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Old 2017-07-07, 12:44   #1
Dr Sardonicus
 
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Feb 2017
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Default The Fibonacci-Chebyshev connection

Sounds like a lurid headline: "Investigators probe Fibonacci-Chebyshev connection!" Fibonacci (Leonardo of Pisa, or Leonardo Pisano Fibonacci, 1170-1250) is famous for his mathematical work, which predated the printing press. The sequence named for him arose from a question he posed with regard to the reproduction of rabbits.

Pafnuty Lvovich Chebyshev, May 4 [May 16, New Style], 1821, Okatovo, Russia — November 26 [December 8], 1894, St. Petersburg, was a Russian mathematician, known for his work on the distribution of prime numbers, and also in real analysis.

Obviously, the two men never met. They worked in different areas of mathematics. And yet, there is a connection between things named for them. On the one hand, we have the Fibonacci numbers. We can generalize these as follows: If a and b are integers, the quadratic polynomial

x^2 - a*x + b

gives rise to two integer sequences

(*F) Fn: F0 = 0, F1 = 1, Fn+2 = a*Fn+1 - b*Fn

(*L) Ln: L0 = 2, L1 = a, Ln+2 = a*Ln+1 - b*Ln

If the quadratic factors as (x - r)*(x - r') we may write these as

(**) Fn = (r^n - r'^n)/(r - r'). and Ln = r^n + r'^n.

These sequences are generalizations of the Fibonacci and Lucas numbers; they are polynomials in a and b.

Where's the connection to Chebyshev? Well, once upon a time, long long ago, I was thinking about "reciprocal polynomials" F(z), for which

F(z) = z^n*F(1/z), where n is the degree of F(z).

It is easy to see that, if n is odd, then F(-1) = 0; so, with irreducibility in mind, we assume that n is even, n = 2*k. Clearly the lead coefficient is 1. The quadratic case is

F(z) = z^2 - a*z + 1, which is irreducible for integers a other than 2 or -2.

Anyway, it seemed clear to me (as I'm sure has been known for centuries) that

F(z)/z^k

can be expressed as a polynomial in X = z + 1/z. All that is needed is to express z^n + 1/z^n as a polynomial in z + 1/z,

z^n + 1/z^n = Tn(z + 1/z) = Tn(X)

That this is possible is easily shown by induction, using T0(X) = 2, T1(X) = X, and

(z^n + 1/z^n)*(z + 1/z) = z^(n+1) + 1/z^(n+1) + z^(n-1) + 1/z^(n-1), or

Tn+1(X) = X*Tn(X) - Tn-1(X). We have

T0(X) = 2, T1(X) = X, T2(X) = X^2 - 2, T3(X) = X^3 - 3*X, etc

This is a sequence of monic polynomials with integer coefficients. Now, I was sure that these polynomials had been well known for a long time, but how? Then, it suddenly occurred to me: If you take

z = exp(i*t), z + 1/z = 2*cos(t) and z^n + 1/z^n = 2*cos(n*t). That is,

Tn(2*cos(t)) = 2*cos(n*t).

In other words, the Tn(X) are (up to a change of scale and a constant multiplier) "Chebyshev polynomials of the first kind." They give the sequence Ln for the quadratic x^2 - X*x + 1. The sequence Fn obviously gives the polynomials in X = z + 1/z for

(z^n - 1/z^n)/(z - 1/z),

which with z = exp(i*t) becomes sin(n*t)/sin(t). Thus, for the quadratic x^2 - X*x + 1, the sequence Fn expresses sin(n*t)/sin(t) as polynomials in 2*cos(t). These are also monic with integer coefficients. They are (up to a change of scale and a constant multiplier) the "Chebyshev polynomials of the second kind." We have

F0(X) = 0, F1(X) = 1, F2(X) = X, F3(X) = X^2 - 1, etc.

Now, what happens if the constant term 1 is replaced by another constant b? As (*F) and (*L) show, we obtain a sequence of polynomials in two variables. One also has r' = b/r in (**), and we can write

r + r' = sqrt(b)*(r/sqrt(b) + sqrt(b)/r).

The (perhaps) most curious case is b = -1. In this case, the effect is simply to change the (rescaled) Chebyshev polynomials Tn and Fn by making all the coefficients positive. These (particularly the transformed Fn) are called the "Fibonacci polynomials," giving the formulas in (*L) and (*F) with b = -1 for x^2 - a*x - 1.

Notes:

The (rescaled) Chebyshev polynomials above allow very simple proofs by induction, that cos(n*t) is a polynomial with rational coefficients in cos(t), and that sin(n*t) is sin(t) times a polynomial in cos(t).

The coefficients of the (rescaled) Chebyshev polynomials may easily be found using the well known power series method on the second-order differential equation satisfied by the cosine, to develop a recursion relation.

The formulas (*) give (perhaps) interesting composition formulas. Let n = a*b. Then

Tn(X) = Ta(Tb(X)), and Fn(X) = Fa(Tb(X))*Fb(X) F[sub]a[/sub](X*F[sub]b[/sub](X))*F[sub]b[/sub](X).

Last fiddled with by Dr Sardonicus on 2022-01-14 at 21:16 Reason: Fix misstated formula
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