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Math Help - Recursive Formula

  1. #1
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    Recursive Formula

    Given: asub0 = 1 asub1 = 3

    (asubn)^2 - asubn-1 * asubn+1 = (-1)^n for n greater or equal to 1

    find asub3

    i simplified it to (square root (n+(asubn-1)*(asubn+1) )= asubn

    and i got asub2 = 8 and asub3 = 20.666...
    Last edited by stones44; March 10th 2008 at 06:06 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stones44 View Post
    Given: asub0 = 1 asub1 = 3

    (asubn)^2 - asubn-1 * asubn+1 = (-1)^n for n greater or equal to 1

    find asub3

    i simplified it to (square root (n+(asubn-1)*(asubn+1) )= asubn

    and i got asub2 = 8 and asub3 = 20.666...
    a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n
    a_0 = 1 and a_1 = 3

    Am I reading something wrong? I get
    a_n = \frac{a_{n - 1}^2 + (-1)^n}{a_{n - 2}}

    (Hint: Solve the problem for a_{n + 1}, not a_n.)

    -Dan
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  3. #3
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    Hello, stones44!

    It is a rather confusing recurrence.
    It took a while to straighten it out . . .


    Given: . -1)^n\quad\text{for }n \geq 1" alt="a_0 = 1,\;\;\;a_1 = 3.\;\;\;(a_n)^2 - (a_{n-1})\cdot(a_{n+1}) \:= \-1)^n\quad\text{for }n \geq 1" />

    \text{Find }\:a_3
    Solve for a_{n+1}\!:\;\;a_{n+1} \;=\;\frac{(a_n)^2 - (-1)^n}{a_{n-1}}


    For n = 1, we have: . a_2 \;=\;\frac{(a_1)^2 - (-1)^1}{a_0} \;=\;\frac{3^2 +1}{1}\quad\Rightarrow\quad a_2 \:=\:10

    For n=2, we have: . a_3 \;=\;\frac{(a_2)^2 - (-1)^2}{a_1} \;=\;\frac{10^2 - 1}{3} \quad\Rightarrow\quad\boxed{a_3 \:=\:33}

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  4. #4
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    can you write out the algebra from my step to isolating asub n+1
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