American Mathematical Monthly, volume 116, number 1, january 2009

Why Did Lagrange “Prove” the Parallel Postulate? Judith V. Grabiner

1. INTRODUCTION. We begin with an often-told story from the Budget of Paradoxes by Augustus de Morgan: “Lagrange, in one of the later years of his life, imagined” that he had solved the problem of proving Euclid’s parallel postulate. “He went so far as to write a paper, which he took with him to the [Institut de France], and began to read it.” But, De Morgan continues, “something struck him which he had not observed: he muttered ‘Il faut que j’y songe encore’ [I’ve got to think about this some more] and put the paper in his pocket” [8, p. 288]. Is De Morgan’s story true? Not quite in that form. But, as Bernard Cohen used to say, “Truth is more interesting than fiction.” First, according to the published minutes of the Institut for 3 February 1806, “M. Delagrange read an analysis of the theory of parallels” [25, p. 314; italics added]. Those present are listed in the minutes: Lacroix, Cuvier, Bossut, Delambre, Legendre, Jussieu, Lamarck, Charles, Monge, Laplace, Ha¨uy, Berthollet, Fourcroy—a most distinguished audience! Furthermore: Lagrange did not throw his manuscript away. It survives in the library of the Institut de France [32]. There is a title page that says, in what looks to me like Lagrange’s handwriting, “On the theory of parallels: memoir read in 1806,” together with the signatures of yet more distinguished people: Prony and Poisson, along with Legendre and Lacroix. The first page of text says, again in Lagrange’s handwriting, that it was “read at the Institut in the meeting of 3 February 1806.” It is true that Lagrange never did publish it, so he must have realized there was something wrong. In another version of the story, told by Jean-Baptiste Biot, who claims to have been there (though the minutes do not list his name), everybody there could see that something was wrong, so Lagrange’s talk was followed by a moment of complete silence [2, p. 84]. Still, Lagrange kept the manuscript with his papers for posterity to read. This episode raises the three questions I will address in this article. First, what did Lagrange actually say in this paper? Second, once we have seen how he “proved” the parallel postulate, why did he do it the way he did? And last, above all, why did Joseph-Louis Lagrange, the consummate analyst, creator of the Analytical Mechanics, of Lagrange’s theorem in group theory and the Lagrange remainder of the Taylor series, pioneer of the calculus of variations, champion of pure analysis and foe of geometric intuition, why did Lagrange risk trying to prove Euclid’s parallel postulate from the others, a problem that people had been unsuccessfully trying to solve for 2000 years? Why was this particular problem in geometry so important to him? I think that the manuscript is interesting in its own right, but I intend also to use it to show how Lagrange and his contemporaries thought about mathematics, physics, and the universe. As we will see, this was not the way we view these topics today. 2. THE CONTENTS OF LAGRANGE’S 1806 PAPER. First, we look at the contents of the paper Lagrange read in 1806. The manuscript begins by asserting that the theory of parallels is fundamental to all of geometry. Notably, that includes the January 2009]

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facts that the sum of the angles of a triangle is two right angles, and that the sides of similar triangles are proportional. But Lagrange agreed with both the ancients and moderns who thought that the parallel postulate should not be assumed, but needed to be proved. To see why people wanted to prove the parallel postulate, let us recall Euclid’s five geometric postulates [9, pp. 154–155]. The first is that a straight line can be drawn from any point to any other point; the second, that a finite straight line can be produced to any length; the third, that a circle can be drawn with any point as center and any given radius; the fourth, that all right angles are equal; and the fifth, the so-called parallel postulate, which is the one in question. Euclid’s parallel postulate is not, as a number of writers wrongly say (e.g., [5, p. 126]), the statement that only one line can be drawn parallel to a given line through an outside point. Euclid’s postulate states that, if a straight line falls on two straight lines making the sum of the interior angles on the same side of that line less than two right angles, then the two straight lines eventually meet on that side. Euclid used Postulate 5 explicitly only once: to prove that if two lines are parallel, the alternate interior angles are equal. Of course, many later propositions rest on this theorem, and thus presuppose the parallel postulate. Already in antiquity, people were trying to prove Postulate 5 from the others. Why? Of course one wants to assume as little as possible in a demonstrative science, but few questions were raised about Postulates 1–4. The historical focus on the fifth postulate came because it felt more like the kind of thing that gets proved. It is not self-evident, it requires a diagram even to explain, so it might have seemed more as though it should be a theorem. In any case, there is a tradition of attempted proofs throughout the Greek and then Islamic and then eighteenth-century mathematical worlds. Lagrange followed many eighteenth-century mathematicians in seeing the lack of a proof of the fifth postulate as a serious defect in Euclid’s Elements. But Lagrange’s criticism of the postulate in his manuscript is unusual. He said that the assumptions of geometry should be demonstrable “just by the principle of contradiction”—the same way, he said, that we know the axiom that the whole is greater than the part [32, p. 30R]. The theory of parallels rests on something that is not self-evident, he believed, and he wanted to do something about this. Now it had long been known—at least since Proclus in the fifth century—that the “only one parallel” property is an easy consequence of Postulate 5. In the eighteenth century, “only one parallel” was adopted as a postulate by John Playfair in his 1795 textbook Elements of Geometry and by A.-M. Legendre in his highly influential Elements of Geometry [34]. So this equivalent to Postulate 5 had long been around; in the 1790s people focused on it, and so did Lagrange. But Lagrange, unlike Playfair and Legendre, didn’t assume the uniqueness of parallels; he “proved” it. Perhaps now the reader may be eager to know, how did Lagrange prove it? Recall that Lagrange said in this manuscript that axioms should follow from the principle of contradiction. But, he added, besides the principle of contradiction, “There is another principle equally self-evident,” and that is Leibniz’s principle of sufficient reason. That is: nothing is true “unless there is a sufficient reason why it should be so and not otherwise” [42, p. 31; italics added]. This, said Lagrange, gives as solid a basis for mathematical proof as does the principle of contradiction [32, p. 30V]. But is it legitimate to use the principle of sufficient reason in mathematics? Lagrange said that we are justified in doing this, because it has already been done. For example, Archimedes used it to establish that equal weights at equal distances from the fulcrum of a lever balance. Lagrange added that we also use it to show that three equal forces acting on the same point along lines separated by a third of the circumference of a circle are in equilibrium [32, pp. 31R–31V]. 4

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Now we are ready to see how Lagrange deduced the uniqueness of parallels from the principle of sufficient reason. Suppose DE is drawn parallel to the given line AB through the given point C. Now suppose the parallel DE isn’t unique. Then we can also draw FG parallel to AB. (See Figure 1a.) But everything ought to be equal on each side, Lagrange said, so there is no reason that FG should make, with DE, the angle ECG on the right side; why not also on the left side? So the line HCI, making the angle DCH equal to angle ECG, ought also to be parallel to AB. One can see why the argument so far seemed consistent with Lagrange’s views on sufficient reason. F

I C

D

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G

H

B

A Figure 1a. Lagrange’s proof, step 1.

By the same procedure, he continued, we can now make another line KL that makes angle HCK equal to angle ICG, but placed on the other side of the new parallel line HI (see Figure 1b); and we can keep on in this way to make arbitrarily many in this fashion (see Figure 1c), which, as he said, “is evidently absurd” [32, pp. 32V–33R]. L F

I C

D

E

G

H K

B

A Figure 1b. Lagrange’s proof, step 2.

The modern reader may object that Lagrange’s symmetry arguments are, like the uniqueness of parallels, equivalent to Euclid’s postulate. But the logical correctness, or lack thereof, of Lagrange’s proof is not the point. (In this manuscript, by the way, Lagrange went on to give an analogous proof—also by the principle of sufficient reason—that between two points there is just one straight line, because if there were a January 2009]

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N

L

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Figure 1c. Lagrange’s proof, last step.

second straight line on one side of the first, we could then draw a third straight line on the other side, and so on [32, pp. 34R–34V]. Lagrange, then, clearly liked this sort of argument.) 3. WHY DID HE ATTACK THE PROBLEM THIS WAY? It is now time to address the second, and more important question: Why did he do it in the way he did? I want to argue this: Lagrange’s arguments from sufficient reason were shaped by properties of space, space as it was believed to be in the seventeenth and eighteenth centuries. These properties are profoundly Euclidean. To eighteenth-century thinkers, space was infinite, it was exactly the same in all directions, no direction was privileged, it was like the plane in having no curvature, and symmetrical situations were equivalent. Lagrange himself explicitly linked his symmetry arguments to Leibniz’s principle of sufficient reason, but—as we will see—these ideas are also historically linked to Giordano Bruno’s arguments for the infinite universe, Descartes’ view of space as indefinite material extension, the projective geometry used to describe perspective in Renaissance art, various optimization arguments like “light travels in straight lines because that is the shortest path,” and, above all, the Newtonian doctrine of absolute space. As we will soon see, these properties were essential to physical science in the seventeenth and eighteenth centuries: both physics and philosophy promoted the identification of space with its Euclidean structure. This goes along with a shift in emphasis concerning what Euclidean geometry is about. Geometry, in ancient times, was the study of geometric figures: triangles, circles, parallelograms, and the like, but by the eighteenth century it had become the study of space [41, chapter 5]. The space eighteenth-century geometry was about was, in Henri Poincar´e’s words, “continuous, infinite, three-dimensional, homogeneous and isotropic” [44, p. 25]. Bodies moved through it preserving their sizes and shapes. The possible curvature of three-dimensional space did not even occur to eighteenth-century geometers. Their space was Euclidean through and through. Why did philosophers conclude that space had to be infinite, homogeneous, and the same in all directions? Effectively, because of the principle of sufficient reason. For instance, Giordano Bruno in 1600 argued that the universe must be infinite because there is no reason to stop at any point; the existence of an infinity of worlds is no less reasonable than the existence of a finite number of them. Descartes used similar reasoning in his Principles of Philosophy: “We recognize that this world. . . has no 6

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limits in its extension. . . . Wherever we imagine such limits, we . . . imagine beyond them some indefinitely extended space” [28, p. 104]. Similar arguments were used by other seventeenth-century authors, including Newton. Descartes identified space and the extension of matter, so geometry was, for him, about real physical space. But geometric space, for Descartes, had to be Euclidean. This is because the theory of parallel lines is crucial for Descartes’ analytic geometry—not for Cartesian coordinates, which Descartes did not have, but because he needed the theory of similar figures in order to give meaning to expressions of arbitrary powers of x [23, p. 197]. Descartes was the first person to justify using such powers. But an expression like x 4 for Descartes is not the volume of a 4-dimensional figure, but a line, which can be defined as the fourth proportional to the unit line, x, and x 3 . That is, 1/x = x 3 /x 4 . They are all lines, and since all powers of x are lines, they can all be constructed geometrically— but only if we have the theory of similar triangles, for which we need the theory of parallels. Now let us turn from seventeenth-century philosophy to seventeenth-century physics. Descartes, some 50 years before Newton published his first law of motion, was a co-discoverer of what we call linear inertia: that in the absence of external influences a moving body goes in a straight line at a constant speed. Descartes called this the first law of nature, and for him, this law follows from what we now recognize as the principle of sufficient reason. Descartes said, “Nor is there any reason to think that, if [a part of matter] moves. . . and is not impeded by anything, it should ever by itself cease to move with the same force” [30, p. 75]. And the straight-line motion of physical moving bodies obviously requires the indefinite extendability of straight lines and thus indefinitely large, if not infinite, space [23, p. 97]. Descartes’ contemporary Pierre Gassendi, another co-discoverer of linear inertia, used “sufficient reason” to argue for both inertia and the isotropy of space. Gassendi said, “In principle, all directions are of equal worth,” so that in empty spaces, “motion, in whatever direction it occurs. . . will neither accelerate nor retard; and hence will never cease” [29, p. 127]. Artists, too, helped people learn to see space as Euclidean. We see the space created in the paintings and buildings of the Renaissance and later as Euclidean. Renaissance artists liked to portray floors with rectangular tiles and similar symmetric architectural objects—to show how good they were at perspective. These works of art highlight the observations that parallel lines are everywhere equidistant, that two lines perpendicular to a third line are parallel to each other. And our experience of perspective in art and architecture helps us shape the space we believe we live in. (See Figures 2, 3, and 4.) We have seen pictures like these many times, but consider them now as conditioning people to think in a particular way about the space we live in: as Euclidean, symmetric, and indefinitely extendible—going on to infinity [11].

Figure 2. Piero della Francesca (1410/1420–1492), “The Ideal City.”

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Figure 3. Leonardo da Vinci (1452–1519), “The Last Supper.”

Figure 4. Raphael (1483–1520), “The School of Athens.”

Artist-mathematicians like Piero della Francesca began the development of the subject of projective geometry, but the first definitive mathematical treatise on it is that of Girard Desargues in the 1630s. Seventeenth-century projective geometry used the cone (like the artist’s rays of sight or light) to prove properties of all the conic sections as projections of the circle. For instance, geometers treated the ellipse as the circle projected to a plane not perpendicular to the cone. And the parabola, as Kepler pointed 8

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out, behaves projectively like an ellipse with one focus at infinity. So projective geometry explicitly brought infinity into Euclidean geometry: planes and lines go to infinity; parallel lines meet at the point at infinity. And the geometry of perspective and projective geometry reinforced Euclideanness in a wide variety of other ways, from the role of Euclid’s Optics in the humanistic classical tradition to the use of the theory of parallels to draw military fortifications from 2-dimensional battlefield sketches [12, p. 24]. Although these Euclidean views prevailed, perhaps they didn’t have to. There were alternatives suggested even in the eighteenth century [22]. Is visual space Euclidean? Not necessarily. Bishop Berkeley, for instance, said that we don’t “see” distance at all; we merely infer it from the angles we do see. And Thomas Reid pointed out that a straight line right in front of you looks exactly like a circle curved with you at the center—or even a circle curved away from you in the other direction. Reid gave a set of rules for visual space—he called this the “geometry of visibles”—which clearly are not Euclid’s rules; a modern philosopher has called Reid’s geometry of visibles “the geometry of the single point of view” [46, p. 396]. And there are other alternatives to Euclideanness. Cultures other than the western often speak about space differently and order their perceptions differently: particular directions have special connotations, and “closeness” can be cultural as well as metrical. Many cultures do not use the idea of an outside abstract space at all; instead—as Leibniz did—they recognize only the relations between bodies [3], [35], [36]. So, as a matter of empirical fact, abstract Euclidean space is not something that all human thinkers do use, let alone that all humans must use. In the twentieth century, experimental psychologists showed that when people in a dark room are asked to put luminous points into two equidistant lines, or two parallel lines, the people are satisfied when the lines in fact curve away from the observer. As a result, Rudolf Luneburg in the 1940s claimed that visual space is a hyperbolic space of constant curvature; later psychological experiments suggest that visual space is not represented by any consistent geometry [48, pp. 30–31]. Even in the Renaissance, some painters portrayed what we now recognize as 3-dimensional non-Euclidean spaces, using reflections in a convex mirror, notably Parmigianino’s (1524) “Self Portrait in a Convex Mirror,” and, most famously, the “Arnolfini Wedding” by Jan van Eyck (1434). (See Figures 5, 6a, and 6b.) In the spaces in these mirrors, parallel lines are not everywhere equidistant.

Figure 5. Parmigianino (1503–1540), “Self Portrait in a Convex Mirror.”

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Figure 6a. Jan van Eyck (c. 1390–1441), “The Arnolfini Wedding.”

Figure 6b. Detail from “The Arnolfini Wedding.”

A modern physicist, John Barrow, has said that if people had paid more attention to these mirrors, non-Euclidean geometry might have been discovered much sooner [1, p. 176]. But I am not so sure. I think that these artists viewed convex mirrors as presenting an especially difficult problem in portraying 3-dimensional Euclidean space on a 2-dimensional Euclidean canvas; for instance, J. M. W. Turner included such drawings in his strongly Euclidean lectures [45] on perspective. (See Figures 7a, 7b.) The winning view, I think, is that expressed by the Oxford art historian Martin Kemp, who says that from the Renaissance to the nineteenth century, the goal of constructing “a model of the world as it appears to a rational, objective observer” [27, p. 314] was shared by scientists and artists alike. Virtually unanimously, artists, armed with Euclid’s Optics, have long helped teach us to “see” a Euclidean world. 10

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c Tate, London Figure 7a. J. M. W. Turner (1775–1851), “Spheres at Different Distances from the Eye.”  2008.

Figure 7b. J. M. W. Turner, “Reflections in a Single Metal Globe and in a Pair of Polished Metal Globes.” c Tate, London 2008. 

4. THE CRUCIAL ARGUMENT: NEWTONIAN PHYSICS. Let us now return to physics and to the most important seventeenth-century argument of all for the reality of infinite Euclidean space: Newtonian mechanics. Newton needed absolute space as a reference frame, so he could argue that there is a difference between real and apparent accelerations. He wanted this so he could establish that the forces involved with absolute (as opposed to relative) accelerations are real, and thus that gravity is real. Newton’s absolute space is infinite and uniform, “always similar and immovable” [38, p. 6], and he described its properties in Euclidean terms. And it is real; it has a Platonic kind of reality. Leibniz, by contrast, did not believe in absolute space. He not only said that spatial relations were just the relations between bodies, he used the principle of sufficient reason to show this. If there were absolute space, there would have to be a reason to explain why two objects would be related in one way if East is in one direction and West in the opposite direction, and related in another way if East and West were reversed [24, p. 147]. Surely, said Leibniz, the relation between two objects is just one thing! But Leibniz did use arguments about symmetry and sufficient reason— sufficient reason was his principle, after all. Thus, although Descartes and Leibniz did not believe in empty absolute space and Newton did, they all agreed that what I am calling the Euclidean properties of space are essential to physics. In the eighteenth century, the Leibniz-Newton debate on space was adjudicated by one of Lagrange’s major intellectual influences, Leonhard Euler. In his 1748 essay “Reflections on Space and Time,” Euler argued that space must be real; it cannot be January 2009]

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just the relations between bodies as the Leibnizians claim [10]. This is because of the principles of mechanics—that is, Newton’s first and second laws. These laws are beyond doubt, because of the “marvelous” agreement they have with the observed motions of bodies. The inertia of a single body, Euler said, cannot possibly depend on the behavior of other bodies. The conservation of uniform motion in the same direction makes sense, he said, only if measured with respect to immovable space, not to various other bodies. And space is not in our minds, said Euler; how can physics—real physics—depend on something in our minds? So space for Euler is real. The philosopher Immanuel Kant was influenced by Euler’s analysis [14, pp. 29, 207]. Kant agreed that we need space to do Newtonian physics. But in his Critique of Pure Reason of 1781, Kant placed space in the mind nonetheless. We order our perceptions in space, but space itself is in the mind, an intuition of the intellect. Nevertheless, Kant’s space turned out to be Euclidean too. Kant argued that we need the intuition of space to prove theorems in geometry. This is because it is in space that we make the constructions necessary to prove theorems. And what theorem did Kant use as an example? The sum of the angles of a triangle is equal to two right angles, a result whose proof requires the truth of the parallel postulate [26, “Of space,” p. 423]. Even outside of mathematics and physics, explicit appeals to sufficient reason, symmetry, parallels, and infinity pervade eighteenth-century thought, from balancing chemical equations to symmetry in architecture to the balance of powers in the U.S. Constitution. Let me call one last witness from philosophy: Voltaire. Like many thinkers in the eighteenth century, Voltaire said that universal agreement was a marker for truth. Religious sects disagree about many things, he said, so on these topics they are all wrong. But by contrast, they all agree that one should worship God and be just; therefore that must be true. Voltaire pointed out also that “There are no sects in geometry” [47, p. 195]. One does not say, “I’m a Euclidean, I’m an Archimedean.” What everyone agrees on: that is what is true. “There is but one morality,“ said Voltaire, “as there is but one geometry” [47, p. 225]. 5. THE ARGUMENT FROM EIGHTEENTH-CENTURY MATHEMATICS AND SCIENCE. Now let us turn to eighteenth-century mathematics and science. Eighteenth-century geometers tended to go beyond Euclid himself in assuming Euclideanness. As a first example, look at the 1745 El´emens de G´eom´etrie by AlexisClaude Clairaut. Clairaut grounded geometry not on Euclid’s postulates but on the capacity of the mind to understand clear and distinct ideas. For instance, Euclid had defined parallel lines as lines in the same plane that never meet. Clairaut, less interested in proof than in Euclidean plausibility, defined parallel lines as lines that are everywhere equally distant from one another [6, p. 10]. The great French Encyclopedia [7, vol. 11, pp. 905–906] defined parallel lines as “lines that prolonged to infinity never get closer or further from one another, or that meet at an infinite distance” [italics added] assuming, then, a uniform, flat Euclidean space infinitely extended. The “equidistant” definition of parallels is reinforced by ordinary language, as we speak of parallel developments, or, more geometrically, ships on parallel courses, and even of parallels of latitude. And speaking of latitude raises the question of why the fact that the geometry on the surface of a sphere, with great circles serving as “lines,” is not Euclidean—there are no parallels, for example—did not shake mathematicians’ conviction that all of Euclid’s postulates are true and mutually consistent. Lagrange himself is supposed to have said that spherical trigonometry does not need Euclid’s parallel postulate [4, pp. 52–53]. But the surface of a sphere, in the eighteenth-century view, is not non-Euclidean; it 12

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exists in 3-dimensional Euclidean space [20, p. 71]. The example of the sphere helps us see that the eighteenth-century discussion of the parallel postulate’s relationship to the other postulates is not really about what is logically possible, but about what is true of real space. Now, let us turn to eighteenth-century physics. As we will see, Euclideanness, especially the theory of parallels and the principle of sufficient reason, was essential to the science of mechanics in the eighteenth century, not only to its exposition, but to its progress. Johann Heinrich Lambert was one of the mathematicians who worked on the problem of Postulate 5. Lambert explicitly recognized that he had not been able to prove it, and considered that it might always have to remain a postulate. He even briefly suggested a possible geometry on a sphere with an imaginary radius. But Lambert also observed that the parallel postulate is related to the law of the lever [20, p. 75]. He said that a lever with weightless arms and with equal weights at equal distances is balanced by a force in the opposite direction at the center equal to the sum of the weights, and that all these forces are parallel. So either we are using the parallel postulate, or perhaps, Lambert thought, some day we could use this physical result to prove the parallel postulate. Lagrange himself in his Analytical Mechanics [31, pp. 4–5] gave an argument about balancing an isosceles triangle similar to, but much more complex than, Lambert’s discussion. Lagrange himself did not explicitly link the law of the lever to the parallel postulate, but the geometry of the equilibrium situation that Lagrange was describing nonetheless requires it [4, pp. 182–183]. In a similar move, d’Alembert had tried to deduce the general law of conservation of momentum purely from symmetry principles [13, pp. 821–823]. And in the 1820s, J.-B. Fourier, from a very different philosophical point of view, also said that the parallel postulate could be derived from the law of the lever. From this Fourier concluded that geometry follows from statics and so geometry is a physical science [20, pp. 78–79]. But note that it is still Euclidean geometry. Let us now concentrate further on Lagrange’s mechanics. His deepest conviction was that a subject must be seen in its full generality. Like many Enlightenment thinkers only more so, Lagrange wanted to reduce the vast number of laws and principles to a single fundamental general principle, preferably one that is independent of experience. “Sufficient reason” was such a principle. Although he did not explicitly cite Leibniz’s principle in his Analytical Mechanics [43, p. 146], Lagrange used it frequently. For instance, he wrote, “The equilibrium of a straight and horizontal lever with equal weights and with the fulcrum at its midpoint is a self-evident truth because there is no reason that either of the weights should move.” Another key example of Euclideanness as physical argument is the use of parallelograms to find resultant forces. Lagrange, in his Analytical Mechanics, used the principle of sufficient reason, Euclid’s theory of parallels, and the infinity of space and its Euclidean nature to discuss the composition of forces [31, p. 17]. He said that a body which is moved uniformly in two different directions simultaneously must necessarily traverse the diagonal of the parallelogram whose sides it would have followed separately. So parallels are needed. Lagrange continued, “with regard to the direction in the case of two equal forces, it is obvious that there is no reason that the resultant force should be nearer to one than the other of these two equal forces; therefore it must bisect the angle formed by these two forces” [31, p. 21]. Lagrange also used the principle of composition of forces to get the conditions of equilibrium when two parallel forces are applied to the extremities of a straight lever. He suggested that we imagine “that the directions of the forces extend to infinity” and then, on this basis, we can prove that “the resultant force must pass through the point of support.” In effect January 2009]

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this is the parallelogram argument with the corner of the parallelogram at infinity. Lagrange tried to reduce even his own fundamental physical principle—the principle of virtual velocities—to levers and parallelograms of forces, and after him Amp`ere, Carnot, Laplace, and Poisson tried to do the same [39, p. 218]. Pierre-Simon Laplace, too, related a priori arguments, including sufficient reason and Euclid’s theory of parallels, to argue that physical laws had to be the way they were. For instance he said that a particle on a sphere moves in a great circle because “there is no reason why it should deviate to the right rather than the left of that great circle”—notice, “not a word about the forces acting on this particle on this sphere” [15, p. 104]. Laplace also asked why gravitation had to be inverse-square, and gave a geometric answer [4, p. 53], [16, p. 310]. He said that inverse-square gravitation implies that if the size of all bodies and all distances in the whole universe were to decrease proportionally, the bodies would describe the same curves that they do now, so that universe would still look exactly the same. The observer, then, needs to recognize only the ratios. So, Laplace said, even though we haven’t proved Euclid’s fifth postulate, we know it must be true, and so the theorems deduced from it must also be true. For Laplace, then, the idea of space includes the following self-evident property: similar figures have proportional sides [4, p. 54]. The Newtonian physical universe requires similar figures to have proportional sides, and this of course requires the theory of parallels, and thus geometry must be Euclidean [33, p. 472]. These men did not want to do mechanics, as, say, Newton had done. They wanted to show not only that the world was this way, but that it necessarily had to be. A modern philosophical critic, Helmut Pulte, has said that Lagrange’s attempt to “reduce” mechanics to analysis strikes us today as “a misplaced endeavour to mathematize. . . an empirical science, and thus to endow it with infallibility” [39, p. 220]. Lagrange would have responded, “Right! That’s just exactly what we are all doing.” Lagrange thought these two things: Geometry is necessarily true; mechanics is mathematics. He needed them both. 6. WHY DID IT MATTER SO MUCH? And now, we are ready for the last question. Why did actually proving Postulate 5 matter so much to Lagrange and to his contemporaries? I trust I have convinced the reader of the central role of Euclideanness in the eighteenth century. But still, if it were just a matter of simple logic, surely after 2000 years people should have concluded: we have been trying as hard as possible, we cannot imagine how to prove this, so let us just concede defeat. It can’t be done. Euclid was right in deciding that it had to be assumed as a postulate. Why did eighteenth-century geometers not settle for this, and, in particular, why didn’t Lagrange? Because there was so much at stake. Because space, for Newtonian physics, has to be uniform, infinite, and Euclidean, and because metaphysical principles like that of sufficient reason and optimality were seen both as Euclidean and as essential to eighteenth-century thought. How could all of this rest on a mere assumption? So, many eighteenth-century thinkers believed that it was crucial to shore up the foundations of Euclid’s geometry, and we can place Lagrange’s manuscript in the historical context of the many attempts in the eighteenth century to cure this “blemish” in Euclid by proving Postulate 5. Also, Lagrange was not just any eighteenth-century mathematician. Lagrange was, mathematically speaking, a Cartesian and a Leibnizian. His overall philosophy of mathematics was to reduce each subject to the most general possible principle. In calculus, as I have argued at length in two books [17], [18], Lagrange wanted to reduce all the ideas of limits and infinites and infinitesimals and rates of change or fluxions to 14

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“the algebraic analysis of finite quantities.” In algebra, Lagrange said that even Newton’s idea of algebra as “universal arithmetic” wasn’t general enough; algebra was the study of systems of operations. In mechanics, his goal was to reduce everything to the principle of virtual velocities—and then to use “only algebraic operations subject to a regular and uniform procedure” [31, preface]. Lagrange even composed his Analytical Mechanics without a single diagram, precisely so he could show he had reduced physics to pure analysis. Geometry, then, ought also to be reducible to self-evident principles, to clear, distinct, and general ideas. Finally, there are social causes to be considered. First, the social background will help answer this question: Why was Lagrange doing this in 1806, as opposed, say, to the 1760s when he taught mathematics at the military school in Turin or in the 1770s and 1780s when he was the leading light of the Berlin Academy of Sciences? One reason is that in about 1800 there was a revival of interest in synthetic geometry in France. There was a Parisian school in synthetic geometry including Monge, Servois, Biot, Lacroix, Argand, Lazare Carnot and his students, and Legendre. Important reasons for this were partly practical, partly ideological [40, p. 450]. The practical needs are related to Monge’s championing of descriptive geometry, so clearly useful in architecture and in military planning. Monge also helped directly to pique Lagrange’s interest, writing two letters to him in the early 1790s soliciting his assistance on problems involving the geometry of perspective [37]. As for the ideology promoting geometry in France after the Revolution, as Joan Richards has written, “the quintessentially reasonable study of universally known space had a central role to play in educating a rational populace” [40, p. 454]. Lagrange himself articulated such views throughout his lifetime, writing as early as 1775 that synthetic geometry was sometimes better than analytic because of “the luminous clarity that accompanies it” [19, p. 135], and, near the end of his life, telling his friend Fr´ed´eric Maurice that “geometric considerations give force and clarity to judgement” [19, p. 1295]. So the French geometers would not have been favorably disposed to inventing a non-Euclidean geometry. It is no wonder that only comparative outsiders like the Hungarian Janos Bolyai and the Russian Nikolai Ivanovich Lobachevsky were the first to publish on this topic. Even Gauss, who out of fear of criticism did not publish his own invention of the subject that he christened “non-Euclidean geometry,” was somewhat outside the French mathematical mainstream. The British would not have found inventing non-Euclidean geometry enticing either. Even William Rowan Hamilton, who in the 1840s was to devise the first noncommutative algebra, wrote in 1837, “No candid and intelligent person can doubt the truth of the chief properties of Parallel Lines, as set forth by Euclid in his Elements, two thousand years ago. . . . The doctrine involves no obscurity nor confusion of thought and leaves in the mind no reasonable ground for doubt” [21, p. 354]. In fact, even after Hermann von Helmholtz and W. K. Clifford had introduced non-Euclidean geometry into Victorian Britain, some British thinkers continued to maintain that real space had to be Euclidean. There was a great deal at stake in Britain: the doctrine of the unity of truth, the established educational program based on the Euclidean model of reason, and the attitudes toward authority that this entailed. The authority and rigor of Euclid, both in Britain and in France, were part and parcel of the established intellectual order. And—one last social point—non-Euclidean geometry even in the twentieth century was culturally seen as anti-establishment, partly through its association with relativity theory. For instance, surrealist artists used it that way: misunderstood, perhaps, but still explicitly part of their assault on traditional artistic canons. Two examples are Yves Tanguy’s “Le Rendez-vous des parall`eles” January 2009]

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c 2008 The M.C. Escher CompanyFigure 8. M. C. Escher (1898–1972), “Hand with Reflecting Sphere.”  Holland: All rights reserved.

