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.
Best Answer
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.