Vol. 42, No. 1, January 2011
THE COLLEGE MATHEMATICS JOURNAL
Anne Burns: Iterated M¨ obius In this issue: •
Augustus De Morgan Behind the Scenes
•
Computing Determinants by Double-Crossing
•
Two-Person Pie-Cutting
•
Chutes and Ladders for the Impatient An Official Publication of the Mathematical Association of America
EDITORIAL POLICY The College Mathematics Journal serves all who are interested in college mathematics. CMJ seeks lively, well-motivated articles that will enrich undergraduate instruction and enhance classroom learning, as well as expository papers that stimulate the thinking and broaden the perspectives of those who teach undergraduate-level mathematics. Articles involving all aspects of mathematics are welcome, including history, philosophy, problem solving, pedagogy, applications, computation, and so on. Emphasis is placed on topics taught in the first two years of college. Contributions from teachers in high schools, two-year colleges, four-year colleges, and universities are welcome. In addition to articles, CMJ publishes short pieces containing material suitable for immediate classroom use, problems, solutions to published problems, student research projects, media reviews, and all sorts of mathematical ephemera: proofs without words, arresting examples of fallacious proofs, mistakes and/or other mathematical anomalies, quotations, poetry, humor, cartoons, and doodles. Letters to the Editor on any topic are also welcome, as are all kinds of comments, criticisms, and suggestions for making CMJ more lively, entertaining, and informative.
SUBMITTING AN ARTICLE Manuscripts should be submitted electronically. Please send a pdf file attached to an email to
[email protected]. The author’s name(s) must not appear in this file, but should be listed in the accompanying e-mail, along with the postal address of a designated corresponding author. In lieu of a pdf, an author may submit a single paper copy of the manuscript, printed on only one side of the paper. Note: The only electronic file format accepted is pdf. Authors who use TEX are asked to use a standard environment with no custom formatting. A TEX file will be requested only if a manuscript is accepted for publication. Contributions to the sections Classroom Capsules, Problems, Solutions, Student Research Projects, and Media Highlights should be submitted to the appropriate section editor at the address listed inside the journal. All other submissions should be sent to
[email protected]. For more information about writing for CMJ, please visit the web site at http://www.maa.org/pubs/ cmj.html and follow the links to Information for Authors and Electronic Production Information.
The COLLEGE MATHEMATICS JOURNAL (ISSN 0746-8342) is published by the Mathematical Association of America at 1529 Eighteenth Street, N.W., Washington, D.C. 20036 and Lancaster, PA, monthly in January, March, May, September, and November. Subscription correspondence and notice of change of address should be sent to the Membership/ Subscriptions Department, Mathematical Association of America, at the address above. Microfilmed issues may be obtained from University Microfilms International, Serials Bid Coordinator, 300 North Zeeb Road, Ann Arbor, MI 48106. Advertising correspondence should be addressed to MAA Advertising 1529 Eighteenth St. NW Washington DC 20036 Phone: (866) 821-1221 Fax: (866) 387-1208 E-mail:
[email protected] Further advertising information can be found online at www.maa.org. Permission to make copies of individual articles, in paper or electronic form, including posting on personal and class web pages, for educational and scientific use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the following copyright notice: Copyright the Mathematical Association of America 2011. All rights reserved. Abstracting with credit is permitted. To copy otherwise, or to republish, requires specific permission of the MAA’s Director of Publications and possibly a fee. Periodicals postage paid at Washington, DC, and additional mailing offices. Postmaster: Send address changes to Membership/ Subscription Department, Mathematical Association of America, 1529 Eighteenth Street, N.W., Washington, D.C. 20036-1385 Printed in the United States of America
¨ ABOUT ITERATED MOBIUS Anne Burns is a professor of mathematics at Long Island University, C. W. Post Campus. She began her studies as an art major but later switched to mathematics. Programming a computer showed her that she could combine her interests in art and mathematics. She is fascinated by the recursive properties evident in nature and spends her spare time recreating them. There is no end to the possibilities in using mathematics to create art. How lucky we are to be able to participate in this exciting new field! ¨ Iterated Mobius is the result of applying an iterated function (IFS) system to ¨ a seed map four times. The IFS incorporates a Mobius transformation which scatters the various parts of the image along distorted lines. For more about ¨ how Iterated Mobius was constructed see page 14.
Vol. 42, No. 1, January 2011
THE COLLEGE MATHEMATICS JOURNAL Editor Michael Henle, Oberlin College, Oberlin OH 44074 Sally Moffitt, Editorial Assistant
Board of Editors Ricardo Alfaro, University of Michigan–Flint, Flint MI 48502 Ed Barbeau, University of Toronto, Toronto, Ontario, Canada M5S 3G3 sarah-marie belcastro, Sarah Lawrence College, Bronxville NY 10708 Susan Jane Colley, Oberlin College, Oberlin OH 44074 Curtis Cooper, University of Central Missouri, Warrensburg MO 64093 Susan Goldstine, St. Mary’s College of Maryland, St. Mary’s City MD 20686 Lixing Han, University of Michigan–Flint, Flint MI 48502 Reuben Hersh, University of New Mexico, Albuquerque NM 87131 Donald E. Hooley, Bluffton University, Bluffton OH 45817 Heather A. Hulett, University of Wisconsin–La Crosse, La Crosse WI 54601 Michael A. Jones, Mathematical Reviews, Ann Arbor MI 48103 Gary Kennedy, The Ohio State University-Mansfield, Mansfield OH 44906 Dan King, Sarah Lawrence College, Bronxville NY 10708 Warren Page, 30 Bonnie Way, Larchmont NY 10538 Cecil Rousseau, University of Memphis, Memphis TN 38152 Kenneth E. Schilling, University of Michigan–Flint, Flint MI 48502 Brigitte Servatius, Worcester Polytechnic Institute, Worcester MA 01609 Shing So, University of Central Missouri, Warrensburg MO 64093 Todd G. Will, University of Wisconsin–La Crosse, La Crosse WI 54601 Harry Waldman, Managing Editor
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
Chutes and Ladders for the Impatient Leslie A. Cheteyan, Stewart Hengeveld, and Michael A. Jones
Leslie A. Cheteyan (
[email protected]) received her B.S. from Montclair State University in 2008. She is currently working towards her master’s degree in pure and applied mathematics at Montclair. From an early age, she has had a love for math and its implications. Besides math, Leslie enjoys playing all types of sports in which basketball is her favorite. Her competitive nature helps to fuel her motivation in different areas of mathematics. Stewart Hengeveld (
[email protected] ) is currently working on his master’s degree at Montclair State University. He has spent the past few years teaching developmental mathematics at Bergen Community College and tutoring in Bergen’s Henry and Edith Cerullo Learning Assistance Center. He believes that tutoring is an integral part of modern education, and should be made available and known to all students. Since 2008, he has been President of the Buehler Columbia Amateur Astronomy Association, located at Bergen. He organizes public viewing nights at which he shows various celestial objects on one of the college’s 16 Meade telescopes. Michael A. Jones (
[email protected]) is an Associate Editor at Mathematical Reviews in Ann Arbor, MI. Leslie and Stewart worked with him on this article as a project for an undergraduate discrete mathematical modeling course that he taught while he was a professor at Montclair State University. He really does play Chutes and Ladders with his son (Stephen), his daughter (Anneliese), and his wife (Carrie).
My 4-year old son likes to play games, but he doesn’t have the longest attention span. His younger sister and my wife are less patient. I’m even worse! Recently, while playing Chutes and Ladders, I wondered how changing the spinner might shorten the game so that we could complete a game without any of us giving up! To find out, two students and I extended the well-known Markov chain model of Chutes and Ladders (e.g., see [1] and [6]) to investigate the relationship between spinner range and the expected number of turns it takes to complete a game. doi:10.4169/college.math.j.42.1.002
2
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 2
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
page 3
A review of Chutes and Ladders Chutes and Ladders (known as Snakes and Ladders in some countries) is a board game popular among the 4-year old set. The game board consists of 100 squares arranged in numerical order, 1 to 100. On each player’s turn, the player spins the spinner and moves his or her token the number of squares indicated, if possible. (For example, a player does not move her token if she spins a 2 when on square 99.) The players begin the game off the board, as if there were a square 0. The first player to reach square 100 wins the game. The chutes and ladders complicate matters. If at the end of a move a player lands at the base of a ladder, then the player climbs to the top of the ladder, thereby getting closer to the goal of square 100. But, if at the end of a move a player lands at the top of a chute, then the player slides down the chute, moving farther from the goal. The game comes with a spinner which returns the numbers 1 through 6 with equal probability. It would be easy to use a different spinner or to modify the given spinner so that each of the numbers 1 to n is equally likely to occur. We wanted to know which value of n results in the shortest expected number of turns for a player to finish the game. In the next section, we develop a Markov chain model that includes the positions of the chutes and ladders and takes into account the spinner range. At first blush, it may seem that the expected number of turns to complete the game will decrease as n increases, but the requirement that a player land exactly on square 100 to end the game (or land on the ladder that goes from square 80 to square 100) means that high values of n result in frequent turns in which a player does not move.
A spinner-range dependent Markov chain for Chutes and Ladders For i = 1 to 100, let state i of a Markov chain represent square i in Chutes and Ladders. Each player begins the game off the board at state 0. We build a 101 × 101 transition matrix P for a Markov chain to model n-spinner Chutes and Ladders by first assuming that there are no chutes or ladders. Then, the addition of each chute or ladder modifies the transition matrix P by a sequence of simple operations in such a way that the order in which the chutes and ladders are added does not matter. The result is the transformed matrix, P ∗ , which models the play of the game. This process is demonstrated with a smaller example below. Recall that entry (i, j) of a transition matrix indicates the probability of transition from state i to state j. For ease of notation, we use the state names, 0 to 100, to represent the row and column numbers, instead of 1 to 101. We assume the spinner range is less than or equal to the number of squares on the board. Hence, for the real game, we consider n ≤ 100. Because a player can no longer move once he or she has completed the board by landing on square 100, the only entry in row 100 is in column 100. That is, p100,100 = 1. Once reached, state 100 cannot be left, making it an absorbing state. Further, because state 0 is off the board, column 0 is all 0’s, i.e., it is impossible to take a turn and land on state 0. Ignoring the effect of the chutes and ladders, the n-spinner defines the following non-zero transition probabilities or entries in the transition matrix: •
For state j = 0 to (100 − n), let p j, j +i =
1 for i = 1 to n; n n−100+ j p j, j = ; and n 1 p j, j +i = n for i = 1 to (100 −
•
For j = (101 − n) to 100, let
•
For j = (101 − n) to 100, let
j).
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
3
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
The resulting matrix contains a diagonal block of entries n1 , until a spin can produce an outcome which overshoots square 100, in which case the player does not move on his or her turn. Because it is awkward to display a 101 × 101 transition matrix, we demonstrate the analysis with the following 10-state variation of Chutes and Ladders. .................................................................................................................. ... ... ... .. ... ... .... ... ... ... ... ... ... ... ... ... ... .. .. ... ... ... ... .... ... .... ... .. . .......................................................................................................................... ... .. .. ... ... ... ... ... ... ... ... ... ... .... ... . ... ... ... .............. ..... .... ... . ... .... .. ...... . ....... .. . ............................... . ... ... ............. .... .......... .......... .. ... . .. . . . ....................................................................................................................................................... . ............ ........ . . . ... .. ... .... . ....... ..... ........ . . ... . . . .. ... ... . .. ... ... .. .. ... .... ................................ .... ........ ... .... .. ... ... ... .... .. .. ... .... ... ... .. ... ... ... ...............................................................................................................
7
8
6
5
1
2
9
4
3
Figure 1. Example game board.
Example. Consider the smaller game board of 9 squares in Figure 1. As in the real game, a player begins in state 0. Assume to begin that the spinner has outcomes 1 through 4 that occur with equal probability. For the time being, ignore any chutes or ladders. A spinner range of 4 results in the 10 × 10 transition matrix P (see Figure 2) where the game ends when the player reaches square 9. Recall that we denote the rows and columns by numbers from 0 to 9. Rows 5 and 6 of the matrix are identical. From state 5, a spin of 1, 2, 3, or 4 results in the player moving to states 6, 7, 8, or 9, respectively, with equal probability. From state 6, a spin of 1, 2, or 3 results in the player moving to states 7, 8, or 9, respectively, with equal probability of 14 . But, a spin of 4 means that the player cannot move (as there is no state 10); hence, the player stays at state 6 with probability 14 . ⎡ 0 ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢0 ⎢ P =⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢ ⎣0 0
1 4
0
1 4 1 4
0
0
1 4 1 4 1 4
0
0
0
1 4 1 4 1 4 1 4
0
0
0
0
1 4 1 4 1 4 1 4
0
0
0
0
0
0
0
0
0
0
1 4 1 4 1 4 1 4 1 4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1 4 1 4 1 4 1 4 2 4
0
0
0
0
0
0
0
0
1 4 1 4 1 4 1 4 3 4
0
0
0
0
0
0
0
0
0
⎤
0⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ 1⎥ ⎥ 4⎥ 1⎥ 4⎥ ⎥ 1⎥ 4⎥ 1⎥ 4⎦ 1
Figure 2. Example matrix P.
To account for the chutes and ladders, P needs to be modified. Recall that when a player lands at the foot of a ladder, then the player climbs to the top of the ladder. Similarly, when a player lands at the top of a chute, then the player slides to the bottom of the chute. Denote ladders and chutes by [i, j], meaning that if a player lands on square i at the end of the turn, then the player moves to square j (by a ladder if i < j and by a chute if j < i). The chutes and ladders from the real game board are: [1, 38], [4, 14], [9, 31], [16, 6], [21, 42], [28, 84], [36, 44], [47, 26], [49, 11], 4
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 4
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
page 5
[51, 67], [56, 53], [62, 19], [64, 60], [71, 91], [80, 100], [87, 24], [93, 73], [95, 75], and [98, 78]; there are 19 chutes or ladders. In general, we construct the transition matrix P ∗ from P as follows: if [i, j] is a chute or ladder, then ∗ = 0 for all k (row i is filled with zeros), 1. set pi,k ∗ 2. set pk, j = pk, j + pk,i for k = i (landing on square i is equated with landing on square j), and ∗ 3. set pk,i = 0 for all k (column i is filled with zeros). ∗ = pk,l . Because row i and column i are For all other entries (k, l) in P ∗ , set pk,l ∗ replaced by zeros, the matrix P can be reduced (by eliminating the rows and columns consisting of all zeros). For the real Chutes and Ladders game, this reduction would result in an 82 × 82 transition matrix (101 − 19 = 82). However, we find it easier not to reduce the matrix in this way, as it makes keeping track of the states more difficult. Let’s examine what happens for the smaller board of Figure 1.
Example (continued). The 9-square version of Chutes and Ladders has only a single ladder, [2, 6], and a single chute, [8, 3]. Take the ladder first. When a player lands on square 2, she climbs the ladder to square 6. To account for this in the transition matrix P ∗ , do the following operations in sequence to P: 1. replace row 2 with zeros (because a player is never on square 2), 2. replace column 6 with the sum of column 2 (the probabilities of landing on square 2) and column 6 (the probabilities of landing on square 6), and 3. replace column 2 with zeros (because a player cannot land on square 2). The chute is handled in a similar fashion. If a player lands on square 8, then he slides down the chute to square 3. To account for this in the transition matrix P ∗ , do the following operations in sequence to the matrix previously modified by the ladder: 1. replace row 8 with zeros (because a player is never on square 8), 2. replace column 3 with the sum of column 8 (the probabilities of landing on square 8) and column 3 (the probabilities of landing on square 3), and 3. replace column 8 with zeros (because a player cannot land on square 8). The resulting matrix P ∗ appears in Figure 3. ⎡
0
1 4
0
1 4 1 4
0
0
1 4 1 4
0
0
0
1 4
1 4 1 4
0
0
0
0
⎤
⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢0 ⎢ ∗ P =⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢ ⎣0
0
0
0
0
0
0
0
0
0
0
1 4
0
0
0
0
0
0
0
0
0
0
0
0
1 4 1 4 1 4 1 4
1 4 1 4 1 4 1 4
0
0
0
1 4 1 4 1 4 1 4 2 4
0
0
1 4 1 4
0
0
0
0
0
0
0
0
0⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ 1⎥ ⎥ 4⎥ 1⎥ 4⎥ ⎥ 1⎥ 4⎥ ⎥ 0⎦
0
0
0
0
0
0
0
0
0
1
0 0 0 0
Figure 3. Example matrix P ∗ .
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
5
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
Fundamental matrix and game analysis The Markov chain of Chutes and Ladders is absorbing because it has an absorbing state and, by a sequence of moves, it is possible to reach the absorbing state from every other state. An absorbing Markov chain can have its states re-ordered so that it is in block form, called fundamental form, with an identity matrix as a lower right block:
Q
R
0
I
Because the absorbing state is the last square on the game board (resulting in the last row and column), Pn∗ is already in fundamental form. For the real Chutes and Ladders, Q n is a 100 × 100 matrix, Rn is a 100 × 1 matrix, 0 is the 1 × 100 matrix of all zeros, and I is a 1 × 1 identity matrix. The block Q n can be used to determine the expected number of turns to complete the game with an n-spinner. Although we provide a short description of how (and why) this analysis works, more details can be found in [5, 7, 8]. Johnson [7] used board games to teach Markov chains and provides a description of the fundamental matrix in the context of board games, while Kemeny and Snell [8] have a well-written text on Markov chains. The book by Grinstead and Snell also covers Markov chains [5, Chapter 11] and is available electronically for free! The matrix Nn = (I100 − Q n )−1 is called the fundamental matrix, where I100 is the 100 × 100 identity matrix. Its entries (i, j) indicate the number of times the player is in state j, given that the player starts in state i. The curious form of Nn is reminiscent of the well-known formula for geometric series: 1 + r + r 2 + · · · = (1 + r )−1 for |r | < 1. The geometric series of the matrix Q n converges because the Markov chain is absorbing, so that Nn = I100 + Q n + Q 2n + · · · = (I100 − Q n )−1 . Let 1100 be the 100 × 1 vector of all 1’s. Then, Nn 1100 is a 100 × 1 vector in which the entry in row i gives the expected number of turns necessary to be absorbed (that is, to transition to state 100) beginning in state i. The entry in row 0 gives the expected number of turns it takes for a player to complete Chutes and Ladders, because the player always starts in state 0. Example (continued). For the 9-square example with a 4-spinner, the matrix in Figure 3 is already in fundamental form. It follows that Q 4 and N4 are the 9 × 9 matrices given in Figure 4. ⎡
0
⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢ Q4 = ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢ ⎣0 0
1 4
0
0
0
1 4
1 4 1 4
0
0
0
0
1 4
0
1 4 1 4
0
0
0
0
1 4 1 4 1 4 1 4
0
0
0
1 4 1 4 1 4 1 4 2 4
⎤ 0 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎦
0
0
0
0
0
1 4 1 4
0
0
1 4 1 4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1 4 1 4 1 4 1 4
0
0
0
0
⎡
1
⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢ N4 = ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢ ⎣0 0
89 60 7 5
41 60 3 5
29 48 3 4
1
0
0
0
0
0
0
0
32 15 4 3 16 15 16 15 16 15
8 15 4 3 4 15 4 15 4 15
2 3 2 3 4 3 1 3 1 3
10 9 10 9 8 9 17 9 5 9
20 9 20 9 16 9 16 9 28 9
⎤ 0 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎦
0
0
0
0
0
0
1
1 4
0
1
0
0 0 0
0
0
0
0
0
0 0
193 144 5 4
37 18
2
Figure 4.
6
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 6
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
page 7
Multiplying N4 by the 9 × 1 vector of all 1’s yields
89 12
7 1
20 3
20 3
16 3
16 3
16 3
1
T
.
Hence, the expected number of turns to complete the 9-square version of the game with a 4-spinner beginning at square 0 is 89 . The expected duration is computed for 12 spinner ranges from 1 to 9 in Table 1. Notice that there is more than one local minimum. Increasing the range of the spinner until the expected duration decreases is not sufficient to determine the global minimum. In the example, the expected number of spins to complete the game decreases as the spinner range increases from 2 to 4. Then, a spinner range of 5 increases the expected duration, while a spinner range of 6 results in the (global) minimum expected length of 43 turns to complete the game. The ex6 pected length of the game then increases as the spinner range increases from 7 to 9. Notice that the expected duration is the same for spinners of range 8 and 9. Table 1. The expected number of turns E for one player to complete the 9-square version, as a function of spinner range n. n
1
E
∞
2 243 22
≈ 11.05
3 173 23
4
≈ 7.52
89 12
5
≈ 7.42
7.75
6 43 6
7
≈ 7.17
57 7
≈ 8.14
8
9
9
9
We use the same approach to determine the expected number of turns to complete the real Chutes and Ladders (see Table 2). The minimum expected duration occurs with a spinner range of 15. Notice that this data set is close to being single-troughed. The expected number of turns to complete the game decreases as the spinner range is increased from 3 to 15 and then increases as the spinner range increases from 16 to Table 2. The expected number E of turns for one player to complete Chutes and Ladders, as a function of spinner range n. n
E
n
E
n
E
n
E
n
E
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
∞ 60.76 65.90 54.49 45.56 39.23 34.70 31.85 30.30 28.77 27.43 27.02 26.22 25.98 25.81 25.84 25.97 26.39 26.71 27.21
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
27.53 27.58 28.12 28.78 29.10 29.72 30.07 30.64 31.46 32.14 32.90 33.61 34.32 35.05 35.77 36.50 37.26 38.05 38.83 39.54
41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
40.30 41.05 41.84 42.74 43.54 44.35 45.15 45.98 46.85 47.67 48.53 49.38 50.25 51.03 51.87 52.55 53.40 54.24 55.12 55.98
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
56.81 57.67 58.53 59.45 60.33 61.19 62.07 62.94 63.77 64.66 65.59 66.50 67.39 68.22 69.12 69.88 70.76 71.65 72.56 72.55
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
73.46 74.40 75.34 76.29 77.23 78.16 79.05 79.99 80.91 81.85 82.80 83.74 84.64 85.57 86.46 87.43 88.39 89.28 90.28 90.28
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
7
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:07 a.m.
jones.tex
79. From 79 to 80, there is a slight decrease in the expected length, but then the game increases for spinner ranges of 81 to 99. As with the 9-square game, interesting behavior happens for spinner ranges equal to or one less than the board size: the expected duration is the same. Is it always the case for an n-square game that the expected lengths are the same for spinner ranges of n − 1 and n? Additional calculations suggest that this is the case, but can you prove it or find a counterexample? This is an unsolved problem.
Your turn! We used the discrete uniform distribution on {1, 2, . . . , n} and considered how changing n affects the expected number of turns to complete Chutes and Ladders. There are other natural questions about how to shorten the game: considering non-uniform spinner distributions or different arrangements and lengths of chutes and ladders (see [1]; a more abstract question regarding the effect of changing the transitions to a state in a Markov chain is considered in [3]). There are endless further questions that can be addressed using Markov chains. You are invited to modify the Markov chain in this paper and have fun playing— both Chutes and Ladders, and with Markov chains. To get started, consider downloading the modifiable Maple file used to compute the expected values in this paper [2]. Of course, Chutes and Ladders is only one game that can be analyzed by Markov chains. Johnson [7] considers Hi Ho! Cherry-O as well as Chutes and Ladders and provides references to other ways to use Markov chains to analyze games. The Maple file can be edited to help explore the relationship between Markov chains and other board games, too. Summary. In this paper, we review the rules and game board for Chutes and Ladders, define a Markov chain to model the game regardless of the spinner range, and describe how properties of Markov chains are used to determine that an optimal spinner range of 15 minimizes the expected number of turns for a player to complete the game. Because the Markov chain consists of 101 states, we demonstrate the analysis with a 10-state variation with a single chute and single ladder. The resulting 10 × 10 transition matrix is easier to display and the manipulations are comparable. We conclude with an unsolved problem about expected lengths for generalized Chutes and Ladders games.
References 1. S. C. Althoen, L. King, and K. Schilling, How long is a game of Snakes and Ladders? Mathematical Gazette 77 (1993) 71–76. doi:10.2307/3619261 2. L. Cheteyan, S. Hengeveld, and M. A. Jones, College Mathematics Journal Supplements, documented Maple code to replicate calculations in this article (2010); available at http://www.maa.org/pubs/cmj_ supplements/index.html. 3. P. Diaconis and R. Durrett, Chutes and Ladders in Markov chains, Journal of Theoretical Probability 14 (2001) 891–926. doi:10.1023/A:1017509611178 4. S. Gadbois, Mr. Markov plays Chutes and Ladders, UMAP Journal 14 (1993) 31–38. 5. C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., American Mathematical Society, Providence RI, 1997; available at http://www.dartmouth.edu/~chance/teaching_aids/books_ articles/probability_book/amsbook.mac.pdf. 6. J. Humpherys, Chutes and Ladders, use of Markov chains to determine probability distribution of game length (2009); available at http://math.byu.edu/~jeffh/mathematics/games/chutes/chutes.html. 7. R. W. Johnson, Using games to teach Markov chains, PRIMUS: Problems, Resources and Issues in Mathematics Undergraduate Studies XIII (2003) 337–348. doi:10.1080/10511970308984067 8. J. G. Kemeny and J. L. Snell, Finite Markov Chains, Van Nostrand, New York, 1960.
8
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 8
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:10 a.m.
jones.tex
page 9
Probability 1/e Reginald Koo and Martin L. Jones
Reginald Koo (
[email protected]) received his B.Sc. in mathematics from the University of the West Indies, his M.Sc. in pure mathematics from the University of Newcastle upon Tyne, and his Ph.D. in analysis from the University of South Carolina. He is an associate professor at the University of South Carolina Aiken. His primary interests lie in the broad area of analysis. Martin Jones (
[email protected]) received his B.A. in mathematics and chemistry from Warren Wilson College, his M.Sc. in mathematics from the University of South Carolina, and his Ph.D. in mathematics from the Georgia Institute of Technology. He is a professor of mathematics at the College of Charleston and his interests are in the areas of probability and statistics.
Seasoned probability students know that when the professor asks for the probability of some event, good guesses are zero and one! Without a doubt, these are excellent numbers to call out when caught napping in class, yet another probability occurs with amazing frequency, namely 1/e ≈ 0.367879. In this article, we give three interesting and somewhat challenging probability problems that have events with probability 1/e, and then attempt to explain why 1/e occurs so frequently.
The wise man The story is told of a group of troubled people who went to seek the counsel of a wise man. Upon arrival, each person is instructed to write his trouble on a slip of paper and place it in a hat on the floor. The hat is shaken and each person randomly selects a slip of paper. As the members of the group read the slips, the wise man tells them “by all the laws of probability, you each have a new problem to worry about.” At the end of the story each person is content that his trouble is not as grave as the one on their slip and so the entire group returns home happy. Problem solved. Or is it? How wise is our wise man? The problem appears in [3] under the name “The Matching Problem,” but it is also known as the “Hat Check Problem,” and the “Dinner Problem.” Here we are concerned with the Matching Problem, not the various personal problems. Suppose that there are n persons in the group and let A j represent the event that person j selects his original problem from the hat (we will assume that the n problems are distinct, even though that is probably not true in real life!). The event ofinterest, A, that no person selects his or her original problem can be expressed A = nj =1 Acj . doi:10.4169/college.math.j.42.1.009
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
9
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:10 a.m.
jones.tex
Using De Morgan’s laws, we may write n
A=
Acj =
j =1
n
c ,
Aj
j =1
and then by complementation P(A) = P
n
c Aj
=1− P
j =1
n
Aj .
