Combinational Mathematics Part 2

Transcription

1 j1 Combinational Mathematics Part 2 Jon T. Butler Naval Postgraduate School, Monterey, CA, USA We are here I live here Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 1 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 2 Naval Postgraduate School San Francisco Monterey Home Los Angeles Monterey Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 3 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 4 High School & University Rensselaer Polytechnic Institute Troy 7000 Dallas students B.E.E. & M.Engr. High School Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 5 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 6 1

2 Slide 1 j1 jbutler, 2010/03/24

3 Ph.D. & First Job Ohio State Univ., Evanston Columbus 60,000 students Ph.D. Northwestern Univ., Evanston Columbus 16,000 students 15 years Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 7 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 8 Naval Postgraduate School, Monterey 2000 students 24 years Monterey Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 9 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 10 A Few Puzzles 1. Children Paradox Mr. Yamada has two children. At least one is a boy. What is the probability that the other is a boy also? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 11 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

4 There are four equally likely outcomes Older Younger Boy Boy Boy Girl Girl Boy Girl Girl (impossible) Since only one of three possible outcomes consists of two boys, the probability is 1/3! Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Playing Keno Keno is a game in which you choose 6 numbers from 1 to 80 and the casino chooses 20 numbers from 1 to 80. You enter your choice of numbers for 100 yen. The casino pays you back depending on how many matches occur. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 14 Playing Keno For example, if all 6 of your numbers match numbers chosen by the casino, you win 100,000 yen! Question: What is the probability P(k) that k of the numbers you chose match numbers chosen by the casino? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 15 Playing Keno Number of ways k numbers can match. P(k) = C(80,k)C(80-k,20-k)C(60,6-k) C(80,20)C(80,6) Number of ways casino can choose 20 numbers. Number of ways casino can choose the rest of its numbers. Number of ways person can choose the rest of its numbers. Number of ways person can choose 6 numbers. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 16 Playing Keno Playing Keno Probability of matching k numbers Exp. the results of an experiment done in a class of 40 students. Each student chose 6 numbers and the teacher choose 20. k P(k) Exp Probability k Probability Experiment Exact Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 17 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

5 How to Become Rich by Playing Keno You cannot. Spend your money somewhere else. 3. Yen Paradox A coin has two sides Head (H) and Tails (T). In this game, a coin is flipped producing H or T. To play, you pay 100 yen. If it is T, the coin is flipped again. If it is H, you receive some money, as shown on the next slide. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 19 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 20 Yen Paradox Outcome H TH TTH TTTH TTTTH TTTTTH You receive 100 yen 200 yen 400 yen 800 yen 1600 yen 3200 yen How much should you pay to play this game? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 21 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 22 Yen Paradox Of course, if you pay 100 yen, you should play because you will always receive at least 100 yen. However, should you play if it costs 10,000 yen instead of 100 yen? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 23 Yen Paradox You can expect to receive 1_ 1_ You should pay ANY amount to play!! Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

6 4.Paths in a Chessboard You are here. Go here. Only go EAST or NORTH. How many paths are there? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 25 Paths in a Chessboard Make 7 EAST moves and 7 NORTH moves. Once you make the EAST moves, the rest are NORTH moves. There are C(14,7) = 3,432 ways to choose the EAST moves. Thus, there are 3,432 different paths. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Goats and Car Behind three doors are a car and two goats. You win what is behind the door you choose. The car is more valuable than a goat; you want the car. However, once you choose, the host opens a door behind which is a goat and asks if you want to change your answer. Should you? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 27 There are three possibilities Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 28 Without any information, your chances of choosing the car is 1/3. However, if you change, your chances are 2/3. That is, if you choose 1 and change, you lose. But if you choose 2 or 3 and change, you win. That is, in 2 out of 3 equally likely possibilities, you win. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Impossible Theorem Theorem: Let D = A + Δ. Then, D = A for all values of Δ. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

