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Math Help - 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 s be a real number, s \ne 0. Find a sequence a such that a_n=s \Delta a_n and a_0=1.

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

    Changing indexes:

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

    Is that really all they're looking for?
    Thanks.
    Last edited by MSUMathStdnt; March 15th 2011 at 09:51 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MSUMathStdnt View Post
    I feel like I am missing something to this question.

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

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

    Changing indexes:

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

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

    . . Hence: . 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 0 to n, of a_n. The sum could then be replaced by:

    \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; March 16th 2011 at 12:49 AM.
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