David Gwyn J. answered • 11/26/20
Highly Experienced Tutor (Oxbridge graduate and former tech CEO)
A recurrence relation is an equation which describes a series recursively (i.e. in relation to prior terms):
xn = 2xn-1 + xn-2 - 3xn-3
We are given x0 = 2, x1 = 1, x2 = 4
Then x3 = 2(4) + 1 - 3(2) = 8 + 1 - 6 = 3
And x4 = 2(3) + 4 - 3(1) = 6 + 4 - 3 = 7
And x5 = 2(7) + 3 - 3(4) = 14 + 3 - 12 = 5
And x6 = 2(5) + 7 - 3(3) = 10 + 7 - 9 = 8
And x7 = 2(8) + 5 - 3(7) = 16 + 5 - 21 = 0
etc
and we have 2, 1, 4, 3, 7, 5, 8, 0...
We can also go backwards to find the previous term...
so to find xn-1 we can say that 4 = 2(1) + (2) - 3(xn-1) => 4 = 4 - 3(xn-1) => xn-1 = 0
and for xn-2 then 1 = 2(2) + (0) - 3(xn-2) => 1 = 4 - 3(xn-2) => xn-2 = 1
for xn-3 then 2 = 2(0) + (1) - 3(xn-3) => 2 = 1 - 3(xn-3) => xn-3 = -1/3
for xn-4 then 0 = 2(1) + (-1/3) - 3(xn-4 ) => 0 = 2 - 1/3 - 3(xn-4) => 0 = 5/3 - 3(xn-4) => xn-4 = 5/9
etc
and we have 5/9, -1/3, 1, 0, 2, 1, 4, 3, 7, 5, 8, 0...
We can double-check with recursive calculators online.
David Gwyn J.
Yup, my bad. I assumed that "solve" was not really solve in a question on this board. You are absolutely right, of course, you can solve this by finding the non-recursive polynomial which is equivalent to the above recurrence relation. But that's some university maths, wayyyy above my high school paygrade. :-) But there are some ninja mathematicians here, hopefully one of them can help you out.11/27/20
Lemur L.
don't we need to first find the characteristic equation since it is a third-order recurrence relation?11/26/20