Register to reply

Roots of linear sum of Fibonacci polynomials

by ekkilop
Tags: fibonacci, linear, polynomials, roots
Share this thread:
Nov30-12, 01:08 PM
P: 28
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
Phys.Org News Partner Mathematics news on
Researcher figures out how sharks manage to act like math geniuses
Math journal puts Rauzy fractcal image on the cover
Heat distributions help researchers to understand curved space
Nov30-12, 04:18 PM
P: 12,081
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.
Nov30-12, 04:53 PM
P: 28
Thanks for your reply!

Interesting observations. Yes, you're right! The product of the solutions will be 2 since G can be written as the characteristic polynomial of a matrix with determinant 2. Since the solutions come in complex conjugated pairs this suggests some pretty strict bounds.

The solutions also add up to 2 as it seems.

Have you got any clues to why the real part of the solutions are larger than zero?

Register to reply

Related Discussions
Roots of polynomials over GF(p) Linear & Abstract Algebra 3
Roots of polynomials Calculus & Beyond Homework 2
Roots of polynomials... Precalculus Mathematics Homework 7
Polynomials do or don't have integer roots? Precalculus Mathematics Homework 12
Roots of complex polynomials Linear & Abstract Algebra 7