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Fibonacci polynomials
Polynomial sequence considered as a generalization of Fibbonacci numbers

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.

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Definition

These Fibonacci polynomials are defined by a recurrence relation:1

F n ( x ) = { 0 , if  n = 0 1 , if  n = 1 x F n − 1 ( x ) + F n − 2 ( x ) , if  n ≥ 2 {\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

L n ( x ) = { 2 , if  n = 0 x , if  n = 1 x L n − 1 ( x ) + L n − 2 ( x ) , if  n ≥ 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 by3

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

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

Examples

The first few Fibonacci polynomials are:

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

The first few Lucas polynomials are:

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

Properties

  • The degree of Fn is n − 1 and the degree of Ln is n.
  • 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 ordinary generating functions for the sequences are:4 ∑ n = 0 ∞ F n ( x ) t n = t 1 − x t − t 2 {\displaystyle \sum _{n=0}^{\infty }F_{n}(x)t^{n}={\frac {t}{1-xt-t^{2}}}} ∑ n = 0 ∞ L n ( x ) t n = 2 − x t 1 − x t − 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 F n ( x ) = U n ( x , − 1 ) , {\displaystyle F_{n}(x)=U_{n}(x,-1),\,} L n ( x ) = V n ( x , − 1 ) . {\displaystyle L_{n}(x)=V_{n}(x,-1).\,}
  • They can also be expressed in terms of Chebyshev polynomials T n ( x ) {\displaystyle {\mathcal {T}}_{n}(x)} and U n ( x ) {\displaystyle {\mathcal {U}}_{n}(x)} as F n ( x ) = i n − 1 ⋅ U n − 1 ( − i x 2 ) , {\displaystyle F_{n}(x)=i^{n-1}\cdot {\mathcal {U}}_{n-1}({\tfrac {-ix}{2}}),\,} L n ( x ) = 2 ⋅ i n ⋅ T n ( − i x 2 ) , {\displaystyle L_{n}(x)=2\cdot i^{n}\cdot {\mathcal {T}}_{n}({\tfrac {-ix}{2}}),\,}
where i {\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 as5

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

Closed form expressions, similar to Binet's formula are:6

F n ( x ) = α ( x ) n − β ( x ) n α ( x ) − β ( x ) , L n ( x ) = α ( x ) n + β ( x ) n , {\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

α ( x ) = x + x 2 + 4 2 , β ( x ) = x − x 2 + 4 2 {\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

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

For Lucas Polynomials n > 0, we have

L n ( x ) = ∑ k = 0 ⌊ n / 2 ⌋ n n − k ( n − k k ) x n − 2 k . {\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 by7

x n = F n + 1 ( x ) + ∑ k = 1 ⌊ n / 2 ⌋ ( − 1 ) k [ ( n k ) − ( n k − 1 ) ] F n + 1 − 2 k ( x ) . {\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,

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

Combinatorial interpretation

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

F n ( x ) = ∑ k = 0 n F ( n , k ) x k , {\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.8 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 ) = { ( 1 2 ( n + k − 1 ) k ) if  n ≢ k ( mod 2 ) , 0 else . {\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.

Further reading

  • Hoggatt, V. E.; Bicknell, Marjorie (1973). "Roots of Fibonacci polynomials". Fibonacci Quarterly. 11: 271–274. ISSN 0015-0517. MR 0332645.
  • Hoggatt, V. E.; Long, Calvin T. (1974). "Divisibility properties of generalized Fibonacci Polynomials". Fibonacci Quarterly. 12: 113. MR 0352034.
  • Ricci, Paolo Emilio (1995). "Generalized Lucas polynomials and Fibonacci polynomials". Rivista di Matematica della Università di Parma. V. Ser. 4: 137–146. MR 1395332.
  • Yuan, Yi; Zhang, Wenpeng (2002). "Some identities involving the Fibonacci Polynomials". Fibonacci Quarterly. 40 (4): 314. MR 1920571.
  • Cigler, Johann (2003). "q-Fibonacci polynomials". Fibonacci Quarterly (41): 31–40. MR 1962279.

References

  1. Benjamin & Quinn p. 141 ↩

  2. Benjamin & Quinn p. 142 ↩

  3. Springer ↩

  4. Weisstein, Eric W. "Fibonacci Polynomial". MathWorld. /wiki/Eric_W._Weisstein ↩

  5. Springer ↩

  6. Springer ↩

  7. A proof starts from page 5 in Algebra Solutions Packet (no author). https://web.archive.org/web/20170202051159/http://cmimc.org/Documents/Archive/AlgebraSolutions_2016.pdf ↩

  8. Benjamin & Quinn p. 141 ↩