Roots of linear sum of Fibonacci polynomials

1. Nov 30, 2012

ekkilop

For what complex numbers, x, is

Gn = fn-1(x) - 2fn(x) + fn+1(x) = 0

where the terms are consecutive Fibonacci polynomials?

Here's what I know:

1) Each individual polynomial, fm, has roots x=2icos(kπ/m), k=1,...,m-1.

2) The problem can be rewritten recursively as
Gn+2 = xGn+1 + Gn,
G1 = x-2,
G2 = x2 - 2x + 2
with characteristic equation Y2 - xY - 1.
If a and b are the roots of the characteristic equation, then
Gn = an + bn - 2(an - bn)/(a-b)
Choosing x=-2icosh(z) is an option that leads to an expression in terms of cosh(nx), sinh(nx) and sinh(x) but it doesn't get me any further.

Has anyone got an idea on an alternative approach to this problem?
Does anyone know of previous studies of this type of problem?

Thank you

2. Nov 30, 2012

Staff: Mentor

Some approach to guess solutions:

G1 has a single solution $x=2$
G2 has two solutions $x=1\pm i$
$G_3 = x^3-2x^2+3x-2$ has three solutions $x=1$, $x=\frac{1}{2}(1\pm i\sqrt{7})$
$G_4 = x^4-2x^3+4x^2-4x+2$ has four solutions $x=\frac{1}{2}(1\pm i)-\sqrt{-1\mp \frac{i}{2}}$ and $x=\frac{1}{2}(1\pm i)+\sqrt{-1\mp \frac{i}{2}}$
The product of all solutions is 2 in all tested cases, and looking at the recursive definition and the first expressions I think this will be true for all n.

G_5 gives an interesting graph for the roots. Looks a bit like a christmas tree.
Same thing for G_6, but without simplification the expression is quite long.

Last edited: Nov 30, 2012
3. Nov 30, 2012