# Recurrence relation

• May 7th 2008, 11:08 PM
ctwi001
Recurrence relation
I have the following problem to solve:

a_n = 4a_{n-1} - 4a_{n-2}
where a_0 = -5 and a_1 = 3

which I have attempted to solve via substitution, ie a_n = r^n

When I do this, I end up with the result (r-2)^2 = 0
hence, r = 2.

When I put this back into the equation, I then get
A + B = -5
2A + 2B = 2(A + B) = 3
=> 0 = 13.

What am I fumbling in here??
• May 7th 2008, 11:45 PM
Isomorphism
Quote:

Originally Posted by ctwi001
I have the following problem to solve:

a_n = 4a_{n-1} - 4a_{n-2}
where a_0 = -5 and a_1 = 3

which I have attempted to solve via substitution, ie a_n = r^n

When I do this, I end up with the result (r-2)^2 = 0
hence, r = 2.

When I put this back into the equation, I then get
A + B = -5
2A + 2B = 2(A + B) = 3
=> 0 = 13.

What am I fumbling in here??

When there are two distinct roots $r_1$ and $r_2$, the general solution has the form $Ar_1 ^n + Br_2 ^n$.

But when $r_1 = r_2=r$ ,the general solution is $Ar^n + Bnr^n$.
• May 8th 2008, 12:06 AM
ctwi001
Thanks
Thank you so much - I had been trying to come up with a pattern like that; and the text books etc didn't give that example of non-distinct roots.

(Clapping)