THE AMERICAN MATHEMATICAL MONTHLY

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THE AMERICAN MATHEMATICAL MONTHLY VOLUME 119, NO JANUARY 2012 A Letter from the Editor Scott Chapman Invariant Histograms Daniel Brinkman and Peter J Olver Zariski Decomposition: A New (Old) Chapter of Linear Algebra 25 Thomas Bauer, Mirel Caib˘r, and Gary Kennedy a Another Way to Sum a Series: Generating Functions, Euler, and the Dilog Function 42 Dan Kalman and Mark McKinzie NOTES A Class of Periodic Continued Radicals 52 Costas J Efthimiou A Geometric Interpretation of Pascal’s Formula for Sums of Powers of Integers 58 Parames Laosinchai and Bhinyo Panijpan Covering Numbers in Linear Algebra 65 Pete L Clark PROBLEMS AND SOLUTIONS 68 REVIEWS An Introduction to the Mathematics of Money By David Lovelock, Marilou Mendel, and A Larry Wright Alan Durfee An Official Publication of the Mathematical Association of America 76 New title from the MAA! Rediscovering Mathematics: You Do the Math Shai Simonson Rediscovering Mathematics is an eclectic collection of mathematical topics and puzzles aimed at talented youngsters and inquisitive adults who want to expand their view of mathematics By focusing on problem solving, and discouraging rote memorization, the book shows how to learn and teach mathematics through investigation, experimentation, and discovery Rediscovering Mathematics is also an excellent text for training math teachers at all levels Topics range in difficulty and cover a wide range of historical periods, with some examples demonstrating how to uncover mathematics in everyday life, including: • • • number theory and its application to secure communication over the Internet, the algebraic and combinatorial work of a medieval mathematician Rabbi, and applications of probability to sports, casinos, and everyday life Rediscovering Mathematics provides a fresh view of mathematics for those who already like the subject, and offers a second chance for those who think they don’t To order call 1-800-331-1622 or visit us online at www.maa.org! THE AMERICAN MATHEMATICAL MONTHLY Volume 119, No January 2012 EDITOR Scott T Chapman Sam Houston State University NOTES EDITOR Sergei Tabachnikov Pennsylvania State University Douglas B West University of Illinois BOOK REVIEW EDITOR Jeffrey Nunemacher Ohio Wesleyan University PROBLEM SECTION EDITORS Gerald Edgar Ohio State University Doug Hensley Texas A&M University ASSOCIATE EDITORS William Adkins Louisiana State University David Aldous University of California, Berkeley Elizabeth Allman University of Alaska, Fairbanks Jonathan M Borwein University of Newcastle Jason Boynton North Dakota State University Edward B Burger Williams College Minerva Cordero-Epperson University of Texas, Arlington Beverly Diamond College of Charleston Allan Donsig University of Nebraska, Lincoln Michael Dorff Brigham Young University Daniela Ferrero Texas State University Luis David Garcia-Puente Sam Houston State University Sidney Graham Central Michigan University Tara Holm Cornell University Roger A Horn University of Utah Lea Jenkins Clemson University Daniel Krashen University of Georgia Ulrich Krause Universităt Bremen a Jeffrey Lawson Western Carolina University C Dwight Lahr Dartmouth College Susan Loepp Williams College Irina Mitrea Temple University Bruce P Palka National Science Foundation Vadim Ponomarenko San Diego State University Catherine A Roberts College of the Holy Cross Rachel Roberts Washington University, St Louis Ivelisse M Rubio Universidad de Puerto Rico, Rio Piedras Adriana Salerno Bates College Edward Scheinerman Johns Hopkins University Susan G Staples Texas Christian University Dennis Stowe Idaho State University Daniel Ullman George Washington University Daniel Velleman Amherst College EDITORIAL ASSISTANT Bonnie K Ponce NOTICE TO AUTHORS The MONTHLY publishes articles, as well as notes and other features, about mathematics and the profession Its readers span a broad spectrum of mathematical interests, and include professional mathematicians as well as students of mathematics at all collegiate levels Authors are invited to submit articles and notes that bring interesting mathematical ideas to a wide audience of MONTHLY readers The MONTHLY’s readers expect a high standard of exposition; 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it may be searched online in a variety of ways for any specified keyword(s) MAA members whose institutions not provide JSTOR access may obtain individual access for a modest annual fee; call 800-3311622 See the MONTHLY section of MAA Online for current information such as contents of issues and descriptive summaries of forthcoming articles: http://www.maa.org/ Proposed problems or solutions should be sent to: DOUG HENSLEY, MONTHLY Problems Department of Mathematics Texas A&M University 3368 TAMU College Station, TX 77843-3368 In lieu of duplicate hardcopy, authors may submit pdfs to monthlyproblems@math.tamu.edu Advertising Correspondence: MAA Advertising 1529 Eighteenth St NW Washington DC 20036 Phone: (877) 622-2373 E-mail: tmarmor@maa.org Further advertising information can be found online at www.maa.org Change of address, missing issue inquiries, and other subscription correspondence: MAA Service Center, maahq@maa.org All at the address: The Mathematical Association of America 1529 Eighteenth Street, N.W Washington, DC 20036 Recent copies of the MONTHLY are available for purchase through the MAA Service Center maahq@maa.org, 1-800-331-1622 Microfilm Editions: University Microfilms International, Serial Bid coordinator, 300 North Zeeb Road, Ann Arbor, MI 48106 The AMERICAN MATHEMATICAL MONTHLY (ISSN 0002-9890) is published monthly except bimonthly June-July and August-September by the Mathematical Association of America at 1529 Eighteenth Street, N.W., Washington, DC 20036 and Lancaster, PA, and copyrighted by the Mathematical Association of America (Incorporated), 2012, including rights to this journal issue as a whole and, except where otherwise noted, rights to each individual contribution 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 2012 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 the American Mathematical Monthly, Membership/Subscription Department, MAA, 1529 Eighteenth Street, N.W., Washington, DC, 20036-1385 A Letter from the Editor Scott Chapman Time is always marching forward We have again reached the quinquennial changing of the guard at the Monthly It was a great pleasure for me to serve during 2011 as the Monthly’s Editor-Elect This marks my first of 50 issues as Editor, and I wish to use this opportunity to take a look both backwards and forwards The Monthly, its readers, and the MAA owe a huge debt of gratitude to my predecessor, Professor Daniel J Velleman of Amherst College Few people understand the daunting task of managing this publication The Monthly receives somewhere between 800 and 1000 submissions annually, and we are able to publish less than 10% of the manuscripts we receive The process of juggling submissions, Associate Editors’ reports, requests for referees, referees’ reports, and sometimes multiple revisions, can be mind boggling As many Monthly authors would profess, Dan was a master at whipping an accepted article into publishable shape While many authors were frustrated by being asked to revise a paper as many as three times after it was accepted, the end result was unmistakably an article of the highest expository quality I thank Dan for his unending help during the transition and am glad he has agreed to remain on the Editorial Board for another 5-year term The Monthly remains the world’s most-read mathematics journal; this is in no small part due to the efforts of Dan during his Editorship It would be a terrible omission for me to not also thank Dan’s Editorial Assistant of years, Nancy Board We wish her well as she heads to New Mexico to begin her well-deserved retirement In the ever-changing world of academic publication, the