- Thread starter
- #1

#### LLand314

##### Member

- Mar 26, 2013

- 77

everytime I get 550 for a1 a2 a3 and a4

is that correct or am I doing it wrong.

- Thread starter LLand314
- Start date

- Thread starter
- #1

- Mar 26, 2013

- 77

everytime I get 550 for a1 a2 a3 and a4

is that correct or am I doing it wrong.

- Thread starter
- #3

- Mar 26, 2013

- 77

It just seemed to easy for Calculus thats all.

Thanks

Thanks

- Admin
- #4

\(\displaystyle a_{n+1}=0.4a_{n}+330\)

\(\displaystyle a_{n+2}=0.4a_{n+1}+330\)

We find the homogenous recursion via differencing:

\(\displaystyle a_{n+2}=1.4a_{n+1}-0.4a_{n}\)

whose associated characteristic roots are:

\(\displaystyle r=\frac{2}{5},\,1\)

and hence the closed form is:

\(\displaystyle A_n=k_1+k_2\left(\frac{2}{5} \right)^n\)

Now, using:

\(\displaystyle A_0=A_1=550\), we may determine:

\(\displaystyle k_1=550,\,k_2=0\) and so:

\(\displaystyle A_n=550\)

- Feb 13, 2012

- 1,704

Using the procedure described in...

everytime I get 550 for a1 a2 a3 and a4

is that correct or am I doing it wrong.

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... the difference equation can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n} = 330 - .6\ a_{n} = f(a_{n})$ (1)

... and the function f(*) is represented here...

There is only one attractive fixed point in x=550 and, because the linearity of f(*) the stable point is met at the first step...

Kind regards

$\chi$ $\sigma$