Nov 2, 2010 - heads and hearts who taught by way of love and devotion. .... 73. 7 Moments and centres of mass: preliminaries. 75. 7.1 Informal definitions . ...... Cleary, the same argument will not work for a block set such as c1,100.
Optimal block stacking and combinatorial identities via Archimedes’ method David Treeby (BScAdvHons)
A thesis submitted for the degree of Doctor of Philosophy at Monash University in 2017 Mathematical Sciences
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Copyright Notice © David Treeby 2017
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Abstract What is the maximum overhang that can be obtained when a set of blocks of variable width are stacked so that there is one block at each level? We determine the configuration of blocks that maximises the overhang when these blocks are stacked in a prescribed order. We also describe infinite sets of blocks of variable width that can be stacked to yield towers with arbitrarily large overhang. This investigation is a generalisation of a classic problem, which asks how n identical blocks should be stacked to create a tower of maximal overhang. The solution to this problem is well-known: one can achieve an overhang that is proportional to the nth harmonic number. Since the harmonic numbers diverge, one can theoretically construct a balanced tower of any conceivable overhang, given a sufficient quantity of blocks. When considering blocks of variable width, we show that one can obtain arbitrarily large overhang if and only if a particular series diverges. The series in question has not been studied to date. Using this result, we find characterisations of sets of blocks that can be stacked to yield towers with arbitrarily large overhang. We will obtain a counterpart to the Abel-Dini Theorem, which, informally, states that there is no slowest diverging series of positive terms. The block stacking counterpart asserts that there is no stack whose overhang diverges more slowly than all others. We then allow the blocks to be permuted, obtaining a partial description of how a finite set of blocks should be stacked, and in what order, to maximise the stack’s iii
overhang. For small sets of blocks, this partial description will allow us to determine the maximal overhang in exponential time. In an additional line of inquiry, we will consider various symmetric configurations of blocks and point masses in the plane. By considering the centres of mass of such configurations we establish various well-known summation formulas, and discover others that are new and appealing. The method that we employ can be regarded as a discrete counterpart to the physical method conceived by Archimedes to determine the areas and volumes of various geometric figures. For instance, Archimedes used his Law of the Lever to establish the area of a parabolic segment, while we use similar arguments to establish a formula for the sum of consecutive squares. We extend the investigation by finding essentially physical arguments to find formulas for other sums of powers, combinatorial identities and Fibonacci summation formulas.
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Declaration This thesis contains no material that has been accepted for the award of any other degree or diploma at any university or equivalent institution and that, to the best of my knowledge and belief, this thesis contains no material previously published or written by another person, except where due reference is made in the text of the thesis.
Signature: Name: Date:
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Publications during enrolment Many of the results in this thesis have been published, or accepted for publication. The results in Chapters 2 to 6 are due to appear in a condensed form in [V]. An anonymous referee for this article suggested the proof of Theorem 5.6.1, which appears in this thesis. The contents of Chapters 8 and 9 contributed to articles [III] and [IV]. Finally, the results in Chapters 10 and 11 first appeared in articles [I] and [II], respectively. [I] D. Treeby. Hidden formulas in Fibonacci tilings. Fibonacci Quart., 54(1):23– 30, 2016. [II] D. Treeby. Physical derivations of Fibonacci summations. Fibonacci Quart., 54(4):327–334, 2016. [III] D. Treeby. A moment’s thought: Centres of mass and combinatorial identities. Math. Mag., 90(1):19–25, 2017. [IV] D. Treeby. Finding sums of powers using physical arguments. Math. Gaz., 101(551):227–235, 2017. [V] D. Treeby. Further thoughts on a paradoxical tower. Amer. Math. Monthly, Forthcoming.
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Acknowledgments The production of this thesis was not a sole e↵ort. It could not have been written without the help of those who have supported me in a variety of ways. First and foremost, thanks must go to my wife, Renae Bishop, who su↵ered a tired and distracted husband during the time it took to write this thesis. And, despite her understanding little of what was written, I’m grateful that she survived reading the entire thing. Thanks must also be extended to my supervisor, Burkard Polster, who, apart from being a sublime communicator of mathematics, is so capable of contending with bureaucracy that I scarcely encountered single administrator during the course of my candidature. In short, he allowed me to work on what really matters. Thanks also goes to my co-supervisor, Heiko Dietrich, who provided a steady and invaluable supply of stylistic advice and computational assistance. His feedback was always prompt and detailed. I hope that I capitalised all of the right words. A special thanks goes to Marty Ross, my honorary supervisor and friend, who has supported me since my undergraduate days, and who has taught me a thing or two about mathematical culture, intellectual passion and determination. I must also mention Maria Athanassenas, my mathematical mother, the filler of heads and hearts who taught by way of love and devotion. And it was Michelle Liu and her doting mother Ruth Chiu who kept me and my dependants generously fed over the past few years. Finally thanks goes to my brother and sister, of whom I’m proud, and for whom I’m ever thankful. vii
Contents 1 Introduction
1
1.1
Thesis overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Line of inquiry I: Generalisations of a classic block stacking problem
2
1.2.1
Historical notes . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2.2
Generalisations . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2.3
Variable block widths . . . . . . . . . . . . . . . . . . . . . .
4
1.2.4
Diverging block sequences . . . . . . . . . . . . . . . . . . .
4
1.2.5
Permutations of blocks . . . . . . . . . . . . . . . . . . . . .
5
Line of inquiry II: Centres of mass and integer summations . . . . .
7
1.3.1
Historical remarks on Archimedes’ Method . . . . . . . . . .
7
1.3.2
Archimedes’ Method applied discretely . . . . . . . . . . . .
8
1.3.3
Sums and alternating sums of powers . . . . . . . . . . . . .
8
1.3.4
Combinatorial identities . . . . . . . . . . . . . . . . . . . .
9
1.3.5
Fibonacci summations . . . . . . . . . . . . . . . . . . . . .
10
1.3
2 Block stacking preliminaries 2.1
12
Key definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2.1.1
Balanced stacks . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.1.2
Right-aligned stacks . . . . . . . . . . . . . . . . . . . . . .
13
2.1.3
Overhang and divergence . . . . . . . . . . . . . . . . . . . .
14
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3 Right-aligned stacks
16
3.1
A formula for the right-aligned overhang . . . . . . . . . . . . . . .
16
3.2
The harmonic stack . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
3.2.1
Optimality of the harmonic stack . . . . . . . . . . . . . . .
18
3.2.2
A harmonically suspended elephant . . . . . . . . . . . . . .
20
3.3
Stacking cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
3.4
Reordering right-aligned stacks . . . . . . . . . . . . . . . . . . . .
21
3.4.1
3.5
Block sequences with divergent overhang whose terms tend to zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
Another sufficient condition . . . . . . . . . . . . . . . . . . . . . .
25
4 Configurations of maximal overhang 4.1
28
A formula for the maximal overhang . . . . . . . . . . . . . . . . .
32
4.1.1
A word on notation . . . . . . . . . . . . . . . . . . . . . . .
33
4.1.2
The maximal overhang of monotone decreasing n-block sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Characterising block sequences with divergent overhang
34 37
5.1
Asymptotic Notation . . . . . . . . . . . . . . . . . . . . . . . . . .
38
5.2
Characterisation I . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
5.2.1
Sums of block sequences . . . . . . . . . . . . . . . . . . . .
43
Characterisation II . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
5.3.1
Cubes sequences with divergent overhang . . . . . . . . . . .
47
Comments on unbounded block sequences . . . . . . . . . . . . . .
48
5.4.1
50
5.3
5.4
A surprising block sequence . . . . . . . . . . . . . . . . . .
5.5
Slowly diverging block sequences
. . . . . . . . . . . . . . . . . . .
52
5.6
Strongly divergent block sequences . . . . . . . . . . . . . . . . . .
54
6 Permuting block sets 6.1
57
Optimal partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
57
6.2
Estimates for counterbalance blocks . . . . . . . . . . . . . . . . . .
61
6.3
Bounding the size of the counterbalance . . . . . . . . . . . . . . .
63
6.4
The Subset-Sum Property . . . . . . . . . . . . . . . . . . . . . . .
65
6.4.1
Proof of the Subset-Sum Property . . . . . . . . . . . . . . .
66
6.5
Optimal partitions for consecutive integers . . . . . . . . . . . . . .
70
6.6
Conjectures pertaining to optimal partitions . . . . . . . . . . . . .
73
7 Moments and centres of mass: preliminaries
75
7.1
Informal definitions . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
7.2
Formal definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
7.2.1
Fundamental properties of the centre of mass . . . . . . . .
78
7.2.2
Fundamental properties of the first moment . . . . . . . . .
81
7.2.3
The centre of mass of a planar region . . . . . . . . . . . . .
82
8 Sum of powers using physical arguments 8.1
8.2
8.3
84
On sums and alternating sums of integers . . . . . . . . . . . . . . .
85
8.1.1
Sums of consecutive integers . . . . . . . . . . . . . . . . . .
85
8.1.2
Alternating sums of consecutive integers . . . . . . . . . . .
86
On sums and alternating sums of squares . . . . . . . . . . . . . . .
87
8.2.1
Sums of squares: An alternative configuration . . . . . . . .
89
8.2.2
Sums of triangular numbers . . . . . . . . . . . . . . . . . .
90
8.2.3
Alternating sums of squares . . . . . . . . . . . . . . . . . .
91
8.2.4
A demonstration with particles of positive mass . . . . . . .
93
8.2.5
Alternating sums of triangular numbers . . . . . . . . . . . .
96
On sums and alternating sums of cubes . . . . . . . . . . . . . . . .
96
8.3.1
Sums of cubes . . . . . . . . . . . . . . . . . . . . . . . . . .
97
8.3.2
Alternating sums of cubes . . . . . . . . . . . . . . . . . . .
98
8.3.3
A demonstration with positive masses . . . . . . . . . . . . . 104
8.3.4
A generalisation . . . . . . . . . . . . . . . . . . . . . . . . . 107
x
8.4
The sum and alternating sum of powers of four . . . . . . . . . . . 108 8.4.1
The sum of powers of four . . . . . . . . . . . . . . . . . . . 109
8.4.2
The alternating sum of powers of four . . . . . . . . . . . . . 110
8.5
The sum of powers of five . . . . . . . . . . . . . . . . . . . . . . . 114
8.6
Sums of odd powers of consecutive integers . . . . . . . . . . . . . . 116
8.7
A continuous counterpart . . . . . . . . . . . . . . . . . . . . . . . . 118
9 Combinatorial Identities
120
9.1
Moments of a symmetric system . . . . . . . . . . . . . . . . . . . . 120
9.2
Pairs of symmetric sequences
. . . . . . . . . . . . . . . . . . . . . 122
10 Summations of Fibonacci numbers
125
10.1 Hidden identities in geometric tilings . . . . . . . . . . . . . . . . . 125 10.2 The method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 10.3 A Simple Fibonacci tiling . . . . . . . . . . . . . . . . . . . . . . . 129 10.4 A preliminary result . . . . . . . . . . . . . . . . . . . . . . . . . . 130 10.5 The sum of products of three consecutive Fibonacci numbers . . . . 132 10.6 The sum of cubes of Fibonacci numbers . . . . . . . . . . . . . . . . 133 11 Balancing finite sums of Fibonacci numbers
136
11.1 The method and the main result . . . . . . . . . . . . . . . . . . . . 137 11.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 11.3 Further comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Bibliography
145
1
2
Chapter 1 Introduction
Giving an address on the teaching of mathematics, Vladimir Arnold [8] claimed that “Mathematics is a part of physics. Physics is an experimental science, a part of natural science. Mathematics is the part of physics where experiments are cheap”. This thesis will explore the intersection of physics and mathematics. On one hand, we will look at a problem of physics that leads to a series of mathematically interesting problems. On the other, we look at how certain mathematical problems have physical interpretations that lead to simple solutions.
1.1
Thesis overview
This thesis will consider two lines of inquiry, each of which is connected by the need to carefully study the centres of mass of various systems of particles. Line of inquiry I (Chapters 2–6). How should a set of blocks of variable width be stacked, one on top of another, to maximise their overhang? Line of inquiry II (Chapters 7–11). How can sets of blocks and systems of particles be carefully arranged to establish and discover formulas relating to wellknown integer sequences? 1
1.2
Line of inquiry I: Generalisations of a classic block stacking problem
What is the maximum overhang that can be achieved when n identical blocks of width 2 are stacked one on top of another? Using a simple argument that utilises Archimedes’ Law of the Lever [7, p. 192], It is easily shown (see, e.g., [18], [35] and [64]) that one can achieve an overhang of at least Hn = 1 +
1 1 1 + + ··· + , 2 3 n
(1.2.1)
where Hn is the nth harmonic number. This fact is undoubtably the source of the problem’s enduring appeal; divergence of the harmonic series allows one to achieve an arbitrarily large overhang given a sufficiently large quantity of blocks. One natural configuration that achieves this overhang is called the harmonic stack, and this is shown below in Figure 1.1. Harmonic stacks are constructed from top to bottom: the top block is placed so that its centre of mass is placed on the right edge of the second block. The top two blocks are then placed together so that their combined centre of mass is placed on the right edge of the third block, and so on. When sets of blocks are stacked this way, we call them right-aligned. We will formally define this notion in Chapter 2. 2
1 1 5 4
table
1 3
1 2
1 1
Figure 1.1: Identical blocks stacked one on top of another.