future of the Monthly remains bright The beginning of my term as Editor-Elect saw the adoption by the Monthly of the Editorial Manager System for manuscript management I thank the authors of submitted papers and referees for their patience with the system during its initial months of operation While isolated problems and glitches occur, we are confident that Editorial Manager has allowed us improve our administrative function As the year unfolds, I not think you will notice many changes in the Monthly, but as with all organizations, there will be some The incoming Editorial Board consists of 38 members Of these, 19 are new appointments Representation on the Board by members of most underrepresented groups has increased drastically Moreover, the number of Board members with expertise in Applied Mathematics has increased from to The members of the Board hail from 22 states (including the District of Columbia and Alaska), Puerto Rico, Australia, and Germany Of note in the group are Sergei Tabachnikov (Notes Editor), Jeffery Nunemacher (Book Reviews Editor), and Doug Hensley, Gerald Edgar, and Doug West (Problems Editors) I wish to thank Sam Houston State University, most notably Provost Jaime Hebert, for providing the funds to renovate a set of offices we will use over the next years Bonnie Ponce has joined our staff as my Editorial Assistant Please not hesitate to contact us at monthly@shsu.edu when you have questions or concerns I look forward to serving you over the next years http://dx.doi.org/10.4169/amer.math.monthly.119.01.003 January 2012] A LETTER FROM THE EDITOR Invariant Histograms Daniel Brinkman and Peter J Olver Abstract We introduce and study a Euclidean-invariant distance histogram function for curves For a sufficiently regular plane curve, we prove that the cumulative distance histograms based on discretizing the curve by either uniformly spaced or randomly chosen sample points converge to our histogram function We argue that the histogram function serves as a simple, noise-resistant shape classifier for regular curves under the Euclidean group of rigid motions Extensions of the underlying ideas to higher-dimensional submanifolds, as well as to area histogram functions invariant under the group of planar area-preserving affine transformations, are discussed INTRODUCTION Given a finite set of points contained in R n , equipped with the usual Euclidean metric, consider the histogram formed by the mutual distances between all distinct pairs of points An interesting question, first studied in depth by Boutin and Kemper [4, 5], is to what extent the distance histogram uniquely determines the point set Clearly, if the point set is subjected to a rigid motion—a combination of translations, rotations, and reflections—the interpoint distances will not change, and so two rigidly equivalent finite point sets have identical distance histograms However, there exist sets that have identical histograms but are not rigidly equivalent (The reader new to the subject may enjoy trying to find an example before proceeding further.) Nevertheless, Boutin and Kemper proved that, in a wide range of situations, the set of such counterexamples is “small”—more precisely, it forms an algebraic subvariety of lower dimension in the space of all point configurations Thus, one can say that, generally, the distance histogram uniquely determines a finite point set up to rigid equivalence This motivates the use of the distance histogram as a simple, robust, noise-resistant signature that can be used to distinguish most rigidly inequivalent finite point sets, particularly those that arise as landmark points on an object in a digital image The goal of this paper is to develop a comparable distance histogram function for continua—specifically curves, surfaces, and higher-dimensional submanifolds of Euclidean spaces Most of the paper, including all proofs, will concentrate on the simplest scenario: a “regular” bounded plane curve Regularity, as defined below, does allow corners, and so, in particular, includes polygons We will approach this problem using the following strategy We first sample the curve using a finite number of points, and then compute the distance histogram of the sampled point set Stated loosely, our main result is that, as the curve becomes more and more densely sampled, the appropriately scaled cumulative distance histograms converge to an explicit function that we name the global curve distance histogram function Alternatively, computing the histogram of distances from a fixed point on the curve to the sample points leads, in the limit, to a local curve distance histogram function, from which the global version can be obtained by averaging over the curve Convergence of both local and global histograms is rigorously established, first for uniformly sampled points separated by a common arc length distance, and then for points randomly sampled with respect to the uniform arc length distribution http://dx.doi.org/10.4169/amer.math.monthly.119.01.004 MSC: Primary 53A04, Secondary 68U10 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 The global curve distance histogram function can be computed directly through an explicit arc length integral By construction, it is invariant under rigid motions Hence, a basic question arises: does the histogram function uniquely determine the curve up to rigid motion? While there is ample evidence that, under suitably mild hypotheses, such a result is true, we have been unable to establish a complete proof, and so must state it as an open conjecture A proof would imply that the global curve histogram function, as approximated by its sampled point histograms, can be unambiguously employed as an elementary, readily computed classifier for distinguishing shapes in digital images, and thus serve as a much simpler alternative to the joint invariant signatures proposed in [15] Extensions of these ideas to subsets of higher-dimensional Euclidean spaces, or even general metric spaces, are immediate Moreover, convergence in sufficiently regular situations can be established along the same lines as the planar curve case treated here Following Boutin and Kemper [4], we also consider area histograms formed by triangles whose corners lie in a finite point set In two dimensions, area histograms are invariant under the group of equi-affine (meaning area-preserving affine) transformations We exhibit a limiting area histogram function for plane curves that is also equi-affine invariant, and propose a similar conjecture Generalizations to other transformation groups, e.g., similarity, projective, conformal, etc., of interest in image processing and elsewhere [9, 16], are worth developing The corresponding discrete histograms will be based on suitable joint invariants—for example, area and volume cross ratios in the projective case—which can be systematically classified by the equivariant method of moving frames [15] Analysis of the corresponding limiting histograms will be pursued elsewhere Our study of invariant histogram functions has been motivated in large part by the potential applications to object recognition, shape classification, and geometric modeling Discrete histograms appear in a broad range of powerful image processing algorithms: shape representation and classification [1, 23], image enhancement [21, 23], the scale-invariant feature transform (SIFT) [10, 18], object-based query methods [22], and as integral invariants [11, 19] They provide lower bounds for and hence establish stability of Gromov–Hausdorff and Gromov–Wasserstein distances, underlying an emerging new approach to shape theory [12, 13] Local distance histograms underly the method of shape contexts [2] The method of shape distributions [17] for distinguishing three-dimensional objects relies on a variety of invariant histograms, including local and global distance histograms, based on the fact that objects with different Euclidean-invariant histograms cannot