(1935) and Max Ernst’s “Young Man Intrigued by the Flight of a Non-Euclidean Fly” (1942–1947). And Figure 8 shows an example from an artist who really did understand what a 3-dimensional non-Euclidean space might look like. 7. CONCLUSION. I cannot explain why Lagrange initially thought that his proof was a good one, but I hope it is clear why he thought he needed to prove the parallel postulate, and why he tried to prove it using the techniques that he used. The story I have told reminds us that, although the great eighteenth-century mathematicians are our illustrious forbears, our world is not theirs. We no longer live in a world of certainty, symmetry, and universal agreement. But it was only in such a world that the work of Lagrange and Laplace, Fourier and Kant, Euler and d’Alembert could flourish. That space must be Euclidean was part of the Cartesian, Leibnizian, Newtonian, symmetric, economical, and totally rationalistic world view that underlies all of Lagrange’s mathematics and classical mechanics—ideas that, from Newton and Leibniz through Kant and Laplace, buttressed the whole eighteenth-century view of the universe and the laws that govern it. And the certainty of Euclidean geometry was the model for the whole Enlightenment program of finding universally-agreed-upon truth through reason. 16

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Thus, though Lagrange’s illustrious audience in Paris may have realized that his proof was wrong, their world-view made them unable to imagine that the parallel postulate couldn’t be proved, much less to imagine that the world itself might be otherwise. ACKNOWLEDGMENTS. I thank the Department of the History and Philosophy of Science, University of Leeds, England, for its hospitality and for vigorous discussions of this research. I also thank the Biblioth`eque de l’Institut de France for permission to study Lagrange’s manuscripts, the donors of the Flora Sanborn Pitzer Professorship at Pitzer College for their generous support, the Mathematical Association of America for inviting me to talk about this topic at MathFest 2007, and Miss Kranz, my trigonometry teacher at Fairfax High School in Los Angeles, who once on a slow day in class revealed to us all that there was such a thing as non-Euclidean geometry. For the right to reproduce the works of art in this paper, I thank the following: Figures 2, 3, 4, 5, 6a, 6b: Erich Lesser/Art Resource; Figures 7a and 7b: The Tate Gallery, London; c 2008 The M.C. Escher Company-Holland: All Figure 8: M.C. Escher’s “Hand with Reflecting Sphere”  rights reserved. www.mcescher.com

REFERENCES 1. J. D. Barrow, Outer space, in Space: In Science, Art and Society, F. Penz, G. Radick, and R. Howell, eds., Cambridge University Press, Cambridge, 2004, 172–200. 2. J.-B. Biot, Note historique sur M. Lagrange, in M´elanges Scientifiques et Litt´eraires, vol. III, Michel L´evy Fr`eres, Paris, 1858, 117–124. 3. P. Bloom et al., Language and Space, MIT Press, Cambridge, MA, 1996. 4. R. Bonola, Non-Euclidean Geometry, Dover, New York, 1955. 5. R. Carnap, An Introduction to the Philosophy of Science, Dover, New York, 1995; reprint of Basic Books, New York, 1966. ´ em´ens de G´eom´etrie, Par la Compagnie des Libraires, Paris, 1765. 6. A.-C. Clairaut, El´ 7. J. D’Alembert and D. Diderot, eds., Encyclop´edie, ou Dictionnaire Raisonn´e des Sciences, des Arts et des M´etiers, Briasson, Paris, 1762–1772. 8. A. De Morgan, A Budget of Paradoxes, Longmans Green, London, 1872. 9. Euclid, The Thirteen Books of Euclid’s Elements, T. L. Heath, ed., Cambridge University Press, Cambridge, 1956. 10. L. Euler, Reflexions sur l’espace et le tems, M´emoires de l’acad´emie de Berlin 4 (1750) 324–333. 11. J. V. Field, The Invention of Infinity: Mathematics and Art in the Renaissance, Oxford University Press, Oxford, 1997. 12. J. V. Field and J. J. Gray, The Geometrical Work of Girard Desargues, Springer, New York, 1987. 13. J. Franklin, Artifice and the natural world: Mathematics, logic, technology, in Cambridge History of Eighteenth Century Philosophy, K. Haakonssen, ed., Cambridge University Press, Cambridge, 2006, 815–853. 14. M. Friedman, Kant and the Exact Sciences, Harvard University Press, Cambridge, MA 1992. 15. E. Garber, The Language of Physics: The Calculus and the Development of Theoretical Physics in Europe, 1750–1914, Birkh¨auser, Boston, 1999. 16. C. C. Gillispie, Pierre-Simon Laplace, 1749–1827: A Life in Exact Science, in collaboration with R. Fox and I. Grattan-Guinness, Princeton University Press, Princeton, 1997. 17. J. V. Grabiner, The Calculus as Algebra: J.-L. Lagrange, 1736–1813, Garland, New York, 1990. , The Origins of Cauchy’s Rigorous Calculus, MIT Press, Cambridge, MA, 1981. 18. 19. I. Grattan-Guinness, Convolutions in French Mathematics, 1800–1840, 3 vols., Birkh¨auser, Basel, 1990. 20. J. Gray, Ideas of Space: Euclidean, Non-Euclidean and Relativistic, 2nd ed., Clarendon Press, Oxford, 1989. 21. T. L. Hankins, Algebra as pure Time: William Rowan Hamilton and the foundations of algebra, in Motion and Time, Space and Matter, P. Machamer and R. Turnbull, eds., Ohio State University Press, Columbus, OH, 1976, 327–359. 22. R. J. Herrnstein and E. G. Boring, A Source Book in the History of Psychology, Harvard University Press, Cambridge, MA, 1966. 23. L. Hodgkin, A History of Mathematics, Oxford University Press, Oxford, 2005.

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24. N. Huggett, ed., Space from Zeno to Einstein: Classic Readings with a Contemporary Commentary, MIT Press, Cambridge, MA, 1999. 25. Institut de France, Acad´emie des Sciences, Proc`es-Verbaux des S´eances de l’Acad´emie, 1804–1807, vol. III, Acad´emie des Sciences, Hendaye, 1913. 26. I. Kant, Critique of Pure Reason (trans. F. M. M¨uller), Macmillan, New York, 1961. 27. M. Kemp, The Science of Art: Optical Themes in Western Art from Brunelleschi to Seurat, Yale University Press, New Haven, CT, 1990. 28. A. Koyr´e, From the Closed World to the Infinite Universe, Johns Hopkins University Press, Baltimore, MD, 1957. , Metaphysics and Measurement, Harvard University Press, Cambridge, MA, 1968. 29. , Newtonian Studies, University of Chicago Press, Chicago, IL, 1965. 30. 31. J.-L. Lagrange, Analytical Mechanics, A. Boissonnade and V. N. Vagliente, trans. and eds., Kluwer, Dordrecht, 1997; from J.-L. Lagrange, M´ecanique analytique, 2nd ed., Courcier, Paris, 1811–1815. In Oeuvres de Lagrange, Gauthier-Villars, Paris, 1867–1892, vol. XI. 32. , Sur la Th´eorie des Parall`eles, M´emoire lu en 1806. Unpublished manuscript in the Biblioth`eque de l’Institut de France, Inst MS 909, ff 18–35. 33. P.-S. Laplace, Exposition du syst`eme du monde, Cercle-Social l’An IV, Paris, 1796, in Ouevres compl`etes de Laplace, Gauthier-Villars, Paris, 1878–1912, vol. VI. 34. A.-M. Legendre, El´ements de G´eom´etrie, Didot, Paris, 1794. 35. S. C. Levinson, Language and mind: Let’s get the issues straight, in Language in Mind: Advances in the Study of Language and Thought, D. Gertner and S. Goldin-Meadow, eds., MIT Press, Cambridge, MA, 2003, 25–46. , Frames of reference and Molyneux’s question: Crosslinguistic evidence, in Language and 36. Space, MIT Press, Cambridge, MA, 1996, 109–170. 37. G. Monge, Two letters to Lagrange, n. d., Oeuvres de Lagrange, Gauthier-Villars, Paris, 1867–1892, vol. XIV, 308–310, 311–314. 38. I. Newton, Sir Isaac Newton’s Mathematical Principles of Natural Philosophy and His System of the World, trans. A. Motte, rev. F. Cajori, University of California Press, Berkeley, 1960. 39. H. Pulte, 1788: Joseph Louis Lagrange, Mechanique analitique, in Landmark Writings in Western Mathematics, 1640–1940, I. Grattan-Guinness, ed., Elsevier, Amsterdam, 2005, 208–224. 40. J. L. Richards, The Geometrical Tradition: Mathematics, Space, and Reason in the Nineteenth Century, in The Modern Physical and Mathematical Sciences, M. J. Nye, ed., Vol. 5 of the Cambridge History of Science, Cambridge University Press, Cambridge, 2003, 449–467. 41. B. A. Rosenfeld, A History of Non-Euclidean Geometry: Evolution of the Concept of a Geometric Space, Springer, Berlin and Heidelberg, 1988. 42. B. Russell, A Critical Exposition of the Philosophy of Leibniz, Allen and Unwin, London, 1937. 43. R. Taton, Lagrange et Leibniz: De la th´eorie des functions au principe de raison suffisante, in Beitr¨age zur Wirkungs- und Rezeptionsgeschichte von Gottfried Wilhelm Leibniz, A. Heinekamp, ed., Franz Steiner Verlag, Stuttgart, 1986, 139–147. 44. R. Torretti, Philosophy of Geometry from Riemann to Poincar´e, D. Reidel, Boston, 1978. 45. J. M. W. Turner, drawings, in A. Fredericksen, Vanishing Point: The Perspective Drawings of J. M. W. Turner, Tate, London, 2004. 46. J. Van Cleve, Thomas Reid’s geometry of visibles, Philosophical Review 111 (2002) 373–416. 47. F. M. Arouet de Voltaire, The Portable Voltaire, B. R. Redman, ed., Viking, New York, 1949. 48. M. Wagner, The Geometries of Visual Space, Erlbaum, Mahwah, NJ, 2006.

JUDITH V. GRABINER received her B.S. in Mathematics from the University of Chicago in 1960 and her Ph.D. in the History of Science from Harvard in 1966. She is the author of The Origins of Cauchy’s Rigorous Calculus, MIT Press, Cambridge, MA, 1981, and The Calculus as Algebra, Garland, New York and London, 1990, to be reprinted, along with a number of her articles, by the MAA. For articles published in MAA journals, she has won the Carl Allendoerfer award three times and the Lester Ford award three times. In 2003 she was awarded the Deborah and Franklin Tepper Haimo Award for Distinguished College or University Teaching. She is the Flora Sanborn Pitzer Professor of Mathematics at Pitzer College, where her students keep her continually aware of the beauty of mathematics and the excitement of seeing its relationship with the world. Department of Mathematics, Pitzer College, Claremont, CA 91711 [email protected]

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Overhang∗ Mike Paterson and Uri Zwick

1. INTRODUCTION. How far off the edge of the table can we reach by stacking n identical, homogeneous, frictionless blocks of length 1? A classical solution achieves an overhang asymptotic to 12 ln n. This solution is widely believed to be optimal. We show, however, that it is exponentially far from optimality by constructing simple nblock stacks that achieve an overhang of cn 1/3 , for some constant c > 0. The problem of stacking a set of objects, such as bricks, books, or cards, on a tabletop to maximize the overhang is an attractive problem with a long history. J. G. Coffin [2] posed the problem in the “Problems and Solutions” section of this M ONTHLY, but no solution was given there. The problem recurred from time to time over subsequent years, e.g., [18], [19], [12], [4]. Either deliberately or inadvertently, these authors all seem to have introduced the further restriction that there can be at most one object resting on top of another. Under this restriction, the harmonic stacks, described below, are easily seen to be optimal.

Figure 1. A harmonic stack with 10 blocks.

The classical harmonic stack of size n is composed of n blocks stacked one on top of the other, with the ith block from the top extending by 2i1 beyond the block below it. (We assume that the length of each is 1.) The overhang achieved by the block n 1 construction is clearly 12 Hn , where Hn = i=1 ∼ ln n is the nth harmonic number. i Both a 3D and a 2D view of the harmonic stack of size 10 are given in Figure 1. The harmonic stack of size n is balanced since, for every i < n, the center of mass of the topmost i blocks lies exactly above the right-hand edge of the (i + 1)st block, as can be easily verified by induction. Similarly, the center of mass of all the n blocks lies exactly above the right edge of the table. A formal definition of “balanced” is given in Definition 2.1. A perhaps surprising and counterintuitive consequence of the harmonic stacks construction is that, given sufficiently many blocks, it is possible to obtain an arbitrarily large overhang! ∗ A preliminary version of this paper [14] appeared in the Proceedings of the 17th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA’06), pages 231–240.

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Harmonic stacks became widely known in the recreational math community as a result of their appearance in the Puzzle-Math book of Gamow and Stern [5, BuildingBlocks, pp. 90–93] and in Martin Gardner’s “Mathematical Games” section of the November 1964 issue of Scientific American [6] (see also [7, Chapter 17: Limits of Infinite Series, p. 167]). Gardner refers to the fact that an arbitrarily large overhang can be achieved, using sufficiently many blocks, as the infinite-offset paradox. Harmonic stacks were subsequently used by countless authors as an introduction to recurrence relations, the harmonic series, and simple optimization problems; see, e.g., [8, pp. 258– 260]. Hall [9] notes that harmonic stacks started to appear in textbooks on physics and engineering mechanics as early as the mid-19th century (see, e.g., [13, p. 341], [16, pp. 140–141], [21, p. 183]). It is perhaps surprising that none of the sources cited above realizes how limiting is the one-on-one restriction under which the harmonic stacks are optimal. Without this restriction, blocks can be used as counterweights, to balance other blocks. The problem then becomes vastly more interesting, and an exponentially larger overhang can be obtained. Stacks with a specific small number of blocks that do not satisfy the one-on-one restriction were considered before by several other authors. Sutton [20], for example, considered the case of three blocks. One of us set a stacking problem with three uniform thin planks of lengths 2, 3, and 4 for the Archimedeans Problems Drive in 1964√ [10]. Ainley [1] found the maximum overhang achievable with four blocks to be 15−4 2 ∼ 1.16789. The optimal stacks with 3 and 4 blocks are shown, together with 8 the corresponding harmonic stacks, in Figure 2.

Figure 2. Optimal stacks with 3 and 4 blocks compared with the corresponding harmonic stacks.

Very recently, and independently of our work, Hall [9] explicitly raises the problem of finding stacks of blocks that maximize the overhang without the one-on-one restriction. (Hall calls such stacks multiwide stacks.) Hall gives a sequence of stacks which he claims, without proof, to be optimal. We show, however, that the stacks suggested by him are optimal only for n ≤ 19. The stacks claimed by Hall to be optimal fall into a natural class that we call spinal stacks. We show in Section 3 that the maximum overhang achievable using such stacks is only ln n + O(1). Thus, although spinal stacks achieve, asymptotically, an overhang which is roughly twice the overhang achieved by harmonic stacks, they are still exponentially far from being optimal. Optimal stacks with up to 19 blocks are shown in Figures 3 and 4. The lightly shaded blocks in these stacks form the support set, while the darker blocks form the balancing set. The principal block of a stack is defined to be the block which achieves 20

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blocks  2 overhang  0.75

blocks  3 overhang  1

blocks  4 overhang  1.16789

blocks  5 overhang  1.30455

blocks  6 overhang  1.4367

blocks  7 overhang  1.53005

blocks  8 overhang  1.63151

blocks  9 overhang  1.71527

blocks  10 overhang  1.78713

Figure 3. Optimal stacks with 2 up to 10 blocks.

blocks  11 overhang  1.85878

blocks  12 overhang  1.92509

blocks  13 overhang  1.98451

blocks  14 overhang  2.03822

blocks  15 overhang  2.0929

blocks  16 overhang  2.14384

blocks  17 overhang  2.1909

blocks  18 overhang  2.23457

blocks  19 overhang  2.27713

Figure 4. Optimal stacks with 11 up to 19 blocks.

the maximum overhang. (If several blocks achieve the maximum overhang, the lowest one is chosen.) The support set of a stack is defined recursively as follows: the principal block is in the support set, and if a block is in the support set then any block on which this block rests is also in the support set. The balancing set consists of all the blocks that do not belong to the support set. A stack is said to be spinal if its support set has a single block in each level, up to the level of the principal block. All the stacks shown in Figures 3 and 4 are thus spinal. It is very tempting to conclude, as done by Hall [9], that the optimal stacks are spinal. Surprisingly, the optimal stacks for n ≥ 20 are not spinal! Optimal stacks conJanuary 2009]

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blocks  20 overhang  2.32014

blocks  30 overhang  2.70909

Figure 5. Optimal stacks with 20 and 30 blocks.

taining 20 and 30 blocks are shown in Figure 5. Note that the right-hand contours of these stacks are not monotone, which is somewhat counterintuitive. For all n ≤ 30, we have searched exhaustively through all combinatorially distinct arrangements of n blocks and found optimal displacements numerically for each of these. Some of the resulting stacks were shown in Figures 3, 4, and 5. We are confident of their optimality, though we have no formal optimality proofs, as numerical techniques were used. While there seems to be a unique optimal placement of the blocks that belong to the support set of an optimal stack, there is usually a lot of freedom in the placement of the balancing blocks. Optimal stacks seem not to be unique for n ≥ 4. In view of the non-uniqueness and added complications caused by balancing blocks, it is natural to consider loaded stacks, which consist only of a support set with some external forces (or point weights) attached to some of their blocks. We will take the weight of each block to be 1; the size, or weight, of a loaded stack is defined to be the number of blocks contained in it plus the sum of all the point weights attached to it. The point weights are not required to be integral. Loaded stacks of weight 40, 60, 80, and 100, which are believed to be close to optimal, are shown in Figure 6. The stack of weight 100, for example, contains 49 blocks in its support set. The sum of all the external forces applied to these blocks is 51. As can be seen, the stacks become more and more non-spinal. It is also interesting to note that the stacks of Figure 6 contain small gaps that seem to occur at irregular positions. (There is also a scarcely visible gap between the two blocks at the second level of the 20-block stack of Figure 5.) That harmonic stacks are balanced can be verified using simple center-of-mass considerations. These considerations, however, are not enough to verify the balance of more complicated stacks, such as those in Figures 3, 4, 5, and 6. A formal mathematical definition of “balanced” is given in the next section. Briefly, a stack is said to be balanced if there is an appropriate set of forces acting between the blocks of the stacks, and between the blocks at the lowest level and the table, under which all blocks are in equilibrium. A block is in equilibrium if the sum of the forces and the sum of the moments acting upon it are both 0. As shown in the next section, the balance of a given stack can be determined by checking whether a given set of linear inequalities has a feasible solution. Given the fact that the 3-block stack that achieves the maximum overhang is an inverted 2-triangle (see Figure 2), it is natural to enquire whether larger inverted triangles are also balanced. Unfortunately, the next inverted triangle is already unbalanced and 22

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Figure 6. Loaded stacks, believed to be close to optimal, of weight 40, 60, 80, and 100.

would collapse in the way indicated in Figure 7. Inverted triangles show that simple center-of-mass considerations are not enough to determine the balance of stacks. As an indication that balance issues are not always intuitive, we note that inverted triangles are falsely claimed by Jargodzki and Potter [11, Challenge 271: A staircase to infinity, p. 246] to be balanced. Another appealing structure, the m-diamond, illustrated for m = 4 and 5 in Figure 8, consists of a symmetric diamond shape with rows of length 1, 2, . . . , m − 1, m,

Figure 7. The balanced inverted 2-triangle and the unbalanced inverted 3-triangle.

Figure 8. The balanced 4-diamond and the unbalanced 5-diamond.

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m − 1, . . . , 2, 1. Small diamonds were considered by Drummond [3]. The m-diamond uses m 2 blocks and would give an overhang of m/2, but unfortunately it is unbalanced for m ≥ 5. A 5-diamond would collapse in the way indicated in the figure. An mdiamond could be made balanced by adding a column of sufficiently many blocks resting on the top block. The methodology introduced in Section 3 can be used to show that, for m ≥ 5, a column of at least 2m − m 2 − 1 blocks would be needed. We can show that this number of blocks is also sufficient, giving a stack of 2m − 1 blocks with an overhang of m/2. It is interesting to note that these stacks are already better than the classical harmonic stacks, as with n = 2m − 1 blocks they give an overhang of 12 log2 (n + 1)  0.693 ln n. Determining the exact overhang achievable using n blocks, for large values of n, seems to be a formidable task. Our main goal in this paper is to determine the asymptotic growth of this quantity. Our main result is that there exists a constant c > 0 such that an overhang of cn 1/3 is achievable using n blocks. Note that this is an exponential improvement over the 12 ln n + O(1) overhang of harmonic stacks and the ln n + O(1) overhang of the best spinal stacks! In a subsequent paper [15], with three additional coauthors, we show that our improved stacks are asymptotically optimal, i.e., there exists a constant C > 0 such that the overhang achievable using n blocks is at most Cn 1/3 . Our stacks that achieve an asymptotic overhang of cn 1/3 , for some c > 0, are quite simple. We construct an explicit sequence of stacks, called parabolic stacks, with the r th stack in the sequence containing about 2r 3 /3 blocks and achieving an overhang of r/2. One stack in this sequence is shown in Figure 9. The balance of the parabolic stacks is established using an inductive argument.

Figure 9. A parabolic stack consisting of 111 blocks and giving an overhang of 3.

The remainder of this paper is organized as follows. In the next section we give formal definitions of all the notions used in this paper. In Section 3 we analyze spinal stacks. In Section 4, which contains our main results, we introduce and analyze our parabolic stacks. In Section 5 we describe some experimental results with stacks that seem to improve, by a constant factor, the overhang achieved by parabolic stacks. We end in Section 6 with some open problems. 2. STACKS AND THEIR BALANCE. As the maximum overhang problem is physical in nature, our first task is to formulate it mathematically. We consider a 2-dimensional version of the problem. This version captures essentially all the interesting features of the overhang problem. 24

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A block is a rectangle of length 1 and height h with uniform density and unit weight. (We shall see shortly that the height h is unimportant.) We assume that the table occupies the quadrant x, y ≤ 0 of the 2-dimensional plane. A stack is a collection of blocks. We consider only orthogonal stacks in which the sides of the blocks are parallel to the axes, with the length of each block parallel to the x-axis. The position of a block is then determined by the coordinate (x, y) of its lower left corner. Such a block occupies the box [x, x + 1] × [y, y + h]. A stack composed of n blocks is specified by the sequence (x1 , y1 ), . . . , (xn , yn ) of the coordinates of the lower left corners of its blocks. We require each yi to be a nonnegative integral multiple of h, the height of the blocks. Blocks can touch each other but are not allowed to overlap. The overhang n of the stack is 1 + maxi=1 xi . A block at position (x1 , y1 ) rests on a block in position (x2 , y2 ) if |x1 − x2 | < 1 and y1 − y2 = h. The interval of contact between the two blocks is then [max{x1 , x2 }, 1 + min{x1 , x2 }] × {y1 }. A block placed at position (x, 0) rests on the table if x < 0. The interval of contact between the block and the table is [x, min{x + 1, 0}] × {0}. When block A rests on block B, the two blocks may exert a (possibly infinitesimal) force on each other at every point along their interval of contact. A force is a vector acting at a specified point. By Newton’s third law, forces come in opposing pairs. If a force f is exerted on block A by block B, at (x, y), then a force − f is exerted on block B by block A, again at (x, y). We assume that edges of all the blocks are completely smooth, so that there is no friction between them. All the forces exerted on block A by block B, and vice versa, are therefore vertical forces. Furthermore, as there is nothing that holds the blocks together, blocks A and B can push, but not pull, one another. Thus, if block A rests on block B, then all the forces applied on block A by block B point upward, while all the forces applied on block B by block A point downward, as shown on the left in Figure 10. Similar forces are exerted between the table and the blocks that rest on it.

Figure 10. Equivalent sets of forces acting between two blocks.

The distribution of forces acting between two blocks may be hard to describe explicitly. Since all these forces point in the same direction, they can always be replaced by a single resultant force acting at some point within their interval of contact, as shown in the middle drawing of Figure 10. As an alternative, they may be replaced by two resultant forces that act at the endpoints of the contact interval, as shown on the right in Figure 10. Forces acting between blocks and between the blocks and the table are said to be internal forces. Each block is also subjected to a downward gravitational force of unit size, acting at its center of mass. As the blocks are assumed to be of uniform density, the center of mass of a block whose lower left corner is at (x, y) is at (x + 12 , y + h2 ). A rigid body is said to be in equilibrium if the sum of the forces acting on it, and the sum of the moments they apply on it, are both zero. A 2-dimensional rigid body acted January 2009]

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upon by k vertical forces f 1 , f 2 , . . . , f k at (x1 , y1 ), . . . , (xk , yk ) is in equilibrium if k k and only if i=1 f i = 0 and i=1 xi f i = 0. (Note that f 1 , f 2 , . . . , f k are scalars that represent the magnitudes of vertical forces.) A collection of internal forces acting between the blocks of a stack, and between the blocks and the table, is said to be a balancing set of forces if the forces in this collection satisfy the requirements mentioned above (i.e., all the forces are vertical, they come in opposite pairs, and they act only between blocks that rest on each other) and if, taking into account the gravitational forces acting on the blocks, all the blocks are in equilibrium under this collection of forces. We are now ready for a formal definition of balance. Definition 2.1 (Balance). A stack of blocks is balanced if and only if it admits a balancing set of forces. Static balance problems of the kind considered here are often under-determined, so that the resultants of balancing forces acting between the blocks are usually not uniquely determined. It was the consideration by one of us of balance issues that arise in the game of Jenga [22] which stimulated this current work. The following theorem shows that the balance of a given stack can be checked efficiently. Theorem 2.2. The balance of a stack containing n blocks can be decided by checking the feasibility of a collection of linear equations and inequalities with O(n) variables and constraints. Proof. Let (x1 , y1 ), . . . , (xn , yn ) be the coordinates of the lower left corners of the blocks in the stack. Let Bi , for 1 ≤ i ≤ n, denote the ith block of the stack, and let B0 denote the table. Let Bi /B j , where 0 ≤ i, j ≤ n, signify that Bi rests on B j . If Bi /B j , we let ai j = max{xi , x j } and bi j = min{xi , x j } + 1 be the x-coordinates of the endpoints of the interval of contact between blocks i and j. (If j = 0, then ai0 = xi and bi0 = min{xi + 1, 0}.) For all i and j such that Bi /B j , we introduce two variables f i0j and f i1j that represent the resultant forces that act between Bi and B j at ai j and bi j . By Definition 2.1 and the discussion preceding it, the stack is balanced if and only if there is a feasible solution to the following set of linear equalities and inequalities:   ( f i0j + f i1j ) − ( f ki0 + f ki1 ) = 1, for 1 ≤ i ≤ n; j : Bi /B j



k : Bk /Bi

(ai j f i0j + bi j f i1j ) −

j : Bi /B j



1 (aki f ki0 + bki f ki1 ) = xi + , for 1 ≤ i ≤ n; 2 k : Bk /Bi

f i0j , f i1j ≥ 0,

for i, j such that Bi /B j .