(1)
j =1
Using the Principle of Inclusion-Exclusion we find that n n Aj = P(A j ) − P(Ai ∩ A j ) + P(Ai ∩ A j ∩ Ak ) P j =1
j =1
i< j
i< j
− · · · + (−1)n+1 P(A1 ∩ · · · ∩ An ).
(2)
Now there are n! different arrangements possible for the slip selections by members of the group. There are (n − 1)! arrangements under the restriction that person j selects their original problem. Therefore P(A j ) = (n − 1)!/n! = 1/n for each j = 1, 2, . . . , n. Similarly P(Ai ∩ A j ) = 1/(n(n − 1)) for each i < j and so on. Substituting these expressions into (2) yields n n 1 1 1 P − + · · · + (−1)n+1 Aj = n n(n − 1) n! i< j j =1 j =1 n 1 1 1 + · · · + (−1)n+1 =n· − 2 n(n − 1) n n! =1−
1 1 (−1)n+1 + − ··· + . 2! 3! n!
(3)
Finally substituting (3) into (1) yields
1 (−1)n+1 1 P(A) = 1 − 1 − + − · · · + 2! 3! n! 1 (−1)n 1 − + ··· + 2! 3! n! 1 1 (−1)n 1 = 1 − + − + ··· + 1! 2! 3! n! n j (−1) . = j! j =0
=
(4)
As the number of members of the group n increases we have P(A) →
∞ (−1) j j =0
10
j!
=
1 . e
(5)
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 10
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:10 a.m.
jones.tex
page 11
The coffee shop game You have an hour to kill at a coffee shop before your plane departs, so you decide to play the following game. As people enter the coffee shop one at a time, with each arrival you must decide if the person currently entering will be the last person to enter the shop before the hour ends. (This problem appears in [6, Ex. 2.9, p. 90].) If a person enters and you claim that they will be the last, then you win the game if the hour ends with no other arrivals. If another person enters before the hour is up, then you lose. Similarly, you lose if the hour ends and you have not selected any entering person. You decide, as a strategy, to select a time t and choose the first person (if any) who arrives after that time. A reasonable model for the arrival of customers is a Poisson process. This is a counting process in which the probability of k arrivals in an interval of length s is given by the Poisson distribution: P(k arrivals in (t, t + s]) = (λs)k exp(−λs)/k!,
(6)
where λs is the mean number of arrivals in (t, t + s]. Let A represent the event that you win, and note that A occurs if there is exactly one arrival in the time interval (t, 1]. Now the number of arrivals in the interval (t, 1] has a Poisson distribution with mean λ(1 − t). Therefore, with k = 1, s = 1 − t in (6), P(A) =
λ(1 − t)e−λ(1−t) . 1!
(7)
Your objective is to select t to maximize P(A). To this end, set f (t) = λ(1 − t)e−λ(1−t) , differentiate with respect to t to obtain
f (t) = λe−λ(1−t) λ(1 − t) − 1 , and then set f (t) equal to zero. We find that P(A) is maximized with t = 1 − 1/λ. Substituting this into (7), yields the maximum probability of winning the game using this strategy, namely
1 P(A) = λ 1 − (1 − 1/λ) e−λ(1−(1−1/λ)) = . e
(8)
The marriage problem Consider a woman who has n men willing to marry her. If all the men showed up at once, she could order them and choose the best. Unfortunately for her, the men arrive one at a time and in random order. After dating each man for a short period of time, she must decide, before moving on to the next, whether or not to marry him. If she rejects his marriage proposal, she cannot recall him at a later time and, should she decide to marry, she will have to forego meeting the remaining suitors. What strategy should she adopt if she wants to maximize the probability of marrying the best suitor? This problem appears in [3, pp. 90–95] as the “Beauty Pageant Problem” and is also known as the “Classical Secretary Problem,” “The Fussy Suitor Problem,” and the “Best Choice Problem.” The development here follows [2]. Let X 1 , X 2 , . . . , X n be the relative ranks of men so that X j is the rank of the jth man among the first j men that she dates (with rank 1 being the best). These random variables are independent, and since all of the arrival orderings are equally likely, VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
11
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:10 a.m.
jones.tex
P(X 1 = 1) = 1, P(X 2 = 1) = P(X 2 = 2) = 1/2, . . . . We will call a suitor at stage j a candidate if X j = 1 (called a keeper in [5, pp. 18–20]). Clearly she should reject a suitor who is not a candidate. Note that the probability that the best suitor among the first j is the best overall is just the probability that the best suitor appears among the first j suitors which is j/n. Let W j represent the maximum probability of selecting the best suitor using a strategy that rejects the first j suitors. Then W j ≥ W j +1 since any rule that rejects the first j + 1 suitors rejects the first j suitors. Thus it will be optimal to accept a candidate at stage j if j/n ≥ W j . Moreover, since ( j + 1)/n > j/n ≥ W j ≥ W j +1 , our bride-to-be can use a threshold rule, that is, a rule of the form “reject r − 1 suitors and then accept the next candidate, if any, who is better than all previously appearing suitors.” Here r ≥ 1. Let Pr be the probability of selecting the best suitor using the threshold rule with threshold r . Then Pr =
n
P(suitor k is best and is selected)
k=r
=
n
P(suitor k is best)P(suitor k is selected | suitor k is best)
k=r
=
n 1 P(best of first k − 1 appears before stage r) n k=r
=
n n r −1 1r −1 1 = , n k − 1 n k − 1 k=r k=r
where the first term in the series is interpreted as 1 when r = 1. The optimal value of r is the value that maximizes Pr . Note that Pr ≥ Pr +1 if and only if n n r r −1 1 1 ≥ n k=r k − 1 n k=r +1 k − 1
if and only if 1≥
n
1 . k−1 k=r +1
(9)
1 If f (r ) = nk=r +1 k−1 , then f (r ) is decreasing and, therefore, the optimal value r ∗ is the first r satisfying (9). For large n, f (r ) ≈ ln(n/r ), and so ln(n/r ∗ ) ≈ 1 or r ∗ /n ≈ e−1 . Thus it is optimal to reject roughly 1/e or 37% of the suitors before looking for candidates and the probability of success, Pr , is approximately 1/e.
Why 1/e ? Here is one possible explanation. The binomial distribution is useful in modeling a variety of different probability problems, since it counts the number of “successes” in n independent trials in which each trial has probability p of being a success. The Poisson distribution, in turn, is often a good approximation to the binomial distribution with the mean λ of Poisson(λ) set equal to the mean np of Binomial(n, p) (particularly when n is large and p is small). Therefore, if X ∼ Binomial(n, p) and Y ∼ Poisson(λ = np), then the probability of exactly one success will be P(X = 1) ≈ P(Y = 1) = λe−λ . 12
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 12
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:10 a.m.
jones.tex
page 13
For a Poisson process with rate λ, the probability of exactly one arrival in a time interval of length s is λse−λs . When s = 1/λ, the mean interarrival time, then the probability of exactly one arrival is e−1 . In the Coffee Shop Game, the goal is to have exactly one arrival in the interval (t, 1] and the optimal value of t makes the length of this interval 1/λ. In the Marriage Problem, the goal is to have exactly one candidate in the list of suitors from r ∗ to n. As in the Coffee Shop Game, the cutoff value r ∗ is such that the mean number of candidates in the list {r ∗ , r ∗ + 1, . . . , n} is nearly one. In the story of the Wise Man, the objective is to have the hat become empty before anyone selects their own problem. In this case, the objective is to have no matches and, using the connection between the Poisson and the binomial distributions, we again have that P(X = 0) ≈ P(Y = 0) = e−λ = e−1 , since the mean number of matches in this problem is 1. This is similar to the situation in which n players are playing a lottery. Each has probability 1/n of winning and for the probability that no one wins we have, in the limit, that limn→∞ (1 − 1/n)n = e−1 . Therefore, as we see it, the connection between the Poisson and binomial distributions is the driving force behind the e−1 phenomenon. All that is required is that appropriate choices of λ occur so that the mean of the Poisson distribution is one. Summary. Quite a number of interesting problems in probability feature an event with probability equal to 1/e. This article discusses three such problems and attempts to explain why this probability occurs with such frequency.
References 1. J. N. Brawner, Dinner, dancing, and tennis, anyone? Math. Mag. 73 (2000) 29–36. 2. T. S. Ferguson, Optimal Stopping and Applications, electronic text; available at http://www.math.ucla. edu/~tom/. 3. R. J. Larsen, and M. L. Marx, An Introduction to Probability and Its Applications, Prentice Hall, Englewood Cliffs NJ, 1985. 4. B. H. Margolius, Avoiding your spouse at a bridge party, Math. Mag. 74 (2001) 33–41. 5. K. Merow, A delta for your epsilon: Finding your match mathematically, Math Horizons (September 2009) 18–20. 6. S. Ross, Stochastic Processes, 8th ed., Wiley, New York, 1996.
. . . for all the petty annoyances resulting from society’s impatience with math and science, being a mathematician has some considerable compensating advantages . . . . there is the deference I am given when the conversation turns to topics of math and science (which it often does when I am in the room). That’s rather pleasant. Social conventions being what they are, it is quite rare that my opinion on number-related questions is challenged. Unless, that is, we are discussing the Monty Hall problem. —from The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brainteaser by Jason Rosenhouse [p. 2] Reviewed in this issue on page 71
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
13
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:10 a.m.
jones.tex
¨ Anne Burns explains how Iterated Mobius was constructed We start with an iterated function system that maps the unit circle into 9 smaller circles (see Figure 1(a)). Iterating the maps results in a symmetric pattern (see Figure 1(b)). In order to achieve more interesting designs we compose the maps with a map from a family of functions that depend on three real parameters.
(a) Initial map
(b) After one iteration Figure 1.
The M¨obius transformations of the form U (z) = (uz + v)/(vz + u), with |u|2 − |v|2 = 1, map the unit circle onto itself and the interior onto the interior. Since |v|2 = |u|2 − 1, there are 3 real parameters to be freely chosen. We compose the original transformations of the iterated function system with U . Changing u and v causes a distortion in the placement of the circles which compounds with each iteration. Figure 2 illustrates the effect after one iteration using three distinct choices of u with arg(v) = 0. Iterated M¨obius is the result of iterating this system four times with parameters |u| = 1.25, arg(u) = π/8, arg(v) = 0. For more details, see [1].
One iteration: u = 1.25, arg(v) = 0
One iteration: |u| = 1.25, arg(u) = π/4, arg(v) = 0
One iteration: |u| = 1.25, arg(u) = π/8, arg(v) = 0
Figure 2. Reference 1. A. M. Burns, Iterated M¨obius transformations, J. Math. and the Arts 2 (2008) 171–188.
14
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 14
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
10:24 a.m.
swanson.tex
The Band Around a Convex Body David Swanson
David Swanson (
[email protected]) is an associate professor of mathematics at the University of Louisville, where he has worked since 2003. He has a B.S. from the University of Wisconsin–Madison and a Ph.D. from Indiana University. His research interests include real function theory and differential equations.
For a convex body C in the plane, we denote by Ct (t > 0) the set C surrounded by a band of thickness t as illustrated in Figure 1.
t
C
Ct
Figure 1.
Sekino, in this J OURNAL [3], proved that the perimeter and area of C and Ct are related by the formulas P(Ct ) = P(C) + 2πt
(1)
A(Ct ) = A(C) + P(C)t + πt 2
(2)
and
provided that the boundary of C is a piecewise smooth Jordan curve. However, formulas (1) and (2) are known to be true for arbitrary convex bodies C (and in fact versions of these formulas hold in higher dimensions as well; see, e.g., [2, p. 271]). The purpose of this article is to give an elementary proof of (1) and (2) for general planar convex bodies using an entirely different method from Sekino’s.
Preliminaries A convex body is a set C ⊂ R2 that is convex, compact, and that contains at least three non-collinear points. This includes familiar sets like disks and regular polygons, but excludes sets like half-planes and line segments. doi:10.4169/college.math.j.42.1.015
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
15
page 15
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:24 a.m.
swanson.tex
Given a convex body C and t > 0, the set Ct may be defined precisely by Ct = {x : dist(x, C) ≤ t}, where the distance from a point x to C is defined as the minimum of all distances |x − y| among points y ∈ C. It is not hard to see that Ct is also a convex body. In certain cases, formulas (1) and (2) are easy to verify. For instance, if C is a closed disk of radius r then Ct is a closed disk of radius r + t, and in this case P(C) = 2πr , A(C) = πr 2 , P(Ct ) = 2π(r + t), and A(Ct ) = π(r + t)2 . A little algebra shows that (1) and (2) both hold. On the other hand, if C is a convex polygon as shown in Figure 2, then Ct is the union of C, several rectangles with width t and total length Per(C), and several circular sectors of radius t. It can be shown that formulas (1) and (2) hold in this case as well. Note that formulas (1) and (2) imply that d A(Ct ) = P(Ct ), dt which is a generalization of the fact that the formula for the perimeter of a circle (2πr ) is the derivative of the formula of its area (πr 2 ).
C
Ct
Figure 2.
The primary technical issue in the proof of formulas (1) and (2) is how properly to define the perimeter of a general convex body. If no special assumption is made about the smoothness of its boundary, the classical arclength formula cannot be used. Instead, the perimeter is defined using approximating polygons. The main tool in the proof of these formulas is our Theorem 1, which is an integral formula for the perimeter of a convex body in terms of its projections onto lines. This theorem is a special case of a general arclength formula known as Crofton’s formula, a good discussion of which, for smooth curves, may be found in [4, p. 33] and the references therein.
The perimeter of a convex body In order to define the perimeter of a convex body properly, we recall that the perimeter of a polygon Q is the sum of the lengths of its boundary edges. We will denote this quantity by Per(Q). 16
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 16
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:24 a.m.
swanson.tex
page 17
Now let be a simple curve in the plane parameterized by γ : [a, b] → R2 . If γ is smooth then the length of is given by b () = |∇γ (t)| dt, (3) a
familiar from calculus. If γ is not smooth, but merely continuous, one may define for any partition P = {t0 , t1 , . . . , tn } of the interval [a, b], the sum L(P ) =
n
|γ (t j ) − γ (t j −1 )|.
i=1
This sum is precisely the length of the associated polygonal curve comprised of the segments γ (t1 )γ (t0 ), γ (t2 )γ (t1 ), . . . , γ (tn )γ (tn−1 ). According to the principle that the shortest distance between two points is a segment, the length of should equal or exceed L(P ) for all such P . One may in turn define the length of as L() = sup L(P ),
(4)
P
where the supremum is taken over all partitions P of the interval [a, b]. This is the commonly accepted definition of the length of a continuous curve , and if is smooth then () = L(). See, e.g., [1, p. 153] for details. A curve for which L() < ∞ is called rectifiable. Now let C be a convex body with boundary . We will use (4) to find a formula for the perimeter of C. Let x be a point interior to C. Define a function γ : [0, 2π] → R2 by setting γ (θ) to be the point where intersects the ray anchored at x making the angle θ with the positive x-axis (see Figure 3).
γ (θ )
x
θ
Figure 3.
The fact that C is convex implies that the point γ (θ) is uniquely determined and that γ is continuous (the justification is left as an exercise for the reader). We conclude that is a simple closed curve parameterized by γ and the numbers L(P ) in (4) correspond to the perimeters of the convex polygons Q with vertices on . Formula (4) may be expressed as L() = sup{Per(Q) : Q ⊆ C is an inscribed convex polygon}. What is remarkable about this expression is that it is defined for any convex body C with no assumption regarding the smoothness of its boundary. Motivated by these remarks, we define the perimeter of a convex body C by P(C) = sup{Per(Q) : Q ⊆ C is an inscribed convex polygon}. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
(5) 17
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:24 a.m.
swanson.tex
A small issue of consistency will need to be addressed: we have not demonstrated that P(Q) assigns the correct perimeter to a convex polygon Q since the boundary of Q is not smooth. An immediate consequence of the definition is that Per(Q) ≤ P(Q), and the fact that these quantities are indeed equal will be shown at the end of the next section.
Perimeters of polygons Given an angle 0 ≤ θ < π we define: • •
L θ is the line passing through the origin whose ray in the upper-half plane makes the angle θ with the positive x-axis. pθ (E) is the orthogonal projection of a set E onto L θ (see Figure 4).
Lθ
E θ pθ (E)
Figure 4.
We will make use of the following properties of convex bodies: if C is a convex body, then pθ (C) is a segment on L θ for every θ ∈ [0, π) and ( pθ (Ct )) = ( pθ (C)) + 2t
(6)
for all t > 0. The proof of this fact is left to the reader. The following lemma, which is central to the rest of the paper, shows that the perimeter of a convex polygon Q may be recovered from the lengths of its projections. Lemma. If Q is a convex polygon, then Per(Q) =
π
( pθ (Q)) dθ.
0
Proof. Consider a horizontal segment I in the plane with length (I ). For each angle θ ∈ [0, π), the projection pθ (I ) is a segment on the line L θ with length ( pθ (I )) = (I )|cos θ |, 18
(7)
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 18
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:24 a.m.
swanson.tex
page 19
as shown in Figure 5. This formula includes the degenerate case θ = π/2, where pθ (I ) consists of a single point. Now integrate (7) over [0, π) to obtain π π ( pθ (I )) dθ = (I ) |cos θ| dθ = 2(I ). 0
0
Likewise, if I is a segment which makes an angle α with the positive x-axis, then ( pθ (I )) = (I )|cos(θ − α)|. In this case π π π+α ( pθ (I )) dθ = (I )|cos(θ − α)| dθ = (I ) |cos θ | dθ 0
0
= (I )
α
π
|cos θ| dθ,
0
because |cos θ| is periodic with period of length π. Thus for any segment I in the plane, 1 π (I ) = ( pθ (I )) dθ. (8) 2 0
Lθ
I θ
( pθ (I )) = (I )|cos θ |
pθ (I )
Figure 5.
Now let Q be a convex polygon and let be its boundary. Writing the edges of as I1 , . . . , Ik , we have from (8) that π k 1 k (I j ) = ( pθ (I j )) dθ. (9) Per(Q) = 0 2 j =1 j =1 Since bounds a convex set, pθ () is a segment for every angle θ and the projection pθ : → L θ is two-to-one between the endpoints of pθ () (see Figure 6). It follows that the segments pθ (I j ) form a 2-fold covering of pθ () and consequently k 1 ( pθ (I j )). ( pθ ()) = 2 j =1
(10)
Comparison of (9) with (10) yields Per(Q) =
π
( pθ ()) dθ.
(11)
0
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
19
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
10:24 a.m.
swanson.tex
pθ is two-to-one between the endpoints of pθ ()
Figure 6.
Finally, since Q is convex we have pθ (Q) = pθ () and consequently ( pθ (Q)) = ( pθ ()) for every angle 0 ≤ θ ≤ π. This completes the proof. The lemma has several interesting corollaries. First, if Q and Q are convex polygons with Q ⊆ Q , then P(Q) ≤ P(Q ).
(12)
This follows at once because Q ⊆ Q implies that θ (Q) ≤ θ (Q ) for every angle θ. Second, if Q is a convex polygon, then P(Q) = Per(Q),
(13)
so that P(Q) assigns the correct value to every convex polygon Q. Although this is intuitively clear from the definition, the lemma can be used to give a correct justification. Since any polygon Q is naturally inscribed in itself, the definition of perimeter implies that Per(Q) ≤ P(Q). On the other hand, if R ⊂ Q is a convex polygon inscribed in Q, then inequality (12) states that Per(R) ≤ Per(Q). We take the supremum over all such R to obtain the opposite inequality P(Q) ≤ Per(Q). In light of this fact, from now on we will write P(Q) instead of Per(Q). Third, the word “inscribed” is superfluous in the definition of P(C), so that P(C) = sup{P(Q) : Q ⊆ C is a convex polygon},
(14)
for any convex body C. The definition of perimeter (5) directly implies that P(C) ≤ sup{P(Q) : Q ⊆ C is a convex polygon}. To obtain the opposite inequality let Q be any convex polygon contained in C. We construct from Q an inscribed polygon Q which will be the convex hull of a finite set S. Let I be an edge of the boundary of Q. If I is contained in the boundary of C, place the endpoints of I in S. Otherwise, extend I to a line which intersects the boundary of C at two points and add these points to S (see Figure 7). Repeating this process for each edge I yields a finite set S contained in the boundary of C. The convex hull Q of S is therefore a convex polygon inscribed in C, and since Q contains each edge of Q it contains Q itself. Thus Q ⊆ Q ⊆ C. Since P(Q) ≤ P(Q ) by (12) and Per(Q ) ≤ Per(C) by definition, we have P(Q) ≤ P(C). It follows that sup{P(Q) : Q ⊂ C is a convex polygon} ≤ P(C). 20
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 20
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:24 a.m.
swanson.tex
page 21
P
P
P C
C
Figure 7.
Fourth, if C and C are convex bodies with C ⊆ C , then P(C) ≤ P(C ).
(15)
This is a direct extension of (12) to convex bodies. Note that if Q ⊆ C is a convex polygon then Q ⊆ C , and by (14) we have P(Q) ≤ P(C ). A second application of (14) implies P(C) ≤ P(C ) by taking the supremum over all such Q. Fifth, and finally, every convex body C has finite perimeter. To see this, let S be a square in the plane satisfying C ⊆ S and note that (15) implies P(C) ≤ P(S) < ∞.
An integral formula for the perimeter of a convex body A convex body C may be approximated from the interior by convex polygons in the following sense: for any ε > 0 there exists a convex polygon Q ⊆ C such that P(Q) ≤ P(C) ≤ P(Q) + ε,
(16)
Q ⊆ C ⊆ Qε .
(17)
and
To see why this is possible, select a finite set of points x1 , . . . , xk ∈ C so that C⊆
k j =1
(x j , ε),
where (x, ε) is the closed disk of radius ε centered at x (the existence of such points may be justified for any bounded set C). Since C has finite perimeter, the characterization of P(C) given in (14) implies that there is a convex polygon R ⊆ C satisfying P(C) ≤ P(R) + ε. Denote by y1 , . . . , ym the vertices of R and define Q = convex hull of {x1 , . . . , xk , y1 , . . . , ym }. Then R ⊆ Q ⊆ C, and since P(R) ≤ P(Q) by (12) we obtain P(Q) ≤ P(C) ≤ P(R) + ε ≤ P(Q) + ε. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
21
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
10:24 a.m.
swanson.tex
Moreover, since any point y ∈ (x j , ε) is within distance ε of the point x j ∈ Q we have (x j , ε) ⊆ Q ε and thus Q⊆C ⊆
k j =1
(x j , ε) ⊆ Q ε .
Using this approximation, we extend the lemma of the previous section from convex polygons to convex bodies. Theorem 1. If C is a convex body, then
π
P(C) =
( pθ (C)) dθ.
0
Proof. Let C be a convex body. Let ε > 0 be given and let Q be a convex polygon satisfying (16) and (17). For every 0 ≤ θ < π, the inclusions in (17) imply that ( pθ (Q)) ≤ ( pθ (C)) ≤ ( pθ (Q ε )) = ( pθ (Q)) + 2ε, where the last equality is given by formula (6). We may integrate these inequalities over the interval [0, π) to obtain π π π ( pθ (Q)) dθ ≤ ( pθ (C))dθ ≤ ( pθ (Q)) dθ + 2πε. 0
0
0
The integrals on the left and right sides of this inequality are precisely the perimeter of Q. Thus π ( pθ (C))dθ ≤ P(Q) + 2πε, P(Q) ≤ 0
and, in light of (16), we have
π
P(C) − ε ≤
( pθ (C))dθ ≤ P(C) + 2πε.
0
Since ε > 0 was arbitrary, we obtain Theorem 1 by letting ε → 0+ .
The main result Theorem 2. If C is a convex body and t > 0, then 1. P(Ct ) = P(C) + 2πt, and 2. A(Ct ) = A(C) + P(C)t + πt 2 . Proof. Let C be a convex body and let t > 0. To prove (1) we apply Theorem 1 to get
π
P(Ct ) = 0
22
π
( pθ (Ct )) dθ =
( pθ (C)) + 2t dθ = P(C) + 2πt.
0
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 22
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
10:24 a.m.
swanson.tex
To prove (2) we will use the fact that if Q is a convex polygon then A(Q t ) = A(Q) + P(Q)t + πt 2 , (see the discussion preceding Figure 2 above) and approximate C from the interior by convex polygons. Select ε > 0 and let Q be a convex polygon satisfying (16) and (17) above so that P(Q) ≤ P(C) ≤ P(Q) + ε, and Q ⊆ C ⊆ Qε . These inclusions imply that Q t ⊆ Ct ⊆ Q t+ε as well. A simple comparison of areas shows that A(Ct ) ≤ A(Q t+ε ) = A(Q) + P(Q)(t + ε) + π(t + ε)2 ≤ A(C) + P(C)(t + ε) + π(t + ε)2 ,
(18)
and A(Ct ) ≥ A(Q t ) = A(Q) + P(Q)t + πt 2 . Estimating the last expression requires a little work. Since A(C) ≤ A(Q ε ) = A(Q) + P(Q)ε + πε 2 ≤ A(Q) + P(C)ε + πε 2 , and P(C) ≤ P(Q ε ) = P(Q) + 2πε, we have that A(Q) ≥ A(C) − P(C)ε − πε 2 , and P(Q) ≥ P(C) − 2πε. Returning to the lower bound for A(Ct ), these imply A(Ct ) ≥ A(C) − P(C)ε − πε 2 + (P(C) − 2πε)t + πt 2 = A(C) + P(C)(t − ε) + π(t 2 − ε 2 ) − 2πεt.
(19)
Passing to the limit in (18) and (19) as ε → 0+ , it follows that A(Ct ) = A(C) + P(C)t + πt 2 , as desired. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
23
page 23
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:24 a.m.
swanson.tex
Summary. We give elementary proofs of formulas for the area and perimeter of a planar convex body surrounded by a band of uniform thickness. The primary tool is an integral formula for the perimeter of a convex body which describes the perimeter in terms of the projections of the body onto lines in the plane. References 1. G. Edgar, Measure, Topology, and Fractal Geometry, Undergraduate Texts in Mathematics, Springer-Verlag, 1990. 2. H. Federer, Geometric Measure Theory, Classics in Mathematics, Springer-Verlag, 1969. 3. J. Sekino, The band around a convex set, College Math. J. 32 (2001) 110–114. doi:10.2307/2687115 4. S. Tabachnikov, Geometry and Billiards, Student Mathematical Library 30, American Mathematical Society, Providence RI, 2005.