7 Proof: Multiply both sides by D A. D 2 DA = DA +D A 2 A Move D to left side D 2 DA D = DA A 2 A Factor out D on left and A on right. Proof: Factor out D on left and A on right. D (D A ) = A (D A ) Divide each side by D A D = A! Q.E.D. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 31 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 32 Proof: Factor out D on left and A on right. D (D A ) = A (D A ) Divide each side by D A D = A! D A is 0 We are dividing both sides by 0. Q.E.D. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Geography Puzzle A person walks one kilometer south, one kilometer east, and one kilometer north and arrives at the start point. He sees a bear! What color is the bear? Discuss later. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 34 Outline Combinatorial Math Part 2 Introduction Generating Functions Intro Using Generating Functions Solving Recurrence Relations Fibonacci Numbers in Nature Outline Combinatorial Math Part 2 Introduction Generating Functions Intro Using Generating Functions Solving Recurrence Relations Fibonacci Numbers in Nature Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 35 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

8 Ordinary Generating Functions Example: How many ways are there to make change for 50? Consider a harder problem How many ways are there to make change for X using 1 5 and 10? One way is shown here But, there are other ways. = List all ways, as shown on next slide. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 37 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 38 Consider a harder problem Choose one from each row This means choose none. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 39 Examples Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 40 List the choices using + OR and. AND ( + ( + ( ) + ) + ) Now substitute 1 for, x for 1, x 5 for 5, and x 10 for 10. (1 + x + x x + x x x + ) (1 + x 5 + x 5 x 5 + x 5 x 5 x 5 + ) (1 + x 10 + x 10 x 10 + x 10 x 10 x 10 + ) Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 41 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

9 Now, multiply this out, yielding (1 + x + x 2 + x 3 + x 4 + 2x 5 + 2x 6 + 2x 7 + 2x 8 + 2x 9 + 4x x x x x x x 17 + ) For example, there is only one way to make change for 4 ( ). Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 43 Now, multiply this out, yielding (1 + x + x 2 + x 3 + x 4 + 2x 5 + 2x 6 + 2x 7 + 2x 8 + 2x 9 + 4x x x x x x x 17 + ) There are two ways to make change for 5 ( and 5). Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 44 Now, multiply this out, yielding (1 + x + x 2 + x 3 + x 4 + 2x 5 + 2x 6 + 2x 7 + 2x 8 + 2x 9 + 4x x x x x x x 17 + ) And, there are four ways to make change for 10 ( , , 5+5, and 10). Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 45 Observation (1 + x + x 2 + x 3 + ) = (1-x) -1 (1 + x 5 +x 10 + x 15 + ) = (1-x 5 ) -1 (1 +x 10 +x 20 + x 30 + ) = (1-x 10 ) -1 whenever the left sum converges. P(x) = (1-x) -1 (1-x 5 ) -1 (1-x 10 ) -1. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 46 Here, P(x) = (1-x) -1 (1-x 5 ) -1 (1-x 10 ) -1 is the ordinary generating function for the number of ways to make change using 1, 5, and 10 coin. That is, P(x) generates these solutions. How do we answer the original question: How many ways are there to make change for 50? How many ways are there to make change for 50? Simple answer: Use a symbolic math application like MATLAB, SCILAB, Mathematica, Maple, etc. Answer: 36 For example, in MATLAB, > syms x > f = (1-x)^-1*(1-x^5)^-1*(1-x^10)^-1 > taylor(f, x, 'Order', 51) This yields 36*x^ *x^ *x^ x^4 + x^3 + x^2 + x + 1 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 47 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

