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Terms of a sequence

LLand314

Member
Mar 26, 2013
77
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

Well-known member
MHB Math Helper
Jan 27, 2012
267
550 * 0.4 = 220

220 + 330 = 550

Pretty clear. Why do you doubt?
 

LLand314

Member
Mar 26, 2013
77
It just seemed to easy for Calculus thats all.

Thanks
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If we write the inhomogeneous recursions:

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

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

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:

\(\displaystyle A_n=550\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
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...





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

$\chi$ $\sigma$