The first 2n equations require the forces applied on the blocks to exactly cancel the forces and moments exerted on the blocks by the gravitational forces. (Note that the table is not required to be in equilibrium.) The inequalities f i0j , f i1j ≥ 0, for every i and j such that Bi /B j , require the forces applied on Bi by B j to point upward. As a unit length block can rest on at most two other unit length blocks, the number of variables is at most 4n and the number of constraints is therefore at most 6n. The feasibility of such a system of linear equations and inequalities can be checked using linear programming techniques. (See, e.g., Schrijver [17].) 26

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Definition 2.3 (Maximum overhang, the function D(n)). The maximum overhang that can be achieved using a balanced stack comprising n blocks of length 1 is denoted by D(n). We now repeat the definitions of the principal block, the support set, and the balancing set sketched in the introduction. Definitions 2.4 (Principal block, support set, balancing set). The block of a stack that achieves the maximum overhang is the principal block of the stack. If several blocks achieve the maximum overhang, the lowest one is chosen. The support set of a stack is defined recursively as follows: the principal block is in the support set, and if a block is in the support set then any block on which this block rests is also in the support set. The balancing set consists of any blocks that do not belong to the support set. The blocks of the support sets of the stacks in Figures 3, 4, and 5 are shown in light gray while the blocks in the balancing sets are shown in dark gray. The purpose of blocks in the support set, as its name indicates, is to support the principal block. The blocks in the balancing set, on the other hand, are used to counterbalance the blocks in the support set. As already mentioned, there is usually a lot of freedom in the placement of the blocks of the balancing set. To concentrate on the more important issue of where to place the blocks of the support set, it is useful to introduce the notion of loaded stacks. Definitions 2.5 (Loaded stacks, the function D∗ (w)). A loaded stack consists of a set of blocks with some point weights attached to them. The weight of a loaded stack is the sum of the weights of all the blocks and point weights that participate in it, where the weight of each block is taken to be 1. A loaded stack is said to be balanced if it admits a balancing set of forces, as for unloaded stacks, but now also taking into account the point weights. The maximum overhang that can be achieved using a balanced loaded stack of weight w is denoted by D ∗ (w). Clearly D ∗ (n) ≥ D(n), as a standard stack is a trivially loaded stack with no point weights. When drawing loaded stacks, as in Figure 6, we depict point weights as external forces acting on the blocks of the stack, with the length of the arrow representing the force proportional to the weight of the point weight. (Since forces can be transmitted vertically downwards through any block, we may assume that point weights are applied only to upper edges of blocks outside any interval of contact.) As the next lemma shows, balancing blocks can always be replaced by point weights, yielding loaded stacks in which all blocks belong to the support set. Lemma 2.6. For every balanced stack that contains k blocks in its support set and n − k blocks in its balancing set, there is a balanced loaded stack composed of k blocks, all in the support set, and additional point weights of total weight n − k that achieves the same overhang. Proof. Consider the set of forces exerted on the support set of the stack by the set of balancing blocks. From the definition of the support set, no block of the support set can rest on any balancing block; therefore the effect of the balancing set can be represented by a set of downward vertical forces on the support set, or equivalently by a finite set January 2009]

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of point weights attached to the support set, with the same total weight as the set of balancing blocks. Given a loaded stack of integral weight, it is in many cases possible to replace the set of point weights by a set of appropriately placed balancing blocks. In some cases, however, such a conversion of a loaded stack into a standard stack is not possible. The optimal loaded stacks of weight 3, 5, and 7 cannot be converted into standard stacks without decreasing the overhang, as the number of point weights needed is larger than the number of blocks remaining. (The cases of weights 3 and 5 are shown in √ 11−2 6 ∗ Figure 11.) In particular, we get that D (3) = 6 > D(3) = 1. Experiments with optimal loaded stacks lead us, however, to conjecture that the difference D ∗ (n) − D(n) tends to 0 as n tends to infinity.

Figure 11. Optimal loaded stacks of weight 3 and 5.

Conjecture 2.7. D(n) = D ∗ (n) − o(1). 3. SPINAL STACKS. In this section we focus on a restricted, but quite natural, class of stacks which admits a fairly simple analysis. Definitions 3.1 (Spinal stacks, spine). A stack is spinal if its support set has just a single block at each level. The support set of a spinal stack is referred to as its spine. The optimal stacks with up to 19 blocks, depicted in Figures 3 and 4, are spinal. The stacks of Figure 5 are not spinal. A stack is said to be monotone if the x-coordinates of the rightmost blocks in the various levels, starting from the bottom, form an increasing sequence. It is easy to see that every monotone stack is spinal. Definitions 3.2 (The functions S(n), S∗ (w), and S∗k (w)). Let S(n) be the maximum overhang achievable using a spinal stack of size n. Similarly, let S ∗ (w) be the maximum overhang achievable using a loaded spinal stack of weight w, and let Sk∗ (w) be the maximum overhang achievable using a spinal stack of weight w with exactly k blocks in its spine. It is tempting to make the (false) assumption that optimal stacks are spinal. (As mentioned in the introduction, this assumption is implicit in [9].) The assumption holds, however, only for n ≤ 19. (See the discussion following Theorem 4.4.) As spinal stacks form a very natural class of stacks, it is still interesting to investigate the maximum overhang achievable using stacks of this class. A generic loaded spinal stack with k blocks in its spine is shown in Figure 12. We denote the blocks from top to bottom as B1 , B2 , . . . , Bk , with B1 being the principal 28

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Figure 12. A generic loaded spinal stack.

block. We regard the tabletop as Bk+1 . For 1 ≤ i ≤ k, the weight attached to the left edge of Bi is denoted by wi , and the relative overhang of Bi beyond Bi+1 is denoted  by di . We define ti = ij =1 (1 + w j ), the total downward force exerted upon Bi+1 by block Bi . We also k define t0 = 0. Note that ti = ti−1 + wi + 1, for 1 ≤ i ≤ k, and that wi , the total weight of the loaded stack. tk = w = k + i=1 The assumptions made in Figure 12, that each block is supported by a force that acts along the right-hand edge of the block underneath it and that all point weights are attached to the left-hand ends of blocks, are justified by the following lemma. Lemma 3.3. In an optimal loaded spinal stack: (i) Each block is supported by a force acting along the right-hand edge of the block underneath it. In particular, the stack is monotone. (ii) All point weights are attached to the left-hand ends of blocks. Proof. For (i), suppose there were some block Bi+1 (1 ≤ i ≤ k) where the resultant force exerted on it from Bi does not go through its right-hand end. If i < k then Bi+1 could be shifted some distance to the left and Bi together with all the blocks above it shifted to the right in such a way that the resultant force from Bi+1 on Bi+2 remains unchanged in position and the stack is still balanced. In the case of i = k (where Bk+1 is the tabletop), the whole stack could be moved to the right. The result of any such change is a balanced spinal stack with an increased overhang, a contradiction. As an immediate consequence, we get that optimal spinal stacks are monotone. For (ii), suppose that some block has weights attached other than at its left-hand end. We may replace all such weights by the same total weight concentrated at the left end. The result will be to move the resultant force transmitted to any lower block somewhat to the left. Since the stack is monotone, this change cannot unbalance the stack, and indeed would then allow the overhang to be increased by slightly shifting all blocks to the right; again a contradiction. We next note that for any nonnegative point weights w1 , w2 , . . . , wk ≥ 0, there are appropriate positive displacements d1 , d2 , . . . , dk > 0 for which the generic spinal stack of Figure 12 is balanced. January 2009]

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Lemma 3.4. A loaded spinal stack with k blocks in its spine that satisfies the two conditions of Lemma 3.3 is balanced if and only if di =

wi + ti

1 2

=1−

ti−1 + 12 , ti

for 1 ≤ i ≤ k. Proof. The lemma is verified using a simple calculation. The net downward force acting on Bi is (wi + ti−1 + 1) − ti = 0, by the definition of ti . (Recall that ti = i j =1 (1 + w j ).) The net moment acting on Bi , computed relative to the right-hand edge of Bi , is di ti − ( 12 + wi ), which vanishes if and only if di =

1 2

ti−1 + 12 + wi =1− , ti ti

as required. Note, in particular, that if wi = 0 for 1 ≤ i ≤ k, then ti = i and di = 2i1 , and we are back to the classic harmonic stacks. We can now also justify the claim made in the introduction concerning the unbalanced nature of diamond stacks. Consider the spine of an m-diamond. In this case, di = 12 for all i and so the balance conditions give the equations ti = 2ti−1 + 1 for 1 ≤ i ≤ m. As t0 = 0, we have ti ≥ 2i − 1 for all i and hence tm ≥ 2m − 1. Since tm is the total weight of the stack, the number of extra blocks required to be added for balance is at least 2m − 1 − m 2 , which is positive for m ≥ 5. Next, we characterize the choice of the weights w1 , w2 , . . . , wk , or alternatively of the total loads t1 , t2 , . . . , tk , that maximizes the overhang achieved by a spinal stack of total weight w. (Note that wi = ti − ti−1 − 1, for 1 ≤ i ≤ k.) Lemma 3.5. If a loaded spinal stack with total weight w and with k blocks in its spine achieves the maximal overhang of Sk∗ (w), then for some j (1 ≤ j ≤ k) we have ti2 = (ti−1 + 12 )ti+1 , for 1 ≤ i < j, and wi = 0, for j < i ≤ k. Proof. Let w1 , w2 , . . . , wk be the point weights attached to the blocks of an optimal spinal stack with overhang Sk∗ (w). For some i satisfying 1 ≤ i < k and a small x, consider the stack obtained by increasing the point weight at the left-hand end of block Bi from wi to wi + x, and decreasing the point weight on Bi+1 from wi+1 to wi+1 − x, assuming that wi+1 ≥ x. Note that this small perturbation does not change the total weight of the stack. The overhang of the perturbed stack is    wj + 1 wi+1 − x + 12 ti−1 + 12 2 + V (x) = 1 − + . ti + x ti+1 t j j =i,i+1 The first two terms in the expression above are the new displacements di (x) and di+1 (x). Note that all other displacements are unchanged. Differentiating V (x) we get ti−1 + 12 1 V (x) = − (ti + x)2 ti+1

30

and



V (0) =

ti−1 + ti2

1 2

−

1 . ti+1

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If wi = 0 while wi+1 > 0, then ti−1 = ti − 1 and ti+1 > ti + 1, which in conjunction with ti ≥ 1 implies that V (0) > 0, contradicting the optimality of the stack. Thus, if in an optimal stack we have wi = 0, then also wi+1 = wi+2 = · · · = wk = 0. If wi , wi+1 > 0, then we must have V (0) = 0, or equivalently ti2 = (ti−1 + 12 )ti+1 , as claimed. The optimality equations given in Lemma 3.5 can be solved numerically to obtain the values of Sk∗ (w) for specific values of w and k. The value of S ∗ (w) is then found by optimizing over k. The optimal loaded spinal stacks of weight 3 and 5, which also turn out to be the optimal loaded stacks of these weights, are shown in Figure 11. The optimality equations of Lemma 3.5 were also used to compute the spines of the optimal stacks with up to 19 blocks shown in Figures 3 and 4. The spines of the stacks with 3 and 5 blocks were obtained by adding the requirement that no point weight be attached to the topmost block of the spine. A somewhat larger example is given on the top left of Figure 14 where the optimal loaded spinal stack of weight 100 is shown. It is interesting to note that the point weights in optimal spinal stacks form an almost arithmetical progression. This observation is used in the proof of Theorem 3.8. Numerical experiments suggest that for every w ≥ 1, all the point weights in the spinal stacks with overhang S ∗ (w) are nonzero. There are, however, non-optimal values of k for which some of the bottom blocks in the stack that achieves an overhang of Sk∗ (w) have no point weights attached to them. We next show, without explicitly using the optimality conditions of Lemma 3.5, that S ∗ (w) = ln w + (1). Theorem 3.6. S ∗ (w) < ln w + 1. Proof. Forfixed total weight w = tk and fixed k, the largest possible overhang k Sk∗ (w) = i=1 di is attained when the conditions of Lemmas 3.3 and 3.4 (and 3.5) hold. Thus, as t0 = 0,   k k k    ti−1 + 12 ti−1 1−
ti−1 , ti

we see that Sk∗ (w) < k −

k 

xi

and

i=2

k 

xi =

i=2

t1 1 ≥ . tk w

The minimum sum for a finite set of positive real numbers with fixed product is attained when the numbers are equal, hence 1

Sk∗ (w) < k − (k − 1)w − k−1 . Let z =

k−1 , ln w

1

so that k − 1 = z ln w and w − k−1 = e−1/z . Then Sk∗ (w) < 1 + z ln w(1 − e−1/z ) < 1 + ln w,

as z(1 − e−1/z ) ≤ 1, for every z > 0. Corollary 3.7. S(n) < ln n + 1. January 2009]

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We can now describe a construction of loaded spinal stacks which achieves an overhang agreeing asymptotically with the upper bound proved in Theorem 3.6. Theorem 3.8. S ∗ (w) > ln w − 1.313. √ Proof. We construct a spine with k =  w blocks in it with wi = 2(i − 1), for 1 ≤ i ≤ k. It follows easily by induction that ti = i 2 , for 1 ≤ i ≤ k. In particular, the total weight of the stack is tk = k 2 ≤ w, as required. By Lemma 3.4, we get that di =

wi + ti

1 2

=

2(i − 1) + i2

1 2

=

3 2 − 2. i 2i

Thus, ∗

S (w) ≥

k  i=1

di = 2

k  1 i=1

> ln w + 2γ −

√

 w k 3 1 3  1 √ − = 2H −  w i 2 i=1 i 2 2 i=1 i 2

π2 > ln w − 1.313. 4

In the above inequality, γ  0.5772156 is Euler’s gamma.

Figure 13. A spinal stack with a shield.

We next discuss a technique that can be used to convert loaded spinal stacks into standard stacks. This is of course done by constructing balancing sets that apply the required forces on the left-hand edges of the spine blocks. The first step is the placement of shield blocks on top of the spine blocks, as shown in Figure 13. We let Bi , for 0 ≤ i ≤ k − 1, be the shield block placed on top of spine block Bi+1 and alongside spine block Bi for i > 0. We let yi be the x-coordinate of the left edge of Bi , for 1 ≤ i ≤ k − 1. Note that xi+1 − 1 < yi ≤ xi − 1, where xi is the x-coordinate of the left edge of Bi . 32

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Shield block Bi applies a downward force of wi+1 on Bi+1 . The force is applied at xi+1 , i.e., at the left edge of Bi+1 . Block Bi also applies a downward force of u i+1 on



Bi+1 at z i+1 , where yi ≤ z i+1 ≤ yi+1 + 1. Similarly, block Bi−1 applies a downward

force of u i on Bi at z i . Finally a downward external force of vi is applied on the left edge of Bi . The goal of the shield blocks is to aggregate the forces that should be applied on the spine blocks and to replace them by a set of fewer integral forces that are to be applied on the shield blocks. We will therefore place the shield blocks and choose the forces u i and their positions in such a way that most of the vi will be 0. (This is why we use dashed arrows to represent the vi forces in Figure 13.) The shield blocks are in equilibrium provided that the following balance conditions are satisfied: u i + vi + 1 = u i+1 + wi+1 , z i u i + yi vi + (yi + 12 ) = z i+1 u i+1 + xi+1 wi+1 , for 1 ≤ i ≤ k − 1. (We define u k = 0.) It is easy to see that if u i+1 , wi+1 , xi+1 , yi+1 , and z i+1 are set, then any choice of vi uniquely determines u i and z i . The choice is feasible if u i , vi ≥ 0 and yi−1 ≤ z i . In our constructions, we used the following heuristic to place the shield blocks and specify the forces between them. We start placing the shield blocks from the bottom up. In most cases, we choose yi = xi − 1 and vi = 0, i.e., Bi is adjacent to Bi and no external force is applied to it. Eventually, however, it may happen that z i+1 < xi − 1, which makes it impossible to place Bi adjacent to Bi and still apply the force u i+1 down on Bi+1 at z i+1 . In that case we choose yi = z i+1 . A more significant event, that usually occurs soon after the previous event, is when z i+1 ≤ xi+1 − 1, in which case

no placement of Bi allows it to apply the forces u i+1 and vi+1 on Bi+1 and Bi+1 at the required positions, as they are at least a unit distance apart. In this case, we introduce a nonzero, integral, external force vi+1 as follows. We let vi+1 = (1 − z i+1 + yi+1 )u i+1 and then recompute u i+1 and z i+1 . It is easy to check that u i+1 , vi+1 ≥ 0 and that yi ≤ z i+1 ≤ yi+1 + 1. If we now have z i+1 > xi+1 − 1, then the process can continue. Otherwise we stop. In our experience, we were always able to use this process to place all the shield blocks, except for a very few top ones. The vi forces left behind tend to be few and far apart. When this process is applied, for example, on the optimal loaded spinal stack of weight 100, only one such external force is needed, as shown in the second diagram of Figure 14. The nonzero vi ’s can be easily realized by erecting appropriate towers, as shown at the bottom of Figure 14. The top part of the balancing set is then designed by solving a small linear program. We omit the fairly straightforward details. The overhang achieved by the spinal stack shown at the bottom of Figure 14 is about 3.6979, which is a considerable improvement on the 2.5937 overhang of a 100block harmonic stack, but is also substantially less than the 4.23897 overhang of the non-spinal loaded stack of weight 100 given in Figure 6. Using the heuristic described above we were able to fit appropriate balancing sets for all optimal loaded spinal stacks of integer weight n, for every n ≤ 1000, with the exception of n = 3, 5, 7. We conjecture that the process succeeds for every n = 3, 5, 7. Conjecture 3.9. S(n) = S ∗ (n) for n = 3, 5, or 7. 4. PARABOLIC STACKS. We now give a simple explicit construction of n-block stacks with an overhang of about (3n/16)1/3 , an exponential improvement over the January 2009]

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Figure 14. Optimal loaded spinal stack of weight 100 (top left), with shield added (top right) and with a complete balancing set added (bottom).

O(log n) overhang achievable using spinal stacks in general and the harmonic stacks in particular. Though the stacks of this sequence are not optimal (see the empirical results of the next section), they are within a constant factor of optimality, as will be shown in a subsequent paper [15]. The stacks constructed in this section are what we term brick-wall stacks. The blocks in each row are contiguous, and each is centered over the ends of blocks in the row beneath. This resembles the simple “stretcher-bond” pattern in real-life bricklaying. Overall the stacks have a symmetric roughly parabolic shape, hence the name, with vertical axis at the table edge and a brick-wall structure. An illustration of a 111block parabolic 6-stack with overhang 3 was given in Figure 9. An r -row is a row of r adjacent blocks, symmetrically placed with respect to x = 0. An r -slab, for r ≥ 2, has height 2r − 3 and consists of alternating r -rows and (r − 1)rows, starting and finishing with r -rows. An r -slab therefore contains r (r − 1) + (r − 1)(r − 2) = 2(r − 1)2 blocks. Figure 15 shows r -slabs, for r = 2, 3, . . . , 6. A

Figure 15. A 6-stack composed of r -slabs, for r = 2, 3, . . . , 6, and an additional block.

34

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parabolic d-stack, or just d-stack, for short, is a d-slab on a (d − 1)-slab on . . . on a 2-slab on a single block. The slabs shown in Figure 15 thus compose a 6-stack. Lemma 4.1. A parabolic d-stack contains has an overhang of d2 .

d(d−1)(2d−1) 3

+ 1 blocks and, if balanced,

Proof. The number of blocks contained in a d-stack is 1+

d  r =2

2(r − 1)2 = 1 +

d(d − 1)(2d − 1) . 3

The overhang achieved, if the stack is balanced, is half the width of the top row, i.e., d2 . In preparation for proving the balance of parabolic stacks, we show in the next lemma that a slab can concentrate a set of forces acting on its top together with the weights of its own blocks down into a narrower set of forces acting on the row below it. The lemma is illustrated in Figure 16.

Figure 16. A 6-slab with a grey 5-slab contained in it.

Lemma 4.2. For any g ≥ 0, an r -slab with forces of g, 2g, 2g, . . . , 2g, g acting downwards onto its top row at positions r −2 r −4 r −2 r r ,− ,..., , , − ,− 2 2 2 2 2 respectively, can be balanced by applying a set of upward forces g , 2g , 2g , . . . , 2g , g , r where g = r −1 g + r − 1, on its bottom row at positions −

r −3 r −1 r −1 r −3 ,− ,..., , , 2 2 2 2

respectively. Proof. The proof is by induction on r . For r = 2, a 2-slab is just a 2-row, which is clearly balanced with downward forces of g, 2g, g at −1, 0, 1 and upward forces of 2g + 1, 2g + 1 at − 12 , 12 , when half of the downward force 2g acting at x = 0 is applied January 2009]

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on the right-hand edge of the left block and the other half applied on the left-hand edge of the right block. For the induction step, we first observe that for any r ≥ 2 an (r + 1)-slab can be regarded as an r -slab with an (r + 1)-row added above and below and with an extra block added at each end of the r − 2 rows of length r − 1 of the r -slab. The 5-slab (shaded) contained in a 6-slab together with the added blocks is shown in Figure 16.

Figure 17. The proof of Lemma 4.2.

Suppose the statement of the lemma holds for r -slabs and consider an (r + 1)-slab with the supposed forces acting on its top row. Let f = g/r , so that g = r f . As in the basis of the induction, the top row can be balanced by r + 1 equal forces of 2r f + 1 from below (the 1 is for the weight of the blocks in the top row) acting at positions − r2 , − r −2 , . . . , r −2 , r2 . As 2 2 2r f + 1 = (r − 1) f + ((r + 1) f + 1) = 2(r − 1) f + 2 f + 1, we can express this constant sequence of r + 1 forces as the sum of the following two force sequences: (r − 1) f, (r + 1) f + 1,

2(r − 1) f, 2(r − 1) f, . . . , 2(r − 1) f, 2(r − 1) f, 2 f + 1,

2 f + 1,

... ,

2 f + 1,

2 f + 1,

(r − 1) f (r + 1) f + 1

The forces in the first sequence can be regarded as acting on the r -slab contained in the (r + 1)-slab, which then, by the induction hypothesis, yield downward forces on the bottom row of r f + r − 1, 2r f + 2(r − 1), . . . , 2r f + 2(r − 1), r f + r − 1 at positions − r −1 , − r −3 , . . . , r −3 , r −1 . 2 2 2 2 36

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The forces of the second sequence, together with the weights of the outermost blocks of the (r + 1)-rows, are passed straight down through the rigid structure of the r -slab to the bottom row. The combined forces acting down on the bottom row are now (r + 1) f + r −1, r f + r −1, 2 f + 1, 2r f + 2(r −1), 2 f + 1, . . . , 2 f + 1, r f + r −1, (r + 1) f + r −1 at positions − r2 , − r −1 , . . . , r −1 , r2 . The bottom row is in equilibrium when the se2 2 quence of upward forces (r + 1) f + r, 2(r + 1) f + 2r, 2(r + 1) f + 2r, . . . , 2(r + 1) f + 2r, 2(r + 1) f + 2r, (r + 1) f + r is applied on the bottom row at positions − r2 , − r −2 , . . . , r −2 , r2 , as required. 2 2 Theorem 4.3. For any d ≥ 2, a parabolic d-stack is balanced, contains d(d − 1)(2d − 1) +1 3 blocks, and has an overhang of d2 . Proof. The balance of a parabolic d-stack follows by a repeated application of Lemma 4.2. For 2 ≤ r ≤ d, let g(r ) denote the value of g in Lemma 4.2 for the r -slab in the d-stack. Although the argument does not rely on the specific values that d−1 2 g(r ) assumes, it can be verified that g(r ) = r1 i=r i . Note that g(d) = 0, as no downward forces are exerted on the top row of the d-slab, which is also the top row of r the d-stack, and that g(r − 1) = r −1 g(r ) + r − 1, as required by Lemma 4.2. Theorem 4.4. D(n) ≥ ( 3n )1/3 − 16

1 4

for all n.

Proof. Choose d so that (d − 1)d(2d − 1) d(d + 1)(2d + 1) +1≤n ≤ . 3 3 Then Theorem 4.3 shows that a d-stack yields an overhang of d/2 and can be constructed using n or fewer blocks. Any extra blocks can be just placed in a vertical pile in the center on top of the stack without disturbing balance (or arbitrarily scattered on the table). Hence 2(d + 12 )3 n< 3

 and so

D(n) ≥ d/2 >

3n 16

1/3

1 − . 4

In Section 3 we claimed that optimal stacks are spinal only for n ≤ 19. We can justify this claim for n ≤ 30 by exhaustive search, while comparison of the lower bound from Theorem 4.4 with the upper bound of S(n) < 1 + ln n from Corollary 3.7 deals with the range n ≥ 5000. The intermediate values of n can be covered by combining a few explicit constructions, such as the stack shown in Figure 20, with numerical bounds using Lemma 3.5. January 2009]

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Can parabolic d-stacks be built incrementally by laying one brick at a time? The answer is no, as the bottom three rows of a parabolic stack form an unbalanced inverted 3-triangle. The inverted 3-triangle remains unbalanced when the first block of the fourth row is laid down. Furthermore, the bottom six rows, on their own, are also not balanced. These, however, are the only obstacles to an incremental row-by-row and block-by-block construction of parabolic stacks and they can be overcome by the modified parabolic stacks shown in Figure 18. We simply omit the lowest block and move the whole stack half a block length to the left. The bricks can now be laid row by row, going in each row from the center outward, alternating between the left and right sides, with the left side, which is over the table, taking precedence. The numbers in Figure 18 indicate the order in which the blocks are laid. Thus, unlike with harmonic stacks, it is possible to construct an arbitrarily large overhang using sufficiently many blocks, without knowing the desired overhang in advance.

...

57

59 54 50

56 52 47

48 45

41

39

32

30 27

31

22

11 9

24

14

12 10 7

6

5

3 1

21 17

8

4

33 28

19 15

16

42 37

26

18

13

40

29

23 20

46

35

25

51

49

38 34

60 55

44

43

36

58 53

2

Figure 18. Incremental block-by-block construction of modified parabolic stacks.

5. GENERAL STACKS. We saw in Section 2 that the problem of checking whether a given stack is balanced reduces to checking the feasibility of a system of linear equations and inequalities. Similarly, the minimum total weight of the point weights that are needed to balance a given loaded stack can be found by solving a linear program. Finding a stack with a given number of blocks, or a loaded stack with a given total weight, that achieves maximum overhang seems, however, to be a much harder computational task. To do so, one should, at least in principle, consider all possible combinatorial stack structures and for each of them find an optimal placement of the blocks. The combinatorial structure of a stack specifies the contacts between the blocks of the stack, i.e., which blocks rest on which, and in what order (from left to right), and which rest on the table. The problem of finding a (loaded) stack with a given combinatorial structure with maximum overhang is again not an easy problem. As both the forces and their locations are now unknowns, the problem is not linear, but rather a constrained quadratic 38

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

programming problem. Though there is no general algorithm for efficiently finding the global optimum of such constrained quadratic programs, such problems can still be solved in practice using nonlinear optimization techniques. For stacks with a small number of blocks, we enumerated all possible combinatorial stack structures and numerically optimized each of them. For larger numbers of blocks this approach is clearly not feasible and we had to use various heuristics to cut down the number of combinatorial structures considered. The stacks of Figures 3, 4, 5, and 6 were found using extensive numerical experimentation. The stacks of Figures 3, 4, and 5 are optimal, while the stacks of Figure 6 are either optimal or very close to being so. The collections of forces that balance the loaded stacks of Figure 6 (and the loaded stacks contained in the stacks of Figures 3, 4, and 5) have the following interesting properties. First, the balancing collections of forces of these stacks are unique. Second, almost all downward forces in these collections are applied at the edges of blocks. The only exceptions occur when a downward force is applied on a right-protruding block, i.e., a rightmost block in a level that protrudes beyond the rightmost block of the level above it. In addition, all point weights are placed on the left-hand edges of left-protruding blocks, where left-protruding blocks are defined in an analogous way. The table, of course, supports the (only) block that rests on it at its right-hand edge. A collection of balancing forces that satisfies these conditions is said to be well-behaved. A schematic description of well-behaved collections of balancing forces is given in Figure 19. The two right-protruding blocks are shown with a slightly lighter shading. A right-protruding block is always adjacent to the block on its left. We conjecture that forces that balance optimal loaded stacks are always well-behaved.

