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MASS Colloquium
Ramanujan, Fibonacci numbers, and Continued Fractions or
Why I took Zeckendorf ’s Theorem along my last trip to Canada
Notes of the talk given by George Andrews
By Loren Anderson and Jason Green
Abstract: This talk focuses on the famous Indian genius, Ramanujan. One objective
will be to give some account of his meteoric rise and early death. We shall try to lead
from some simple problems involving Fibonacci numbers to a discussion of some of
Ramanujan’s achievements including some things from his celebrated Lost Notebook.
1
A Guessing Game
To start, let us play a little game of numbers. Pick a number between 1 and 100. Now
write this number down somewhere (so you don’t forget it!) Do not tell me the number!
Now turn to the final section of these notes, titled “Game Cards”. This final section
has a total of 10 cards with many numbers on them. The Guessing Game proceeds as
follows:
1. Look at Card 1
2. Check to see if your number is on this Card
3. If your number is not on this Card, look at the next Card (and go back to Step 2)
4. If your number is on this Card, look at the next Card
5. I will guess whether your number appears on this Card
6. Let me think...no, your number is not on this Card
7. Look at the next Card (and go back to step 2)
Continue this process until you have looked at and checked all 10 cards.
You will find that I was able to correctly guess that if your number appeared on a
Card, that it would not appear on the next Card. Also, I would be able to tell you which
number you picked! But how did I know? Let me show you with an example.
I will pick a number between 1 and 100. Our number will be...72. Let’s check Card 1.
We see that 72 is on Card 1. So if we check Card 2, there is no 72. Let’s check Card 3.
We see that 72 is also on Card 3. So if we check Card 4, there is no 72. Let’s check Card
5. There is no 72, so we check Card 6. We see that 72 is on Card 6. So if we check Card
7, there is no 72. Let’s check Card 8. There is no 72, so we check Card 9. We see that 72
is on Card 9. So if we check Card 10, there is no 72.
1
Now, which Cards had 72 on them? The Cards with 72 were Card 1, Card 3, Card
6, and Card 9. Now take a look at the number in the upper left corner of each of these
Cards. They are 1, 3, 13, and 55. If you take their sum, you get 72! This idea will be
true for any number you pick. Don’t believe me? Give it another try! The reason for this
result is that the numbers in the upper left corner are a part of the sequence of Fibonacci
Numbers.
2
Fibonacci and Zeckendorf
The Fibonacci Numbers are a sequence of numbers defined recursively as follows:
F0 = 0,
F1 = 1,
Fn+1 = Fn + Fn−1 ,
n ∈ N.
So the first few terms of the sequence are
F0 = 0, F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5,
F6 = 8, F7 = 13, F8 = 21, F9 = 34, F10 = 55, · · · .
Zeckendorf’s Theorem: Every positive integer is uniquely a sum of distinct, nonconsecutive Fibonacci Numbers. For example:
72 = 55 + 13 + 3 + 1 = F10 + F7 + F4 + F2
43 = 34 + 8 + 1 = F9 + F6 + F2
54 = 34 + 13 + 5 + 1 = F9 + F7 + F5 + F2 .
So when we played the Guessing Game, each Card was created so that your number
would appear on the Card only if the Fibonacci Number was in the Zeckendorf decomposition of your number. Also, we were able to say that if your number appeared on a Card
then it would not appear on the next Card because Zeckendorf’s Theorem says a positive
integer is the sum of nonconsecutive Fibonacci Numbers.
3
Kilometers to Miles
Imagine driving from the U.S. to Canada on a nice summer’s afternoon. As soon as
you cross the border, the speed limit changes. Not only that, but instead of it being
in miles per hour, the speed is represented in kilometers per hour. Your car might only
display miles per hour on the speedometer. At first you simply take a guess at the speed
in miles per hour and hope that you don’t get pulled over. After a short time however, you
remember a few facts about the Fibonacci numbers that allow you to make a sufficient
conversion. The conversion from kilometers to miles is roughly 0.621 miles per kilometer.
n
is approximately 0.618. Combining this with Zeckendorf’s Theorem
Also, limn→∞ FFn+1
yields an approximation formula to convert kilometers to miles.
2
First, write the number of kilometers we have in terms of its unique sum of nonconsecutive Fibonacci Numbers, thanks to Zeckendorf’s Theorem, say
x = Fi1 + Fi2 + Fi3 + ... + Fik .
Then, multiply each Fk by 0.618 to yield
0.618x = 0.618Fi1 + 0.618Fi2 + 0.618Fi3 + ... + 0.618Fik .
