In mathematics, the Fibonacci polynomials are a polynomial sequence which can be considered as a generalization of the Fibonacci numbers. The polynomials generated in a similar way from the Lucas numbers are called Lucas polynomials.
These Fibonacci polynomials are defined by a recurrence relation:[1]
Fn(x)=\begin{cases} 0,&ifn=0\\ 1,&ifn=1\\ xFn(x)+Fn(x),&ifn\geq2 \end{cases}
The Lucas polynomials use the same recurrence with different starting values:[2]
Ln(x)=\begin{cases} 2,&ifn=0\\ x,&ifn=1\\ xLn(x)+Ln(x),&ifn\geq2. \end{cases}
They can be defined for negative indices by[3]
F-n(x)=(-1)n-1Fn(x),
L-n
nL | |
(x)=(-1) | |
n |
(x).
The Fibonacci polynomials form a sequence of orthogonal polynomials with
An=Cn=1
Bn=0
The first few Fibonacci polynomials are:
F0(x)=0
F1(x)=1
F2(x)=x
2+1 | |
F | |
3(x)=x |
3+2x | |
F | |
4(x)=x |
4+3x | |
F | |
5(x)=x |
2+1
5+4x | |
F | |
6(x)=x |
3+3x
The first few Lucas polynomials are:
L0(x)=2
L1(x)=x
2+2 | |
L | |
2(x)=x |
3+3x | |
L | |
3(x)=x |
4+4x | |
L | |
4(x)=x |
2+2
5+5x | |
L | |
5(x)=x |
3+5x
6+6x | |
L | |
6(x)=x |
4+9x2+2.
infty | |
\sum | |
n=0 |
Fn(x)tn=
t | |
1-xt-t2 |
infty | |
\sum | |
n=0 |
Ln(x)tn=
2-xt | |
1-xt-t2 |
.
Fn(x)=Un(x,-1),
Ln(x)=Vn(x,-1).
l{T}n(x)
l{U}n(x)
Fn(x)=in-1 ⋅ l{U}n-1(\tfrac{-ix}2),
Ln(x)=2 ⋅
n ⋅ l{T} | |
i | |
n(\tfrac{-ix}2), |
where
i
See main article: Lucas sequence. As particular cases of Lucas sequences, Fibonacci polynomials satisfy a number of identities, such as
Fm+n(x)=Fm+1(x)Fn(x)+Fm(x)Fn-1(x)
Lm+n(x)=Lm(x)L
nL | |
m-n |
(x)
Fn+1(x)Fn-1(x)-
2=(-1) | |
F | |
n(x) |
n
F2n(x)=Fn(x)Ln(x).
F | ||||
|
n+\beta(x) | |
,L | |
n(x)=\alpha(x) |
n,
\alpha(x)= | x+\sqrt{x2+4 |
t2-xt-1=0.
Ln(x)=\sum
\lfloorn/2\rfloor | |
k=0 |
n | |
n-k |
\binom{n-k}{k}xn-2k.
A relationship between the Fibonacci polynomials and the standard basis polynomials is given by[4]
n=F | |
x | |
n+1 |
\lfloorn/2\rfloor | |
(x)+\sum | |
k=1 |
(-1)k\left[\binomnk-\binomn{k-1}\right]Fn+1-2k(x).
x4=F5(x)-3F3(x)+2F1(x)
x5=F6(x)-4F4(x)+5F2(x)
x6=F7(x)-5F5(x)+9F3(x)-5F1(x)
x7=F8(x)-6F6(x)+14F4(x)-14F2(x)
If F(n,k) is the coefficient of xk in Fn(x), namely
Fn(x)=\sum
n | |
k=0 |
F(n,k)xk,
F(n,k)=\begin{cases}\displaystyle\binom{
12(n+k-1)}{k} | |
&ifn |
\not\equivk\pmod2,\\[12pt] 0&else. \end{cases}
This gives a way of reading the coefficients from Pascal's triangle as shown on the right.