Figure 19. A schematic description of a well-behaved set of balancing forces.

A useful property of well-behaved collections of balancing forces is that the total weight of the stack and the positions of its blocks uniquely determine all the forces in the collection. This follows from the fact that each block has either two downward forces acting upon it at specified positions, namely at its two edges, or just a single force in an unspecified position. Given the upward forces acting on a block, the downward force or forces acting upon it can be obtained by solving the force and moment January 2009]

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39

equations of the block. All the forces in the collection can therefore be determined in a bottom-up fashion. We conducted most of our experiments, on blocks with more than 30 blocks, on loaded stacks balanced by well-behaved sets of balancing forces. We saw in Section 2 that loaded stacks of total weight 3, 5, and 7 achieve a larger overhang than the corresponding unloaded stacks, simply because the number of blocks available for use in their balancing sets is smaller than the number of point weights to be applied. The loaded stacks of Figure 6 exhibit another trivial impediment to the conversion of loaded stacks into standard ones: the point weight to be applied in the lowest position has magnitude less than 1. Thus, these stacks can be converted into standard ones only after making some small adjustments. These adjustments have only a very small effect on the overhang achieved. Thus, although we believe that the difference between the maximum overhangs achieved by loaded and unloaded stacks is bounded by a small universal constant, we also believe that for most sizes, loaded stacks yield slightly larger overhangs. Although the placements of the blocks in the optimal, or close to optimal, stacks of Figure 6 are somewhat irregular, with some small (essential) gaps between blocks of the same layer, at a high level, these stacks seem to resemble brick-wall stacks, as defined in Section 4. This, and the fact that brick-wall stacks were used to obtain the (n 1/3 ) lower bound on the maximum overhang, indicate that it might be interesting to investigate the maximum overhang that can be achieved using brick-wall stacks. The parabolic brick-wall stacks of Section 4 were designed to enable a simple inductive proof of their balance. Parabolic stacks, however, are far from being optimal brick-wall stacks. The balanced 95-block symmetric brick-wall stack with an overhang of 4 depicted in Figure 20, for example, contains fewer blocks and achieves a larger overhang than that achieved by the 111-block overhang-3 parabolic stack of Figure 9.

Figure 20. A 95-block symmetric brick-wall stack with overhang 4.

Loaded brick-wall stacks are especially easy to experiment with. Empirically, we have again discovered that the minimum weight collections of forces that balance them turn out to be well-behaved, in the formal sense defined above. When the brick-wall stacks are symmetric with respect to the x = 0 axis, and have a flat top, point weights 40

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

are attached only to blocks at the top layer of the stack. Protruding blocks, both on the left and on the right, then simply serve as props, while all other blocks are perfect splitters, i.e., they are supported at the center of their lower edge and they support other blocks at the two ends of their upper edge. In non-symmetric brick-wall stacks it is usually profitable to use the left-protruding blocks as splitters and not as props, attaching point weights to their left ends. A schematic description of well-behaved forces that balance symmetric and asymmetric brick-wall stacks is shown in Figure 21. As can be seen, all forces in such well-behaved collections are linear functions of w, the total weight of the stack. This allows us, in particular, to find the minimum total weight needed to balance a brick-wall loaded stack without solving a linear program. We simply choose the smallest total weight w for which all forces are nonnegative. This observation enabled us to experiment with huge symmetric and asymmetric brickwall stacks.

Figure 21. A schematic description of well-behaved collections of forces that balance symmetric and asymmetric brick-wall stacks.

The best symmetric loaded brick-wall stacks with overhangs 10 and 50 that we have found are shown in Figures 22 and 23. Their total weights are about 1151.76 and 115,467, respectively. The blocks in the larger stack are so small that they are not shown individually. We again believe that these stacks are close to being the optimal stacks of their kind. They were found using a local search approach. In particular, these stacks cannot be improved by widening or narrowing layers, or by adding or removing single layers. Essentially the same symmetric stacks were obtained by starting from almost any initial stack and repeatedly improving it by widening, narrowing, adding, and removing layers. As can be seen from Figures 22 and 23, the shapes of optimal symmetric loaded stacks, after suitable scaling, seem to tend to a limiting curve. This curve, which we have termed the vase, is similar to but different from that of an inverted normal distribution. We have as yet no conjecture for its equation. We have conducted similar experiments with asymmetric loaded brick-wall stacks. The best such stack with overhang 10 that we have found is shown in Figure 24. Its total weight of about 1128.84 is about 3.38% less than the weight of the symmetric stack of Figure 22. The scaled shapes of optimal asymmetric loaded brick-wall stacks seem again to tend to a limiting curve which we have termed the oil lamp. We again have no conjecture for its equation. January 2009]

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41

weight  1151.76 blocks  1043 overhang  10 Figure 22. A symmetric loaded brick-wall stack with an overhang of 10.

weight  115467. blocks  112421 overhang  50 Figure 23. A scaled outline of a loaded brick-wall stack with an overhang of 50.

6. OPEN PROBLEMS. Some intriguing problems still remain open. In a subsequent paper [15], we show that the (n 1/3 ) overhang lower bound presented here is optimal, up to a constant factor, but it would be interesting to determine the largest constant cover for which overhangs of (cover − o(1))n 1/3 are possible. Can this constant cover be achieved using stacks that are simple to describe, e.g., brick-wall stacks, or simple modifications of them, such as brick-wall stacks with adjacent levels having a displacement other than 12 , or small gaps left between the blocks of the same level? What are the limiting vase and oil lamp curves? Do they yield, asymptotically, the maximum overhangs achievable using symmetric and general stacks? Another open problem is the relation between the maximum overhangs achievable using loaded and unloaded stacks. We believe, as expressed in Conjecture 2.7, that 42

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

weight  1112.88 blocks  921 overhang  10 Figure 24. An asymmetric loaded brick-wall stack with an overhang of 10.

the difference between these two quantities tends to 0 as the size of the √ stacks tends to infinity. We also conjecture that D ∗ (n) − D(n) ≤ D ∗ (3) − D(3) = 5−26 6  0.017, for every n ≥ 1. Our notion of balance, as defined formally in Section 2, allows stacks to be precarious: stacks that achieve maximum overhang are always on the verge of collapse. It is not difficult, however, to define more robust notions of balance, where there is some stability. In one such natural definition, a stack is stable if there is a balancing set of forces in which none of the forces acts at the edge of any block. We note in passing that Farkas’ lemma, or the theory of linear programming duality (see [17]), can be used to derive an equivalent definition of stability: a stack is stable if and only if every feasible infinitesimal motion of the blocks of the stack increases the total potential energy of the system. This requirement of stability raises some technical difficulties but does not substantially change the nature of the overhang problem. Our parabolic d-stacks, for example, can be made stable by adding a (d − 1)-row symmetrically placed on top. The proof of this is straightforward but not trivial. We believe that for any n = 3, the loss in the overhang due to this stricter definition is infinitesimal. Our analysis of the overhang problem was made under the no friction assumption. All the forces considered were therefore vertical. The presence of friction introduces horizontal forces and thus changes the picture completely, as also observed by Hall [9]. We can show that there is a fixed coefficient of friction such that the inverted triangles are all balanced, and so achieve overhang of order n 1/2 . REFERENCES 1. S. Ainley, Finely balanced, Mathematical Gazette 63 (1979) 272. 2. J. G. Coffin, Problem 3009, this M ONTHLY 30 (1923) 76. 3. J. E. Drummond, On stacking bricks to achieve a large overhang (Note 65.8), Mathematical Gazette 65 (1981) 40–42. 4. L. Eisner, Leaning tower of the Physical Review, Amer. J. Phys. 27 (1959) 121. 5. G. Gamow and M. Stern, Puzzle-Math, Viking, New York, 1958.

January 2009]

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6. M. Gardner, Mathematical games: Some paradoxes and puzzles involving infinite series and the concept of limit, Scientific American (Nov. 1964) 126–133. , Martin Gardner’s Sixth Book of Mathematical Games from Scientific American, W.H. Freeman, 7. San Francisco, 1971. 8. R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley Longman, Reading, MA, 1988. 9. J. F. Hall, Fun with stacking blocks, Amer. J. Phys. 73 (2005) 1107–1116. 10. J. E. Hearnshaw and M. S. Paterson, Problems drive, Eureka 27 (1964) 6–8 and 39–40. Available at http://archim.org.uk/eureka/27/problems.html and .../27/solutions.html. 11. C. P. Jargodzki and F. Potter, Mad About Physics: Braintwisters, Paradoxes, and Curiosities, John Wiley, New York, 2001. 12. P. B. Johnson, Leaning tower of lire, Amer. J. Phys. 23 (1955) 240. 13. G. M. Minchin, A Treatise on Statics: With Applications to Physics, 6th ed., Clarendon, Oxford, 1907. 14. M. Paterson and U. Zwick, Overhang, in Proceedings of the 17th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA’06), Society for Industrial and Applied Mathematics, Philadelphia, 2006, 231–240. 15. M. Paterson, Y. Peres, M. Thorup, P. Winkler, and U. Zwick, Maximum overhang, to appear in this M ONTHLY; available at http://arXiv.org/pdf/0707.0093. 16. J. B. Phear, Elementary Mechanics, MacMillan, Cambridge, 1850. 17. A. Schrijver, Theory of Linear and Integer Programming, John Wiley, New York, 1998. 18. R. T. Sharp, Problem 52, Pi Mu Epsilon Journal 1 (1953) 322. , Problem 52, Pi Mu Epsilon Journal 2 (1954) 411. 19. 20. R. Sutton, A problem of balancing, Amer. J. Phys. 23 (1955) 547. 21. W. Walton, A Collection of Problems in Illustration of the Principles of Theoretical Mechanics, 2nd ed., Deighton, Bell, Cambridge, 1855. 22. U. Zwick, Jenga, in Proceedings of the 13th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA’02), Association for Computing Machinery, New York, 2002, 243–246. MIKE PATERSON (Ph.D., FRS) took degrees in mathematics at Cambridge. A glimpse of his early interests can be found in the 1964 paper referenced above, before he rose to fame as the co-inventor with John Conway of Sprouts. He evolved from president of the Trinity Mathematical Society to president of the European Association for Theoretical Computer Science, and migrated from MIT to the University of Warwick, where he has been in the Computer Science department for 37 years. Department of Computer Science, University of Warwick, Coventry CV4 7AL, United Kingdom. [email protected]

URI ZWICK received his B.Sc. in Computer Science from the Technion, Israel Institute of Technology, and his M.Sc. and Ph.D. in Computer Science from Tel Aviv University, where he is currently a professor of Computer Science. His main research interests are: algorithms and complexity, combinatorial optimization, mathematical games, and recreational mathematics. In an early collaboration with Mike Paterson he determined the optimal strategy for playing the popular Memory Game, known also as Concentration. His favorite Hawaiian musician is Israel Kamakawiwo’ole. School of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel [email protected]

44

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A Probabilistic Proof of the Lindeberg-Feller Central Limit Theorem Larry Goldstein 1. INTRODUCTION. The Central Limit Theorem, one of the most striking and useful results in probability and statistics, explains why the normal distribution appears in areas as diverse as gambling, measurement error, sampling, and statistical mechanics. In essence, the Central Limit Theorem states that the normal distribution applies whenever one is approximating probabilities for a quantity which is a sum of many independent contributions all of which are roughly the same size. It is the LindebergFeller Theorem [4] which makes this statement precise in providing the sufficient, and in some sense necessary, Lindeberg condition whose satisfaction accounts for the ubiquitous appearance of the bell-shaped normal. Generally the Lindeberg condition is handled using Fourier or analytic methods and is somewhat hard to interpret. Here we provide a simpler, equivalent, and more easily interpretable probabilistic formulation of the Lindeberg condition and demonstrate its sufficiency and partial necessity in the Central Limit Theorem using more elementary means. The seeds of the Central Limit Theorem, or CLT, lie in the work of Abraham de Moivre, who, in 1733, not being able to secure himself an academic appointment, supported himself consulting on problems of probability and gambling. He approximated the limiting probabilities of the binomial distribution, the one which governs the behavior of the number Sn of successes in an experiment which consists of n independent trials, each one having the same probability p ∈ (0, 1) of success. Each individual trial of such an experiment can be modelled by X , a (Bernoulli) random variable which records one for each success and zero for each failure, P(X = 1) = p

and

P(X = 0) = 1 − p,

and has mean E X = p and variance Var(X ) = p(1 − p). The record of successes and failures in n independent trials is then given by an independent sequence X 1 , X 2 , . . . , X n of these Bernoulli variables, and the total number of successes Sn by their sum Sn = X 1 + · · · + X n .

(1)

Exactly, Sn has the binomial distribution, which specifies that   n k p (1 − p)n−k for k = 0, 1, . . . , n. P(Sn = k) = k  For even moderate values of n managing the binomial coefficients nk becomes unwieldy, to say nothing of computing the sum which yields the cumulative probability  n  P(Sn ≤ m) = p k (1 − p)n−k k k≤m that there will be m or fewer successes. January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

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The great utility of the CLT is in providing an easily computable approximation to such probabilities that can be quite accurate even for moderate values of n. To state this result, let Z denote a standard normal variable, that is, one with distribution function P(Z ≤ x) = (x) given by    x 1 1 2 (x) = ϕ(u)du where ϕ(u) = √ exp − u , (2) 2 2π −∞ and recall that we say a sequence of random variables Yn is said to converge in distribution to Y , written Yn →d Y , if lim P(Yn ≤ x) = P(Y ≤ x)

n→∞

for all continuity points x of P(Y ≤ x).

(3)

√ Letting Wn = (Sn − np)/ np(1 − p), the binomial standardized to have mean zero and variance one, obtained from Sn by subtracting its mean and dividing by its standard deviation, the CLT yields Wn →d Z. This result allows us to approximate the cumbersome cumulative binomial probability √ P(Sn ≤ m) by the simpler ((m − np)/ np(1 − p)), noting that (x) is continuous for all x, and therefore that the convergence of distribution functions in (3) here holds everywhere. It was only for the special case of the binomial that normal approximation was first considered. Only many years later with the work of Laplace around 1820 did it begin to be systematically realized that the same normal limit is obtained when the underlying Bernoulli variables are replaced by any variables with a finite variance. The result was the classical Central Limit Theorem, which states that Wn converges in distribution to Z whenever √ Wn = (Sn − nμ)/ nσ 2 is the standardization of a sum Sn , as in (1), of independent and identically distributed random variables each with mean μ and variance σ 2 . From this generalization it now becomes somewhat clearer why various distributions observed in nature, which may not be at all related to the binomial, such as the errors of measurement averages, or the heights of individuals in a sample, take on the bell-shaped form: each observation is the result of summing many small independent factors. A further extension of the classical CLT could yet come. In situations where the summands do not have identical distributions, can the normal curve still govern? For an example, consider the symmetric group S n , the set of all permutations π on the set {1, 2, . . . , n}. We can represent π ∈ S7 , for example, by two-line notation   1 2 3 4 5 6 7 π= 4 3 7 6 5 1 2 from which one can read that π(1) = 4 and π(4) = 6. This permutation can also be represented in the cycle notation π = (1, 4, 6)(2, 3, 7)(5) with the meaning that π maps 1 to 4, 4 to 6, 6 to 1, and so forth. From the cycle representation we see that π has two cycles of length 3 and one of length 1, for a total 46

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

of three cycles. In general, let K n (π) denote the total number of cycles in a permutation π ∈ Sn . If π is chosen uniformly from all the n! permutations in Sn , does the Central Limit Theorem imply that K n (π) is approximately normally distributed for large n? To answer this question we will employ the Feller coupling [3], which constructs a random permutation π uniformly from Sn with the help of n independent Bernoulli variables X 1 , . . . , X n with distributions P(X i = 0) = 1 −

1 i

P(X i = 1) =

and

1 , i

i = 1, . . . , n.

(4)

Begin the first cycle at stage 1 with the element 1. At stage i, i = 1, . . . , n, if X n−i+1 = 1 close the current cycle and begin a new one starting with the smallest number not yet in any cycle, and otherwise choose an element uniformly from those yet unused and place it to the right of the last element in the current cycle. In this way at stage i we complete a cycle with probability 1/(n − i + 1), upon mapping the last element of the current cycle to the one which begins it. As the total number K n (π) of cycles of π is exactly the number of times an element closes the loop upon completing its own cycle, K n (π) = X 1 + · · · + X n ,

(5)

a sum of independent, but not identically distributed, random variables. Hence, despite the similarity of (5) to (1), the hypotheses of the classical Central Limit Theorem do not hold. Nevertheless, in 1922 Lindeberg [8] provided a general condition which can be applied in this case to show that K n (π) is asymptotically normal. To explore Lindeberg’s condition, first consider the proper standardization of K n (π) in our example. As any Bernoulli random variable with success probability p has mean p and variance p(1 − p), the Bernoulli variable X i in (4) has mean i −1 and variance i −1 (1 − i −1 ) for i = 1, . . . , n. Thus, hn =

n  1 i=1

i

and

σn2

=

n   1 i=1

1 − 2 i i

 (6)

are the mean and variance of K n (π), respectively; the mean h n is known as the nth harmonic number. In particular, standardizing K n (π) to have mean zero and variance 1 results in Wn =

K n (π) − h n , σn

which, by absorbing the scaling inside the sum, can be written as Wn =

n  i=1

X i,n

where

X i,n =

X i − i −1 . σn

(7)

In general, it is both more convenient and more encompassing to deal not with a sequence of variables but rather with a triangular array as in (7) which satisfies the following condition. Condition 1.1. For every n = 1, 2, . . . , the random variables making up the collection Xn = {X i,n : 1 ≤ i ≤ n} are independent with mean zero and finite variances January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

47

2 σi,n = Var(X i,n ), standardized so that

Wn =

n 

X i,n

has variance

Var(Wn ) =

i=1

n 

2 σi,n = 1.

i=1

Of course, even under Condition 1.1, some further assumptions must be satisfied by the summand variables for Wn to converge in distribution to the normal. For instance, if the first variable accounts for some nonvanishing fraction of the total variability, it will strongly influence the limiting distribution, possibly resulting in the failure of normal convergence. Ruling out such situations, the Lindeberg-Feller CLT, see [4], says that Wn →d Z under the Lindeberg condition ∀ > 0

lim L n, = 0

n→∞

where

L n, =

n 

2 E{X i,n 1(|X i,n | ≥ )},

(8)

i=1

where for an event A, the ‘indicator’ random variable 1(A) takes on the value 1 if A occurs, and the value 0 otherwise. The appearance of the Lindeberg condition is justified by explanations such as the one given by Feller [4], who roughly says that it requires the individual variances be due mainly to masses in an interval whose length is small in comparison to the overall variance. We present a probabilistic condition which is seemingly simpler, yet equivalent. Our probabilistic approach to the CLT is through the so-called zero bias transformation introduced in [6], which was motivated by Stein’s characterization [10] of N (0, σ 2 ), the mean zero normal distribution with variance σ 2 . In particular, Stein proves that X has distribution N (0, σ 2 ) if and only if σ 2 E[ f (X )] = E[X f (X )]

(9)

for all absolutely continuous functions f for which these expectations exist. To motivate the zero bias transformation, let B be a Bernoulli random variable with success probability p ∈ (0, 1) and let X be the (nonnormal) centered Bernoulli variable B − p, having variance σ 2 = p(1 − p). Then, computing the right-hand side in Stein’s identity (9), we find E[X f (X )] = E[(B − p) f (B − p)] = p(1 − p) f (1 − p) − (1 − p) p f (− p) = σ 2 [ f (1 − p) − f (− p)]  1− p 2 f (u)du =σ −p

= σ E f (U ), 2

for U having uniform density over [− p, 1 − p]. Hence, a version of Stein’s identity holds for this nonnormal variable upon replacing the X appearing on the left-hand side of (9) by a variable with a distribution different from that of the given X . This calculation gives one instance of the zero bias transformation: the centered Bernoulli B − p is transformed to the uniform U over [− p, 1 − p]. With ∗ indicating the transformation and =d the equality of two random variables in distribution, we write (B − p)∗ =d U 48

where U has distribution U [− p, 1 − p].

(10)

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In fact, it turns out [6] that for every X with mean zero and finite, nonzero variance σ 2 , there exists a unique ‘X -zero biased distribution’ on an X ∗ which satisfies σ 2 E f (X ∗ ) = E[X f (X )]

(11)

for all absolutely continuous functions f for which these expectations exist. By Stein’s characterization, X ∗ and X have the same distribution if and only if X has the N (0, σ 2 ) distribution, that is, the normal distribution is the zero bias transformation’s unique fixed point. The Bernoulli example highlights the general fact that the distribution of X ∗ is always absolutely continuous, regardless of the nature of the distribution of X , and that the zero bias transformation moves a nonnormal distribution in some sense closer to normality. One way to see the ‘if’ direction of Stein’s characterization, that is, why the zero bias transformation fixes the normal, is to note that the density function ϕσ 2 (x) = σ −1 ϕ(σ −1 x) of a N (0, σ 2 ) variable, with ϕ(x) given by (2), satisfies the differential equation with a form ‘conjugate’ to (11), σ 2 ϕσ 2 (x) = −xϕσ 2 (x), and now (11), with X ∗ = X , follows for a large class of functions f by integration by parts. It is the uniqueness of the fixed point of the zero bias transformation, that is, the fact that X ∗ has the same distribution as X only when X is normal, that provides the probabilistic reason behind the CLT. This ‘only if’ direction of Stein’s characterization suggests that a distribution which gets mapped to one nearby is close to being a fixed point of the zero bias transformation, and therefore must be close to the transformation’s only fixed point, the normal. Hence the normal approximation should apply whenever the distribution of a random variable is close to that of its zero bias transformation. Moreover, the zero bias transformation has a special property that immediately shows why the distribution of a sum Wn of comparably sized independent random variables is close to that of Wn∗ : a sum of independent terms can be zero biased by randomly selecting a single summand, with each summand chosen with probability proportionally to its variance, and replacing it with one of comparable size. Thus, by differing only in a single summand, the variables Wn and Wn∗ are close, making Wn an approximate fixed point of the zero bias transformation, and therefore approximately normal. This explanation, when given precisely, becomes a probabilistic proof of the Lindeberg-Feller CLT under a condition equivalent to (8) which we call the ‘small zero bias condition’. To set the stage for the introduction of the small zero bias condition we first consider more precisely this special property of the zero bias transformation on independent ∗ sums. Given Xn satisfying Condition 1.1, let X∗n = {X i,n : 1 ≤ i ≤ n} be a collection ∗ of random variables so that X i,n has the X i,n zero bias distribution and is independent of Xn . Further, let In be a random index, independent of Xn and X∗n , with distribution 2 P(In = i) = σi,n ,

(12)

and write the variables X In ,n and X ∗In ,n which are selected by In , that is, the mixtures, using indicator functions as X In ,n =

n  i=1

January 2009]

1(In = i)X i,n

and

X ∗In ,n =

n 

∗ 1(In = i)X i,n .

(13)

i=1

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

49

Then, upon replacing in the sum Wn the variable selected by In by the corresponding variable having its zero biased distribution, we obtain Wn∗ = Wn − X In ,n + X ∗In ,n ,

(14)

a variable which has the Wn zero bias distribution. The proof of this fact is simple. For all absolutely continuous functions f for which the following expectations exist,  n  E[Wn f (Wn )] = E X i,n f (Wn ) i=1

=

n 

E[X i,n f ((Wn − X i,n ) + X i,n )]

i=1

=

n 

2 ∗ E[σi,n f ((Wn − X i,n ) + X i,n )]

i=1



=E

n 

∗ 1(In = i) f (Wn − X i,n + X i,n )

i=1

= E[ f (Wn − X In ,n + X ∗In ,n )], where in the third equality we have used the fact that X n,i and Wn − X i,n are independent, and in the fourth equality that In is independent of Xn and X∗n . In view of (11), we now have the equality of the expectations of Wn∗ and Wn − X In ,n + X ∗In ,n on a large enough class of functions sufficient to guarantee that these two random variables have the same distribution. From (14) we see that the CLT should hold when the random variables X In ,n and X ∗In ,n are both small asymptotically, since then the distribution of Wn is close to that of Wn∗ , making Wn an approximate fixed point of the zero bias transformation. The following theorem shows that properly formalizing the notion of smallness results in a condition equivalent to Lindeberg’s. Recall that we say a sequence of random variables Yn converges in probability to Y , and write Yn → p Y , if lim P(|Yn − Y | ≥ ) = 0 for all  > 0.

n→∞

Theorem 1.1. If Xn , n = 1, 2, . . . is a collection of random variables satisfying Condition 1.1, then the small zero bias condition X ∗In ,n → p 0

(15)

and the Lindeberg condition (8) are equivalent. Our probabilistic proof of the Lindeberg-Feller Theorem develops by first showing that the small zero bias condition implies that X In ,n → p 0, and hence that Wn∗ − Wn = X ∗In ,n − X In ,n → p 0. 50

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Theorem 1.2 confirms that this convergence in probability to zero, mandating that Wn have its own zero bias distribution in the limit, is sufficient to guarantee that Wn converges in distribution to the normal. Theorem 1.2. If Xn , n = 1, 2, . . . satisfies Condition 1.1 and the small zero bias condition X ∗In ,n → p 0, then Wn →d Z, a standard normal random variable. a random permutation in Sn , with We return now to the number K n (π) of cycles

of ∞ mean h n and variance σn2 given by (6). Since i=1 1/i 2 < ∞, by upper and lower bounding the nth harmonic number h n by integrals of 1/x, we have σn2 = 1. (16) n→∞ log n

n In view of (7) and (4) we note that in this case Wn = i=2 X i,n , as X 1 = 1 identically makes X 1,n = 0 for all n. Now by the linearity relation hn =1 n→∞ log n lim

and therefore

(a X )∗ =d a X ∗

lim

for all a = 0,

which follows directly from (11), by (10) we have ∗ =d Ui /σn , where Ui has distribution U [−1/i, 1 − 1/i], i = 2, . . . , n. X i,n

In particular, |Ui | ≤ 1 with probability one for all i = 2, . . . , n, and therefore |X ∗In ,n | ≤ 1/σn → 0

(17)

by (16). Hence the small zero bias condition is satisfied, and Theorem 1.2 may be invoked to show that the number of cycles of a random permutation is asymptotically normal. More generally, the small zero bias condition will hold for an array Xn with elements X i,n = X i /σn whenever the independent mean zero summand variables X 1 , X 2 , . . . satisfy |X i | ≤ C with probability one for some constant C, and the variance σn2 of their sum Sn tends to infinity. In particular, from (11) one can verify that |X i | ≤ C with probability one implies |X i∗ | ≤ C with probability one, and hence (17) holds with C replacing 1. In such a case, the Lindeberg condition (8) is also not difficult to verify: for any  > 0 one has C/σn <  for all n sufficiently large, and all terms in the sum in (8) are identically zero. Next consider the verification of the Lindeberg and small zero bias conditions in the identically distributed case, showing that the classical CLT is a special case of the Lindeberg-Feller theorem. Let X 1 , X 2 , . . . be independent with X i =d X, i = 1, 2, . . . , where X is a random variable with mean μ and variance σ 2 . By replacing X i by (X i − μ)/σ , it suffices to consider the case where μ = 0 and σ 2 = 1. Now set 1 X i,n = √ X i n

and

Wn =

n 

X i,n .

i=1

For the verification of the classical √ Lindeberg condition, first use the fact that the distributions are identical and the n-scaling to obtain √ 2 1(|X 1,n | ≥ )} = E{X 2 1(|X | ≥ n)}. L n, = n E{X 1,n January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

51

√ Now note that X 2 1(|X | ≥ n) tends to zero as n → ∞, and is dominated by the integrable variable X 2 ; hence, the dominated convergence theorem may be invoked to provide the needed convergence of L n, to zero. Verification that the small zero bias condition is satisfied in this case is more mild. Again using that (a X )∗ = a X ∗ , we have 1 X ∗In ,n =d √ X ∗ , n the mixture on the left being of these identical distributions. But now for any  > 0 √ lim P(|X ∗In ,n | ≥ ) = lim P(|X ∗ | ≥ n) = 0, n→∞

n→∞

that is, X ∗In ,n → p 0. Though Theorems 1.1 and 1.2 show that the Lindeberg condition is sufficient for normal convergence, it is easy to see, and well known, that it is not necessary. In particular, consider the case where for all n the first summand X 1,n of Wn has the mean zero normal distribution σ Z with variance σ 2 ∈ (0, 1), and the Lindeberg condition is satisfied for the remaining variables, that is, that the limit is zero when taking the sum in (8) over all i = 1. Since the sum of independent normal variables is again normal, Wn will converge in distribution to Z, but (8) does not hold, since for all  > 0 2 1(|X 1,n | ≥ )} = σ 2 E{Z 2 1(σ |Z| ≥ )} > 0. lim L n, = E{X 1,n

n→∞

Defining 2 m n = max σi,n

(18)

1≤i≤n

to use for excluding such cases, we have the following partial converse to Theorem 1.2 (see also [4, Theorem 2, page 492] proved independently by Feller and L´evy, and also [7] for interesting CLT history). Theorem 1.3. If Xn , n = 1, 2, . . . satisfies Condition 1.1 and lim m n = 0,

(19)

n→∞

then the small zero bias condition is necessary for Wn →d Z. We prove Theorem 1.3 in Section 5 by showing that Wn →d Z implies that Wn∗ →d Z, and that (19) implies X In ,n → p 0. But then Wn + X ∗In ,n = Wn∗ + X In ,n →d Z also, and we can then prove that W n →d Z

and

Wn + X ∗In ,n →d Z

imply that

X ∗In ,n → p 0.