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1.2.1
Historical notes
This classic problem dates back to at least the mid-nineteenth century, when it was presented in two mechanics textbooks, [60, pp. 140–141] and [81, p. 183]. However, its first appearance in the mathematics literature was in 1923, when the problem was posed, without solution, by Coffin [24, p. 76] in The American Mathematical Monthly. In 1953, the problem resurfaced when it was presented by Sharp [70] in the Pi Mu Epsilson Journal. However, it achieved its widest exposure when it was considered by Gardner [31] in the “Mathematical Games” column of Scientific American.
1.2.2
Generalisations
Various generalisation of the classic problem have already been considered. One obvious generalisation is to relax the requirement that the blocks be stacked so that there is only one block per level. The question then becomes: what is the maximum overhang that can be achieved when n identical blocks are stacked in any stable fashion? Ainley [2] and then Drummond [28] investigated specific examples of socalled multi-wide stacks, however Hall [35] provided the first systematic treatment, and demonstrated that by employing blocks as a counterbalance, one can form stacks whose overhang is twice the overhang obtained by harmonic stacks employing the same number of blocks. Despite Hall’s hasty claim that his multi-wide stacks were optimal, Paterson and Zwick [59] subequently improved the result by demonstating that n blocks can be stacked with the use of similar counterbalancing to achieve an overhang p proportional to 3 n. In a follow-up paper, Paterson, Peres, Thorup, Winkler and Zwick [58] then showed that one cannot do any better; an overhang proportional to p 3 n is the greatest possible for any balanced stack of n blocks. A continuous analogue of the classic block stacking problem was considered by Polster, Ross and Treeby [65]. In this paper, the authors investigated various 3
properties of the asymptotic shape of the harmonic stack as blocks are successively added and then rescaled.
1.2.3
Variable block widths
Despite the aforementioned generalisations of the original problem, none of the literature considers stacks for which the block size can vary, and this provides scope for further investigation along these lines. This will be the first line of inquiry for this thesis. We consider infinite sequences of blocks for which the width may vary. These can be described by a so-called block-sequence w = (w1 , w2 , w3 , . . .), where each wi is equal to half of the width of the ith block. These blocks will be stacked in a one-on-one fashion. We will address the following questions: (1) When a finite set of blocks of variable width is stacked in a prescribed order can we describe the configuration of these blocks that achieves the maximal overhang? (2) Can we exhibit infinite sequences of blocks whose widths converge to zero that can nonetheless achieve an arbitrarily large overhang? (3) If such sequences of blocks exist, what are sufficient and necessary conditions for this to occur? (4) How can we maximise the overhang if a finite set of blocks can be stacked in any order?
1.2.4
Diverging block sequences
Much of our analysis will depend on the following preliminary result, which generalises the formula for the right-aligned overhang of the harmonic stack. This will be restated and proved in Chapter 3.
4
Theorem. For any block-sequence w = (wi )1 i=1 let Wi = w1 + · · · + wi . The rightaligned overhang of the truncated n-block sequence wn = (w1 , . . . , wn ) is n X wi2 R(wn ) = . Wi i=1
From the above theorem, we see that the right-aligned overhang diverges if and P wi2 only if the series 1 i=1 Wi diverges. Note that this series bears some resemblance to a well-known series that has received considerable attention in the literature, P wi specifically 1 i=1 W ↵ , The best known result related to this series is the Abel-Dini i
Theorem.
Theorem (Abel-Dini Theorem). Suppose that (di )1 i=1 is an arbitrary sequence of P1 positive terms such that the infinite series i=1 di diverges. If Di = d1 + · · · + di , P di then the series 1 i=1 D ↵ diverges if ↵ 1 and converges if ↵ > 1. i
This result was first proved in a slightly weaker form by Abel in [1], and com-
pleted by Dini in [27]. Various authors have found improvements of the Abel-Dini Theorem (see, e.g., [37], [66] and [69]). However, to the best of our knowledge, none P wi2 have considered series of the form 1 i=1 Wi . As a counterpart to the Abel-Dini Theorem we will prove the following result in
Chapter 5, which states that the divergence of the series in question is tantamount to the divergence of a more tractable series. Proposition. If w = (wi )1 i=1 is any monotone decreasing sequence of positive terms and Wi = w1 + · · · + wi , then n n X X wi2 wi =⇥ Wi i i=1 i=1
1.2.5
!
.
Permutations of blocks
After considering blocks stacked in a prescribed order, in Chapter 6 we allow our blocks to be arranged in any order. The question then becomes: how should a set of 5
blocks of variable width be stacked, and in what order, to maximise their overhang? By combining the results of Proposition 4.0.1 and Theorem 6.1.3, we will prove that every maximal stack has certain characteristics, and these are exemplified in Figure 1.2 shown below. Specifically, we will prove that: (1) the protruding block is the largest block, (2) blocks above the protruding block are placed as a counterbalance on its left edge, (3) blocks beneath the protruding block are right-aligned and arranged in decreasing order of size.
counterbalance set the largest block protrudes decreasing and right-aligned
Figure 1.2: A typical stack of maximal overhang.
This means that in order to achieve the maximal overhang, we need only consider which blocks are to serve as a counterbalance for the largest block. As this appears to be a combinatorially difficult problem, we aim to reduce the range of viable counterbalances by finding upper bounds for the width and number of blocks that may appear in the counterbalance. Such results are obtained in Propositions 6.2.1 and 6.4.1, and Corollary 6.3.1. We will pay particular attention to sets of blocks whose half-widths are consecutive integers, cm,n = {m, m + 1, . . . , n}. We will explain why determining the maximal overhang for a block set such as c1,100 is difficult while determining the maximal overhang for c201,300 is trivial.
6
1.3
Line of inquiry II: Centres of mass and integer summations
Subsequent to stacking blocks one-on-one to construct towers of maximal overhang, we then consider carefully arranged sets of blocks and particles in the plane. By considering the centres of mass of these configurations we develop a method for efficiently finding proofs of various well-known formulas relating to integer sequences, and discovering others.
1.3.1
Historical remarks on Archimedes’ Method
Archimedes was the first to provide a mathematical formulation of the notion of centre of mass. One of Archimedes’ best-known results is his Law of the Lever. Archimedes demonstrated that the torque exerted by various weights resting at di↵erent positions along a lever can be balanced at a single point, which we now call the centre of mass. This result is demonstrated in his book, On the Equilibrium of Planes, which can be found in [7], a collection of Archimedes’ extant works. In this book, Archimedes famously developed a systematic approach that allowed him to determine the centres of mass of a variety of geometric objects (e.g., parabolic segments, spheres and cones). However, this does not appear to be the primary purpose of his investigation; rather, Archimedes used these results to determine the area and volume of such objects. This approach is best exemplified in a more recently discovered book, The Method of Mechanical Theorems, which also features in [7]. This approach was refined by Pappus, and is encapsulated by Pappus’ Centroid Theorems [57]. These are two related theorems for which the centre of mass plays a crucial part in determining the surface area and volume of a surface of revolution. The works of Archimedes and Pappus emphasise the important role played by the concept of centre of mass, not only to physics, but to mathematics more broadly. 7
1.3.2
Archimedes’ Method applied discretely
A recent article by Lord [48] contained an elegant derivation of the formula for the sum of squares using a centre of mass argument, which we recount in Section 8.2. Lord, in e↵ect, applied Archimedes’ Method in a discrete rather than a continuous setting. Lord’s insight provides impetus for further research. Might one be able to prove or discover other identities by similar considerations? This will be the primary focus for the second half of this thesis. Broadly speaking, our approach is to begin with a carefully arranged system of particles whose masses are given by some chosen sequence of numbers. The system should possess either global or internal symmetry that allows for its centre of mass to be calculated two di↵erent ways. Equating the two results yields some desired identity. The technique bears some resemblance to the use of combinatorial arguments to establish similar identities. In this case, identities are proved by enumerating the same set two di↵erent ways. This combinatorial approach is best exemplified by Benjamin and Quinn in [15]. Our first application of the mechanical method is to demonstrate how one can obtain formulas for sums and alternating sums of powers.
1.3.3
Sums and alternating sums of powers
Finding geometric demonstrations of the well-known formulas n(n + 1) , 2 n(n + 1)(2n + 1) 12 + 2 2 + · · · + n2 = , 6 n2 (n + 1)2 13 + 2 3 + · · · + n3 = , 4 1 + 2 + ··· + n =
has become something of a mathematical pastime. The literature abounds with such examples (see, e.g. [4], [30], [32], [40], [45], [53], [63], [71], [80] and [82]). We will continue this tradition in Chapter 8 by demonstrating how these (and other) 8
results can be interpreted physically and in a way that often makes these formulas almost immediately apparent. We will do this by considering the centres of mass of a carefully arranged systems of particles in the plane. Other authors (e.g., [25], [55], [61], [62], and [72]) have geometrically demonstrated the corresponding formulas for alternating sums of powers, 8 > : k, if n = 2k, 8 > : k(2k + 1), if n = 2k, 8 > : k 2 (4k + 3), if n = 2k.
By interpreting each of these (and other) sums as the first moment (i.e., torque) of a system of signed masses (i.e. masses that may be negative) we can also provide a physical demonstration of each of these sums.
1.3.4
Combinatorial identities
Subsequent to our study of sums of powers, in Chapter 9 we will apply a similar method to find physical proofs for a variety of combinatorial identities. Some of these are well-known, such as the simple identity ✓ ◆ n X n i = n2n 1 . i i=0 Others result for which a physical proof is possible are less well-known, such as p ✓ ◆ X p 1 1 1 ( 1)i = . i i+2 p+1 p+2 i=0 We likewise physically prove a variety of other identities that, to the best of our knowledge, appear to be new. 9
1.3.5
Fibonacci summations
In 1972, Brousseau [17], a founding editor of the Fibonacci Quarterly, published an entertaining account of how geometric tilings can be used to generate identities involving Fibonacci numbers. In this thesis, we explain how there is a hidden identity in each of his tilings that can be discovered by considering their centres of mass. To demonstrate the utility of this approach, in Chapter 10 we provide a simple derivation of a formula for the sum of cubes of Fibonacci numbers, n X
2 3Fn+1 Fn
Fi3 =
i=1
3 Fn+1 2
Fn3 + 1
.
This is clearly not as appealing as the formula for the sum of squares of Fibonacci numbers, which is well-known to be n X
Fi2 = Fn Fn+1 .
i=1
Various authors have found similar formulas for a variety of sums involving products of Fibonacci and Lucas numbers (see, e.g., [19], [22], [23], [50], [51] and [20]). The “gold standard” is to obtain a closed formula that can be factored into some product of Fibonacci and Lucas numbers. One such result was discovered by Block [16], n X i=1
1 Fi2 Fi+1 = Fn Fn+1 Fn+2 . 2
(1.3.1)
In Chapter 11 we use a physical argument based on the centre of mass of a configuration of rectangles in the plane to extend Block’s formula. We prove the following theorem. Theorem. Equation (1.3.1) implies (1.3.2) which in turn implies (1.3.3) where n X
1 3 2 2 Fi3 Fi+1 = Fn2 Fn+1 Fn+2 ; 4
(1.3.2)
1 5 4 4 Fi5 Fi+1 F2i+1 = Fn4 Fn+1 Fn+2 . 8
(1.3.3)
i=1
n X i=1
10
Various other identities are established likewise. With our overview complete, we begin our investigation of the block stacking problem that constitutes our first line of inquiry.
11
Chapter 2 Block stacking preliminaries Our thesis begins with a generalisation of the classic block stacking problem. An abridged version of the material presented in Chapters 2 through 6 will soon be published in [79]. Before we begin this study, we first establish our key definitions and the required notation.
2.1
Key definitions
We consider an infinite set of blocks that are uniform in every respect except for their widths. It is helpful to describe each block by its half-width rather than its width. We call a sequence of positive half-widths w = (w1 , w2 , . . .) a block sequence. Of particular importance will be monotone decreasing block sequences, for which wi+1 wi for all i 2 N. We can assume that the mass of each block is equal to its half-width. By truncating the sequence at its nth term we obtain an n-block sequence, wn = (w1 , . . . , wn ). For any given n-block sequence we arrange the blocks on a table sequentially from top to bottom. For each block i we keep track of the horizontal position xi 2 R of its midpoint relative to some chosen origin. We call xn = (x1 , . . . , xn ) a configuration. The pairing {(wi , xi ) : i = 1, 2, . . . , n} of an n-block sequence and a 12
configuration is called an n-stack.
2.1.1
Balanced stacks
An n-stack is said to be balanced if it is in static equilibrium. This is determined by ensuring that, for each i 2 {1, . . . , n
1}, the balancing point of blocks 1 through
i combined is placed along the top edge of block i + 1 as shown in Figure 2.1 (a), where the balancing point is depicted by a small triangle. Since this balancing point is given by
Pi
j=1
we require that xi+1
xi = Pi
xj w j
j=1 wj
,
(2.1.1)
wi+1 xi xi+1 + wi+1 . This is equivalent to xi+1 2 [xi
wi , xi + wi ].
(2.1.2)
We assume that the table edge is always located at the origin. Therefore, we also require that the balancing point xn of blocks 1 through n combined satisfies xn 0.
2.1.2
Right-aligned stacks
We say that an n-stack is right-aligned at block i if the balancing point xi of blocks 1 through i combined is located at the right edge of block i + 1. That is, if xi = xi+1 + wi+1 .
(2.1.3)
This is shown in Figure 2.1 (b). An n-stack is right-aligned if it is right-aligned at block i for every i 2 {1, . . . , n
1}, and if xn = 0. Examples of these right-aligned
n-stacks are seen in Figures 3.2 (a) and 3.3 (a). These configurations are balanced but precariously so; each block is on the verge of tipping over the block beneath.
