# How to solve recurrence relation with one real root and two complex roots ?

1. Aug 24, 2010

### huifei

How to solve recurrence relation with one real root and two complex roots ???

The Example is ;

Solve the recurrence relation a n-1 + a n-3 = 0 where n ≥ 3 and a 1 = 1 a 2 = 1 a 3 = 2
a n = nth order
a n-1 = (n-1)th order.
a n-3 = (n-3)th order.

I've started the solving ;
a n = r^n
so
the equation will be ;

r^n - r^(n-1) - r^(n-3)
r^3 - r^2 - 1 = 0
I could'nt do anything after find the roots ?
What should i do ?
Thanks for helping.

2. Aug 24, 2010

### jasonRF

Re: How to solve recurrence relation with one real root and two complex roots ???

I think you are making this harder than it is. Hint:

you can rewrite you recurrence relation as

$$a_{n} = - a_{n-2}.$$

So what is $$a_4$$? How about $$a_5$$? See the pattern?

jason