Quick observation: I recently noticed that Chebyshev polynomials and Fibonacci numbers have analogous formulas.

The *n*th Chebyshev polynomial satisfies

for |*x*| ≥ 1, and the *n*th Fibonacci number is given by

There’s probably a way to explain the similarity in terms of the recurrence relations that both sequences satisfy.

Both the Fibonacci sequence and the Chebyshev polynomials satisfy a constant coefficient three term recurrence equation.

The solution to a constant coefficient three term recurrence relation whose characteristic equation has two distinct roots r1 and r2 is given by A*r1^n + B*r2^n.

The Fibonacci sequence is closely related to the Chebyshev polynomials of the second kind evaluated at x=i/2.

F(n) = U(n-1, i/2)/i^(n-1)

Yeah, if you let S denote the shift operator (ST)(n) = T(n+1), the the Chebyshev polynomials satisfy

(S^2 – 2xS + I) T = 0,

and when you solve for the roots you get

\frac{2x \plusminus \sqrt{4x^2 – 4}}{2},

and then you divide out the 2 and fit the n = 0 and n =1 terms.

At first this looked like just formal manipulation and in need of a bit more justification, but you can see that it’s 100% rigorous if you just “rotate your mind 90 degrees”. By that I mean: Instead of thinking of {T_n(x)} as an n-indexed sequence of functions, think of it an x-indexed collection of sequences. Then x is just a parameter in defining our sequence, and plays that role in the rest of the argument too!