# linear recurrence sequences

• Feb 13th 2010, 12:26 PM
scouse280
linear recurrence sequences
i am having trouble with this sequence:

x1=3 xn+1=0.8 xn-5 (n=1,2,3....)

how do i find a closed form, also in the tenth term correct to 4SF

Can someone explain how can i describe a long term behaviour of the sequence

many thanks
rich
• Feb 13th 2010, 01:09 PM
TKHunny
One way is to work it out. Write down a few terms and see what emerges in your mind.

3
-2.6
-7.08
...

Okay, that is totally not what I meant.

n=1) 3

n=2) 3*(4/5) - 5

n=3) (4/5)[3*(4/5) - 5] - 5 = 3*(4/5)^2 - 5(4/5) - 5

I'm beginning to see it. Are you? Let's just guess what the next one will be.

n=4) 3*(4/5)^3 - 5*(4/5)^2 - 5(4/5) - 5

And the nth one? (Feel free to prove this by induction.)

n) 3*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5

n) 8*(4/5)^(n-1) - 5*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5

Some algebra

n) 8*(4/5)^(n-1) - 5*[(4/5)^(n-1) + (4/5)^(n-2) + ... + (4/5)^2 + (4/5) + 5]

A little more algebra

n) 8*(4/5)^(n-1) - 5*[(1 - (4/5)^n)/(1-(4/5))

Simplify

n) $x_{n}\;=\;35\left[\frac{4}{5}\right]^{n} - 25$

I think I'm done. Of course, you're not. You still have to prove the formula.

Note: There is a better way. It starts with simply assuming the form of the final result and solving for a few parameters. In this case, there is a hint. Rewrite the formula a little:

$x_{n+1}\;=\;\frac{4x_{n}-25}{5}$

Compare this to the final result. From now on, with a first order recursion and a constant

$x_{n+1}\;=\;\frac{a}{b}x_{n} + c\;=\;\frac{ax_{n}+bc}{b}$

You say to yourself, it must look like this.

$x_{n}\;=\;D\cdot \left[\frac{a}{b}\right]^{n} + bc$

Use your initial values to solve for D and you are rather done.
• Feb 13th 2010, 01:27 PM
scouse280
trying to figure it out meself
i presumed that u1=5*(0.4)n-1

so find the 12tth term would be u12= 5(0.4)12

i guess i am totally off the mark with this one
• Feb 13th 2010, 02:04 PM
scouse280
help
• Feb 13th 2010, 03:53 PM
math2009
$

x_{n+1}=0.8x_{n}-5\rightarrow \vec{y}_{n+1}=
\begin{bmatrix} x_{n+1} \\ 1 \end{bmatrix}=
\begin{bmatrix} 0.8x_{n}-5 \\ 1 \end{bmatrix}=
\begin{bmatrix} 0.8 &-5 \\ 0 &1 \end{bmatrix}
\begin{bmatrix} x_{n} \\ 1 \end{bmatrix}=
A\vec{y}_{n}

$

$

f_{A}(\lambda)=\det\begin{bmatrix} A-\lambda I_{2} \end{bmatrix}=0
\rightarrow \lambda_{1,2}=\{~0.8~,~1~\}

$

$
\vec{v}_1\in \ker(A-\lambda_{1}I_{2})=span(\vec{e}_1)~,~
\vec{v}_2\in \ker(A-\lambda_{2}I_{2})=span(\begin{bmatrix} -25 \\ 1 \end{bmatrix})
$

$

\vec{y}_{1}=\begin{bmatrix} 3 \\ 1 \end{bmatrix}=
\begin{bmatrix} \vec{v}_{1} &\vec{v}_{2} \end{bmatrix}
(\begin{bmatrix} \vec{v}_{1} &\vec{v}_{2} \end{bmatrix}^{-1}\vec{y}_{1})=28\vec{v}_{1} + \vec{v}_{2}

$

$

\vec{y}_{n}=A^{n-1}\vec{y}_{1}=
A^{n-1}(28\vec{v}_{1} + \vec{v}_{2})=
28\lambda_{1}^{n-1}\vec{v}_{1} + \lambda_{1}^{n-1}\vec{v}_{2}=
28(0.8)^{n-1}\vec{e}_{1}+1^{n-1}\begin{bmatrix} -25 \\ 1 \end{bmatrix}

$

$

=\begin{bmatrix} 28(0.8)^{n-1}-25 \\ 1 \end{bmatrix}~,~
x_{n}=28(0.8)^{n-1}-25

$
• Feb 19th 2010, 04:13 PM
sarandor
As this is part of question 2 on the first Assignment with the MST121 course at the open uni I am guessing you are on this course, it would be worth contacting your tutor for support. That is what they are there for..