be rigidly equivalent; the converse, however, was not addressed Indeed, there are strong indications that the distance histogram alone is insufficient to distinguish surfaces, although we not know explicit examples of rigidly inequivalent surfaces that have identical distance histograms DISTANCE HISTOGRAMS Let us first review the results of Boutin and Kemper [4, 5] on distance histograms defined by finite point sets For this purpose, our initial setting is a general metric space V , equipped with a distance function d(z, w) ≥ 0, for z, w ∈ V , satisfying the usual axioms Definition The distance histogram of a finite set of points P = {z , , z n } ⊂ V is the function η = η P : R+ → N defined by η(r ) = #{(i, j) | ≤ i < j ≤ n, d(z i , z j ) = r } (2.1) In this paper, we will restrict our attention to the simplest situation, when V = R m is endowed with the usual Euclidean metric, so d(z, w) = z − w We say that January 2012] INVARIANT HISTOGRAMS two subsets P, Q ⊂ V are rigidly equivalent, written P Q, if we can obtain Q by applying an isometry to P In Euclidean geometry, isometries are rigid motions: the translations, rotations, and reflections generating the Euclidean group [25] Clearly, any two rigidly equivalent finite subsets have identical distance histograms Boutin and Kemper’s main result is that the converse is, in general, false, but is true for a broad range of generic point configurations Theorem Let P (n) = P (n) (R m ) denote the space of finite (unordered) subsets P ⊂ R m of cardinality #P = n If n ≤ or n ≥ m + 2, then there is a Zariski dense open subset R(n) ⊂ P (n) with the following property: if P ∈ R(n) , then Q ∈ P (n) has the same distance histograms, η P = η Q , if and only if the two point configurations are rigidly equivalent: P Q In other words, for the indicated ranges of n, unless the points are constrained by a certain algebraic equation, and so are “nongeneric,” the distance histogram uniquely determines the point configuration up to a rigid motion Interestingly, the simplest counterexample is not provided by the corners of a regular polygon For example, the corners of a unit square have side distances of and diagonal distances of √ √ 2, and so√ distance histogram has values η(1) = 4, η( ) = 2, while η(r ) = its for r = 1, Moreover, this is the only possible way to arrange four points with the given distance histogram A simple nongeneric configuration is provided by the corners of the kite and trapezoid quadrilaterals shown in Figure Although clearly not rigidly equivalent, both point configurations have the same distance histogram, √ √ with nonzero values η( 2) = 2, η(2) = 1, η( 10 ) = 2, η(4) = A striking onedimensional counterexample, discovered in [3], is provided by the two sets of integers P = {0, 1, 4, 10, 12, 17}, Q = {0, 1, 8, 11, 13, 17} ⊂ R, which, as the reader can check, have identical distance histograms, but are clearly not rigidly equivalent √ √ 10 √ √ √ √ 10 √ 10 10 √ 2 2 Figure Kite and trapezoid To proceed, it will be more convenient to introduce the (renormalized) cumulative distance histogram P (r ) = + n n2 η P (s) = s≤r # (i, j) | d(z i , z j ) ≤ r , n2 (2.2) where n = #P We note that we can recover the usual distance histogram (2.1) via η(r ) = n P (r ) − P (r − δ) for sufficiently small δ (2.3) We further introduce a local distance histogram that counts the fraction of points in P that are within a specified distance r of a given point z ∈ R m : λ P (r, z) = c 1 # j | d(z, z j ) ≤ r = #(P ∩ Br (z)), n n THE MATHEMATICAL ASSOCIATION OF AMERICA (2.4) [Monthly 119 where Br (z) = v ∈ V | d(v, z) ≤ r (2.5) denotes the ball (in the plane, the disk) of radius r centered at the point z Observe that we recover the cumulative histogram (2.2) by averaging its localization: P (r ) = n λ P (r, z) = z∈P n2 #(P ∩ Br (z)) (2.6) z∈P In this paper, we are primarily interested in the case when the points lie on a curve Until the final section, we restrict our attention to plane curves: C ⊂ V = R A finite subset P ⊂ C will be called a set of sample points on the curve We will assume throughout that the curve C is bounded, rectifiable, and closed (Extending our results to non-closed curves is straightforward, but we will concentrate on the closed case in order to simplify the exposition.) Further mild regularity conditions will be introduced below We use z(s) to denote the arc length parametrization of C, measured from some base point z(0) ∈ C Let ds < ∞ l(C) = (2.7) C denote the curve’s length, which we always assume to be finite Our aim is to study the limiting behavior of the cumulative histograms constructed from more and more densely chosen sample points It turns out that, under reasonable assumptions, the discrete histograms converge, and the limiting function can be explicitly characterized as follows Definition Given a curve C ⊂ V , the local curve distance histogram function based at a point z ∈ V is h C (r, z) = l(C ∩ Br (z)) , l(C) (2.8) i.e., the fraction of the total length of the curve that is contributed by those parts contained within the disk of radius r centered at z The global curve distance histogram function of C is obtained by averaging the local version over the curve: HC (r ) = l(C) h C (r, z(s)) ds (2.9) C Observe that both the local and global curve distance histogram functions have been normalized to take values in the interval [0, 1] The global function (2.9) is invariant under rigid motions, and hence two curves that are rigidly equivalent have identical global histogram functions An interesting question, which we consider in some detail towards the end of the paper, is whether the global histogram function uniquely characterizes the curve up to rigid equivalence Modulo the definition of “regular,” to be presented in the following section, and details on how “randomly chosen points” are selected, provided in Section 4, our main convergence result can be stated as follows January 2012] INVARIANT HISTOGRAMS Theorem Let C be a regular plane curve Then, for both uniformly spaced and randomly chosen sample points P ⊂ C, the cumulative local and global histograms converge to their continuous counterparts: λ P (r, z) −→ h C (r, z), P (r ) −→ HC (r ), (2.10) as the number of sample points goes to infinity UNIFORMLY SPACED POINTS Our proof of Theorem begins by establishing convergence of the local histograms In this section, we work under the assumption that the sample points are uniformly spaced with respect to arc length along the curve Let us recall some basic terminology concerning plane curves, mostly taken from Guggenheimer’s book [8] We will assume throughout that C ⊂ R has a piecewise C2 arc length parametrization z(s), where s belongs to a bounded closed interval [0, L], with L = l(C) < ∞ being its overall length The curve C is always assumed to be simple, meaning that there are no self-intersections, and closed, so z(0) = z(L), and thus a Jordan curve We use t (s) = z (s) to denote the unit tangent, and1 κ(s) = z (s) ∧ z (s) the signed curvature at the point z(s) Under our assumptions, both t (s) and κ(s) have left- and right-hand limiting values at their finitely many discontinuities A point z(s) ∈ C where either the tangent or curvature is not continuous will be referred to as a corner We will often split C up into a finite number of nonoverlapping curve segments, with distinct endpoints A closed curve is called convex if it bounds a convex region in the plane A curve segment is convex if the region bounded by it and the straight line segment connecting its endpoints is a convex region A curve segment is called a spiral arc if the curvature function κ(s) is continuous, strictly monotone,2 and of one sign, i.e., either κ(s) ≥ or κ(s) ≤ Keep in mind that, by strict monotonicity, κ(s) is only allowed to vanish at one of the endpoints of the spiral arc Definition A plane curve is called regular if it is piecewise C2 and the union of a finite number of convex spiral arcs, circular arcs, and