This argument formalizes the probabilistic reason that the small zero bias condition, or Lindeberg condition, is necessary for normal convergence under (19). Section 2 draws a parallel between the zero bias transformation and the one better known for size biasing, and there we consider its connection to the differential equation method of Stein using test functions. In Section 3 we prove the equivalence of the classical Lindeberg condition and the small zero bias condition, and then, in Sections 4 and 5, we prove its sufficiency and partial necessity for normal convergence. 52

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

Some pains have been taken to keep the treatment as elementary as possible, in particular by avoiding the use of characteristic functions. Though some technical argument is needed, only real functions are involved, and the development remains at a level as basic as the material permits. With the help of two general results in Section 6, the presentation is self-contained with the exception of the existence of the zero bias distribution in the generality already specified, and the following two results stated in Section 2: the bounds (24) to the Stein equation, applied in Theorem 1.2, and the wellknown fact involving convergence in distribution, that (3) implies (21) when C = Cb , used in Theorem 1.3. 2. BIASING AND THE STEIN EQUATION. Relation (11) characterizing the zero bias distribution of a mean zero random variable with finite variance is quite similar to the better known identity which characterizes the size bias distribution of a nonnegative random variable X with finite mean μ. In particular we say that X s has the X -size biased distribution if μE f (X s ) = E[X f (X )]

(20)

for all functions f for which these expectations exist. For example, if X is a nonnegative integer valued random variable, it is easy to verify that P(X s = k) =

k P(X = k) μ

for k = 0, 1, . . . ,

that is, the probability that X s takes on the value k is ‘biased’ by the size k of the value taken. Size biasing can appear unwanted in various sampling contexts, and is also implicated in generating the waiting time paradox ([4, sec. I.4]). We note that the size biasing relation (20) is of the same form as the zero biasing relation (11), but with the mean μ replacing the variance σ 2 , and f rather than f

evaluated on the biased variable. Hence zero biasing is a kind of analog of size biasing on the domain of mean zero random variables. In particular, the two transformations share the property that a sum of independent terms can be biased by replacing a single summand by one having that summand’s biased distribution; in zero biasing the summand is selected with probability proportional to its variance, and in size biasing with probability proportional to its mean. The biasing relations (11) and (20) are in terms of ‘test functions’, while definition (3) of convergence in distribution is expressed using distribution functions. The connection between these frameworks (see [1]) is the fact that convergence in distribution of Yn to Y implies the convergence of expectations of functions of Yn to those of Y , precisely, Yn →d Y implies lim Eh(Yn ) = Eh(Y ) for all h ∈ C

n→∞

(21)

when C = Cb , the collection of all bounded, continuous functions. Clearly, (21) hold∞ ing with C = Cb implies that it holds with C = Cc,0 , the set of all functions with com∞ pact support which integrate to zero and have derivatives of all orders, since Cc,0 ⊂ Cb . In Section 6 we prove the following result, making all three statements equivalent. ∞ Theorem 2.1. If (21) holds with C = Cc,0 , then Yn →d Y .

In light of the characterization (9), a strategy first suggested in [9] (see also [11]) ∞ for proving Wn →d Z is to choose a class of test functions C , such as Cc,0 , which is January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

53

rich enough to guarantee convergence in distribution, and then for given h ∈ C to find a function f which solves the ‘Stein equation’ f (w) − w f (w) = h(w) − Eh(Z).

(22)

Now demonstrating Eh(Wn ) → Eh(Z) can be accomplished by showing  lim E f (Wn ) − Wn f (Wn ) = 0. n→∞

It is easy to verify that when Eh(Z) exists an explicit solution to (22) is given by  w f (w) = ϕ −1 (w) [h(u) − Eh(Z)]ϕ(u)du, (23) −∞

where ϕ(u) is the standard normal density given in (2). Stein [11] showed that when h is a bounded differentiable function with bounded derivative h , the solution f is twice differentiable and satisfies || f || ≤ 2||h|| and

|| f

|| ≤ 2||h ||,

(24)

where for any function g, ||g|| =

sup

−∞<x<∞

|g(x)|.

These bounds will be applied in Section 4. 3. EQUIVALENCE: PROOF OF THEOREM 1.1. Since the random index In is independent of Xn and X∗n , taking expectations in (13) and using (12) yields E f (X In ,n ) =

n 

2 σi,n E f (X i,n ) and

E f (X ∗In ,n ) =

i=1

n 

2 ∗ σi,n E f (X i,n ).

(25)

i=1

Let  > 0 be given and set ⎧ ⎨ x + 0 f (x) = ⎩ x −

if x ≤ −, if − < x < , if x ≥ .

The function f is absolutely continuous with f (x) = 1(|x| ≥ ) almost everywhere. Hence, using the zero bias relation (11) for the second equality, 2 ∗ 2 ∗ P(|X i,n | ≥ ) = σi,n E f (X i,n ) σi,n   2 = E X i,n − |X i,n | 1(|X i,n | ≥ )  2  ≤ E X i,n + |X i,n | 1(|X i,n | ≥ )  2  ≤ 2E X i,n 1(|X i,n | ≥ ) ,

54

(26)

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

and summing over i = 1, . . . , n and applying identity (25) for the indicator function 1(|x| ≥ ) we obtain P(|X ∗In ,n | ≥ ) =

n 

2 ∗ σi,n P(|X i,n | ≥ ) ≤ 2L n, .

i=1

Hence the Lindeberg condition (8) implies the small zero bias condition (15). For the implication in the other direction use that for all x,      . x 2 1(|x| ≥ ) ≤ 2 x 2 − |x| 1 |x| ≥ 2 2 Applying (26) and (25) to obtain the second and third equalities, respectively, we have L n, =

n 

2 E{X i,n 1(|X i,n | ≥ )} ≤ 2

i=1

n  i=1

     2 E X i,n − |X i,n | 1 |X i,n | ≥ 2 2

  2 ∗ σi,n P |X i,n |≥ 2 i=1    , = 2P |X ∗In ,n | ≥ 2

=2

n 

proving that the small zero bias condition (15) implies the Lindeberg condition (8).

4. SUFFICIENCY: PROOF OF THEOREM 1.2. With the help of two lemmas we prove Theorem 1.2, the ‘small zero bias’ version of the Lindeberg-Feller theorem. Lemma 4.1. Let Xn , n = 1, 2, . . . satisfy Condition 1.1 and m n be given by (18). Then X In ,n → p 0 whenever

lim m n = 0.

n→∞

Proof. From (25) with f (x) = x we find E X In ,n = 0, and hence Var(X In ,n ) = E X 2In ,n . Now (25) again, with f (x) = x 2 , yields Var(X In ,n ) =

n 

4 σi,n .

i=1 4 2 2 2 Since for all i, σi,n ≤ σi,n max1≤ j ≤n σ j,n = σi,n m n , for any  > 0 we have

P(|X In ,n | ≥ ) ≤

n n  Var(X In ,n ) 1  1 1 4 2 = σ ≤ m σi,n = 2 mn , n i,n 2 2 2   i=1   i=1

the first inequality being Chebyshev’s, and the last equality following from Condition 1.1. As m n → 0 by hypotheses, the proof is complete. Lemma 4.2. If Xn , n = 1, 2, . . . satisfies Condition 1.1 and the small zero bias condition, then X In ,n → p 0. January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

55

Proof. For all n, 1 ≤ i ≤ n, and  > 0, 2 2 2 σi,n = E(X i,n 1(|X i,n | < )) + E(X i,n 1(|X i,n | ≥ )) ≤  2 + L n, .

Since the upper bound does not depend on i, m n ≤  2 + L n, , and now, since Xn satisfies the small zero bias condition, by Theorem 1.1 we have lim sup m n ≤  2 n→∞

and therefore lim m n = 0.

n→∞

The claim now follows by Lemma 4.1. We are now ready to prove the forward direction of the Lindeberg-Feller CLT. ∞ and let f be the solution to the Stein equation, for Proof of Theorem 1.2. Let h ∈ Cc,0 that h, given by (23). Substituting Wn for w in (22), taking expectation, and using (11) we obtain   E [h(Wn ) − Eh(Z)] = E f (Wn ) − Wn f (Wn ) = E f (Wn ) − f (Wn∗ ) (27)

with Wn∗ given by (14). Since Wn∗ − Wn = X ∗In ,n − X In ,n , the small zero bias condition and Lemma 4.2 imply Wn∗ − Wn → p 0.

(28)

By (24) f is bounded with a bounded derivative f

, hence its global modulus of continuity η(δ) = sup | f (y) − f (x)| |y−x|≤δ

is bounded and satisfies limδ→0 η(δ) = 0. Now, by (28), η(|Wn∗ − Wn |) → p 0,

(29)

and by (27) and the triangle inequality

  |Eh(Wn ) − Eh(Z)| =  E( f (Wn ) − f (Wn∗ ))   ≤ E  f (Wn ) − f (Wn∗ ) ≤ Eη(|Wn − Wn∗ |).

Therefore lim Eh(Wn ) = Eh(Z)

n→∞

by (29) and the bounded convergence theorem. Invoking Theorem 2.1 finishes the proof. 56

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

5. PARTIAL NECESSITY. In this section we prove Theorem 1.3, showing in what sense the Lindeberg condition, in its equivalent zero bias form, is necessary. We begin with Slutsky’s lemma, see [5], which states that Un →d U and Vn → p 0

implies Un + Vn →d U.

(30)

When independence holds, we have the following kind of reverse implication, whose proof is deferred to Section 6. Lemma 5.1. Let Un and Vn , n = 1, 2, . . . be two sequences of random variables such that Un and Vn are independent for every n. Then Un →d U and Un + Vn →d U

implies

Vn → p 0.

Next, we show that the zero bias transformation enjoys the following continuity property. Lemma 5.2. Let Y and Yn , n = 1, 2, . . . be mean zero random variables with finite, nonzero variances σ 2 = Var(Y ) and σn2 = Var(Yn ), respectively. If Y n →d Y

lim σn2 = σ 2 ,

and

n→∞

then Yn∗ →d Y ∗ . y ∞ and F(y) = −∞ f (t)dt. Since Y and Yn have mean zero and Proof. Let f ∈ Cc,0 finite variances, their zero bias distributions exist, so in particular,  σn2 E f (Yn∗ ) = E Yn F(Yn ) for all n. By (21), since y F(y) is in Cb , we obtain σ 2 lim E f (Yn∗ ) = lim σn2 E f (Yn∗ ) = lim E[Yn F(Yn )] = E[Y F(Y )] = σ 2 E f (Y ∗ ). n→∞

n→∞

n→∞

∞ Hence E f (Yn∗ ) → E f (Y ∗ ) for all f ∈ Cc,0 , so Yn∗ →d Y ∗ by Theorem 2.1.

We now provide the proof of the partial converse to the Lindeberg-Feller theorem. Proof of Theorem 1.3. Since Wn →d Z and Var(Wn ) → Var(Z), the sequence of variances and the limit being identically one, Lemma 5.2 implies Wn∗ →d Z ∗ . But Z is a fixed point of the zero bias transformation, hence Wn∗ →d Z. Since m n → 0, Lemma 4.1 yields that X In ,n → p 0, and Slutsky’s lemma (30) now gives that Wn + X ∗In ,n = Wn∗ + X In ,n →d Z. Hence W n →d Z

and

Wn + X ∗In ,n →d Z.

Since Wn is a function of Xn , which is independent of In and X∗n and therefore of X ∗In ,n , invoking Lemma 5.1 yields X ∗In ,n → p 0. January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

57

6. APPENDIX. Here we provide the proofs that convergence of expectations over ∞ the class of functions Cc,0 implies convergence in distribution, and for the converse of Slutsky’s lemma under an additional independence assumption. Proof of Theorem 2.1. Let a < b be continuity points of P(Y ≤ x). Billingsely [1] exhibits an infinitely differentiable function ψ taking values in [0, 1] such that ψ(x) = 1 for x ≤ 0 and ψ(x) = 0 for x ≥ 1. Hence, for all u > 0 the function ψa,b,u (x) = ψ(u(x − b)) − ψ(u(x − a) + 1) is infinitely differentiable, has support in [a − 1/u, b + 1/u], equals 1 for x ∈ [a, b], and takes values in [0, 1] for all x. Furthermore,  ∞ 1 ψa,b,u (x)d x = + (b − a), u −∞ so for every  ∈ (0, 1], letting d=−

  1 1 +  −1 + (b − a) u u

the function ψa,b,u, (x) = ψa,b,u (x) − ψb+2/u,b+2/u+d,u (x) ∞ is an element of Cc,0 . Furthermore, for all u > 0 and  ∈ (0, 1], ψa,b,u, (x) equals 1 on [a, b], lies in [0, 1] for x ∈ [a − 1/u, b + 1/u], and lies in [−, 0] for all other x. Hence

lim sup P(Yn ∈ (a, b]) ≤ lim sup Eψa,b,u, (Yn ) +  n→∞

n→∞

= Eψa,b,u, (Y ) +     1 1 ≤ P Y ∈ a − ,b + + . u u Letting  tend to zero and u to infinity, since a and b are continuity points, lim sup P(Yn ∈ (a, b]) ≤ P(Y ∈ (a, b]). n→∞

A similar argument using ψa+1/u,b−1/u,u, (x) shows that the reverse inequality holds with lim inf replacing lim sup, so for all continuity points a and b, lim P(Yn ∈ (a, b]) = P(Y ∈ (a, b]).

n→∞

For b any continuity point and  ∈ (0, 1], there exist continuity points a < b < c with P(Y ∈ (a, c]) < . Since P(Y ≤ a) ≤ P(Y ∈ (a, c]) < , for all n sufficiently large P(Yn ≤ a) ≤ P(Yn ∈ (a, c]) ≤ , and we have |P(Yn ≤ b) − P(Y ≤ b)| ≤ |P(Yn ∈ (a, b]) − P(Y ∈ (a, b])| + 2, 58

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

yielding lim P(Yn ≤ b) = P(Y ≤ b).

n→∞

Proof of Lemma 5.1. We first prove Lemma 5.1 for the special case where Un =d U for all n, that is, we prove that if U + Vn →d U

with U independent of Vn

(31)

then Vn → p 0. By adding to U an absolutely continuous random variable A, independent of U and Vn , (31) holds with U replaced by the absolutely continuous variable U + A; we may therefore assume without loss of generality that U possesses a density function. If Vn does not tend to zero in probability, there exist positive  and p such that for infinitely many n 2 p ≤ P(|Vn | ≥ ) = P(Vn ≥ ) + P(−Vn ≥ ), so either Vn or −Vn is at least  with probability at least p. Assume that there exists an infinite subsequence K such that P(Vn ≥ ) ≥ p

for all n ∈ K,

(32)

a similar argument holding in the opposite case. Since U has a density the function s(x) = P(x ≤ U ≤ x + 1) is continuous, and as the limits of s(x) at plus and minus infinity are zero, s(x) attains its maximum value, say s, in a bounded region. In particular, y = inf{x : s(x) = s} is finite, and, by definition of y and the continuity of s(x), sup s(x) = r < s.

x≤y−

Since U and Vn are independent P(y ≤ U + Vn ≤ y + 1|Vn ) = s(y − Vn ) for all n.

(33)

Therefore, on the one hand we have P(y ≤ U + Vn ≤ y + 1|Vn ≥ ) ≤ r

for all n ∈ K,

but by conditioning on Vn ≥  and its complement, using (33), (32), (31), and the fact that U is absolutely continuous, we obtain the contradiction lim inf P(y ≤ U + Vn ≤ y + 1) ≤ r p + s(1 − p) n→∞

< s = P(y ≤ U ≤ y + 1) = lim P(y ≤ U + Vn ≤ y + 1). n→∞

To generalize to the situation where Un →d U through a sequence of distributions which may depend on n, we use Skorohod’s construction (see Theorem 11.7.2 of [2]), January 2009]

THE LINDEBERG-FELLER CENTRAL LIMIT THEOREM

59

which implies that whenever Yn →d Y , there exist Y n and Y on the same space with Y n =d Yn and Y =d Y such that Y n → p Y . In particular, Y n and Y can be taken to be the inverse distribution functions of Yn and Y , respectively, evaluated on the same uniform random variable. In this way we may construct U n and U on the same (but now perhaps enlarged) space as Vn . Then, by the hypotheses and Slutsky’s lemma (30), we obtain U + Vn = (U n + Vn ) + (U − U n ) →d U

with U independent of Vn .

Since this is the situation of (31), we conclude Vn → p 0. ACKNOWLEDGMENT. We enthusiastically thank the generous efforts of an anonymous reviewer, whose helpful comments, of both a technical and nontechnical nature, greatly improved this work in all respects.

REFERENCES 1. 2. 3. 4. 5. 6.

P. Billingsley, Convergence of Probability Measures, Wiley, New York, 1968. R. Dudley, Real Analysis and Probability, Cambridge University Press, Cambridge, 1989. W. Feller, The fundamental limit theorems in probability, Bull. Amer. Math. Soc. 51 (1945) 800–832. , An Introduction to Probability Theory and Its Applications, vol. 2, Wiley, New York, 1967. T. Ferguson, A Course in Large Sample Theory, Chapman & Hall, New York, 1996. L. Goldstein and G. Reinert, Stein’s method and the zero bias transformation with application to simple random sampling, Ann. Appl. Probab. 7 (1997) 935–952. 7. L. Le Cam, The central limit theorem around 1935, Statist. Sci. 1 (1986) 78–96. 8. J. Lindeberg, Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung, Math. Z. 15 (1922) 211–225. 9. C. Stein, A bound for the error in the normal approximation to the distribution of a sum of dependent random variables, Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability, vol. 2, University of California Press, Berkeley, (1972) 583–602. , Estimation of the mean of a multivariate normal distribution, Ann. Statist. 9 (1981) 1135–1151. 10. 11. , Approximate Computation of Expectations, Institute of Mathematical Statistics, Hayward, CA, 1986. LARRY GOLDSTEIN received his M.A. in Mathematics (1979), M.S. in Electrical Engineering (1982), and Ph.D. in Mathematics (1984), all from the University of California, San Diego. Since 1984 he has been in the mathematics department at the University of Southern California, where he is now full professor. His interests lie in probability and statistics, presently in Stein’s method and cohort sampling schemes in epidemiology. He has applied the Central Limit Theorem while serving as a shipboard scientist on an exploration of the MidAtlantic Ridge, and as a consultant for VISA’s Grinch sweepstakes to estimate the odds of winning and the average prize payout, which it did to surprising accuracy. Department of Mathematics, KAP 108, University of Southern California, Los Angeles, CA 90089-2532 [email protected]

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Does Lipschitz with Respect to x Imply Uniqueness for the Differential Equation y = f (x, y)? ´ Jos´e Angel Cid and Rodrigo L´opez Pouso 1. INTRODUCTION. Of course, the answer to the question posed in the title is no, in general, but, surprisingly enough, yes in a significant situation to be detailed later. This paper aims to bring to the attention of the widest possible mathematical audience an elementary result, based on the theorem of differentiation of inverse functions, which extends the applicability of uniqueness theorems. Loosely speaking, it transforms every uniqueness theorem into an alternative version of it with the roles of the dependent and the independent variables interchanged in the assumptions. A remarkable example of such a theorem is the version of Lipschitz’s theorem alluded to in the first paragraph, but there are at least as many new possibilities as old uniqueness theorems. Next we introduce some notation to discuss properly the question considered. Let N be a neighborhood of a point (x0 , y0 ) ∈ R2 , f : N → R a given mapping, and consider the scalar initial value problem y  = f (x, y),

y(x0 ) = y0 .

(1.1)

We recall that a solution of (1.1) is a function Y : I → R which is continuously differentiable on the interval I such that x0 ∈ I , Y (x0 ) = y0 , and for all x ∈ I we have (x, Y (x)) ∈ N and Y  (x) = f (x, Y (x)). As usual, we say that (1.1) has a unique solution if there exists α > 0 with the property that (1.1) has one solution defined on [x0 − α, x0 + α] and any other solution Y : I → R coincides with it on I ∩ [x0 − α, x0 + α]. Here and henceforth, we assume that f is continuous on N ; thus Peano’s theorem guarantees that there exists ε > 0 such that (1.1) has at least one solution defined on the interval [x0 − ε, x0 + ε]. The scalar version of Peano’s theorem was published in 1886, see [11], and was extended to systems in 1890, see [12].1 At that time Peano was already aware of the fact that some continuous functions f allow (1.1) to have more than one solution. He presented the following example in [12]: Example 1.1. The problem (1.1) for f (x, y) = 3y 2/3 and x0 = 0 = y0 has more than one solution. Indeed, one can check by direct computation that Y1 (x) = 0 and Y2 (x) = x 3 are both solutions defined on the whole real line. Lavrentieff constructed a more dramatic example in 1925, which consisted in a continuous function on a rectangle such that uniqueness fails for (1.1) at every initial condition (x0 , y0 ) in the rectangle’s interior, see [6]. Later, in 1963, Hartman published in this M ONTHLY a simpler example of that type with a function defined on the whole plane, see [5]. 1 Peano wrote that paper using his own logical symbols, which makes it rather difficult to read. Three years later, G. Mie wrote in German a more understandable version of it, see [8].

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The main usefulness of differential equations is that they serve as models that describe mathematically many real phenomena and processes. Especially in those cases the existence of more than one solution is disturbing and misleading, because it produces uncertainty about the behavior of the object that we are studying. Moreover, uniqueness and nonuniqueness also have a number of theoretical implications as, for example, in the study of the qualitative behavior at infinity of global solutions, see [13]. Therefore it is of fundamental importance to have adequate tools to decide whether a concrete problem has a unique solution. Almost every textbook on ordinary differential equations contains a version of the well-known uniqueness theorem published by Lipschitz in 1877, see [7] and, for instance, the monographs [4, 14]. The following suffices for our purposes in this article: Theorem 1.1 (Lipschitz’s uniqueness theorem). Let N be a neighborhood of a point (x0 , y0 ) ∈ R2 and let f : N → R be continuous on N . If f satisfies a Lipschitz condition with respect to the second variable on N , i.e., ∃ L > 0 such that (x, y), (x, z) ∈ N ⇒ | f (x, y) − f (x, z)| ≤ L|y − z|,

(1.2)

then (1.1) has a unique solution. Now, does a Lipschitz condition with respect to the first variable imply uniqueness for (1.1)? Example 1.1 shows that this is not true in general, and this is often the end of the question, but we urge the reader to go through the remaining few pages to find out that the answer to our question is positive provided that f (x0 , y0 ) = 0. 2. DOUBLE THE UNIQUENESS THEOREMS THAT YOU KNOW (WITH LITTLE EFFORT!). We begin this section with the following simple technical remark: if N  ⊂ N is another neighborhood of (x0 , y0 ) and the problem y  = f |N  (x, y),

y(x0 ) = y0

has a unique solution, then (1.1) has a unique solution (here, f |N  stands for the restriction of f to N  ). This observation guarantees that we can pass, without losing generality, to more convenient smaller neighborhoods when studying uniqueness. In doing so, we avoid some technicalities in the proofs. The next theorem is the core of the present article and establishes the equivalence between uniqueness for (1.1) and uniqueness for a related reciprocal problem, thus doubling the applicability of uniqueness theorems. Theorem 2.1. Let N be a neighborhood of a point (x0 , y0 ) ∈ R2 and let f : N → R be continuous on N . If f (x0 , y0 ) = 0 then (1.1) has a unique solution if and only if the problem x =

1 , f (x, y)

x(y0 ) = x0

(2.3)

has a unique solution. Proof. Since f (x0 , y0 ) = 0 and f is continuous at (x0 , y0 ), there exists a neighborhood of (x0 , y0 ) where f has constant sign and is bounded. For simplicity, we assume that f and 1/ f have constant sign and are bounded on N . 62

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To establish the result we will use the following claim, which is interesting in its own right: Claim. If Y is a solution of (1.1) then Y −1 is a solution of (2.3) and, conversely, if X is a solution of (2.3) then X −1 is a solution of (1.1). Let Y : I → R be a solution of (1.1); for all x ∈ I we have (x, Y (x)) ∈ N and Y  (x) = f (x, Y (x)), so Y  has constant sign on I and, in particular, it has an inverse Y −1 : Y (I ) → R. Let us show that X = Y −1 solves (2.3). First, Y (I ) is an interval that contains y0 and X (y0 ) = x0 ; second, we use the theorem of differentiation of inverse functions for all y ∈ Y (I ) to obtain that X  (y) = (Y −1 ) (y) =

1 Y  (Y −1 (y))

=

1 . f (X (y), y)

The proof of the converse is analogous, so we omit it, and the claim is proven. Finally, suppose that Y is the unique solution to (1.1) on the interval I = [x0 − α, x0 + α], for some α > 0. We are going to prove that Y −1 is the unique solution to (2.3) on the interval J = [y0 − β, y0 + β] provided that β > 0 is so small that J ⊂ Y (I ) and if X is any solution to (2.3) defined on an interval J˜ ⊂ J then X ( J˜) ⊂ I (such a choice of β is possible because 1/ f is bounded on N ). Let X be a solution to (2.3) on an interval J˜ ⊂ J . Since X −1 is a solution to (1.1) on X ( J˜) and X ( J˜) ⊂ I , we conclude that X −1 = Y on X ( J˜). Hence X = Y −1 on J˜. Analogous arguments show that uniqueness for (2.3) implies uniqueness for (1.1). The main theoretical importance in the previous equivalence lies in the fact that the dependent and the independent variables interchange their roles when passing from (1.1) to (2.3), and this has the consequence that assumptions are transferred from one argument to the other. The plan for generating new uniqueness theorems from old ones in case f (x0 , y0 ) = 0 is very simple now: look for appropriate assumptions on f which imply that (2.3) falls inside the scope of the uniqueness theorem that you choose. We carry out this plan with Lipschitz’s theorem in the next section. 3. LIPSCHITZ’S UNIQUENESS THEOREM REVISITED. This section is devoted to the following version of Lipschitz’s uniqueness theorem, which is a straightforward consequence of Theorems 1.1 and 2.1. Theorem 3.1. Let N be a neighborhood of a point (x0 , y0 ) ∈ R2 and let f : N → R be continuous on N . If f (x0 , y0 ) = 0 and, moreover, f satisfies a Lipschitz condition with respect to the first variable on N , i.e., ∃ L > 0 such that (s, y), (x, y) ∈ N ⇒ | f (s, y) − f (x, y)| ≤ L|s − x|,

(3.4)

then (1.1) has a unique solution. Proof. Theorem 2.1 applies because f (x0 , y0 ) = 0, so it suffices to prove that (2.3) has a unique solution. To do so, note that the continuity of f implies that there exists a neighborhood of (x0 , y0 ) where | f | ≥ | f (x0 , y0 )|/2 =: r > 0. For simplicity we January 2009]

