2
$\begingroup$

I have been playing around with Fibonacci-type of sequence that involve complex numbers. I have stumbled upon the following sequence, which seemed interesting to me: $$0,1,2i,-3,-4i,5,6i,...$$ so $F_n = 2iF_{n-1} + F_{n-2}$. These look like a sequence of natural numbers (except for $0$) where every other is multiplied by $i$ and the signs change after two sequences.

I understand the algebra behind the above sequence, but I have been wondering whether there is an intuition behind why the sequence looks like a "modified" sequence of natural numbers.

$\endgroup$
3
  • 1
    $\begingroup$ $(X-i)^2=X^2-(2iX+1)$. $\endgroup$ – Angina Seng Jul 20 '18 at 14:50
  • $\begingroup$ Isn't this just $n\cdot i^{n-1}$? $\endgroup$ – mechanodroid Jul 20 '18 at 14:54
  • $\begingroup$ Also related: (1), (2), (3), (4), and so on. $\endgroup$ – Xander Henderson Jul 24 '18 at 20:27
2
$\begingroup$

The characteristic polynomial for your recursion is $$x^2-2ix-1=(x-i)^2$$

Visibly, this has a double root at $x=i$. Thus the general form of the solution to the recursion is $$F_n=Ai^n+Bni^n$$ Using your initial conditions it is easy to specify the solution to your case.

$\endgroup$

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