Mathematical Jeopardy? During a recent airing of that ever-popular TV show Mathematical Jeopardy, contestants had to define the following terms: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m)
a cancellation law; a cyclic group; a discontinuous function; indirect proof; an integral domain; an irreducible element; max-flow, min-cut; a mean value; a non-homogeneous case; a planar graph; prime factors; quadratic residues; a recurrence relation.
Match the correct answers above with the appropriate questions below. What is a biker gang? What is circumstantial evidence? What is a country which has repelled all invasions? What is a flight plan? What are health and wealth? What is hooking up with an old flame? What is an interrupted party? What is the last stronghold? What is a loose blouse and a short skirt? What is an odd ball? What is a refund policy? What are semi-finalists? What is viciousness? —Andy Liu, The University of Alberta (For more Mathematical Jeopardy see Word Ways)
24
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 24
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
page 25
Two-Person Pie-Cutting: The Fairest Cuts Julius B. Barbanel and Steven J. Brams
Julius B. Barbanel (
[email protected]) received his B.S. from Case Western Reserve University in 1973 and his Ph.D. in mathematics from the State University of New York at Buffalo in 1979. He is a Professor of Mathematics at Union College, has published articles in both Set Theory and Fair Division, and has authored a book on Fair Division entitled The Geometry of Efficient Fair Division (Cambridge University Press 2005). He has interests in ancient mathematics and also enjoys bicycling and cross-country skiing. Steven J. Brams is Professor of Politics at New York University and the author, co-author, or co-editor of 17 books and about 250 articles. His most recent book is Mathematics and Democracy: Designing Better Voting and Fair-Division Procedures (Princeton University Press, 2008); Game Theory and the Humanities is forthcoming (MIT Press, 2011). He has applied game theory and social-choice theory to voting and elections, bargaining and fairness, international relations, and the Bible and theology. He is a former president of the Peace Science Society (1990–1991) and of the Public Choice Society (2004–2006). Brams is a Fellow of the American Association for the Advancement of Science (1986), a Guggenheim Fellow (1986–1987), and was a Visiting Scholar at the Russell Sage Foundation (1998–1999).
Pie-cutting is different from cake-cutting. For one thing, a cake is usually rectangular, a pie circular. Cutting a cake into n ≥ 2 pieces typically involves making n − 1 parallel, vertical cuts across the cake, whereas cutting a pie usually means cutting wedge-shaped pieces from the center, which requires n cuts. We can think of a cake as being obtained from a pie by making some initial cut, which determines the two ends of the cake, so its remaining division requires only n − 1 additional cuts. It is the initial cut that essentially differentiates pie-cutting from cake-cutting. We focus on pie-cutting here, although we return to cake-cutting to compare it with pie-cutting. We restrict our analysis to two players, because the properties we impose on the allocation of pie pieces are demanding and cannot all be satisfied if there are three or more players.
Assumptions and properties We assume that the two players, player 1 and player 2, wish to divide a pie into two pieces, using two cuts from the center, with one piece allocated to each player. Before doi:10.4169/college.math.j.42.1.025
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
25
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
stating the properties that we want the allocation to satisfy, it is helpful to provide some mathematical formalism. We assume that the cuts of the pie are radial, so the pie is mathematically equivalent to a circle. Player 1 and player 2 use additive measures m 1 and m 2 , respectively, to assess the value of a piece of pie. These measures are nonatomic (any piece of pie of positive measure is divisible into two pieces, each of positive measure, and therefore any single point of pie has measure 0) and absolutely continuous with respect to each other (any piece of pie that has measure 0 according to one player’s measure has measure 0 according to the other player’s measure). We lose no generality by assuming that any interval on the circle of positive length has positive measure because, if this were not so, we could eliminate this interval, join its endpoints, and redefine the measures on this new circle. We also assume that each player assigns measure 1 to the whole pie. We wish an allocation of the pie to have three natural properties: • • •
be envy-free, that is, neither player prefers the other player’s piece to his or her own piece; be undominated, that is, no other allocation gives one player more, and the other player at least as much, value in their own eyes; and be equitable, meaning both players assign exactly the same value to the pieces they receive, so neither player envies the other’s “degree of happiness”—that is, thinks the other player did better in his or her own eyes than one did in one’s own eyes.
Barbanel, Brams, and Stromquist [3] showed there always exists an allocation that is envy-free, undominated, and equitable when there are two players, but this is not true for more than two players. No algorithm for obtaining such an allocation was given, however; in fact, it was an open question whether or not there is a “moving-knife procedure” (we illustrate such procedures later) that produces such an allocation.
Approximate pie-cutting: the gap procedure In this section we describe a moving-knife procedure that is approximate, in a sense to be made precise, whereas in the next section we show how it can be made exact, given additional assumptions. The approximate procedure gives an allocation that is envy-free, as close to undominated as desired, and as close to equitable as desired. To make these notions precise, fix ε > 0, which will bound (in a way that we specify) the degree to which our allocation may fall short of being undominated and equitable. Let A, B denote an allocation of the pie, A to player 1 and B to player 2. Then A, B is (i) ε-undominated iff for no allocation A , B is it true that m 1 (A) + ε < m 1 (A ) and m 2 (B) + ε < m 2 (B ), (ii) ε-equitable iff |m 1 (A) − m 2 (B)| < ε. In other words, (i) there is no other allocation that is better for both players by ε or more; and (ii) the valuations of the players are within ε. We next describe a procedure that produces an allocation that is ε-undominated and ε-equitable, provided both players are truthful. Furthermore, no truthful player will envy the other player, whereas an untruthful player may be envious. The procedure Choose an even positive integer k so that 1/k < ε. A referee holds a radial knife at some arbitrarily chosen position on the circle and slowly moves the 26
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 26
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
page 27
knife clockwise around the circle. Meanwhile, the two players each move k − 1 knives around the circle in such a way that these knives, together with the referee’s knife, partition the pie into k pieces that, according to each player’s measure, are of equal value. Assume, for now, that the players’ placements of their knives accurately reflect their valuations. Later we consider what happens if a player lies. For each position of the referee’s knife, this knife—together with the k − 1 knives of each player—determine two partitions of the pie into k pieces. We require that the movement of the players’ knives be consistent: Assume that when the referee’s knife is at position p, one player’s k − 1 knives are at positions p1 , p2 , . . . , pk−1 (going clockwise from p). Then, when the referee’s knife moves to position p1 , this player’s knives (again going clockwise, this time from p1 ) are at positions p2 , p3 , . . . , pk−1 , p. Note that while a player’s truthfulness cannot be observed by the referee, his or her consistency can be. We assume that each player cannot see the other player’s moving knives. For example, the players may go through this procedure separately, or be screened from seeing the other player’s knives. As the players move their knives, the referee makes a movie of the players’ movements so as to show the two sets of knives moving simultaneously. For any integer i, and for each of the two players, we refer to that player’s clockwise i-piece or counterclockwise i-piece of pie, counting from the referee’s knife clockwise or counterclockwise, respectively, to that player’s knife i. At each point in the process, there is either an overlap or a gap between these i-pieces. (If they meet at a point, we consider this a gap, since a single point has measure zero.) First, let i = k/2 (and recall that k is even, so i is an integer). As the referee’s knife moves clockwise, we observe player 1’s clockwise i-piece and player 2’s counterclockwise i-piece. (Since we are presently assuming that the players are truthful, each player values his or her i-piece at i/k = 1/2.) We claim that there must be some time in this process when there is a gap between the two players’ i-pieces. Consider, for example, the starting position. If there is not a gap at this time, then there is an overlap. It is not hard to see that when the referee’s knife moves around to any point in this initial overlap, there will now be a gap, and the initial position of the referee’s knife will be in this gap. This follows because each measure is additive, and so the complement of any piece having measure 1/2 also has measure 1/2. We repeat this process with i = (k/2) + 1. If there is no gap at any point in the referee’s 12-hour rotation, we stop. If there is a gap, we repeat the rotation with i = (k/2) + 2. When finally i = k, there will be no gap at any point in the rotation, since the i = k rotation involves pieces with value 1, i.e., the whole pie. Hence, for some j = 1, 2, . . . , (k/2), the i = (k/2) + j − 1 rotation has a gap at some point, but the i = (k/2) + j rotation does not. For convenience, set s = (k/2) + j − 1, the last rotation showing a gap. We call this the s-rotation. Arbitrarily choose a position p for the referee’s knife so that there is a gap in the srotation when the referee’s knife is at position p. We refer to this gap in the s-rotation as the s-gap. In the (s + 1)-rotation, there are no gaps. In particular, when the referee’s knife is at p in the (s + 1)-rotation, there is an overlap between the players’ (s + 1)-pieces (specifically, an overlap between player 1’s clockwise (s + 1)-piece and player 2’s counterclockwise (s + 1)-piece). We refer to this overlap as the (s + 1)-overlap. Note that it is not possible for the s-gap and the (s + 1)-overlap to be disjoint. We illustrate this in Figure 1. The four diagrams correspond to the four possible relative VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
27
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
1
9:37 a.m.
barbanel.tex
1
2
2 1
1
2
2 1
1
2 1
2 1
2
2
Figure 1. Arrows indicate the interval from knife s to knife (s + 1) for each player. The s-gap is dark, and the (s + 1)-overlap is grey.
positions of the players’ knives s and (s + 1), and the corresponding four relative positions of the s-gap and the (s + 1)-overlap. Pick a point q that is in both the s-gap and the (s + 1)-overlap. Let A be the piece of pie obtained by going clockwise from p to q, and let B be the complement. We will show that A, B is the desired allocation. Because A and B are each obtained by starting at the point p and going to the point q (clockwise for player 1, counterclockwise for player 2), and q is at or beyond each player’s knife s, it follows that any player who truthfully indicates the position of his or her knife in the s-rotation will get a piece of pie of size at least s/k. Since j ≥ 1, it follows that each player receives a piece of pie whose value is at least s/k = [(k/2) + j − 1]/k ≥ [(k/2)/k] = 1/2. Therefore, if a player is truthful, he or she will not envy the other player. What if a player is not truthful? Suppose, for example, that player 1 and player 2 have identical measures, that player 1 is truthful, and that player 2 is not. This means that at some time in the s-rotation, some pair of corresponding knives of the players does not coincide. This, together with our consistency assumption, implies that there will be a gap of more than a single point at some time in the s-rotation. If the point q is chosen from within (i.e., not at an endpoint of) this gap, player 1 will get strictly more than half the pie, since he or she was truthful, but because player 1 and player 2 have identical measures, player 2 will get strictly less than half the pie and, hence, will envy player 1. If both players are truthful, we prove allocation A, B is ε-undominated and εequitable by first supposing, by way of contradiction, that A, B is not ε-undominated. Then for some allocation A , B , m 1 (A) + ε < m 1 (A ) and m 2 (B) + ε < m 2 (B ). Our construction and the players’ truthfulness imply that m 1 (A) ≥ s/k and m 2 (B) ≥ s/k. This, plus the fact that (1/k) < ε, implies that m 1 (A ) > m 1 (A) + ε ≥ (s/k) + ε > (s/k) + (1/k) = (s + 1)/k. Similarly, m 2 (B ) > (s + 1)/k. But this contradicts the fact that the (s + 1)-rotation has no gaps. To show that A, B is ε-equitable, we simply observe that each player’s piece runs from the referee’s knife to somewhere between (but not necessarily strictly between) 28
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 28
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
page 29
each player’s knife s and knife s + 1. Hence, if each player is truthful, then each player values his or her piece at between s/k and (s + 1)/k. The difference between these two values is at most 1/k. Since (1/k) < ε, it follows that the allocation A, B is ε-equitable. We call this procedure the gap procedure, not to be confused with a different gap procedure for the fair division of both divisible and indivisible goods (Brams and Kilgour [7], Brams [4] gives updated references). Because truthfulness maximizes the minimum (i.e., 1/2 the value of the cake) that a player can guarantee for himself or herself—irrespective of what the other player does—truthfulness is a maximin strategy for the gap procedure. There is nothing in our procedure that prevents there from being a gap, not chosen by the referee, that gives both players more value (i.e., is closer to being undominated), narrows the difference in their valuations (i.e., is more equitable), or both. However, such a preferred allocation cannot yield both players ε or more in value, or give one player ε or more in value over what the other player receives. Thus, the allocation that the gap procedure produces is approximately undominated and approximately equitable while being entirely envy-free. We close this section by noting that while it was convenient to present the idea of the procedure in stages (i.e., as a sequence of rotations), in truth the same actions are being performed by the referee and by the players at every stage. The crucial stage is that in which there are no longer gaps, so the referee chooses a point for a second cut in the intersection of the s-gap and the (s + 1)-overlap.
An exact procedure? One approach to rendering the approximate allocation of the gap procedure exact is to allow the referee to ask for more information from the players. In particular, assume that the referee begins by asking each player to report, unbeknownst to the other player, his or her measure. (For now we assume that the players are truthful in their reports. Shortly we show that a player risks ending up envious by lying.) Knowing the players’ measures, the referee (R) can, in principle, find the maximum gap in a single 12-hour rotation of two knives. R does so by placing one knife at some starting point (say, 12 o’clock), and another knife at another starting point (say, 5 o’clock), that equitably divides the pie: Player 1’s valuation of the pie between 12 and 5 is the same as player 2’s valuation between 5 and 12. R rotates the two knives simultaneously clockwise (or counterclockwise), maintaining the equitable valuations of the two players through 12 hours, after which R determines a pair of cutpoints that maximize the players’ equitable valuations. The resulting allocation is equitable (by construction), undominated (because it is maximal), and, necessarily, envy-free (equitability and undominatedness imply envy-freeness for two players). Suppose one player lies about his or her measure, underestimating value during some parts of the rotation and overestimating it during others. This player has no assurance that the equitable cutpoint found by the referee gives him or her a piece whose value is underestimated and that he or she could not have done better by being truthful. In particular, a lying player has no assurance that he or she will receive a piece with value at least 1/2 the pie, and thus no assurance of not being envious of the other player. Only truthfulness on the part of each player maximizes the minimum that each can guarantee for himself or herself, and only truthfulness on the part of both players guarantees that the allocation is equitable, envy-free, and undominated. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
29
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
To guarantee these properties, R must continuously pay attention to both players’ measures to ensure that the two continuously moving cutpoints equalize the players’ valuations at every instant, a big task. On the other hand, is it less demanding to ask the players to respond instantaneously to R’s moving knife under the gap procedure? This also requires an infinite number of calculations. Whether one should allow a calculating R—rather than one who just moves a knife—is a philosophical issue more than a mathematical one. The gap procedure requires only that the players make calculations at every point of the circle, whereas the exact procedure requires that a referee calculate equitable cutpoints and then find a pair that maximizes the players’ valuations. In either case, the mathematical calculation required is clear; it is less clear which calculation should be allowed. To complicate matters, there is a third philosophical position: Rule out entirely any possibility of instantaneous calculations and continuously moving knives. Instead, allow only discrete procedures and cuts at discrete times. The argument for this position is the well-known definition of an algorithm as a finite procedure that a Turing machine can execute. This disallows continuous, and hence infinitely many, calculations. To be sure, continuous calculations can be “discretized,” but this makes them only approximate. Moving-knife procedures have long been part of the fair-division literature (Brams, Taylor, and Zwicker [9], Brams and Taylor [8], Robertson and Webb [11], Barbanel and Brams [2]); in our view, they should not be banned for two reasons. One is that they can be used to prove the existence of an allocation with specified properties; the other is that such procedures are intuitively appealing, even if they presume instantaneous calculations. Existence proofs do not say how players can achieve a particular allocation but, instead, show only that such allocations exist (see, for example, Barbanel [1]). Algorithms are finite, step-by-step procedures that enable players to implement certain allocations, which we illustrate for cake-cutting in the next section. Continuous procedures involving moving knives, like the gap procedure, are not finite, but they enable players to implement allocations, with the help of a referee, who uses information that he or she can directly observe. For example, while a referee can observe and verify that knives are moved and cuts are made according to his or her instructions, he or she has no access to the private information held by players and so cannot determine that they are acting according to their stated preferences. By contrast, when the players report their measures to the referee under the exact procedure, the referee can move their knives and make cuts as if he or she were acting for them. This kind of procedure has been allowed in the recent pie-cutting literature (Brams, Jones, and Klamler [6]).
Relationship to cake-cutting Whether discrete or continuous, with or without a calculating referee, we insist that a procedure not prevent a truthful player from obtaining an allocation with specified properties. We illustrate this with a discrete cake-cutting procedure, somewhat like the gap procedure, which gives players the incentive to be truthful about their 5050 points but fails to guarantee that the resulting envy-free allocation is equitable. Indeed, we conjecture that there is no two-person cake-cutting procedure—discrete or continuous—in which truth-telling by both players is the only strategy that guarantees them an envy-free and equitable allocation of a cake. (All envy-free allocations of a cake are undominated; see Gale [10], and Barbanel, Brams, and Stromquist [3].) 30
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 30
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
page 31
Assume that each of two players is asked to indicate, unbeknownst to the other, where they would make their 50-50 cut in a cake. If these are not the same point and so leave a gap, begin by giving each player the side of the gap that gives him or her 1/2 the cake, as he or she values it. This is equivalent to choosing i = k/2 with the gap procedure but, because this is a cake, not a pie, without the need for a referee to move a knife around a circle. As with pie-cutting, a player has good reason to be truthful about his or her 50-50 point on the cake (Brams, Jones, and Klamler [5]). If the player is not, he or she may end up with less than 1/2 cake, even adding in the entire gap. This can be seen below, where 1 and 2 are the true 50-50 points of players 1 and 2, respectively, and B and E are the beginning and end boundaries of the cake. B
1
2
1
E
If player 1 misrepresents his or her 50-50 point as 1 , then giving him or her all of the gap plus the remainder of the right side yields the player [2, E], which he or she values at less than 1/2, whereas player 2 obtains 1/2 from [B, 2). If the players are truthful about their 50-50 points, how can they divide the gap [1, 2]? Unlike pie-cutting, the gap procedure, applied to cake-cutting, is vulnerable to strategizing. If the players are asked to cut the cake into k > 2 equal pieces instead of two, they have no incentive, after accurately indicating their 50-50 cutpoints, to tell the referee their true measures. Instead, they should pretend to value cake near its endpoints at almost 50 percent each, and then very little until it reaches exactly 50 percent at their 50-50 points. More specifically, if a player pretends to value the cake around his or her 50-50 point as very low, he or she will need more of the interval that separates the 50-50 points to obtain the same value as the other player receives from it, giving both players an incentive to lie. The players’ claims would cause the interval to disappear (when j = 2), so the referee’s choice would be an arbitrary cutpoint in [1, 2]. While this would ensure each player an envy-free piece of at least 1/2 the cake, any guarantees about equitability fall by the wayside. However, Brams, Jones, and Klamler [5] provide a partial solution to this problem by giving a procedure that produces a “proportionally equitable” allocation of a cake. With pie-cutting, on the other hand, there will, in general, be an infinite number of gaps, which could be anywhere around the circle. Unlike cake-cutting, the players are at a loss in cutting a pie to formulate a strategy of lying that can increase their shares, beyond what truthfulness guarantees. To be sure, a player may do better by lying, but as we have demonstrated, a player also may do worse, because only truthfulness is a maximin strategy (i.e., maximizes one’s minimum allocation).
Conclusion In [3], Barbanel, Brams, and Stromquist suggest that pie-cutting is harder than cakecutting. One reason is that cake allocations can have stronger properties. For example, it is always possible to find an undominated, envy-free allocation of a cake, but there are pies for which this is not possible for three or more players. Ironically, for two players, pie-cutting is easier, as we have shown—at least using the gap or the exact procedure—because the players have no incentive to lie, lest they do worse. But this is not true of cake-cutting. Thus, envy-free, undominated, and equitable allocations—or at least approximations to the latter two properties—are easier to achieve for pies than for cakes. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
31
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
9:37 a.m.
barbanel.tex
Acknowledgment. We are grateful to the editor and a reviewer for their valuable contributions to this article. Summary. Barbanel, Brams, and Stromquist (in 2009) asked whether there exists a twoperson moving-knife procedure that yields an envy-free, undominated, and equitable allocation of a pie. We present two procedures: One yields an envy-free, almost undominated, and almost equitable allocation, whereas the second yields an allocation with the two “almosts” removed. The latter, however, requires broadening the definition of a “procedure,” which raises philosophical, as opposed to mathematical, issues. An analogous approach for cakes fails because of problems in eliciting truthful preferences.
References 1. J. B. Barbanel, The Geometry of Efficient Fair Division, Cambridge University Press, Cambridge, 2005. 2. J. B. Barbanel and S. J. Brams, Cake division with minimal cuts: envy-free procedures for 3 persons, 4 persons, and beyond, Math. Social Sci. 48 (2004) 251–269. doi:10.1016/j.mathsocsci.2004.03.006 3. J. B. Barbanel, S. J. Brams, and W. Stromquist, Cutting a pie is not a piece of cake, Amer. Math. Monthly 116 (2009) 496–514. doi:10.4169/193009709X470407 4. S. J. Brams, Mathematics and Democracy: Designing Better Voting and Fair-Division Procedures, Princeton University Press, Princeton, 2008. 5. S. J. Brams, M. A. Jones, and C. Klamler, Better ways to cut a cake, Notices Amer. Math. Soc. 35 (2006) 1314–1321. 6. , Proportional pie-cutting, Internat. J. Game Theory 36 (2008) 353–367. doi:10.1007/ s00182-007-0108-z 7. S. J. Brams and D. M. Kilgour, Competitive fair division, Journal of Political Economy 109 (2001) 418–443. doi:10.1086/319550 8. S. J. Brams and A. D. Taylor, Fair Division: From Cake-Cutting to Dispute Resolution, Cambridge University Press, Cambridge, 1996. 9. S. J. Brams, A. D. Taylor, and W. S. Zwicker, Old and new moving-knife schemes, Math. Intelligencer 17 (1995) 547–554. doi:10.1007/BF03024785 10. D. Gale, Mathematical entertainments, Math. Intelligencer 15 (1993) 48–52. doi:10.1007/BF03025257 11. J. Robertson and W. Webb, Cake-Cutting Algorithms: Be Fair If You Can, A K Peters, Natick MA, 1998.
. . . I discovered that the professional literature on the problem was far more vast than I had ever imagined. Even more so than mathematicians, cognitive scientists and psychologists have been making a good living trying to explain why people find this particular problem so maddeningly difficult. Philosophers have used it directly in exploring the nature of probability, and indirectly in shedding light on seemingly unrelated philosophical questions. Economists have made their own contributions to the literature, studying what the Monty Hall problem tells us about human decision making. This barely scratches the surface. —from The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brainteaser by Jason Rosenhouse [p. x] Reviewed in this issue on page 71
32
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 32
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
Augustus De Morgan Behind the Scenes Charlotte Simmons
Charlotte Simmons (
[email protected]) is a professor of mathematics and department chair at the University of Central Oklahoma in Edmond, Oklahoma. She earned her Ph.D. at the University of Oklahoma. Her research interests include finite hyperbolic geometry, cryptology, and the history of mathematics.
“[I]t will ever be much more pleasing to grant even more praise than is actually due, than to pluck the laurel from the deserving brow.” —Benjamin Gompertz [10] Augustus De Morgan (1806–1871) was a nineteenth century mathematician and prolific writer, author of more than 160 papers and 18 textbooks on algebra, arithmetic, trigonometry, probability, logic, and calculus, plus 850 articles in the popular, working-class oriented Penny Cyclopedia [3]. Here, however, we explore his contributions from behind the scenes, as a mentor to other mathematicians. Both Sir William Rowan Hamilton and George Boole, for example, two of the greatest algebraists of the nineteenth century, were close friends of De Morgan. During the period in which they produced some of their greatest work, both were in regular correspondence with De Morgan, who provided them with encouragement and reassurance in both their professional and personal lives. It is doubtful that either would have attained the level of success that they ultimately achieved without his help. De Morgan also provided support for the actuarial pioneer Benjamin Gompertz and the Indian mathematician Ramchundra, who has been called “De Morgan’s Ramanujan” [13]. If De Morgan had done nothing else noteworthy in mathematics besides supporting the efforts of these four men and championing their work, we would still owe him a great debt.
Who was De Morgan? De Morgan entered Trinity College, Cambridge at age 16, where he developed a lifelong love of mathematics. He obtained his bachelor’s degree in 1827, and was offered the first Professorship of Mathematics at University College London the following year. Though he had no teaching experience, no publications, and was only twentyone, he was selected from 30 applicants “after the most distinguished competition that there has been for any chair” according to a letter from Thomas Coates [19]. The selection committee chose wisely. De Morgan became one of the longest serving and highly respected professors at the university. Professor M. J. M. Hill (Astor Chair of Mathematics at London College from 1884–1924) said of him that “amongst the great doi:10.4169/college.math.j.42.1.033
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
33
page 33
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
men who have lectured within the walls of the College he was probably by reason of his scholarship, by the profundity of his work, and by his personal character, one of the greatest, if not the greatest of them all” [19]. By his retirement in 1867, De Morgan was one of the most distinguished mathematicians in Britain. In 1887, the London Mathematical Society established the De Morgan Medal in his honor. De Morgan prided himself on being a champion of the underdog [18]. He is the only professor in the history of University College London to have resigned twice on matters of principle. His first resignation, in 1831, was in defense of a colleague who he felt had been unfairly dismissed. De Morgan returned five years later, after his replacement drowned, to teach an additional 30 years. Then, when the most qualified candidate was denied a position at the College because he was a controversial Unitarian minister, De Morgan resigned again, never to return. He refused even to sit for a bust for the college library, explaining that as far as he was concerned his old college no longer existed [21].
Hamilton The Irish mathematician Sir William Rowan Hamilton (1805–1865) is famous for constructing the first working algebraic system to break away from the real and complex numbers. Most importantly, Hamilton took the unheard of step of abandoning the commutative law in formulating the quaternions (1843). In his Lectures on Quaternions, Hamilton credits De Morgan’s attempts to extend the geometrical representation of complex numbers as having helped to lead him to this discovery. Hamilton enjoyed a fast rise to the top. Born in Dublin at the stroke of midnight, “like Christ and Newton,” by age three he had begun a rigid curriculum designed to prepare him for Trinity College [16]. At age 17, he was proclaimed a possible second Newton by Brinkley, Astronomer Royal of Ireland. He published his first paper while still an undergraduate; as a result of this work, he was unanimously elected to fill Brinkley’s recently vacated position as Astronomer Royal and was appointed Professor of Astronomy at Trinity College just prior to completing his undergraduate degree. At age 30, he became the first person in Ireland ever to be knighted for scientific merit.
