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Math Help - Generating Function (Finding An)

  1. #1
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    Generating Function (Finding An)

    Discrete Mathematics with Graph Theory
    by Goodaire and Parmenter

    Page 182 # 10b

    an = -10an-1 - 25an-2, n ≥ 2, given a0=1, a1 = 25

    So far I have:

    f(x) - a0 -a1x = -10x(f(x) - a0) - 25x2f(x)
    f(x) - 1 -25x = -10x(f(x) - 1) - 25x2f(x)

    Then solving for f(x)

    I get:
    f(x) = (1+35x)/(1+10X+25x2)
    f(x) = (1+35x)/(1+5x)2

    Then when I do partial fractions I get:
    (1+35x) = (A/(5x+1)^2) + (B/(5x+1))

    Then solving for A & B I get:
    A = 7
    B = -6

    And now I'm having trouble finding my an coefficient.
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  2. #2
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    Re: Generating Function (Finding An)

    Hello, ThatPinkSock!

    I am unfamiliar with your method.
    I'll show you the procedure I was taught.


    a_n \:=\:-10a_{n-1}- 25a_{n-2},\;\; n \ge2,\;\;a_0 = 1,\;a_1 = 25

    We have: . a_n + 10a_{n-1} + 25a_{n-2} \:=\:0

    Let: x^n = a_n

    Substitute: . x^n +10x^{n-1} + 25x^{n-2} \:=\:0

    Divide by x^{n-2}\!:\;\;x^2 + 10x + 25 \:=\:0 \quad\Rightarrow\quad (x+5)(x+5) \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}5,\,\text{-}5

    The function has the form: . f(n) \:=\:A(\text{-}5)^n + Bn(\text{-}5)^n


    Substitute the first two terms:

    . . f(0) \,=\, 1\!:\;\;A(\text{-}5)^0 + B(0)(\text{-}5)^0 \:=\:1 \quad\Rightarrow\quad A \:=\:1

    . . f(1)\,=\,25\!:\;\;A(\text{-}5)^1 + B(1)(\text{-}5)^1 \:=\:25 \Rightarrow\quad \text{-}5A - 5B \:=\:25

    . . . . . . . . . . . \text{-}5(1) - 5B \:=\:25 \quad\Rightarrow\quad \text{-}5B \:=\:30 \quad\Rightarrow\quad B \:=\:\text{-}6



    Therefore: . f(n) \:=\:(\text{-}5)^n - 6n(\text{-}5)^n

    . . . . . . . . . . a_n \;=\;(\text{-}5)^n(1 - 6n)
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  3. #3
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    Re: Generating Function (Finding An)

    Hi ThatPinkSock,

    For the generating function approach, use the series
     \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n to expand  \frac{B}{5x-1}
    and use the series  \frac{1}{(1-x)^2} = \sum_{n=0}^\infty (n+1) x^n to expand   \frac{A}{(5x-1)^2} .

    Here's one part:  \frac{B}{5x-1} = \frac{-B}{1-5x} = -B \sum_{n=0}^\infty (5x)^n

    I'll leave the rest for you...
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