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Math Help - recursive sequence

  1. #1
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    recursive sequence

    Hi

    i have the recursive sequence which is defined this way:

    A(1) = p
    A(2) = q

    A(n) = 0.5[A(n-1) + A(n-2)]

    and i need to prove convergence and find the limit

    it is very easy to prove convergence of the sequence from cantor's theorem

    it is also possible to see that the limit is L=[p+2q]/3 but i cant manage to prove that this is the limit

    please help im going crazy
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  2. #2
    MHF Contributor
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    Quote Originally Posted by omer.jack View Post
    Hi

    i have the recursive sequence which is defined this way:

    A(1) = p
    A(2) = q

    A(n) = 0.5[A(n-1) + A(n-2)]

    and i need to prove convergence and find the limit

    it is very easy to prove convergence of the sequence from cantor's theorem

    it is also possible to see that the limit is L=[p+2q]/3 but i cant manage to prove that this is the limit

    please help im going crazy
    If we assume the solution of the difference equation is of the form

    a_n = c \rho^n

    then substituting into

    2a_n - a_{n-1} - a_{n-2} = 0

    gives

    2 \rho^2 - \rho - 1 = 0

    which has two solutions: \rho = 1,\;\; -\frac{1}{2}

    so the general solution of your difference equation is

    a_n = c_1 + c_2 \left( - \frac{1}{2} \right)^n

    Using the fact that a_1 = p,\;\;\;a_2 = q gives c_1 = \frac{1}{3}(p+3q)\;\;\;c_2 = - \frac{4}{3}(p-q) so

    a_n = \frac{1}{3}(p+3q) - \frac{4}{3}(p-q) \left( - \frac{1}{2} \right)^n

    Now as n\, \rightarrow\, \infty the last term drops giving a_n \rightarrow \frac{1}{3}(p+3q)
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  3. #3
    Super Member PaulRS's Avatar
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    Okay, we have 2a_{n+2}=a_{n+1}+a_n thus: 2(a_{n+2}-a_{n+1})=a_n-a_{n+1}

    So define b_n=a_{n+1}-a_{n} to get: 2b_{n+1}=-b_{n}

    And that can be easily solved. b_{n}=\left(- \tfrac{1}{2}\right)\cdot{b_{n-1}}=...=\left(- \tfrac{1}{2}\right)^n\cdot{b_{0}}

    Now sum: a_n-a_0=\sum_{k=0}^{n-1}{\left(a_{k+1}-a_k\right)}= \sum_{k=0}^{n-1}b_k=b_0 \cdot \sum_{k=0}^{n-1}\left(- \tfrac{1}{2}\right)^k

    And now this is just a geometric sum

    In fact we have: \lim_{n\to +\infty}a_n=a_0+b_0 \cdot \sum_{k=0}^{\infty}\left(- \tfrac{1}{2}\right)^k=a_0+b_0 \cdot\frac{1}{1+\tfrac{1}{2}}
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  4. #4
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    thanks guys that was very helpful
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