
#1
Nov3012, 01:08 PM

P: 25

For what complex numbers, x, is
G_{n} = f_{n1}(x)  2f_{n}(x) + f_{n+1}(x) = 0 where the terms are consecutive Fibonacci polynomials? Here's what I know: 1) Each individual polynomial, f_{m}, has roots x=2icos(kπ/m), k=1,...,m1. 2) The problem can be rewritten recursively as G_{n+2} = xG_{n+1} + G_{n}, G_{1} = x2, G_{2} = x^{2}  2x + 2 with characteristic equation Y^{2}  xY  1. If a and b are the roots of the characteristic equation, then G_{n} = a^{n} + b^{n}  2(a^{n}  b^{n})/(ab) 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
Nov3012, 04:18 PM

Mentor
P: 10,791

Some approach to guess solutions:
G_{1} has a single solution ##x=2## G_{2} has two solutions ##x=1\pm i## ##G_3 = x^32x^2+3x2## has three solutions ##x=1##, ##x=\frac{1}{2}(1\pm i\sqrt{7})## ##G_4 = x^42x^3+4x^24x+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. 



#3
Nov3012, 04:53 PM

P: 25

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? 


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