13
xi+1
wi+1
xi+1 xi
xi+1 + wi+1
xi+1
i
i
i+1
i+1
overhang
xi = xi+1 + wi+1
overhang
(a) Stack is balanced at block i.
(b) Stack is right-aligned at block i.
Figure 2.1
2.1.3
Overhang and divergence
The overhang of an n-stack is the horizontal distance from the edge of the table to the right-most edge of the blocks; that is, the overhang is max1in (xi + wi ). We say that block j protrudes if it is in the maximum overhang position. For each n-block sequence wn , the aim is to find the balanced configurations that yield the maximal overhang, which we denote by M (wn ). The existence of these configurations is ensured by the following proposition. Proposition 2.1.1. For each n-block sequence wn there exists a balanced configuration that maximises the overhang. Proof. By (2.1.1) and (2.1.2), the set of balanced configurations is a compact subset of Rn . For any balanced configuration xn , the overhang is given by f (xn ) = max1in [(xi + wi )]. As f is a continuous function on a compact set, there is some configuration that maximises f . If an n-block sequence is right-aligned, then its overhang will be denoted by R(wn ). Right-aligned stacks are obvious candidates for configurations of maximal overhang. However, there are instances when M (wn ) > R(wn ).
14
The block sequence w is said to have divergent overhang if M (wn ) ! 1 as n ! 1; otherwise we say that it has convergent overhang. We already know that the constant block sequence has divergent overhang. Generalising this, we ask if there are block sequences with divergent overhang whose terms tend to zero monotonically. In Chapter 3 we will see that such block sequences are not hard to come by and in Chapter 5 we find characterisations of such block sequences.
15
Chapter 3 Right-aligned stacks Having established our notational conventions and definitions, we commence our study by investigating right-aligned n-stacks. Right-aligning a given n-block sequence will often maximise its overhang. Even when this is not the case, we will see that any stack of maximal overhang will be right-aligned at all blocks beneath the protruding block. Therefore, the ideas and formulas developed in relation to right-aligned stacks will ultimately prove useful for addressing the general problem of maximal overhang. Right-aligned n-stacks will also provide us with the means to analyse block sequences with divergent overhang as they often provide a useful lower bound for the maximal overhang.
3.1
A formula for the right-aligned overhang
We first establish a formula for the overhang of a right-aligned stack. This requires a little elementary physics and, if required, the reader can find the necessary details in Chapter 7.
16
Theorem 3.1.1. For any block sequence w = (wi )1 i=1 define Wi = w1 +w2 +· · ·+wi . The right-aligned overhang of the truncated n-block sequence wn = (w1 , . . . , wn ) is n X wi2 R(wn ) = . Wi i=1
(3.1.1)
Proof. Let di be the horizontal distance from the right edge of block i to the right edge of block i + 1, as shown in Figure 3.1. Then R(wn ) = d1 + · · · + dn . Recall that the mass of each block is equal to its half-width. Blocks 1 through i total mass w1 + · · · + wi
1
1 have
and, by assumption, their balancing point is located at
distance di from the right edge of block i + 1. Balancing these is the single block of mass wi located at distance wi
di from the right edge of block i + 1. We equate
moments about the right edge of block i + 1 to see that (wi
di )wi = di (w1 + · · · + wi 1 ).
Solving for di gives di = which completes the proof. mass = w1 + · · · + wi
wi2 , w1 + · · · + wi
(3.1.2)
1
mass = wi
i wi
di di
1
i i+1
Figure 3.1: Finding di by taking moments about the right edge of block i + 1. Example 3.1.2. If we right-align w3 = (1, 1, 4), as depicted in Figure 3.3 (a), then its right-aligned overhang is R(1, 1, 4) =
12 12 42 25 + + = . 1 1+1 1+1+4 6 17
4
3.2
The harmonic stack
If the constant n-block sequence hn = (1, 1, . . . , 1) is right-aligned, then we call it the harmonic n-stack. Equation (3.1.1) yields the expected right-aligned overhang for this stack. Corollary 3.2.1. If c > 0 and chn = (c, c, . . . , c), then R(chn ) = cHn where Hn is the nth harmonic number.
3.2.1
Optimality of the harmonic stack
We see that the right-aligned overhang of the harmonic n-stack is equal to the nth harmonic number. However, despite the significant historical attention devoted to harmonic stacks, our reading of the relevant literature suggests that it has not been definitively proved that Hn gives the maximal overhang. This is contrary to the assurance given in articles [58], [59] and [65]. Each of these refer to Hall [35], who provides what appears to be the closest approximation to a proof. A careful perusal of his argument indicates that he has only provided a lower bound for the maximal overhang. Indeed, Hall admits that his argument is premised on the assumption that the top block overhangs most. However, he later observes that if the vertical position of the top two blocks is switched, then the new configuration is balanced and the second block now protrudes furthest from the edge of the table. This is shown in Figure 3.2 (b). He then states that “[t]his case is believed to be the only way that the top block does not set the maximum overhang”.
18
2
11 1 1 54 3 2
table
2
1 1
11 1 1 54 3 2
table
(a)
1 1
(b)
Figure 3.2: Switching the vertical position of the top two blocks does not alter the overhang.
Might there be some other configuration that yields a greater overhang? Indeed, when the three irregular blocks in Figure 3.3 (a) are right-aligned, the overhang is readily shown to be
25 6
(see Example 3.1.2). However, in Figure 3.3 (b) observe that
a greater overhang is attained when the top two blocks are employed as a counterbalance:
32 6
(see Example 4.1.3). Here, the third block from the top overhangs
most. What precludes something similar from happening for blocks of equal width? 2
2
8
8 25 6
32 6
(a)
(b)
Figure 3.3: A greater overhang can be achieved when the top two blocks are employed as a counterbalance.
We settle this problem in Corollary 4.1.4. We prove that the maximal overhang is indeed Hn , and that this is only obtained when the blocks have one of the two configurations depicted in Figure 3.2.
19
3.2.2
A harmonically suspended elephant
The harmonic n-stack su↵ers from the following impracticality: even the tiniest mass resting upon its topmost black will cause the n-stack to topple. Fans of suspended elephants need not despair; a sufficiently large set of blocks of halfwidth 1 can be configured to suspend any given mass m over any desired overhang. Indeed, suppose mass m is placed upon the middle of the top-most block and that the n-stack is right-aligned. By an argument identical to that seen in the proof of Theorem 3.1.1, the n-stack will have a right-aligned overhang equal to 1 1 1 + + ··· + . m+1 m+2 m+n This clearly diverges as n ! 1, for any m > 0.
Figure 3.4: Any given mass can be suspended over any desired distance.
3.3
Stacking cubes
Much of the preceding work can also be generalised in obvious directions. For instance, take an infinite set of cubes of uniform density whose half-widths are given by a so-called cube sequence, w = (wi )1 i=1 . We can assume that the mass of 20
the ith cube is wi3 . In this case, the argument given in the proof of Theorem 3.1.1 can be readily modified to give the following result. Proposition 3.3.1. For any cube sequence w = (wi )1 i=1 , the right-aligned overhang of the truncated n-cube sequence wn = (w1 , . . . , wn ) is R(wn ) =
n X i=1
wi4 . w13 + · · · + wi3
(3.3.1)
We will characterise the divergence of (3.3.1) in Section 5.3.1.
3.4
Reordering right-aligned stacks
It is clear from the formula for R(wn ) that its value depends on the order of the elements in wn . We list the right-aligned overhang of each of the six permutations of the 3-block sequence w3 = (1, 2, 3).
R (3, 2, 1) =
238 60
R (3, 1, 2) =
235 60
R (1, 2, 3) =
230 60
R (2, 3, 1) =
238 60
R (1, 3, 2) =
235 60
R (2, 1, 3) =
230 60
Table 3.1: The right-aligned overhang for all permutations of w3 = (1, 2, 3). Note that swapping the top two blocks has no e↵ect on the overhang. Aside from this, the overhang is greatest when the blocks are arranged in order of decreasing size. Lemmas 3.4.1 and 3.4.2 generalise these two observations. Lemma 3.4.1. Interchanging the first two entries of the n-block sequence wn = (w1 , w2 , . . . , wn ) does not change the right-aligned overhang. Proof. Let wn0 = (w2 , w1 , . . . , wn ) be the n-block sequence obtained from wn by interchanging the first two entries. The result follows from Equation (3.1.1) and 21
the fact that w12 w22 w2 + w1 w2 + w22 w2 w12 + = 1 = 2+ . w1 w1 + w2 w1 + w2 w2 w2 + w1 Lemma 3.4.2. The permutation vn = (v1 , . . . , vn ) of wn = (w1 , . . . , wn ) that yields the maximal [minimal ] right-aligned overhang satisfies vi+1 vi [ vi+1 i 2 {2, . . . , n
vi ] for all
1}.
Proof. This will be a proof by contradiction. Suppose that vn = (v1 , . . . , vn ) is the permutation of wn = (w1 , . . . , wn ) that yields the maximal right-aligned overhang and that vi+1 > vi for some i 2 {2, . . . , n
1}. If we obtain vn0 from vn by
interchanging vi and vi+1 , then this increases the right-aligned overhang: R(vn0 ) > R(vn ). To prove this we first set m = v1 + · · · + vi 1 . As i
2 we know that m > 0.
Therefore R(vn0 )
2 2 vi+1 vi+1 vi2 vi2 + m + vi+1 m + vi+1 + vi m + vi m + vi+1 + vi mvi vi+1 (vi+1 vi ) = > 0, (m + vi+1 )(m + vi )(m + vi + vi+1 )
R(vn ) =
which gives us the desired contradiction. The corresponding result for the permutation that minimises the overhang is proved likewise. The following lemma is physically obvious; if we increase the width of the bottom block n, then the right-aligned overhang will increase. The extra width of the bottom block serves two purposes: it further extends the stack that it supports and provides the counterbalance required to do so. Lemma 3.4.3. For fixed w1 , . . . , wn
1
the function f (x) = R(w1 , . . . , wn 1 , x) is
strictly increasing for all x > 0. Proof. By Equation (3.1.1) we have f (x) =
w12 w22 x2 + + ··· + w1 w1 + w2 w1 + · · · + wn 22
1
+x
.
By setting a = w1 + · · · + wn
1
> 0 and then di↵erentiating, we see that
f 0 (x) =
x(2a + x) >0 (a + x)2
for all x > 0. Remark 3.4.4. Note that unilaterally increasing the half-width of any other block may not improve the right-aligned overhang. For instance, a simple calculation 4
using Equation (3.1.1) shows that R(2, 20, 20) < R(1, 20, 20).
By utilising Lemmas 3.4.2 and 3.4.3 we obtain the following result, which gives a simple estimate for the right-aligned overhang of any n-stack. Theorem 3.4.5. Consider any n-block sequence wn = (wi )ni=1 such that 0 < d wi D for all i 2 {1, . . . , n}. Then dHn R(wn ) DHn , where each inequality is strict unless wn = dhn or wn = Dhn . Proof. Permute the n-block sequence wn to obtain the decreasing n-block sequence vn . Then as vi D for all i 2 {1, . . . , n} we have R(w1 , . . . , wn ) R(v1 , . . . , vn )
(by Lemma 3.4.2)
R(v1 , . . . , D)
(by Lemma 3.4.3)
R(D, v1 , . . . , vn 1 ) (by Lemma 3.4.2). By repeating this argument inductively and then applying Corollary 3.2.1, we see that R(wn ) = R(w1 , . . . , wn ) R(D, . . . , D) = R(Dhn ) = DHn , with equality if and only if wn = Dhn . The proof that R(wn )
dHn is similar.
This result warrants a few comments. First, we see that the constant block sequence Dh has the greatest right-aligned overhang amongst all block sequences
23
bounded above by D. Second, a na¨ıve application of formula (3.1.1) yields weaker results. For instance: R(wn ) =
n X i=1
wi2 w1 + · · · + wi
We can likewise show that R(wn )
n X i=1
n d2 d2 X 1 d2 = = Hn . D · · + D} D i=1 i D | + ·{z
D2 Hn . d
i times
Moreover, Theorem 3.4.5 also gives a
simple sufficient condition for a block sequence to have divergent overhang. Corollary 3.4.6. If the block sequence w = (wi )1 i=1 is bounded below by d > 0, then it has divergent overhang. Proof. By Theorem 3.4.5, M (wn )
dHn ! 1 as n ! 1.
R(wn )
Example 3.4.7. The block sequence w = (a for all a > 1 by Corollary 3.4.6, since a
sin(i))1 i=1 has divergent overhang
sin(i)
a
1 > 0. If a = 1, then
Corollary 3.4.6 is of no use. To see this, we note that w is dense in [0, 2], as shown by Demos and Staib [26]. Therefore, there is no d > 0 such that 1
sin(i)
d for
all i. In Example 3.5.4 we will also prove that this block sequence also has divergent 4
overhang when a = 1.
One might hope to improve Corollary 3.4.6 by asking if the result still holds if we simply assume that w = (wi )1 i=1 has a subsequence bounded below by some d > 0. In Example 5.4.3 we show that no such general result can be proved.
3.4.1
Block sequences with divergent overhang whose terms tend to zero
Our next corollary will yield an example of a block sequence with divergent overhang whose half-widths tend to zero. This will show that the converse of Corollary 3.4.6 is not true. Corollary 3.4.8. If w = (wi )1 i=1 is a monotone decreasing block sequence such that wn log n ! 1 as n ! 1, then w has divergent overhang. 24
Proof. We use the well-known estimate, Hn
log n. Since the truncated n-block
sequence wn is monotone decreasing we have that wi
wn = d for all 1 i n.