De Morgan and Hamilton Though his reputation was established long before his correspondence with De Morgan began, De Morgan still provided valuable assistance to Hamilton. When Hamilton asked De Morgan to review his text, Lectures on Quaternions, he graciously agreed: “There is a pleasure in reading while any thing that strikes may do service” [8]. De Morgan never complained once though the text grew to be 700 pages long, and took five years to complete. De Morgan’s task was thus no small undertaking. Though Hamilton expressed his “most sincere and unaffected conviction” that one who had had the benefit of a scientific education at a good university and attempted to study this text would find it “almost light reading,” Herschel described the same work as one which would “take any man a twelvemonth to read, and near a lifetime to digest” [7]. As Hamilton once said, “[H]ow deeply man desires in intellectual things themselves the sympathy of man” [7]. De Morgan provided this sympathetic ear. Hamilton’s “monstrous innovations” drew intense criticism at the time, but De Morgan provided support when Hamilton’s quaternions were being attacked: “As to people ridiculing quaternions, let them do it; but do not let them succeed in making you feel it” [8]. 34
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 34
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
Besides mathematics, their letters (only a portion fills 400 pages [8]) were filled with accounts of their children’s activities, social events, poetry, politics, religion, and other details of their personal lives. Hamilton had this to say of their correspondence: “[H]e and I have exchanged a great number of pleasant letters, partly no doubt on mathematics, but we are not afraid to write nonsense to each other; at least I send him nonsense at times, and he sends me back wit in return, rising occasionally to humour” [8]. An example of this wit is the following, which De Morgan wrote to Hamilton after a lengthy break in their correspondence [8]: If you are dead and buried, why do you not say so at once, like a man, instead of insinuating it in this roundabout way by solemn silence? What has become of the Manual of Quaternions? I write to you because I want to know something about you . . . If you do not write I shall circulate a report that you have shipped yourself to fight for the Pope. De Morgan also did his best to console Hamilton during times when “the earth seemed to him draped in black” [7]. When Catherine, the only woman he ever loved, married someone else, Hamilton contemplated suicide. He never really recovered emotionally and suffered from bouts of depression throughout his life. When she resurfaced years later, De Morgan received many letters detailing how the subject of Catherine “continued to agitate him to a degree beyond what is rational” [9]. Upon Hamilton’s death, De Morgan wrote: “I have called him one of my dearest friends, and most truly; for I know now how much longer than twenty-five years we have been in intimate correspondence, of most friendly agreement or disagreement, of most cordial interest in each other. And yet we did not know each other’s faces. I met him about 1830 at Babbage’s breakfast table, and there for the only time in our lives we conversed . . . . And this is all I ever saw, so it has pleased God, all I shall see in this world of a man whose friendly communications were amongst my greatest social enjoyments, and greatest intellectual treats” [8].
Boole George Boole (1815–1864) was almost entirely self-educated. Though one of his fellow-pupils at the primary school he attended reports, “This George Boole was a sort of prodigy among us and we looked upon him as a star of the first magnitude,” his parents were so poor that they could not afford to send him to secondary school, let alone college [11]. At 16, Boole had to seek employment as an assistant teacher to support his parents and siblings. It didn’t go well. He was terminated for reading mathematics on Sunday and working math problems in chapel. Ultimately, he secured a professorship at a college in Ireland, where he taught until his death. Boole’s first major contribution to mathematics was the paper, “Exposition of a General Theory of Linear Transformations,” which marked the beginning of algebraic invariant theory; his next was “On a General Method in Analysis,” which helped pave the way for operator theory. But his greatest gift to mathematics was in logic. His first book on the subject, The Mathematical Analysis of Logic, is recognized as the beginning of modern symbolic logic. Here, Boole showed that logic is actually a branch of mathematics, and introduced the notion that symbols need not always be interpreted quantitatively but can be used to represent objects as well. In addition, he developed a method for expressing the established rules of syllogistic reasoning algebraically, which others before him (including Leibniz) had attempted but failed. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
35
page 35
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
De Morgan and Boole Within mathematics, De Morgan is also best known for his work in logic. He created the first logic of relations, and promoted a symbolic approach to the subject. Unfortunately, as De Morgan noted in his review of the work, Boole’s masterpiece, An Investigation of the Laws of Thought (1854) appeared “on the very day” as his own and soon overshadowed it [22]. Although Boole’s work is of greater depth than De Morgan’s, his initial publications on the subject were inspired, at least in part, by De Morgan’s work as he notes in the preface. Boole introduced himself to De Morgan by letter in 1842, when not yet widely known. Excessively modest and self-conscious of the fact that he had no degree, Boole said he had no “claim to [the] notice” of those mathematicians who did [11]. De Morgan encouraged him to submit his work to the Royal Society, assuring him that he need not hesitate a moment to publish. This award-winning paper immediately brought him to the notice of the mathematical community. As Boole’s reputation grew, he turned to De Morgan on numerous occasions to proofread his papers and arrange their publication. (Some ninety letters from the Boole-De Morgan correspondence have survived [23].) De Morgan always supplied positive reinforcement, even when busy with his own work or he didn’t understand Boole’s train of thought: “To say how far I agree with you would be difficult at this time, as it is my busiest time . . . But I must urge on you to continue and publish.” On another occasion, De Morgan writes, “I hope you will expand your views of probabilities—which I am not sure I understand” [23]. De Morgan secured a pass to the library of the British Museum for Boole, and was instrumental in securing a position for him at Queen’s College, Cork, no small accomplishment given that the last line of Boole’s application read “I am not a member of any University and have never studied at a college.” De Morgan writes in his recommendation, “I can speak confidently to the fact of his being not only well-versed in the highest branches of Mathematics, but possessed of original power” [11]. Though he enjoyed his position, Boole was not happy in Ireland and troubles within the college worried him. With no wife or family nearby, Boole turned to De Morgan for advice and support [23]: Now this is what I would not say to any one in whose good feeling and discretion I could not place entire confidence. What I ask of you is not to mention these circumstances but to inform me at any future period of what you suppose might suit me in England. No one else knows of my present views and feelings. By 1855, Boole’s mood had improved and he reported to De Morgan: “My objections to Ireland are however growing less and less and I have really very little to complain of besides the smallness of the remuneration which I receive” [23]. The reason for his cheerfulness was apparent when he revealed that he had married four months prior. At the request of a friend, Boole had agreed in 1850 to tutor Mary Everest. Boole had initially warned Mary that he was too old ever to think of marrying, but her father’s death in 1855 left her ill and destitute. At the age of 40, Boole proposed to the 23-year old, cautioning that he had reservations about “imprisoning a young girl’s life” [11]. By all accounts, their marriage was an extremely happy one: Mary referred to their nine years together as “a sunny dream” [2]. Boole’s correspondence with De Morgan clearly indicates the depth of respect and trust he had for him. As mathematician Robert Graves, a contemporary of De Morgan, puts it, “the sterling truthfulness” of De Morgan’s nature nurtured confidence in his friends and they confided in him their most private thoughts and feelings [7]. When 36
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 36
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
Boole died unexpectedly at the age of forty-nine, De Morgan worked hard to secure a government pension for his wife and five small children. In a letter to Hamilton dated December 13, 1864, De Morgan writes, “There will be no need to tell you that you must be aiding and assisting in getting a pension for Boole’s wife and daughters. An application will be made and must be well-backed” [8]. Significantly De Morgan used distinctly different tones in his letters to Boole and to Hamilton. With Hamilton, De Morgan often assumed a bossy, even reprimanding voice. Sarcastic remarks were common: “Ink must be cheap in Ireland if you can afford to waste it on such a supposition as that” [8]. With the shy and insecure Boole, De Morgan was much gentler. His perceptiveness is remarkable, given that he was not well acquainted with either of them in person.
Gompertz Benjamin Gompertz (1779–1865) was the son of a distinguished diamond merchant. Like Boole, he was forced to teach himself mathematics for, as a Jew, he was barred from a college education. He studied Newton and Maclaurin, stealing out into the garden and pursuing his investigations by moonlight after his parents removed his candles fearing he would stay up late studying. His first significant mathematics paper described an application of a method of differences to series. It was submitted to the Royal Society in 1806, but the two papers that established his reputation as a brilliant mathematician and won him an invitation to join the Royal Society appeared in 1817 and 1818. When the Royal Society rejected the first for being too profound to be understood, Gompertz published them at his own expense [1]. Gompertz was, as De Morgan wrote in an obituary [6], in a certain sense, ‘the last of the learned Newtonians.’ He was the last who adhered to the old language of fluxions, which [as of 1865] has been obsolete in the English mathematical world for nearly half a century. It is adherence to fluxional notation that apparently prevented wide recognition of his work at the time. Today Gompertz is regarded as a pioneer in actuarial science. Gompertz’s wife came from a wealthy Jewish family with strong links to the stock exchange. Gompertz’s brother-in-law set up the Alliance Assurance Company in 1824. Gompertz served as actuary from then until his retirement from active life. In 1825 he introduced the ‘law of mortality’ which became a fundamental tool of the life insurance industry. (See page 40, where an application of Gompertz’s law to insurance is described.)
De Morgan and Gompertz Though his primary job was at University College, De Morgan frequently accepted work as an actuarial consultant. In 1838 he published An Essay on Probabilities, and on Their Applications to Life Contingencies and Insurance Offices, which applied probabilistic methods to insurance problems. It was widely used for over a generation. A close personal friend and correspondent of Gompertz, De Morgan supported Gompertz when his law of mortality did not immediately capture the attention of those in the field. De Morgan wrote that it was “not by any means so well known as it ought to be, even by actuaries” [4]. Worse, it was attacked, in 1832, by T. R. Edmonds, VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
37
page 37
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
claiming that he had discovered the same law “independent of the imperfect one of Mr. Gompertz” [10]. De Morgan defended his friend writing, “All of the points of Mr. Edmonds’s alleged discovery had been published by Mr. Gompertz,” and whereas Edmonds claims “that the discovery of Mr. Gompertz is imperfect—meaning, of course, as compared with that claimed by Mr. Edmonds . . . there is no difference between the two” [5]. A fifteen-page response from Edmonds followed, but De Morgan’s defense of Gompertz was convincing.
Ramchundra Born about fifty miles from Delhi, Yesudas Ramchundra (1821–1880) was the son of a revenue collector for the East India Company. De Morgan reports that he attended a school where “no particular attention was paid to mathematics,” and “studied [mathematics] at home with such books as he could procure” [17]. One year after his father died, Ramchundra was married at the age of eleven to the daughter of a wealthy man of Delhi. While her dowry eased the burden of supporting his mother and five siblings somewhat, his wife was deaf and mute as Ramchundra only learned after the wedding. He was forced to drop out of school for three years to care for his new wife and allow his siblings to get an education. During this period, he worked as a journalist. Returning to Delhi College on scholarship and subsequently completing his studies in 1844, Ramchundra was hired by his alma mater to teach science and mathematics. Ramchundra was born during a period in which the cultural and literary life of Delhi was vibrant. He was responsible for Delhi College publication during the 1840s and 50s, making Western developments in science and technology available to the literate public of North India and advocating openness to knowledge independent of its origin. Praised by scholars for a straightforward and conversational style of writing, he wrote on many topics apart from mathematics, including water mills, banyon trees, steamboats, railways, balloons, mirages, deception, irrigation, Confucious, the circulation of blood, and the education of girls [12]. During the Indian revolt of 1857, the principal of Delhi College was killed and the college collapsed. In a letter to De Morgan, Ramchundra detailed his own narrow escape when mutineers raided his village [17]: [A] very prudent Brahmin zemindar advised me and my servant to fly to the jungles before the mutineers could arrive. We did so; but before we could run three quarters of a mile, we heard a great noise in the village, bullets were whistling about us, and horsemen appeared to be in our pursuit, for the noise of galloping was distinctly heard. I then rushed into a thorny little bush, not minding the thorns that went into my flesh . . . the mutineers, after plundering and giving a good beating to the zemindars, &c. with whom I lived in the village, did not penetrate into the jungle, but went their way towards Delhi. After a brief appointment as head master of what is today the Indian Institute of Technology-Roorkee, he was appointed in 1858 as head master of a newly organized school in Delhi. He retired at the age of 45 due to poor health, but was later appointed director of education in 1870 and honored for his contribution to the development of education in the state of Patiala. 38
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 38
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
De Morgan and Ramchundra In 1850, De Morgan received a copy of a work by a twenty-nine year old self-taught Indian mathematician. “[M]y own birth and descent having always given me a lively interest in all that relates to India, I took up the work of Ramchundra with a mingled feeling of satisfaction and curiosity: a few minutes of perusal added much to both” [17]. Like Gompertz, Ramchundra had published his treatise in Calcutta at his own expense. Endorsing the work as being worthy of encouragement, De Morgan recommended that the treatise be reprinted in London for circulation in Europe and India and offered his services as editor. A Treatise on Problems of Maxima and Minima appeared nine years later with a twenty-three page preface by De Morgan. Upon De Morgan’s recommendation, the Indian government authorized a reward of 2,000 rupees to a very grateful Ramchundra. De Morgan had hoped that the reprint would bring Ramchundra to the notice of scientific men in Europe and promote a revival of interest in India. He pointed out that mathematics is one of the sciences for which Europe is indebted to India. Unfortunately, things didn’t turn out as De Morgan had planned and Ramchundra’s name is not widely known today even within mathematics. Niven [15], for example, contains no mention of Ramchundra, although Ramchundra’s A Treatise on Problems of Maxima and Minima, Solved by Algebra (1859) addresses the same ideas and even some of the same problems. Rice [20] characterizes De Morgan’s labor on Ramchundra’s behalf as “maximum effort, minimum effect.” On the other hand, since the appearance of Mus`es paper [13], a biography of Ramchundra has been added to various online encyclopedias (e.g., Wikepedia and Answers.com), and a 2007 publication by Nahin [14] has a section titled “Apollonius Pursuit and Ramchundra’s Intercept Problem.” Perhaps seeds sown by De Morgan more than a century ago finally have taken root. Acknowledgments. The author would like to thank the editor and the reviewers for their excellent suggestions, and M. Meo for bringing De Morgan’s relationship with Ramchundra to her attention. She would also like to acknowledge support received from the UCO Office of Research & Grants while conducting research for this manuscript. Summary. Augustus De Morgan’s support was crucial to the achievements of mathematicians whose work is considered greater than his own. This article explores the contributions he made to mathematics from behind the scenes by supporting the work of Hamilton, Boole, Gompertz, and Ramchundra.
References 1. M. N. Adler, Memoirs of the late Benjamin Gompertz, Journal of the Institute of Actuaries 13 (1866) 1–20. 2. M. E. Boole, Home-side of a scientific mind, Mary Everest Boole: Collected Works vol. 1, C.W. Daniel Company, London, 1931. 3. R. Corrie, Penny Magazine Online, Electronic Historical Publications; available at http://www.history. rochester.edu/pennymag/, accessed August 23, 2009. 4. A. De Morgan, On a property of Mr. Gompertz’s law of mortality, Journal of the Institute of Actuaries 8 (1858–1860) 181-184. , On an unfair suppression of due acknowledgment to the writings of Mr. Benjamin Gompertz, Jour5. nal of the Institute of Actuaries 9 (1860–1861) 86–89. , Benjamin Gompertz, The Athenaeum, 22 July 1865, 117. 6. 7. R.P. Graves, Life of Sir William Rowan Hamilton 1882–1889 vol. 2, Arno P., New York, 1975. , Life of Sir William Rowan Hamilton 1882–1889 vol 3, Arno P., New York, 1975. 8. 9. T. L. Hankins, Sir William Rowan Hamilton, John Hopkins UP, Baltimore, 1980. 10. P. F. Hooker, Benjamin Gompertz, Journal of the Institute of Actuaries 91 (1965) 203–212.
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
39
page 39
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
11. D. MacHale, George Boole: His Life and Work, Boole P., Dublin, Ireland, 1985. 12. G. Minault, Master Ramchandra of Delhi College: teacher, journalist, and cultural intermediary, The Annual of Urdu Studies 18 (2003) 95–104; available at http://www.urdustudies.com/pdf/18/ 10MinaultRamchandra.pdf, accessed 23 August, 2009. 13. C. Mus`es, De Morgan’s Ramanujan: an incident in recovering our endangered cultural memory of mathematics, Math. Intelligencer 20(3) (1998) 47–51. doi:10.1007/BF03024806 14. P. J. Nahin, In Chases and Escapes: The Mathematics of Pursuit and Evasion, Princeton UP, New Jersey, 2007. 15. I. Niven, Maxima and Minima Without Calculus, Mathematical Association of America, Washington DC, 1981. 16. S. O’Donnell, William Rowan Hamilton: Portrait of a Prodigy, Boole P., Dublin, Ireland, 1983. 17. Ramchundra, A Treatise on Problems of Maxima and Minima, Solved By Algebra, Wm. H. Allen, London, 1859. 18. A. Rice, Augustus De Morgan: historian of science, Hist. Sci. 34 (1996) 201–240. 19. , Inspiration or desperation? Augustus De Morgan’s appointment to the chair of mathematics at London University in 1828, British J. Hist. Sci. 30 (1997) 257–274. doi:10.1017/S0007087497003075 20. , Maximum effort, minimum effect: De Morgan and his Indian prot´eg´e, History of Mathematics: Mathematics in the Americas and the Far East, 1800–1940, conference presentation, October, 1998; abstract available at http://www.mfo.de/programme/schedule/1998/43/Report_41_98.pdf, accessed 23 August, 2009. 21. , What makes a great mathematics teacher? the case of Augustus De Morgan, American Mathematical Monthly 106 (1999) 534–552. doi:10.2307/2589465 22. , Everybody makes errors: the intersection of De Morgan’s logic and probability, 1837–1847, Hist. Philos. Logic 24 (2003) 289–305. doi:10.1080/01445340310001599579 23. G. C. Smith, The Boole-De Morgan Correspondence 1842–1864, Oxford, New York, 1982.
Teaching Tip: Actuarial Science and Gompertz’s Law of Mortality Jesse Byrne (
[email protected]) Actuaries use a unique combination of mathematical, statistical, and business skills to solve a variety of financial and social problems, such as the computation of premiums for insurance policies. A highly stratified profession, actuarial science dates back to the 17th century and welcomes mathematics majors. This short primer gives students and their teachers some idea of what actuaries do as well as describing Gompertz’s Law of Mortality, a major contribution to actuarial science. (For more about Gompertz, see Charlotte Simmons’ article on p. 33.) Pricing life insurance If an insurance company knew exactly when the person buying insurance would die, this would be simple. Unfortunately (from the actuary’s point of view), this date is unknown and the company must use a simple but lengthy process based on expected values and the Law of Large Numbers. Specifically, actuaries create a table giving the number of years the insured might live from the date the insurance policy was written, the present value of the benefit being purchased for each of these corresponding years of life, and the probability that the insured will die in the corresponding year. Table 1 illustrates with a benefit of $1,000. The expected value of the policy is computed by multiplying the values from columns two and three and then summing the results. For example, using a $1,000 benefit and a 6% interest rate, the premium turns out to be $817.65, the sum of the entries in the fourth column.
40
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 40
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:00 a.m.
simmons.tex
Table 1. Number of Years
Present Value of $1000 at 6% interest
Probability of Mortality
Expected Value Computation
1
1,000(1.06)−1 = 943.40
.10
$94.34
2
1,000(1.06)−2 = 890.00
.15
$133.50
3
1,000(1.06)−3
= 839.62
.20
$167.92
4
1,000(1.06)−4
= 792.09
.25
$198.02
5
1,000(1.06)−5 = 747.26
.30
$224.18
Total
$817.65
After receiving the premium, the insurance company invests the $817.65 at 6% interest in order to accumulate the $1,000 benefit it must pay when the insured dies. Note that the insurance company loses money if the insured actually dies within one year of buying the policy; it should have charged a premium of $943.40 instead of $817.65. On the other hand, the insurance company makes a profit if the insured lives until year five; then a premium of $747.26 would have sufficed. It is essential the company sell a large number of policies in order to break even. Gompertz’s Law of Mortality A natural question is how the values in the third column of Table 1 are computed. For this, actuaries use a life table, a list of values giving the proportion of x-year olds that survive to the following year. Specifically, let d(x, k) be the probability that an x-year old dies in year k (at age x + k). Let p(x) be the probability that an x-year old survives one year (to age x + 1). Then q(x) = 1 − p(x) is the probability that an x-year old dies within a year and, d(x, k) = p(x) p(x + 1) · · · p(x + k − 1)q(x + k)
(1)
Prior to Gompertz’s Law of Mortality, actuaries attempted to determine such values by gathering census information from the population under consideration, quite a difficult task in the mid-1800’s. Gompertz discovered a simple analytic function that accurately approximates the probability distribution p(x) for the future lifetimes of a given population. Let S(x) be the probability that a newborn survives to the age of x. Gompertz modeled S by S(x) = exp(−m(c x − 1)),
(2)
where m and c are constants depending on the specific population under consideration (e.g., humans, elephants, insects, etc.). Using S, we obtain p(x) = S(x + 1)/S(x), and from p(x), using (1), we obtain d(x, k), which is the third column of Table 1. Gompertz’s key observation is that the rate of mortality of a given population increases in geometric proportion. That is, if L(x) is the total population of x-
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
41
page 41
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
1:27 p.m.
simmons.tex
year olds, then L (x) = −Bc x . L(x) This differential equation is separable and the solution has the form −B x L(x) = exp (c − 1) . L(0) ln(c) Therefore, since S(x) is the percentage of newborns that survive to age x, i.e., L(x)/L(0), it follows that S has the form (2). This is Gompertz’s Law of Mortality. This law greatly simplified the computation of life insurance premiums since life tables were now much easier to determine. Rather than finding accurate census information for each age group in the table, actuaries now only needed to find the data for two age groups, because the survival function (2) only has two parameters. Conclusion Gompertz’s law had great impact on the insurance industry but it has applications outside human populations. It has been broadly studied and used to predict maximum life expectancies for many animals and the growth of tumors as well.
101. Mr Jeavons said that I liked maths because it was safe. He said I liked maths because it meant solving problems and these problems were difficult and interesting but there was always a straightforward answer at the end. And what he meant was that maths wasn’t like life because in life there are no straightforward answers at the end. I know he meant this because this is what he said. This is because Mr. Jeavons doesn’t understand numbers. Here is a famous story called The Monty Hall Problem which I have included in this book because it illustrates what I mean. There used to be a column called Ask Marilyn . . .
··· ··· ··· ··· And this shows that intuition can sometimes get things wrong. And intuition is what people use in life to make decisions. But logic can help you work out the right answer. It also shows that Mr. Jeavons was wrong and numbers are sometimes very complicated and not very straightforward at all. And that is why I like The Monty Hall Problem. —from The Curious Incident of the Dog in the Night-time by Mark Haddon [pp. 61–65], For more about the Monty Hall problem see page 71.
42
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 42
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
Computing Determinants by Double-Crossing Deanna Leggett, John Perry, and Eve Torrence
Deanna Leggett (
[email protected]) received her B.S. in mathematics from the University of Southern Mississippi in 2009. She is currently pursuing a M.S. in mathematics at the same institution, continuing the research that she began as an undergraduate. When she is not studying, she enjoys running, playing guitar, and spending time with her family.
John Perry (
[email protected]) earned a B.S. from Marymount University, an M.S. from Northern Arizona University, and a Ph.D. from North Carolina State University. He is now an assistant professor at the University of Southern Mississippi, where he pursues research in algorithmic algebra. Outside the office, he spends time with his wife and children, who are patiently teaching him a new language. Eve Torrence (
[email protected]) is a professor of mathematics at Randolph-Macon College. Her current research interests include mathematical origami and sculpture. She has served as the Chair of the Maryland-DC-Virginia Section of the MAA and is currently President-Elect of Pi Mu Epsilon Mathematics Honor Society.
Students learn this simple pattern for the determinant of a 2 × 2 matrix: a b = ad − bc. c d A similar pattern exists for 3 × 3 matrices, but to compute the determinant of larger matrices, the student must learn something a little more complicated. Most students learn expansion by minors, first developed by Laplace. Some also learn to triangularize the matrix. Both methods are effective, and triangularization is efficient; but hand computations frequently lead to many mistakes, expansion by minors is tedious, and triangularization can turn a matrix of integers into a matrix of fractions. In 1866, the Rev. Charles Lutwidge Dodgson (better known as Lewis Carroll) described a quick and conceptually simple method to compute determinants [2] by condensation. Given an n × n matrix A, a condensation method sets A(n) = A, then uses doi:10.4169/college.math.j.42.1.043
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
43
page 43
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
each A(k) to compute the (k − 1) × (k − 1) matrix A(k−1) , until the sole entry of A(1) turns out to be det A. (Other condensation methods appear in [1].) If Dodgson’s method terminates successfully, it computes the determinant of an n × n matrix using O(n 3 ) integer operations—not bad, especially considering that all divisions are exact, so no fractions are introduced. There’s a rub, though: division by zero can happen even though no zeros appear in the interior of A itself. The workaround described in this J OURNAL in [2] swaps rows of the original matrix in such a way that zeros are moved out of the interior. Unfortunately, this also has drawbacks. Is there another workaround? Yes, and this paper presents it. The first two sections describe Dodgson’s method and explain how it works. This suggests a new, doublecrossing, method that we describe and explain in the two following sections.
Dodgson’s condensation method We present Dodgson’s method, followed by an example. Let A(n) be the given n × n matrix. Then for each k = n − 1, n − 2, . . . , 1: • •
•
Let B (k) be the k × k matrix of determinants of contiguous 2 × 2 submatrices of A(k+1) . We refer to this as condensing A(k+1) . If k = n − 1, let A(k) = B (k) . If k ≤ n − 2, let A(k) be the k × k matrix whose (i, j)th element is the (i, j)th element of B (k) divided by the (i, j)th element of the interior of A(k+2) . (The interior of an n × n matrix M is the (n − 2) × (n − 2) submatrix whose (i, j)th element is the (i + 1, j + 1)th element of M). The single element of A(1) is the determinant of A.
Example 1.
Given the matrix ⎞ 2 −3 1 2 5 1 −2 −3 2⎟ ⎜ 4 ⎟ ⎜ 2 2 −3 ⎟ , A = ⎜ 5 −4 ⎝ 3 −1 5 2 1⎠ −4 1 5 −1 2 ⎛
we set A(5) = A. For A(4) , compute the matrix B (4) of condensations of A(5) :
A(4) = B (4)
⎛ ⎞ 2 −3 −3 1 ··· ⎞ ⎜4 ⎟ ⎛ 14 1 1 −2 5 1 19 ⎜ ⎟ ⎜ ⎟ ⎜ −21 −6 2 5 ⎟ ⎟=⎝ =⎜ 1 1 −2 ⎠. ⎜4 ⎟ 7 −18 −6 8 · · · ⎜ 5 −4 −4 ⎟ 2 ⎝ ⎠ −1 −10 −15 5 .. .. .. . . .
Condense A(4) to obtain ⎞ 21 16 −33 46 ⎠ , = ⎝ 420 72 −88 210 90 ⎛
B (3)
44
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 44
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
and divide the entries of B (3) by the corresponding entries of the interior of A(5) to obtain ⎞ ⎛ 21 −8 11 A(3) = ⎝ −105 36 23 ⎠ . 88 42 45 Condensing A(3) , we obtain B
(2)
=
−84 −580 −7578 654
and
A
(2)
=
14 −290 421 −109
.
