How does the tribonacci sequence have anything to do with hyperbolic functions?

Question

The Fibonacci sequence has always fascinated me because of its beauty. It was in high school that I was able to understand how the ratio between 2 consecutive terms of a purely integer sequence came to be a beautiful irrational number.

So I wondered yesterday if instead of 2 terms, we kept 3 terms. So I wrote a python program to calculate the ratio. At the 10000th term it came to be close to 1.839...

After some research on OEIS and Wikipedia, I found that the series is popular and is known as the tribonacci sequence. But what surprised me the most was the exact ratio given on this link.

The tribonacci constant $$\frac{1+\sqrt[3]{19+3\sqrt{33}} + \sqrt[3]{19-3\sqrt{33}}}{3} = \frac{1+4\cosh\left(\frac{1}{3}\cosh^{-1}\frac{19}{8}\right)}{3} \approx 1.83928675$$ (sequence A058265 in the OEIS)

I wonder how a sequence with nothing but natural numbers leads us to non-Euclidian geometry. I wonder if someone could tell me how these two are related.

Note: I don't actually want the exact solution which would be extremely difficult to understand for a high schooler like me, I just want to know if there is a way to connect number theory and non-Euclidian geometry.

Answer 1

Similar to De Moivre's formula:

$$\cos nx \pm i\sin nx = (\cos x\pm i\sin x)^n$$

there is the hyperbolic De Moivre formula:

$$\cosh nx \pm \sinh nx = (\cosh x\pm\sinh x)^n$$

which means this: if you can represent a real number $a$ as $a=\cosh x\pm\sinh x$, then $\sqrt[n]{a}=\cosh (x/n)\pm\sinh (x/n)$. In other words, hyperbolic trigonometric functions can help us exponentiate and take roots. (Note the "ordinary" trigonometric functions can do the same - for roots of complex numbers.)

In this case, let's take $x=\pm\cosh^{-1}\left(2+\frac{3}{8}\right)$ so that $\cosh x=2\frac{3}{8}=\frac{19}{8}$. This (from well-known identity $\cosh^2x-\sinh^2x=1$) gives $\sinh x=\pm\frac{3\sqrt{33}}{8}$. Now, take $a=\frac{1}{8}(19\pm 3\sqrt{33})=\cosh x\pm \sinh x$. All that is left is to apply the hyperbolic De Moivre's formula with $n=3$ to take the cube root and prove that the formula from the Wikipedia article you have cited is correct.

Answer 2

I wonder how a sequence with nothing but natural numbers leads us to trigonometry. I wonder if someone would say me how these two are related.

Briefly, the ratio between consecutive terms of the Tribonacci sequence

approaches the real root of $x^3-x^2-x=1$

(like that of the Fibonacci sequence approaches the positive root of $x^2-x=1$),

and the solution of that cubic equation can be expressed in terms of hyperbolic functions.

Answer 3

1. The terms of an $n$-th order linear recurrence are calculated from roots of a degree $n$ polynomial equation (the characteristic equation of the recurrence, the coefficients of the polynomial are the same as the coefficients of the recurrence)

2. Here $n=3$ so we get a cubic equation to solve

3. The roots of a cubic equation can be expressed using cube roots, but this involves complex numbers, even if the solutions are all real. Equivalent results that stay within real-valued expressions involve the cube-root-like operation of passing from $\cos(t)$ to $\cos(t/3)$ (angle trisection) when there are 3 real solutions, or from $\cosh(t)$ to $\cosh(t/3)$ when there is 1 real solution, with the (ordinary or hyperbolic) cosine with argument $t$ having values that are rational (or maybe square root) functions of the coefficients of the cubic equation. This is the "trigonometric solution of the cubic" used to avoid expressions with complex numbers.

The hyperbolic angle trisection is just another way of saying "the $n$-th term of the recurrence is $p\alpha^n + q\beta^n + r\gamma^n$ for $\alpha,\beta,\gamma$ the three roots of the characteristic equation", the hyperbolic trisection being one way to express the operation of extracting the roots. The largest root ~ 1.839 being the Tribonacci constant.