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Math Help - Difference Equations.. Again :D

  1. #1
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    Difference Equations.. Again :D

    Given that the following sequence, Q_{n} , satisfies a homogeneous second order difference equation with constant coefficients find Q_{n} :

     Q_{1} = 1, Q_{2} = 3, Q_{3} = 4, Q_{4} = 7, Q_{5} = 11, Q_{6} = 18, ...

    I reckon that it would have to start with  AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0 and do i then just put in the various numbers and solve them simultaneiously?
    Last edited by brd_7; January 26th 2008 at 06:40 AM.
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  2. #2
    Senior Member Peritus's Avatar
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    exactly, and after you get the constant coefficients you can solve the homogeneous equation to find Qn.
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  3. #3
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    just a quick question, when solving them simultaneously, would i just have

    18A + 11B + 7C = 0
    4A + 3B + C = 0
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  4. #4
    Senior Member Peritus's Avatar
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    you have to have 3 equations in order to find an unambiguous solution for the constants.
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  5. #5
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    Ok, so ill add 7A +4B +3C to that, but how do i re-arrange for solving?
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  6. #6
    Senior Member Peritus's Avatar
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    Quote Originally Posted by brd_7 View Post
    Ok, so ill add 7A +4B +3C to that, but how do i re-arrange for solving?
    I'm sorry if I sound rude, but if you deal with a topic in "advanced" algebra such as difference equations you must at least know how to solve a system of linear equations. If you don't know or forgot, there are plenty of resources on the internet such as:

    System of linear equations - Wikipedia, the free encyclopedia
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  7. #7
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    Nah you wern't rude by any means, its just i get given work that hasnt been explained properly, i then look at the notes and the examples are really poor, and so out of frustration i forget how to do the simplest of things.

    But, through simultaneous equations i keep getting -

    -5A - 5B = 0
    -5A - 5B = 0

    Which gets me nowhere.. Or at least i think..
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  8. #8
    Senior Member Peritus's Avatar
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    <br /> <br />
AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0<br />

    to get the first equation substitute n=3:


    AQ_3  + BQ_2  + CQ_1  = 0

    4A + 3B + C = 0<br />

    now n=4:


    \begin{gathered}<br />
  AQ_4  + BQ_3  + CQ_2  = 0 \hfill \\<br />
  7A + 4B + 3C = 0 \hfill \\ <br />
\end{gathered}

    now n=5:

    \begin{gathered}<br />
  AQ_5  + BQ_4  + CQ_3  = 0 \hfill \\<br />
  11A + 7B + 4C = 0 \hfill \\ <br />
\end{gathered}
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  9. #9
    Senior Member Peritus's Avatar
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    <br /> <br />
AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0<br />

    to get the first equation substitute n=3:


    AQ_3  + BQ_2  + CQ_1  = 0

    4A + 3B + C = 0<br />

    now n=4:


    \begin{gathered}<br />
  AQ_4  + BQ_3  + CQ_2  = 0 \hfill \\<br />
  7A + 4B + 3C = 0 \hfill \\ <br />
\end{gathered}

    now n=5:

    \begin{gathered}<br />
  AQ_5  + BQ_4  + CQ_3  = 0 \hfill \\<br />
  11A + 7B + 4C = 0 \hfill \\ <br />
\end{gathered}

    <br />
\left\{ {\begin{array}{*{20}c}<br />
   {4A + 3B + C = 0}  \\<br />
   {7A + 4B + 3C = 0}  \\<br />
   {11A + 7B + 4C = 0}  \\<br /> <br />
 \end{array} } \right.<br /> <br />

    as you can see the third equation is the sum of the forst two, so the equations are linearly dependent, that's why you had a problem.

    after solving the system we get that the general solution is:

    A = -B = -C

    let's choose A=1, B=C=-1

    thus:

    <br />
Q_n  - Q_{n - 1}  - Q_{n - 2}  = 0<br />

    you can notice that we've got the fibonacci sequence.
    Last edited by Peritus; January 26th 2008 at 09:42 AM.
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  10. #10
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    Ok, i should be able to do the rest then. Thanks for the help
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