41
$\begingroup$

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.

$\endgroup$
10
  • 14
    $\begingroup$ it's the root of a cubic equation $\endgroup$ – J. W. Tanner Apr 11 at 18:19
  • 14
    $\begingroup$ What do you mean by I wonder how a sequence with nothing but natural numbers leads us to trigonometry? It is like that, that's the beauty of math! $\endgroup$ – Giuppox Apr 11 at 18:19
  • 2
    $\begingroup$ See the link I added to my comment above $\endgroup$ – J. W. Tanner Apr 11 at 18:24
  • 17
    $\begingroup$ The usual Fibonacci sequence also connects to trigonometry. In particular, to the geometry of the regular pentagon. $2\cos(72°)=\phi-1$ $\endgroup$ – PM 2Ring Apr 11 at 18:58
  • 5
    $\begingroup$ Does it necessarily need to be trigonometry? I'd thought that "tribonacci" came from smashing together "triple" and "Fibonacci". Saying that they're related to trigonometry because they're related to hyperbolic functions which are related to trigonometry seems to be more of a stretch. $\endgroup$ – Teepeemm Apr 12 at 18:24
45
$\begingroup$

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.

$\endgroup$
4
  • 14
    $\begingroup$ +1. While the switch between circular and hyperbolic functions is simple (replace $x$ by $ix$) and well known (to me), it never occurred to me to apply it on DeMoivre theorem. Really nice! $\endgroup$ – Paramanand Singh Apr 12 at 7:21
  • 2
    $\begingroup$ And the reason why trigonometric functions can help us exponentiate is because they’re thinly veiled $e^x$ ($\cosh x = (e^x+e^{-x})/2, \cos x = \cosh ix$). Why should anyone think the $\cosh$-based expression is an improvement over using radicals is a different question. $\endgroup$ – Roman Odaisky Apr 12 at 23:01
  • 1
    $\begingroup$ There are trigonometric functions and there are hyperbolic functions. But there is no such thing like hyperbolic trigonometric functions. $\endgroup$ – CiaPan Apr 13 at 19:29
  • $\begingroup$ @RomanOdaisky: Maybe because it is 'cleaner'? It doesn't have repeated numbers unlike the radical version? And also, if it's outside the domain of the real cosh, you can use the real cos instead, so you don't have to deal with complex numbers for real roots of the cubic? $\endgroup$ – user21820 May 9 at 4:29
33
$\begingroup$

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.

$\endgroup$
8
  • 4
    $\begingroup$ I think, now I am able to understand why one would land in such a cubic equation, thanks a lot. $\endgroup$ – Aryan Raj Apr 11 at 20:23
  • 6
    $\begingroup$ I wonder if this applies to n-Fibonacci sequences, where the sum of the first n terms is taken... does it approach (x^n - x^(n-1) - ... - x = 1)? $\endgroup$ – icedwater Apr 12 at 10:36
  • 9
    $\begingroup$ @icedwater Yes, it does! One can show this with generating functions. $\endgroup$ – Michael Engen Apr 12 at 12:23
  • 1
    $\begingroup$ Thanks @MichaelEngen, I'll poke around at this on the weekend :) $\endgroup$ – icedwater Apr 13 at 2:45
  • 2
    $\begingroup$ @icedwater: 1. Make the ansatz $a_k = x^k$. Note that the recurrence relation is satisfied if x is a root of the polynomial that you give. 2. To satisfy the initial value conditions, one can take a linear combination of these sequences for all the roots of the polynomial. (Here one would need some linear independence argument for the existence of a solution, at the moment I don't quite know how to do that in general.) 3. Asymptotically, the root with the largest absolute value wins. $\endgroup$ – Carsten S Apr 13 at 13:19
1
$\begingroup$
  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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.