10 How many ways are there to make change for 50? You can also go online. Checkout: olynomials. Here, you can type in (1-x)^-1*(1-x^5)^-1*(1-x^10)^-1 and click on the expand button, which shows the series expansion about x = 0, which is 1 +x +x 2 + x 3 + x 4 + 2x x x Answer: 36 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 49 How many ways are there to make change for 50? Long answer (by hand). 1) (1-x) -1 = a 0 +a 1 x+a 2 x 2 + 2) (1-x) -1 (1-x 5 ) -1 = b 0 +b 1 x+b 2 x 2 + 3) (1-x) -1 (1-x 5 ) -1 (1-x 10 ) -1 = c 0 +c 1 x+c 2 x 2 + Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 50 How many ways are there to make change for 50? For the first step, we know 1) (1-x) -1 = 1+x+x 2 +x 3 For the second step, we know 2)1+x+x 2 +x 3 = (1-x 5 ) (b 0 +b 1 x+b 2 x 2 + ) or a i = b i b i-5. or b i = a i +b i-5 =1+b i-5 How many ways are there to make change for 50? For the third step, we know 3)(1-x) -1 (1-x 5 ) -1 (1-x 10 ) -1 =(c 0 +c 1 x+c 2 x 2 + ) and from this (c 0 +c 1 x+c 2 x 2 + )(1-x 10 ) = (1-x) -1 (1-x 5 ) -1 or c i = b i +c i-10. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 51 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 52 a i =1 i = a i b i c i b i =a i +b i-5 c i =b i +c i-10 Computing c 50 x 50, the number of ways to make change for 50 c 50 =36 How many ways are there to make change for 50 using 0, 1, 2, 3, 4, and 5 10 coin? Answer: Instead of using (1+x 10 +x 20 + ), use (1+yx 10 +y 2 x 20 + ). That is, instead of (1-x 10 ) -1, use (1- yx 10 ) -1. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 53 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

11 The generating function is P(x,y) =(1-x) -1 (1-x 5 ) -1 (1-yx 10 ) -1. The coefficient of x 50 is (d 0 + d 1 y +d 2 y 2 + d 3 y 3 + d 4 y 4 + d 5 y 5 ), where d i = number of 10 coin to make change for 50. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 55 a i =1 Computing c 50 x 50, using 10 coin i = a i b i y 4 y 3 y 2 y 1 y 0 c i b i =a i +b i-5 c i =b i +c i-10 y 5 c 50 =36 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 56 From the table on the previous page [11 + 9y +7y 2 + 5y 3 + 3y 4 + y 5 ] x 50 In making change for 50, there are 11 ways using no 10 yen coin 11 9 ways using 1 10 yen coin 9y 7 ways using 2 10 yen coins 7y 2 5 ways using 3 10 yen coins 5y 3 3 ways using 4 10 yen coins 3y 4 You can also use MATLAB, Mathematica, Maple, and Wolfram Alpha ( alpha.com/examples/polynomials) to evaluate a generating function on two variables. That generating function is (1-x) -1 (1-x 5 ) -1 (1-yx 10 ) way using 5 10 yen coins y 5 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 58 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 57 Outline Combinatorial Math Part 2 Introduction Generating Functions Intro Using Generating Functions Solving Recurrence Relations Fibonacci Numbers in Nature Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 59 Other Generating Functions (1+x) 3 =(1 + 3x + 3x 2 + x 3 ) Object #1 Object #2 Object #3 (1+x) 3 = (1+x) (1+x) (1+x) This represents ways to choose from 3 objects. For each object, we can either choose it (x) or not choose it (1). Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

12 Other Generating Functions (1 + 3x + 3x 2 + x 3 ) Here, we can choose NO object in 1 way, ONE object in 3 ways, TWO objects in 3 ways, and THREE objects in 1 way. Other Generating Functions In general, (1+x) n = 1+C(n,1)x+C(n,2)x 2 + +C(n,n)x n, where C(n,i) = number of ways to choose i objects from n. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 61 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 62 Prove: 2 n = 1+C(n,1)+C(n,2)+ +C(n,n) Substitute 1 for x in (1+x) n = 1+C(n,1)x+C(n,2)x 2 + +C(n,n)x n Other Generating Functions How many ways are there to select from 3 red balls, 2 green balls and 1 blue ball? Prove: 0 = 1-C(n,1)+C(n,2)- +C(n,n) Substitute -1 for x in (1+x) n = 1+C(n,1)x+C(n,2)x 2 + +C(n,n)x n Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 63 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 64 How many ways are there to select from 3 red balls, 2 green balls and 1 blue ball? r r r g g (1+rx+r 2 x 2 +r 3 x 3 ) (1+gx+g 2 x 2 ) (1+bx) Choose none Choose 1 Choose 2 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 65 g Choose 3 How many ways are there to select from 3 red balls, 2 green balls and 1 blue ball? (1+rx+r 2 x 2 +r 3 x 3 ) (1+gx+g 2 x 2 ) (1+bx) = 1 + (r+g+b)x + (rg+rb+gb+r 2 +g 2 )x 2 + (rgb +rg 2 +r 2 g+r 2 b+r 3 +g 2 b)x 3 + (rg 2 b +r 2 gb+r 2 g 2 +r 3 g+r 3 b)x 4 + (r 2 g 2 b +r 3 gb +r 3 g 2 )x 5 + (r 3 g 2 b)x 6. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

