Fibonacci Polynomial Calculator

F_{n+1}(x) = xF_n(x)+F_{n-1}(x)

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About Fibonacci Polynomial Calculator

A Fibonacci polynomial calculator computing F_n(x) with F_1=1, F_2=x, F_{n+1}=xF_n+F_{n-1} and Lucas polynomials L_n(x) with L_1=x, L_2=x²+2. At x=1: F_n(1)=Fibonacci numbers. Connections to Chebyshev polynomials and continued fractions. Client-side.

Fibonacci Polynomial Calculator Features

  • F_n(x) value
  • L_n(x) value
  • Coefficients
  • Special values
  • Sequence
Fibonacci polynomials: F_1=1, F_2=x, F_{n+1}=xF_n+F_{n-1}. So F_3=x²+1, F_4=x³+2x, F_5=x⁴+3x²+1. At x=1: 1,1,2,3,5,8... (Fibonacci!). Coefficients are binomial-like. Related to Chebyshev: F_n(x)=i^{1-n}U_{n-1}(ix/2).

How to Use

Enter n and x:

  • F_n(x): Fibonacci poly
  • L_n(x): Lucas poly
  • x=1: Classic numbers

Coefficient Pattern

F_n(x) = Σ C(n-1-k,k)·x^{n-1-2k}. Coefficients are entries from the 'shallow diagonal' of Pascal's triangle. This gives the beautiful connection: Fibonacci numbers = shallow diagonal sums of Pascal's triangle.

Chebyshev Connection

F_n(x) = i^{1-n}·U_{n-1}(ix/2) where U is Chebyshev second kind. This means Fibonacci polynomials are 'hyperbolic Chebyshev' polynomials — the same recurrence but with + instead of − in the argument.

Step-by-Step Instructions

  1. 1Enter n.
  2. 2Enter x.
  3. 3Compute F_n(x).
  4. 4Compute L_n(x).
  5. 5Try x=1 for classic.

Fibonacci Polynomial Calculator — Frequently Asked Questions

How are Fibonacci polynomials related to Fibonacci numbers?+

F_n(1) = F_n (nth Fibonacci number). F_n(2) = Pell numbers. F_n(3) gives another sequence. The polynomial F_n(x) encodes ALL these sequences: just substitute x. It's a 'master formula' generating infinitely many Fibonacci-like sequences.

What's the connection to continued fractions?+

The nth convergent of the continued fraction [0; x, x, x, ...] is F_n(x)/F_{n+1}(x). As n→∞, this converges to (−1+√(x²+4))/(2x), generalizing the golden ratio (x=1 gives φ−1).

What are the coefficients of F_n(x)?+

F_n(x) = Σ_{k=0}^{⌊(n-1)/2⌋} C(n-1-k,k)·x^{n-1-2k}. These are the 'rising diagonal' sums of Pascal's triangle. For F_5: C(4,0)x⁴+C(3,1)x²+C(2,2)=x⁴+3x²+1.

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