straight lines Thus, any regular curve has only finitely many corners, finitely many inflection points, where the curvature has an isolated zero, and finitely many vertices, meaning points where the curvature has a local maximum or minimum, but is not locally constant In particular, polygons are regular, as are piecewise circular curves, also known as biarcs [14] (But keep in mind that our terminological convention is that polygons and biarcs have corners, not vertices!) Examples of irregular curves include the graph of the infinitely oscillating function y = x sin 1/x near x = 0, and the nonconvex spiral arc r = e− θ for ≤ θ < ∞, expressed in polar coordinates Theorem If C is a regular plane curve, then there is a positive integer m C such that the curve’s intersection with any disk having center z ∈ C and radius r > 0, namely C ∩ Br (z), consists of at most m C connected segments The minimal value of m C will be called the circular index of C The symbol ∧ denotes the two-dimensional cross product, which is the scalar v ∧ w = v w − v w for 2 v = (v1 , v2 ), w = (w1 , w2 ) Guggenheimer [8] only requires monotonicity, allowing spiral arcs to contain circular subarcs, which we exclude Our subsequent definition of regularity includes curves containing finitely many circular arcs and straight line segments c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 r s dr dsdθdφ = |E| = E π/2 = 4π 0 π π −π −π 1 − s2 s dθ dφds − s sin2 (θ − φ) 1−s s dθ ds, − s sin2 θ by periodicity and symmetry Note that < − s ≤ − s sin2 (θ − φ) in the interior of E , so the integrand reflects the bound r ≤ To perform the integration over θ , substitute u = s tan θ with s held constant, so that dθ = s du/(s + u ) and sin2 θ = u /(s + u ) For < s < 1, this yields π/2 ∞ − s2 s dθ = − s sin2 θ 0 s (1 − s ) du πs = s + (1 − s )u 2 − s2 Therefore, the required volume is |E| = 2π s − s ds = 2 π Also solved by N Caro (Brazil), R Chapman (U K.), W J Cowieson, E A Herman, O Kouba (Syria), J H Lindsey II, W Nuij (Netherlands), P Perfetti (Italy), J Simons (U K.), W Song, R Stong, M Wildon (U K.), L Zhou, Barclays Capital Quantitative Analytics Group (U K.), and the proposer The Short Vector Problem 11524 [2010, 741] Proposed by H A ShahAli, Tehran, Iran A vector v in Rn is short if v ≤ (a) Given six short vectors in R2 that sum to zero, show that some three of them have a short sum (b)∗ Let f (n) be the least M such that, for any finite set T of short vectors in Rn that sum to 0, and any integer k with ≤ k ≤ |T |, there is a k-element subset S of T such that v∈S v ≤ M The result of part (a) suggests f (2) = Find f (n) for n ≥ Solution by the proposer We need a preliminary result Lemma Given a collection of two or more short vectors in R2 that sum to zero, some two of them have a short sum Proof If one of the vectors is zero, then together with any other we have a short sum and are done Now assume we have m vectors, all nonzero We show that the angle θ between some two of them is at least 2π/3, so the cosine of their angle is at most −1/2 and their sum is short If the two vectors are u and w, with u ≤ w , then u+w = u = w 2 + w + u + cos θ u u − w w ≤ u ≤ w 2 + w − u w ≤ Write v1 , , vm for the given vectors, with numbering to be determined We may rotate coordinates so that one of the vectors, v1 , lies on the positive x-axis; let each vector v j make angle θ j with the positive x-axis Thus ≤ θ j < 2π, with θ1 = If all vectors lie on the x-axis, then (since their sum is 0) one of them (say v2 ) lies on the negative x-axis, so v1 + v2 is short, and we are done Now assume not all the vectors are on the x-axis Because the sum is 0, at least one vector is in the upper half plane Among these, let v2 be one with the largest angle Thus < θ2 < π If θ2 ≥ 2π/3, then v1 + v2 is short and we are done, so we may assume < θ2 < 2π/3 72 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 The sum of all the vectors is zero, and one of them, v1 , is on one side of the line through and v2 , so another of them, say v3 , is on the other side of that line, hence θ2 < θ3 < θ2 + π Now θ2 < θ3 < π is ruled out by the maximal choice for θ2 If 2π/3 ≤ θ3 ≤ 4π/3, then v1 + v3 is short and we are done Thus we have θ3 > 4π/3, so θ3 − θ2 < π and θ3 − θ2 > 4π/3 − 2π/3 = 2π/3, and so v2 + v3 is short (a) If one of the vectors is 0, then apply the Lemma to the remaining vectors to get two with short sum; with the zero vector we then have three with short sum Now assume the given vectors are all nonzero We write v1 , , v6 for the given vectors, with numbering to be determined Apply the Lemma to {v1 , , v6 } to conclude that some two have short sum, say v1 and v2 Now apply the Lemma to {v1 + v2 , v3 , v4 , v5 , v6 } to conclude that some two have short sum If v1 + v2 + v j is short for some j with ≤ j ≤ 6, we are done Therefore, we may assume some two from {v3 , , v6 } have short sum, say v3 and v4 Now apply the Lemma to {v1 + v2 , v3 + v4 , v5 , v6 } to conclude that some two have short sum If that choice of two is one of v1 + v2 , v3 + v4 and one of v5 , v6 , then we are done If v1 + v2 + v3 + v4 is short, then so is v5 + v6 So we may assume v5 + v6 is short Now among the three short vectors u 12 , u 34 , and u 56 given by u 12 = v1 + v2 , u 34 = v3 + v4 , and u 56 = v5 + v6 , there are two such that the angle θ between them satisfies θ ≤ 2π/3, since the entire circle has circumference 2π We consider two cases: First assume θ > π/3 We rotate again to put the u jk ’s into the x-y plane, and, taking θ jk to be the signed angle of u jk with the positive x-axis,where −π ≤ θ jk < π , we spin them so that θ12 = −θ/2 and θ34 = θ/2 The sum is 0, and u 12 and u 34 are both in the right half-plane, while u 56 is in the left half-plane Thus at least one of v5 and v6 is in the left half-plane Say v5 is in the left half-plane If v5 is in the upper half-plane, then the angle between u 12 and v5 is greater than 2π − θ/2 − π , which is more than 2π/3, and hence u 12 + v5 is short Similarly, if v5 is in the lower half-plane, then u 34 + v5 is short The other case is θ ≤ π/3, so in particular θ < π/2 and the dot product u 12 · u 34 is positive Compute v5 + u 12 + v5 + u 34 + v6 + u 12 + v6 + u 34 = v5 + v6 + u 12 + u 34 + (v5 + v6 ) · (u 12 + u 34 ) = v5 + v6 + u 12 + u 34 + u 56 · (u 12 + u 34 ) = v5 + v6 + u 12 + u 34 − (u 12 + u 34 ) · (u 12 + u 34 ) = v5 + v6 − u 12 · u 34 ≤ Therefore at least one vector in {v5 + u 12 , v5 + u 34 , v6 + u 12 , v6 + u 34 } is short Editorial comment The other solution for (a) also involves taking cases No solution for (b) was received The proposer conjectures that f (n) = − max{2,n} , which is achieved when k = and the vectors are the vertices of the regular simplex centered at the origin Part (a) also solved by Barclays Capital Quantitative Analytics Group (U K.) Plane Geometric Arrangements 11525 [2010, 741] Proposed by Grigory Galperin, Eastern Illinois University, Charleston, IL, and Yury Ionin, Central Michigan University, Mount Pleasant, MI January 2012] PROBLEMS AND SOLUTIONS 73 (a) Prove that for each n ≥ there is a set of regular n-gons in the plane such that every line contains a side of exactly one polygon from this set (b) Is there a set of circles in the plane such that every line in the plane is tangent to exactly one circle from the set? (c) Is there a set of circles in the plane such that every line in the plane is tangent to exactly two circles from the set? (d) Is there a set of circles in the plane such that every line in the plane is tangent to exactly three circles from the set? Composite Solution by Jim Simons, Cheltenham, U K., and Barclays Capital Quantitative Analytics Group, London, U K (a) For odd n, consider any regular n-gon W Choose a direction d that is not parallel to any side of W , and consider the set of all translations of W in the direction d Clearly, if a line l is parallel to a side of W , then it is a side of exactly one of these n-gons Now take the n-gons