Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Spring 2004 MATH025.1001
The Fibonacci Sequence
The Fibonacci sequence is the sequence of whole numbers 0, 1, 1, 2, 3, 5, 8,
13, 21, 34, 55, 89, 144, and so on.
Combinations of the numbers in this sequence occur in nature, see the book on
page 48.
Let us present these Fibonacci numbers as a function F(n).
When we put the function in a table, the result looks like
n
F(n)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
....
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
....
So F(3) = 2 and F(11) = 89.
So far we understand the definition of the sequence of Fibonacci numbers
somewhat informally.
An example of a precise mathematical definition is:
F(0) = 0 and F(1) = 1, the initial conditions part
of the definition.
F(n) = F(n-1) + F(n-2), for all n greater than or equal 2
the recursive part of the definition.
With these conditions, the Fibonacci sequence is completely and uniquely
determined.
With the definition given, the calculation of a term like F(100),
requires the calculation of the first 100 terms F(0), F(1), F(2), and so on
all the way to get to F(100).
Let us experiment by starting at the other end:
F(100) = F(99) + F(98) =
2F(98) + F(97) =
3F(97) + 2F(96) =
5F(96) + 3F(95) =
8F(95) + 5F(94) =
13F(94) + 8F(93) = and so on
We observe a pattern, like F(100) = F(7)F(94) + F(6)F(93).
Good mathematics requires that we convert this into a general pattern:
F(100) = F(n+1)F(100-n) + F(n)F(100-n-1)
and then try to prove that it is valid (it is).
So, for example, F(100) = F(31)F(70) + F(30)F(69).
A good mathematician makes other observations like: In the calculations above
there is nothing special about the number 100.
We could start with another number say k, instead of 100, and so derive the
still more general pattern
F(k) = F(n+1)F(k-n) + F(n)F(k-n-1)
In fact, we could replace the function F by another function satisfying
H(n) = H(n-1) + H(n-2) for all n greater than or equal 2 and still get
If H(n) = H(n-1) + H(n-2) for all n greater than or equal 2, then H(k)
= F(n+1)H(k-n) + F(n)H(k-n-1)
Let us look for a very special G(n) satisfying G(n) = G(n-1) + G(n-2)
for all n greater than or equal 2.
Namely, try G(n) = g^n for some (positive) value of g.
Recall that
g^0 equals 1
g^1 equals g
g^2 equals g times g
g^3 equals g times g times g
and so on.
We have not yet found g, but its table would look like
n
G(n)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
....
1
g
g^2
g^3
g^4
g^5
g^6
g^7
g^8
g^9
g^10
g^11
g^12
g^13
g^14
g^15
g^16
....
A number g will make the equation G(n) = G(n-1) + G(n-2) work for all n greater
than or equal to 2, exactly when g^2 = g + 1 or, equivalently
g^2 - g - 1 = 0
When we solve this quadratic equation, we get two solutions
g_1 = (1 + sqrt(5))/2 as positive root (approximately 1.618033989)
g_2 = (1 - sqrt(5))/2 as negative root (approximately -0.618033989)
The positive root g= g_1 is known as the golden ratio.
Additionally, the second root equals g_2 = -1/g.
Now we have, for example, g^28 = g^27 + g^26 and so on.
Bu also (-1/g)^54 = (-1/g)^53 + (-1/g)^52, and so on.
A Fibonacci style sequence K(0), K(1), K(2), K(3), K(4), and so on,
satisfies the recursive equation
K(n) = K(n-1) + K(n-2), for all n greater than or equal to 2.
So if, additionally, we pick values for K(0) and K(1), then the whole sequence
is uniquely determined.
Here is an example:
Using the recursive relations, we get K(2) = g^2 + (-1/g)^2; and K(3) = g^3 +
(-1/g)^3; and K(4) = g^4 + (-1/g)^4; and K(5) = g^5 + (-1/g)^5; and so on.
Now -1/g = g_2 is approximately equal to -0.618033989, which is less than 1
from 0.
So higher powers (-1/g)^2, or (-1/g)^3, or (-1/g)^4, must get smaller and
smaller.
Thus if n is big, then K(n) is approximately equal to g^n.
For example, K(5) = 11, while g^5 = 11.090167 approximately.
And K(8) = 47, while g^8 = 46.978714 approximately.
The accuracy gets better with higher powers.
For example, K(14) = 843, while g^14 = 842.9988 approximately.
We started off with the Fibonacci sequence F(0), F(1), F(2), ... near
the top of the page.
We sidetracked and ended up with the numbers g_1 = g (the golden ratio) and g_2
= -1/g.
In good mathematics it frequently occurs that, in the end, one gets back to the
original work.
In this case, we can show that g^0 / sqrt(5) - (-1/g)^0 / sqrt(5) equals 0,
and g^1 / sqrt(5) - (-1/g)^1 / sqrt(5) equals 1.
Let us start a Fibonacci style sequence with
M(0) = g^0 / sqrt(5) - (-1/g)^0 / sqrt(5) = 0
M(1) = g^1 / sqrt(5) - (-1/g)^1 / sqrt(5) = 1
satisfying M(n) = M(n-1) + M(n-2) for all n greater than or equal 2.
With the recursive relations for g_1 and g_2, we get
M(2) = g^2 / sqrt(5) - (-1/g)^2 / sqrt(5)
M(3) = g^3 / sqrt(5) - (-1/g)^3 / sqrt(5)
M(4) = g^4 / sqrt(5) - (-1/g)^4 / sqrt(5)
M(5) = g^5 / sqrt(5) - (-1/g)^5 / sqrt(5)
M(6) = g^6 / sqrt(5) - (-1/g)^6 / sqrt(5)
and so on
Here is a little surprise: Since M(0) = F(0) and M(1) = F(1), the new Fibonacci
style sequence M(0), M(1), M(2), ... and the original Fibonacci sequence F(0),
F(1), F(2), ... must be the same.
So
F(n) = g^n / sqrt(5) - (-1/g)^n / sqrt(5), for all n
Since (-1/g)^n is very tiny for big n, we have F(n) approximately equals g^n /
sqrt(5), for bigger n.
The approximations are better when n is bigger.
For example, F(14) = 377, while g^14 / sqrt(5) = 377.00053 approximately.
Last updated: February 2004
Comments & suggestions:
wimr@mscs.mu.edu