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Math Help - Solve the following inhomogeneous recurrence relation...

  1. #1
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    Solve the following inhomogeneous recurrence relation...

    Solve the inhomogeneous recurrence relation

    an=3an-1+n2-3 with a0=1

    Any help please? I tried a few different cases which got me nowhere.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Solve the following inhomogeneous recurrence relation...

    You may use the process of symbolic differencing to get a homogeneous recurrence. We are given:

    (1) a_{n}=3a_{n-1}+n^2-3

    Now, replace n with n+1 to get:

    (2) a_{n+1}=3a_{n}+(n+1)^2-3

    Now, subtract (1) from (2) to get another recurrence. Repeat this process until you have a homogeneous recurrence, from which you may determine the closed form from the characteristic roots, and then determine the parameters from the initial values.
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  3. #3
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    Re: Solve the following inhomogeneous recurrence relation...

    Think of recurrence equations as being like a discrete type of differential equation.

    1) Solve the homogeneous case: \tilde{a}_n = 3\tilde{a}_{n-1} to get \tilde{a}_n = c3^n.

    2) Now use variation of parameters. Try: a_n = 3^nb_n \ (\text{with }1 = a_0 = 3^0b_0 = b_0).

    a_n = 3a_{n-1} + n^2-3 \text{ becomes } 3^nb_n = 3(3^{n-1}b_{n-1}) +n^2-3 = 3^nb_{n-1}+ n^2-3,

    \text{giving } b_n = b_{n-1}+ \frac{n^2-3}{3^n}, \text{ with }b_0 = 1.

    \text{That's solveable as a sum: } b_n = b_0 + \sum_{k=1}^n \frac{k^2-3}{3^k} \text{ for } n \ge 1.

    You can further simply that.
    Last edited by johnsomeone; October 29th 2012 at 12:25 PM.
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