described above and rotate each of them clockwise about the origin by all possible angles in the range [0, 2π ) Every line l can be made parallel to n a side of W by rotating it counterclockwise around the origin by exactly one angle in this range Thus, every line is the side of exactly one of these n-gons For even n, the construction above does not quite work Since opposite sides of W are parallel, the construction above would produce a set of polygons with every line containing a side of two of them We modify the construction by choosing the direction d more carefully and allowing only certain translates Suppose the initial polygon has width w between two parallel sides and the direction d makes an angle α with these sides The two polygons that differ by a translation in the d direction through a distance w csc α have sides that lie on the same line Suppose we choose a set X ⊂ R of translation amounts such that for all x exactly one of {x, x + w csc α} is in X Every line parallel to this pair of sides will contain a side of exactly one translate, as required Let m = n/2 The m pairs of parallel sides all have the same width w and make angles of α + 2π k/n with d, where ≤ k ≤ m − Thus we need only show that there is a subset X ⊂ R such that for every x and every k exactly one of {x, x + w csc(α + 2πk/n)} lies in X For ≤ k ≤ m − 1, let Dk = w csc(α + 2π k/n) If these Dk are linearly dependent over Q, then there are integers rk , not all zero, such that m−1 rk csc(α + 2π k/n) = The left side of this equation written in terms of k=0 complex exponentials is a rational function of eiα If rk = 0, then this rational function has a pole at eiα = e−2πik/n and hence is non-trivial Hence the equation has only finitely many solutions Thus there are only countably many α for which the Dk are linearly dependent over Q Choose the direction d so that the distances Dk are linearly independent over Q Choose a Hamel basis of R containing these distances, and for a ∈ R let pk (a) be the coefficient of Dk in the expansion of a in this basis Now one can take m−1 pk (a) ≡ (mod 2) X= a: k=0 as the set of translations This set has the required property (b) There is no such set of circles If two circles are not nested, then there is a line tangent to both Therefore in any such set the circles would have to be nested, totally ordered by radius The intersection of the compact circular discs defined by these circles would be a non-empty closed set F and any line intersecting F would not be tangent to any of the circles 74 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 (c, d) Using the Axiom of Choice, we will show that for any k > 1, there is a set of circles in the plane such that every line in the plane is tangent to exactly k circles from the set Let c be the cardinality of the real numbers, and let ωc be the first ordinal of this cardinality There are c lines in the plane, so they can be indexed as {lα : α < ωc } Using transfinite induction, we construct the set C of circles and simultaneously the set L of lines in the plane tangent to precisely k of the circles in C Initially C and L are empty Throughout the induction we will have no line in the plane tangent to more than k of the circles in C, and |C| < c Note that for each pair of circles there are at most lines tangent to both of them, and therefore |L| ≤ C < c For the inductive step, let α be the smallest ordinal for which lα ∈ L At any given / point of lα there are c circles tangent to lα , but fewer than c of these are already in C or are tangent to a line in L Hence we can choose a circle, not already in C, that is tangent to lα , and whose addition does not produce a line tangent to more than k circles Repeating this construction at most k times, we shall be adding lα to L At the point when we are up to step α in the induction, we have |C| ≤ k|α| < c as required for the induction step The construction ends when we reach ordinal ωc , and at this point L is all lines in the plane Editorial comment The proposers showed that (c) holds without requiring the Axiom of Choice Simply take the set to consist of all circles whose radius is an odd integer and whose center is on the unit circle We did not count a single point as a degenerate polygon or circle Also solved by GCHQ Problem Solving Group (U K.)(part b), and the proposers (parts a–c) Expanders Increase Dimension 11526 [2010, 742] Proposed by Marius Cavachi, “Ovidius” University of Constanta, Constanta, Romania Prove that there is no function f from R3 to R2 with the property that f (x) − f (y) ≥ x − y for all x, y ∈ R3 Solution by Ralph Howard, University of South Carolina, Columbia, SC When (X, d) k is a metric space and k ∈ (0, ∞), write H(X,d) for the k-dimensional Hausdorff outer measure defined on the subsets of X The Hausdorff dimension of (X, d) is k dimH (X, d) = inf{k ∈ (0, ∞) : H(X,d) (X ) = 0} Recall that, with the usual metric, dimH (Rn ) = n If (X, d X ) and (Y, dY ) are metric spaces, call a map f : X → Y an expanding map if dY ( f (x1 ), f (x2 )) ≥ d X (x1 , x2 ) for all x1 , x2 ∈ X Such a map need not be continuous, but it is clearly injective It suffices to prove the following: if there is an expanding map from (X, d X ) to (Y, dY ), then dimH (X, d X ) ≤ dimH (Y, dY ) Let f : X → Y be an expanding map The image f [X ] is a subset of Y , so dimH ( f [X ]) ≤ dimH (Y ) Without loss of generality, we may replace Y by f [X ] and assume that f is surjective and thus bijective Now f has an inverse g : Y → X As f is an expanding map, g is a contraction; that is, d X (g(y1 ), g(y2 )) ≤ dY (y1 , y2 ) for all y1 , y2 ∈ Y For k > and S ⊆ Y , it follows directly from the definition of the k k Hausdorff outer measures that H(X,d X ) (g[S]) ≤ H(Y,dY ) (S) Thus, since g is surjective, we have dimH (X, d X ) ≤ dimH (Y, dY ) Also solved by N Eldredge, O Geupel (Germany), J Grivaux (France), E A Herman, O P Lossers (Netherlands), K Schilling, J Simons (U K.), R Stong, Barclays Capital Quantitative Analytics Group (U K.), and the proposer January 2012] PROBLEMS AND SOLUTIONS 75 REVIEWS Edited by Jeffrey Nunemacher Mathematics and Computer Science, Ohio Wesleyan University, Delaware, OH 43015 An Introduction to the Mathematics of Money By David Lovelock, Marilou Mendel, and A Larry Wright Springer, New York, 2007, xii + 300 pp., ISBN 978-0-387-34432-4 $79.95 Reviewed by Alan Durfee For the last few years I have been teaching an advanced course in mathematical finance, despite the fact that my field of research, the interface of topology and algebraic geometry, is on the opposite side of the mathematical spectrum I started teaching it simply because there was an upper-level course in applied mathematics listed in the catalogue, but it hadn’t been taught for a while So I decided to teach it (this is the advantage of being at a small college) I made a quick guess at what I thought would be two interesting topics, cryptology and mathematical finance We did some cryptology in the first rather large half of the course, and the students loved it In the remaining time we did some mathematical finance, in particular the pricing of options, and their reaction was less enthusiastic In retrospect I can see why; the central ideas of mathematical finance are more difficult and we didn’t have enough time to treat them properly In addition I barely knew the subject In the next few years cryptology became a lower level course, the kind that mathphobes take to satisfy a distribution requirement A MAA mini-course helped me learn mathematical finance, which stayed at the upper level I’ve now taught the course every other year for the last ten years, and each time through it I learn a bit more So it’s been working well Also it’s been fun to teach The students tend to be upper level math majors and math-oriented economics majors The last time I taught it I actually had more students than in any other advanced course in the department There are many fine textbooks for mathematical