# Fibonacci polynomials

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]

${\displaystyle F_{n}(x)={\begin{cases}0,&{\mbox{if }}n=0\\1,&{\mbox{if }}n=1\\xF_{n-1}(x)+F_{n-2}(x),&{\mbox{if }}n\geq 2\end{cases}}}$

The first few Fibonacci polynomials are:

${\displaystyle F_{0}(x)=0\,}$
${\displaystyle F_{1}(x)=1\,}$
${\displaystyle F_{2}(x)=x\,}$
${\displaystyle F_{3}(x)=x^{2}+1\,}$
${\displaystyle F_{4}(x)=x^{3}+2x\,}$
${\displaystyle F_{5}(x)=x^{4}+3x^{2}+1\,}$
${\displaystyle F_{6}(x)=x^{5}+4x^{3}+3x\,}$

The Lucas polynomials use the same recurrence with different starting values:[2] ${\displaystyle L_{n}(x)={\begin{cases}2,&{\mbox{if }}n=0\\x,&{\mbox{if }}n=1\\xL_{n-1}(x)+L_{n-2}(x),&{\mbox{if }}n\geq 2.\end{cases}}}$

The first few Lucas polynomials are:

${\displaystyle L_{0}(x)=2\,}$
${\displaystyle L_{1}(x)=x\,}$
${\displaystyle L_{2}(x)=x^{2}+2\,}$
${\displaystyle L_{3}(x)=x^{3}+3x\,}$
${\displaystyle L_{4}(x)=x^{4}+4x^{2}+2\,}$
${\displaystyle L_{5}(x)=x^{5}+5x^{3}+5x\,}$
${\displaystyle L_{6}(x)=x^{6}+6x^{4}+9x^{2}+2.\,}$

The Fibonacci and Lucas numbers are recovered by evaluating the polynomials at x = 1; Pell numbers are recovered by evaluating Fn at x = 2. The degrees of Fn is n − 1 and the degree of Ln is n. The ordinary generating function for the sequences are:[3]

${\displaystyle \sum _{n=0}^{\infty }F_{n}(x)t^{n}={\frac {t}{1-xt-t^{2}}}}$
${\displaystyle \sum _{n=0}^{\infty }L_{n}(x)t^{n}={\frac {2-xt}{1-xt-t^{2}}}.}$

The polynomials can be expressed in terms of Lucas sequences as

${\displaystyle F_{n}(x)=U_{n}(x,-1),\,}$
${\displaystyle L_{n}(x)=V_{n}(x,-1).\,}$

## Identities

Main article: Lucas sequence

As particular cases of Lucas sequences, Fibonacci polynomials satisfy a number of identities.

First, they can be defined for negative indices by[4]

${\displaystyle F_{-n}(x)=(-1)^{n-1}F_{n}(x),\,L_{-n}(x)=(-1)^{n}L_{n}(x).}$

Other identities include:[4]

${\displaystyle F_{m+n}(x)=F_{m+1}(x)F_{n}(x)+F_{m}(x)F_{n-1}(x)\,}$
${\displaystyle L_{m+n}(x)=L_{m}(x)L_{n}(x)-(-1)^{n}L_{m-n}(x)\,}$
${\displaystyle F_{n+1}(x)F_{n-1}(x)-F_{n}(x)^{2}=(-1)^{n}\,}$
${\displaystyle F_{2n}(x)=F_{n}(x)L_{n}(x).\,}$

Closed form expressions, similar to Binet's formula are:[4]

${\displaystyle F_{n}(x)={\frac {\alpha (x)^{n}-\beta (x)^{n}}{\alpha (x)-\beta (x)}},\,L_{n}(x)=\alpha (x)^{n}+\beta (x)^{n},}$

where

${\displaystyle \alpha (x)={\frac {x+{\sqrt {x^{2}+4}}}{2}},\,\beta (x)={\frac {x-{\sqrt {x^{2}+4}}}{2}}}$

are the solutions (in t) of

${\displaystyle t^{2}-xt-1=0.\,}$

A relationship between the Fibonacci polynomials and the standard basis polynomials is given by

${\displaystyle x^{n}=F_{n+1}(x)+\sum _{k=1}^{\lfloor n/2\rfloor }(-1)^{k}\left[{\binom {n}{k}}-{\binom {n}{k-1}}\right]F_{n+1-2k}(x).}$

For example,

${\displaystyle x^{4}=F_{5}(x)-3F_{3}(x)+2F_{1}(x)\,}$
${\displaystyle x^{5}=F_{6}(x)-4F_{4}(x)+4F_{2}(x)\,}$
${\displaystyle x^{6}=F_{7}(x)-5F_{5}(x)+9F_{3}(x)-5F_{1}(x)\,}$
${\displaystyle x^{7}=F_{8}(x)-6F_{6}(x)+14F_{4}(x)-14F_{2}(x)\,}$

A proof of this fact is given starting from page 5 here.

## Combinatorial interpretation

The coefficients of the Fibonacci polynomials can be read off from Pascal's triangle following the "shallow" diagonals (shown in red). The sums of the coefficients are the Fibonacci numbers.

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

${\displaystyle F_{n}(x)=\sum _{k=0}^{n}F(n,k)x^{k},\,}$

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) is equal to the binomial coefficient

${\displaystyle F(n,k)={\binom {\tfrac {n+k-1}{2}}{k}}}$

when n and k have opposite parity. This gives a way of reading the coefficients from Pascal's triangle as shown on the right.

## References

1. ^ a b Benjamin & Quinn p. 141
2. ^ Benjamin & Quinn p. 142
3. ^
4. ^ a b c Springer