So, by Theorem 3.4.5, as n ! 1, M (wn )
R(wn )
w n Hn
wn log n ! 1.
Example 3.4.9. Here, and often later, it is convenient for our block sequence to begin with a di↵erent index. For ↵ 2 (0, 1) we define w = (1/(log i)↵ )1 i=2 . Then w is a decreasing block sequence whose half-widths tend to zero. Moreover, wn log n = (log n)1
↵
! 1 as n ! 1. Therefore, w has divergent overhang by
Corollary 3.4.8. In Example 5.3.5 we will prove that this block sequence also has divergent overhang when ↵ = 1 but convergent overhang if ↵ > 1.
3.5
4
Another sufficient condition
Corollaries 3.4.6 and 3.4.8 provided sufficient conditions for a block sequence to have divergent overhang. The first of these is applicable to block sequences whose terms are bounded below by some d > 0 while the second of these is applicable to blocks sequences whose terms may tend to zero, but not too quickly. We now find a sufficient condition applicable to block sequences that may have some subsequence whose half-widths tend to zero. We show that if the half-widths of a block sequence do not go to zero on average, then the block sequence will have divergent overhang. To make the meaning of this statement clear, we say that w 2 R is the Ces` aro mean of the sequence (wi )1 i=1 if w = limi!1
Wi . i
Our proof of the following condition will
require two ingredients. The first of these is Abel’s Lemma (see, e.g., [73, p. 388]). Lemma 3.5.1 (Abel’s Lemma). For real numbers a1 , . . . , an and b1 , . . . , bn we define Ai = a1 + · · · + ai . Then n X i=1
ai bi = An bn +
n 1 X i=1
25
Ai (bi
bi+1 ).
We also require the well-known and so-called RMS-AM Inequality, which compares the root mean square (RMS) )with the arithmetic mean (AM) of a finite sequence of non-negative real numbers [34]. Lemma 3.5.2 (RMS-AM Inequality). For non-negative real numbers a1 , . . . , an , r a21 + · · · + a2n a1 + · · · + an , n n with equality if and only if ai = aj for all i, j 2 {1, . . . , n}. Proposition 3.5.3. If w > 0 is the Ces` aro mean of w = (wi )1 i=1 , then w has divergent overhang. Proof. We will show that the right-aligned overhang diverges. Since Wi = w > 0, i!1 i lim
there exists N such that for each i
N,
w Wi 3w . 2 i 2
(3.5.1)
Therefore, wi2 Wi
2 wi2 . (3.5.2) 3w i P wi2 By Equation (3.5.2), it suffices to show that 1 i=1 i diverges. To this end, we let Pn Tn = i=1 wi2 . By an application of Abel’s Lemma 3.5.1 we obtain ◆ n 1✓ n X X wi2 1 1 1 = Tn + Ti i n i i+1 i=1 i=1 n 1
X 1 Ti 1 = Tn + n i+1 i i=1 n X i=1
1 Ti . i+1 i
By employing the RMS-AM Inequality 3.5.2 and then Estimate (3.5.1) we obtain !2 ✓ ◆ i i 2 Ti 1X 2 1X Wi w2 = wj wj = , i i j=1 i j=1 i 4 26
provided i > N . Therefore n X w2 i
i=1
i
n X i=1
1 Ti i+1 i
n X i=N
1 Ti i+1 i
n w2 X 1 !1 4 i=N i + 1
as n ! 1. Example 3.5.4. We return to the block sequence w = (a
sin(i))1 i=1 , but now let
a = 1. By Proposition 3.5.3, w has divergent overhang since its Ces`aro mean can be readily shown to be equal to Wi i + csc(1/2) sin(i/2 + 1/2) sin(i/2) = lim = 1 > 0. i!1 i i!1 i lim
The right-aligned 20-block sequence w20 is shown in Figure 3.5 below. Despite its 4
precarious appearance, the n-stack is balanced.
Figure 3.5: The right-aligned n-block sequence w20 = (1
sin(i))20 i=1 .
Remark 3.5.5. Note that a block sequence w = (wi )1 i=1 need not have a positive Ces`aro mean to have divergent overhang. To see this, the block sequence defined in Example 3.4.9 has divergent overhang and its terms tend to zero. It will therefore have a Ces`aro mean of zero. This is simple consequence of Theorem 5.3.1, which 4
follows.
27
Chapter 4 Configurations of maximal overhang It is important to stress here that our blocks are stacked in a prescribed sequential order from top to bottom. Permutations are not (currently) permitted. Having established a formula for the right-aligned overhang of an n-block sequence, we now find a corresponding formula for its maximal overhang. To obtain such a formula we require a description of the configurations that yield this maximum. We will show that stacks of maximal overhang are right-aligned at all blocks beneath the protruding block. Any block above the protruding block serves as a counterbalance for the protruding block. One such configuration is shown below in Figure 4.1.
counterbalance protruding block right-aligned section
Figure 4.1: The first three blocks counterbalance the fourth block. Beneath this, the stack is right-aligned.
28
Proposition 4.0.1. Any maximal configuration xn = (xi )ni=1 of the n-block sequence wn = (wi )ni=1 satisfies the following two conditions: (a) If block j protrudes, then for every i
j the stack is right-aligned at block i.
(b) The balancing point of blocks 1 through j
1 combined is aligned with the left
edge of block j. Informal proof of (a). Suppose that the n-stack of maximal overhang is not rightaligned at block i
j. Then the balancing point of blocks 1 through i is some
distance d > 0 to the left of the right edge of block i + 1. This is shown in Figure 4.2. Let M = w1 + · · · + wi be the mass of blocks 1 through i and let m = wi+1 be the mass of block i + 1. Now pick ✏, 2 R such that 0 < ✏,
0 to the right of the left edge of the protruding block j. By part (a), every block beneath block j is right-aligned. Therefore, we can shift blocks 1 though j
1 to
the left by some sufficiently small 0 < ✏ < d so that the stack is still balanced. Now the balancing point xn of the entire n-stack is some distance
> 0 to the left of the
table edge, as shown in Figure 4.3. Therefore, we can shift the entire n-stack by distance
to the right, yielding an n-stack with greater overhang and our desired
contradiction.
30
✏
d j
j
Figure 4.3: By shifting the counterbalance to the left, the balancing point of the entire stack is now to the left of the table edge.
Formal proof of (b). Again, the reader may skip the following details if the previous argument was sufficiently convincing. By part (a), before shifting any blocks, the n-stack is right-aligned at block i for each j i n
1. Therefore, blocks 1
through i combined are balanced at xi = xi+1 + wi+1 . When blocks 1 through j
1
are shifted by ✏ to the left, the n-stack is configured so that x0n = (x1
✏, x2
✏, . . . , xj
1
✏, xj , . . . , xn ).
Therefore, the balancing point of blocks 1 through i combined is Pj 1 P ✏)wk + ik=j xk wk k=1 (xk 0 xi = Pi k=j wk Pi P ✏ jk=11 wk k=1 xk wk = Pi k=j wk Pj 1 wk = xi ✏ Pk=1 i k=j wk = xi+1 + wi+1
where
✏bi
(4.0.1) (4.0.2)
Pj 1 wk bi := Pk=1 > 0. i k=j wk
For the new configuration to be balanced, we need to ensure that the balancing point of blocks 1 through i is still to the right of the left edge of block i + 1, for each j i n
1. This occurs if and only if x0i 31
xi+1
wi+1 and, by (4.0.2), this
occurs if and only if ✏ 2wi+1 /bi for all j i n
1. Therefore, if we set ✏ = min {2wi+1 /bi : j i n
1} ,
then by (4.0.1) the balancing point of the entire n-stack has shifted to the left by x0n = ✏bn > 0. The entire stack can then be right-shifted by
distance
= xn
distance
> 0 to yield an improved overhang, which contradicts the maximality of
the initial n-stack.
4.1
A formula for the maximal overhang
Henceforth we will assume that every n-block sequence wn is configured according to the description provided in Proposition 4.0.1. Using this description, we can now find a formula for the maximal overhang M (wn ) corresponding to such configurations. Suppose that blocks 1, . . . , j
1 serve as a counterbalance for block j. Then, by
Proposition 4.0.1, we can assume that the balancing point of each of these blocks is aligned with the left edge of block j, as shown below in Figure 4.4. The balancing point of blocks 1 through j will then be some distance d measured from the right edge of block j. As the mass of each block is equal to its half-width, we find d as a mass-weighted average, d=
wj2 + 2wj (w1 + · · · + wj 1 ) = 2wj w1 + · · · + wj
32
wj2 . w1 + · · · + wj
(4.1.1)
mass = w1 + · · · + wj
1
wj
wj
mass = wj d
Figure 4.4: Blocks 1, . . . , j
1 counterbalance block j.
As the n-stack is right-aligned at every block beneath block j, the overhang contributed by the right-aligned section is n X
wi2 . w + · · · + wi i=j+1 1
(4.1.2)
The relevant calculation can be found in the proof of Proposition 3.1.1. By piecing together (4.1.1) and (4.1.2) obtain Proposition 4.1.1. Proposition 4.1.1. For any n-block sequence wn , the overhang Mj (wn ) obtained when block j protrudes is given by n X wj2 wi2 + . w1 + · · · + wj i=j+1 w1 + · · · + wi
Mj (wn ) = 2wj
(4.1.3)
To determine the maximal overhang we compare the overhang when block j protrudes for each j 2 {1, . . . , n} so that M (wn ) = max Mj (wn ). 1jn
4.1.1
(4.1.4)
A word on notation
If wn = (w1 , . . . , wn ) is configured so that block j protrudes, then we write wn = (w1 , . . . , wj 1 )(wj )(wj+1 , . . . , wn ). The first pair of parentheses indicate those blocks that are in the counterbalance, the second indicate the protruding block and the third indicate those blocks beneath the protruding block; that is, the right-aligned section. If the first pair of parentheses are empty, then this indicates that block 33
1 protrudes, while if the final pair are empty, then this indicates that block n protrudes. Example 4.1.2. If we configure w4 = (5, 6, 7, 8) so that its third block protrudes, then the corresponding overhang is 72 82 3215 + = . 5+6+7 5+6+7+8 234
M3 [(5, 6)(7)(8)] = 2 · 7
4
Example 4.1.3. We can now determine the maximal overhang of the 3-block sequence w3 = (1, 1, 4) depicted earlier in Figure 3.3. We compare the overhang when blocks j = 1, 2, 3 protrude, yielding Table 4.1 shown below. The maximal overhang 4
is achieved when j = 3.
j
1
2
3
w3
()(1)(1, 4)
(1)(1)(4)
(1, 1)(4)()
Mj (w3 )
25 6
25 6
32 6
Table 4.1: The overhang of w3 = (1, 1, 4) depends on which block protrudes.
4.1.2
The maximal overhang of monotone decreasing n-block sequences
Although Equation (4.1.4) for the maximal overhang is somewhat unpleasant, there is still much that one can do with it. First, we can now settle the open question related to the optimality of harmonic stacks discussed in Section 3.2.1. Corollary 4.1.4. If the constant n-block sequence hn = (1)ni=1 is configured to achieve the maximal overhang, then either block 1 or block 2 protrudes. Moreover, M (hn ) = R(hn ).
34
Proof. If block j protrudes, then, since wi = 1 for all 1 i n, Equation (4.1.3) gives n X wj2 wi2 + w1 + · · · + wj i=j+1 w1 + · · · + wi
M (hn ) = 2wj =2
=2 =2 Since M (hn ) 2
n X 1 1 + j i=j+1 i
1 j 1 j
j X 1 i=1 j
i
X1 i=1
i
+
n X 1 i=1
i
+ R(hn ).
R(hn ), 1 j
j X 1 i=1
i
0 () j = 1 or j = 2 () 2
1 j
j X 1 i=1
i
= 0.
Therefore, either block 1 or block 2 protrudes and in either case M (hn ) = R(hn ). More generally, we can prove the same for any monotone decreasing n-block sequence. Proposition 4.1.5. Suppose that wn = (wi )ni=1 is a monotone decreasing n-block sequence and that wn is configured to achieve the maximal overhang. If w1 > w2 , then block 1 protrudes, otherwise either block 1 or block 2 may protrude. Proof. Suppose that the overhang is maximised when block j protrudes and that j
3. Let b = wj and a = wj 1 . Moreover, by setting c = w1 + · · · + wj
2
> 0,
we can assume that j = 3. We will show that the overhang obtained when block 2 protrudes is greater than the overhang obtained when block 3 protrudes. To see this, note that a M2 [(c)(a)(b, . . .)]
b
0 and 3a
2b > 0. Therefore, by Equation (4.1.3),
M3 [(c, a)(b)(. . .)] =
a2 (a
35
b) + 2c2 (a b) + ac(3a (c + a)(c + a + b)
2b)
> 0.
This contradicts the maximality of the overhang. Therefore, either block 1 or block 2 protrudes and c = 0. We then observe that M1 [()(a)(b, . . .)]
