# Thread:

1. ## 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?

2. exactly, and after you get the constant coefficients you can solve the homogeneous equation to find Qn.

3. just a quick question, when solving them simultaneously, would i just have

18A + 11B + 7C = 0
4A + 3B + C = 0

4. you have to have 3 equations in order to find an unambiguous solution for the constants.

5. Ok, so ill add 7A +4B +3C to that, but how do i re-arrange for solving?

6. Originally Posted by brd_7
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

7. 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..

8. $<br /> <br /> AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0<br />$

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:

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

9. $<br /> <br /> AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0<br />$

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.

10. Ok, i should be able to do the rest then. Thanks for the help
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