Fibonacci polynomials

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Template:Short description 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)={0,if n=01,if n=1xFn1(x)+Fn2(x),if n2

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

Ln(x)={2,if n=0x,if n=1xLn1(x)+Ln2(x),if n2.

They can be defined for negative indices by[3]

Fn(x)=(1)n1Fn(x),
Ln(x)=(1)nLn(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
F3(x)=x2+1
F4(x)=x3+2x
F5(x)=x4+3x2+1
F6(x)=x5+4x3+3x

The first few Lucas polynomials are:

L0(x)=2
L1(x)=x
L2(x)=x2+2
L3(x)=x3+3x
L4(x)=x4+4x2+2
L5(x)=x5+5x3+5x
L6(x)=x6+6x4+9x2+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 polynomials can be expressed in terms of Lucas sequences as
    Fn(x)=Un(x,1),
    Ln(x)=Vn(x,1).
  • They can also be expressed in terms of Chebyshev polynomials 𝒯n(x) and 𝒰n(x) as
    Fn(x)=in1𝒰n1(ix2),
    Ln(x)=2in𝒯n(ix2),
where i is the imaginary unit.

Identities

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

Fm+n(x)=Fm+1(x)Fn(x)+Fm(x)Fn1(x)
Lm+n(x)=Lm(x)Ln(x)(1)nLmn(x)
Fn+1(x)Fn1(x)Fn(x)2=(1)n
F2n(x)=Fn(x)Ln(x).

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

Fn(x)=α(x)nβ(x)nα(x)β(x),Ln(x)=α(x)n+β(x)n,

where

α(x)=x+x2+42,β(x)=xx2+42

are the solutions (in t) of

t2xt1=0.

For Lucas Polynomials n > 0, we have

Ln(x)=k=0n/2nnk(nkk)xn2k.

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

xn=Fn+1(x)+k=1n/2(1)k[(nk)(nk1)]Fn+12k(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

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

Fn(x)=k=0nF(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)={(12(n+k1)k)if n≢k(mod2),0else.

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

References

Template:Reflist

Further reading

  1. 1.0 1.1 Benjamin & Quinn p. 141
  2. Benjamin & Quinn p. 142
  3. 3.0 3.1 3.2 Springer
  4. Template:MathWorld
  5. A proof starts from page 5 in Algebra Solutions Packet (no author).