M2 [(a)(b)(. . .)] =
a2 (a b) . a(a + b)
(4.1.5)
If a > b, then (4.1.5) shows that block 1 must protrude to maximise the overhang. On the other hand, if a = b, then (4.1.5) shows that either block 1 or block 2 may protrude, since the overhang is the same in either case. Next, the following proposition states that a block sequence that is bounded above has divergent overhang if and only if its right-aligned overhang diverges. This is due to the fact that we can easily estimate the limited extent to which a counterbalance may improve the maximal overhang compared with when the block sequence is fully right-aligned. Proposition 4.1.6. If the block sequence w = (wi )1 i=1 is bounded above, then w has divergent overhang if and only if its right-aligned overhang diverges. Proof. We need to prove that M (wn ) ! 1 if and only if R(wn ) ! 1, as n ! 1. First, it is clear that R(wn ) M (wn ). Now suppose that M (wn ) is maximised when block j protrudes and that wi D for all i. Then by (4.1.3) we have that M (wn ) = 2wj 2D +
n X wj2 wi2 + w1 + · · · + wj i=j+1 w1 + · · · + wi
n X i=1
wi2 w1 + · · · + wi
2D + R(wn ). Therefore, R(wn ) M (wn ) 2D + R(wn ). The result follows. The requirement that the block sequences be bounded above is essential to Proposition 4.1.6. Later, we will prove the following surprising result: there are unbounded block sequences with divergent overhang and convergent right-aligned overhang (see Example 5.4.3). 36
Chapter 5 Characterising block sequences with divergent overhang I think that it was Harald Bohr who remarked to me that all analysts spend half their time hunting through the literature for inequalities which they want to use and cannot prove. G. H. Hardy [36]
Proposition 4.1.6 asserts that a block sequence that is bounded above will have divergent overhang if and only if its right-aligned overhang diverges. However, this in itself is not entirely helpful since Equation (3.1.1) for the right-aligned overhang is not particularly tractable. Might we account for this difficulty by finding a clean characterisation of block sequences whose overhang diverges?
37
5.1
Asymptotic Notation
We will show that, for certain natural classes of block sequences, the divergence of the right-aligned overhang is tantamount to the divergence of a more pliable series. In doing so, it will be convenient to utilise some of the asymptotic notation first employed by Bachmann [10] and Landau [44], and then modified by Knuth [42]. The notation that we require is listed below in Table 5.1.
Notation
Description
Definition
f (n) = O (g(n))
f is asymptotically bounded
There exists n0 2 N and k >
above by g
0 such that f (n) kg(n) for all n
f (n) = ⌦ (g(n))
f is asymptotically bounded
There exists n0 2 N and k >
below by g
0 such that f (n) all n
f (n) = ⇥ (g(n))
f (n) ⇠ g(n)
n0
kg(n) for
n0
f is asymptotically bounded
f (n) = O (g(n)) and f (n) =
above and below by g
⌦ (g(n))
f is asymptotically equal to g
limn!1
f (n) g(n)
=1
Table 5.1: The required asymptotic notation.
5.2
Characterisation I
We will now establish a characterisation of monotone decreasing blocks sequences whose overhang diverges. This is a natural class of block sequences since, in Lemma 38
3.4.2, we proved that the right-aligned overhang of an n-block sequence is maximised when the blocks are arranged in decreasing order of size. Our proof of this characterisation will require two ingredients. We have already encountered the first of these: Abel’s Lemma 3.5.1. Second, and most essentially, we require an inequality that appears in the literature as Titu’s Lemma, the elegant proof of which was first given by Titu Andreescu cite[p. 8]Titu. Lemma 5.2.1 (Titu’s Lemma). For real numbers a1 , . . . , an and positive real numbers b1 , . . . , bn ,
a21 a22 a2 + + ··· + n b1 b2 bn
Equality occurs if and only if
ai bi
aj bj
=
(a1 + a2 + · · · + an )2 . b1 + b2 + · · · + bn
for all i, j 2 {1, . . . , n}.
Proposition 5.2.2. If w = (wi )1 i=1 is any sequence of positive terms and Wi = w1 + · · · + wi , then
n n X X wi wi2 =⌦ Wi i i=k i=k
!
.
(5.2.1)
Proof. By an application of Abel’s Lemma we have that n ✓ X
◆ 1 = wi · i i i=1 ✓ n 1 Wn X 1 = + Wi n i i=1
n X wi i=1
1 i+1
◆
n 1
=
Wn X Wi + n i(i + 1) i=1
n X i=1
Wi i(i + 1)
n
1 X Wi . 2 i=1 i2
Therefore
n X Wi i=1
i2
2
39
n X wi i=1
i
.
(5.2.2)
Now, by Titu’s Lemma 5.2.1 and then Estimate (5.2.2) we have n n X wi2 X wi2 /i2 = Wi i=1 Wi /i2 i=1 P 2 ( ni=1 wi /i) Pn 2 i=1 Wi /i Pn 2 ( i=1 wi /i) P 2 ni=1 wi /i n 1 X wi = , 2 i=1 i
(5.2.3)
which proves the result. P 1 wi So, for any sequence w = (wi )1 i=1 of positive terms, if i=1 i diverges, then so P wi2 too does 1 i=1 Wi . However, the converse does not hold. To see this, consider the following example.
Example 5.2.3. Define a sequence of positive terms w = (wi )1 i=1 by 8 > : 1i , otherwise. 2 We first note that the series n X wi i=1
i
=
P1
wi i=1 i
n X wi
i6=2j
i
On the other hand, the series j
W 2j =
2 X i=1
j
+
converges since, for any n 2 N,
1 1 X X 1 j < + = log 2 + 2. i i i2 2j i=1 j=1
n X wi
i=2j
P1
wi2 i=1 Wi
diverges. To see this, we first note that
j
2 2 X 1 X j(j + 1) wi = + k < 1 + (1 + 2 + · · · + j) = 1 + 2j 2 . i 2 2 k k i6=2
i=2
Therefore, n
2 X wi2 Wi i=1
n
2 n X X wi2 j2 n > = ! 1, 2 Wi 2j 2 j j=1
i=2
as n ! 1.
4 40
It is also possible to find block sequences that are not monotone decreasing for P wi2 P 1 wi which 1 i=1 Wi and i=1 i both diverge, but at di↵erent rates. Such behaviour is illustrated in Example 5.2.4 below.
Example 5.2.4. Let w = (i2i )1 i=1 . Since Wi = 1 · 21 + 2 · 22 + · · · + i · 2i ⇠ i2i+1 we can readily show that n n X X wi2 i2 22i = ⇠ n2n . 1 2 i Wi 1 · 2 + 2 · 2 + ··· + i · 2 i=1 i=1
On the other hand,
n X wi i=1
i
=
n X i2i i=1
i
⇠ 2n+1 .
Note that in this instance, the right-aligned overhang is, asymptotically, no better than the overhang that one could achieve by simply stacking the blocks vertically. We will have more to say about unbounded block sequences in Section 5.4.
4
Notwithstanding Examples 5.2.3 and 5.2.4, we can improve Proposition 5.2.2 if we simply demand that our sequence be monotone decreasing. Proposition 5.2.5. If w = (wi )1 i=1 is any monotone decreasing sequence of positive terms and Wi = w1 + · · · + wi , then n n X X wi2 wi =O Wi i i=1 i=1
!
.
Proof. As w = (wi )1 i=1 is monotone decreasing we have that Wi
(5.2.4) iwi . Therefore,
n n n X X X wi2 wi2 wi = . W iw i i i i=1 i=1 i=1
(5.2.5)
This gives the result. By combining Propositions 5.2.2 and 5.2.5 we obtain the following characterisation of monotone decreasing block sequences with divergent overhang. 41
Theorem 5.2.6 (Characterisation I). If w = (wi )1 i=1 is any monotone decreasing block sequence and Wi = w1 + · · · + wi , then n n X X wi2 wi =⇥ Wi i i=1 i=1
!
.
Moreover, the block sequence w has divergent overhang if and only if verges.
(5.2.6) P1
wi i=1 i
di-
Proof. From Propositions 5.2.2 and 5.2.5 we know that n
n
n
X wi 1 X wi X wi2 . 2 i=1 i Wi i i=1 i=1
(5.2.7)
This establishes (5.2.6). Finally, since w is bounded above by w1 , we can employ Propositions 3.1.1 and 4.1.6 to see that w = (wi )1 i=1 has divergent overhang if and P1 wi2 P wi only if i=1 Wi diverges. That this is equivalent to the divergence of 1 i=1 i follows from (5.2.6).
We conclude this section with a range of examples that illustrate some typical applications of Theorem 5.2.6. The first of these confirms that if a decreasing block sequence w = (wi )1 i=1 is to have divergent overhang, then its terms must tend to zero very slowly. Example 5.2.7. The monotone decreasing block sequence w = (1/ip )1 i=1 has convergent overhang for all p > 0. To prove this, we simply note that 1 X wi i=1
i
=
1 X 1 p+1 i i=1
4
converges for all p > 0.
Example 5.2.8. We return to the block monotone decreasing block sequence w = (1/(log i)↵ )1 i=2 . We can now show that this has divergent overhang if 0 ↵ 1 and convergent overhang if ↵ > 1. To prove this, we simply note that the integral P 1 test indicates that 1 i=2 i(log i)↵ diverges if 0 ↵ 1 and converges if ↵ > 1. The 4
result then follows from Theorem 5.2.6.
42
We can improve Example 5.2.8 by giving a family of blocks sequences {w(j)}j2N
with divergent overhang such that w(j + 1) = (wi (j + 1))1 i=1 has much smaller terms than w(j) = (wi (j))1 i=1 in the sense that wi (j + 1)/wi (j) ! 0 as i ! 1, for all j 2 N.
Example 5.2.9. Define log1 (x) = log (x) and logk (x) = log(logk 1 (x)) for k
2.
For each j 2 N we can find a suitably large m 2 N such that logj (m) > 1. For any 0 ↵ 1 we define a monotone decreasing block sequence w(j) = (wi (j))1 i=m by wi (j) =
1 . log(i) log2 (i) · · · logj 1 (i)(logj (i))↵
(5.2.8)
Note that m is chosen so that each factor in the denominator of (5.2.8) is greater P wi than 1. It is well-known that 1 i=m i diverges (see, e.g. [41, p. 293]). Therefore, Theorem 5.2.6 ensures that the block sequence w(j) also has divergent overhang, for each j 2 N.
4
Remark 5.2.10. Theorem 5.2.6 can also be used to simply demonstrate the convergence or divergence of some rather intimidating series. For instance, if we define wi i! a monotone decreasing sequence by (wi )1 i=1 by i = ii+1 , then, by the ratio test, we P wi P1 i! know that 1 i=1 i = i=1 ii+1 converges. Theorem 5.2.6 then ensures that 1 X
1! i=1 11
i! 2 ii
+ ··· +
4
also converges.
5.2.1
i! ii
Sums of block sequences
P P1 For any pair of positive term series, 1 i=1 ai and i=1 bi , it is an elementary fact P1 P1 P1 that i=1 (ai + bi ) converges if and only if i=1 ai and i=1 bi both converge. We will now prove the corresponding result for blocks sequences. If w = (wi )1 i=1
and v = (vi )1 i=1 are block sequences, then we define their sum in the natural way: w + v = (wi + vi )1 i=1 . 43
1 Proposition 5.2.11. If w = (wi )1 i=1 and v = (vi )i=1 are monotone decreasing block
sequences, then their sum v + w has convergent overhang if and only if v and w have convergent overhang. Proof. If v and w are monotone decreasing, then so too is v + w. By Proposition 4.1.6, it suffices to consider the right-aligned overhang of each block sequence. By repeatedly applying Estimate (5.2.7) we have ! n 2 X w i i + V W i i i=1 i=1 ! n n X vi X w i + i i i=1 i=1 ! n X vi + w i i i=1
n X v2
1 2
1 2
=
1 2
n X (vi + wi )2
=
i=1 n X i=1
V i + Wi
w i + vi i
n X vi i=1
2
i
+
n X wi i=1
i
n X v2 i=1
n X wi2 i + Vi Wi i=1
!
.
Therefore, 1 2
n n X wi2 X vi2 + Wi i=1 Vi i=1
!
n X (wi + vi )2 i=1
Wi + V i
n n X wi2 X vi2 2 + Wi i=1 Vi i=1
!
.
The proposition follows obviously. 1 Example 5.2.12. Let v = (1/(log i)↵ )1 i=2 and w = (1/(log i) )i=2 where ↵,
0.
Then, by Proposition 5.2.11 and Example 5.2.8, v + w has convergent overhang if ↵ > 1 and
> 1 and divergent overhang if ↵ 1 or
44
1.
4
5.3
Characterisation II
Theorem 5.2.6 does not hold if we relax the assumption that our block sequences be monotone decreasing, as shown in Examples 5.2.3 and 5.2.4. However, if we do relax this assumption, then we can find a similar characterisation provided we impose a new demand that the ratios of consecutive half-widths approach unity sufficiently fast, in a sense that we will soon make clear. This second characterisation will also yield a precise asymptotic identity rather than the bounds seen in Theorem 5.2.6. For this modified characterisation, we also require two crucial ingredients. The first of these is the Stolz-Ces`aro Theorem, whose proof is given by Muresan [52, p. 85]. Theorem 5.3.1 (Stolz-Ces`aro Theorem). If (ai )1 i=1 is a sequence of real numbers and (bi )1 i=1 is a strictly increasing and divergent sequence, then ai ai+1 ai lim sup ; bi+1 bi i!1 bi i!1 ai ai+1 ai lim inf lim inf . i!1 bi i!1 bi+1 bi
lim sup
In particular, ai i!1 bi lim
ai bi
1
ai = L. i!1 bi
= L =) lim
1
Our characterisation requires one further ingredient: Raabe-Duhamel’s Test of Divergence, whose proof can also be found in Muresan [52, p. 120]. Theorem 5.3.2 (Raabe-Duhamel). If (ai )1 i=1 is a sequence of positive terms for ⇣ ⌘ P ai which limi!1 i ai+1 1 < 1, then 1 i=1 ai diverges. We use each of Theorems 5.3.1 and 5.3.2 in our proof of the following lemma.