13 The following shows, in detail, all ways to choose. If you care only about the number of ways, set r = g = b = 1. (1+rx+r 2 x 2 +r 3 x 3 ) (1+gx+g 2 x 2 ) (1+bx) = 1 + (r+g+b)x + (rg+rb+gb+r 2 +g 2 )x 2 +(rgb +rg 2 +r 2 g+r 2 b+r 3 +g 2 b)x 3 +(rg 2 b +r 2 gb+r 2 g 2 +r 3 g+r 3 b)x 4 +(1 + r 2 g 2 b +r 3 gb +r 3 g 2 )x 5 +(r 3 g 2 b)x 6. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 67 The following shows, in detail, all ways to choose. If you want just a number of ways without detail, set r = g = b = 1. This yields (1+rx+r 2 x 2 +r 3 x 3 ) (1+gx+g 2 x 2 ) (1+bx) = 1 + 3x + 5x 2 + 6x 3 + 5x 4 + 3x 5 + x 6, from r = g = b = 1. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 68 Other Generating Functions From before, (1+x) n = 1+C(n,1)x+C(n,2)x 2 + +C(n,n)x n. So, (1+x) n is the generating function for choosing r objects from n without repetition. Is there a generating function for choice WITH repetition? Choice With Repetition YES! It is (1+x+x 2 +x 3 + +x k + ) n = (1-x) -n Compare this with choice without repetition, which is (1 + x) n Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 69 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 70 Choice With Repetition For choice without repetition, (1 + x) n = (1 + +C(n,r) x r + ) and with repetition (1 - x) -n = (1 + +C(-n,r) (-x) r + ) Rewrite this term. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 71 Choice With Repetition From (1 - x) -n = (1 + +C(-n,r) (-x) r + ) Rewrite this term. C(-n,r)(-x) r = (-n)!/r!(- n-r)! = (-1) r (n+r-1) (n+1)(n)/r! (-x) r = (n+r- 1)!/((n-1)!r!) x r = C(n+r-1,r) x r. From before, we know that C(n+r-1,r) is the number of ways to choose r objects from n with repetition. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

14 Prove: m = m(m+1)/2. Solution: Use generating functions. We have: 1 + x + x x m + = (1-x) -1. Differentiate both sides with respect to x. 1+2x +3x mx m-1 + =(1-x) -2 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 73 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 74 Multiply both sides by x Differentiate both sides with respect to x. x+2x 2 +3x 3 + +mx m + =x(1-x) -2 Note: If A(x) = a 0 + a 1 x + a 2 x 2 +a 3 x 3 + Then, (1-x) -1 A(x) = (a 0 )+ (a 0 +a 1 )x + (a 0 +a 1 +a 2 )x 2 + Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 75 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 76 Therefore, (1-x) -1 (x+2x 2 +3x 3 + +mx m + )= (1)x + (1+2)x 2 + (1+2+3) x 3 +. So, the coefficient of x m in x(1-x) - 3 is ( m). Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 77 But, This is just the coefficient of x m-1 in (1-x) -3. From before, this is the number of ways to choose m-1 objects from 3 with repetition or C(3+(m-1)-1,m-1) = C(m+1,m-1) = C(m+1,2) = (m+1)m/2. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

15 Another way to compute We seek the number of squares in this triangle. m m Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part m Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 80 This is called a look-see proof. That is, you look at it and immediately see it. Unfortunately, most theorems do NOT have a look-see proof. Outline Combinatorial Math Part 2 Introduction Generating Functions Intro Using Generating Functions Solving Recurrence Relations Fibonacci Numbers in Nature Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 81 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 82 Solving Recurrence Relations Generating functions can be used to solve recurrence relations. The most famous recurrence relation is the Fibonacci recurrence relation. n a n Born: Pisa, Italy m We seek the number of squares in this triangle. m+1 It is 1/2 of an m x m+1 rectangle or (m+1)m/2. This originated in 1202 by Leonardo Fibonacci. He was educated in North Africa, where his father held a diplomatic post. He published Liber abaci in 1202, which introduced the Hindu-Arabic placedvalued decimal system and Arabic numerals in Europe. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 83 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