Approximating 0.618 by
0.618x ≈
Fk−1
Fk
yields
Fi1 −1
Fi −1
Fi −1
Fi −1
Fi1 + 2 Fi2 + 3 Fi3 + ... + k Fik .
F i1
F i2
F i3
F ik
Simplifying yields
0.618x ≈ Fi1 −1 + Fi2 −1 + Fi3 −1 + ... + Fik −1 .
This is reasonably close to the true value of 0.621x. (and in fact, it is a bit smaller
than the value, ensuring that you don’t speed!).
For example, suppose you need to convert 80 kilometers into miles. Then by Zeckendorf’s Theorem, one can represent 80 as a unique sum of nonconsecutive Fibonacci
Numbers. Here, 80 = 55 + 21 + 3 + 1 = F10 + F8 + F4 + F2 . From above, we see that we
can find the number of miles by replacing each Fibonacci number with the previous one.
So F10 + F8 + F4 + F2 becomes
F9 + F7 + F3 + F1 = 34 + 13 + 2 + 1 = 50.
Therefore, 80 kilometers is approximately 50 miles. This allows one to estimate that
traveling 80 kilometers per hour is nearly the same as driving 50 miles per hour.
4
Continued Fractions 1
n
, under the assumption that the limit actually
Here, we evaluate x = limn→∞ FFn+1
exists. Since Fn+1 = Fn + Fn−1 , we have:
x≈
Fn
Fn
=
.
Fn+1
Fn + Fn−1
Dividing by Fn yields
x≈
1
1
=
.
Fn−1
Fn−1
1+
1+
Fn
Fn−1 + Fn−2
Dividing the nested fraction by Fn−1 yields
3
x≈
1
1
1+
Fn−2
1+
Fn−1
1
=
1
1+
Fn−2
1+
Fn−2 + Fn−3
.
Letting n → ∞,
1
x=
(∗)
1
1+
1+
1
..
.
We can make a clever observation to yield
x=
1
.
1+x
√
Solving this equation yields x2 + x − 1 =√0, so x = −1±2 5 . The sequence is clearly
positive, so we take the positive solution, −1+2 5 , which is approximately 0.618.
5
Continued Fractions 2
Continued fractions have been studied by many mathematicians. People considered
the quantity
1
1+
1+
|q| < 1.
,
q
q2
q3
1+
.
1 + ..
Notice that when q = 1, it is equal to fraction (*) above.
Ramanujan saw this problem and considered
C(x, q) =
x
.
xq
1+
xq 2
1+
xq 3
1+
1 + ...
This can be rewritten as
C(x, q) =
x
1 + C(xq, q)
4
(1)
Using this equation, Ramanujan deduced some of the following equations.
C(x, q) = x − x2 q + x3 q 2 + (−x4 + x3 )q 3 + (x5 − 2x4 )q 4 + (−x6 + 3x5 − x4 )q 5 + ...
−C(−x, q)
= 1 + xq + x2 q 2 + (x3 + x2 )q 3 + (x4 + 2x3 )q 4 + (x5 + 3x4 + x3 )q 5 + ...
x
Also,
−C(−x, 1)
1
2x
2
3
4
5
6
7
= 1 + x + 2x + 5x + 14x + 42x + 132x + 429x + ... =
.
x
x+1 x
This shows that −C(−x,1)
is the generating function for the Catalan Numbers.
x
Rogers, Schur, and Ramanujan showed that the initial continued fraction (1) is equal
to this product:
∞
Y
(1 − q 5n−4 )(1 − q 5n−1 )
n=1
(1 − q 5n−3 )(1 − q 5n−2 )
.
As q approaches 1 from the left, the product approaches
∞
Y
(5n − 4)(5n − 1)
(5n − 3)(5n − 2)
which converges to the value
6
n=1
√
−1+ 5
,
2
which we showed previously.
Brief History
Srinivasa Ramanujan was born in 1887 in India to a poor Brahmin family. He was
immediately realized as a math prodigy and won many awards and contests in high school.