finance, a few of which are [5] and [7] Another one, which contains many practical details, is the highly regarded (and expensive) [3] (we use this in our course) The book by Lovelock, Mendel and Wright under review here is good for the more elementary topics, but I don’t recommend it for more advanced topics, which are all crammed together into the last 50 pages of this 300 page book In the other books I mentioned, they form the bulk of the material The book by Lovelock et al starts with interest rates, a basic and absolutely necessary topic Simple and compound interest are discussed in some detail, and results are often proved in two ways, using both induction and recurrence relations Continuously compounded interest is then introduced (the book uses i (∞) for this interest rate, a rather awkward notation; it’s simpler to just use r ) This is used exclusively for the more advanced topics discussed in the last chapter There are many useful homework problems in the book, both easy (“walking”) and harder (“running”) http://dx.doi.org/10.4169/amer.math.monthly.119.01.076 76 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 The book goes on to discuss inflation (and includes a table of the consumer price index for every year since 1913), taxes, annuities, loans, amortization (mortgages) and credit cards These topics are useful in understanding the basic ways in which money works I was interested to learn how credit card numbers are formed; the first digit represents the type of card, the next few digits the issuing financial institution, followed by the customer’s number and last of all a checksum digit Chapters eight and nine explain the basics of stocks and bonds Chapter ten, at a higher level, first discusses stock market indices like the Dow-Jones and the S&P 500, then goes on to portfolios and the Capital Asset Pricing Model Chapter eleven, the last chapter in the book, is about options, their basic properties and the Black-Scholes formulas for pricing them The mathematical level of this chapter is much higher than the previous ones Options and their basic properties are discussed in the first four sections of the last chapter The simplest type of options are puts and calls.1 A call on a particular stock is an agreement between two parties It gives the holder (who pays the “premium”) the right of buying on a specified future date (the “expiration date”) one share of the stock for an agreed price (the “strike price”) The person who writes the call receives the premium but is then under the obligation to sell, if asked, the stock to the other person at the strike price The holder has a “long” call, the writer a “short” one If on the expiration date the current stock price is higher than the strike price, then the holder of the call will buy the stock from the other person at the strike price and then sell it on the market at its current value, thus making a profit On the other hand, if on the expiration date the current stock price is lower than the strike price, then the holder does nothing and the writer keeps the premium, thus making a profit More specifically, if St is the price of the stock at time t, the expiration time is T and the strike price is X , then the payoff of the call is max(ST − X, 0) A put is similar except that it gives the buyer the right to sell the share of stock This is a form of insurance against the stock falling in value There are many types of options besides the standard puts and calls; for instance an Asian call has payoff determined by the average stock price between time and time T Put-call parity is proved in the next section This shows that if the value of a call is known, then the value of the corresponding put can be easily computed, and vice-versa The next four sections present the rather sophisticated mathematics behind the motion of stock prices and the value of puts and calls I found these sections rather condensed and hard to follow, so instead I will give my own summary of this theory First we need to talk about the assumptions used in basic mathematical finance Here are some of them: There are no arbitrage opportunities; there is no way to earn more on an investment than putting the money in the bank without taking a risk Another way of saying this is that any risk-free portfolio earns the same as money in the bank This assumption is absolutely fundamental, and will be illustrated below with the simple example of pricing a call using a one-step binomial tree There are no transaction costs; an item can be bought or sold at the same price Also there is just one risk-free interest rate, which is both earned on deposits and charged on loans (this assumption is rather like what happens in an elementary physics course, where one assumes that there is no friction) Actually there are two common types of options “European” options can be exercised only on a specific date, whereas “American” options can be exercised any time up to that date The results described in the book and this review are for European options They are mathematically easier to handle, and the Black-Scholes theory applies to them January 2012] REVIEWS 77 It is possible to buy or sell an arbitrary number (large or small, even not integral) of shares of a stock For example, in the one-step binomial tree described below, one may need to buy a fractional number of shares None of the above is true in the real world There are arbitrage opportunities (though they disappear quickly) if an asset is priced differently in two different markets There are always transaction costs, as can be seen, for instance, in the buy and sell rates at a currency booth in an airport And of course it is not possible to purchase an arbitrarily large number, and especially a fractional amount, of most things Next I will describe three methods for pricing simple options like puts and calls The first method uses binomial trees These are simple to describe but only produce an approximation of the exact price The second method uses geometric Brownian motion, and the third method takes place in the risk-neutral world We’ll start with the first method, and look at the simplest case, a one-step binomial tree (Figure 1) uS0 fu S0 f dS0 fd Figure A one-step binomial tree There is a single node on the left (at time zero), which opens up to two nodes on the right (at time T ) The stock has value S0 at time zero, and this value either goes up by a factor u on the top branch, or down by a factor d on the bottom one We also have an option which has value f at time zero, and whose value at time T is f u if the stock goes up, and f d if the stock goes down We would like to find f To this, we start off at time with a portfolio which contains shares of the stock and one short option.2 The value of the portfolio at time is S0 − f At time T the value of the portfolio is either u( S0 ) − f u if the stock goes up, or d( S0 ) − f d if the stock goes down Choose so the portfolio has the same value, whether it goes up or down (we are using assumption (3) here) This makes the portfolio risk-free Since we know its value at time T we now know its value at time (here we use assumption (1); all risk-free investments earn the same as money in the bank) With some computation we then find that the value of the option at time is f = e−r T p f u + (1 − p) f d (1) where p= e−r T − d u−d (2) and r is the risk-free interest rate.3 The The 78 here is a constant It is not the used in calculus, which will appear a few paragraphs below notation for most of the equations in this review is taken from [3] c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 Note that the above formula contains nothing related to a particular company This is settled later when the u and d are defined in terms of the volatility σ (see below) of the company involved; see [2] The multi-step binomial tree is a generalization of the one-step tree It starts on the left-hand side with a single node, and as one moves from the left to the