Solving a recurrence relation with no real roots?

361 Views Asked by At

$a_n - 2a_{n-1} -12a_{n-2} - 14a_{n-3} -5a_{n-4} = 0 $

I've tried a few method:

  1. setting the denominator to $1 -2x -12x^2 -14x^3 -5x^4$ and then finding the numerator by multiplying $(1 -2x -12x^2 -14x^3 -5x^4)*\sum a_nx^{n}$

  2. finding the roots of C(y) = $y^4 -4y^3 -12y^2 -14y -5$

What else can I do?

1

There are 1 best solutions below

4
On BEST ANSWER

A start: The given recurrence has characteristic equation $x^4-2x^3-12x^2-14x-5=0$. This has $x=-1$ as a solution.

Dividing our polynomial by $x+1$, we get $x^3-3x^2-9x-5$. Note that $-1$ is a zero of this polynomial. Continue.