In recognition of his intellectual prowess, he won a scholarship for Government College,
yet he failed other classes and lost it. Due to his bleak financial situation, he applied
for a dead-end job in Madras Port Trust. Ramanujan became an accounting clerk in
1912 earning 30 pounds per year. In 1913, he wrote letters to Hobson and Baker to
convince them of his talent, but he was ignored. He wrote another letter addressed to
Hardy. Hardy had an advantage: he recognized various theorems that he himself had
trouble proving. Hardy decided to invest time in Ramanujan, urging Ramanujan to come
to England. After about a year of contemplating a trip to a foreign land, Ramanujan
set sail for England to work with Hardy at Cambridge University. Together, they wrote
papers on the following topics: circle method, probabilistic number theory, applications
of modular forms to number theory. It is regarded as a truly great mathematical event
of the 20th century. In 1918, Ramanujan contracted an illness that was believed to be
5
tuberculosis, but no one, even to this day, is truly sure. He returned to India and lived
for a year before passing away at the age of 32. Ramanujan’s contributions are lauded to
this day. His birthday, December 22, is celebrated as National Mathematics Day in India.
Also, they declared the year 2012 as National Mathematics Year.
7
Lost Notebook
In January of 1920, Ramanujan wrote a final letter to Hardy. Within the letter he made
no mention of his declining health, but did mention what he referred to as “mock-theta
functions”. These functions are similar to the theta functions widely used in number
theory and analysis. Ramanujan wrote about wanting to extend the idea of the theta
functions. Dr. Andrews studied mock-theta functions during his time in graduate school,
eventually writing his thesis about them. 1976, Dr. Andrews attended a conference in
Europe. The cost of visiting Europe for more than 3 weeks was , while the cost of visiting
for less than that was 1 for small . After the conference, Dr. Andrews used some of his
remaining time to visit Cambridge University to view some old documents. Whilst there,
he came upon a notebook with equations like the following:
q
1 − 2q cos 2nπ
+ q2
5
q4
+
+ ···
2 )(1 − 2q cos 2nπ + q 4 )
(1 − 2q cos 2nπ
+
q
5
5
F (q) = 1 +
6
q5
1
+
1 − q (1 − q)(1 − q 4 )(1 − q 6 )
q 20
+
+ ···
(1 − q)(1 − q 4 )(1 − q 6 )(1 − q 9 )(1 − q 11 )
1
q5
ψ(q) =
+
1 − q 2 (1 − q 2 )(1 − q 5 )(1 − q 7 )
q 20
+ ···
+
(1 − q 2 )(1 − q 3 )(1 − q 7 )(1 − q 8 )(1 − q 12 )
φ(q) =
and immediately recognised the handwriting as that of Ramanujan. Dr. Andrews had
found what is known today as the “Lost Notebook” of Ramanujan. The formulas and
equations of Ramanujan are still being explored today.
8
Game Cards
1
22
43
64
85
4
25
46
67
88
6
27
48
69
90
9
30
51
72
93
12
33
53
74
95
14 17 19
35 38 40
56 59 61
77 80 82
98 101 103
Card 1
2
36
70
104
138
7
10 15 20 23 28 31
41 44 49 54 57 62 65
75 78 83 86 91 96 99
109 112 117 120 125 130 133
143 146 151 154 159 164 167
Card 2
3
32
58
87
113
4
11 12 16 17 24 25
33 37 38 45 46 50 51
59 66 67 71 72 79 80
88 92 93 100 101 105 106
114 121 122 126 127 134 135
Card 3
5
28
61
94
117
6
7
18 19 20 26 27
39 40 41 52 53 54 60
62 73 74 75 81 82 83
95 96 107 108 109 115 116
128 129 130 141 142 143 149
7
Card 4
8
32
64
88
120
9
10 11 12 29 30 31
33 42 43 44 45 46 63
65 66 67 84 85 86 87
97 98 99 100 101 118 119
121 122 131 132 133 134 135
Card 5
13
47
68
102
136
14 15 16 17 18 19 20
48 49 50 51 52 53 54
69 70 71 72 73 74 75
103 104 105 106 107 108 109
137 138 139 140 141 142 143
Card 6
21
29
79
87
116
22 23 24 25 26 27 28
30 31 32 33 76 77 78
80 81 82 83 84 85 86
88 110 111 112 113 114 115
117 118 119 120 121 122 165
Card 7
34
42
50
126
134
35 36 37 38 39 40 41
43 44 45 46 47 48 49
51 52 53 54 123 124 125
127 128 129 130 131 132 133
135 136 137 138 139 140 141
Card 8
55
63
71
79
87
56 57 58 59 60 61 62
64 65 66 67 68 69 70
72 73 74 75 76 77 78
80 81 82 83 84 85 86
88 199 200 201 202 203 204
Card 9
89
97
105
113
121
90 91 92 93
98 99 100 101
106 107 108 109
114 115 116 117
122 123 124 125
Card 10
8
94
102
110
118
126
95
103
111
119
127
96
104
112
120
128