Lemma 5.3.3. Suppose m 2 N and ↵ 2 R. If (wi )1 i=1 is a sequence of positive terms for which L := lim i i!1
✓
wi wi+1 45
1
◆
< 1,
(5.3.1)
then
P1
wim+↵ i=1 w1m +···+wim
diverges if and only if n X i=1
P1
i=1
wim+↵ ⇠ (1 w1m + · · · + wim
wi↵ i
diverges. In particular,
L)
n X w↵ i
i
i=1
.
(5.3.2)
Proof. First note that n X i=1
n
X w↵ wim+↵ iwim i = . w1m + · · · + wim i w1m + · · · + wim i=1
(5.3.3)
Let ✏ > 0. We will show that for all sufficiently large i, 1
L+✏
iwim 1.
When ↵ 1, we can determine the rate of divergence using (5.3.6) followed by a standard integral comparison to show that n n X X (1/ log i)2↵ 1 1 ⇠ ⇠ (log n)1 ↵ ↵ ↵ (1/ log 2) + · · · + (1/ log i) i(log i) 1 ↵ i=2 i=2
5.3.1
↵
.
4
Cubes sequences with divergent overhang
We can find a similar characterisation for right-aligned stacks made from cubes whose half-widths are given by the cube sequence w = (wi )1 i=1 . In Proposition 3.3.1 47
we indicated that the right-aligned overhang will diverge if and only if
P1
wi4 i=1 w13 +···+wi3
diverges. By letting m = 3 and ↵ = 1 in Lemma 5.3.3 we obtain the following result. Proposition 5.3.6. If w = (wi )1 i=1 is a cube sequence for which ✓ ◆ wi L := lim i 1 < 1, i!1 wi+1 then
P1
wi4 i=1 w13 +···+wi3
diverges if and only if n X i=1
5.4
P1
wi i=1 i
wi4 ⇠ (1 w13 + · · · + wi3
diverges. In particular, n X wi
L)
i=1
i
.
Comments on unbounded block sequences
As Theorem (5.3.4) does not require that our block sequences be decreasing, we can easily check its veracity for simple examples of block sequences with divergent overhang whose terms are increasing. Example 5.4.1. Consider the block sequence w = (ip )1 i=1 where p > 0 and the partial sum for its right-aligned overhang, n X i=1
i2p . 1 p + · · · + ip
By an integral comparison one can readily show that 1 p + · · · + ip ⇠
ip+1 . p+1
Therefore, without recourse to Theorem 5.3.4, one can show directly that the rightaligned overhang is asymptotically equal to n n X X i2p (p + 1) = (p + 1) ip p+1 i i=1 i=1
1
⇠
p+1 p n. p
To confirm that Theorem 5.3.4 also gives the expected result we first note that i [ip /(i + 1)p
⇥ 1] = i (1 + 1/i)
p
⇤ ⇥ 1 =i 1 48
p/i + O 1/i2
⇤ 1 !
p
as i ! 1. Therefore, 1
L = 1 + p. It follows from (5.3.6) that the right-aligned
overhang satisfies, n n n n X X X X wi2 i2p ip = ⇠ (1 + p) = (p + 1) ip p + · · · + ip W 1 i i i=1 i=1 i=1 i=1
1
⇠
p+1 p n, p
as expected. Note that if the first n blocks were simply stacked vertically, then we would achieve an overhang equal to np . Right-aligning the stack only improves this 4
(asymptotically) by a factor of (p + 1)/p.
In Examples 5.2.4 and 5.4.1 we observed increasing n-block sequences whose P w2 right-aligned overhang ni=1 Wii was asymptotically no less than the half-width wn of the nth block. Proposition 5.4.2 generalises this observation.
Proposition 5.4.2. Let w = (wi )1 i=1 be a block sequence for which Wn = w1 + · · · + wn ! 1 as n ! 1 and lim supn!1 wn /wn+1 < 1. Then n X wi2 = ⌦(wn ). Wi i=1
Proof. The result will follow at once if we can establish that lim inf n!1
n 1 X wi2 > 0. wn i=1 Wi
(5.4.1)
First, note that Wn2
w n Wn .
(5.4.2)
We now apply Titu’s Lemma 5.2.1 followed by Estimate (5.4.2) to show that n 1 X wi2 lim inf n!1 wn Wi i=1
lim inf n!1
1 W2 Pn n wn i=1 Wi
lim inf n!1
1 w n Wn Wn Pn = lim inf Pn . n!1 wn i=1 Wi i=1 Wi
We estimate the right-hand side of this with two successive applications of the StolzCes`aro Theorem 5.3.1. We first let an = Wn and bn = W1 + · · · + Wn . Note that bn ! 1 as n ! 1. Therefore Wn an lim inf Pn = lim inf n!1 n!1 bn i=1 Wi
lim inf n!1
an+1 bn+1 49
an wn+1 wn = lim inf = lim inf . n!1 Wn+1 n!1 Wn bn
Now define cn = wn and dn = Wn . Since dn ! 1 as n ! 1, we can apply the Stolz-Ces`aro Theorem 5.3.1 once again to give lim inf n!1
wn cn = lim inf n!1 dn Wn
lim inf n!1
cn+1 dn+1
cn wn+1 wn = lim inf . n!1 dn wn+1
By assumption, lim supn!1 wn /wn+1 < 1. Therefore, we finally see that ✓ ◆ wn+1 wn wn wn lim inf = 1 + lim inf = 1 lim sup > 0, n!1 n!1 wn+1 wn+1 w n!1 n+1 which establishes (5.4.1).
5.4.1
A surprising block sequence
In Section 4.1.2 we indicated that there are block sequences whose maximal overhang diverges even though their right-aligned overhang converges. That is, we can show that Proposition 4.1.6 is not true if we drop the requirement that our block sequence be bounded above. To see this, we first recall that any block sequence that is not bounded above has divergent overhang; we can achieve any desired overhang by simply stacking the blocks vertically. Remarkably, this does not guarantee that the right-aligned overhang diverges. We have concocted the following example to illustrate this fact. The idea is to begin with the block sequence w = (1/i)1 i=1 , whose right-aligned overhang converges (see Example 5.2.7). We note that Wi =
1 1
+
1 2
+ ··· +
1 i
! 1 as i ! 1. That is, the total mass of the first
i blocks increases without bound. We will now lightly intersperse this hefty block sequence with another sequence that tends to infinity so slowly that its contribution to the right-aligned overhang is meagre. Example 5.4.3. We define a block sequence w = (wi )1 i=1 by 8 p > < k, if i = 22k for some k 2 N, wi = > :1, otherwise. i 50
We will show that 1 1
Therefore, Wi
P1
wi2 i=1 wi
1 i
converges. First, note that wi
+ 12 + · · · +
1 i
for all i 2 N. Now consider,
for all i 2 N.
1 1 1 X X X wi2 wi2 wi2 = + . Wi Wi Wi k k i=1 i=22
(5.4.3)
i6=22
The second summation in (5.4.3) converges since 1 1 1 X X X wi2 (1/i)2 1 ⇡2 = . 1 1 1 Wi i2 6 + + · · · + 1 2 i k k i=1 2 2 i6=2
i6=2
The first summation in (5.4.3) also converges. To see this, we note that 1 1 1 + + · · · + 2 k = H 2 2k 1 2 2 Therefore, 1 X
1 X
⇣ p ⌘2 k
k
log 22 = 2k log 2.
1 X
1
X k k 2 = = . 1 1 1 1 1 1 k 2 log 2 log 2 + 2 + ··· + i + 2 + · · · + 2 2k 1 k k 1 k=1 k=1 2 2 i=2 i=2 p Notice that when the block of half-width k is added, the substantial weight of wi2 Wi
blocks that it supports makes its contribution to the overhang insubstantial.
4
Notwithstanding the pathological example above, the right-aligned overhang of a block sequence is sure to diverge provided that it has some subsequence whose terms grow sufficiently fast. 1 Proposition 5.4.4. If the block sequence w = (wi )1 i=1 has a subsequence (wnk )k=1 P wi2 w such that nnkk ! 1 as k ! 1, then 1 i=1 Wi diverges.
Proof. For any n 2 N we have that
Wn2 = (w1 + w2 + · · · + wn
1
+ wn )2
wn (w1 + w2 + · · · + wn 1 + wn ) ◆ ✓ n n 1 2 1 wn w1 + w2 + · · · + wn 1 + wn n n n n wn = (nw1 + (n 1)w2 + · · · + 2wn 1 + wn ) n n wn X = Wi . n i=1 51
(5.4.4)
We now apply Titu’s Lemma 5.2.1 and then Estimate (5.4.4) to show that n X wi2 Wi i=1
It follows that
W2 Pn n i=1 Wi
nk 1 X X wi2 wi2 = lim Wi k!1 i=1 Wi i=1
wn . n
w nk = 1. k!1 nk lim
Example 5.4.5. Take any sequence of positive terms (ai )1 i=1 and then define a block sequence w = (wi )1 i=1 such that 8 > :ai , otherwise.
The block sequence w has divergent overhang by Proposition 5.4.4 since wk2 /k 2 = k 3 /k 2 = k ! 1 as k ! 1.
5.5
4
Slowly diverging block sequences
In Section 5.2 we studied block sequences whose half-widths tend to zero monotonically whose overhang can nonetheless diverge. Might there be some block sequence with divergent overhang whose half-widths tend to zero faster than any other block sequence? For those familiar with the Abel-Dini Theorem, it will be no surprise that the answer is a resounding no. Loosely speaking, the Abel-Dini Theorem states that for any divergent series there is a corresponding series with much smaller terms that still diverges. A simple proof of the Abel-Dini Theorem can be found in Knopp [41, p. 290]. Theorem 5.5.1 (Abel-Dini Theorem). Suppose that (di )1 i=1 is an arbitrary sequence P of positive terms such that the infinite series 1 i=1 di diverges. If Di = d1 + · · · + di , P1 di then the series i=1 D↵ diverges if ↵ 1 and converges if ↵ > 1. i
52
For our purposes we require a quantitative form of the Abel-Dini Theorem in the special case where ↵ = 1. The proof Theorem 5.5.2 below was first given by Ces`aro [21] and is reproduced in English by Knopp [41, p. 292]. An appealing and geometrically motivated proof has also been given by Ash [9]. Theorem 5.5.2 (Ces`aro). Let (di )1 i=1 be any sequence of positive terms such that P1 the infinite series i=1 di diverges. If Di = d1 + · · · + di and Ddii ! 0 as i ! 1, then
n X di ⇠ log Dn . Di i=1
(5.5.1)
Theorem 5.5.2 implies that for every divergent series of positive and monotone decreasing terms, there is a corresponding divergent series whose partial sums grow essentially like the logarithms of the original partial sums. We now prove a corresponding result for block sequences, utilising Characterisation I, which is given as Theorem 5.2.6. Proposition 5.5.3. If w = (wi )1 i=1 is any monotone decreasing block sequence with divergent overhang, then there exists a monotone decreasing block sequence v = (vi ) with divergent overhang whose terms are smaller in that limi!1 n X v2
n X wi2 = ⇥ log Vi Wi i=1 i
i=1
!!
vi wi
.
= 0. Moreover, (5.5.2)
Proof. As w = (wi )1 i=1 is monotone decreasing and has divergent overhang, TheP wi wi orem 5.2.6 ensures that 1 i=1 i diverges. Let di = i and Di = d1 + · · · + di so P di that, by the Abel-Dini Theorem 5.5.1, we know that 1 i=1 Di diverges. Now define a new a decreasing block sequence v = (vi ) by vi =
wi . Di
vi 1 = lim = 0. i!1 wi i!1 Di lim
To prove (5.5.2) we first note that vi wi /Di wi /i di = = = . i i Di Di 53
Note that
Therefore, n X v2 i
i=1
Vi
=⇥
n X vi i=1
i
!
n X di =⇥ Di i=1
= ⇥ log = ⇥ log
(by Theorem 5.2.6)
!
(since vi /i = di /Di )
n X wi i=1 n X i=1
i
!!
wi2 Wi
!!
(by Theorem 5.5.2) (by Theorem 5.2.6).
This proves the result.
5.6
Strongly divergent block sequences
Suppose that we now wish to actually construct a stack, beginning with the bottom block and working our way up. And suppose we also want the overhang to exceed some given value. To achieve this, we could take any block sequence whose rightaligned overhang diverges. We could then calculate the number of blocks required to achieve the desired overhang. Each block would then be placed at the appropriate position, as determined by Equation (3.1.2). However, if we then required a stack with greater overhang, we would be forced to begin anew as any additional block added to the top of the stack would cause it to topple. Is this limitation simply due to the way in which we have decided to configure the block sequence? Or is it a more fundamental problem? With this question in mind, we provide the following definition. Given the block sequence w = (wi )1 i=1 , we wish to construct a stack from the bottom up whose overhang diverges in a stronger sense by satisfying two conditions: (a) the stack is balanced at every level as each block is added, and (b) for every D > 0 there is some block with overhang exceeding D. 54
Any block sequence for which this is possible is called strongly divergent. How might we characterise such sequences? To determine this, it is now convenient to number the blocks from the bottom up. We locate the table edge at the origin. As before, xi is the horizontal displacement of the midpoint of block i and wi is its half-width. Therefore, block i extends from xi
wi to xi + wi . Let w[i, j] be the
sum of half-widths from block i to block j inclusive and let x[i, j] be the centre of mass of these blocks. For each block j added to the tower, Condition (a) demands that x[i + 1, j] 2 [xi
wi , xi + wi ] for every 1 i < j and x[1, j] 0. This is
illustrated below in Figure 5.1. xi
wi
xi x[i + 1, j]
x i + wi
j .. . i+1 i .. . 1
Figure 5.1: The stack is balanced at every level as block j is added to the stack, for all j 2 N. Unfortunately, our hopes of constructing an improbable stack are dashed by the next theorem. We prove that the only strongly divergent block sequences are those that are obviously strongly divergent: unbounded block sequences. A variation of the following proof was contributed by an anonymous referee of an article that is soon to be published [79]. Theorem 5.6.1. A block sequence w is strongly divergent if and only if it is unbounded.
