Fibonacci polynomials explained

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.

Definition

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

and

Bn=0

.

Examples

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.

Properties

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)

and

l{U}n(x)

as

Fn(x)=in-1l{U}n-1(\tfrac{-ix}2),

Ln(x)=2 ⋅

nl{T}
i
n(\tfrac{-ix}2),

where

i

is the imaginary unit.

Identities

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).

Closed form expressions, similar to Binet's formula are:
F
n(x)=\alpha(x)n-\beta(x)n
\alpha(x)-\beta(x)
n+\beta(x)
,L
n(x)=\alpha(x)

n,

where
\alpha(x)=x+\sqrt{x2+4
},\,\beta(x)=\fracare the solutions (in t) of

t2-xt-1=0.

For Lucas Polynomials n > 0, we have

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).

For example,

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)

Combinatorial interpretation

If F(n,k) is the coefficient of xk in Fn(x), namely

Fn(x)=\sum

n
k=0

F(n,k)xk,

then F(n,k) is the number of ways an n−1 by 1 rectangle can be tiled with 2 by 1 dominoes and 1 by 1 squares so that exactly k squares are used.[1] Equivalently, F(n,k) is the number of ways of writing n−1 as an ordered sum involving only 1 and 2, so that 1 is used exactly k times. For example F(6,3)=4 and 5 can be written in 4 ways, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, as a sum involving only 1 and 2 with 1 used 3 times. By counting the number of times 1 and 2 are both used in such a sum, it is evident that

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.

References

Further reading

Notes and References

  1. Benjamin & Quinn p. 141
  2. Benjamin & Quinn p. 142
  3. Springer
  4. A proof starts from page 5 in Algebra Solutions Packet (no author).