The discussion revolves around solving a linear homogeneous recurrence relation defined by the equation an - 5an-1 + 6an-2 = 0, with initial conditions a0 = 1 and a1 = 3. Participants explore the methods for finding a general solution to this recurrence relation.
Some participants have provided guidance on factoring and interpreting the roots of the characteristic equation. There is an ongoing exploration of how to apply these roots to the original problem, with some participants expressing confusion about the process and the significance of the initial conditions.
Participants note that the recurrence relation includes specific initial conditions and a domain restriction (n ≥ 2), which may affect the interpretation of the solution. There is also mention of a similar example from a textbook, which raises questions about the consistency of the approach taken in this problem.
Office_Shredder said:Do you see something you can factor out of that polynomial?
pavel329 said:I'm not sure how to factor anything with (n-1) or (n-2) as an exponent.
The book turned a similar equation into t2-t-6.
Which made no sense to me.
Dick said:First factor out t^(n-2). It divides all three terms, right?
pavel329 said:Ok so I now have it down to
t=2,3.
Now I assume i need to put that into an equation for the original problem.
Which I have no clue how to do.
But where does the n≥2 stand?
Does that affect my final outcome?
Dick said:Are you sure you've seen examples of this before? They should have given you one before they gave you this problem. The t=2,3 tells you the solution is a_n=A*2^n+B*3^n. You know a0 and a1. Use those to find A and B.
pavel329 said:I've only seen one example.
sn = sn-1 + 6sn-2, s0 = 4 s1 = 7
However this equation put a 0 in place of the an and added the n≥2.
Which maybe I shouldn't have paid attention to.
But my final answer is: A=1 B=0. So an = 3n. Is this correct?
Dick said:It sure is.