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Thread: Recurrence Relation

  1. #1
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    Recurrence Relation

    Could anyone direct me on how to solve these? Im stuck. (n-1 are subscripts)

    An=2An-1 -3, a0 = -1
    An=(n+1)An-1, a0=2
    An=2nAn-1, a0=3
    An=-An-1 + n - 1, a0 = 7

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  2. #2
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    Even a general method would help! Thanks
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  3. #3
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    Hello, vexiked!

    I'll walk through the first one.


    Could anyone direct me on how to solve these?

    $\displaystyle (1)\;A_n\:=\:2A_{n-1} -3,\quad a_0 = -1$

    We conjecture that $\displaystyle A_n$ is of the form: .$\displaystyle X^n$, an exponential expression.

    $\displaystyle \begin{array}{ccccc}\text{Then we have:} & X^n &=& 2X^{n-1} - 3 & {\color{blue}[1]} \\
    \text{Write the "next" term:} & X^{n+1} &=& 2X^n - 3 & {\color{blue}[2]}\end{array}$

    Subtract [2] - [1]: .$\displaystyle X^{n+1} - X^n \:=\:2X^n - 2X^{n-1} \quad\Rightarrow\quad X^{n+1} - 3X^n + 2X^{n-1} \:=\:0 $

    Divide by $\displaystyle X^{n-1}\!:\;\;X^2 - 3X + 2 \:=\:0 \quad\Rightarrow\quad (X-1)(X-2) \:=\:0 \quad\Rightarrow\quad X \:=\:1,\:2$

    . . Hence, the function has: .$\displaystyle 1^n\text{ and }2^n$


    Form a linear combination of the two roots: .$\displaystyle f(n) \;=\;A\cdot1^n + B\cdot2^n$

    . . and we have: .$\displaystyle f(n) \;=\;A + B\!\cdot\!2^n$


    Use the first two terms of the sequence:

    . . $\displaystyle \begin{array}{ccccccccc}f(0) = \text{-}1\!: & A + B\!\cdot\!2^0 &=& \text{-}1 & \Rightarrow & A + B &=& \text{-}1 & {\color{blue}[3]} \\
    f(1) = \text{-}5\!: & A + B\!\cdot\!2^1 &=&\text{-}5 & \Rightarrow & A + 2B &=& \text{-}5 & {\color{blue}[4]} \end{array}$

    Subtract [4] - [3]: .$\displaystyle B \:=\:\text{-}4$

    Substitute into [3]: .$\displaystyle A - 4 \:=\:\text{-}1 \quad\Rightarrow\quad A \:=\:3$


    Therefore: .$\displaystyle f(n) \:=\:3 - 4\!\cdot\!2^n \quad\Rightarrow\quad\boxed{ f(n) \:=\:3 - 2^{n+2}}$

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