Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Fall 2016 MATH 1300-101
Last updated: November 2016
Comments and suggestions: Email wim.ruitenburg@marquette.edu
Book, chapter 13 on 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 page 393.
The Fibonacci sequence started as a simple example or game.
Mathematicians discovered many interesting properties about Fibonacci
numbers.
Fibonacci numbers occur in art, biology, computer science.
Use and understanding partly flows from the mathematics done before.
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.
Mathematicians look for patterns
Here is a pattern from page 396.
Following the book, consider the growth rate of the terms of the
Fibonacci sequence.
It lists
F(2) / F(1) = 1/1 = 1
F(3) / F(2) = 2/1 = 2
F(4) / F(3) = 3/2 = 1.5
F(5) / F(4) = 5/3 = 1.66666667 approximately
F(6) / F(5) = 8/5 = 1.6
F(7) / F(6) = 13/8 = 1.625
F(8) / F(7) = 21/13 = 1.61538462 approximately
As this pattern suggests, the consecutive quotients approach some constant
number we call φ (Greek letter phi) or G, for Golden Ratio.
When we keep extending the table above by more quotients, we see that
φ = G = 1.618033988749894848204586834 approximately
When we stop our investigations just because we can approximate φ with
arbitrary precision, we potentially lose new patterns.
So we search for equations involving φ = G.
If F(100000) = N for some big N, then F(100001) = N φ and F(100002) = N
φ^2 approximately.
Since F(100002) = F(100001) + F(100000), we get φ^2 = φ + 1.
So φ is root of X^2 - X - 1 = 0.
Solve this and we get φ = (1 + 5^(1/2)) / 2.
Here is another way to get this same Golden Ratio.
One easily shows that the quotients in the list above can also be written as
These nested quotients have a convergence property by which it is reasonable to
write
G = 1 + 1/G (or: φ = 1 + 1/φ)
Multiply both sides by G.
So G^2 = G + 1. So G^2 - G - 1 = 0. When we solve this quadratic equation, we
get roots (1 + 5^(1/2)) / 2 and (1 - 5^(1/2)) / 2.
The correct solution is
φ = G = (1 + 5^(1/2)) / 2
Note the presence of 5^(1/2), also written sqrt(5), the square root of
5.
Mathematicians look for more patterns
Here is a pattern involving divisibility: Each third term is even.
So F(0) = 0 is even, F(3) = 2 is even, F(6) = 8 is even, F(9) = 34 is even,
F(12) = 144 is even, and so on.
These are the only even terms.
Mathematicians often write m = 0 modulo 2 to indicate that m is even.
Similarly, mathematicians often write m = 0 modulo 3 to indicate that m is
divisible by 3.
And so on.
With this notation, the pattern we discovered, can be written as
F(n) = 0 modulo 2, exactly when n = 0 modulo 3
Mathematicians can easily prove this statement.
Once mathematicians see one pattern, they look for more of the same
kind.
Here is one: Each 4th term is divisible by 3.
So F(0), F(4), F(8), F(12), F(16), F(20), and so on, are divisible by 3, and
no other F(n) are divisible by 3.
Each 5th one is divisible by 5.
Each 8th one is divisible by 7.
Here is a list of some of these patterns.
F(n) = 0 modulo 2, exactly when n = 0 modulo 3
F(n) = 0 modulo 3, exactly when n = 0 modulo 4
F(n) = 0 modulo 5, exactly when n = 0 modulo 5
F(n) = 0 modulo 7, exactly when n = 0 modulo 8
F(n) = 0 modulo 11, exactly when n = 0 modulo 10
F(n) = 0 modulo 13, exactly when n = 0 modulo 7
F(n) = 0 modulo 17, exactly when n = 0 modulo 9
F(n) = 0 modulo 19, exactly when n = 0 modulo 18
F(n) = 0 modulo 23, exactly when n = 0 modulo 24
F(n) = 0 modulo 29, exactly when n = 0 modulo 14
Now we have a pattern of patterns.
For each prime number p we find a number k such that
F(n) = 0 modulo p, exactly when n = 0 modulo k
Mathematicians can prove that this pattern holds.
Mathematicians now look for patterns inside this pattern of
patterns.
Here is one:
F(3) = F(2+1) = 0 modulo 2
F(4) = F(3+1) = 0 modulo 3
F(5) = F(5+0) = 0 module 5. An exception in the pattern
F(8) = F(7+1) = 0 modulo 7
F(10) = F(11-1) = 0 modulo 11
F(14) = F(13+1) = 0 modulo 13
F(18) = F(17+1) = 0 modulo 17
F(18) = F(19-1) = 0 modulo 19
F(24) = F(23+1) = 0 modulo 23
F(28) = F(29-1) = 0 modulo 29
So a pattern we see is: For all primes p with the exception of p = 5, we have
that
Either F(p-1) = 0 modulo p, or F(p+1) = 0 modulo p
For example, choose p = 13.
Then either F(12) = 144 is divisible by 13 (it isn't) or F(14) = 377 is
divisible by 13 (it is).
Mathematicians can prove these patterns.
Why is 5 so special?
There is a connection with the occurrence of sqrt(5), also written 5^(1/2), in
the formula for the golden ratio φ = (1 + sqrt(5))/2.
Fibonacci Nim
We are given a pile of two or more pennies.
Two players alternatingly must take pennies from the pile, subject to the
following rules:
Whoever moves first, must take at least one penny, but not all.
In response to the previous person's move of some pennies from the pile,
the other player must take at least one penny, but at most twice as much as the
other player just took.
You win when you empty the pile so that the (losing) player can not make a next
move.
Winning strategy:
In class we hinted at writing the number of pennies as a sum of non-consecutive
Fibonacci numbers.
Fibonacci on the Web
Some sample websites on Fibonacci numbers, collected with Phillip van Hoven:
Recommended problems from Chapter 13 of the book:
1, 3
Recommended problems from Chapter 13 of the book:
7, 14c, 18ab
We are given that the Golden Ratio φ = 1.618033988749894848204586834
approximately.
Compute its square φ^2 to 10 decimal places without using a calculator.
One of the Fibonacci numbers F(1001), F(1002), F(1003), and F(1004) is
divisible by 3.
Which one and why?
You are given that 89 is a prime number.
Either F(88) or F(90) is divisible by 89.
Which one of these two Fibonacci numbers is divisible by 89, and why?
Suppose we play Fibonacci Nim starting with a pile of 88 pennies.
I begin by taking 33 pennies.
Do you have a winning countermove, and if so what is it?
Suppose we start a game of Fibonacci Nim with a pile of 13 pennies.
You decide whether you are the beginning player or the second player in trying
to win the game.
What is the best choice, and why?
Suppose we start a game of Fibonacci Nim with a pile of 20 pennies.
You decide whether you are the beginning player or the second player in trying
to win the game.
What is the best choice, and why?
We play Fibonacci Nim.
With my turn I take 12 pennies from a pile of 88 pennies, and leave it with 76
pennies.
Do you have a countermove that guarantees that you can ultimately win?
If so, what is that countermove?