Condensing A(2) , we obtain B (1) = (120564) and A(1) = (3349). Compute det A using your favorite method (expansion of cofactors, triangularization, etc.). You will find that det A = 3349. However, Dodgson’s method fails to compute the determinant of a matrix A whenever a zero appears in the interior of A(k) for any k > 2. One workaround swaps rows and columns of A so as to move zeros out of the interior. Unfortunately, this technique does not always succeed. First, we can only swap rows or columns of the original matrix; we cannot do this for an intermediate matrix. Second, swapping may remove certain zeros from the interior, but may introduce other zeros. Since zeros can appear in an intermediate matrix, this isn’t easy to control. For some matrices, it won’t work at all. Example 2. No combination of row or column swaps allows Dodgson’s method to compute the determinant of ⎞ 1 0 3 0 ⎜ 0 −1 0 1 ⎟ , A=⎝ 1 1 2 0⎠ 0 2 0 1 ⎛
because there will always be a zero in the interior of A.
Why does it work? Dodgson’s method relies on the following theorem of Jacobi. Jacobi’s Theorem. Let L be an n × n matrix and M an m × m minor of L chosen from rows i 1 , i 2 , . . . , i m and columns j1 , j2 , . . . , jm . Let M be the corresponding m × m minor of L , the matrix of cofactors of L, and let M ∗ be the complementary (n − m) × (n − m) minor of M in L. Then det M = (det L)m−1 · det M ∗ · (−1)
m
=1 i + j
.
A proof of Jacobi’s Theorem is sketched in [2]. To apply it, we adopt the following notation. If A is a matrix, then A is its matrix of cofactors, and Ai... j,k... denotes the submatrix composed of rows i, i + 1, . . . , j and columns k, k + 1, . . . , of A. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
45
page 45
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
Example 1 (continued). Jacobi’s Theorem can be used alone to compute determinants. Let L = A, the matrix from Example 1. Choose for M the 3 × 3 submatrix in the upper left corner of L, so ⎞ 2 −3 1 1 −2 ⎠ . = ⎝4 5 −4 2 ⎛
M = L 1...3,1...3
We have M∗ =
2 1 −1 2
⎞ −109 −490 −144 61 ⎠ . M = ⎝ 302 −25 −26 −885 150 ⎛
and
Then by Jacobi’s Theorem, det M = (det L)3−1 · det M ∗ · (−1)1+1+2+2+3+3 . Since det M = 56,079,005 and det M ∗ = 5, we find that (det L)2 = 11,215,801, so that det L = 3349. Dodgson’s method relies on Jacobi’s Theorem. For a non-trivial illustration, let A be a generic 4 × 4 matrix. The first two condensations of Dodgson’s method are ⎞ |A1...2,1...2| |A1...2,2...3| |A1...2,3...4| ⎟ ⎜ = ⎝ |A2...3,1...2| |A2...3,2...3| |A2...3,3...4| ⎠ , |A3...4,1...2| |A3...4,2...3| |A3...4,3...4| ⎛
A(3)
and ⎛ A(2)
⎜ ⎜ ⎜ =⎜ ⎜ ⎜ ⎝
|A 1...2,1...2 | |A 1...2,2...3 | |A 2...3,1...2 | |A 2...3,2...3 | a2,2 |A 2...3,1...2 | |A 2...3,2...3 | |A 3...4,1...2 | |A 3...4,2...3 |
|A 1...2,2...3 | |A 1...2,3...4 | |A 2...3,2...3 | |A 2...3,3...4 | a2,3 |A 2...3,2...3 | |A 2...3,3...4 | |A 3...4,2...3 | |A 3...4,3...4 |
a3,2
a3,3
⎞ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎠
(1)
To relate this to Jacobi’s Theorem, consider the upper left 3 × 3 submatrix of A(4) ⎞ a1,1 a1,2 a1,3 = ⎝ a2,1 a2,2 a2,3 ⎠ . a3,1 a3,2 a3,3 ⎛
L = A1...3,1...3
Cross out row 2 and column 2 of L, obtaining M= 46
a1,1 a1,3 a3,1 a3,3
;
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 46
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
its complement in L is cofactors of M in L is ⎛ a2,2 ⎜ a3,2 ⎜ M = ⎜ ⎝ a1,2 a2,2
November 17, 2010
10:28 a.m.
torrence.tex
M ∗ = (a2,2 ). The 2 × 2 submatrix of L corresponding to the a2,3 a3,3 a1,3 a2,3
a2,1 a3,1 a1,1 a2,1
⎞ a2,2
a3,2 ⎟ |A2...3,2...3 | |A2...3,1...2 | ⎟ . ⎟ = |A1...2,2...3 | |A1...2,1...2 | a1,2 ⎠ a2,2
Using Jacobi’s Theorem and a column and row swap, we obtain det M = (det L)2−1 · det M ∗ · (−1)1+1+3+3 det M = det L det M ∗ |A1...2,1...2 | |A1...2,2...3 | |A2...3,1...2 | |A2...3,2...3 | = det L = det A1...3,1...3 . a2,2 This is equivalent to the element in the upper left corner of A(2) in equation (1)! Likewise, the upper right corner of A(2) has the value |A1...3,2...4 |; the lower left corner has the value |A2...4,1...3 |; and the lower right corner has the value |A2...4,2...4 |. So Jacobi’s Theorem gives us
|A1...3,1...3 | |A1...3,2...4 | (2) A = . |A2...4,1...3 | |A2...4,2...4 | The next (and final) condensation in Dodgson’s method is
|A(2) | |A1...3,1...3 ||A2...4,2...4 | − |A2...4,1...3 ||A1...3,2...4 | (1) . = A = (3) |A2...3,2...3 | a2,2
(2)
Now we apply Jacobi’s Theorem with L = A. Select the 2 × 2 minor of the corners,
a1,1 a1,4 M= . a4,1 a4,4 Its complementary minor is M ∗ = (A2...3,2...3 ). The matrix of cofactors is
|A2...4,2...4 | −|A2...4,1...3 | M = . |A1...3,1...3 | −|A1...3,2...4 | By Jacobi’s Theorem, det M = (det A)2−1 · det M ∗ · (−1)1+1+4+4 det M = det A det M ∗ |A1...3,1...3 ||A2...4,2...4 | − |A2...4,1...3 ||A1...3,2...4 | = det A. |A2...3,2...3 | This is precisely the singleton element of A(1) in equation (2)! Generalizing this observation leads to the following theorem: VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
47
page 47
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
Dodgson’s Condensation Theorem. Let A be an n × n matrix. After k successful condensations, Dodgson’s method produces the matrix ⎞ ⎛ |A1...k+1,1...k+1 | |A1...k+1,2...k+2 | · · · |A1...k+1,n−k...n | ⎟ ⎜ |A ⎜ 2...k+2,1...k+1 | |A2...k+2,2...k+2 | · · · |A2...k+2,n−k...n | ⎟ A(n−k) = ⎜ ⎟, .. .. .. .. ⎠ ⎝ . . . . |An−k...n,1...k+1 | |An−k...n,2...k+2 | · · · |An−k...n,n−k...n | whose entries are the determinants of all (k + 1) × (k + 1) submatrices of A. Proof. By “successful” condensations, we mean not encountering division by zero. The proof is by induction on k. Base case. When k = 1, the theorem is trivial: the first condensation is ⎞ ⎛ |A1...2,2...3 | ··· |A1...2,n−1...n | |A1...2,1...2 | ⎜ |A |A2...i+2,2...i+2 | · · · |A2...3,n−1...n | ⎟ 2...3,1...2 | ⎟ ⎜ A(n−1) = ⎜ ⎟. . . .. .. .. .. ⎠ ⎝ . . |An−1...n,1...2 | |An−i...n,2...i+2 | · · · |An−1...n,n−1...n | Inductive hypothesis. Fix k ≥ 1. Assume that for all = 1, . . . , k, the th condensation gives us A(n−) where for all 1 ≤ i, j ≤ n = |Ai...i+(n−), j ... j +(n−) |. ai,(n−) j Inductive step. Let i, j ∈ {1, . . . , n − (k + 1)}. The next condensation in Dodgson’s method gives us ai,(n−(k+1)) = j
(n−k) (n−k) (n−k) ai+1, ai,(n−k) j j +1 − ai+1, j ai, j +1 (n−(k−1)) ai+1, j +1
.
From the inductive hypothesis, we know that
|Ai...i+k, j ... j +k ||Ai+1...i+k+1, j +1... j +k+1 | − |Ai+1...i+k+1, j ... j +k ||Ai...i+k, j +1... j +k+1 | . = ai,(n−(k+1)) j |Ai+1...i+k, j +1... j +k | Let L = Ai...i+k+1, j ... j +k+1 and M be the 2 × 2 minor made up of the corners of L. Let M be the corresponding 2 × 2 minor of L , and M ∗ be the interior of L (that is, the complementary k × k minor of M in L). By Jacobi’s Theorem, det M = (det L)2−1 · det M ∗ · (−1)i+ j +(i+k+1)+( j +k+1), or det L =
det M = det M ∗
|Ai...i+k, j ... j +k ||Ai+1...i+k+1, j +1... j +k+1 | − |Ai+1...i+k+1, j ... j +k ||Ai...i+k, j +1... j +k+1 | |Ai+1...i+k, j +1... j +k |
= ai,n−(k+1) , j (3)
48
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 48
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
so long as the denominator is not zero. Hence = det L = |Ai...i+k+1, j ... , j +k+1 |, ai,(n−(k+1)) j as claimed.
The double-crossing condensation method The double-crossing method solves some of the problems caused by zeros in the interior of a matrix. The method does not work in all cases; however, it does work in many interesting cases and preserves the spirit of Dodgson’s method. The Double-Crossing Theorem. Let A be an n × n matrix. Suppose that after k successful condensations, we encounter a zero in the interior of A(n−k) , say in row i and column j. Let r, s ∈ {−1, 0, 1}. If the element α in row i + r and column j + s of A(n−k) is non-zero, then we compute A(n−(k+1)) and A(n−(k+2)) as usual, with the exception of the element in row i − 1 and column j − 1 of A(n−(k+2)) . Let = k + 1. Then (a) identify the ( + 2) × ( + 2) submatrix L whose upper left corner is the element in row i − 1 and column j − 1 of A(n) ; (b) identify the complementary minor M ∗ by crossing out the × submatrix M of L whose upper left corner is the element in row r + 2 and column s + 2 of L; (c) compute the matrix M of determinants of minors of M ∗ in L; (d) compute the element in row i − 1 and column j − 1 of A(n−(k+2)) by dividing the determinant of M by α. The resulting matrix A(n−(k+2)) has the form described by Dodgson’s Condensation = |Ai...i+(k+2), j ... j +(k+2) |, and the condensation can proTheorem; that is, ai,(n−(k+2)) j ceed. The Double-Crossing Theorem applies whenever the non-zero element is immediately above, below, left, right, or catty-corner to the zero; that is, the non-zero element is adjacent to the zero. If the zero appears in a 3 × 3 block of zeros, then the doublecrossing method fails. By a successful condensation, we mean either using Dodgson’s method or this Double-Crossing Theorem. Example 3. Let ⎞ 1 3 1 = ⎝1 0 1⎠ 0 1 1 ⎛
A=A
(3)
and
A
(2)
=
−3 3 1 −1
.
The zero in the interior of A(3) causes Dodgson’s method to fail when computing A(1) . So we apply the Double-Crossing Theorem with i = j = 2. Above the zero is (3) a non-zero element, a1,2 = 3. We divide by this element instead, but this requires us (2) to compute A in a slightly different manner. Now in this example k = 0, r = −1, s = 0, and = k + 1 = 1, so the ( + 2) × ( + 2) submatrix L is A itself. Since VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
49
page 49
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
r + 2 = 1 and s + 2 = 2, cross out the first row and second column of L = A (the row and column containing 3). We are left with the 2 × 2 complementary matrix ∗
M =
1 1 0 1
.
Recall that we compute the minor of an element of a matrix by (again) crossing out the row and column containing that element, then taking the determinant of the remaining submatrix. Hence the matrix M of minors of elements of L that correspond to the elements of M ∗ is ⎞ ⎛ 3 1 1 3 ⎜1 1 0 1⎟ ⎜ ⎟ M = ⎜ ⎟. ⎝3 1 1 3⎠ 0 1 1 0 Put A
(2)
=M =
2 1 3 −3
.
Now that we have our new A(2) , we can compute A(1) by dividing det A(2) by the non(3) zero a1,2 : A(1) =
|A(2) | 3
=
−6 − 3 3
9 = − = (−3), 3
and in fact, det A = −3. This method preserves the “spirit” Dodgson’s method, inasmuch as we compute all determinants by condensing 2 × 2 contiguous submatrices. We call it “doublecrossing” because we find a non-zero element adjacent to the zero element, and cross out its row and column, obtaining the complementary matrix M ∗ . For each element in M ∗ , we compute its minor by (again) crossing out the row and column of its location in A(3) , and use the determinant of the remaining matrix to compute the elements of A(2) . When a zero appears in an intermediate matrix A(k) (where n > k ≥ 3), we cross out rows and columns that make up a submatrix of A(n) . Now we turn to the matrix of Example 2, which cannot be resolved by swaps. Example 2 (continued). Let A(4) be the matrix A of Example 2. We have one zero element in the interior, at (i, j) = (2, 3). As set forth in the Double-Crossing Theorem, this affects element (1, 2) of A(2) . The other three elements of A(2) can be computed as usual: ⎞ ⎛
−1 3 3 1 ? (3) (2) ⎠ ⎝ 1 −2 −2 −→ A = , A = 0 −6 2 −4 2 but the double-crossing method is needed for the last element of A(2) . The interior zero appears in the original matrix, so = 1. We choose the 3 × 3 submatrix whose upper 50
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 50
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
left corner is the element in row i − 1 = 1, column j − 1 = 2 of A(4) , ⎞ 0 3 0 L = ⎝ −1 0 1 ⎠ . 1 2 0 ⎛
There is a non-zero element in row 1, column 2, so r = −1 and s = 0. Cross out the 1 × 1 submatrix in row 1, column 2 of L; identify the complementary matrix, and compute the corresponding matrix of minors in L, ∗
M =
−1 1 1 0
⎛ 3 ⎜2 ⎜ M = ⎜ ⎝3 0
and
0 0 0 1
⎞ 0 3
1 2 ⎟ 0 −3 ⎟ . ⎟ = 3 3 0 3⎠ −1 0
(4) (2) = 3 we have a2,1 = 3 so The determinant of M is 9; after dividing by the nonzero a1,3
A
(2)
=
1 3 0 −6
.
We conclude by computing A(1)
⎞ ⎛ 1 3 = ⎝ 0 −6 ⎠ = (3), −2
and, in fact, the determinant of A is 3. Finally, let’s look at a matrix without an interior zero, but where zero appears in the interior of an intermediate matrix. Example 4. Let ⎛
1 ⎜ 0 ⎜ A=⎜ 1 ⎝ −1 0
0 1 2 1 1
1 1 1 1 0
0 1 1 2 1
⎞ 1 1⎟ ⎟ 2⎟. 1⎠ 0
Using Dodgson’s method, we compute A(5) , and then (horror of horrors!) ⎞ 1 −1 1 −1 1⎟ ⎜ −1 −1 0 . =⎝ 3 1 1 −3 ⎠ −1 −1 1 −1 ⎛
A(4)
(4) There’s a zero in the interior! But the element to its southeast is non-zero, a3,4 = −3. Double-crossing to the rescue!
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
51
page 51
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
Note that k = 1, i = 2, j = 3 and r = s = 1. We compute A(3) as usual, and all of (2) A(2) as usual except for a1,2 : ⎞ −2 1 1 = ⎝ 1 −1 −1 ⎠ −2 2 1 ⎛
A(3)
A(2) =
and
−1 ? 0 1
.
Let = k + 1 = 2. Identify the 4 × 4 submatrix L of A whose upper left corner is in row 1 and column 2. In this case, it turns out that L = A1...4,2...5 . Identify the complementary matrix M ∗ by crossing out the 2 × 2 submatrix of L whose upper left corner is in row r + 2 = 3, column s + 2 = 3. That gives M ∗ = L 1...2,1...2 . Compute the matrix M of determinants of minors of elements of M ∗ in L. That gives ⎛ 1 ⎜1 ⎜ ⎜1 ⎜ M = ⎜ ⎜1 ⎜ ⎝1 1
1 1 1 1 1 2 2 1 2 1 1 2 0 1 0 0 1 2 2 1 2 1 1 2
⎞ 1 2 ⎟ ⎟
1⎟ −1 0 ⎟ . ⎟ = −2 3 1 ⎟ ⎟ 2 ⎠ 1
(2) (4) by dividing det M = −3 by a3,4 = −3. So Compute a1,2
A
(2)
=
−1 1 0 1
and
A
(1)
=
−1 −1
= (1).
In fact, det A = 1.
Why does it work? The double-crossing method is based on Dodgson’s Condensation Theorem. The goal in step k of the algorithm is to compute each determinant |Ai...i+k, j ... j +k |. Dodgson’s method fails when the corresponding denominator of (3) is zero. However, (3) is not the only way to apply Jacobi’s Theorem. As long as some (k − 1) × (k − 1) minor of Ai...i+k, j ... j +k has non-zero determinant, we can recover, reusing most of the computations already performed in previous steps, and re-calculating only a few new minors using the same approach as Dodgson’s method. For example, the proof of the DoubleCrossing Theorem can be summarized by applying Jacobi’s Theorem with L of the Double-Crossing Theorem standing in for L of Jacobi’s Theorem; the 2 × 2 submatrix M ∗ of the Double-Crossing Theorem standing in for M ∗ of Jacobi’s Theorem; and the matrix of determinants of minors of M of the Double-Crossing Theorem standing in for M of Jacobi’s Theorem. One difference does require investigation: the negatives in the matrix of cofactors are now in different places! With Dodgson’s method, we choose the central minor for M, which has two important consequences in Jacobi’s Theorem: • •
m
m
i = j for each = 1, 2, . . . , m, so that (−1) =1 i + j = (−1) =1 2i = 1; and if negatives appear in the 2 × 2 matrix of cofactors, they appear off the main diagonal, cancelling each other out.
52
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 52
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:51 a.m.
torrence.tex
The double-crossing method works in a similar way in Example 4, but Examples 2 and 3 are a little more complicated. We chose the top-middle minor for M, so: • •
m
(−1) =1 i + j = (−1)(1+2)+(2+3)+···+[(k+1)+(k+2)] = (−1)k+1 ; and the determinant of the 2 × 2 matrix M of cofactors is (−1)(k+2)+2(k+3)+1 · D, where D is the determinant of the minors. The multiple of D simplifies to (−1)k+1 ; thus the signs cancel, and we can consider only the determinants of the minors.
(2) To clarify this, let’s revisit Example 2. The problem in computing a1,2 was the zero in (2) position a2,3 . Since a1,2 = det A1...3,2...4 we identified the complementary minor M ∗ = a1,3 = 3. In this case, the matrix of cofactors is
a1,3 a1,4 2+3 a1,2 ⎜ (−1) a3,3 a3,4 (−1) a3,2 ⎜ M = ⎜ ⎝ 3+1 a1,3 a1,4 3+3 a1,2 (−1) (−1) a2,3 a2,4 a2,2 ⎞ ⎛ 3 0 −0 3 − ⎜ 2 0 1 2⎟ −1 · 0 ⎜ ⎟ =⎜ ⎟ = 3 ⎝ 3 0 0 3⎠ 0 1 −1 0 ⎛
2+1
⎞ a1,3 a3,3 ⎟ ⎟ ⎟ a1,3 ⎠ a2,3 (−1)(−3) 3
.
By Jacobi’s Theorem, det L =
(−1)1 · 9 det M = = 3. det M ∗ · (−1)2+1+3+3 3 · (−1)1
Conclusion Dodgson’s method is a special case of the double-crossing method. The most computationally intensive part of the double-crossing method is that of computing M . Depending on the choice of M, one or two of its elements are in a corner of L, and Dodgson’s method has already computed their minors. So to successfully resolve an interior zero for Dodgson’s method, the double-crossing method must compute two or three new determinants. Double-crossing is not guaranteed to succeed; it fails if an intermediate matrix contains a 3 × 3 block of zeros. Sparse matrices are an excellent example where the double-crossing method is an abject failure (Dodgson’s too, for that matter). When double-crossing fails, one can still preserve the computations that work, and adapt a hybrid with another method. The Double-Crossing Theorem tells us the precise minor L whose determinant we need; we can compute only the determinant of this minor using another method, substitute its value into row i − 1 and column j − 1 of A(n−(k+2)) , and proceed. Summary. Dodgson’s method of computing determinants is attractive, but fails if an interior entry of an intermediate matrix is zero. This paper reviews Dodgson’s method and introduces a generalization, the double-crossing method, that provides a workaround for many interesting cases. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
53
page 53
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
December 8, 2010
9:27 a.m.
torrence.tex
References 1. A. Aitken, Determinants and Matrices, Interscience Publishers, New York, 1951. 2. A. Rice and E. Torrence, “Shutting up like a telescope”: Lewis Carroll’s “curious” condensation method for evaluating determinants, Coll. Math. J. 38 (2007) 85–95. 3. J. von zur Gathen and J. Gerhard, Modern Computer Algebra, Cambridge University Press, Cambridge, 1999. 4. C. Yap, Fundamental Problems of Algorithmic Algebra, Oxford University Press, Oxford, 2000.
Teaching Tip: Correcting Cramer’s Rule Vagarshak Vardanyan (
[email protected]) There is a misunderstanding about Cramer’s rule that examination of more than a dozen intermediate algebra texts (including [1], [2], and [3]), suggests few have avoided. The error also appeared in Wikipedia [4]. It has been corrected by the author. Let D be the determinant of a linear system of equations and Dx , Dy , Dz , etc. be determinants formed by replacing in turn each column of D by the constants of the system. The sources cited all contain the following incorrect statement (with some unimportant differences in wording): for a 3 3 system of linear equations, if all determinants are zero, then the system has infinitely many solutions. What this statement doesn’t take into account is that besides infinitely many solutions, the system may be inconsistent and have no solution, for example 8 < x C 2y C 3z D 4 x C 2y C 3z D 5 : x C 2y C 3z D 6: Geometrically this system represents three parallel planes. Table 1. System Solution Kronecker-Capelli Theorem Cramer’s Rule
22
US
r D r0 D 2
NS
r D 1,
r0
D2
D D 0, Dx ¤ 0, Dy ¤ 0
IMS
r D 1,
r0
D1
D D 0, Dx D 0, Dy D 0
US
r D r0 D 3
NS 33
r D 1; 2; r D 1,
IMS
rD
r0
D¤0
D¤0 r0 D 3
r0 D 2
x D Dx =D, y D Dy =D
x D Dx =D, y D Dy =D, z D Dz =D
2 C D2 C D2 > 0 D D 0, Dx y z
D2
D D 0, Dx D 0, Dy D 0, Dz D 0
r D r0 D 1
The situation is clarified by the Kronecker-Capelli theorem. Let A be the matrix of coefficients of a system of linear equations; let A0 be its augmented matrix; and let r and r 0 be the ranks of these matrices. The Kronecker-Capelli theorem states that the system has a solution only if r D r 0 . The system has no solution if r < r 0 . The extension of the theorem is that the solution is unique if r D r 0 D n, and there are infinitely many solutions if r D r 0 < n.
54
c
THE MATHEMATICAL ASSOCIATION OF AMERICA
page 54
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
December 8, 2010
9:27 a.m.
torrence.tex
page 55
Ignoring the case r D 0, Table 1 lists all cases (US D unique solution, NS D no solution, and IMS D infinitely many solutions) and their connection with Cramer’s Rule. The error-prone, ambiguous case is marked by an asterisk. References 1. M. L. Bittinger, Intermediate Algebra, Addison Wesley, Boston, 2007. 2. C. P. McKeague, Intermediate Algebra, 8th ed., Brooks-Cole, Belmont CA, 2008. 3. M. Sullivan and K. R. Struve, Intermediate Algebra, 2nd ed., Prentice Hall, Upper Saddle River NJ, 2010. 4. Wikipedia, Cramer’s Rule; available at http://en.wikipedia.org/wiki/Cramer’s_rule; accessed 1 December 2010; corrected 2 October 2010.
Flaws, Fallacies, and Flimflam: Who’s Right? The following items appeared in the Atlantic Monthly for March, 2010 in the Letters Column (page 17). Is either one of these contributions correct? Misleading Math? Megan McArdle attempts to illustrate her point that survey respondents are unreliable (“Misleading Indicator,” November Atlantic) by telling us that it is mathematically impossible for men to report an average number of female sexual partners that is much higher than the average number of male partners reported by women. I agree that survey respondents are unreliable, but so is McArdle’s math. It may be unlikely that the number of partners reported by honest males would be higher, but it is not mathematically impossible. —Fred Graf, Concord, NH Megan McArdle replies: We are talking about two different meanings of the word average. True, in Harvey Levin’s scenario, it is more common for women to be monogamous than men. But I was using average to describe the mean number of sexual partners, which is to say, the sum of everyone’s number of partners, divided by the number of people. If there are 10 men and 10 women, and one of the women has slept with all 10 men, while the other women are monogamous, the average number of sexual partners is the same for the men and the women. Nine of the men have had two sexual partners, while one of the men has had one partner, for an average of 19=10 D 1:9. Meanwhile, one of the women has had 10 sexual partners while the other nine women have had one partner, for an average of 19=10 D 1:9. The numbers come out the same, no matter how you vary the particulars. Yet surveys can generate differences of three- or fourfold between the mean numbers of sexual partners that women and men report. —Ed Barbeau (
[email protected]), University of Toronto
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
55
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:55 a.m.
whitcher.tex
Boundary Conditions Ursula Whitcher
Ursula Whitcher (
[email protected]) is a Teaching and Research Postdoctoral Fellow at Harvey Mudd College. She received her M.S. and Ph.D. in mathematics from the University of Washington; as an undergraduate at Swarthmore College, she studied mathematics, physics, and Latin. Her extracurricular interests include rock climbing and medieval history.
Royal Academy of Science, Paris, 1823 This is her moment of triumph: a seat at the center, a node. Mademoiselle Germain sits silent, head upright, chaperoned. Academy members rise or dip; the speaker drones. Steel plate hums to the bow like silk stretched tight. Who grasps the edge controls— she claims—the waves inside. She makes her hands unfold. Her lips taste dry.
Sophie Germain was awarded the Royal Academy of Science’s prix extraordinaire for research in elasticity, in absentia, in 1816. Her friend Joseph Fourier extended a formal invitation to the Academy’s meetings in 1823, after his November 1822 election as Permanent Secretary. Summary. A poem about the nineteenth-century mathematician Sophie Germain.
doi:10.4169/college.math.j.42.1.056
56
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 56
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:58 a.m.
classroom.tex
CLASSROOM CAPSULES EDITORS
Ricardo Alfaro
Lixing Han
Kenneth Schilling
University of Michigan–Flint Flint MI 48502
[email protected]
University of Michigan–Flint Flint MI 48502
[email protected]
University of Michigan–Flint Flint MI 48502
[email protected]
Classroom Capsules are short (1–3 page) notes that contain new mathematical insights on a topic from undergraduate mathematics, preferably something that can be directly introduced into a college classroom as an effective teaching strategy or tool. Classroom Capsules should be prepared according to the guidelines on the inside front cover and sent to any of the above editors.