right each node branches into more nodes Given a price for the stock at the left, this price increases and decreases at each node as for a one-step tree Hence at the ending time we know the value of the stock at each node on the far right Suppose we want to find the price of a call in this situation We know its payoff at the ending time, which is on the far right Then going backwards we can find the value of the call at each step as we did above, eventually ending up at the left with the value we are looking for This gives an approximate value for the option The exact value is determined by letting the number of steps in the tree go to infinity; see [2] Now let’s describe the second (more mathematically advanced) method of pricing an option We’ll start with the discrete version of the Geometric Brownian Motion model for the stock price S as time changes, which is S =µ t +σ z S (3) √ with z = t Here t is a small time interval and S is the change in the stock price during that time interval Hence S/S is the return on one share of stock On the right hand side of the equation there are two constants; µ is the growth rate of the stock and σ its volatility These will be different for different companies The variable has a standard normal distribution; we assume that these are independent for different times This adds randomness to the equation; after all, we don’t know where the stock is going The equation thus says that the percentage returns are normally distributed with mean µ t and variance σ t Do real stocks actually follow this model? Can we estimate the constants µ and σ ? It’s easy to find the daily prices of stocks, some for many years.4 For instance the price of GE stock is available back to 1962, giving more than twelve thousand lines of data This data can be statistically analyzed and the model above can be tested to see how well it holds (here we take t to be one day) There is general agreement on the answers here, as one can find by reading the references below, in particular [1] I have turned these questions into homework assignments for my students Each chooses a different company, takes about a year of daily stock prices, calculates the daily returns, and then statistically analyzes them.5 After reading 200 or so homework papers I too can say that the answers tend to be the same Are the returns normally distributed? The answer is yes, more or less When a normal distribution is superimposed on the histogram of the returns one sees that it approximately fits, though the histogram has fat tails This means that stock prices have more extreme activity than expected Even though the normal distribution doesn’t work that well in this situation, abandoning it would be difficult since so much of the theory depends on it, for instance, all of the Black-Scholes-Merton results discussed below Are the daily returns independent of each other? The answer again is yes, more or less The scatterplots of today’s returns against yesterday’s show no patterns, and the autocorrelations are not statistically significant See We for example finance.yahoo.com use SPSS January 2012] REVIEWS 79 There is also a theoretical answer to this question: If we knew the stock was going up tomorrow then we would buy it Indeed everybody would buy it, so the stock price would have already arrived at that value (this property is called the “efficient market hypothesis”) Can the constants µ and σ can be estimated from the data? The growth rate µ cannot be estimated accurately; in fact the 95% confidence interval is so large that we can’t tell whether the stock is going up or down On the other hand the volatility σ can be estimated fairly well This is fortunate since the volatility plays an important role in the Black-Scholes model, but the growth rate is not needed at all Now let’s continue with the mathematical aspects of the GBM model (3) above If we let t → 0, then, as we all know from Calc I, we just replace with d, so S/ t becomes d S/dt The problem here though is that S is not a function, but instead a random process For each time t we don’t know exactly what value S will have, but only a general idea where it should be In fact this is exactly what happens with stock prices So d S/dt doesn’t belong in elementary calculus, but rather in stochastic calculus The infinitesimal version of Equation (3) above is dS = µdt + σ dz S (4) Here z is Brownian motion, which can be thought of as an infinitely fast random walk, appropriately scaled, and dz is an infinitesimal piece of it Itˆ ’s lemma is a fundamental result of stochastic calculus; it is the analogue of the o chain rule in ordinary calculus Applying Itˆ ’s lemma gives o √ ln ST − ln S0 = µ − σ T + σ T (5) Here the stock starts at time and goes to time T (Itˆ ’s lemma accounts for the extra o term in the expression for the mean) Another way to express this equation is to say that ln ST − ln S0 is normally distributed with mean µ − (1/2)σ T and standard √ deviation σ T Note that when σ = (so there is no volatility), then the equation becomes ST = S0 exp(µT ), the equation for continuously compounded money in the bank Now suppose we have an option on a stock whose value f depends on the time t and the price S of the stock (European puts and calls satisfy this criterion; Asian calls not) The value of the stock is governed by Equation (4) above, and the value of the option has a similar expression These two expressions can be combined It is rather surprising that in this step the growth factor µ and the randomness factor dz drop out The result is the famous Black-Scholes partial differential equation (where r is the interest rate); ∂f ∂f ∂2 f + rS + σ S 2 = r f ∂t ∂S ∂S The value of a particular option can be found by specifying the boundary conditions for the PDE and then solving the equation For example if we have a call with strike price X and expiration time T , then the boundary conditions are: 80 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 (a) if t = T then f = max(S − X, 0), (b) if S = then the option has no value, so f = 0, and (c) if S → ∞ then f → S Solving the PDE gives the value of the call at time zero (see for instance [6, Ch 5]) The resulting formula for the value of the call (which can be found in any textbook) is a function of the stock price at time zero, the strike price, the interest rate, the volatility and the time T The whole process of finding the PDE and deriving this formula is non-trivial; after all, Merton and Scholes were awarded the Nobel prize for their work (unfortunately Black had died) Before this formula traders made intelligent guesses for the price of an option; afterwards they simply punched buttons on a calculator The third method for finding the value of a call is to the calculations in the “risk-neutral world” In this world the average growth rate of a stock is the same as money in the bank Hence the expected returns of all stocks are the same, independent of whether they are conservative or risky investments A result obtained in the riskneutral world also holds in the real world Why this method works requires some serious explaination The derivation of the Black-Scholes formula using this method can be found, for instance, in [3, §13.8] A simple case of risk-neutral pricing can already be seen for the one-step binomial tree above [3, §11.2]: If we think of the p defined by Equation (2) as a probability, then the price of the option is its discounted expected value as in Equation (1) Furthermore this definition of p is equivalent to assuming that the expected value of ST is S0 er T I find that the book by Lovelock et al confuses the real and risk-neutral worlds In fact the latter is never mentioned Geometric Brownian motion is used in §11.6 to model the price of a stock with growth rate µ; this is essentially Equation (3) above However later in the section µ becomes the risk-free interest rate i (∞) Of course in the real world growth rates µ are different for different stocks; they