Hi.

I'm simply curious about one thing. Is it possible to make a Fibonacci... whatever it's called again.... a Fibonacci list which converges? I myself haven't found it, still, I just want to know.

Can you give an example of what you mean?

The Fibonacci sequence is:

1,1,2,3,5,8,13,21,34,55,....

If you take the ratios of successive terms, the ratio converges to the golden ratio as you take larger and larger pairs of terms.

But maybe you're talking about something else.

What kind of list are you referring to?

Perhaps he was asking about generalized Fibonacci sequence, i.e. sequences with the same recurrence relation f(x) = f(x-1) + f(x-2), but with other starting numbers. If so, the answer is no. Any sequence with that recurrence relation will always grow exponentially, either positive or negative.

I actually meant about a sequence (thanks for the term), I mean, Fibonacci sequence which starts from any number but 0, and grows to 0 (impossible as far as I know.. once it hits 0, it will start to diverge), or at least oscillating between positive and negative.... is it possible?

f(x) = f(x-1) + f(x-2)

For the defenition of f(x) to be complete, you must define a constant value for f(0). For example; f(0)=1 (which gives the fibonacci serie)

There are no values for f(0) that will make this serie converge (other than 0).

It can oscillate between positive and negative, though. Try starting with f(0)=1 and f(1)=(-2/(1+squareroot[5])).
By the way, if you start with f(1)=1 and f(1)=(1+sqrrt[5])/2, which is the golden ratio - call it tau, then f(x+1) will always equal tau times f(x), proving that 1+tau=tau^2, tau+tau^2=tau^3, etc.
Aaron

Ah okay. I'll do the math if I have time. Thank you :) So the point is that FIbonacci sequence can never converge, though it may oscilate, right?