An Elementary Treatment of General Inner Products Jack E. Graver (
[email protected]), Department of Mathematics, Syracuse University, Syracuse NY 13244-1150 A typical first course on linear algebra is usually restricted to vector spaces over the reals and the usual positive-definite inner product. Hence, the proof of the result dim(S ) + dim(S ⊥ ) = dim(V ) is not presented in a way that is generalizable to non positive-definite inner products or to to inner products on vector spaces over other fields. In [1], this author made a case for proving this result in a way that does generalize to an arbitrary symmetric, nonsingular, bilinear form for vector spaces over any field. He then went on to describe just how that could be done. The case is based on the fact that there are many useful inner products that are not positive-definite: the usual inner n product for linear codes over finite fields (x1 , x2 , . . . , xn ), (y1 , y2 , . . . , yn ) = i=1 xi yi and the 4-dimensional real vector space with the inner product of special relativity (t, x, y, z), (t , x , y , z ) = tt − x x − yy − zz , to name two. The dimension theorem is still valid for arbitrary inner products, but the usual proof will not work since the stronger condition S ⊕ S ⊥ = V may not hold. Unfortunately, the remedy proposed in [1] involves a rather extensive alteration of the usual sequencing of the material in a typical first or second linear algebra course. In this note, a much simpler adjustment is proposed. What follows is an outline of the definitions and lemmas leading to a proof of the dimension theorem in its full generality. To start, we develop the necessary terminology including an extra definition. Given a finite-dimensional vector space V over an arbitrary field, a function v, w mapping the pairs of vectors into the scalars such that (i) (symmetric) v, w = w, v for all v, w ∈ V , (ii) (bilinear) v, (αu + βw) = αv, u + βv, w, for all u, v, w ∈ V and all scalars α, β, is a symmetric bilinear form on V . If in addition, (iii) (non-singular) v, w = 0, for all w, implies that v is the zero vector, doi:10.4169/college.math.j.42.1.057
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
57
page 57
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:58 a.m.
classroom.tex
the symmetric bilinear form is called an inner product for V . In an elementary course, the field is the reals and the non-singular condition is replaced by (iii ) (positive-definite) v, v > 0, for all non-zero vectors v. The plan here is to show how the dimension theorem can be presented in a traditional course in a way that can be generalized to an arbitrary inner product over an arbitrary field. To do this we prove as much of the dimension theorem as possible without invoking (iii) or the positive-definite condition (iii ). Many standard results, like Lemmas 1 and 2 below, use only (i) and (ii), and we have changed their statements to reflect this. For any symmetric bilinear form, we say v is orthogonal to w and write v ⊥ w whenever v, w = 0. Lemma 1. Consider a vector space with a symmetric bilinear form. If v is orthogonal to w1 , . . . , wk , then v is orthogonal to every vector in the subspace spanned by w1 , . . . , wk . Lemma 2. Consider a vector space V with a symmetric bilinear form, and let S be any subspace of V . Then
S ⊥ = {v : v ⊥ w, f or all w ∈ S } is also a subspace of V . We need just one new lemma to prove the dimension theorem. Lemma 3. S and T .
Consider a vector space with a symmetric bilinear form and subspaces
(i) If dim(S ) > dim(T ), then S contains a nonzero vector that is orthogonal to every vector in T . (ii) If S contains no nonzero vector orthogonal to every vector in T and T contains no nonzero vector orthogonal to every vector in S , then dim(S ) = dim(T ). Proof. Denote the bilinear form by , . Let b1 , . . . , bk be a basis for S and d1 , . . . , dh a basis for T , where k > h. Consider the h-tuples ti = (bi , d1 , . . . , k bi , dh ), for i = 1, . . . , k. Since k > h, {t1 , . . . , tk } is dependent and i=1 αi ti = (0, . . . , 0) for some set of scalars α1 , . . . , αk not all of which are zero. But then k v = i=1 αi bi is a nonzero vector of S , and one easily sees that (v, d1 , . . . , v, dh ) =
k
αi ti = (0, . . . , 0).
i=1
Thus v ⊥ d j , for j = 1, . . . , h, and v is a nonzero vector of S orthogonal to every vector in T . Part (i) is proved, and Part (ii) follows. Arbitrary symmetric bilinear forms may admit troublesome vectors of two types: vectors orthogonal to every vector in the space and vectors orthogonal to themselves. A vector v that is orthogonal to every vector in the space is called a null vector. Clearly, the only null vector in an inner-product space is the zero vector. Also, the only self-orthogonal vector in a positive-definite inner product space is the zero vector. However, nonzero null vectors and nonzero self-orthogonal vectors may exist for an arbitrary symmetric bilinear form. Indeed, self-orthogonal vectors do exist in the 58
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 58
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
10:31 a.m.
classroom.tex
spaces mentioned above corresponding to codes and special relativity. It is the existence of these two types of vectors that prevent one from extending the traditional proof of the dimension theorem to the general case. To give a general proof, we simply work around these troublesome vectors. Theorem. Consider a vector space V with a symmetric bilinear form and let S be a subspace that contains no nonzero null vector. Then dim(S ) + dim(S ⊥ ) = dim(V ). Proof. Let b1 , . . . , bk be a basis for S ⊥ , let b1 , . . . , bn be an extension of that basis for S ⊥ to a basis for V , and let T be the subspace spanned by bk+1 , . . . , bn . Clearly, dim(T ) + dim(S ⊥ ) = dim(V ). We need only show that dim(S ) = dim(T ). Suppose that v ∈ T is orthogonal to every vector in S . Then v ∈ S ⊥ ∩ T and, therefore, must be the zero vector. Suppose that v ∈ S is orthogonal to every vector in T . Then v ⊥ bi , for all i. Thus v is a null vector and, by hypothesis, must be the zero vector. The result now follows by Part (ii) of Lemma 3. Corollary. Consider a vector space V with an inner product. Then, for all subspaces S of V : (i) dim(S ) + dim(S ⊥ ) = dim(V ), (ii) (S ⊥ )⊥ = S . And, if S contains no nonzero self-orthogonal vectors, in particular if the inner product is positive definite, (iii) S ⊕ S ⊥ = V . Proof. Part (i) follows at once from the theorem. To prove (ii), we note that S ⊆ (S ⊥ )⊥ and that, by the theorem, dim(S ) = dim(V ) − dim(S ⊥ ) = dim((S ⊥ )⊥ ). Part (iii) follows from (i) and the observation that vectors in S ∩ S ⊥ are selforthogonal. Summary. A typical first course on linear algebra is usually restricted to vector spaces over the real numbers and the usual positive-definite inner product. Hence, the proof that dim(S ) + dim(S ⊥ ) = dim(V ) is not presented in a way that generalizes to non-positive–definite inner products or to vector spaces over other fields. In this note we give such a proof.
Reference 1. J. E. Graver, An alternative approach to the dimension theorem for inner product spaces, Amer. Math. Monthly 94 (1987) 671–679. doi:10.2307/2322224
◦
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
59
page 59
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:58 a.m.
classroom.tex
Cantor Groups Ben Mathes (
[email protected]), Colby College, Waterville ME 04901; Chris Dow (
[email protected]), University of Chicago, Chicago IL 60637; and Leo Livshits (
[email protected]), Colby College, Waterville ME 04901 The Cantor set C is simultaneously small and large. Its cardinality is the same as the cardinality of all real numbers, and every element of C is a limit of other elements of C (it is perfect), but C is nowhere dense and it is a nullset. Being nowhere dense is a kind of topological smallness, meaning that there is no interval in which the set is dense, or equivalently, that its complement is a dense open subset of [0, 1]. Being a nullset is a measure theoretic concept of smallness, meaning that, for every > 0, it is possible to cover the set with a sequence of intervals In whose lengths add to a sum less than . It is possible to tweak the construction of C that results in a set H that remains perfect (and consequently has the cardinality of the continuum), is still nowhere dense, but the measure can be made to attain any number α in the interval [0, 1). The construction is a special case of Hewitt and Stromberg’s Cantor-like set [1, p. 70], which goes like this: let E 0 denote [0, 1], let E 1 denote what is left after deleting an open interval from the center of E 0 . If E n−1 has been constructed, obtain E n by deleting open intervals of the same length from the center of each of the disjoint closed intervals that comprise E n−1 . The Cantor-like set is then defined to be the intersection of all the E n . An example of such a construction might begin by removing the middle open subinterval of [0, 1] of length 14 . You will be left with the two intervals [0, 38 ] and [ 58 , 1], from which you remove open intervals of length 412 from the centers of both. Continuing inductively, you will have removed subintervals whose lengths add to 1 1 1 + 2 2 + 22 3 + . . . , 4 4 4 resulting in a Cantor-like set with measure 12 , a set that is large in the sense of cardinality and measure, but retains its topological smallness. Let us consider an algebraic measure of size. The unit interval is a group under addition modulo 1, and algebraically speaking, a subset of this group is small when it generates a proper subgroup of [0, 1). The traditional Cantor set C is algebraically large. Not only is [0, 1) the smallest group containing C , we now show that 2C is already all of [0, 1), where n C denotes the set of all n-fold sums (mod 1) α1 + · · · + αn with αi ∈ C (1 ≤ i ≤ n). Let w ∈ [0, 1) be given, and let z = w/2. Note that one can write z as z = x + y where the ternary (base 3) expansions of x and y contain only 0’s and 1’s. Indeed, if z=
∞ i=1
zi
i 1 , 3
with z i ∈ {0, 1, 2}, then when z i = 0 we take both xi and yi equal to 0, when z i = 1 we take xi = 1 and yi = 0, and when z i = 2 we take both xi and yi equal to 1. Letting 1 i 1 i x= xi 3 and y = yi 3 then gives us z = x + y, and w = 2x + 2y. Thus with c1 = 2x and c2 = 2y we have two elements of C for which w = c1 + c2 . We now build a Hewitt and Stromberg Cantor-like set G that is algebraically tiny. Let μ(E) denote the outer measure of E, so if E is an interval, then μ(E) is just the doi:10.4169/college.math.j.42.1.060
60
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 60
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
10:58 a.m.
classroom.tex
length of the interval. For a general set E, the outer measure μ(E) is the infimum of the set of all sums ∞ μ(Ii ), i=1
for which Ii (i ∈ N) are intervals whose union contains E. The nullsets are those E with μ(E) = 0. In the general construction of a Cantor-like set described above, upon deleting open intervals of the same length from the center of each of the disjoint closed intervals that comprise E n−1 , one is left with E n , which is a disjoint union of exactly 2n closed intervals, each with the same length, which we denote rn . Note that for any two subintervals I and J of [0, 1), the set I + J = {x + y | x ∈ I, y ∈ J } satisfies μ(I + J ) ≤ μ(I ) + μ(J ). Now E n = I1 ∪ · · · ∪ I2n and a crude estimate gives us 2E n ⊂
2n 2n
Ik + I j ,
k=1 j =1
so that μ(2E n ) ≤
2n 2n
μ(Ik + I j ) ≤ (2n )2 (2) rn .
k=1 j =1
Similarly we have μ(m E n ) ≤
2n i 1 =1
···
2n
μ(Ii1 + · · · + Iim ) ≤ (2n )m (m) rn ,
i m =1
for every natural number m. We now choose rn to force limn→∞ μ(m E n ) = 0 for every m; e.g., rn = n!1 is sufficient. Let G denote the Cantor-like set determined by the sequence rn , and let Gˆ denote the group generated by G . Thus G = ∩E n and, since E n contains the inverse of each of its elements, we have that G contains the inverse of each of its elements. It follows that ∞ Gˆ = {x1 + · · · + xm | m ∈ N, xi ∈ G , 1 ≤ i ≤ m}, and Gˆ ⊂ ∪∞ m=0 ∩n=0 m E n . Since ˆ μ(m E n ) → 0 for every m, we see that ∩∞ n=0 m E n is a nullset. This shows that G is a nullset, and it also shows, for those who know about Baire category, that Gˆ is first category, since each set ∩∞ n=0 m E n is closed with empty interior. Summary. The Cantor subset of the unit interval [0, 1) is large in cardinality and also large algebraically, that is, the smallest subgroup of [0, 1) generated by the Cantor set (using addition mod 1 as the group operation) is the whole of [0, 1). In this paper, we show how to construct Cantor-like sets which are large in cardinality but small algebraically. In fact for the set we construct, the subgroup of [0, 1) that it generates is, like the Cantor set itself, nowhere dense in [0, 1).
Reference 1. E. Hewitt and K. Stromberg, Real and Abstract Analysis, Springer-Verlag, Berlin, 1965.
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
61
page 61
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
PROBLEMS AND SOLUTIONS EDITORS
Curtis Cooper
Shing S. So
CMJ Problems Department of Mathematics and Computer Science University of Central Missouri Warrensburg MO 64093 email:
[email protected]
CMJ Solutions Department of Mathematics and Computer Science University of Central Missouri Warrensburg MO 64093 email:
[email protected]
This section contains problems intended to challenge students and teachers of college mathematics. We urge you to participate actively BOTH by submitting solutions and by proposing problems that are new and interesting. To promote variety, the editors welcome problem proposals that span the entire undergraduate curriculum. Proposed problems should be sent to Curtis Cooper, either by email as a pdf, TEX, or Word attachment (preferred) or by mail to the address provided above. Whenever possible, a proposed problem should be accompanied by a solution, appropriate references, and any other material that would be helpful to the editors. Proposers should submit problems only if the proposed problem is not under consideration by another journal. Solutions to the problems in this issue should be sent to Shing So, either by email as a pdf, TEX, or Word attachment (preferred) or by mail to the address provided above, no later than April 15, 2011.
PROBLEMS 941. Proposed by Ovidiu Furdui, Cluj, Romania. Let f : [a, b] → [0, 1] be an increasing function which is differentiable at b with f (b) = 1 and f (b) = 0 and let g be a bounded function on [a, b] with limx→b− g(x) = L. Find the value of b lim n f n (x)g(x) d x. n→∞
a
942. Proposed by Geoffrey Kandall, Hamden CT. Suppose ABC is any triangle with angle bisectors A A , B B , and CC meeting at the incenter I . Let A = I A ∩ B C , B = I B ∩ A C , and C = I C ∩ A B . Prove that I B I C I A + + = 1. A A B B C C 943. Proposed by Juan-Bosco Romero M´arquez, Universidad de Valladolid, Valladolid, Spain and Francisco Javier Garc´ıa Capit´an, Spain. Given the triangle ABC, let D, E be points on line BC such that lines AD and AE are isogonal. Let the internal bisectors of angles ADB and AEC meet lines AB and doi:10.4169/college.math.j.42.1.062
62
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 62
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
AC at F and G, respectively. Show that the points X = BG ∩ CF, Y = DG ∩ EF and Z = DF ∩ EG are on the internal bisector la of angle BAC and satisfy the cross ratio (AXYZ) =
AB AC
2 BD 2 : . DC
A α
α G
E X Y B
L
D
C
E
Z
944. Proposed by Duong Viet Thong, Ha Noi University of Science, Vietnam. 1 Let f : [0, 1] → R be a continuously differentiable function such that 0 f (x) d x = 0 and for all x ∈ [0, 1] we have m ≤ f (x) ≤ M. Show that m ≤ 12
1
x f (x) d x ≤ 0
M . 12
945. Proposed by Erwin Just (Emeritus), Bronx Community College of the City University of New York, Bronx NY. Assume that R is a ring with more than one element, and there exists a positive integer n ≥ 3, such that for each x ∈ R, x + x 9 + x n = 0. Prove that (a) n must be odd. (b) For each x ∈ R, x = x 9 . (c) If n ≡ 3, 5, or 7 mod 8, then for each x ∈ R, x = x 3 .
SOLUTIONS A definite integral and the Glaisher-Kinkelin constant 916. Proposed by Kim McInturff, Santa Barbara CA. Let A denote the Glaisher-Kinkelin constant defined by A = lim n −(6n n→∞
2 +6n+1)/12 n 2 /4
e
n
k k = 1.2824 . . . .
k=1
Prove that
1
−1
x ln (x + 2) d x =
3 − 2 ln A. 4
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
63
page 63
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
Solution by Hongwei Chen, Christopher Newport University, Newport News VA and Khristo Boyadzhiev, Ohio Northern University, Ada OH (independently). Let I denote the desired definite integral. Since (1 + t) = t(t), I =
−1
= =
1
x ln (x + 2) d x +
x ln (x + 2) d x 0
0
−1
0
+
1
x ln(1 + x) +
x[ln(1 + x) + ln x] d x 0
0
−1
x[ln(1 + x) + ln x + ln (x)] d x 0
0
−1
1
x[ln(1 + x) + ln (1 + x)] d x +
1
x ln (1 + x) d x +
x ln (x) d x. 0
Integration by parts yields
0
−1
x ln(1 + x) d x =
3 4
and
1
x[ln(1 + x) + ln x] d x = 0. 0
Substituting t = 1 + x gives
0
−1
1
x ln (1 + x) d x =
(t − 1) ln (t) dt. 0
Thus, I =
3 + 4
1
(2t − 1) ln (t) dt. 0
We shall show
1
(2x − 1) ln (x)d x = −2 ln A.
(1)
0
Since, for 0 < x < 1, ln (x) =
1 1 1 ln(2π) − ln(2 sin(π x)) + (γ + ln(2π))(1 − 2x) 2 2 2 ∞ ln k 1 sin(2πkx), + π k=1 k
where γ is Euler’s constant. This yields
1 0
64
∞ 1 ln k 1 (2x − 1) ln (x)d x = − (γ + ln(2π) − 2 . 6 π k=1 k 2
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 64
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
Since, from the Glaisher product, ∞ ln k
=
k=1
k2
∞
1/k 2
k
π2 (12 ln A − γ − ln(2π)), 6
=
k=1
A12 2πeγ
π 2 /6 ,
we establish (1) as desired. Editor’s Note. Jet Foncannon of Philadelphia PA solved this problem by the following p+1 general approach: Let p ≥ 0 be an integer and define H p = p x ln (x) d x, U p = p+1 ln (x) d x, and C = 12 ln(2π). Then p
p+1
H p+1 − H p =
(x + 1) ln x d x + U p p
and
H p − H1 =
1 p2 ln p − ( p + 5)( p − 1) + C( p − 1) p+ 2 4 p( p − 1) + j ln j. 2 j =1 p−1
−
From the asymptotic expansions p−1 j =1
and
p2 + j ln j ∼ ln A − 4
p+1
Hp ∼
x
x−
p
p2 p 1 − + ln p 2 2 2
1 1 ln x − x + C + d x, 2 12x
5 3 H1 = C − ln A − 2 4 and hence, for p ≥ 2, Hp =
p2 p+ 2
ln p − ln A + C
1 p+ 2
3 p2 + 2 p + j ln j. 4 j =1 p−1
−
Thus,
1
3 x ln (x + 2)d x = H1 + H2 − 2(U1 + U2 ) = −2 ln A + . 4 −1
Also solved by M ICHEL B ATAILLE , Rouen, France; J ET F ONCANNON, Philadelphia PA; G. C. G REUBEL, ´ NGEL P LAZA (jointly), U. of Las Palmas de Gran ´ (student) and A Newport News VA; S ANTIAGO DE L UX AN Canaria, Spain; PAOLO P ERFETTI, Dipartimento di Matematica, Universit`a degli studi di Tor Vergata Roma, Rome, Italy; W ILLIAM S EAMAN, Albright C.; P ETER S IMONE , U. of Nebraska Medical Center; J OHN S UMNER and A IDA K ADIC -G ALEB (jointly), U. of Tampa; and the proposer.
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
65
page 65
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
An inequality of a triangle with its circumradius and inradius 917. Proposed by Jos´e Luis D´ıaz-Barrero, Universitat Polit´ecnica de Catalunya, Barcelona, Spain. Let a, b, c be the lengths of the sides of triangle ABC with inradius r and circumradius R. Prove that
r a b c 1 3 ≤ < . 2r + R b + c + 2a c + a + 2b a + b + 2c 3 Solution by Jos`e Nieto, Universidad del Zulia, Maracaibo, Venezuela and Peter N¨usch, Lausanne, Switzerland (independently). We prove the following stronger result.
r ≤ 2r + R
3
abc 1 ≤ , (b + c + 2a)(c + a + 2b)(a + b + 2c) 4
(1)
with equality if and only if a = b = c. √ 4 Since by the Arithmetic-Geometric mean inequality b + c + 2a ≥ 4 bca 2 , √ 4 (b + c + 2a)(c + a + 2b)(a + b + 2c) ≥ 43 bca 2 cab2 abc2 = 43 abc, and hence
3
abc ≤ (b + c + 2a)(c + a + 2b)(a + b + 2c)
3
abc 1 = , 43 abc 4
with equality if and only if a = b = c. Therefore, the right inequality in (1) is established. Let denote the area of ABC and s = a+b+c . From the well-known results 2 abc = sr = s(s − a)(s − b)(s − c), 4R R abcs 2abc abc = = . = 2 r 4 4(s − a)(s − b)(s − c) (−a + b + c)(a − b + c)(a + b − c)
=
(2)
Since 6abc − (−a + b + c)(a − b + c)(a + b − c)
b+c a+c a+b + + a b c
1 1 (a + b)(a − b)2 (a + b − c) + (a + c)(a − c)2 (a + c − b) c b 1 + (b + c)(b − c)2 (b + c − a) a ≥ 0,
=
66
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 66
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
we find that
November 11, 2010
11:00 a.m.
problems.tex
b+c a+c a+b (−a + b + c)(a − b + c)(a + b − c) + + ≤ 6abc. a b c
(3)
By (2) and (3), 2abc 1 b+c a+c a+b R = ≥ + + . r (−a + b + c)(a − b + c)(a + b − c) 3 a b c By the Arithmetic-Geometric mean inequality, 3 (b + c + 2a)(c + a + 2b)(a + b + 2c) abc c + a + 2b a + b + 2c 1 b + c + 2a + + ≤ 3 a b c 1 b+c c+a a+b R + + =2+ ≤2+ . 3 a b c r Therefore, the left inequality in (1) holds with equality if and only a = b = c. Also solved by A RKADY A LT (2 solutions), San Jose CA; G EORGE A POSTOLOPOULOS, Messolonghi, Greece; H ABIB FAR, Lone Star C.; J OHN H EUVER, Grande Prairie, Canada; S EON W OO K IM, Inst. of Sci. Ed. for the Gifted and Talented, Yonsei U., Seoul, Korea; K EE -WAI L AU, Hong Kong, China; D OUGLAS M AGOMO, Northland C.; G EORGE M ATTHEWS, Indianapolis IN; W ILLIAM S EAMAN, Albright C.; M ICHAEL VOWE , Therwil, Switzerland; and the proposer.
Conditions for a perfect cube 918. Proposed by Cezar Lupu (student), University of Bucharest, Bucharest, Romania. Find all integers a, b, c such that for every positive integer n, the number a · 2n + b · 3n + c is a perfect cube. Solution by Carlo Pagano, Universit`a di Roma “Tor Vergata,” Rome, Italy. We prove the following general result: √ Let A, B be integers which are not perfect cubes such that 1 < A3 < B 2 . If xn = 3 a An + bB n + c is an integer for every n ∈ N, then a = b = 0 and c is perfect cube. Assume b = 0. Since B > A, n A a A n xn = b1/3 B n/3 1 + +o . 3b B B Hence, aB xn+3 − Bxn ∼ 2/3 3b because 0 < This means
A B 2/3
A B 2/3
n
A3 −1 →0 B3
< 1. Since xn+3 − Bxn is an integer for all n, xn+3 = Bxn eventually. a A3 An + bB 3 B n + c = a B 3 An + bB 3 B n + B 3 c,
and hence a(A3 − B 3 )An is eventually constant, which implies a = 0. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
67
page 67
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
√ Next, assume a = 0. Then xn = 3 a An + c, and hence n 1 c 1 n +o xn = a 1/3 An/3 1 + . 3a A A It follows that xn+3 − Axn ∼
cA 3a 2/3
1 A2/3
n
1 − 1 →0 A3
because 0 < A1 < 1. Since xn+3 − Axn is an integer for all n, xn+3 = Axn eventually. for a An to be a perfect cube This means that (A3 − 1)c = 0, and hence c = 0. In order √ 3 for all n, a = 0, which is a contradiction. Therefore, if a An + bB n + c is an integer for every n ∈ N, then a = b = 0 and c is a perfect cube. Also solved by G EORGE A POSTOLOPOULOS, Messoloughi, Greece and the proposer.
Two incomplete solutions were received.
Two triangle identities 919. Proposed by Michel Bataille, Rouen, France. Let a, b, c be the lengths of the sides of triangle ABC with inradius r and radii of excircles ra , rb , rc at angles A, B, C, respectively. Prove that (1) (rb − rc ) cos A + (rc − ra ) cos B + (ra − rb ) cos C = 0 and . (2) (rb + rc ) csc A + (rc + ra ) csc B + (ra + rb ) csc C = abc 2r 2 Solution by Chip Curtis, Missouri Southern State University, Joplin MO and Habib Far, Lone Star College, Montgomery, Conroe TX (independently). (1) Let R be the circumradius of ABC. From the known formula ra = 4R sin
B C A cos cos , 2 2 2
with analogous formulas for rb and rc , we have C A C A B B rb − rc = 4R sin cos cos − sin cos cos 2 2 2 2 2 2 = 4R cos
B −C A sin . 2 2
Thus, B −C A (rb − rc ) cos A = 4R cos A cos sin 2 2 A− B +C A+ B −C − sin = 2R cos A sin 2 2 π
π − C − sin −B = 2R cos A sin 2 2 = 2R cos A(cos C − cos B). 68
(1)
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 68
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:00 a.m.
problems.tex
Similarly, (rc − ra ) cos B = 2R cos B(cos A − cos C)
(2)
(ra − rb ) cos C = 2R cos C(cos B − cos A).
