# Difference Equations.. Again :D

• January 26th 2008, 07:27 AM
brd_7
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?
• January 26th 2008, 07:43 AM
Peritus
exactly, and after you get the constant coefficients you can solve the homogeneous equation to find Qn.
• January 26th 2008, 07:50 AM
brd_7
just a quick question, when solving them simultaneously, would i just have

18A + 11B + 7C = 0
4A + 3B + C = 0
• January 26th 2008, 07:54 AM
Peritus
you have to have 3 equations in order to find an unambiguous solution for the constants.
• January 26th 2008, 07:56 AM
brd_7
Ok, so ill add 7A +4B +3C to that, but how do i re-arrange for solving?
• January 26th 2008, 08:04 AM
Peritus
Quote:

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
• January 26th 2008, 08:16 AM
brd_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..
• January 26th 2008, 08:33 AM
Peritus
$

AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0
$

to get the first equation substitute n=3:

$AQ_3 + BQ_2 + CQ_1 = 0$

$4A + 3B + C = 0
$

now n=4:

$\begin{gathered}
AQ_4 + BQ_3 + CQ_2 = 0 \hfill \\
7A + 4B + 3C = 0 \hfill \\
\end{gathered}$

now n=5:

$\begin{gathered}
AQ_5 + BQ_4 + CQ_3 = 0 \hfill \\
11A + 7B + 4C = 0 \hfill \\
\end{gathered}$
• January 26th 2008, 08:51 AM
Peritus
$

AQ_{n} + BQ_{n - 1} + CQ_{n - 2} = 0
$

to get the first equation substitute n=3:

$AQ_3 + BQ_2 + CQ_1 = 0$

$4A + 3B + C = 0
$

now n=4:

$\begin{gathered}
AQ_4 + BQ_3 + CQ_2 = 0 \hfill \\
7A + 4B + 3C = 0 \hfill \\
\end{gathered}$

now n=5:

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

$
\left\{ {\begin{array}{*{20}c}
{4A + 3B + C = 0} \\
{7A + 4B + 3C = 0} \\
{11A + 7B + 4C = 0} \\

\end{array} } \right.

$

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:

$
Q_n - Q_{n - 1} - Q_{n - 2} = 0
$

you can notice that we've got the fibonacci sequence.
• January 26th 2008, 08:52 AM
brd_7
Ok, i should be able to do the rest then. Thanks for the help