6
$\begingroup$

Pick four integers $a,b,c$ and $d$. Then we get a corresponding sequence given by $$t_{n+2} = at_{n+1} +bt_n, \; t_1 = c, \;t_2 = d.$$

From what I can tell, we seem to get an especially rich theory when we choose $a=1,b=1,c=1,d=1$, thereby obtaining the Fibonacci sequence. Just take a look at the relevant wikipedia page; it's simply huge, and full of interesting-looking identities and connections.

Question. Why is this? What is about these four numbers that gives such a rich theory for the corresponding sequence?

A good answer should either:

  • Explain that most of the results about the Fibonacci sequence have analogs that work for any $a,b,c$ and $d$ satisfying some weak conditions, so really the Fibonacci sequence isn't that special, or:

  • Specify a very strong constraint on the relationship between $a,b,c$ and $d$ and explain why this constraint makes this particular sequence and the (few) others like it to have a very rich theory.

$\endgroup$
  • 2
    $\begingroup$ Your question is biased. The Fibonacci sequence is not only special as a linear recurrence of degree $2$. $\endgroup$ – J.-E. Pin Jun 22 '18 at 4:20
  • 1
    $\begingroup$ In mathematics there is a "sporadic phenomenon" that makes it so provoking, so interesting. For example, there are two Mathieu groups of 4 and 5-transitivity, no other so transitive. Why? nobody knows. We have to live with it . $\endgroup$ – Boyku Jun 22 '18 at 4:27
  • 1
    $\begingroup$ In order to stay within the realm of Fibonacci and Lucas numbers, or any of the generalized forms of them, then a, b, c, and d are limited. If choosing any a, b, c, d values then one can find the Pell, Pell-Lucas, Jacobsthal, Jacobsthal-Lucas, Fermat, etc numbers. One will find that these numbers are all "solutions" of second order difference equations. Fibonacci, and Lucas, are more widely known due to there values being found in all fields of science. $\endgroup$ – Leucippus Jun 22 '18 at 4:31
  • 1
    $\begingroup$ There is also the Tribonacci Sequence. You can go here (from MSE) for some info about it, and here (from Brilliant). For some extra detail and more formulae go here. Also, a general "Ratio Formula" I will put it as, can be found by watching this YouTube video. $\endgroup$ – Mr Pie Jun 22 '18 at 4:40
  • $\begingroup$ @J.-E.Pin, I don't understand your comment. Can you elaborate a bit? $\endgroup$ – goblin Jun 22 '18 at 5:10
6
$\begingroup$

In calculus and differential equations, the most important functions are those which are their own derivatives: $e^x, \sin x, \cosh x, $ etc. The most fundamental is $e^x$ and the others can be expressed in terms of it. It satisfies the equation $y' = y.$

The Fibonacci sequence is its own difference sequence. (Write down the sequence and then write the differences between each pair of successive terms, and you'll get another copy of the Fibonacci sequence.) So it satisfies the equation $F = \Delta F.$ So it's not a surprise that solutions to difference equations can be expressed in terms of the Fibonacci sequence. Your recursive definition involving $a,b, c$ and $d$ is really a difference equation, as is the definition of the Tribonacci sequence and other more general things.

$\endgroup$
  • $\begingroup$ "Write down the sequence and then write the differences between each pair of successive terms, and you'll get another copy of the Fibonacci sequence" -- this doesn't seem to be literally true. In particular, the first value of Fibonacci seems to change when you do this. For instance, if $x$ is defined to be $1,1,2,3,5,\ldots$, then $\Delta x$ seems to be $0,1,1,2,3,5,\ldots$. If we apply $\Delta$ again we seem to get $1,0,1,1,2,3,5,\ldots$. If we do it again we seem to pick up a negative term at the beginning. $\endgroup$ – goblin Jun 27 '18 at 4:45
  • $\begingroup$ The closest I think we can get to making this claim literally true is this: if we extend the Fibonacci sequence into the negatives, then $\Delta$ seems to have the same effect as shifting the sequence right by $1$. That is, $(\Delta F)_n = F_{n-1}$. $\endgroup$ – goblin Jun 27 '18 at 4:51
  • $\begingroup$ @goblin That's what's normally done. $F_{-n} = (-1)^nF_n$. $\endgroup$ – B. Goddard Jun 27 '18 at 11:03
  • $\begingroup$ Okay, but it's still not the case that it's its own difference sequence, right? Differencing it seems to shift it right one step. So it's not really an eigenvector in the usual sense of the word. I think we can view real sequences as forming an R[x] module where x is a right shift. Then the fibonacci sequence will be an eigenvector with eigenvalue x. But you have to move beyond vector spaces to modules over a PID for this to work $\endgroup$ – goblin Jun 27 '18 at 12:52
  • $\begingroup$ @goblin Write the differences of $\ldots,-8, +5, -3, +2, -1, 0, 1, 1, 2, 3, 5, 8,\ldots$ and see what you get. I had a typo:$ F_{-n} = (-1)^{n+1}f_n$. $\endgroup$ – B. Goddard Jun 27 '18 at 14:40

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.