(3)
and
The desired identity follows by adding equations (1), (2), and (3). (2) Let s = a+b+c and denote the area of ABC. Then 2
=
s(s − a)(s − b)(s − c) = r s.
bc Since s − a = b+c−a , ra = s−a , and csc A = 2 , with analogous formulas for 2 s − b, s − c, rb , rc , csc B, and csc C, 1 1 bc + (rb + rc ) csc A = 2 a + c − b a + b − c 2
=
2abc (a + c − b)(a + b − c)
with analogous results for (ra + rc ) csc B and (ra + rb ) csc C. It follows that (rb + rc ) csc A + (ra + rc ) csc B + (ra + rb ) csc C =
abcs 2abc(a + b + c) = (a + c − b)(a + b − c)(b + c − a) 2(s − b)(s − c)(s − a)
=
abcs 2 abcs 2 abcs 2 abc = = = 2. 2 2 2 2s(s − a)(s − b)(s − c) 2 2r s 2r
Also solved by G EORGE A POSTOLOPOULOS, Messoloughi, Greece; A RKADY A LT , San Jose CA; S COTT B ROWN, Auburn U. Montgomery; D MITRY F LEISCHMAN, Santa Monica CA; M ICHAEL G OLDENBERG, The Ingenuity Project, Baltimore Poly. Inst. and M ARK K APLAN, C.C. of Baltimore County (jointly); J OHN H EUVER, Grande Prairie, Canada; S EON W OO K IM, Inst. of Sci. Ed. for the Gifted and Talented, Yonsei U., Seoul, Korea; K EE -WAI L AU, Hong Kong, China; S UPAWAN L ERTSKRAI, Harford C.C.; G EORGE M ATTHEWS, Indianapolis ¨ IN; K IM M C I NTURFF, Santa Barbara CA; P ETER N USCH , Lausanne, Switzerland; RUTHVEN M URGATROYD; J OS E´ N IETO, Universidad del Zulia, Maracaibo, Venezuela; J OEL S CHLOSBERG, Bayside NY; W ILLIAM S EAMAN, Albright C.; J OHN S UMNER and A IDA K ADIC -G ALEB (jointly), U. of Tampa; E RCOLE S UPPA, Liceo Scientifico Statale E. Einstein, Teramo, Italy; M ARIA T RIPOLI and J OHN Z ACHARIAS (jointly), Melbourne FL; M ICHAEL VOWE , Therwil, Switzerland; and the proposer.
An infinite sum with the power of a product of cosines 920. Proposed by Ovidiu Furdui, Cluj, Romania. Let α > 0 and β ≥ 0. Prove that the series ∞
n β (cos 1 cos 2 · · · cos n)α
n=1
is absolutely convergent. VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
69
page 69
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
10:32 a.m.
problems.tex
Solution by Eugene Herman, Grinnell College, Grinnel IA. More generally, we show that ∞ n=1
nβ
n
α cos(kp + q)
(1)
k=1
is absolutely convergent whenever α > 0, β ≥ 0, 0 < p < π, q ∈ R. The continuous function f (x) = |cos(x) cos(x + p)| has an absolute maximum M ≤ 1 on [−π, π], and hence has the same maximum on R by the periodicity of f . Since 0 < p < π, f (x) never equals 1, and so M < 1. Thus, for even values of n, n cos(kp + q) = |cos( p + q) cos( p + q + p)||cos(3 p + q) cos(3 p + q + p)| k=1 · · · |cos((n − 1) p + q) cos((n − 1) p + q + p)| =
n/2
f ((2 j − 1) p + q) ≤ M n/2 .
j =1
When n is odd, n
k=1
(n−1)/2 cos(kp + q) ≤ n−1 . Thus, for all n, k=1 cos(kp + q) ≤ M α n β cos(kp + q) ≤ n β M (n−1)α/2 . n k=1
β (n−1)α/2 converges by the ratio test, and therefore the series (1) The series ∞ n=1 n M converges absolutely. Also solved by G EORGE A POSTOLOPOULOS, Messoloughi, Greece; H ERB B AILEY, Rose-Hulman Inst. Tech.; PAUL B UDNEY, Sunderland MA; DAVID G ETLING, Berlin, Germany; M ARTY G ETZ and D IXON J ONES (jointly), U. of Alaska Fairbanks; A. B ATHI K ASTURIARACHI, Kent State U. at Stark; K EE -WAI L AU, Hong Kong, China; M ICHAEL N ATHANSON, St. Mary’s C. of CA; N ORTHWESTERN U NIVERSITY M ATH P ROBLEM S OLVING G ROUP, Northwestern U.; PAOLO P ERFETTI, Dipartimento di Matematica, Universit`a degli studi di Tor Vergata Roma, Rome, Italy; J OEL S CHLOSBERG, Bayside, NY; W ILLIAM S EAMAN, Albright C.; J OHN S UMNER and A IDA K ADIC -G ALEB (jointly), U. of Tampa; T ONY TAM, Calexico CA; A LEXANDER WALKER (student), Boston C.; A LBERT W HITCOMB, Castle Shannon PA; and the proposer.
Editor’s Note. Paul Budney of Sunderland MA noted that the result in this problem can be generalized as follows: Suppose {ti } is a real arithmetic progression with common difference d such that |sin(d)|, |cos(d)| = 1. Furthermore, suppose { f i } is a sequence such that, for each i ∈ N, f i = sin or f i = cos. Then n of functions ∞ β α n=1 n ( i=1 f i (ti )) is absolutely convergent, where α > 0 and β ≥ 0.
70
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 70
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:09 a.m.
review.tex
page 71
BOOK REVIEW The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brain Teaser by Jason Rosenhouse. Oxford University Press, 2009, xii + 194 pp. ISBN 9780-19-536789-8, $24.95. Reviewed by Edward J. Barbeau (
[email protected]), University of Toronto, Toronto, ON M5S 2E4 Every once in a while, a problem captures the imagination and is widely circulated within the mathematical community. Occasionally, it breaks out of the professional circle into the public domain. A notorious example is the Monty Hall problem, inspired by the television game show Let’s Make a Deal: Monty Hall, a television host, shows a contestant three identical doors, behind one of which is a car and behind the other two are goats. He asks the contestant to select one of the doors and will give the contestant the prize behind it. After the contestant makes her selection, Monty opens one of the remaining doors to reveal a goat. He then offers the contestant the opportunity to switch her choice to the remaining unopened door. Should the contestant, preferring the car, stick with her first choice, or switch? This problem first came to my attention in a newsletter from Washington State University [9] and was reported on in this J OURNAL in 1993 [1], with follow-up comments in [2–6]. Earlier, in 1990, Marilyn Vos Savant dealt with the problem in her weekly column, “Ask Marilyn,” in Parade. Public exposure resulted in a flurry of controversy in which opposing answers were offered with great passion and tenacity. Mathematicians began to produce papers purporting to sort things out; some of these came to the editor of the College Mathematics Journal. Before these were refereed, it seemed useful to find out what was already known about the problem, and with the help of several correspondents, I found several antecedents, notably a problem about three prisoners awaiting execution popularized by Martin Gardner in the 1950s: Three prisoners, A, B, C are to be executed. The governor decides, at random, to pardon one of them and informs the prison warden. Prisoner A, aware of this, convinces the warden to tell him which of B and C will be executed (selecting at random if both are to be executed). What is the chance now that A will be pardoned? In essence, the problem dated back to at least 1889 when Joseph Bertrand published a box problem in his book Calcul des Probabilit´es. Some solvers of the Monty Hall problem were of the opinion the two unopened doors were equally likely to conceal the car. Others were equally sure that the door initially chosen gave a probability 1/3 of success and the remaining door 2/3; Monty just told you which of the doors might hide the car should you switch. The resolution you favour turns on whether you feel that Monty has provided any pertinent information, and if so, how this governs the eventual probability. It is not surprising that this sort of probability problem engenders dispute. Since, in any given play of the game, the positions of the car and goats are given, probability is an artefact to handle the uncertainty. So we have to consider how this abstraction connects with reality. If we are to come up with useful advice as to whether one ought doi:10.4169/college.math.j.42.1.071
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
71
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:09 a.m.
review.tex
to stick or switch, a lot of spadework is needed to interpret the situation and decide what probability actually measures. The book under review makes this analysis. The author delves into the problem with zest, considers it from different angles and treats generalizations. This book can be regarded as a treatise on probability in which the Monty Hall problem is a vehicle for introducing and clarifying the concepts. Indeed, the author tells us that “it is a recurring theme of this book that you can teach an entire course in probability theory using nothing more than variations on the Monty Hall problem” (p. 105). It is not quite selfcontained in this respect, although students in a first probability course could study it with profit. The opening chapter sets the stage with a brief historical and philsophical sketch of probability, followed by an account of the origin and early renditions of the Monty Hall situation. Quite a bit of space is devoted to details of the controversy between Marilyn Vos Savant and her critics. Quotations from Parade and the American Statistician give a flavour of the acrimony in the exchange of opinions. The second chapter eases us into the problem. It opens with a statement of its “canonical version,” where Monty is assumed always to open a door concealing a goat and does so with equal probability when he has a choice. There is an informal discussion of reasons that one might stick or switch. To work through the fog, we need a mathematical model, so the author provides a brief primer on probability before applying it to the canonical problem. A Monte Carlo simulation validates the conclusion that it is better to switch. An important tool for treating the problem and its variants is Bayes’ Theorem. This is the burden of the third chapter, where the reader is treated to a clear exposition of independent events, conditional probability, the law of total probability and the theorem itself. As an example, this result is applied to a concern of every air traveller: after I have waited for a long time at the baggage carousel, what are the chances that the airline has lost my case? (There is a small error in the middle of page 72, where one should consider the smallest integer x for which P(A | B) is larger than 12 .) Now we can play with various scenarios and determine the best strategy when, for example, Monty opens one of the remaining doors at random (possibly revealing a car); Monty points to a door and says, possibly incorrectly, that it conceals a goat; the car is behind the doors with unequal probabilities; or Monty always reveals a goat, but opens the doors with unequal probabilities when given a choice. The next two chapters introduce more variants and demand close attention from the reader to negotiate the mathematics. We can have more doors, more cars, more kinds of prize, more players or more hosts. For example, with many doors and one car, we can have Monty opening doors one by one until only the door chosen by the contestant and one other door remain closed. If the contestant is offered a chance to switch anywhere along the way, is it best to stick with the original choice, switch only at the very end, or follow some particular regime of sticking and switching? (At the bottom of page 103, the last inequality should read P(C˜i ) ≤ n−1 ; at the top of page n 116, two occurrences of the word “probability” should be replaced by “event”; on page 125, several occurrences of m should be replaced by n, the number of doors.) There is some confusion at the beginning of Section 5.3 due to a couple of minor mistakes that do not affect the overall correctness of the analysis. However, as the author uses an alternative to the Baysian approach that was unfamiliar to me, it is worth describing the situation in detail, correcting the mistakes, using Bayes theorem as an example of how he treated cases elsewhere in the book, and then describing his treatment. The setting is similar to the standard problem, except that there are two hosts available, Coin-Toss Monty (CTM) and Three-Obsessed Monty (TOM). The presiding 72
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 72
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:09 a.m.
review.tex
page 73
host is selected by a coin toss. The contestant starts by selecting Door 1. If the car is behind this door, Monty is assumed to open Door 3 with probability q; q = 12 for CTM and q = 1 for TOM. We suppose that Monty opens Door 3 to reveal a goat. For i = 1, 2, 3, let Mi be the event that Monty opens Door i and Ci the event that the car is behind Door i. With p = P(C1 | M3 ) and q = P(M3 | C1 ), since P(C1 | M3 )P(M3 ) = P(M3 | C1 )P(C1 ) and P(M3 ) = q P(C1 ) + P(C2 ) = 13 (1 + q), we have that p(1 + q) = q, so that ( p, q) = ( 13 , 12 ) for CTM and ( p, q) = ( 12 , 1) for TOM. Turning to the situation at hand, the chances are even that the presiding host is CTM and TOM. In any case, P(M1 ) = 0. Since CTM will open Door 3 with probability 1 7 and TOB will open Door 3 with probability 23 , then P(M3 ) = 12 · 12 + 12 · 23 = 12 2 5 7 and P(M2 ) = 12 . Similarly, P(M3 | C1 ) = 12 · 12 + 12 · 1 = 34 . Then p · 12 = 34 · 13 so 3 4 p = 7 . Thus the probability of winning by switching is 7 . The author’s treatment relies on a “proportionality principle”: If various alternatives are equally likely, and then some event is observed, the updated probabilities for the alternatives are proportional to the probabilities that the observed event would have occurred under those alternatives” (p. 82). Basically, it is argued that if we see the host open Door 3, then the host is more likely to be TOM than CTM. Since the respective values of P(M3 ) are 23 and 12 for the two hosts, we conclude that we are 43 more likely to see the host open Door 3 if he is TOM rather than CTM. “Thus we conclude that we have drawn Coin-Toss Monty with probability 37 and Three-Obsessed Monty with probability 47 . Our probability of winning by switching will therefore be the following weighted average 47 · 12 + 37 · 23 = 47 ” (pp. 117–118). The apparent irrationality of people tackling this problem has caught the attention of psychologists. Even if you believe that it does not matter whether you switch, why, according to some investigators, do fewer than 20% of players elect to switch? It seems that switching from a winning position weighs more heavily than failing to switch from a losing position. A study by Bar-Hillel and Falk [8] aims to understand why intuition may be an unreliable guide. In his next chapter, the author reports in detail on one of their examples: Mr. Smith, a father of two, is seen walking along the street with a lad who turns out to be his son. What is the probability that Mr. Smith’s other child is also a boy? Before this question can be answered, according to Rosenhouse, one has to consider not just the data but “the precise statistical experiment that led you to the data.” The answer depends on whether you learn that it is the elder who is the boy, that a child chosen at random is a boy, or that one of the children is a boy. Using the work of Falk, and of Shimojo and Ichikawa [10], the author discusses how the solver’s implicit subjective principles may militate against a cogent analysis. He concludes that “it seems that there is something in our cognitive architecture that leads us to make fools of ourselves when discussing problems of this sort” (p. 145). Whatever this is seems to cut across cultural boundaries, as this behaviour is common to people the world over. Other research on cognitive issues is briefly sketched. Finally, it is the turn of the philosophers. Two issues are discussed. The first is whether, as philosophers Paul Moser and D. Hudson Mulder hold, you can justifiably follow one strategy in a single play and a quite different one over many repetitions. The second turns on a variant with two contestants, where philosopher Peter Baumann claims that the solution violates the principle that two rational people with the same information who fully determine their rational degree of belief in a proposition, should assign the same probability to the proposition. The author, convincingly in my view, argues against both positions.
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
73
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:09 a.m.
review.tex
The author has not only thought deeply about this problem. He has gone carefully through the literature and provides us with an entertaining, multifaceted and thoughtprovoking study of this challenging problem. References 1. E. J. Barbeau (ed.), Fallacies, flaws, and flimflam, College Math. J. 24 (1993) 149–153. 2. , Fallacies, flaws, and flimflam, College Math. J. 26 (1995) 132–134. 3. , Fallacies, flaws, and flimflam, College Math. J. 27 (1996) 46. 4. , Fallacies, flaws, and flimflam, College Math. J. 27 (1996) 205. 5. , Fallacies, flaws, and flimflam, College Math. J. 28 (1997) 44. 6. , Fallacies, flaws, and flimflam, College Math. J. 29 (1998) 136. 7. , Mathematical Fallacies, Flaws, and Flimflam, Mathematical Association of America, Washington DC, 2000, pp. 86–90. 8. M. Bar-Hillel and R. Falk, Some teasers concerning conditional probabilities, Cognition 11(2) (1982) 109– 122. doi:10.1016/0010-0277(82)90021-X 9. S. C. Saunders, Math. Notes 32(2) (whole no. 129), April, 1990, Dept. of Mathematics, Washington State University, Pullman WA. 10. S. Shimojo and S. Ichikawa, Intuitive reasoning about probability: theoretical and experimental analysis of three prisoners, Cognition 32 (1989) 1–24. doi:10.1016/0010-0277(89)90012-7
A Jungian Nightmare School came to bore me. It took up far too much time which I would rather have spent drawing battles and playing with fire. Divinity classes were unspeakably dull, and I felt a downright fear of the mathematics class. The teacher pretended that algebra was a perfectly natural affair, to be taken for granted, whereas I didn’t even know what numbers really were. They were not flowers, not animals, not fossils; they were nothing that could be imagined, mere quantities that resulted from counting. To my confusion these quantities were now represented by letters, which signified sounds, so that it became possible to hear them, so to speak. Oddly enough, my classmates could handle these things and found them self-evident. No one could tell me what numbers were, and I was unable even to formulate the question. To my horror I found that no one understood my difficulty. The teacher, I must admit, went to great lengths to explain to me the purpose of this curious operation of translating understandable quantities into sounds. I finally grasped that what was aimed at was a kind of system of abbreviation, with the help of which many quantities could be put in a short formula. But this did not interest me in the least. I thought the whole business was entirely arbitrary. Why should numbers be expressed by sounds? One might just as well express a by apple tree, b by box, and x by a question mark. a, b, c, x, y, z were not concrete and did not explain to me anything about the essence of numbers, any more than an apple tree did. But the thing that exasperated me most of all was the proposition: If a = b and b = c, then, a = c, even though by definition a meant something other than b, and, being different, could therefore not be equated with b, let alone with c. Whenever it was a question of an equivalence, then it was said that a = a, b = b, and so on. This I could accept, whereas a = b seemed to me a downright lie or a fraud. I was equally outraged when the teacher stated, in the teeth of his own definition of parallel lines, that they met at infinity. This seemed to me no better than a stupid trick to catch peasants with, and I could not and would not have anything to do with it. My intellectual morality
74
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 74
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 11, 2010
11:09 a.m.
review.tex
page 75
fought against these whimsical inconsistencies, which have forever debarred me from understanding mathematics. Right into old age I have had the incorrigible feeling that if, like my schoolmates, I could have accepted without a struggle the proposition that a = b, or that sun = moon, dog = cat, then mathematics might have fooled me endlessly—just how much I only began to realize at the age of eighty-four. All my life it remained a puzzle to me why it was that I never managed to get my bearings in mathematics when there was no doubt whatever that I could calculate properly. Least of all did I understand my own doubts concerning mathematics. Equations I could comprehend only by inserting specific numerical values in place of the letters and verifying the meaning of the operation by actual calculation. As we went on in mathematics I was able to get along, more or less, by copying out algebraic formulas whose meaning I did not understand, and by memorizing where a particular combination of letters had stood on the blackboard. I could no longer make headway by substituting numbers, for from time to time the teacher would say, “Here we put the expression so-and-so,” and then he would scribble a few letters on the blackboard. I had no idea where he got them and why he did it—the only reason I could see was that it enabled him to bring the procedure to what he felt was a satisfactory conclusion. I was so intimidated by my incomprehension that I did not dare to ask any questions. Mathematics classes became sheer terror and torture to me. —from Memories, Dreams, Reflections by C. G. Jung. Recorded and edited by Aniela Jaffe. Translated by Richard and Clara Winston —suggested by Steven C. Althoen
The ideal audience? A theatre full of mathematicians, philosophers and scientists. Ask any conjuror. The intelligent—the more rational—a spectator, the more readily he will be deceived. Because what you are doing in magic is creating the semblance of a chain of cause and effect. I do this, then this happens: I do that, then that happens. Basic logic. And a logical mind, receptive to a connection between each apparent cause and its apparent effect, is more prone to surprise when an illusion reaches its ‘illogical’ climax. —from The Houdini Girl by Martyn Bedford
It wasn’t the best of times; it wasn’t the worst of times; it was the times you’d get if you arranged all possible times (including even fictional times in which the nights were usually dark and stormy) in order from worst to best on the real number line from 0.0 inclusive to 1.0 exclusive and then used a really good uniform random number generator to pick a value in that range thus choosing the corresponding times—that’s the times it was. —from A Tale of Two Statisticians by Dale Dellutri
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
75
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
MEDIA HIGHLIGHTS EDITOR
Warren Page 30 Bonnie Way Larchmont NY 10538
[email protected] with
Tanya Leise
Cecil Rousseau
Amherst College Amherst MA 01002
University of Memphis Memphis TN 38152
Media Highlights are short, approximately half-page, reviews intended to help CMJ readers monitor a broad spectrum of publications, web materials, professional activities, and instructional resources. Readers are encouraged to submit items that will be of interest to colleagues in the mathematical community. Media Highlights should be sent to Warren Page at
[email protected].
The Last Boat from Lisbon: Conversations with Peter D. Lax, Istvan Hargittai. The Mathematical Intelligencer 23:3 (Fall 2010) 24–30. With his family, Peter Lax emigrated from Hungary to the United States in 1941, on the last boat to leave Lisbon before the United States entered World War II. He bore a letter from Rozsa Peter to John von Neumann that praised the 15-year-old Lax as her most talented student, and ending, “I would like to see him in good hands out there because I am convinced that he may amount to something.” Lax fulfilled her prophecy, becoming one of our most distinguished applied mathematicians, the Director of the Courant Institute, and winning the 2005 Abel Prize “for his groundbreaking contributions to the theory and application of partial differential equations and to the computation of their solutions.” In this interview, he talks candidly about his education, his career, and other great Hungarian-born mathematicians and scientists. Here are a few samples: •
• •
•
On schools: “In America, the teachers were friends. In the European school, you recognized who your enemy was—the teachers. That could explain why European schools were so efficient. You had to fight for your life.” (Rozsa Peter tutored him outside of school.) On Paul Halmos’ famously-titled article “Applied Mathematics is Bad Mathematics”: “The fact is that he knew nothing about it, so whatever he said was irrelevant.” On Paul Erd˝os: “Erd˝os did some very great things, but . . . what I found strange was that he was willing to work on anything. It was partly kindness: when people came to him with problems, he was very willing to do it. And partly it was just that he was interested in everything.” On the “five Martians” of Hungarian science—Theodore von Karman, Leo Szilard, Eugene Wigner, John von Neumann, and Edward Teller: “Szilard perhaps had the most fantastic imagination. Perhaps he was the most remarkable among them. But von Neumann had a mind which was, in its power, unlike anybody else’s.”
