Hey there,
I've been having a bit of trouble solving the following recurrence relation.
Given:
Any ideas would be greatly appreciated! Thanks in advance.
Hello, kevi555!
When there is a constant term, I was taught to do it like this . . .Solve the following recurrence relation.
Given: .
We have: . .[1]
Consider the next term:
. . . . . . . .[2]
Subtract [1] from [2]: .
. . and we have: .
Divide by
. . and this cubic factors: .
Can you finish it now?
My problem with this is that I need one more condition to get a particular solution? I'm getting a similar form to galactus', but I'm getting an unknown constant that I can't get rid of.
My solution so far is
This reproduces both initial conditions.
-Dan
I'll pick it up from this point. (Odd, I had thought that Soroban had the solution to the cubic correct earlier.)
The cubic factors to
so we have solutions x = 1, -4, and -4.
Thus the general solution to the recursion equation will be
Now all you need to do is fit and to this. I found I needed a third condition to solve the system completely, and discovered that if I calculated from the original recursion relation, I could set n = 2 in my solution and thus get a third relationship from which I could eliminate A.
So the problem essentially boils down to solving the system:
-Dan
Notice the Maple generated solution is the same as topquarks, only in an expanded form. Can't you use when you have three terms?. I tried and kept getting the incorrect solution. So, either I done it wrong or I am misconstrued. What are your thoughts?.
I thought with recursions, when you have k terms, you could set it up as
If you have a root z of your auxiliary equation repeated k times, then representing that root we have terms
in the solution to the recursion relation.
So in this case we have the root -4 repeated twice, so we have terms in the solution.
(This is all very similar to solving a linear differential equation with constant coefficients and having repeated roots to the auxiliary equation producing terms in the solution.)
-Dan