16 The French mathematician, Edouard Lucas, used Fibonacci numbers to prove is prime. This 39-digit number was the largest known until 1952 when a computer was used to find five higher primes, the largest being The Lucas sequence is ,... Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 85 The Fibonacci recurrence relation is a n = a n-1 + a n-2 Solving this by generating function, multiply both sides by x n. a n x n = a n-1 x n + a n-2 x n This yields an infinite number of equalities, like so. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 86 This yields an infinite number of equalities, like so. a 2 x 2 = a 1 x 2 + a 0 x 2 a 3 x 3 = a 2 x 3 + a 1 x 3 a 4 x 4 = a 3 x 4 + a 2 x 4 Summing both sides yields a 2 x 2 + a 3 x 3 + a 4 x 4 + = (a 1 x 2 +a 2 x 3 +a 3 x 4 + )+ (a 0 x 2 +a 1 x 3 +a 2 x 4 + ) (1) Let the Fibonacci generating function be A(x) = a 0 x 0 + a 1 x 1 + a 2 x 2 + where a 0 =1 and a 1 =1 From (1), we have A(x) a 0 a 1 x = x(a(x)-a 0 ) + x 2 A(x) (2) Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 87 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 88 Solving (2) yields or A(x)= 1/(1-x-x 2 ) A(x) = 1+ x + 2x 2 + 3x 3 + 5x 4 +8x 5 + This can be obtained using a polynomial manipulation package. Rewriting this A(x)= -1/(x 2 +x-1) or A(x)= -1/[(x-a)(x-b)], where a = (-sqrt(5)-1)/2 and b = (sqrt(5)-1)/2. a = and b = Expand as a partial fraction A(x) = (1/sqrt(5))/(x-a) + (-1/sqrt(5))/(x-b) Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 89 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

17 Expand as a partial fraction A(x) = (1/sqrt(5))/(x-a) + (-1/sqrt(5))/(x-b) Rewriting yields A(x) = (1/(sqrt(5)b))/[1-(1/b)x] + (- 1/(sqrt(5)a))/[1-(1/a)x] Solving yields a n = (1.618) n (-0.618) n Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 91 a n = (1.618) n (-0.618) n Note that the second term approaches 0 as n increases. Thus, a n is approximated by a n ~ (1.618) n The number is famous. the golden ratio. It is Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 92 Outline Combinatorial Math Part 2 Introduction Generating Functions Intro Using Generating Functions Solving Recurrence Relations Fibonacci Numbers in Nature In sunflowers, there are two rows of seeds. Each row has a Fibonacci number of seeds. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 93 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 94 Fibonacci Numbers in the Solar System Planet Aphelian Dis. From Sun (x 10 6 km) Fibonacci Number Dis. Proport. to Fibonacci Number Mercury Venus Earth Mars Asteroids 8 Jupiter Saturn Uranus Neptune Pluto For example, 7375/51.2 = Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 95 Fibonacci tiling places fxf squares together, where f is a Fibonacci number. Now, draw an arc between opposite corners, like so. Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

18 The result is the golden spiral Shells Galaxies which often occurs in nature. Typhoons Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 97 Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Geography Puzzle A person walks one kilometer south, one kilometer east, and one kilometer north and arrives at the start point. He sees a bear! What color is the bear? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part 2 99 White Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Is there another place on earth where a person walks one kilometer south, one kilometer east, and one kilometer north and arrives at the start point? Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Near the South Pole 1 kilometer About 1 + 1/(2π)=1.159 kilometers South Pole There are other circles corresponding to twice, three times, etc., around the South Pole! Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part

19 And, finally Merry Christmas and Happy New Year Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part Meiji Univ. 10:30-12:00 December 18, 2015 J. T. Butler Combinatorial Mathematics Part