On 1/16/2012 11:05 AM, Invisible wrote:
> To many people, this is a seemingly magical property. BUT WAIT! It turns
> out that if you apply the same rules but starting with ANY NUMBERS YOU
> WANT, the same thing happens!
>
> Pic[0] = 31
> Pic[1] = 42
> Pic[n] = Pic[n-1] + Pic[n-2]
>
> 31 + 42 = 73
> 42 + 73 = 115
> 73 + 115 = 188
> 115 + 188 = 303
> 188 + 303 = 491
> 303 + 491 = 794
>
> 42 / 31 = 1.354 838 709 677...
> 73 / 42 = 1.738 095 238 095...
> 115 / 73 = 1.575 342 465 753...
> 188 / 115 = 1.634 782 608 695...
> 303 / 188 = 1.611 702 127 659...
> 303 / 491 = 1.620 462 046 204...
> φ ≈ 1.618 033 988 749...

If you start with any two integers, at least one of which is not zero, 
and proceed, the series that follows can be constructed from some 
combination of the original Fibonacci series, with some of the input 
series being shifted, and some being multiplied by some factor.

For instance, if you start with 1 and 3, the series you get is:

  1 3 4 7 11 18 29 47 76 123

Now subtract from that the same series beginning with 0 and 1, and you get:

  1 3 4 7 11 18 29 47 76 123
  0 1 1 2  3  5  8 13 21  34
---------------------------
  1 2 3 5  8 13 21 34 55  89

Which is the (0,1) series, shifted to the left by two members.

Since the two component series have member-to-member ratios that 
converge to phi, intuition suggests that the sums of individual series 
members will also have a ratio converging to this same value.

The rigorous proof that all such series will have a member-to-member 
ratio approaching phi will entail showing that for any two non-negative 
integers, A and B, with B>A, that the value of B/(A+B) is closer to phi 
(1.618~) than the value A/B.

|A/B - phi| > |B/(A+B) - phi| for all A and B, 0 < A < B

The full proof is left as an exercise for the interested student.

Regards,
John

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