Calculating Terms of Sequence: a1,a2,a3,a4

[Logan Land]

write the terms a1,a2,a3,a4 of the following sequence. an+1=0.4an+330, a0=550

everytime I get 550 for a1 a2 a3 and a4
is that correct or am I doing it wrong.

 

[tkhunny]

550 * 0.4 = 220

220 + 330 = 550

Pretty clear. Why do you doubt?

 

[Logan Land]

It just seemed to easy for Calculus that's all.

Thanks

 

[MarkFL]

If we write the inhomogeneous recursions:

$$a_{n+1}=0.4a_{n}+330$$

$$a_{n+2}=0.4a_{n+1}+330$$

We find the homogenous recursion via differencing:

$$a_{n+2}=1.4a_{n+1}-0.4a_{n}$$

whose associated characteristic roots are:

$$r=\frac{2}{5},\,1$$

and hence the closed form is:

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

Now, using:

$$A_0=A_1=550$$, we may determine:

$$k_1=550,\,k_2=0$$ and so:

$$A_n=550$$

 

[chisigma]

write the terms a1,a2,a3,a4 of the following sequence. an+1=0.4an+330, a0=550

everytime I get 550 for a1 a2 a3 and a4
is that correct or am I doing it wrong.

Using the procedure described in...

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...

http://www.123homepage.it/u/i68681865._szw380h285_.jpg.jfif
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

$\chi$ $\sigma$