- #1
ekkilop
- 29
- 0
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
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