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assume that this holds on N , so for (x, y), (s, y) ∈ N we have      1 1   f (s, y) − f (x, y)  L   f (x, y) − f (s, y)  =  f (x, y) f (s, y)  ≤ r 2 |s − x|, and therefore Theorem 1.1 guarantees that (2.3) has a unique solution (remember that x is the dependent variable in (2.3)). A very useful consequence of Theorem 3.1 concerns differential equations with continuously differentiable right-hand sides. Corollary 3.1. Let N be a neighborhood of a point (x0 , y0 ) ∈ R2 and let f : N → R be continuous on N . If f (x0 , y0 ) = 0 and, moreover, ∂ f /∂ x is continuous on N , then (1.1) has a unique solution. Proof. Let N  be a compact neighborhood of (x0 , y0 ) such that N  ⊂ N , and let L > 0 be an upper bound of |∂ f /∂ x| on N  . Now for (x, y), (s, y) ∈ N  , x = s, the mean value theorem guarantees the existence of r , strictly between x and s, such that   ∂ f  | f (x, y) − f (s, y)| =  (r, y) |x − s| ≤ L|x − s|. ∂x Hence Theorem 3.1 implies that (1.1) has a unique solution. Example 3.1. The nonlinear problem y  = cos x + x

√ 3

y,

y(0) = 0

has a unique solution by virtue of Corollary 3.1. Notice that the right-hand side of the differential equation does not satisfy the assumptions of Lipschitz’s uniqueness theorem on any neighborhood of the initial condition. An important particular case of the preceding corollary is that of autonomous differential equations. The following classical uniqueness result, which goes back to Peano, see [12], follows immediately from Corollary 3.1. Corollary 3.2. Let ε > 0, g : (y0 − ε, y0 + ε) → R continuous, and x0 ∈ R. If g(y0 ) = 0 then the autonomous problem y  = g(y),

y(x0 ) = y0

(3.5)

has a unique solution. Example 3.2. In the nonautonomous case the condition f (x0 , y0 ) = 0 alone is not sufficient for uniqueness for (1.1). As an example observe that the change of variable y = z + x transforms the autonomous problem dz/d x = 3z 2/3 , z(0) = 0, which appears in Example 1.1, into the nonautonomous one y  = 3(y − x)2/3 + 1,

y(0) = 0,

for which the right-hand side does not vanish at the initial data and has Y1 (x) = x and Y2 (x) = x 3 + x as solutions for all x ∈ R. In this case the right-hand side does not satisfy either (1.2) or (3.4) on any neighborhood of the initial condition. 64

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4. CONCLUDING REMARKS. 1. The claim in the proof of Theorem 2.1 also provides us with an integration method, as (2.3) may be integrable even though (1.1) is not. In fact, it is a well-known trick in the field of differential equations to try and solve d x/dy = 1/ f (x, y) instead of dy/d x = f (x, y) whenever the last differential equation is not solvable by elementary methods. It seems however that its underlying theoretical implications in connection with uniqueness were not fully exploited until the last decade. The claim in the proof of Theorem 2.1 was established in a more general form by the authors in [3], and it was used there to derive a number of consequences, including Theorem 3.1 (Theorem 2.7 in [3]). Later, Cid extended Theorem 3.1 to systems of differential equations in [2]. It was after the publication of [2, 3] that the authors became aware of the work done by Mortici in [9, 10]. As far as we know, Mortici was the first author who deduced Theorem 3.1. 2. Most uniqueness criteria are based on generalizations of the standard Lipschitz condition (1.2), see [1], or require some monotonicity assumptions such as those in [15, 16]. Theorem 2.1 is the key to establishing alternative versions of all of them with assumptions “transferred from y to x”. This yields a lot of new uniqueness theorems that readers may find useful in different situations. A complete account of even the most relevant of them exceeds the objectives of the present article, but we point out as a final example the analog of the so-called Peano uniqueness criterion (which can be looked up in [1]): Theorem 4.1. Let N be a neighborhood of a point (x0 , y0 ) ∈ R2 and let f : N → R be continuous on N . If f (x0 , y0 ) = 0 and, moreover, f is nondecreasing with respect to its first variable (i.e., f (s, y) ≤ f (x, y) whenever (s, y), (x, y) ∈ N and s ≤ x), then (1.1) has a unique solution. ACKNOWLEDGMENTS. We would like to thank the referees for their valuable suggestions. The research for this paper was partially supported by Ministerio de Educaci´on y Ciencia, Spain, project MTM2007-61724, and by Xunta de Galicia, Spain, project PGIDIT06PXIB207023PR.

REFERENCES 1. R. P. Agarwal and V. Lakshmikantham, Uniqueness and Nonuniqueness Criteria for Ordinary Differential Equations, Series in Real Analysis, vol. 6, World Scientific, Singapore, 1993. ´ Cid, On uniqueness criteria for systems of ordinary differential equations, J. Math. Anal. Appl. 281 2. J. A. (2003) 264–275. ´ Cid and R. L. Pouso, On first order ordinary differential equations with non-negative right-hand 3. J. A. sides, Nonlinear Anal. 52 (2003) 1961–1977. 4. E. A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill, New York, 1955. 5. P. Hartman, A differential equation with non-unique solutions, this M ONTHLY 70 (1963) 255–259. 6. M. Lavrentieff, Sur une e´ quation diff´erentielle du premier ordre, Math. Z. 23 (1925) 197–209. 7. R. Lipschitz, Sur la possibilit´e d’int´egrer compl`etement un syst`eme donn´e d’´equations diff´erentielles, Darboux Bull. X. (1877) 149–159. 8. G. Mie, Beweis der integrirbarkeit gew¨ohnlicher differentialgleichungssysteme nach Peano, Math. Ann. 43 (1893) 553–568. 9. C. Mortici, On the solvability of the Cauchy problem, Nieuw Arch. Wiskd. IV. Ser. 17 (1999) 21–23. , Approximation methods for solving the Cauchy problem, Czechoslovak Math. J. 55 (2005) 10. 709–718. 11. G. Peano, Sull’integrabilit`a delle equazioni differenzialli di primo ordine, Atti. Accad. Sci. Torino 21 (1886) 677–685.

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12. 13. 14. 15. 16.

, D´emonstration de l’int´egrabilit´e des e´ quations diff´erentielles ordinaires, Math. Ann. 37 (1890) 182–228. W. Rudin, Nonuniqueness and growth in first-order differential equations, this M ONTHLY 89 (1982) 241–244. W. Walter, Ordinary Differential Equations, Graduate Texts in Mathematics, vol. 182, Springer-Verlag, New York, 1998. D. V. V. Wend, Uniqueness of solutions of ordinary differential equations, this M ONTHLY 74 (1967) 948–950. , Existence and uniqueness of solutions of ordinary differential equations, Proc. Amer. Math. Soc. 23 (1969) 27–33.

´ ´ ANGEL JOSE CID received his Ph.D. in mathematics from the Universidade de Santiago de Compostela under the advice of Alberto Cabada and the second author. He currently teaches at Universidad de Ja´en. He enjoys reading, bicycling, and travelling. Departamento de Matem´aticas, Universidad de Ja´en, Campus Las Lagunillas, 23701, Ja´en, Spain [email protected] ´ RODRIGO LOPEZ POUSO received his Ph.D. in mathematics from the Universidade de Santiago de Compostela under the advice of Alberto Cabada and Eduardo Liz. He currently teaches at the same university. He enjoys reading and cinema. Departamento de An´alise Matem´atica, Universidade de Santiago de Compostela, 15782, Santiago de Compostela, Spain [email protected]

From the Letters of Lady Mary Wortley Montagu “You have given me a great deal of Satisfaction by your account of your eldest Daughter. I am particularly pleas’d to hear she is a good Arithmetician; it is the best proofe of understanding. The knowledge of Numbers is one of the cheif distinctions between us and Brutes.” Lady Mary Wortley Montagu, letter to her daughter, Lady Bute, Jan. 28, 1753, in R. Halsband, ed., The Complete Letters of Lady Mary Wortley Montagu, Vol. III: 1752–1762, Oxford University Press, London, 1967, p. 20. —Submitted by Robert Haas, Cleveland Heights, OH

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NOTES Edited by Ed Scheinerman

A Simple Complex Analysis and an Advanced Calculus Proof of the Fundamental Theorem of Algebra Anton R. Schep It is hard not to have Ray Redheffer’s title of [2] as a reaction to another article on the Fundamental Theorem of Algebra. In fact at least 28 notes have appeared in this M ONTHLY about this theorem. In this note we present nevertheless two proofs of the Fundamental Theorem of Algebra which do not seem to have been observed before and which we think are worth recording. The first one uses Cauchy’s integral theorem and is, in the author’s opinion, as simple as the most popular complex analysis proof based on Liouville’s theorem (see [3] for this and three other proofs using complex analysis). Problem 5 on p. 126 of [1] gives a proof of the Fundamental Theorem of Algebra based on a complex contour integral that is similar to the one used here, but the details are not quite the same. The second one considers the integral obtained by parameterizing the contour integral from the first proof and uses only results from advanced calculus. This proof is similar to the proof of [4], where the same ideas were used to prove the nonemptiness of the spectrum of an element in a complex Banach algebra. There the companion matrix of a polynomial was then used to derive the Fundamental Theorem of Algebra. Theorem (Fundamental Theorem of Algebra). Every polynomial of degree n ≥ 1 with complex coefficients has a zero in C. Proof. Let p(z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a polynomial of degree n ≥ 1 and assume that p(z)  = 0 for all z ∈ C. First Proof. By Cauchy’s integral theorem we have  2πi dz =  = 0, zp(z) p(0) |z|=r where the circle is traversed counterclockwise. On the other hand     2π dz   ≤ 2πr · max 1 = → 0 as r → ∞   |z|=r |zp(z)| min|z|=r | p(z)| |z|=r zp(z) (since | p(z)| ≥ |z|n |(1 − |an−1 |/|z| − · · · − |a0 |/|z n |)|), which is a contradiction. Second Proof. Define g : [0, ∞) × [0, 2π] → C by g(r, θ) = 1/ p(r eiθ ). Then the function g is continuous on [0, ∞) × [0, 2π] and has continuous partials on (0, ∞) × ∂g (0, 2π) satisfying ∂θ = ir · ∂g . ∂r January 2009]

NOTES

67

 2π Define now F : [0, ∞) → C by F(r ) = 0 g(r, θ) dθ. Then by Leibniz’s rule for differentiation under the integral sign we have for all r > 0 



ir F (r ) = ir 0

2π

∂g dθ = ∂r

 0

2π

∂g dθ = g(r, 2π) − g(r, 0) = 0. ∂θ

Hence F  (r ) = 0 for all r > 0. This implies that F is constant on [0, ∞) with F(r ) = 2π F(0) = p(0)  = 0. On the other hand | p(z)| → ∞ as |z| → ∞ implies that g(r, θ) → 0 as r → ∞ uniformly in θ. Therefore F(r ) → 0 as r → ∞, which is a contradiction.

REFERENCES 1. N. Levinson and R. M. Redheffer, Complex Variables, Holden-Day, San Francisco, CA, 1970. 2. R. M. Redheffer, What! Another note just on the fundamental theorem of algebra?, this M ONTHLY 71 (1964) 180–185. 3. R. Remmert, Theory of Complex Functions (trans. R. B. Burckel), Graduate Texts in Mathematics, vol. 122, Springer-Verlag, New York, 1991. 4. D. Singh, The spectrum in a Banach algebra, this M ONTHLY 113 (2006) 756–758. Department of Mathematics, University of South Carolina, Columbia, SC 29208 [email protected]

√ Yet Another Proof of the Irrationality of 2 ˜ Natalia Cas´as Ferreno √ The classical Greek proof of the irrationality of 2 is an example of a first-class theorem, simple enough to be presented to a wide audience but deep in essence, as has been written by the British mathematician G. H. Hardy [7]. Nowadays many proofs √ of the irrationality of 2 are known. Indeed, on the web page [3] fourteen different proofs appear, and several have been published in this M ONTHLY (see [1, 2, 4, 5, 6, 9]). The aim of this note is to present a new √ proof of this fact. Let us consider the linear mapping f : R → R given by f (x) = ( 2 − 1)x as a one-dimensional discrete dynamical system. For each point x0 ∈ R let us define its orbit O(x0 ) as the sequence of iterates of f starting at x0 , namely

O(x0 ) =

√ n 2 − 1 x0 : n ∈ {0} ∪ N .

√ Since 0 < 2 − 1 < 1√it is clear that for each x0 ∈ R the orbit O(x0 ) converges to zero. Now suppose that 2 = p/q is a rational number with p, q ∈ N. Then for all n ∈ N it holds that √ n 2 q= 68



2n/2 q, if n is even, 2(n−1)/2 p, if n is odd.

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

 2π Define now F : [0, ∞) → C by F(r ) = 0 g(r, θ) dθ. Then by Leibniz’s rule for differentiation under the integral sign we have for all r > 0 



ir F (r ) = ir 0

2π

∂g dθ = ∂r

 0

2π

∂g dθ = g(r, 2π) − g(r, 0) = 0. ∂θ

Hence F  (r ) = 0 for all r > 0. This implies that F is constant on [0, ∞) with F(r ) = 2π F(0) = p(0)  = 0. On the other hand | p(z)| → ∞ as |z| → ∞ implies that g(r, θ) → 0 as r → ∞ uniformly in θ. Therefore F(r ) → 0 as r → ∞, which is a contradiction.

REFERENCES 1. N. Levinson and R. M. Redheffer, Complex Variables, Holden-Day, San Francisco, CA, 1970. 2. R. M. Redheffer, What! Another note just on the fundamental theorem of algebra?, this M ONTHLY 71 (1964) 180–185. 3. R. Remmert, Theory of Complex Functions (trans. R. B. Burckel), Graduate Texts in Mathematics, vol. 122, Springer-Verlag, New York, 1991. 4. D. Singh, The spectrum in a Banach algebra, this M ONTHLY 113 (2006) 756–758. Department of Mathematics, University of South Carolina, Columbia, SC 29208 [email protected]

√ Yet Another Proof of the Irrationality of 2 ˜ Natalia Cas´as Ferreno √ The classical Greek proof of the irrationality of 2 is an example of a first-class theorem, simple enough to be presented to a wide audience but deep in essence, as has been written by the British mathematician G. H. Hardy [7]. Nowadays many proofs √ of the irrationality of 2 are known. Indeed, on the web page [3] fourteen different proofs appear, and several have been published in this M ONTHLY (see [1, 2, 4, 5, 6, 9]). The aim of this note is to present a new √ proof of this fact. Let us consider the linear mapping f : R → R given by f (x) = ( 2 − 1)x as a one-dimensional discrete dynamical system. For each point x0 ∈ R let us define its orbit O(x0 ) as the sequence of iterates of f starting at x0 , namely

O(x0 ) =

√ n 2 − 1 x0 : n ∈ {0} ∪ N .

√ Since 0 < 2 − 1 < 1√it is clear that for each x0 ∈ R the orbit O(x0 ) converges to zero. Now suppose that 2 = p/q is a rational number with p, q ∈ N. Then for all n ∈ N it holds that √ n 2 q= 68



2n/2 q, if n is even, 2(n−1)/2 p, if n is odd.

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√ Thus in any case ( 2)n q is a natural number, and from the binomial theorem it follows that  n  √  √ n n  k n−k 0< 2−1 q = 2 (−1) q ∈ Z. k k=1 Therefore O(q) ⊂ N which contradicts the fact that O(q) goes to zero. We remark that if α ∈ R is an algebraic integer of degree k (i.e., α is a real root of a monic polynomial of degree k with integer coefficients) then a similar argument applied to the mapping f (x) = (α − [α])x, where [·] denotes the integer part, using q k−1 as starting point, implies that α is either an integer or an irrational number. Finally we note that our proof is related to that of since f (x) represents the  [8], a 2 dynamics of the two dimensional map ( ab ) → −1 restricted to the eigenspace ( ) 1 −1√ b √ a = 2b which is associated to the √ eigenvalue λ = 2 − 1, and to the proof given in [10], where it is pointed out that if 2 = p/q then O(1) is bounded below by 1/q > 0, contradicting the fact that O(1) converges to zero. ACKNOWLEDGMENTS. One of the anonymous referees suggested the following alternative proof: Sup√ pose 2 is rational and choose the least q ∈ N such that f (q) ∈ N. Clearly f (q) < q, and since f ( f (q)) ∈ N, as an easy computation shows, we get a contradiction. This is essentially Niven’s proof presented in [2, Proposition 1]. The first remark after the proof is also based on a comment of the same referee, and moreover he/she called my attention to [4]. I am very grateful for his/her valuable comments that certainly have improved this note.

REFERENCES 1. T. M. Apostol, Irrationality of the square root of two—A geometric proof, this M ONTHLY 107 (2000) 841–842. 2. R. Beigel, Irrationality without number theory, this M ONTHLY 98 (1991) 332–335. 3. A. Bogomolny, Square Root of 2 is Irrational, Interactive Mathematics Miscellany and Puzzles, available at http://www.cut-the-knot.org/proofs/sq_root.shtml/. 4. R. W. Floyd, What else Pythagoras √ could have done (letter to the editor), this M ONTHLY 96 (1989) 67. 5. R. Gauntt, The irrationality of√ 2, this M ONTHLY 63 (1956) 247. 6. E. Halfar, The irrationality of 2, this M ONTHLY 62 (1955) 437. 7. G. H. Hardy, A Mathematician’s Apology, Foreword by C. P. Snow, Cambridge University Press, Cambridge, 1992. 8. D. Kalman, R. Mena, and S. Shahriari, Variations on an irrational theme—Geometry, dynamics, algebra, Math. Mag. 70 (1997) 93–104. 9. Y. Sagher, What Pythagoras could have done, this M ONTHLY 95 (1988) 117. 10. P. Ungar, Irrationality of square roots, Math. Mag. 79 (2006) 147–148. Department of Mathematics, IES Ingenio, Gran Canaria, Las Palmas, Spain [email protected]

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A General Lagrange Theorem Giovanni Panti 1. INTRODUCTION The ordinary continued fractions expansion of a real number is based on the Euclidean algorithm. Variants of the latter yield variants of the former, all encompassed by a more general dynamical systems framework. For all these variants the Lagrange theorem holds: a number has an eventually periodic expansion if and only if it is a quadratic irrational. This fact is surely known for specific expansions, but the only proof for the general case that I could trace in the literature follows as an implicit corollary from much deeper results by Boshernitzan and Carroll on interval exchange transformations [2]. It may then be useful to have at hand a simple and virtually computation-free proof of a general Lagrange theorem. Let D be any unimodular partition of the real unit interval. By this we mean that D is a family (finite or countable, of cardinality at least 2) of half-open intervals a = ( p/q, r/s], with a varying in a fixed index set I ⊆ Z, such that: (i) each interval a ∈ D is unimodular, i.e., its extrema 0 ≤ p/q < r/s ≤ 1 are rational numbers (always written in reduced form), and  qp rs  = −1 (this amounts to saying that the column vectors ( p q)t and (r s)t constitute a Zbasis for Z2 ); (ii) two distinct intervals in D have empty intersection;  (iii) the set X = D \ {1} contains all irrational numbers in [0, 1]. Fix an arbitrary function ε : I → {−1, +1}, let a = ( p/q, r/s] ∈ D, and let

Ga =

  ε(a)+1

−1 2 0 1 0 1 p r . 1 1 1 0 q s

  Any matrix T = vt wu in GL2 Z induces a fractional-linear homeomorphism φT of the projective real line R ∪ {∞} onto itself via φT (x) = (t x + u)/(vx + w). Using projective coordinates (x y)t to represent the real number x/y (with (1 0)t representing ∞), φT amounts to multiplication on the left by T . We apply the above to T = G a , observing that G a ( p q)t and G a (r s)t are (0 1)t and (1 1)t (in one order or the other, depending on the value of ε(a)). We can then check easily that φG a  a is monotonically increasing with range (0, 1] if ε(a) = −1, and monotonically decreasing with range [0, 1) if ε(a) = +1. The resulting piecewise-fractional map G : X → [0, 1], defined by Gx = φG a (x) for x ∈ a , is the Gauss map determined by D and ε. Denote by ψa = φG −1  [0, 1] the ath inverse branch of G; the range of ψa is the a topological closure of a . Computing G −1 a , we obtain explicitly ⎧ (r − p)x + p ⎪ ⎪ , if ε(a) = −1; ⎨ (s − q)x + q ψa (x) = ⎪ ( p − r )x + r ⎪ ⎩ , if ε(a) = +1. (q − s)x + s If x, Gx, G 2 x, . . . , G n−1 x are all in X then, letting G t−1 x ∈ at for 1 ≤ t ≤ n, we have the identity x = ψa1 ψa2 · · · ψan (G n x). 70

(1)

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In dynamical systems language, the (finite or infinite) sequence a1 , a2 , . . . is the symbolic sequence of x. By (1), if x has a finite symbolic sequence (i.e., G n x ∈ [0, 1] \ X ⊂ Q for some n), then x is rational. Conversely, let x = u/v ∈ a be rational, and let p/q < r/s be the extrema of a . Then (u v)t = l( p q)t + m(r s)t for some nonnegative integers l, m with m ≥ 1. Multiplying by G a to the left, we see that the denominator of Gx equals l + m, which is strictly less than the denominator v = lq + ms of x. Hence the sequence of the denominators along the G-orbit of x is strictly decreasing, and x must eventually leave X (note that 0 and 1 are the only rational points having denominator 1, and neither of them is in X ). Example 1. The first and main given by ordinary continued frac example is of course  tions: I = {1, 2, . . . }, a = 1/(a + 1), 1/a , X = (0, 1), and ε = +1 throughout. The Gauss map is Gx = 1/x − 1/x, and its graph is shown in Figure 1.

Figure 1.

We have ψa (x) = 1/(a + x), and hence (1) assumes the familiar shape 1

x=

.

1

a1 +

1 ..

a2 + an−1 +

.

1 an + G n x

Example 2. In odd continued fractions we have I , X , and the a ’s as above, and ε(a) = (−1)a+1 . The Gauss map is Gx = |1/x − b|, where b is the odd integer in {1/x, 1/x + 1}, and its graph is shown in Figure 2.

Figure 2.

January 2009]

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71

Setting (bt , εt ) equal to (at , +1) for at = 1/G t−1 x odd, and to (at + 1, −1) for at even, we have ψat (x) = 1/(bt + εt x), and therefore 1

x=

.

ε1

b1 +

ε2 ..

b2 + bn−1 +

.

εn−1 bn + εn G n x

Among the continued fractions algorithms covered by our framework are the even continued fractions, the nearest integer fractions, the Farey fractions, and so on; see [1], [7] for more examples and a detailed analysis of the ergodic properties of the corresponding Gauss maps. Other algorithms—such as various versions of the cyclic method discovered by the Indians in the 12th century for solving the Pell equation— are not covered; see [8, §2-3] and references therein for these algorithms. 2. CONVERGENCE . . . Unimodular intervals have a curious property. Observation 3. Let 1 ⊃ 2 ⊃ · · · be a nested sequence of closed unimodular intervals, each one strictly contained in the previous one. Then lim length(n ) = 0.

n→∞

Proof. The length of n = [ pn /qn , rn /sn ] is (qn sn )−1 , which is less than or equal to [max(qn , sn )]−1 . If u/v is a rational in the topological interior of n , then v is strictly greater than max(qn , sn ) (again, because (u v)t = l( pn qn )t + m(rn sn )t , for certain positive l, m ∈ Z). It follows that the sequence max(q1 , s1 ), max(q2 , s2 ), . . . is strictly increasing and goes to infinity, and therefore the sequence of the reciprocals, which bounds above the sequence of the lengths, goes to 0. As a consequence, we have: (i) the map k : [0, 1] \ Q → I N that associates to each irrational its symbolic sequence is injective, continuous, and open; (ii) if x is irrational with symbolic sequence a1 , a2 , . . . , and y ∈ [0, 1], then lim ψa1 · · · ψan (y) = x.

n→∞

Indeed, let k(x) = a1 , a2 , . . . ; by §1(1), the set of all numbers whose symbolic sequence agrees with k(x) up to an is contained in the closed unimodular interval n whose extrema are ψa1 · · · ψan (0) and ψa1 · · · ψan (1). Since each a ∈ D is properly contained in (0, 1], the inclusions 1 ⊃ 2 ⊃ · · · are proper, and Observation 3 implies (ii) and the injectivity of k. We leave the continuity and openness of k as an exercise for the reader; k may be surjective (e.g., in the case of ordinary continued fractions), but is never so if I is finite (because then I N is compact, while [0, 1] \ Q is not). 72

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3. . . . AND PERIODICITY. On August 25th, 1769, Joseph Louis Lagrange read at the Royal Berlin Academy of Science a M´emoire on the resolution of algebraic equations via continued fractions [5]. He had been in Berlin since 1766, as director of the mathematical section of the same Academy, on the recommendation of his predecessor, Leonhard Euler. In this M´emoire he proved that the irrational x has an eventually periodic continued fractions expansion if and only if it has degree two over the rationals. The “only if” implication is clear even in our general setting. If x has symbolic sequence a1 , . . . , at , at+1 , . . . , at+r under any Gauss map G, then G t x and G t+r x have the same symbolic sequence, namely at+1 , . . . , at+r , and hence are equal. This implies that the vectors

 x G at · · · G a1 1 and G at+r

 x · · · G at+1 G at · · · G a1 1

are projectively equal, i.e., differ by a nonzero multiplicative constant. But then (x 1)t is an eigenvector for the integer matrix (G at · · · G a1 )−1 G at+r · · · G at+1 G at · · · G a1 , and hence x is quadratic. The “if” direction is trickier. The usual proof [4], [6] is basically Lagrange’s. One considers the minimal polynomial cn X 2 + dn X + en ∈ Z[X ] of 1/G n x, and shows that, from some n on, the coefficients cn , dn , en must satisfy certain inequalities. Sometimes this proof is reworded by writing 1/G n x in reduced form √ 1 Pn + D = , Gn x Qn and again bounding Pn and Q n in terms of the common discriminant D of the above polynomials, for n large enough. All of this is tightly related to Gauss’s theory of reduced quadratic forms; see [3, §5.7]. It is not clear how to adapt the above proof to the case of arbitrary piecewisefractional expansions. I will switch to a more geometric vein by formulating another observation. Observation 4. Let x be a quadratic irrational. Then there exists a matrix H with integer entries having (x 1)t as an eigenvector, with corresponding eigenvalue λ > λ¯ > 0 (λ¯ being the other eigenvalue). Proof. Let cX 2 + d X + e ∈ Z[X ] be the minimal polynomial of x, and let x¯ be the algebraic conjugate. Plainly

  

x x −d −e . = cx 1 1 c 0 Let K be the sum of the matrix displayed above with t times the 2 × 2 identity matrix. For t a sufficiently large positive integer, the eigenvalues cx + t and c x¯ + t of K are both positive. If c x¯ + t < cx + t, we take H = K and we are through. Otherwise, we take H = |K |K −1 and observe that 0 < |K |(c x¯ + t)−1 = cx + t < c x¯ + t = |K |(cx + t)−1 . January 2009]

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73

We may now prove the Lagrange theorem for arbitrary piecewise-fractional expansions. Let x ∈ [0, 1] be a quadratic irrational, with symbolic sequence a1 , a2 , . . . ¯ x¯ be as above. Let h : R2 → R2 be the under some Gauss map G, and let H, λ, λ, linear transformation whose matrix is H , with respect to the standard basis. By §1(1), x belongs to the interval n whose extrema are ψa1 · · · ψan (0) and ψa1 · · · ψan (1), for every n ≥ 0. The vectors (ψa1 · · · ψan (0) 1)t and (ψa1 · · · ψan (1) 1)t are positively proportional to the columns of the matrix

