Thread: Recurrence Relation HELP

1. Recurrence Relation HELP

1. In predicting future sales of a product, one assumption is to say that the amount sold next year will be the average of the amount sold this year and last year. Suppose that
an is the amount sold in year n.

(a) Find a recurrence relation for an.
(b) Solve the recurrence relation if a0 = a1 = 1.

What i have found so far is that "an = 2an+1 - an-1"

2. Originally Posted by sanorita_belle
1. In predicting future sales of a product, one assumption is to say that the amount sold next year will be the average of the amount sold this year and last year. Suppose that
an is the amount sold in year n.

(a) Find a recurrence relation for an.
(b) Solve the recurrence relation if a0 = a1 = 1.

What i have found so far is that "an = 2an+1 - an-1"

It is normal for recurences to be causal (that is the $n$-th term depends only on earlier terms:

$a_{n+1}=(a_n+a_{n-1})/2$

so:

$a_{n}=(a_{n-1}+a_{n-2})/2$

CB

3. Originally Posted by CaptainBlack
It is normal for recurences to be causal (that is the $n$-th term depends only on earlier terms:

$a_{n+1}=(a_n+a_{n-1})/2$

so:

$a_{n}=(a_{n-1}+a_{n-2})/2$

CB

How did you get $a_{n-2}$? Can you please explain a bit more? Thanks in advance!

4. Originally Posted by sanorita_belle
How did you get $a_{n-2}$? Can you please explain a bit more? Thanks in advance!
n is a dummy variable it can be replaced with anything you want so replace n by n-1

CB

5. I tried solving this recurrence relation, and i got two roots x=1, x= -1/2

then i wrote it in the general form of the solution as:

$a_{n} = A (1)^ n + B (-1/2)^ n$ for some constants A and B.

Then by substituting the given initial values and solving equations, i got A= 1, B=0

So according to this the solution should be, $a_{n} = 1^n$.

Does it make sense? Can i have a solution of the form $a_{n} = 1^n$ which means that solution is always 1?