Fiddling With Fibonacci

While fiddling with his calculator, Tony Buckland noticed that the 
eighth root of 2207, minus 2, is almost equal to (sqrt(5) - 1)/2, 
to seven significant figures, and asked if this was merely a 
coincidence.  To understand why this gives such a close approximation
to the Fibonacci ratio, recall that the characteristic polynomial of
the Fibonacci sequence is  x^2 - x - 1, which has the roots 

          u  = (1+sqr(5))/2 =  1.618...

          v  = (1-sqr(5))/2 = -0.618...

The Lucas sequence consists of "Newton's sums" for this polynomial,
i.e., the nth Newton sum is the sum of the nth powers of the roots,
and since these are symmetric functions they are necessarily integers.
The first several Lucas numbers are

    2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 
  521, 843, 1364, 2207, 3571, 5778, ...

(Naturally these numbers satisfy the Fibonacci recurrence relation.)
Notice that 2207 is L(16), so we have 2207 = u^16 + v^16.  However,
since |v| is less than 1, it's clear than raising v to a high power
will yield an extremely small contribution, and the value of v^n goes
to zero as n increases.  Thus we can say that L(n) ~ u^n  for
sufficiently large n.  Also, remember that u^2 = u+1  (from the 
characteristic polynomial), so u^16 can be written as (1+u)^8.
Consequently we have 2207 ~ (1+u)^8 which, as you observed on your
calculator, implies 2207^(1/8) ~ 1+u = -v+2.  The next such case 
is 5778^(1/9) = 2.618033997... and so on.

You might wonder about the odd-order Lucas numbers, such as L(15) = 
3571.  Remember this is approximately u^15, so we can't make the 
substitution for u^2 and cut the exponent in half.  So our 
observation here is simply the fact that the *fifteenth* root 
of 3571 is 1.618033931...

You might also notice the relation between the continued fraction 
convergents of sqrt(5) and the expansions of (2+sqrt(5))^n.  The
continued fraction of sqrt(5) is 

                   1
          2 + -----------
                      1
               4 + ---------
                         1
                    4 + --------
                              1
                         4 + ------
                              etc.

so the convergents are

 2,  9/4,  38/17,  161/72,  682/305,  2889/1292,  12238/5473, ...

If we denote the nth convergent as  A_n / B_n,  it's easy to show
that
          (2+sqrt(5))^n  =  A_n + B_n sqrt(5)

Notice that 2A_n = L(3n), showing that the previous approximations
are essentially based on the continued fraction of sqrt(5).