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Math Help - Finding the generating function of a recurrence

  1. #1
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    Arrow Finding the generating function of a recurrence

    a_n = 5a_(n-1) - 6a_(n-2) for n>=2, a_0 = 1 and a_1 = c

    I got A(x) = 1 + (c+5)x / 1-5x+6x^2 and this is wrong.

    Help?
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  2. #2
    ynj
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    Quote Originally Posted by DaRush19 View Post
    a_n = 5a_(n-1) - 6a_(n-2) for n>=2, a_0 = 1 and a_1 = c

    I got A(x) = 1 + (c+5)x / 1-5x+6x^2 and this is wrong.

    Help?
    G(x)=\frac{1+(c-5)x}{1-5x+6x^2}
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  3. #3
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    Hello, DaRush19!

    Your function does not proeduce: a_1 \,=\, c


    a_n \;=\; 5a_{n-1} - 6a_{n-2}\quad\text{for }n\geq 2,\;\; a_0 \,=\, 1,\;\;a_1 \,=\, c
    I don't know what method you used . . .


    We have: . a(n) \;=\;5\cdot a(n\text{-}1) - 6\cdot a(n\text{-}2)

    Let X^n \:=\:a(n) . . . That is, assume a(n) is an exponential function.

    Then we have: . X^n \;=\;5X^{n-1} - 6X^{n-2} \quad\Rightarrow\quad X^n - 5X^{n-1} + 6X^{n-2} \;=\;0

    Divide by X^{n-2}\!:\quad X^2 - 5X + 6 \:=\:0 \quad\Rightarrow \quad (X - 2)(X - 3) \:=\:0 \quad\Rightarrow\quad X \:=\:2,3


    Form a linear combination of these roots: . a(n) \;=\;2^n\!\cdot\!A + 3^n\!\cdot\!B


    From the first two values, we have: . \begin{array}{ccccc}a(0) &=& A + B &=& 1 \\ a(1) &=& 2A + 3B &=& c \end{array}

    Solve the system of equations: . \begin{array}{c}A \;=\;3-c \\ B \;=\;c-2 \end{array}


    . . Therefore: . a(n) \;=\;(3-c)\!\cdot\!2^n + (c-2)\!\cdot\!3^n

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  4. #4
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    Quote Originally Posted by ynj View Post
    G(x)=\frac{1+(c-5)x}{1-5x+6x^2}
    Could you write down your working out for me please? That is indeed the correct answer, thanks!
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  5. #5
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    Actually I've got it. It was a ridiculous sign error.
    Thank you both for you help, much appreciated!
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