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Math Help - General solution to 3rd order recurrence relation

  1. #1
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    General solution to 3rd order recurrence relation

    Hi, I'm a bit stuck on this question:

    For a fixed r, show that the general solution to the recurrence relation

    x_{n} = rx_{n-1} + r^2x_{n-2} - r^3x_{n-3}

    is given by

     x_{n} = Ar^n + Bnr^n + C(-r)^n for constants A, B, C \in \mathbb{R}

    I've done this so far:

    Let x_{n-1} = a^n

    a^n = ra^{n-1} +r^2na^{n-2} -r^3xa^{n-3}

    Divide through by a^{n-3}

    a^3 = ra^2 + r^2na - r^3

    But I'm stuck now on getting those powers back to n. Any pointers most welcome!
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  2. #2
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    Quote Originally Posted by Salome View Post
    Hi, I'm a bit stuck on this question:

    For a fixed r, show that the general solution to the recurrence relation

    x_{n} = rx_{n-1} + r^2x_{n-2} - r^3x_{n-3}

    is given by

     x_{n} = Ar^n + Bnr^n + C(-r)^n for constants A, B, C \in \mathbb{R}

    I've done this so far:

    Let x_{n-1} = a^n

    a^n = ra^{n-1} +r^2na^{n-2} -r^3xa^{n-3}

    Divide through by a^{n-3}

    a^3 = ra^2 + r^2na - r^3

    But I'm stuck now on getting those powers back to n. Any pointers most welcome!
    Not quite sure where you got your, what is it called, a characteristic equation? Anyway, rewrite your recursion as
    x_{n + 3} - rx_{n + 2} - r^2x_{n + 1} + r^3x_n = 0

    So your characteristic (or whatever) is
    y^3 - ry^2 - r^2y + r^3 = 0

    You can guess that y = r is a solution (use a variant on the rational root theorem. Possible roots in terms of r are y = (+/-) (1, r, r^2, r^3). )

    So eventually we can show that
    y^3 - ry^2 - r^2y + r^3 = (y - r)^2(y + r) = 0

    Can you go from there?

    -Dan
    Last edited by topsquark; May 11th 2011 at 04:40 AM. Reason: Fixed some exponents
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  3. #3
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    Gotcha, thanks so much! Though I think here:

    y^2 - ry^2 - r^2y + r^3 = (y - r)^2(y + r) = 0

    the first y^2 should be y^3?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Salome View Post
    Gotcha, thanks so much! Though I think here:

    y^2 - ry^2 - r^2y + r^3 = (y - r)^2(y + r) = 0

    the first y^2 should be y^3?
    Yes, I have fixed that in my original post. Thanks for the catch!

    -Dan
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