0
$\begingroup$

This answer says that any casual sequence of the kind $y_n = y_{n-1} + y_{n-2} + y_{n-3} + \ldots $ will stay constant-0 because $y_0$ is a sum of zeroes, so is $y_1$ and the rest of the sequence. I can also put it the way $Y(z) = z^{-1}Y(z)+z^{-2}Y(z)+z^{-3}Y(z) + \ldots$, which either means that $z^{-1}+z^{-2}+\ldots+z^{-n} = 1$, which is unlikely, or simply $Y(z) = 0$, which is more plausible. But, Fibonacci is also of this kind. Does it mean that it is non-casual or what?

$\endgroup$
  • $\begingroup$ What is the definition of casual sequences? $\endgroup$ – 5xum Jul 31 '14 at 11:00
  • $\begingroup$ @mxum Read the referenced answer. As far as I could understand, these are sequences which are 0 at n < 0. I would be happy to see more discussion on this topic though. $\endgroup$ – Val Jul 31 '14 at 11:02
  • $\begingroup$ Even in the referenced answer, you are the only one to use the word "casual", and you do not properly define it. Can you please define it here to make this question self-contained? $\endgroup$ – 5xum Jul 31 '14 at 11:05
  • $\begingroup$ It says A sequence is called causal **iff** $x_n = 0$ for all $n <0$. $\endgroup$ – Val Jul 31 '14 at 11:08
  • $\begingroup$ Ah, so casual was a misspelling. sorry. $\endgroup$ – 5xum Jul 31 '14 at 11:10
1
$\begingroup$

Fibonacci is defined as $$F_1 = 1, F_2 = 1\\F_{n+1} = F_n + F_{n-1}$$

and never hits zero since it is obviously increasing. If you would take $$G_1=0,G_2=0\\G_{n+1} = G_n + G_{n-1},$$

then of course the sequence will be constantly zero.


EDIT:

According to your link, a sequence is causal if $x_n=0$ for $n<0$, however, this means that the definition of causal sequences only applies to sequences where $x_n$ is defined both for positive and negative values of $n$. In the case of the Fibonacci sequence, however, $F_n$ is defined only for $n\geq 0$, so it is, by definition, not causal.

$\endgroup$
  • $\begingroup$ It seems there is a way to make the Fibonacci causal, math.stackexchange.com/questions/279868 $\endgroup$ – Val Jul 31 '14 at 11:38
  • $\begingroup$ @val There is a simpler way of proving that: Any causal sequence with a recursive formula $a_n=\alpha_1a_{n-1} +\dots + \alpha_k a_{n-k}$ is obviously constantly equal to $0$. This means Fibonacci is not causal. $\endgroup$ – 5xum Jul 31 '14 at 11:40

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.