Lax tells the story of when he was Director of the Computing Center at the Courant Institute at the height of student unrest in 1970, when students occupied the Institute and threatened to destroy its computer unless the University put up $100,000 bail for the Black Panthers. The University didn’t give in, and after two days the students left. doi:10.4169/college.math.j.42.1.076
76
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 76
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
page 77
When Lax and his colleagues re-entered the building, they smelled smoke and found a burning fuse leading to bottles of flammable liquids tied to the computer. They defused the bomb and saved the computer. “Afterwards, [my wife] Anneli asked how I could be so crazy to run toward a burning bomb. I told her that I was so angry that I didn’t think.” The interview ends with this exchange: On a lighter note, do you count in English or in Hungarian? “In Hungarian.” And you curse in Hungarian. “Sorry about that. I do that out of tact.” PDS The Pattern Collector, Julie Rehmeyer. Science News’ Math Trek (August 6, 2010), http://www.sciencenews.org/view/generic/id/61870. This article outlines the personal and professional history of the encyclopedia of integer sequences founded by Neal Sloane. The encyclopedia started with a set of cards bearing sequences Sloane had encountered, evolved through two books, a website, and is now a wiki. Sloane’s favorite sequence, the Recam´an sequence, is pointed to as one that might “wreck your life.” He spent months working with it. Seemingly “just a curiosity, unconnected to other mathematical questions,” Sloane views it as a wonderful discovery that “provides a welcome escape from the troubles of our planet.” NS What Is the Probability That the Final Person on the Aircraft Sits in His Own Seat?, Paul Belcher. Mathematical Spectrum 42:3 (2009/2010) 107–110. The following problem has been passed around mathematical circles lately: n passengers board a plane with n seats. The first passenger to board has lost his boarding pass and sits in a seat at random. The other n − 1 passengers sit in their seats if they’re unoccupied, otherwise they chose a seat at random. What is the probability that the nth passenger gets his own seat? The surprising answer is one-half, independent of the number n. The article shows this by using a tree diagram for the special-but-suitablygeneral case when n is 4. The eight paths pair up into four pairs, where both paths in a pair have the same probability. Each pair includes a path where the nth passenger gets his own seat and a path where he doesn’t. So for n = 4, the probability that the last passenger gets his own seat must be 12 . The article cleverly shows that the pairing generalizes for any n, by looking at the last person among the first n − 1 passengers who does not get his own seat. It also shows why the tree diagram for n passengers has exactly 2n−1 different paths. The article gives two additional proofs that the answer is always 12 . Both are by mathematical induction, a short one by strong induction and a much longer one by ordinary (or weak) induction. PR The Spread of Behavior in an Online Social Network Experiment, Damon Centola. Science 329 (September 2010) 1194–1196. The author conducted a controlled experimental study of how the topology of a real social network is related to the diffusion of information. Organizing the participants of a community into a clustered lattice network in one set of trials, and into a random network in the other set of trials, Centola explored the time it took for a particular piece of information, in this case a particular health practice, to be acted upon by members of the community. The study also allowed Centola to view the actions of a given actor in the community after receiving the message from multiple other actors. The results show that social reinforcement through multiple, redundant ties is sufficient to greatly reduce the time for diffusion. In all experimental trials, the clustered network spread the message faster than the random network. This stands in contrast to the theory that networks with long-range ties, rather than redundancies, will result in VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
77
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
faster diffusion. The participants in these experiments were significantly more likely to exhibit increased participation in the social network after receiving multiple copies of messages. While the study was highly controlled and idealized, the results suggest that there is still a great deal to learn about modeling real-world social networks. KHG Some Properties of Cyclic Compositions, Arnold Knopfmacher and Neville Robbins. The Fibonacci Quarterly 48:3 (August 2010) 249–255. A composition of n into k parts is a sequence (a1 , a2 , . . . , ak ) of k positive integers whose sum is n. There are n−1 such compositions. Say that two of the corresponding k−1 circular sequences are equivalent if they differ only by a circular shift. The number of equivalence classes is denoted by nk . The table of values of nk for n ≥ 2 and 1 ≤ k ≤ n − 1, 1 1 1 1 1 1 1 1
1 2 2 3 3 4 4
1 2 4 5 7 10
1 3 5 10 14
1 3 7 14
1 4 10
1 4
1,
shares some features with Pascal’s triangle. The appearance of the Fibonacci sequence in Pascal’s triangle is well known. Here we see that every other row of the nk table has a repeated Catalan number: 1, 2, 5, 14, . . . n n The authors prove the symmetry property n−k = k for 1 ≤ k ≤ n − 1 and that 2n is divisible by 2. Analytic and combinatorial methods are used to obtain the conn nection between cyclic composition numbers and binomial coefficients, nk = n1 nk if (n, k) = 1. The analytic approach is based on the bivariate generating function
−1 φ( j) n zju j n k z u = log 1 − , C(z, u) = k j 1 − zj n>0 k≥0 j ≥1 which has as a consequence the formula
n/j 1 = [z u ]C(z, u) = φ( j) , k/j k n j |(n,k)
n
n k
where φ is the Euler totient function. The authors give two proofs of the symmetry property: (i) by means of the explicit formula, and (ii) using a bijection between the necklaces with n beads, k of which are black and n − k white, and the cyclic compositions of n into k parts. CR Impact of Proof Validation on Proof Writing in Abstract Algebra, Robert A. Powers, Cathleen Craviotto, and Richard M. Grassi. International Journal of Mathematical Education in Science and Technology 41:4 (2010) 501–514. The authors report an attempt to improve students’ proof-writing skills in an undergraduate abstract algebra course by having them validate mathematical arguments. Forty students, many of them preservice secondary teachers, were divided into two sections, one taught with a proof-validation activity once a week, and the other served as a control group. Both sections were taught in a semi-interactive manner, incorpo78
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 78
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
page 79
rating lectures “enhanced by questions and answers” and cooperative group work. In the proof validation section, approximately 20–25 minutes were spent every Friday on validation activities during which students worked in small groups to analyze a mathematical argument, decide whether it was a proof, justify their claim, and report back to the class. (All 15 validation activities are given in an appendix.) Statistical tests done on the three exams and the final (also included in the article) showed that those in the validation section improved modestly more than those in the control section. Nevertheless, the authors feel that “proof validation is a valuable objective . . . in and of itself” and they have decided to incorporate validation activities in sections of linear algebra and real analysis. A&JS Massively Collaborative Mathematics, Julie Rehmeyer. SIAM News 43:3 (April 2010) 1–3. This article tells a remarkable story involving Cambridge University mathematician and Fields Medalist Timothy Gowers. Gowers used his blog, http://gowers. wordpress.com, to direct more than two dozen mathematicians in collaborative proof of the density Hales-Jewett theorem. This theorem addresses the question of how many squares would have to be removed from a multidimensional tic-tac-toe board of arbitrary size to make it impossible for either player to win. The theorem has widespread importance in mathematics and theoretical computer science. It plays a particularly important role in Ramsey theory, additive combinatorics, and the “parallel repetition problem” in computer science. Gowers had an approach in mind for proving the theorem, but he had not yet been able to make the approach work. He therefore invited the mathematical world to collaborate through his blog, where he wrote, “The hard thought would be done by a sort of super-mathematician whose brain is distributed amongst bits of the brains of lots of interlinked people.” Within days, mathematicians began to post suggestions. And within six weeks, a beautiful proof was completed that went further than Gowers had expected. The proof will be published under the name “D.H.J. Polymath.” RNG Adding Nesting Structure to Words, Rajeev Alur and P. Madhusudan. Journal of the ACM 56:3 (May 2009) 1–43. The authors introduce a new formalism, called nested words, that can be used to model data with both linear and hierarchical structures, such as arithmetic expressions, computer programs, web documents, and various types of binary trees. In addition to a linear sequence of symbols, a nested word contains nested edges, which are noncrossing directed edges that connect pairs of symbols. Edges with the source or target vertex missing are also allowed. In the nested word below, a linear sequence of eight symbols (not shown) is enhanced by five nested edges (−∞, 1), (−∞, 4), (2, 3), (5, ∞), (7, ∞). 4
5
7 1
2
3
6
8
VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
79
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
The authors establish a bijection between nested words and strings enhanced with the tags “” and “”, where each symbol a has three flavors: a, a, and a. The source vertex a of a nested edge corresponds to a and the target vertex to a. For example, assuming all symbols are a, the corresponding linear encoding of the above nested word is aaaaaaaa. Since the edges are non-crossing, the corresponding tags are properly nested, that is, for any two pairs of , one must be completely inside or completely outside the other. The article extends the classical concept of regular languages to nested words by using finite automata enhanced with nested edges along which additional states may propagate. As in the classical setting, regular languages of nested words have equivalent formulations in terms of formal grammars and monadic second-order logic. Linearly encoded regular languages are called visibly pushdown languages. The set of all properly nested tagged words (as described above) is such an example. The article’s main results show that visibly pushdown languages form a proper subclass of classical context-free languages. Visibly pushdown languages are closed under Boolean operations, homomorphism, concatenation, reversal, and other word operations, and there exist efficient algorithms to determine their emptiness, equivalence, and inclusion decision problems. These properties lead to new efficient algorithms for software verification that were not possible previously. Similar closure and decision results are shown for infinite nested words. The authors also explore relationships between nested words and other models (e.g., regular nested words, which are exponentially more concise over tree automata in describing languages). NT Moody’s Mega Math Challenge Winning Paper: Making Sense of the 2010 Census, A. Das Sarma, J. Hurwitz, D. Tolnay, and S. Yu (Montgomery Blair High School, Silver Spring, MD). SIAM Undergraduate Research Online (SIURO) 3 (August 4, 2010) 159–175. Sponsored by The Moody’s Foundation and organized by SIAM, The M3 Challenge is an annual mathematical modeling competition for teams of 11th and 12th grade students. These students are given 14 hours to analyze an applied problem (with no outside help except for the Internet) and submit a solution subject to certain criteria. The 2010 M3 Challenge problem required the students to analyze U.S. Census Bureau data and methodology, and propose mathematical recommendations for undercount adjustment, the best apportionment method for the U.S. House of Representatives, and the fairest way to draw Congressional districts. After several rounds of judging by professional mathematicians, a tentative rank of the best six papers was made from the 531 initial submissions. The final ranking of the teams was based on their oral presentation of their paper to a panel of five judges at the Moody Foundation’s headquarters. The above cited article is an enhanced version of the winning paper written during the 14-hour period of the competition. For the undercount problem, the winning team proposed examining public records to estimate the values of missing data from different segments of the population. For the House apportionment problem, the team evaluated six historical methods (Hill, Dean, Webster, Adams, Jefferson, and Hamilton-Vinton) and indicated why the Hamilton-Vinton method was best. In their analysis of Congressional districting, the team recommended that states divide their congressional districts impartially according to population density. Complete information on the competition is available at http://m3challenge. siam.org. A description of the 2008 M3 Challenge and its winning paper “Ethanol: Not All It Seems To Be” appeared in the January 2008 issue of this J OURNAL. HJR 80
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 80
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
page 81
Does Sea Level Change When a Floating Iceberg Melts?, Boon Leong Lan. The Physics Teacher 48 (May 2010) 328–329. Archimedes’ Principle states that an object immersed in water is buoyed up by a force that is equal to the weight of water it displaces. This implies that when an ice cube floating in a glass of water melts, the height of the water in the glass will not change. However, there is an implicit assumption here that the densities of the water and the object are equal. In the case of an iceberg melting in the ocean, the density of seawater (1024 kg/m3 ) and of an iceberg (1000 kg/m3 are sufficiently different (icebergs are fresh water with no salt) so the sea level will not remain constant. This article offers a simple analysis, using algebra, to show that Archimedes’ Principle predicts a slight rise in sea level when an iceberg melts. Archimedes’ Principle says that the density of seawater times the volume of iceberg that is under water equals the density of the iceberg times the volume of the entire iceberg. When the iceberg melts, the mass of the resulting water is the same as the mass of the original iceberg. After the iceberg melts, sea level will have a rise of 0.021 times the volume of the original iceberg divided by the surface area of the ocean. The factor 0.021 depends only on the ratio between the densities of seawater and iceberg. TL An Unexpected Use of Primes: Solving Sudokus by Calculator, Mark Spahn, Ron Lancaster, Deborah Moore-Russo, and Gerald Rising. The Mathematical Gazette 94:530 (July 2010) 224–232. The authors show how Sudoku puzzles can be solved by programs written for the TI 84 calculator, which they make available free to the public at: http://www. ascu.buffalo.edu/~insrisg/InsideYourCalculator.htm. The user enters the starting numbers into a program that stores them in a matrix, putting a 0 in all blank cells. A second program, which searches for a solution, changes each 0 to 1, and each number n from 1 to 9 into the nth prime. Checking whether a number k between 1 and 9 is already in any ennead (a row, column, or 3 by 3 box) is accomplished by checking whether the product of the numbers in that ennead is divisible by k. The program contains three algorithms and an optional fourth. The first algorithm goes through all 81 cells to see if the contents of a blank cell can be determined. The second algorithm goes through each ennead to see if any cell in that ennead can be determined. The third algorithm looks for cells that exclude all but two possible digits, noting what happens with the first two algorithms if either of those digits is entered into the cell. The fourth algorithm is usually turned off because it is rarely needed and slows down the execution. For the most difficult Sudokus, in which the third algorithm fails, the program displays the message “HVPS FAIL’D. TOO BAD.” On the rare occasion when this happens, the fourth algorithm can then be implemented to test pairs of alternatives for enneads. This will solve all but the most pathological Sudokus. RNG An Integrated Modeling Environment to Study the Coevolution of Networks, Individual Behavior, and Epidemics, Chris Barrett, Keith Bisset, Jonathan Leidig, Achla Marathe, and Madhav Marathe. AI Magazine (Spring 2010) 75–87. Agent-based modeling (ABM) involves simulating the interactions of many individuals in order to study the emergent patterns. ABM works at the individual level, tracking the actions and interactions of each individual (“agent”) in the population. It allows each agent to act according to its own set of rules (and even to alter those rules in response to interactions). Differential equations-based modeling works at the population level, treating the number of people in each subpopulation (infected vs susceptible, VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
81
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
for example) as variables, and typically assuming uniform behavior and universal contact. Differential equations-based models have the advantage of being mathematically tractable, whereas agent-base models can be very difficult to analyze and typically involve numerical simulation that requires massive computing power. However, agentbased models offer realistic simulations to study, as for example, the effect of interventions, public policy, and heterogeneous individual characteristics on an HIV or H1N1 epidemic. This article describes ABM software that accepts up to 300 million individuals who are geographically located, and whose behaviors are based on real data. It also incorporates various networks, such as social contact and communication networks, that govern interactions between agents. To illustrate the promise of ABM for improving public policy, the authors use a case study of an influenza-like illness and examine the efficacy of interventions such as vaccination of certain subpopulations (children and health care workers), school closure, and self-isolation of exposed adults. Given the extreme complexity of agent-based models, comparing simulations to field data and assessing the structural validity of the model is a crucial and challenging task. The article discusses and analyzes the advances and challenges in ABM. Although agent-based modeling is in an early state of development, rapid progress is being made. It will likely prove to be a valuable tool for improving public policy decisions in complicated situations, such as a looming epidemic. TL Winning Odds, Yutaka Nishiyama and Steve Humble. Plus 55 (June 2010). A television show’s magician game inspired this lucid account of nontransitivity in an elementary coin toss game. About the simplest example of nontransitivity is the scissors-rock-paper game: Scissors cuts paper (beating paper), paper covers rock (beating rock), and rock breaks scissors (beating scissors), yet transitivity would imply scissors beats rocks. In the coin game, each player selects a triplet of heads and tails, and the player whose triplet first comes up in repeated tosses is the winner. No matter which triplet one player selects, the second player can choose a triplet that is more likely to occur first. For an extreme case where one player chooses HHH, the other player can choose THH and will be far more likely to win. Less obviously, the choice THH is beaten by TTH. In fact, every one of the eight possible choices of a triplet can be beaten by an appropriate choice for the second player. Sample probabilities are derived and algorithms are given for choosing the superior triplet. There are historical references, especially to Martin Gardner, who discussed nontransitivity in a Scientific American column and, later, in his book Time Travel and Other Mathematical Bewilderments. This enticing and lively account is at http://plus.maths.org/content/issue/55. NS Recognizing Graph Theoretic Properties with Polynomial Ideals, J. De Loera, C. Hillar, P. Malkin, and M. Omar. The Electronic Journal of Combinatorics 17:1 (2010). The authors are interested in expanding on the polynomial method (a term coined by Noga Alon) to study three notoriously difficult problems in graph theory: kcolorability, Hamiltonicity, and determining the automorphism group of a graph. In each case, the strategy is to associate with a combinatorial question (for instance: “Is this graph 3-colorable?”) a system of polynomial equations so that the answer to the combinatorial question is equivalent to determining if the system of polynomial equations has a solution. There is a (previously known) way to translate 3-colorability into a polynomial formulation. 82
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 82
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
page 83
A graph G is 3-colorable if and only if the system of equations described by JG = {xi3 − 1 = 0, xi2 + xi x j + x 2j = 0 : i is a vertex of G, {i, j} is an edge of G} has a common zero over the algebraic closure over a field K with characteristic relatively prime to 3. If this system of equations does not have a common zero over the finite field of order 2, then the graph is not 3-colorable. The authors are able fully to translate from the lack of a common zero to a combinatorial condition on graphs given by an existence condition for oriented 3- and 4-cycles in the graph G that satisfy certain properties. Furthermore, their combinatorial existence condition can be checked in polynomial time, giving an efficient running time for a solution to a hard combinatorial recognition problem. This leap from system of equations back to a combinatorial characterization that can be verified in polynomial time is a result of the power of this nonlinear polynomial method. The flavor of the results in the rest of the paper are similar to the result about non3-colorability. The article is a nice blend of theory and practical computation, and it contains a wealth of information and ideas. The article is available for free online at http://www.combinatorics.org/Volume_17/PDF/v17i1r114.pdf CS Sets Versus Trial Sequences, Hausdorff Versus von Mises: “Pure” Mathematics Prevails in the Foundation of Probability Around 1920, Reinhard SiegmundSchultze. Historia Mathematica 37:2 (May 2010) 204–241. In 2004, the author discovered two letters in the Richard von Mises Papers in the Harvard University Archives that were written in late 1919, from “pure” mathematician Felix Hausdorff to von Mises. Von Mises had been trained as a mechanical engineer, but it was as an applied mathematician that he became the first director of the Institute for Applied Mathematics at the University of Berlin in 1920. In the previous year, von Mises’s famous article “Foundations of the Calculus of Probability” criticized earlier attempts by Hausdorff and others who, ignoring intuitive notions of empirical probability like randomness, defined probability as “the quotient of the measure of a point set divided by the measure of the set in which that point set is contained.” Von Mises proposed his own alternative foundation for a theory of probability, basing it not on sets and their Lebesgue measure but on limits of sequences of relative frequencies. In his two letters, Hausdorff criticized von Mises’s attempt, in part because limits of relative sequences, when interpreted as set functions, lacked properties such as countable additivity (σ -additivity). Von Mises had used Carath´eodory’s notion of an outer measure and a measure function, that appeared in a book Carath´eodory published in 1918, in the belief that a measure could be assigned to any set. But Hausdorff produced an infinite family of counterexamples to this claim. Siegmund-Schultze asserts, “The subtlety of Carath´eodory’s notion probably escaped von Mises’s attention, or he found it irrevelant for applications.” The author also notes, “von Mises’s controversy with P´olya on the proof of the central limit theorem, which took place at about the same time as his discussion with Hausdorff . . . showed that von Mises did not have full technical command of theorems on sequences of monotonic functions.” Siegmund-Schultze discusses this controversy in his article “Probability in 1919/20: the von Mises-P´olya Controversy,” Archive for History of Exact Sciences 60:5 (September 2006) 431–515. According to the author, Hausdorff had a pure mathematician’s misunderstanding of von Mises’s intent, and, “Generally speaking, Hausdorff seems to have taken von Mises too literally by his words (such as ‘measure function’) . . . ” As a result of Hausdorff’s second letter, von Mises published a “correction” in 1920, that dropped VOL. 42, NO. 1, JANUARY 2011 THE COLLEGE MATHEMATICS JOURNAL
83
Integre Technical Publishing Co., Inc.
College Mathematics Journal 42:1
November 17, 2010
2:50 p.m.
media.tex
his claim of σ -additivity. But von Mises stuck to his two postulates for probability theory, even as the measure-theoretic view gained prevalence in the 1920s and 1930s. This view culminated in 1933 with Kolmorgoroff’s booklet axiomatizing probability theory based on abstract measure theory. John von Neumann originally adopted von Mises’s approach to probability in his quantum mechanics, but “von Mises’s empiricist standpoint with respect to probability theory did not go down well with the G¨ottingen mathematicians, among them famously Hermann Weyl.” It was only late in his life, in the 1950s, that von Mises admitted the failure of his frequentist attempt to rigorously base probability theory on sequences instead of sets. This provides an example of the complicated, historical relation between “real world probability” and “mathematical probability.” For more information about this relation, the author refers to J. L. Doob’s article “The Development of Rigor in Mathematical Probability, (1900–1950),” pp 157–170 in Development of Mathematics 1900–1950, Pier, J. P. (Ed.), Birkh¨auser, Basel, Boston, Berlin, 1994. PR Media Correspondents KHG Kris H. Green; RNG Raymond N. Greenwell; TL Tanya Leise; PR Peter Ross; CR Cecil Rousseau; HJR Henry J. Ricardo; A&JS Annie and John Selden; NS Norton Starr; CS Chris Storm; PDS Philip D. Straffin; NT Nicholas Tran.
Editor’s Note. In July 2010, Rokicki, Kociemba, Morley Davidson, and John Dethridge used the methods described in “Twenty-Two Moves Suffice for Rubik’s Cube,” (September 2010 Media Highlights) to prove that twenty moves suffice, thereby solving the problem completely. See http://www.cube20.org.
84
© THE MATHEMATICAL ASSOCIATION OF AMERICA
page 84
. . . for all the petty annoyances resulting from society’s impatience with math and science, being a mathematician has some considerable compensating advantages. . . . there is the deference I am given when the conversation turns to topics of math and science (which it often does when I am in the room). That’s rather pleasant. Social conventions being what they are, it is quite rare that my opinion on number-related questions is challenged. Unless, that is, we are discussing the Monty Hall problem. —from The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brainteaser by Jason Rosenhouse [p. 2] Reviewed in this issue on page 71
Mathematical Jeopardy? During a recent airing of that ever-popular TV show Mathematical Jeopardy, contestants had to define the following terms: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m)
a cancellation law; a cyclic group; a discontinuous function; indirect proof; an integral domain; an irreducible element; max-?ow, min-cut; a mean value; a non-homogeneous case; a planar graph; prime factors; quadratic residues; a recurrence relation.
Match the correct answers above with the appropriate questions below. What is a biker gang? What is circumstantial evidence? What is a country which has repelled all invasions? What is a flight plan? What are health and wealth? What is hooking up with an old flame? What is an interrupted party? What is the last stronghold? What is a loose blouse and a short skirt? What is an odd ball? What is a refund policy? What are semi-finalists? What is viciousness? —Andy Liu, The University of Alberta (For more Mathematical Jeopardy see Word Ways)
. . . I discovered that the professional literature on the problem was far more vast than I had ever imagined. Even more so than mathematicians, cognitive scientists and psychologists have been making a good living trying to explain why people find this particular problem so maddeningly difficult. Philosophers have used it directly in exploring the nature of probability, and indirectly in shedding light on seemingly unrelated philosophical questions. Economists have made their own contributions to the literature, studying what the Monty Hall problem tells us about human decision making. This barely scratches the surface. —from The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brainteaser by Jason Rosenhouse [p. x] Reviewed in this issue on page 71
101. Mr. Jeavons said that I liked maths because it was safe. He said I liked maths because it meant solving problems and these problems were difficult and interesting but there was always a straightforward answer at the end. And what he meant was that maths wasn’t like life because in life there are no straightforward answers at the end. I know he meant this because this is what he said. This is because Mr. Jeavons doesn’t understand numbers. Here is a famous story called The Monty Hall Problem which I have included in this book because it illustrates what I mean. There used to be a column called Ask Marilyn . . . ... ... ... ... And this shows that intuition can sometimes get things wrong. And intuition is what people use in life to make decisions. But logic can help you work out the right answer. It also shows that Mr. Jeavons was wrong and numbers are sometimes very complicated and not very straightforward at all. And what is why I like The Monty Hall Problem. —from The Curious Incident of the Dog in the Night-time by Mark Haddon [pp. 61–65], For more about the Monty Hall problem see page 71.
Flaws, Fallacies, and Flimflam: Who’s Right? The following items appeared in the Atlantic Monthly for March, 2010 in the Letters Column (page 17). Is either one of these contributions correct? Misleading Math? Megan McArdle attempts to illustrate her point that survey respondents are unreliable (“Misleading Indicator,” November Atlantic) by telling us that it is mathematically impossible for men to report an average number of female sexual partners that is much higher than the average number of male partners reported by women. I agree that survey respondents are unreliable, but so is McArdle’s math. It may be unlikely that the number of partners reported by honest males would be higher, but it is not mathematically impossible. —Fred Graf, Concord, NH Megan McArdle replies: We are talking about two different meanings of the word average. True, in Harvey Levin’s scenario, it is more common for women to be monogamous than men. But I was using average to describe the mean number of sexual partners, which is to say, the sum of everyone’s number of partners, divided by the number of people. If there are 10 men and 10 women, and one of the women has slept with all 10 men, while the other women are monogamous, the average number of sexual partners is the same for the men and the women. Nine of the men have had two sexual partners, while one of the men has had one partner, for an average of 19/10 = 1.9. Meanwhile, one of the women has had 10 sexual partners while the other nine women have had one partner, for an average of 19/10 = 1.9. The numbers come out the same, no matter how you vary the particulars. Yet surveys can generate differences of three- or fourfold between the mean numbers of sexual partners that women and men report. —Ed Barbeau (
[email protected]), University of Toronto
A Jungian Nightmare School came to bore me. It took up far too much time which I would rather have spent drawing battles and playing with fire. Divinity classes were unspeakably dull, and I felt a downright fear of the mathematics class. The teacher pretended that algebra was a perfectly natural affair, to be taken for granted, whereas I didn’t even know what numbers really were. They were not flowers, not animals, not fossils; they were nothing that could be imagined, mere quantities that resulted from counting. To my confusion these quantities were now represented by letters, which signified sounds, so that it became possible to hear them, so to speak. Oddly enough, my classmates could handle these things and found them self-evident. No one could tell me what numbers were, and I was unable even to formulate the question. To my horror I found that no one understood my difficulty. The teacher, I must admit, went to great lengths to explain to me the purpose of this curious operation of translating understandable quantities into sounds. I finally grasped that what was aimed at was a kind of system of abbreviation, with the help of which many quantities could be put in a short formula. But this did not interest me in the least. I thought the whole business was entirely arbitrary. Why should numbers be expressed by sounds? One might just as well express a by apple tree, b by box, and x by a question mark. a, b, c, x, y, z were not concrete and did not explain to me anything about the essence of numbers, any more than an apple tree did. But the thing that exasperated me most of all was the proposition: If a = b and b = c, then a = c, even though by definition a meant something other than b, and, being different, could therefore not be equated with b, let alone with c. Whenever it was a question of an equivalence, then it was said that a = a, b = b, and so on. This I could accept, whereas a = b seemed to me a downright lie or a fraud. I was equally outraged when the teacher stated, in the teeth of his own definition of parallel lines, that they met at infinity. This seemed to me no better than a stupid trick to catch peasants with, and I could not and would not have anything to do with it. My intellectual morality
fought against these whimsical inconsistencies, which have forever debarred me from understanding mathematics. Right into old age I have had the incorrigible feeling that if, like my schoolmates, I could have accepted without a struggle the proposition that a = b, or that sun = moon, dog = cat, then mathematics might have fooled me endlessly—just how much I only began to realize at the age of eighty-four. All my life it remained a puzzle to me why it was that I never managed to get my bearings in mathematics when there was no doubt whatever that I could calculate properly. Least of all did I understand my own doubts concerning mathematics. Equations I could comprehend only by inserting specific numerical values in place of the letters and verifying the meaning of the operation by actual calculation. As we went on in mathematics I was able to get along, more or less, by copying out algebraic formulas whose meaning I did not understand, and by memorizing where a particular combination of letters had stood on the blackboard. I could no longer make headway by substituting numbers, for from time to time the teacher would say, “Here we put the expression so-and-so,” and then he would scribble a few letters on the blackboard. I had no idea where he got them and why he did it—the only reason I could see was that it enabled him to bring the procedure to what he felt was a satisfactory conclusion. I was so intimidated by my incomprehension that I did not dare to ask any questions. Mathematics classes became sheer terror and torture to me. —from Memories, Dreams, Reflections by C. G. Jung. Recorded and edited by Aniela Jaffe. Translated from the German by Richard and Clara Winston —suggested by Steven C. Althoen
The ideal audience? A theatre full of mathematicians, philosophers and scientists. Ask any conjuror. The intelligent—the more rational—a spectator, the more readily he will be deceived. Because what you are doing in magic is creating the semblance of a chain of cause and effect. I do this, then his happens: I do that, then that happens. Basic logic. And a logical mind, receptive to a connection between each apparent cause and its apparent effect, is more prone to surprise when an illusion reaches its ‘illogical’ climax. —from The Houdini Girl by Martyn Bedford
It wasnt the best of times; it wasnt the worst of times; it was the times youd get if you arranged all possible times (including even fictional times in which the nights were usually dark and stormy) in order from worst to best on the real number line from 0.0 inclusive to 1.0 exclusive and then used a really good uniform random number generator to pick a value in that range thus choosing the corresponding times–thats the times it was. —from A Tale of Two Statisticians by Dale Dellutri
New from the MAA A Historian Looks Back The Calculus as Algebra and Selected Writings Judith V. Grabiner Judith Grabiner, the author of A Historian Looks Back, has long been interested in investigating what mathematicians actually do, and how mathematics actually has developed. She addresses the results of her investigations not principally to other historians, but to mathematicians and teachers of mathematics. This book brings together much of what she has had to say to this audience. The centerpiece of the book is The Calculus as Algebra: J.-L. Lagrange, 17361813. The book describes the achievements, setbacks, and influence of Lagrange’s pioneering attempt to reduce the calculus to algebra. Nine additional articles round out the book describing the history of the derivative; the origin of delta-epsilon proofs; Descartes and problem solving; the contrast between the calculus of Newton and Maclaurin, and that of Lagrange; Maclaurin’s way of doing mathematics and science and his surprisingly important influence; some widely held “myths” about the history of mathematics; Lagrange’s attempt to prove Euclid’s parallel postulate; and the central role that mathematics has played throughout the history of western civilization. The development of mathematics cannot be programmed or predicted. Still, seeing how ideas have been formed over time and what the difficulties were can help teachers find new ways to explain mathematics. Appreciating its cultural background can humanize mathematics for students. And famous mathematicians’ struggles and successes should interest—and perhaps inspire—researchers. Readers will see not only what the mathematical past was like, but also how important parts of the mathematical present came to be.
To order visit us online at www.maa.org or call us at 1-800-331-1622. 282 pp., Hardbound, 2010 List: $62.95
Catalog Code: CAGH MAA Member: $49.95
ISBN 978-0-88385-572-0
MATHEMATICAL ASSOCIATION OF AMERICA
1529 Eighteenth St., NW • Washington, DC 20036
CONTENTS 2–56
ARTICLES
2
Chutes and Ladders for the Impatient, By Leslie A. Cheteyan, Stewart Hengeveld, and Michael A. Jones
9
Probability 1/e, By Reginald Koo and Martin L. Jones
15
The Band Around a Convex Body, By David Swanson
25
Two-Person Pie-Cutting: The Fairest Cuts, By Julius B. Barbanel and Steven J. Brams
33
Augustus De Morgan Behind the Scenes, By Charlotte Simmons
40
Teaching Tip: Actuarial Science and Gompertz’s Law of Mortality, By Jesse Byrne
43
Computing Determinants by Double-Crossing, By Deanna Leggett, John Perry, and Eve Torrence
54
Teaching Tip: Correcting Cramer’s Rule, By Vagarshak Vardanyan
56
Boundary Conditions, By Ursula Whitcher
57–61
CLASSROOM CAPSULES
57
An Elementary Treatment of General Inner Products, By Jack E. Graver
60
Cantor Groups, By Ben Mathes
62–70
PROBLEMS AND SOLUTIONS
71–75
BOOK REVIEW
71
76–84
The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brain Teaser by Jason Rosenhouse Reviewed by Edward J. Barbeau
MEDIA HIGHLIGHTS
