Fibonacci sequences and phi

[SLD]

We all know that if you take the Fibonacci sequence out far enough you come to the point where the ratio of the next sequential number is phi, the golden ratio. (1+sqr(5))/2 or about 1.618.

But I wondered, why start with 0 and 1? Why not start with any two random numbers, say 7 and 3? Which creates a different set of numbers than 3 and 7. I tried it several times and I always got phi!

Can this be proven though? For any two random numbers will the Ratios always converge to phi?

SLD

 

[Kharakov]

I recall that the generating function for fibs (with a=1, n>=1, n integer for fibs) is:

fixed... after looking at wiki and checking in Maxima... lol...

\(f(n) \,\, = \,\,\frac{ \left( \frac{1+\sqrt{4a +1}}{2} \right ) ^n - \left( \frac {1-\sqrt{4a +1} } {2} \right ) ^n } {\sqrt{4a+1}}\)


You can get other integer sequences with other a. You can see that with large enough n, the second part approaches 0 (it is less than one, so as n approaches infinity, it approaches 0).

This means (because the sqrt(4a+1) parts on the bottom cancel out):

\( \lim_{n\to\infty} \,\, \frac{f(n+1)} {f(n)} \,\,= \,\, \frac{1+\sqrt{4a+1} } {2}\)


I've no clue about other integer sequences approaching phi. I totally forget.... so am going to cheat and read about it. I would think that they would have to have a generating function that had successive powers of phi, and another part that approached zero, or some combination of generating parts that eventually have parts that approach phi with others canceling out.


duhh... I googled it... and it's obvious once you see it... if you do a bit of algebra, you'll get it. Start with any numbers a and b.....