 −1 −1 0 1 G · · · G . Bn = G −1 a1 a2 an 1 1 Let Cn be the cone spanned positively by these columns. By §2(ii), there exists m such that x¯ ∈ / n , for every n ≥ m. Since 0 < λ¯ < λ, (x 1)t ∈ Cn , and (x¯ 1)t ∈ / Cn ∪ −Cn , we easily see, by writing an arbitrary vector in Cn as a linear combination of the eigenvectors (x 1)t and (x¯ 1)t , that h[Cn ] \ {0} is contained in the topological interior of Cn for n ≥ m. This implies that the matrix Hn = Bn−1 H Bn (i.e., the matrix of h with respect to the basis given by the columns of Bn ) has positive entries for n ≥ m. Hence Hm , Hm+1 , . . . are positive integer matrices, all conjugate to each other via matrices in GL2 Z. Since the determinant and the trace of a matrix are invariant under conjugation, these matrices are finite in number. Therefore Ht = Ht+r , for some t ≥ m and r > 0. The λ-eigenspace of Ht is 1-dimensional with basis Bt−1 (x 1)t , and analogously for 0 1 Ht+r . Multiplying on the left by 1 1 , we see that the vectors G at · · · G a1

 x 1

and G at+r · · · G at+1 G at · · · G a1

 x 1

are projectively equal. This means G t x = G t+r x; hence x is preperiodic under the Gauss map, and its symbolic sequence is eventually periodic. REFERENCES 1. V. Baladi and B. Vall´ee, Euclidean algorithms are Gaussian, J. Number Theory 110 (2005) 331–386. 2. M. D. Boshernitzan and C. R. Carroll, An extension of Lagrange’s theorem to interval exchange transformations over quadratic fields, J. Anal. Math. 72 (1997) 21–44. 3. H. Cohen, A Course in Computational Algebraic Number Theory, Graduate Texts in Mathematics, vol. 138, Springer-Verlag, Berlin, 1993. 4. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, Oxford, 1985. ´ 5. J. L. Lagrange, Additions au M´emorie sur la R´esolution des Equations Numer´eriques, M´emoires de l’Acad´emie royale des Sciences et Belles-Lettres de Berlin XXIV (1770) 581–652; also available at the G¨ottinger Digitalisierungszentrum, http://www.gdz-cms.de. 6. A. M. Rockett and P. Sz¨usz, Continued Fractions, World Scientific, River Edge, NJ, 1992. 7. B. Vall´ee, Euclidean dynamics, Discrete Contin. Dyn. Syst. 15 (2006) 281–352. 8. H. C. Williams, Solving the Pell equation, in Number Theory for the Millennium, III, M. A. Bennett et al., eds., A K Peters, Natick, MA, 2002, 397–435. Department of Mathematics, University of Udine, via delle Scienze 208, 33100 Udine, Italy [email protected]

74

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A Note  nd on the  n Congruence md ≡ m (mod q) Romeo Meˇstrovi´c As noticed in [5], many great mathematicians of the nineteenth century considered problems involving binomial coefficients modulo a prime power (for instance Babbage, Cauchy, Cayley, Gauss, Hensel, Hermite, Kummer, Legendre, Lucas, and Stickelberger). They discovered a variety of elegant and surprising theorems which are often easy to prove. For more information on these classical results, their extensions, and new results about this subject, see Dickson [2], Granville [5], and Singmaster [10]. Suppose that a prime p and pair of integers n ≥ m ≥ 0 are given. A beautiful theorem of E. Kummer (see [6] and [2, p. 270]) states that if pr is the highest power  n of p dividing m , then r is equal to the number of carries when adding m and n − m in base p arithmetic. If n = n 0 + n 1 p + · · · + n s p s and m = m 0 + m 1 p + · · · + m s p s are the p-adic expansions of n and m (so that 0 ≤ m i , n i ≤ p − 1 for each i), then by Lucas’s theorem ([7]; also see [2, p. 271] and [5]),

   s n ni ≡ (mod p). m mi i=0 This immediately yields

  np n ≡ (mod p), mp m

(1)

since the same products of binomial coefficients   are formed on the right side of Lucas’s theorem in both cases, other than an extra 00 = 1. A direct proof of the congruence (1), based on a polynomial method, is given in [11, Solution of Problem A-5, p. 173]. In this note we observe that a partial converse theorem is true: Theorem. If d, q > 1 are integers such that

  nd n ≡ (mod q) md m for every pair of integers n ≥ m ≥ 0, then d and q are powers of the same prime p. Proof. Let p be a prime dividing q and write d = p k s for some integer k ≥ 0, where the integer s is not divisible by p. By our hypothesis and Lucas’s theorem, again, we see that for every integer n ≥ 1,

    n nd nsp k ns n= ≡ = ≡ (mod p). (2) 1 d s sp k Write s = s0 + s1 p + · · · + sk p k with 0 ≤ si ≤ p − 1 for each i. Supposing that there exists j for which s j ≥ 2, choose j so that s j ≥ 2 and si ∈ {0, 1} for 0 ≤ i ≤ j − 1. Let n be the smallest integer for which ns j ≥ p. Write ns = January 2009]

NOTES

75

t0 + t1 p + · · · + tl pl with 0 ≤ ti ≤ p − 1 for each i. Since 2 ≤ n ≤ p − 1 we see that 0 ≤ nsi ≤ p − 1 for 0 ≤ i ≤ j − 1, so ti = nsi for such an i. This implies that t j = ns j − p, and we know that 0 ≤ t j < s j , as (n − 1)s j < p by the definition of n. Therefore, when we use Lucas’s theorem to determine nss (mod p), our product   contains the term st jj , which is ≡ 0 (mod p) since t j < s j . This contradicts (2). We are now left with the case that si = 0 or 1 for every i, and we know that s0 = 1 by the definition of s. Suppose that j exists, the smallest integer ≥ 1 for which s j = 1. Therefore s ≡ p j + 1 (mod p j +1 ). Let n = p j + ( p − 1), so that ns ≡ p − 1 l (mod p j +1 ). Hence, if ns = nst0 + t1 p + · · · + tl p with 0 ≤ ti ≤ p − 1 for each i, then t j = 0 and s j = 1; so that s ≡ 0 (mod p), contradicting (2). Therefore s = 1, that is d = p k . But then p can be the only prime dividing q. Remarks. We do not know which are the possible prime powers p k and pl such that

np k mp k



 n ≡ (mod pl ) m

for all integers n ≥ m ≥ 1. Evidently we wish to determine l p (k), the maximum such l for each given k ≥ 1. By Jacobsthal’s theorem (see, e.g., [5]), l p (1) ≥ 3 for all primes p ≥ 5, and equals 3 unless p divides the numerator of B p−3 , the ( p − 3)rd Bernoulli number. By Wolstenholme’s theorem (see, e.g., [1, Theorem 1]), if p is a prime greater than 1 3, then the numerator of the fraction H ( p − 1) := 1 + 12 + 13 + · · · + p−1 is divisible by p 2 . Now we define w p < p 2 to be the unique nonnegative integer such that the numerator of the fraction w p − H ( p − 1)/ p 2 written in reduced form is divisible by p 2 . It is well known that (see [4]) 1 w p ≡ − B p−3 (mod p) 3 in the sense that p divides the numerator of the fraction w p + 13 B p−3 written in reduced form. Furthermore, by a recent result of J. Zhao [12, congruence (10) of Theorem 3.2], for a given prime p ≥ 7, l p (1) ≥ 5 if and only if w p = 0.  p−1 ≡ A prime p is said to be a Wolstenholme prime if it satisfies the congruence 2p−1 4 1 (mod p ), or equivalently,

 2p ≡ 2 (mod p 4 ). p Then by Glaisher’s congruence ([3, p. 21]; also cf. [8, Corollary, p. 386]), a prime p is a Wolstenholme prime if and only if p divides the numerator of B p−3 . The only two known such primes are 16843 and 2124679, and by a recent result of McIntosh and Roettger from [9], these primes are the only two Wolstenholme primes less than 109 . Finally, note that by the above mentioned results of Jacobsthal and Glaisher, a prime p is a Wolstenholme prime if and only if l p (1) ≥ 4. By using an argument based on the prime number theorem, McIntosh [8, p. 387] conjectured that there are infinitely  p−1 many Wolstenholme primes, and that no prime satisfies the congruence 2p−1 ≡1 5 (mod p ). 76

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ACKNOWLEDGMENTS. The author gratefully acknowledges valuable suggestions from the anonymous referees.

REFERENCES 1. E. Alkan, Variations on Wolstenholme’s theorem, this M ONTHLY 101 (1994) 1001–1004. 2. L. E. Dickson, The History of the Theory of Numbers, vol. 1, Chelsea Publishing, New York, 1966. 3. J. W. L. Glaisher, Congruences relating to the sums and products of the first n numbers and to other sums of products, Q. J. Math. 31 (1900) 1–35. , On the residues of the sums of the inverse powers of numbers in arithmetical progression, Q. J. 4. Math. 32 (1900) 271–288. 5. A. Granville, Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers, in Organic Mathematics–Burnaby, BC 1995, CMS Conf. Proc., vol. 20, American Mathematical Society, Providence, RI, 1997, 253–276. ¨ 6. E. E. Kummer, Uber die erg¨anzungss¨atze zu den allgemeinen reciprocit¨atsgesetzen, J. Reine Angew. Math. 44 (1852) 93–146. 7. E. Lucas, Sur les congruences des nombres eul´eriens et des coefficients diff´erentiels des fonctions trigonom´etriques, suivant un module premier, Bull. Soc. Math. France 6 (1877–1878) 49–54. 8. R. J. McIntosh, On the converse of Wolstenholme’s theorem, Acta Arith. 71 (1995) 381–389. 9. R. J. McIntosh and E. L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Math. Comp. 76 (2007) 2087–2094. 10. D. Singmaster, Notes on binomial coefficients, J. London Math. Soc. 8 (1974) 545–560. 11. The William Lowell Putnam Mathematical Competition, Problem A-5, this M ONTHLY 86 (1979) 171– 173. 12. J. Zhao, Bernoulli numbers, Wolstenholme’s theorem, and p5 variations of Lucas’ theorem, J. Number Theory 123 (2007) 18–26; also available at http://arxiv.org/abs/math.NT/0303332. Department of Mathematics, Maritime Faculty Kotor, University of Montenegro, Dobrota 36, 85330 Kotor, Montenegro [email protected]

A Simple Solution to a Multiple Player Gambler’s Ruin Problem Sheldon M. Ross 1. PROBLEM. Consider a gambler’s ruin problem involving r players, with player i initially having n i units, n i > 0, i = 1, . . . , r . At each stage, two of the players are chosen to play a game, with the winner of the game receiving 1 unit from the loser. Any player whose  fortune drops to 0 is eliminated, and this continues until a single player has all n ≡ ri=1 n i units, with that player designated as the victor. Assuming that the results of successive games are independent and that each game is equally likely to be won by either of its two players, among other results we find (a) the probability that player i is the victor; (b) the expected number of stages until one of the players has all the money; (c) the expected number of games played between two specified players. Moreover we show that none of the preceding quantities depend on the rule for choosing the players in each stage. January 2009]

NOTES

77

ACKNOWLEDGMENTS. The author gratefully acknowledges valuable suggestions from the anonymous referees.

REFERENCES 1. E. Alkan, Variations on Wolstenholme’s theorem, this M ONTHLY 101 (1994) 1001–1004. 2. L. E. Dickson, The History of the Theory of Numbers, vol. 1, Chelsea Publishing, New York, 1966. 3. J. W. L. Glaisher, Congruences relating to the sums and products of the first n numbers and to other sums of products, Q. J. Math. 31 (1900) 1–35. , On the residues of the sums of the inverse powers of numbers in arithmetical progression, Q. J. 4. Math. 32 (1900) 271–288. 5. A. Granville, Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers, in Organic Mathematics–Burnaby, BC 1995, CMS Conf. Proc., vol. 20, American Mathematical Society, Providence, RI, 1997, 253–276. ¨ 6. E. E. Kummer, Uber die erg¨anzungss¨atze zu den allgemeinen reciprocit¨atsgesetzen, J. Reine Angew. Math. 44 (1852) 93–146. 7. E. Lucas, Sur les congruences des nombres eul´eriens et des coefficients diff´erentiels des fonctions trigonom´etriques, suivant un module premier, Bull. Soc. Math. France 6 (1877–1878) 49–54. 8. R. J. McIntosh, On the converse of Wolstenholme’s theorem, Acta Arith. 71 (1995) 381–389. 9. R. J. McIntosh and E. L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Math. Comp. 76 (2007) 2087–2094. 10. D. Singmaster, Notes on binomial coefficients, J. London Math. Soc. 8 (1974) 545–560. 11. The William Lowell Putnam Mathematical Competition, Problem A-5, this M ONTHLY 86 (1979) 171– 173. 12. J. Zhao, Bernoulli numbers, Wolstenholme’s theorem, and p5 variations of Lucas’ theorem, J. Number Theory 123 (2007) 18–26; also available at http://arxiv.org/abs/math.NT/0303332. Department of Mathematics, Maritime Faculty Kotor, University of Montenegro, Dobrota 36, 85330 Kotor, Montenegro [email protected]

A Simple Solution to a Multiple Player Gambler’s Ruin Problem Sheldon M. Ross 1. PROBLEM. Consider a gambler’s ruin problem involving r players, with player i initially having n i units, n i > 0, i = 1, . . . , r . At each stage, two of the players are chosen to play a game, with the winner of the game receiving 1 unit from the loser. Any player whose  fortune drops to 0 is eliminated, and this continues until a single player has all n ≡ ri=1 n i units, with that player designated as the victor. Assuming that the results of successive games are independent and that each game is equally likely to be won by either of its two players, among other results we find (a) the probability that player i is the victor; (b) the expected number of stages until one of the players has all the money; (c) the expected number of games played between two specified players. Moreover we show that none of the preceding quantities depend on the rule for choosing the players in each stage. January 2009]

NOTES

77

2. SOLUTION. We first argue that the expected number of games played is finite. Let X i denote the number of games that involve player i, and let X=

r 1 Xi 2 i=1

denote the total number of games played. Lemma 1. No matter how the choices of the players in each stage are made, E[X ] < ∞. Proof. Fix i, and let L j equal 1 either if player i loses the jth game she plays or if she plays less than j games, and let it equal 0 otherwise. Also, for k ≥ 1, let Ak be the event that L (k−1)n+1 = L (k−1)n+2 = · · · = L kn = 1. Then, with F = min(k : Ak occurs) we have that X i ≤ n F, implying that E[X i ] ≤ n E[F]  P(F ≥ j) =n j ≥1

=n

 j ≥1

P(Ac1 Ac2 · · · Acj −1 )

 ≤n (1 − (1/2)n ) j −1 j ≥1

< ∞. Consequently, E[X ] =

r 

E[X i ] < ∞.

i=1

Proposition 2. No matter how the choices of the players in each stage are made, the probability that player i is the victor is n i /n. Proof. To begin, suppose that there are n players, with each player initially having 1 unit. Consider player i. She starts with 1 and each stage she plays will be equally likely to result in her either winning or losing 1 unit, with the results from each stage being independent. In addition, she will continue to play stages until her fortune becomes either 0 or n. Because this is the same for all players, it follows that each player has the same chance of being the victor. Consequently, no matter how the choices of the players in each stage are made, each player has probability 1/n of being the victor. Now, suppose these n players are partitioned into r teams, with team i containing n i players, i = 1, . . . , r. Then the probability that the victor is a member of team i is n i /n. But because team i initially has a total fortune of n i units, i = 1, . . . , r, and each game played by members of different teams results in the fortune of the winning team increasing by 1 and that of the losing team decreasing by 1, it follows 78

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

that the probability that the victor is from team i is exactly the probability asked for. Consequently, no matter how the choices of the players in each stage are made, P(i is victor) = n i /n. To find E[X i ], we will make use of the following lemma. Lemma 3. Let m j denote the expected number of games needed when there are only 2 players with initial fortunes j and n − j. Then m j = j (n − j). Proof. Conditioning on the outcome of the first game yields that 1 1 m j = 1 + m j +1 + m j −1 , 2 2

j = 1, . . . , n − 1.

(1)

Consequently, m j +1 = 2m j − m j −1 − 2,

j = 1, . . . , n − 1.

(2)

Using that m 0 = 0, the preceding yields that m 2 = 2m 1 − 2 m 3 = 2m 2 − m 1 − 2 = 3m 1 − 6 = 3(m 1 − 2) m 4 = 2m 3 − m 2 − 2 = 4m 1 − 12 = 4(m 1 − 3) suggesting that m i = i(m 1 − i + 1),

i = 1, . . . , n.

(3)

Using (2), the preceding is easily shown by mathematical induction. Letting i = n in (3), and using that m n = 0, gives that m 1 = n − 1, and yields the result m i = i(n − i). Proposition 4. X i , the number of games played by player i, has the same distribution no matter how the choices of the players in each stage are made. Also, E[X i ] = n i (n − n i ). Proof. From the perspective of player i, starting with n i he will continue to play stages, independently being equally likely to win or lose each one, until his fortune is either n or 0. Thus, the number of stages he plays is exactly the same as when he has a single opponent with an initial fortune of n − n i , and the result follows from Lemma 3. Corollary 5. No matter how the choices of the players in each stage are made, the expected number of games played is given by   r r  1 1 E[X ] = n i (n − n i ) = n i2 . n2 − 2 i=1 2 i=1 January 2009]

NOTES

79

Proof. This follows from Proposition 4 upon using the identity X =

1 2

n i=1

Xi .

Remark. Whereas the expected number of games played does not depend on the rule used for choosing players, its distribution (as opposed to the distribution of X i ) does depend on the rule. To see this, suppose r = 3, n 1 = n 2 = 1, and n 3 = 2. Then if players 1 and 2 are chosen in the first stage, then it takes at least three stages to determine a victor, whereas if player 3 is in the first stage then it is possible for there to be only two stages. Now, for any set of players S ⊂ {1, . . . , r }, let X (S) denote the number of games involving only members of S. Also, for disjoint subsets of players A and B, let X (A, B) denote the number of games in which one of the players is in A and the other is in B. Proposition 6. No matter how the choices of the players in each stage are made,  E[X (S)] = ni n j . i< j, {i, j }⊂S

Proof. Let S c be the complement of S. By regarding the players in S as constituting one team and those in S c as constituting a second team, it follows that the changes of a team’s total fortune  will move exactly as if there were 2 players with initial fortunes of n and n − i∈S i i∈S n i . Because there will continue to be games between members of S and S c until the cumulative fortune of one of these teams hits 0, it follows from Lemma 3 that   2       c n− ni ni = n ni − ni . E[X (S, S )] = i∈S

i∈S

i∈S

i∈S

Now, imagine that each player earns 1 point whenever she plays in a game. Then the total number of points earned by players in team S is i∈S X i . But since team S earns 1 point for each game between a member of S and one of S c , and the team earns 2 points for each game between members of S, it follows that  X i = X (S, S c ) + 2X (S). i∈S

Taking expectations yields the result 2E[X (S)] =

 i∈S

n i (n − n i ) − n

 i∈S

 ni +



2 ni

.

i∈S

Corollary 7. For disjoint sets of players A and B, no matter how the choices of the players in each stage are made,  E[X (A, B)] = ni n j . i∈A j ∈B

Proof. For i  = j, let X (i, j) denote the number of games between players i and j. Proposition 6 yields that E[X (i, j)] = E[X ({i, j})] = n i n j . 80

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

The result now follows by taking expectations of both sides of the identity  X (A, B) = X (i, j). i∈A j ∈B

The problem of this paper had previously been considered in [1] and [3]. These papers assumed that r = 3 and also that each choice of the pair of players to contest a game was randomly made in the sense that each pair of remaining players had equal probability of being the contestants. These papers used, respectively, recursive equations and martingale techniques to solve for the mean time E[X ] (see [2] for an introduction to martingales). Neither recognized that E[X ] can be so immediately obtained using the r = 2 result, nor that the result is independent of the manner in which the pairs are selected. Also, neither of these papers considered questions related to the random variables X (S) and X (A, B). REFERENCES 1. A. Engel, The computer solves the three tower problem, this M ONTHLY 100 (1993) 62–64. 2. S. M. Ross and E. Pek¨oz, A Second Course in Probability, ProbabilityBookstore.com, Boston, MA, 2007. 3. D. Stirzaker, Tower problems and martingales, Math. Sci. 19 (1994) 52–59. Epstein Department of Industrial and Systems Engineering, University of Southern California [email protected]

Mathematics Is . . . “Mathematics is to nature as Sherlock Holmes is to evidence.” Ian Stewart, Nature’s Numbers: The Unreal Reality of Mathematical Imagination, Basic Books, New York, 1995, p. 2. —Submitted by Carl C. Gaither, Killeen, TX

January 2009]

NOTES

81

PROBLEMS AND SOLUTIONS Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West with the collaboration of Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall Dougherty, Tam´as Erd´elyi, Zachary Franco, Christian Friesen, Ira M. Gessel, L´aszl´o Lipt´ak, Frederick W. Luttmann, Vania Mascioni, Frank B. Miles, Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ullman, Charles Vanden Eynden, Sam Vandervelde, and Fuzhen Zhang.

Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before May 31, 2009. Additional information, such as generalizations and references, is welcome. The problem number and the solver’s name and address should appear on each solution. An asterisk (*) after the number of a problem or a part of a problem indicates that no solution is currently available.

PROBLEMS 11404. Proposed by Raimond Struble, North Carolina State at Raleigh, Raleigh, NC. Any three non-concurrent cevians of a triangle create a subtriangle. Identify the sets of non-concurrent cevians which create a subtriangle whose incenter coincides with the incenter of the primary triangle. (A cevian of a triangle is a line segment joining a vertex to an interior point of the opposite edge.) 11405. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let P be an interior point of a tetrahedron ABC D. When X is a vertex, let X  be the intersection of the opposite face with the line through X and P. Let X P denote the length of the line segment from X to P. (a) Show that P A · P B · PC · P D ≥ 81P A · P B  · PC  · P D  , with equality if and only if P is the centroid of ABC D. (b) When X is a vertex, let X  be the foot of the perpendicular from P to the plane of the face opposite X . Show that P A · P B · PC · P D = 81P A · P B  · PC  · P D  if and only if the tetrahedron is regular and P is its centroid. 11406. Proposed by A. A. Dzhumadil’daeva, Almaty, Republics Physics and Mathematics School, Almaty, Kazakhstan. Let n!! denote the product of all positive integers not greater than n and congruent to n mod 2, and let 0!! = (−1)!! = 1. Thus, 7!! = 105 and 8!! = 384. For positive integer n, find n    n (2i − 1)!! (2(n − i) − 1)!! i i=0 in closed form. 11407. Proposed by Erwin Just (Emeritus), Bronx Community College of the City University of New York, New York, NY. Let p be prime greater than 3. Does there exists a ring with more than one element necessarily having a multiplicative identity)  p (not x 2i−1 = 0? such that for all x in the ring, i=1 82

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

11408. Proposed by Marius Cavachi, “Ovidius” University of Constant¸a, Constant¸a, Romania. Let k be a fixed integer greater than 1. Prove that there exists an integer n greater than 1, distinct integers a1 , a2 , . . . , an , all greater than 1, such that both  and n n a and φ(a j j ) are kth powers of a positive integer. Here φ denotes Euler’s j =1 j =1 totient function. 11409. Proposed by Paolo Perfetti, Dept. Math, University “Tor Vergata”, Rome, Italy. For positive real α and β, let S(α, β, N ) =

N  n=2

n log(n)(−1)n

n  k=2

α + k log k . β + (k + 1) log(k + 1)

Show that if β > α, then lim N →∞ S(α, β, N ) exists. 11410. Proposed by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. For 0 < φ < π/2, find   ∞  1 (−1)n−1 sin2 (nx) 2 −2 log cos φ + lim x sin (nφ) . x→0 2 n (nx)2 n=1

SOLUTIONS A Solid with the Rupert Property 11291 [2007, 451]. Proposed by Richard Jerrard and John Wetzel, University of Illinois at Urbana-Champaign, Urbana, IL. z The base of a solid P symmetric in the x z-plane is the unit disk x 2 + y 2 ≤ 1 in the x y-plane. The portion of P in the half-space y ≥ 0 is bounded by the surface swept by a segment P Q as P moves uniformly from (1, 0, 2) to (−1, 0, 2) while Q moves uniformly around the unit semicircle from (1, 0, 0) to (−1, 0, 0). The solid is completed y by reflecting this half through the x z-plane (see sketch). x

The projections of P are the unit disk in the x y-plane, a square region of side two in the x z-plane, and an isosceles triangular region in the yz-plane with base and altitude two. Show that P has the Rupert property, that is, it is possible to cut a tunnel through P through which a second copy of P can be passed. Solution by Mark D. Meyerson, U.S. Naval Academy, Annapolis,√MD. The circular base of P is inscribed in the square of side 2 with vertices at ± √2 on the√x and y axes. Let S denote this square. The four points of tangency are (± 2/2, ± 2/2). If we rotate the base disk about the x-axis by less than 90◦ and orthogonally project the result to the x y-plane, we get a ellipsoidal region that lies strictly inside S. Such a rotation applied to the top edge of P will move its projection away from the x-axis; if the rotation is small this projection will continue to lie strictly inside S. And so, for such a small rotation, since S is convex, all segments connecting the top edge of P to the boundary of the base of P will project into the interior of S. Since these segments, together with the base, form the boundary of P, the orthogonal projection to the x yplane of this rotation of P lies strictly inside S. Since S is congruent to the x z-plane slice of P, we can drill a slightly smaller square tunnel in the y-axis direction through which P can pass. January 2009]

PROBLEMS AND SOLUTIONS

83

As an example, after a rotation through 11.537◦ about the x-axis, all points of P will project to points of the x y-plane more than 0.01 units from S. So a square tunnel of side length 1.98 (leaving a border of 0.01) will allow P to pass (with a clearance of about 0.00005). Also solved by R. Bagby, D. Chakerian, O. P. Lossers (Netherlands), J. Schaer (Canada), V. Schindler (Germany), GCHQ Problem Solving Group (U. K.), and the proposer.

An Infinite Product 11299 [2007, 547]. Proposed by Pablo Fern`andez Refolio, UAM, Spain. Show that   √ n2 −1  ∞  1 n2 e e . = e n2 − 1 2π n=2 Solution by Timothy Achenbach, Hillsborough Community College, Plant City, FL. Let   n2 −1  k  1 n2 Pk = e (n − 1)(n + 1) n=2 be the partial product. For 2 ≤ j ≤ k − 1, the factor j appears in the numerator with exponent 2( j 2 − 1) and in the denominator with exponents j ( j + 2) and j ( j − 2). After cancelling, only a factor of j 2 will remain in the denominator. The factor k occurs in the numerator with exponent 2(k 2 − 1) and in the denominator with exponent k(k − 2), hence after cancelling k occurs with exponent k 2 + 2k − 2. The factor k + 1 occurs only in the denominator with exponent k 2 − 1. Hence 2

Pk =

k k +2k . ek−1 (k + 1)k 2 −1 (k!)2

The desired infinite product is thus    √ 2 2 ek−1 k k −1 k k 2πk e2 lim Pk = · lim · lim . k→∞ k→∞ 2π k→∞ (k + 1)k 2 −1 ek k! The second limit is 1 by Stirling’s formula. For the first limit, take logarithms and use the Taylor series for log(1 + z) to get     1 1 1 k − 1 − (k 2 − 1) log 1 + − 2 + O(k −3 ) , = k − 1 − (k 2 − 1) k k 2k which simplifies to −1/2 + O(1/k). Hence the first limit is e−1/2 and the infinite prod√ uct is e e/(2π). Also solved by T. Amdeberhan & V. Moll, S. Amghibech (Canada), R. Bagby, D. Beckwith, B. Brandie, B. S. Burdick, R. Chapman (U. K.), K. Dale (Norway), P. P. D´alyay (Hungary), M. Goldenberg & M. Kaplan, J. Grivaux (France), E. A. Herman, C. Hill, G. Keselman, O. Kouba (Syria), K.-W. Lau (China), O. P. Lossers (Netherlands), K. McInturff, M. Omarjee (France), A. Plaza (Spain), G. T. Prajitura, M. A. Prasad (India), O. G. Ruehr, H.-J. Seiffert (Germany), N. C. Singer, A. Stadler (Switzerland), R. Tauraso (Italy), M. Tetiva (Romania), M. Vowe (Switzerland), F. Wang (China), BSI Problems Group (Germany), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, and the proposer.