55
Proof. Any unbounded block sequence is strongly divergent as the blocks can be stably stacked vertically from the bottom up to achieve any overhang. Now suppose that some bounded block sequence w = (wi )1 i=1 is strongly divergent. Let block k be the lowest block for which no part lies over the table, that is, k = min {i : xi
wi > 0}. As w is bounded and strongly divergent, Condition (b) en-
sures that such a k exists. Note that x[1, k] 0. Now set ✏ = xk
wk and then
find j > k such that w[k + 1, j] >
w[1, k]x[1, k] . ✏
(5.6.1)
Such a j must exist since a strongly divergent stack requires unlimited mass by Condition (b). Note that Condition (a) ensures that x[k + 1, j]
xk
wk = ✏.
(5.6.2)
We now find x[1, j] as a mass-weighted average (see Property 7.2.12), and then employ (5.6.1) and (5.6.2) to give x[1, j] =
w[1, k]x[1, k] + w[k + 1, j]x[k + 1, j] > 0. w[1, j]
This is a contradiction as Condition (a) requires that x[1, j] 0. Therefore, no bounded block sequence is strongly divergent.
56
Chapter 6 Permuting block sets In Section 4.1 we determined the maximal overhang when an n-block sequence is stacked in a prescribed order. What if we now allow for the blocks to be permuted? How should a finite set of blocks be stacked, and in what order, to maximise their overhang? We will see that the answer is not always entirely obvious or easily found.
6.1
Optimal partitions
We will now refer to wn = {wi }ni=1 as a block set. As we want to allow for repeated elements, a block set is actually a multi-set. Of particular interest will be block sets of consecutive integers, defined so that cm,n = {m, m + 1, . . . , n}. In Section 4.1 we established that if wn is stacked in the prescribed order, then we can determine the maximal overhang by first finding the overhang when block j protrudes for each j 2 {1, . . . , n}; the maximal overhang will then be the largest of these values. What if we now allow for the blocks to be permuted? To answer this, we let G(wn ) be the set of permutations of the block set wn . For
2 G(wn ) let (wn ) = ( (w1 ), . . . , (wn )).
By then adapting Equation (4.1.4) we see that the absolute maximal overhang of wn is given by
57
A(wn ) = max max Mj ( (wn )). 2G(wn ) 1jn
(6.1.1)
Example 6.1.1. For c1,6 = {1, 2, 3, 4, 5, 6} we evaluate Mj ( (c1,6 )) numerically for every
2 G(c1,6 ) and every 1 j 6. We find that the absolute maximal
overhang of c1,6 is A(c1,6 ) = 4010/357 and that this obtained when
(c1,6 ) =
(3, 2, 1)(6)(5, 4). This is shown in Figure 6.1. The blocks of half-width 1, 2 and 3 serve as a counterbalance to the protruding block of half-width 6. The order of the counterbalance blocks clearly has no bearing on the overhang.
4
Figure 6.1: The absolute maximal overhang of c1,6 . Note that if each element in wn is unique, then G(wn ) has n! elements. Moreover, the protruding block can be in any one of n di↵erent positions. Therefore, to determine the absolute maximal overhang of wn , Equation (6.1.1) requires as many as n! ⇥ n computations of Mj ( (wn )). This is computationally unfeasible for all but small values of n. Indeed, for an arbitrary block set w25 of size 25, establishing A(w25 ) by a direct application of (6.1.1) would exhaust today’s most powerful computers. Might we improve our application of Equation (6.1.1) by excluding certain unfeasible permutations? For instance, in Example 6.1.1, we note that each block in the optimal counterbalance is smaller than every block not in the counterbalance. Regrettably, the following example shows that this observation does not hold more generally. The following carefully chosen block set was discovered with computational assistance.
58
Example 6.1.2. Consider the block set w8 = {6, 11, 20, 20, 20, 20, 20, 20}. We numerically evaluate Mj ( (w8 )) for every
2 G(wn ) and every 1 j 8. The abso-
lute maximal overhang of w8 is obtained when (w8 ) = (11)(20)(20, 20, 20, 20, 20, 6). Note that the single block in the counterbalance is not the smallest block. It is conjectured that if a block set has less than eight blocks, then the absolute maximal overhang can only be obtained if each block in the counterbalance is smaller than every block not in the counterbalance. To date, a complete proof of this conjecture 4
has eluded us.
Despite the unfortunate potential for irregularity demonstrated in Example 6.1.2, we are able to provide a partial description of the permutations that yield the absolute maximal overhang. In Theorem 6.1.3 we prove that the protruding block is always the largest block and that blocks beneath this are arranged in order of decreasing half-width. Theorem 6.1.3. Consider the block set wn and suppose that the absolute maximal overhang is achieved when block j protrudes in permutation (wn ) = vn = (v1 , . . . , vj 1 )(vj )(vj+1 , . . . , vn ). Then: (a) vj
vj+1
···
vn , and
(b) vj = max{v1 , v2 , · · · , vn }. Proof. Each part will be a proof by contradiction. To prove (a) we would ordinarily first show that vi
vi+1 for each i > j. However, proving this requires precisely the
same argument as given in the proof of Lemma 3.4.2. Therefore, it suffices to prove that vj
vj+1 . Suppose instead that vj < vj+1 . We then switch blocks vj and vj+1
in vn so that block vj+1 now protrudes in position j. This gives the n-stack vn0 = (v1 , . . . , vj 1 )(vj+1 )(vj , vj+2 . . . , vn ). 59
We show that this switch has improved the overhang: Mj (vn0 ) > Mj (vn ). To prove this, let m = v1 + · · · + vj 1 . A routine calculation using Equation (4.1.3) shows that Mj (vn0 )
Mj (vn ) = (vj+1
vj )
vj vj+1 + 2m(vj + vj+1 ) + 2m2 > 0. (vj + m)(vj+1 + m)(vj + vj+1 + m)
This contradicts the assumed optimality. To prove (b), we already know from part (a) that vj if suffices to show that vj
vi for all 1 i j
vi for all i
j. Therefore,
1. In other words, it is enough
to show that the protruding block is larger than each block in the counterbalance. This will be a proof by contradiction. We can assume that v1 > vj . Switch blocks v1 and vj in vn so that block v1 now protrudes in position j. This gives the modified n-stack vn0 = (vj , v2 , . . . , vj 1 )(v1 )(vj+1 , . . . , vn ). We obtain a contradiction by showing that this switch has improved the overhang: Mj (vn0 ) > Mj (vn ). To prove this, we now let m = v2 + · · · + vj
1
0. Equation
(4.1.3) then gives Mj (vn0 )
Mj (vn ) = (v1
✓
vj ) 2
v1 + vj v1 + vj + m
◆
> 0.
When combined, Proposition 4.0.1 and Theorem 6.1.3 permit us to be altogether unconcerned with the order and configuration of the blocks in a stack. Instead, we must simply consider the best way to partition the blocks into two sets: those in the counterbalance, and those not in the counterbalance. Once the partition is chosen, the optimal order and configuration is already determined. The particular partition that yields the absolute maximal overhang of wn is called the optimal partition of wn , and the set of blocks belonging to the counterbalance of the optimal partition is called the optimal counterbalance of wn . Corollary 6.1.4. Determining the optimal partition of the block set wn requires no more than 2n
1
computations of Equation (4.1.3). 60
Proof. By Theorem 6.1.3 (b), the protruding block is the largest block. Each of the other n
1 blocks is either in the counterbalance or not. Therefore, the counter-
balance can be chosen in at most 2n
6.2
1
di↵erent ways.
Estimates for counterbalance blocks
For any given block set, the field of contenders vying for the crown of optimal partition can be further reduced by employing the following proposition. This provides an upper bound for the half-width of each block in the counterbalance. Proposition 6.2.1. Let w be the half-width of some block in an optimal counterbalance of a block set wn . Let m be the sum of half-widths of every other block in the counterbalance, and suppose b is the half-width of the protruding block (see Figure 6.2). Then w b2 /(b + m).
(6.2.1)
m w b
Figure 6.2: Bounding the half-width w of a counterbalance block.
Proof. We can assume that the counterbalance contains just two blocks: one of half-width w and another with half-width m. By assumption, wn = (m, w)(b)(. . .) is optimal. By way of contradiction we assume that w > b2 /(b+m). We now form a new stack wn0 = (m)(b)(w, . . .) by shifting the block of half-width w to the position immediately beneath the protruding block. By a routine a calculation we can show that this improves the overhang, since M2 [(m)(b)(w, . . .)]
M3 [(w, m)(b)(. . .)] = 61
w(w(m + b) b2 ) > 0. (m + b)(w + m + b)
The improvement contradicts the assumed optimality. For any given block set wn , we call any counterbalance that adheres to Proposition 6.2.1 a viable counterbalance for wn . Example 6.2.2. In Example 6.1.2 we numerically determined that the optimal counterbalance of w8 = {6, 11, 20, 20, 20, 20, 20, 20} is {11}. We can now demonstrate this theoretically. One block of half-width b = 20 must protrude. Suppose that there are two or more blocks in the counterbalance and that one of these has half-width w. If m is the sum of half-widths of every other block in the counterbalance, then m
6. By Proposition 6.2.1, we conclude that w b2 /(b + m) 202 /(20 + 6) < 20.
Therefore, the only viable counterbalance containing two or more elements is {6, 11}. A viable counterbalance may also be empty, or contain just one block. Each viable counterbalance is listed below in Table 6.1 along with its respective maximal overhang, calculated to two decimal places using Equation (4.1.3). The optimal counterbalance is confirmed to be {11}. counterbalance {} maximal overhang 50.19
4 {6} 51.85
{11} 51.89
{20} 50.19
{6, 11} 51.86
Table 6.1: The viable counterbalance sets for w8 . For larger sets of blocks, Proposition 6.2.1 can be employed to construct the set of viable counterbalances for an arbitrary block set wn . Beginning with the second largest block, and working through to the smallest block, we form each counterbalance by choosing to either include or omit each block. However each block can only be included if its addition does not make the resulting set unviable. An example of pseudocode that describes this procedure is given in Algorithm 6.2.1. 62
1:
function Counterbalance(w1 , w2 , . . . , wn )
Precondition: 0 < w1 w2 · · · wn 2:
Let HoldingBin[0, 1, . . . , n
3:
Let w0 be a real number
4:
w0
5:
HoldingBin[n
6:
for i
1] be an array with n entries
0
n
1]
{?}
1 to 1 step
1 do
for each set in HoldingBin[i] do
7: 8:
Append set to HoldingBin[i
9:
Append wi to set
10:
m
11:
BlockBound
12:
k
13:
Append set to HoldingBin[k]
1]
the sum of elements in set bwn2 /(wn + m)c
max{0 j < i : wj BlockBound}
end for
14:
end for
15: 16:
return HoldingBin[0]
17:
end function Algorithm 6.2.1: Constructing viable counterbalances.
6.3
Bounding the size of the counterbalance
From Proposition 6.2.1 we can obtain an upper bound for the number of elements in an optimal counterbalance. We can also find an upper bound for the size of each block in the counterbalance. Corollary 6.3.1. Take any block set wn whose smallest and largest blocks have half-widths a and b, respectively. If the optimal counterbalance comprises k blocks
63
whose half-widths are x1 x2 · · · xk , then ✓ ◆2 b b k + 1. a a Moreover, for 1 j < k, xj < p
(6.3.1)
b . k j
(6.3.2)
Proof. Let x be the half-width of any block in the optimal counterbalance. As the smallest block in wn has half-width a, the sum of half-widths of every other block in the counterbalance is m
(k
ax
1)a. By Proposition 6.2.1, b2 b2 . b+m b + (k 1)a
Solving for k gives (6.3.1). Likewise, as the block of half-width xj is in the counterbalance, the sum of half-widths of every other block in the counterbalance is m
mj+1 + mj+2 + · · · + mk
(k
j)xj .
Therefore, by Proposition 6.2.1, xj
b2 b2 < , b + (k j)xj (k j)xj
from which we obtain (6.3.2). Example 6.3.2. Consider any block set of consecutive integers cm,n for which m n
< 32 . Then, since (3/2)2
3/2 + 1 = 7/4 < 2, we can apply Proposition 6.3.1 to
deduce that the optimal counterbalance contains at most one block. For instance, c201,300 has at most one block in its optimal counterbalance.
