pi^{4}/10-9>0 yields y(n)\to\infty as n\to\infty.
    Last edited by bkarpuz; March 30th 2011 at 12:46 AM.
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