Sequence of polynomials defined recursively
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]
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 by[ 3]
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 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] ∑ 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 As particular cases of Lucas sequences, Fibonacci polynomials satisfy a number of identities, such as[ 3]
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:[ 3]
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 by[ 5]
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 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 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.[ 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 ) = { ( 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.
References Benjamin, Arthur T. ; Quinn, Jennifer J. (2003). "Fibonacci and Lucas Polynomial". Proofs that Really Count: The Art of Combinatorial Proof . Dolciani Mathematical Expositions. Vol. 27. Mathematical Association of America . p. 141 . ISBN 978-0-88385-333-7 . Philippou, Andreas N. (2001) [1994], "Fibonacci polynomials" , Encyclopedia of Mathematics , EMS Press Philippou, Andreas N. (2001) [1994], "Lucas polynomials" , Encyclopedia of Mathematics , EMS Press Weisstein, Eric W. "Lucas Polynomial" . MathWorld . Jin, Z. On the Lucas polynomials and some of their new identities. Advances in Differential Equations 2018, 126 (2018). https://doi.org/10.1186/s13662-018-1527-9
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 .
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