# Recurrence relation

• May 7th 2008, 10:08 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, 10:45 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 \$\displaystyle r_1\$ and \$\displaystyle r_2\$, the general solution has the form \$\displaystyle Ar_1 ^n + Br_2 ^n\$.

But when \$\displaystyle r_1 = r_2=r\$ ,the general solution is \$\displaystyle Ar^n + Bnr^n\$.
• May 7th 2008, 11:06 PM
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)