- #1

SYoungblood

- 64

- 1

## Homework Statement

: [/B]I am just learning recurrence relations, and they are proving challenging.

## Homework Equations

-- [/B]Let's have x_{n}= 3x

_{n-1}+ 6x

_{n-2}.

I wanted to look at it with two scenarios. The first is x

_{0}= 1 and x

_{1}= 3. The second is x

_{1}=3 and x

_{2}= 4

## The Attempt at a Solution

Is there a difference between having a staring point of x

_{0}and x

_{1}? That is the sticking point.

As far as the solution is concerned, if I wanted to go to the 5th term, I can do the x

_{1}=3 and x

_{2}= 4 scenario -- x

_{1}= 1, x

_{2}= 4, x

_{3}= x

_{3}=30, x

_{4}= 114, and x

_{5}= 522.

The scenario with x

_{0}is challenging to me. How can you have a recurrence relation where the n-1 term leaves you a negative number? How do you know where to start? Is it understood that, in the equation that is above, I would start with 3(x

_{0}(or 3 * 1) + 6(x

_{1}(6 * 3), with a sum of 21?

Also, how would you turn this into a formula? How do you determine the degree? For this last question, I just need a point in the right direction, I'm not asking for someone to do the questions for me.