A Bernstein Polynomial Integral 11300 [2007, 547]. Proposed by Ulrich Abel, University of Applied Sciences GiessenFriedberg, Friedberg, Germany. For integers k and n with 0 ≤ k ≤ n, let pn,k (t) = 84

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

n

 t k (1 − t)n−k . Let K n (x, y) = nk=0 (y − k/n) pn,k (x) pn,k (y). Prove that for 0 ≤ u u ≤ 1 and 0 ≤ y ≤ 1, x=0 K n (x, y) d x ≥ 0. k

Solution by the BSI Problems Group, Bonn, Germany. We use generating functions. Write [s k ]P(s) for “the coefficient of s k in the polynomial P(s).” Letting f n (s, x) = (x + s(1 − x))n , we have pn,k (x) = [s k ](1 − x + sx)n = [s n−k ](x + s(1 − x))n = [s n−k ] f n (s, x). Thus n 

pn,k (x) pn,k (y) = [s n ](1 − (1 − s)x)n f n (s, y).

k=0

Compute

 y−

Thus K n (x, y) =

n   k=0

 =

k n

 pn,k (y) =

k y− n

−y(1 − y)  pn,k (y). n

 pn,k (x) pn,k (y)

−y(1 − y) ∂ n ∂y





[s n ] (1 − (1 − s)x)n f n (s, y)

= −y(1 − y)[s n ](1 − s)(1 − (1 − s)x)n f n−1 (s, y) = and



u

y(1 − y) n ∂ [s ] (1 − (1 − s)x)n+1 f n−1 (s, y) n+1 ∂x

y(1 − y) n [s ] (1 − (1 − s)u)n+1 − 1 f n−1 (s, y) n+1    n+1  n+1 j j y(1 − y) n  n+1− j [s ] f n−1 (s, y). = s u (1 − u) j n+1 j =1

K n (x, y) d x = 0

For 0 ≤ u ≤ 1 and 0 ≤ y ≤ 1, all coefficients of the two polynomials in s are nonnegative, so all coefficients of their product are also nonnegative. Also solved by R. Chapman (U. K.), P. P. D´alyay (Hungary), A. Stadler (Switzerland), GCHQ Problem Solving Group (U. K.), and Microsoft Research Problems Group.

A Quartic Inequality 11301 [2007, 547]. Proposed by Finbarr Holland, University College Cork, Ireland. Find the least real number M such that, for all complex a, b, and c,

ab(a 2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a 2 ) ≤ M(|a|2 + |b|2 + |c|2 )2 . Solution by Byoung Tae Bae, Institute of Science Education, Yonsei University, Seoul, Korea. First note that ab(a 2 − b2 ) + bc(b2 − c2 ) + ac(c2 − a 2 ) = (b − c)(a − c)(a − b)(a + b + c). It is required to find the smallest real number M such that I = January 2009]

|(a − b)(b − c)(c − a)(a + b + c)| ≤ M. (|a|2 + |b|2 + |c|2 )2 PROBLEMS AND SOLUTIONS

85

Let a = a1 + ia2 , b = b1 + ib2 , c = c1 + ic2 , where a j , b j , and c j are real numbers for j ∈ {1, 2}. Now I =

(a12

+

a22

+

b12

K , + b22 + c12 + c22 )2

where K is given by    K = (a1 − b1 )2 + (a2 − b2 )2 (b1 − c1 )2 + (b2 − c2 )2 (c1 − a1 )2 + (c2 − a2 )2  · (a1 + b1 + c1 )2 + (a2 + b2 + c2 )2 . Now the AM-GM inequality states that for nonnegative numbers x1 , x2 , x3 , and x4 , √ 4

1 (x1 + x2 + x3 + x4 ), 4 with equality if and only if x1 = x2 = x3 = x4 . Applying this to the case where x1 = |a − b|2 , x2 = |b − c|2 , x3 = |c − a|2 , and x4 = |a + b + c|2 gives K ≤

x1 x2 x3 x4 ≤

1 (a1 − b1 )2 + (a2 − b2 )2 + (b1 − c1 )2 + (b2 − c2 )2 16

2 + (c1 − a1 )2 + (c2 − a2 )2 + (a1 + b1 + c1 )2 + (a2 + b2 + c2 )2 ,

with equality if and only if (a1 − b1 )2 + (a2 − b2 )2 = (b1 − c1 )2 + (b2 − c2 )2 = (c1 − a1 )2 + (c2 − a2 )2 = (a1 + b1 + c1 )2 + (a2 + b2 + c2 )2 . However, since (ai − bi )2 + (bi − ci )2 + (ci − ai )2 + (ai + bi + ci )2 = 3(ai2 + bi2 + ci2 ), holds for i = 1, 2, it follows that 9 9 2 (a1 + b12 + c12 + a22 + b22 + c22 )2 = (|a|2 + |b|2 + |c|2 )2 . 16 16 Therefore, I ≤ 9/16. To conclude that M = 9/16, it suffices that there be complex a, b, and c for which = 9/16. Breaking it down by √ real and imaginary parts, we √ I √ √ √ √ take (a1 , a2 )√= (− 3 + √ 2/3, 1 + 2/3), (b1 , b2 ) = ( 3 + 2/3, 1 + 2/3), and (c1 , c2 ) = ( 2/3, −2 + 2/3). K ≤

Also solved by R. Bagby, M. Bataille (France), D. Beckwith, D. R. Bridges, R. Chapman (U. K.), P. P. D´alyay (Hungary), P. De (India), O. Kouba (Syria), O. P. Lossers (Netherlands), J. H. Nieto (Venezuela), T. L. Radolescu (Romania), H.-J. Seiffert (Germany), N. C. Singer, A. Stadler (Switzerland), F. Wang (China), H. Widmer (Switzerland), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, Northeastern University Math Problem Solving Group, and the proposer.

A Series with Harmonic Numbers 11302 [2007, 547]. Proposed by Horst Alzer, Waldbr¨ol, Germany. Find ∞ 

(2k + 1)Hk2 , (k − 1)k(k + 1)(k + 2) k=2  where Hk is the kth harmonic number, defined to be kj =1 1/j. Solution by Michael Vowe, Fachschule Nordwestschweiz, Muttenz, Switzerland. Denote the proposed sum by S. Since 1 1 2k + 1 = − , (k − 1)k(k + 1)(k + 2) (k − 1)k(k + 1) k(k + 2) 86

c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116 

summation by parts gives S= Since

∞ 2

1 H22  + Hk+1 − Hk2 . 3 k(k + 2) k=2



2 Hk+1

−

Hk2

1 = (Hk+1 + Hk )(Hk+1 − Hk ) = 2Hk + k+1



1 , k+1

we obtain ∞ ∞  2Hk 1 3  + + 4 k=2 k(k + 1)(k + 2) k=2 k(k + 1)2 (k + 2)  ∞  ∞  1 1 1 3  − = + Hk + 4 k=2 k(k + 1) (k + 1)(k + 2) k(k + 1)2 (k + 2) k=2

S=

∞ ∞  H2  1 1 3 + + (Hk+1 − Hk ) + 4 6 (k + 1)(k + 2) k(k + 1)2 (k + 2) k=2 k=1  ∞   1 1 + =1+ (k + 1)2 (k + 2) k(k + 1)2 (k + 2) k=2  ∞   1 1 13 1 − = . =1+ =1+ 2k(k + 1) 2(k + 1)(k + 2) 12 12 k=2

=

Also solved by T. Amdeberhan & T. V. Angelis, B. T. Bae (Korea), R. Bagby, M. Bataille (France), D. Beckwith, P. Bracken, M. A. Carlton, R. Chapman (U. K.), K. Dale (Norway), P. P. D´alyay (Hungary), M. N. Deshpande, C. R. Diminnie, M. J. Englefield (Australia), W. Fosheng (China), M. Goldenberg and M. Kaplan, J. Grivaux (France), E. A. Herman, G. Keselman, O. Kouba (Syria), K.-W. Lau (China), O. P. Lossers (Netherlands), M. Omarjee (France), P. Perfetti (Italy), G. T. Prajitura, M. A. Prasad (India), R. Pratt, O. G. Ruehr, H.-J. Seiffert (Germany), N. C. Singer, A. Stadler (Switzerland), R. Tauraso (Italy), M. Tetiva (Romania), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, NSA Problems Group, and the proposer.

Signed Series Terms 11304 [2007, 548]. Proposed by Teodora-Liliana R˘adulescu, Frat¸ii Buzes¸ti College, Craiova, and Vicent¸iu R˘adulescu, University of Craiova, Romania. (a) Find a sequence z n of distinct complex numbers, and a sequence αn of nonzero real numbers, such that for almost all complex numbers z (excluding a set of measure  −1 zero), ∞ diverges to +∞ yet not all α n are positive. n=1 αn |z − z n |  ∞ (b) Let βn be a sequence of real numbers such that n=1 |βn | is finite and such that, ∞ for almost all z in C, n=1 βn |z − z n |−1 converges to a nonnegative real number. Prove that βn ≥ 0 for all n. (c∗ ) Can there be a sequence αn of real numbers, not all positive, and n  a sequence z−1 of distinct complex numbers, such that for almost all complex z, ∞ n=1 αn |z − z n | converges to a positive real number? Solution by the GCHQ Problem Solving Group, Cheltenham, U. K. The problem statement uses metric and measure properties of C but not the algebraic ones, so the setting can be thought of as R 2 rather than C. It is then natural to extend it to R d . Our example for part (a) works for d ≥ 1 and our proof for part (b) works for d ≥ 2. January 2009]

PROBLEMS AND SOLUTIONS

87

(a) Let d ≥ 1. Let z n be any convergent sequence in R d whose limit is, say, z 0 , and let αn be the sequence 2, −1, 2, −1, · · · . Let z ∈ R d be any point except one of the z n . There exists n 0 such that for all n ≥ n 0 , (4/5)|z − z 0 | < |z − z n | < (4/3)|z − z 0 |. When n ≥ n 0 is even, αn |z − z n |−1 + αn+1 |z − z n+1 |−1 ≥ (1/4)|z − z 0 |−1 , so ∞ −1 diverges to +∞. n=1 αn |z − z n | (b) Let d ≥ 2, and let S be the d − 1 dimensional surface area of the unit sphere in  R d . If r > 0 and D = {z ∈ R d : |z| ≤ r }, then D |z|−1 dz = Sr d−1 /(d − 1), where dz is d-dimensional Lebesgue measure. The volume of D is Sr d /d, so the mean value of |z|−1 on D is d/(r (d − 1)). It follows (on pairing off points in the symmetric difference of D and D  ) that the mean value of |z|−1 on any spherical domain D  of radius r is at most d/(r (d − 1)).  If βk < 0 for some k, then there exists n 0 with n 0 > k such that ∞ n=n 0 |βn | < −βk (d − 1)/(2d). There exists r > 0 such that for all z with 0 < |z − z k | < r , n 0 −1 n=1 n=k

−βk |βn | < . |z − z n | 2|z − z k |

N For N > n 0 , the mean value of n=n |βn | |z − z n |−1 on D = {z : |z − z k | ≤ r } is at 0 N most n=n0 |βn |d/(r (d − 1)) ≤ −βk /(2r ). It follows that there is a subdomain D1 of D, with D1 having positive measure, on which ∞  n=n 0

−βk −βk |βn | ≤ ≤ . |z − z n | 2r 2|z − z k |

 −1 converges to a negative Therefore, on the subdomain D1 the series ∞ n=1 βn |z − z n | number. The result follows from this contradiction. Editorial comment. More than one solver noted that part (c) is trivial unless it is required that αn  = 0, but no solution was received for that nontrivial problem. Parts (a) and (b) also solved by J. H. Lindsey II, A. Stadler (Switzerland), and the proposer. Part (a) also solved by the Microsoft Research Problems Group.

An Inequality for Triangles 11306 [2007, 640]. Proposed by Alexandru Rosoiu, University of Bucharest, Bucharest, Romania. Let a, b, and c be the lengths of the sides of a nondegenerate triangle, let p = (a + b + c)/2, and let r and R be the inradius and circumradius of the triangle, respectively. Show that a 4r − R  a · ≤ ( p − b)( p − c) ≤ , 2 R 2 and determine the cases of equality. Solution by Victor Pambuccian, Arizona State University, Pheonix, AZ. Write S for the area of the triangle. The following are well known: S 2 = p( p − a)( p − b)( p − c),

S=

abc = r p. 4R

Let x, y, and z denote the lengths of the tangent segments to the incircle from the vertices opposite c, a, and b, respectively. Now a = z + x, b = x + y, c = y + z, 88

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p = x + y + z, and the inequalities to be proved become: x + z 14x yz − x 2 (y + z) − z 2 (x + y) − y 2 (x + z) √ x +z · ≤ zx ≤ . 2 (x + y)(y + z)(z + x) 2 The second inequality follows from the AM–GM inequality, with equality when x = z (that is, b = c). The first inequality may be rewritten √ 2(x + y)(y + z) zx + x 2 (y + z) + z 2 (x + y) + y 2 (x + z) ≥ 14x yz. (1) √ √ Now by the AM–GM inequality, x + y ≥ 2 x y and y + z ≥ 2 yz, so √ (2) 2(x + y)(y + z) zx ≥ 8x yz. Further applications of the AM–GM inequality yield

x 2 (y + z) + y 2 (z + x) + z 2 (x + y) ≥ 2 x 2 (yz)1/2 + y 2 (zx)1/2 + z 2 (x y)1/2

= 2(x yz)1/2 x 3/2 + y 3/2 + z 3/2

1/3 ≥ 6(x yz)1/2 x 3/2 y 3/2 z 3/2 = 6x yz. (3) Inequality (1) is the sum of (2) and (3), and is thus established. Equality holds in (1) if and only if it holds in both (2) and (3). Since these were based on AM–GM inequalities for the pairs {x, y}, {y, z}, and {z, x}, equality occurs if and only if x = y = z, i.e., if and only if a = b = c and the triangle is equilateral. Editorial comment. Li Zhou proved a slightly stronger result in place of the first inequality, namely a 4r − R 2( p − b)( p − c)  · ≤ ≤ ( p − b)( p − c). 2 R a V. V. Garc´ıa rediscovered and used a lemma that strengthens the inequality R ≥ 2r of Euler: If m and h are the lengths of the median and altitude from the same vertex, respectively, then R/(2r ) ≥ m/ h. (See IX.10.22, p 216, in D. Mitrinovi´c, J. Peˇcari´c, and V. Volonec’s Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, Boston, 1989.) Also solved by S. Amghibech (Canada), M. Bataille (France), D. Beckwith, E. Braune (Austria), R. Chapman (U. K.), P. P. D´alyay (Hungary), A. & A. Darbinyan (Armenia), P. De (India), J. Fabrykowski & T. Smotzer, W. Fosheng (China), V. V. Garc´ıa (Spain), C. Grosu (Romania), J. G. Heuver (Canada), R. A. Kopas, K.-W. Lau ´ Plaza (Spain), J. Posch, C. R. Pranesachar (India), (China), J. Minkus, D. J. Moore, J. H. Nieto (Venezuela), A. M. A. Prasad (India), H.-J. Seiffert (Germany), R. Stong, R. Tauraso (Italy), M. Tetiva (Romania), N. T. Tuan (Vietnam), M. Vowe (Switzerland), J. B. Zacharias, L. Zhou, BSI Problems Group (Germany), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, Northwestern University Problem Solving Group, Princeton Problem Solving Group, and the proposer.

January 2009]

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REVIEWS Edited by Jeffrey Nunemacher Mathematics and Computer Science, Ohio Wesleyan University, Delaware, OH 43015

A Garden of Integrals. By Frank E. Burk. Mathematical Association of America, Washington, DC, 2007, xiv + 281 pp., ISBN 978-0-88385-337-5, $51.95.

Reviewed by Erik Talvila Review dedicated to the memory of Ralph Henstock (1923–2007). Riemann, Lebesgue, Denjoy, Henstock–Kurzweil, McShane, Feynman, Bochner. There are well over 100 named integrals. Why so many? Some are of historical interest and have been superseded by better, newer ones. The Harnack integral is subsumed by the Denjoy. Some are equivalent, as are McShane and Lebesgue in Rn , and Denjoy, Perron, and Henstock–Kurzweil in R. Some are designed to work in special spaces: Feynman for path integrals, where the domain is the set of all continuous functions on [0, 1], Bochner for Banach space–valued functions. Others are designed to invert special derivatives, such as the symmetric derivative or the distributional derivative. And, we continue to keep the Riemann integral around because it is so easy to define. Can’t we have just one super-integral that does everything? The problem is that if an integral is defined to work, for example, on all Banach spaces, then it will probably be unnecessarily complicated when restricted to a simple setting such as the real line. There is also a pedagogical issue. We need to build up mathematics from simple bits before we can define the most abstract structures (Bourbaki notwithstanding). So it seems we will have to live with this plethora of integrals. First there were Newton and Leibniz, for whom integration was anti-differentiation. In the 1800s there were Cauchy and Riemann, who defined integrals in terms of approximating sums formed by partitioning compact intervals on the real line. There are two main themes of integration in the 20th century. The first is the Lebesgue theory. The Lebesgue integral of 1905 immediately revolutionised analysis. Although more complicated than the Riemann integral, it brought with it Borel’s theory of measure. The two concepts of measure and integral led to a very robust theory of integration with powerful limit theorems (such as the Lebesgue dominated convergence theorem), the ability to extend the integral to abstract settings, and a Banach space of integrable functions. These ideas allowed the flowering of such fields as probability and potential theory. In the first decades of the 1900s, while the Lebesgue theory was being refined by mathematicians such as Fatou and Radon, there was a parallel development by Denjoy, Lusin, and Perron. The Lebesgue integral is absolute: a function f is integrable if and only if | f | is integrable. But functions such as g(x) = x 2 sin(x −3 ) with g(0) = 0 have a derivative that exists everywhere and for which |g  | is not integrable in any neighbourhood of the origin. Thus the Lebesgue integral cannot integrate all derivatives. Integrals introduced by Denjoy (1912) and Perron (1914) are able to integrate all derivatives. But these have rather cumbersome definitions, and it is not easy to define them in settings other than the real line. By World War II work on Denjoy–Perron integration had 90

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more or less ceased. In the late 1950s Henstock and Kurzweil independently defined an integral using a modified type of Riemann sum. This integral is equivalent to that of Denjoy and Perron and is much easier to work with. In fact, it is probably more transparent than the improper Riemann integral, has reasonable limit theorems, and integrates all derivatives. Working with measures is also  ∞easy. Whereas the Lebesgue integral has difficulty dealing with integrals such as 0 sin(x 2 ) d x, which exists as a conditionally convergent improper Riemann integral, these are not a problem in the Henstock–Kurzweil theory. It was for these kinds of reasons that the Henstock– Kurzweil integral won many adherents, and there are quite a number of textbooks that discuss it at the undergraduate or graduate level, such as [1] and [5]. But the Henstock– Kurzweil integral lacks a Banach space structure for the space of integrable functions, and there is no canonical extension to Rn . So its development won’t end the search for the perfect integral. One way of defining integrals is via their primitives. A function f is Lebesgue integrable if there is an absolutely continuous function F so that F  = f almost everywhere. This can be done for the Henstock–Kurzweil integral, but the function space of primitives is more complicated. Instead, we could say that a distribution f is integrable if there is a continuous function F so that F  = f , where we now mean the distributional derivative. This simple definition gives a very general integral that includes the Henstock–Kurzweil integral. We get a Banach space of integrable distributions that is isomorphic to C 0 . For a complete but elementary description of this integral see [10]. If one is interested in the history of integration there are several works to consider. An excellent short summary of integration theory is given by Henstock in the otherwise unreadable [7]. See Hawkins [6] for a history of the Lebesgue integral. For a history of non-absolute integration, see Bullen’s survey [2]. The handbook by Zwillinger [11] gives a brief description, an example, and a few references for most types of integrals as well as many techniques of integration, both analytical and numerical. It is an excellent starting point for anyone interested in integration. With this background behind us, let’s get down to the business of A Garden of Integrals. This book was written by Frank E. Burk, who is the author of the successful text Lebesgue Measure and Integration: An Introduction [3]. The author is now deceased, which is a pity as the book should have good legs. Not everything I’m going to say about the book is positive and I feel a bit bad that the author won’t have the chance to respond publicly. This book surveys several types of integrals. There are chapters on the Cauchy, Riemann, Riemann–Stieltjes, Lebesgue, Lebesgue–Stieltjes, Henstock–Kurzweil, Wiener, and Feynman integrals. There is also an historical overview and a chapter on Lebesgue measure. The emphasis is purely theoretical with very little in the way of applications. Throughout the book the author assumes a thorough grounding in undergraduate real analysis. Thus the book is pitched at the senior undergraduate or beginning graduate level. The book is essentially a string of exercises, implicit or explicit, in which the author asks the reader to work out the details of the various theories. So, this is definitely not a coffee table book. To read it you’ll have to turn the TV off, sit down, shut up, and get to work. Working through it will teach you a lot of integration. The writing style is attractive. Sentences flow well, and the subject is laid out logically. In this sense it’s a very good book. What’s not good is that it is not clear for whom it was written. As any writing instructor will tell you, when you sit down to write, the first item to have straight is: Who is your audience? I doubt that was done in this case. There are initial chapters on Cauchy and Riemann integrals. The Cauchy integral is January 2009]

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like the Riemann except that in your Riemann sum you always use the left endpoint of each subinterval when evaluating the integrand. These are pretty basic integrals, but to read these chapters you need to know the following: the trapezoid rule, integration by parts, the improper Riemann integral, Cauchy sequences, uniform convergence, the Weierstrass M-test, Fourier series, separation of variables for the Laplace equation, the Cantor set, and nowhere dense sets. That’s a tall order. I’m wondering just who would know all this stuff but still need to learn something about the Cauchy and Riemann integrals. Then on page 65, in the chapter on the Riemann integral, the author lets the cat out of the bag: “I dare say we can all remember our first encounter with [the Cantor set].” So, to read the book you need to have many results of real analysis at your fingertips. That said, these chapters have some interesting material and some good examples, such as the construction of a continuous nowhere differentiable function and the construction of a Riemann integrable function with a dense set of discontinuities. As Cauchy did, the author defines the Cauchy integral only for continuous functions. Burk is correct in proclaiming that in this sense, the Riemann integral properly includes the Cauchy integral. However, the space of all Cauchy integrable functions (dropping the continuity assumption) is the same as the space of Riemann integrable functions. This was proved by Gillespie [4]. Burk also states that indefinite Cauchy and Riemann integrals are absolutely continuous, but absolute continuity isn’t defined for another 65 pages. It would have been better to give the easy proof that these integrals are Lipshitz continuous. Preceding these chapters is an historical survey. It has some discussion of the geometric methods of the ancients and then brief descriptions of the integrals covered in later chapters. These are too vague to be of much interest, and this chapter should probably have been omitted. A better list of historical sources would have been appreciated. The chapters on Lebesgue measure and Lebesgue integration are much better. Voltera’s example of a function with a bounded derivative that is not Riemann integrable makes good motivation for introducing the Lebesgue integral. These chapters introduce many of the standard topics in measure and integration on the real line. Most concepts here are followed by useful exercises, although the coverage of sigma algebras and Borel sets is very brief. Bounded variation and absolute continuity are defined, but there is so little discussion of them I’m not sure if the uninitiated would get the whole picture. Again, you need to know your real analysis to make your way through this book. Thorough familiarity with topics such as inf, sup, and the Heine–Borel theorem are assumed here. The short chapters on Riemann–Stieltjes and Lebesgue–Stieltjes integrals don’t add much and could well have been omitted. One of the themes of the book is the fundamental theorem of calculus. At the end of each chapter is a summary that consists of the fundamental theorem as it applies to that integral. This is all motivation for the Henstock–Kurzweil integral, since it has the best version of this theorem on the real line. The chapter on Henstock–Kurzweil integration is good. Initially it proceeds at a much more gentle pace than the Lebesgue chapters. The author takes the time to motivate this integral and then works out its properties in some detail. The final two chapters are a tour de force. In treating the Wiener and Feynman integrals in a book of this level and length, Burk is taking on a big task and breaking new ground. The domain of integration is now infinite dimensional: the set of continuous real-valued functions on [0, 1]. These chapters require Lebesgue measure and integration as well as some familiarity with probability, the heat equation, complex variables, and linear operators on a Hilbert space. They are written at a higher level than the pre92

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ceding chapters. The Wiener integral is described reasonably well. The examples and exercises should help the reader understand what’s going on. I was especially impressed with the Feynman integral chapter. It is long and difficult and packed with lots of material: the Schr¨odinger equation, the Fourier transform in the Schwartz space, the Trotter product formula, and semigroups of linear operators. Many of these topics are dealt with quite briefly, so working out the details and filling the gaps will take time and consultation of outside references. I can’t say I understood it all on a first reading, but that gives me plenty to go back to. A note regarding peripherals: The typesetting is generally good and the book is easy to read, with lots of white space and well-displayed formulas. Each chapter begins with an irrelevant quotation by some famous person. I suppose these are meant to inspire, but they didn’t do anything for me. For witty and inspiring beginning-ofchapter quotes, I much prefer [9]. There are about 50 illustrations, mostly in the historical overview, Wiener, and Feynman chapters. These are quite good. The index is thin. There aren’t many typos, although a recurring one is ”hey there” for “hey there”, which the editor should have caught. The binding is good quality; the book is bound in signatures. At $51.95 (cheaper for MAA members) the book has excellent value. Mathematics books usually include a preface that tells the reader roughly what topics will be covered, the chapter dependencies, the level of depth, and the required prerequisites. A Garden of Integrals has no preface. It has a short and completely uninformative foreword written by the author. This is unfortunate, as some unsuspecting souls may try to read this attractive-looking book and find it pretty tough going. It is a major failing of the author/editor not to have told the prospective reader what they are expected to know in order to make sense of any given chapter of this book. Topics I would have liked to have seen are, first and foremost, some discussion of the integral on sets other than compact intervals on the real line. Improper Riemann integrals are a glaring omission. The discussion of the Riemann integral confines itself to [a, b], but later in the chapter on the Wiener integral, improper Riemann integrals are assumed. Integration in Rn , Stokes’ theorem, and some applications to probability would also have been nice. Another thing that bothered me is the references. There is a list at the end of each chapter. These may have been the works (mostly textbooks) the author used to write the book, but a more comprehensive list suitable for further reading should have been compiled. Here are two books similar in scope to the book under review. Theories of Integration by Kurtz and Swartz [8] is an undergraduate text that develops Riemann, Lebesgue, and McShane integration. It assumes some background in analysis; thus the reader is expected to know what countable means, but concepts such as limsup are defined. It has lots of exercises and is a well-crafted textbook. At a slightly more advanced level is Gordon’s The Integrals of Lebesgue, Denjoy, Perron, and Henstock [5]. It covers the integrals in the title as well as such topics as Darboux and Baire class one functions. It has good exercises and is a very carefully written, well-crafted text at the graduate level. One shortcoming is that it discusses only integrals on [0, 1]. Nonetheless, it has become a standard reference. How does Burk’s book compare with these two? Except for the chapters on Wiener and Feynman integration the book is comparable to Kurtz and Swartz and Gordon, although these authors do a better job of incorporating material from the real analysis course. The books cited above are pretty formal. Burk strikes a nice balance between the formal definition, theorem, proof style and chatty, informal discussion. His book is more fun and engaging because he’s constantly forcing you to fill in little details. This is a poor book to use as a reference to look up something like the Lebesgue January 2009]

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dominated convergence theorem, but if you work through the book you’ll have a good time learning the proof. And his chapters on Wiener and Feynman integration set him apart from the rest. Who can benefit from this book? I would be happy to give it to good students with the necessary background for use in an undergraduate reading course or seminar. Such a course could be on Lebesgue or Henstock–Kurzweil integration. Since the book is pretty much just a long set of exercises, this study could work well. However, many topics are dealt with in a cursory manner, so outside reading would have to be prescribed. A graduate seminar could be set up for the Wiener and Feynman chapters. And, of course, the book is ideal for self study. Will I recommend that my college library purchase this book? Definitely. REFERENCES 1. R. G. Bartle and D. R. Sherbert, Introduction to Real Analysis, 3rd ed., Wiley, New York, 2000. 2. P. S. Bullen, Nonabsolute integrals in the twentieth century, in AMS Special Session on Nonabsolute Integration, P. Muldowney and E. Talvila, eds., University of Toronto, Toronto (2000), available at http: //www.emis.de/proceedings/index.html. 3. F. Burk, Lebesgue Measure and Integration: An Introduction, Wiley, New York, 1998. 4. D. C. Gillespie, The Cauchy definition of a definite integral, Ann. Math. 17 (1915) 61–63. 5. R. A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, American Mathematical Society, Providence, RI, 1994. 6. T. Hawkins, Lebesgue’s Theory of Integration, American Mathematical Society, Providence, RI, 2001. 7. R. Henstock, The General Theory of Integration, Oxford University Press, Oxford, 1991. 8. D. S. Kurtz and D. W. Swartz, Theories of Integration, World Scientific, Singapore, 2004. 9. M. Reed and B. Simon, Methods of Mathematical Physics, vol. I, Functional Analysis, Academic Press, New York, 1980. 10. E. Talvila, The distributional Denjoy integral, Real Anal. Exchange 33 (2008) 51–82. 11. D. Zwillinger, Handbook of Integration, Jones and Bartlett, Boston, 1992. University of the Fraser Valley, Abbotsford, BC Canada V2S 7M8 [email protected]

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