# Fibonacci, Lucas, and the square root of five

Discussion in 'Recreational Math' started by walt, Apr 3, 2005.

1. ### waltGuest

I have little formal training in mathematics, but I would guess that
the answer to this question is already well known. Kindly clue me
in if you understand the following:

The square-root of five can be calculated to any arbitrary precision
on a common spreadsheet by doing the following:

1) In column one of a spreadsheet create the Fibonacci numbers
by adding the previous two numbers to make the next in the series.
Start with 1 + 1 = 2, and continue with 1 + 2 = 3, etc.

2) In column two, create the so-called Lucas numbers, which are
closely related to the Fibonacci numbers except that the two
starting numbers are 1 and 3 instead of 1 and 1.

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

and so on...

Now, the part that really puzzles me is this: in the third column,
divide column 2 by column 1 and you will see that the result converges
on the square root of five (2.23606........)

My question, of course, is WTF is special about the square root of five?

The Fibonacci ratio is given by phi = (Sqrt(5) + 1)/2 so I'm not too
surprised to see it appear in my puzzle, but I just can't figure out
why the Lucas numbers should be related to the Fibonacci numbers by
exactly that ratio.

walt, Apr 3, 2005

2. ### Stan LiouGuest

The reccurence relation is F(n+2) = F(n+1) + F(n+1) in both
cases, only subject to different initial values of F(0) and
F(1), so the characteristic polynomial is the same in both
cases: x^2 - x - 1. The roots of this determine the closed-
form solution.

One can also think of it in terms of limits of n->infinity:
L = lim[F(n+1)/F(n) ]
= lim[F(n+2)/F(n+1) ]
= lim[F(n+1)/F(n+1) + F(n)/F(n+1)]
= 1 + L
thus L^2 - L - 1 = 0. Notice that the initial values of
F(0), F(1) are completely irrelevant here. Any sequence
defined by this recurrence relation, be it Fibonacci or
Lucas or whatever, has the same ratio limit.

Stan Liou, Apr 3, 2005

3. ### Stan LiouGuest

Pardon, L = 1 + 1/L.

Stan Liou, Apr 3, 2005
4. ### ProginoskesGuest

That should be: F(n+2) = F(n+1) + F(n).
Any sequence which satisfies F(n+2) = F(n+1) + F(n) is of the
form F(n) = A ((1 + sqrt(5))/2)^n + B ((1 - sqrt(5))/2)^n,
for certain constants A and B.
That last line should be 1 + 1/L.
--- Christopher Heckman

Proginoskes, Apr 3, 2005
5. ### nightGuest

I don't know, but here's something interesting.

If you start the first sequence with 1, 1, ... and the second sequence
with 1, N, ..., then the limit of the ratio is a root of the quadratic:

x^2 + (N-3) x + (1+N-N^2) = 0

The positive root of which is:

((sqrt(5) - 1)/2) N + (3 - sqrt(5))/2

Or, where phi = (sqrt(5) - 1)/2:

(N-1)phi + 1

When the second sequence is the Lucas numbers, N=3, so the limit of the
ratio is 2phi + 1 = sqrt(5).

night, Apr 3, 2005
6. ### Travis WillseGuest

-
Expanding some:

The Fibonacci and Lucas sequences are given by the rule

f_n = f_{n-1}+f_{n-2}

This formula has the characteristic equation

r^2-r-1 = 0

Solving gives (1/2)*(+/-1+Sqrt). Setting

A = (1/2)*(1+Sqrt) (the Golden Ratio)
B = (1/2)*(1-Sqrt) (the negative of the reciprocal of the Golden Ratio),

any sequences defined by the above rule is given by

c_1*A^n+c_2*B^n for some choice of c_1, c_2.

For the Fibonacci sequence, we set f_1 = f_2 = 1 and find have c_1 =
1/Sqrt, c_2 = -1/Sqrt. For the Lucas sequence, we have c_1 = c_2 = 1.

Let's consider the ratio of a given Lucas number L_n and a given
Fibonacci number F_n. Explicitly,

L_n/F_n = (A^n+B^n)/(1/Sqrt*A^n-1/Sqrt*B^n)

|B| < 1, so as n gets very large, B^n approaches zero. So for large n,
we have

L_n/F_n ~ A^n/(1/Sqrt*A^n) = Sqrt

Cheers,
Travis

Travis Willse, Apr 3, 2005
7. 