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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