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 Lucas polynomials use the same recurrence with different starting values:^[2]

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

They can be defined for negative indices by^[3]

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

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

The Fibonacci polynomials form a sequence of orthogonal polynomials with ${\displaystyle A_{n}=C_{n}=1}$ and ${\displaystyle B_{n}=0}$.

Examples

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 first few Lucas polynomials are:

${\displaystyle L_{0}(x)=2\,}$

${\displaystyle L_{1}(x)=x\,}$

${\displaystyle L_{2}(x)=x^{2}+2\,}$

${\displaystyle L_{3}(x)=x^{3}+3x\,}$

${\displaystyle L_{4}(x)=x^{4}+4x^{2}+2\,}$

${\displaystyle L_{5}(x)=x^{5}+5x^{3}+5x\,}$

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

Properties

• The degree of F_n is n − 1 and the degree of L_n is n.

• The Fibonacci and Lucas numbers are recovered by evaluating the polynomials at x = 1; Pell numbers are recovered by evaluating F_n at x = 2.

• The ordinary generating functions for the sequences are:^[4]

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

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

• The polynomials can be expressed in terms of Lucas sequences as

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

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

• They can also be expressed in terms of Chebyshev polynomials ${\displaystyle {\mathcal {T}}_{n}(x)}$ and ${\displaystyle {\mathcal {U}}_{n}(x)}$ as

  ${\displaystyle F_{n}(x)=i^{n-1}\cdot {\mathcal {U}}_{n-1}({\tfrac {-ix}{2}}),\,}$

  ${\displaystyle L_{n}(x)=2\cdot i^{n}\cdot {\mathcal {T}}_{n}({\tfrac {-ix}{2}}),\,}$

where ${\displaystyle i}$ is the imaginary unit.

Identities

Main article: Lucas sequence

As particular cases of Lucas sequences, Fibonacci polynomials satisfy a number of identities, such as^[3]

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

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

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

${\displaystyle F_{2n}(x)=F_{n}(x)L_{n}(x).\,}$

Closed form expressions, similar to Binet's formula are:^[3]

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

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

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

For Lucas Polynomials n > 0, we have

${\displaystyle L_{n}(x)=\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {n}{n-k}}{\binom {n-k}{k}}x^{n-2k}.}$

A relationship between the Fibonacci polynomials and the standard basis polynomials is given by^[5]

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

For example,

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

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

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

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

Combinatorial interpretation

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

If F(n,k) is the coefficient of x^k in F_n(x), namely

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

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 ${\displaystyle F(n,k)={\begin{cases}\displaystyle {\binom {{\frac {1}{2}}(n+k-1)}{k}}&{\text{if }}n\not \equiv k{\pmod {2}},\\[12pt]0&{\text{else}}.\end{cases}}}$

This gives a way of reading the coefficients from Pascal's triangle as shown on the right.