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Math Help - Find solution to a recurrence relation...

  1. #1
    Junior Member
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    Find solution to a recurrence relation...

    I am stuck on this question, and do not know where to head with it.

    Find the solution to An= 2An-1 + An-2 - 2An-3 for n=2,3,4,5......, with a0=3, a1=6 and a2=0

    (Note that n-1, n-2, n-3 are subscripts)
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  2. #2
    MHF Contributor
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    Hi

    Well it depends by your level in maths
    The characteristic polynomial is x^3-2x^2-x+2 whose roots are -1, 1 and 2

    -1)^n + \beta \: 1^n + \gamma \: 2^n" alt="a_n = \alpha \-1)^n + \beta \: 1^n + \gamma \: 2^n" />

    \alpha, \beta, \gamma values are found by the values of a_0, a_1, a_2

    -1)^n + 6 \:\: 1^n - 2^n" alt="a_n = -2 \-1)^n + 6 \:\: 1^n - 2^n" />
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  3. #3
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    Thank you

    Would I approach An = An-1 -n, a0=4 the same way?
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  4. #4
    MHF Contributor
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    No
    A_n - A_{n-1} = -n

    ....

    A_1 - A_0 = -1

    By summation
    A_n - A_0 = -\sum_{k=1}^n\:k = -\frac{n(n+1)}{2}

    A_n = 4-\frac{n(n+1)}{2}
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