are not all the same So assuming that they are the same is moving into the risk-neutral world The main result of §11.6 is to show that trees with an increasing number of steps converge to the lognormal model (5) This result was originally proved in [2] and credit should be given The formula relating the parameters of the tree to the volatility of the stock simply slips into the exposition, though it is actually is a rather important step Also the σ used in §11.6 is different from the σ of §10.3; in fact the former is the variance of the log return, and the latter is apparently the variance of the percentage return Though approximately equal, they are different and this should be noted Section 11.7 finds the price of an option using a binomial tree as explained above The main result of §11.8 is the pricing formula for calls, which is elegantly proved by doing the calculation in the risk-neutral world Once again, this assumes that expected growth rate of a stock is the same as money in the bank, and it should be explained why this result holds in the real world I don’t recommend this book for the more advanced topics connected with stocks and options; the coverage is too brief and rather confused However the earlier parts of the book are fine, with their detailed explanations, examples and homework problems REFERENCES Campbell, J.; Lo, A.; MacKinlay, A C., The econometrics of financial markets Princeton University Press, Princeton, NJ, 1997 Cox, J.; Ross S.; Rubinstein, M., Option pricing: a simplified approach, Journal of Financial Economics (1979), 229-263, available at http://dx.doi.org/10.1016/0304-405X(79)90015-1 Hull, J., Options, futures and other derivatives, seventh edition Pearson, Upper Saddle River, NJ, 2008 January 2012] REVIEWS 81 Lowenstein, R., When genius failed Random House, New York, NY, 2000 Ross, S., An elementary introduction to mathematical finance, third edition Cambridge University Press, New York, NY, 2011 Wilmott, P.; Howison, S.; Dewynne, J., The mathematics of financial derivatives Cambridge University Press, New York, NY, 1995 Wilmott, P., Paul Wilmott on quantitative finance, second edition Wiley, Chichester, England, 2006 Mount Holyoke College, South Hadley, MA 01075 adurfee@mtholyoke.edu A New Problem What is the limit lim n→∞ 1+ 1 1 + ··· + − − − ··· − 2 n n+1 n+2 n equal to? 1 Solution Let Hn = + + · · · + n Then Hn − log n → γ (EulerMascheroni constant) and Hn − log n → γ Hence, limn→∞ (1 + + · · · + 1 − n+1 − n+2 − · · · − n12 = Hn − (Hn − Hn ) = 2Hn − Hn2 = 2(Hn − log n) − n (Hn − log n ) → 2γ − γ = γ So we have a curious definition γ = lim n→∞ 1+ 1 1 + ··· + − − − ··· 2 n n+1 n+2 n —Submitted by Juozas Juvencijus Ma˘ ys c 82 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119 New from the MAA The Hungarian Problem Book IV Edited and Translated by Robert Barrington Leigh and Andy Liu The Eötvös Mathematics Competition is the oldest high school mathematics competition in the world, dating back to 1894 This book is a continuation of Hungarian Problem Book III and takes the contest through 1963 Forty-eight problems in all are presented in this volume Problems are classified under combinatorics, graph theory, number theory, divisibility, sums and differences, algebra, geometry, tangent lines and circles, geometric inequalities, combinatorial geometry, trigonometry and solid geometry Multiple solutions to the problems are presented along with background material There is a substantial chapter entitled "Looking Back," which provides additional insights into the problems Hungarian Problem Book IV is intended for beginners, although the experienced student will find much here Beginners are encouraged to work the problems in each section, and then to compare their results against the solutions presented in the book They will find ample material in each section to help them improve their problem-solving techniques 114 pp., Paperbound, 2011 ISBN 978-0-88385-831-8 Catalog Code: HP4 List: $40.95 Member: $33.95 To order visit us online at www.maa.org or call us at 1-800331-1622 New title by the MAA Counterexamples in Calculus Sergiy Klymchuk As a robust repertoire of examples is essential for students to learn the practice of mathematics, so a mental library of counterexamples is critical for students to grasp the logic of mathematics Counterexamples are tools that reveal incorrect beliefs Without such tools, learners’ natural misconceptions gradually harden into convictions that seriously impede further learning This slim volume brings the power of counterexamples to bear on one of the largest and most important courses in the mathematics curriculum —Professor Lynn Arthur Steen, St Olaf College, Minnesota, USA, Co-author of Counterexamples in Topology Counterexamples in Calculus serves as a supplementary resource to enhance the learning experience in single variable calculus courses This book features carefully constructed incorrect mathematical statements that require students to create counterexamples to disprove them Methods of producing these incorrect statements vary At times the converse of a well-known theorem is presented In other instances crucial conditions are ments are grouped topically with sections devoted to: Functions, Limits, Continuity, Differential Calculus and Integral Calculus using counterexamples as a pedagogical tool in the study of introductory calculus In that light it may well be useful for Catalog Code: CXC 101pp., Paperbound, 2010 ISBN: 978-0-88385-756-6 List: $45.95 MAA Member: $35.95 Order your copy today! 1.800.331.1622 www.maa.org New from the MAA Randomness and Recurrence in Dynamical Systems Rodney Nillsen Randomness and Recurrence in Dynamical Systems bridges the gap between undergraduate teaching and the research level in mathematical analysis It makes ideas on averaging, randomness, and recurrence, which traditionally require measure theory, accessible at the undergraduate and lower graduate level The author develops new techniques of proof and adapts known proofs to make the material accesible to students with only a background in elementary real analysis Over 60 figures are used to explain proofs, provide alternative viewpoints and elaborate on the main text The final part of the book explains further developments in terms of measure theory The results are presented in the context of dynamical systems, and the quantitative results are related to the underlying qualitative phenomena—chaos, randomness, recurrence, and order The final part of the book introduces and motivates measure theory and the notion of a measurable set, and describes the relationship of Birkhoff’s Individual Ergodic Theorem to the preceding ideas Developments in other dynamical systems are indicated, in particular Lévy’s result on the frequency of occurence of a given digit in the partial fractions expansion of a number Historical notes and comments suggest possible avenues for self-study Catalog Code: CAM-31 ISBN: 978-0-88385-043-5 Hardbound, 2010 List: $62.95 MAA Member: $50.95 To order visit us online at www.maa.org or call 1-800-331-1622 MATHEMATICAL ASSOCIATION OF AMERICA 1529 Eighteenth St., NW • Washington, DC 20036 Looking for a great textbook for your class? The MAA offers affordable textbooks in a variety of subjects View our online textbook catalogue: http://www.maa.org/pubs/2011OnlineTextbookCatalog.pdf ... distinguish the shapes Indeed, the × rectangle and the star appear more similar to each other than they are to a second randomized version of themselves On the other hand, for the star and the circle, the. .. Zeeb Road, Ann Arbor, MI 48106 The AMERICAN MATHEMATICAL MONTHLY (ISSN 0002-9890) is published monthly except bimonthly June-July and August-September by the Mathematical Association of America... retirement In the ever-changing world of academic publication, the future of the Monthly remains bright The beginning of my term as Editor-Elect saw the adoption by the Monthly of the Editorial