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Thread: Solve recurrence: a_n=s*a_{n+1}-s*a_{n}

  1. #1
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    Solve recurrence: a_n=s*a_{n+1}-s*a_{n}

    I feel like I am missing something to this question.

    Let $\displaystyle s$ be a real number, $\displaystyle s \ne 0$. Find a sequence $\displaystyle a$ such that $\displaystyle a_n=s \Delta a_n$ and $\displaystyle a_0=1$.

    My answer (so far):
    $\displaystyle a_n = s \Delta a_n = sa_{n+1}-sa_n\; \Rightarrow\; a_{n+1}=\frac{1+s}{s}a_n$.

    Changing indexes:

    $\displaystyle a_n=\frac{1+s}{s}a_{n-1}$
    $\displaystyle a_0=1$
    $\displaystyle a_1=\left( \frac{1+s}{s} \right) \left( 1 \right) = \frac{1+s}{s}$
    $\displaystyle a_2=\left( \frac{1+s}{s} \right) \left( \frac{1+s}{s} \right) = \left( \frac{1+s}{s} \right)^2$
    $\displaystyle a_3=\left( \frac{1+s}{s} \right) \left( \frac{1+s}{s} \right)^2 = \left( \frac{1+s}{s} \right)^3$
    $\displaystyle ... $
    $\displaystyle \mathbf{a_n= \left( \frac{1+s}{s} \right)^n}$

    Is that really all they're looking for?
    Thanks.
    Last edited by MSUMathStdnt; Mar 15th 2011 at 08:51 PM.
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  2. #2
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    Quote Originally Posted by MSUMathStdnt View Post
    I feel like I am missing something to this question.

    Let $\displaystyle s$ be a real number, $\displaystyle s \ne 0$. Find a sequence $\displaystyle a$ such that $\displaystyle a_n=s \Delta a_n$ and $\displaystyle a_0=1$.

    My answer (so far):
    $\displaystyle a_n = s \Delta a_n = sa_{n+1}-sa_n\; \Rightarrow\; a_{n+1}=\frac{1+s}{s}a_n$.

    Changing indexes:

    $\displaystyle a_n=\frac{1+s}{s}a_{n-1}$
    $\displaystyle a_0=1$
    $\displaystyle a_1=\left( \frac{1+s}{s} \right) \left( 1 \right) = \frac{1+s}{s}$
    $\displaystyle a_2=\left( \frac{1+s}{s} \right) \left( 1 \right) = \left( \frac{1+s}{s} \right)^2$
    $\displaystyle a_3=\left( \frac{1+s}{s} \right) \left( 1 \right)^2 = \left( \frac{1+s}{s} \right)^3$
    $\displaystyle ... $
    $\displaystyle \mathbf{a_n= \left( \frac{1+s}{s} \right)^n}$

    Is that really all they're looking for?
    Thanks.
    Well, your notation for the a values is a little off, but your final formula is correct.

    -Dan
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  3. #3
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    Hello, MSUMathStdnt!

    $\displaystyle a_n \:=\:s\!\cdot\!a_{n+1} - s\!\cdot\!a_n,\;\;a_o = 1$

    If there are no typos in the problem, the solution is elementary.


    We have: .$\displaystyle a_n + s\!\cdot\!a_n \;=\;s\!\cdot\!a_{n+1} \quad\Rightarrow\quad (s+1)a_n \:=\:a_{n+1} $

    . . Hence: .$\displaystyle a_{n+1} \:=\:\frac{s+1}{s}\,a_n$


    Each term is a constant ratio of the preceding term.

    . . We have a geometric series.

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  4. #4
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    Quote Originally Posted by topsquark View Post
    Well, your notation for the a values is a little off, but your final formula is correct.

    -Dan
    I think I fixed the notation you were referring to. I used copy and paste to type that up and forgot to change some lines.


    Quote Originally Posted by Soroban View Post
    Hello, MSUMathStdnt!


    If there are no typos in the problem, the solution is elementary.

    ...

    Each term is a constant ratio of the preceding term.

    . . We have a geometric series.

    Thanks, Soroban.

    My book calls that a geometric sequence. The geometric series would be the sum, from $\displaystyle 0$ to $\displaystyle n$, of $\displaystyle a_n$. The sum could then be replaced by:

    $\displaystyle \frac{1-\left( \frac{1+s}{s} \right)^n}{1-\frac{1+s}{s}}=\left( \frac{1+s}{s} \right)^n-1$
    Last edited by MSUMathStdnt; Mar 15th 2011 at 11:49 PM.
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