4
Example (6.3.1) above shows why it is straightforward to find optimal counterbalance when the ratio of largest to smallest blocks widths is sufficiently small. Cleary, the same argument will not work for a block set such as c1,100 . In the following section we develop an additional tool that will help us to determine the optimal counterbalance when the ratio of largest to smallest block widths is too large for Estimate (6.3.1) to be of any use, at least for sufficiently small sets of blocks. 64
6.4
The Subset-Sum Property
The next proposition gives further cause to regard certain partitions as sub-optimal. Its proof is elementary but intricate. We first complete the substantial ground-work required before presenting its formal proof. Proposition 6.4.1 (The Subset-Sum Property). The optimal counterbalance of a block set contains no block whose half-width equals the sum of two or more halfwidths of blocks not in the counterbalance. We now call a counterbalance viable if it adheres to both Propositions 6.2.1 and 6.4.1. Assuming each of these results, Example 6.4.2 below indicates how we can find the viable counterbalances for a small block set. Example 6.4.2. Consider the block set c1,4 = {1, 2, 3, 4}. The block of half-width b = 4 must protrude. Therefore, the counterbalance could be any one of the eight sets: {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}. However, since 3 = 1 + 2, the Subset Sum Property 6.4.1 dictates that the optimal counterbalance cannot be {3}. Moreover, if the counterbalance contains the block of half-width 2, then it cannot contain the block of half-width 3. To see this, we use Proposition 6.2.1, noting that 42 /(4 + 2) < 3. This leaves just five viable counterbalance sets. Each viable counterbalance is listed in Table 6.2 along its approximate overhang, calculated to two decimal places using Equation (4.1.3). The optimal counterbalance is {1, 2}. counterbalance {} overhang 5.83
4 {1} 6.32
{2} 6.43
{1, 2} {1,3} 6.61 6.40
Table 6.2: Viable counterbalance sets for c1,4 .
65
6.4.1
Proof of the Subset-Sum Property
We now turn our attention to the deferred proof of the Subset-Sum Property 6.4.1. Our proof requires that we sequentially make one of two types of alteration to a given stack. The e↵ect of each alternation will be to increase the overhang by incrementally shifting mass up the stack.
Block-swaps and mass transfers. Suppose that a block of half-width c rests on a block of half-width d, somewhere beneath the protruding block; that is, in the right-aligned section. Move 1: A block-swap. If c < d, then we may swap the two blocks. The block order, including the designated protruding block, is unchanged. Such a move will increase the overhang. To prove this, the relevant calculation is given in Lemma 3.4.2. We call this move a block-swap, and this is illustrated below in Figure 6.3.
c
d
d
c
Figure 6.3: A block swap.
Move 2: A mass-transfer by . Suppose c the upper block’s half-width by
d. For any
2 (0, d] we increase
and decrease the lower block’s half-width by .
The block order, including the designated protruding block, is unchanged. We call this move a mass-transfer by , and this is illustrated below in Figure 6.4. Let m be the total mass of blocks above the block of half-width c. By a long but elementary calculation we can show that the overhang has increased by at least (c + )2 (d )2 + m+c+ m+c+d
c2 m+c 66
d2 > ✏(c m+c+d
d)
0,
for some ✏ > 0. Note that if
= d, then the bottom block attains zero half-width.
In the following proof, we will see that such blocks can simply be eliminated.
c+
c
d
d
Figure 6.4: A mass transfer by .
Remark 6.4.3. We stress that the mass transfer move is only permitted when the upper block is no smaller than the lower block. If this is not the case, then a mass transfer might otherwise decrease the overhang. To see this, consider the 2-stacks w2 = ()(1)(3) and v2 = ()(2)(2) = ()(1 + 1)(3
1). Note that v2 can be obtained
from w2 by a mass transfer of 1. Moreover, M1 [()(1)(3)] =
12 32 13 22 22 + = >3= + = M1 [()(2)(2)]. 1 1+3 4 2 2+2
4
We will soon explain how a sequence of block swaps and mass transfers yields a proof of the Subset-Sum Property 6.4.1. However, Example 6.4.4 will first provide a template for the formal proof of this property. Example 6.4.4. Consider the block set w6 = {2, 3, 3, 4, 6, 7}. Note that 6 = 4 + 2. We will show that the absolute maximal overhang is not given by (w6 ) = (6)(7)(4, 3, 3, 2); that is, we show that the counterbalance cannot contain a block whose half-width (6) is the sum of half-widths (4 + 2) of two blocks that are not in the counterbalance. The following sequence of mass-transfers and block-swaps
67
improves the overhang at each step: M2 [(6)(7)(4, 3, 3, 2)] < M2 [(6)(7)(4, 3, 4, 1)]
(mass-transfer by 1)
< M2 [(6)(7)(4, 4, 3, 1)]
(block-swap)
< M2 [(6)(7)(5, 3, 3, 1)]
(mass-transfer by 1)
< M2 [(6)(7)(5, 3, 4, 0)]
(mass-transfer by 1)
< M2 [(6)(7)(5, 3, 4)]
(delete the zero width block)
< M2 [(6)(7)(5, 4, 3)]
(block-swap)
< M2 [(6)(7)(6, 3, 3)]
(mass-transfer by 1)
= M2 [(6)(7)(4 + 2, 3, 3)] = M3 [(4, 2)(7)(6, 3, 3)]. The last move restores the original block set by switching the block of half-width 6 in the counterbalance with two blocks of half-width 2 and 4. Note that the mass-transfer move is restricted by the requirement that, at each step, mass (i.e., half-width) is transferred up the stack to a block that is no smaller. Also note that the block-swap move would not have been required if all of the blocks had di↵erent 4
half-widths.
The process demonstrated in Example 6.4.4 is now generalised as our proof of the Subset Sum Property 6.4.1. If the reader is already convinced that such a generalisation is possible, then they might like to skip the following detailed argument. Proof of Proposition 6.4.1. By relabelling blocks, we suppose that the absolute maximal overhang is obtained when (w) = (w1 + wm )(w)(wm , . . . , w1 ). We make explicit three assumptions: 1. There is one block in the optimal counterbalance whose half-width w1 + wm 68
is the sum of half-widths of just two blocks not in the counterbalance. The general case is obtained by an obvious induction argument. 2. There are no other blocks in the counterbalance. If there were additional blocks, then these would be irrelevant to the following calculation. 3. The block immediately beneath the protruding block has half-width wm and the bottom block has half-width w1 . If there were additional blocks above or below these blocks, then these would also be irrelevant to the calculation that follows. Since we assume that the partition is optimal, by Theorem 6.1.3 we have that wi
wi
1
for every i 2 {2, . . . , m}. First, if wi = wi
1
for every i 2 {2, . . . , m},
then we can employ one mass-transfer by w1 to improve the overhang since M2 [(w1 + w1 )(w)(w1 , . . . , w1 , w1 )] < M2 [(w1 + w1 )(w)(w1 , . . . , w1 + w1 , 0)] = M3 [(w1 , w1 )(w)(w1 , . . . , w1 + w1 )]. The last step we restores the original block set by deleting the block of zero halfwidth and then switching the single block of half-width w1 +w1 in the counterbalance with two blocks of half-width of w1 . As this move has improved the overhang we obtain a contradiction. If wi > wi
1
for some i 2 {2, . . . , m}, then we set = min {wi 2im
For any 0 < d
wi
1
: wi
wi
1
> 0} > 0.
and all k 2 {2, . . . , m} we will prove that M2 [(wm + w1 )(w)(wm , . . . , wk , . . . , w1 )] < M2 [(wm + w1 )(w)(wm , . . . , wk + d, . . . , w1
d)].
(6.4.1)
We induct on k. The base case follows from a mass-transfer by d from block w1 to block w2 . For the inductive step we first suppose that (6.4.1) holds for some 69
k 2 {2, . . . , m}. Now consider two cases. Case 1. First suppose that wk+1 > wk . Therefore, as wk+1
wk + d, we can
employ a mass-transfer by d to show that M2 [(wm + w1 )(w)(wm . . . , wk+1 , wk + d, . . . , w1 < M2 [(wm + w1 )(w)(wm , . . . , wk+1 + d, wk , . . . , w1
d)] d)].
(6.4.2)
Case 2. Second, if wk+1 = wk , then wk+1 wk + d. Therefore, we can complete a block-swap to see that (6.4.2) still holds. This proves (6.4.1) for any 0 < d
and
all k 2 {2, . . . , m}. We now find the non-negative integer r for which w1 = r + d, where 0 d < . Employing (6.4.1) a finite number of times gives M2 [(wm + w1 )(w)(wm , . . . , w2 , w1 )] < M2 [(wm + w1 )(w)(wm + r , . . . , w2 , w1
r )]
M2 [(wm + w1 )(w)(wm + r + d, . . . , w2 , w1
r
d)]
= M2 [(wm + w1 )(w)(wm + w1 , . . . , w2 , 0)] = M3 [(wm , w1 )(w)(wm + w1 , . . . , w2 )]. The last step restores the original block set by deleting the block of zero half-width and then switching the single block of half-width wm + w1 in the counterbalance with two blocks of half-width of wm and w1 . This again yields a contradiction as we have improved the overhang.
6.5
Optimal partitions for consecutive integers
We now demonstrate how one can systematically and simultaneously employ Propositions 6.2.1 and 6.4.1 to determine the optimal partition for block sets of the form c1,n = {1, . . . , n}. The following example illuminates the approach that will be subsequently generalised. 70
Example 6.5.1. We consider the block set c1,8 = {1, 2, 3, 4, 5, 6, 7, 8}. The largest block of half-width b = 8 must protrude. We now show that the optimal counterbalance cannot contain any block of half-width w
6; for if it did, then the
Subset-Sum Property 6.4.1 ensures that the counterbalance must also contain at least one block from each of the sets {1, w
1} and {2, w
2}. Therefore, if m is the
sum of half-widths of every other block in the counterbalance, then m
1 + 2 = 3.
By Proposition 6.2.1, we have that w b2 /(b + m) 82 /(8 + 3) < 6, which is a contradiction. It is then straight-forward to tabulate the fourteen viable counterbalance sets that adhere to both Propositions 6.2.1 and 6.4.1.
{} {1} {2} {1,2} {1,3} {1,2,3} {1,2,4} {1,2,5} {1,3,4} {1,3,5}
{1,4} {2,3} {2,3,4} {1,2,3,4}.
Table 6.3: The viable counterbalance sets for c1,8 . We can then use Equation (4.1.3) to find the overhang corresponding to each of these counterbalance sets. The optimal counterbalance is easily found to be {1, 2, 3}. 4 By generalising this approach we can improve Corollary 6.1.4 to give a better estimate for the number of viable partitions for block sets of consecutive integers. Proposition 6.5.2. If x is the half-width of the largest block in the optimal counterbalance for c1,n , then
⌅ ⇧ x < 2n2/3 + 2/3 .
(6.5.1)
Proof. The largest block of half-width n must protrude. For convenience, we assume that the half-width of the largest block in the counterbalance is an even integer, 2a. The proof is almost identical when the half-width of the largest block in the counterbalance is an odd integer, and has therefore been omitted.
71
The Subset-Sum Property 6.4.1 ensures that the counterbalance must also contain at least one block from each of the sets {1, 2a
1}, {2, 2a
2}, . . . , {a
1, a + 1}.
Therefore, if m is the sum of half-widths of every other block in the counterbalance, then m
1 + 2 + · · · + (a
1) = a(a
1)/2.
By Proposition 6.2.1, we see that n2 2a () a3 n + a(a 1)/2 Set f (a) = a3
a2 + 2an f 0 (a) = 3a2
a2 + 2an
n2 < 0.
(6.5.2)
n2 . Since 2a + 2n = 3 (a
1/3)2 + 2n
1/3 > 0,
it follows that f (a) = 0 has one real root. By Cardano’s Formula [39, p. 148] we can then show that
q p p 3 27n2 + 3 3 27n4 4n3 18n + 2 + 2/3 p 24 /3 3 2(6n 1) p . p p 3 3 27n2 + 3 3 27n4 4n3 18n + 2
1 2a 22/3 3
As the final term in the above expression is negative, we estimate that q p p 1 2/3 3 2a < 2 27n2 + 3 3 27n4 4n3 18n + 2 + 2/3 3 q p p 1 2/3 3 < 2 27n2 + 3 3 27n4 + 2/3. 3 = 2n2/3 + 2/3.
which gives the desired result. Remark 6.5.3. For the block set c1,75 , Estimate (6.5.1) dictates that the counterbalance has no block whose half-width exceeds 36. Although numerically helpful, this estimate is still clearly not as sharp as we might desire. As seen in Table 6.4, we have determined that the half-width of the largest block in the optimal 4
counterbalance is 9. 72
Estimate (6.5.1) can be used to reduce the computational complexity of Algorithm 6.2.1 when dealing with block sets of consecutive integers. We have produced Table 6.4 with the help of this result. This table records the optimal counterbalance for c1,n , for each 1 n 75. If 1 n 24, then each block in the counterbalance is smaller than every other block not in the counterbalance. The pattern first fails when n = 25, in which case the optimal counterbalance is {1, 2, 3, 4, 5, 6} \ {4}. We have also determined the optimal counterbalance for some larger block sets. Indeed, for c1,100 , a day’s computation was required to determine that the optimal counterbalance is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} \ {6}.
6.6
Conjectures pertaining to optimal partitions
We conclude our investigation of towers of maximal overhang by making two conjectures. The first of these pertains to block sets of consecutive integers. Every example of this type that we studied adhered to the following conjecture. Conjecture 6.6.1. With at most one exception, each block in the optimal counterbalance for cm,n = {m, . . . , n} is smaller than every block not in the counterbalance. If this could be proved, then determining the maximal overhang of cm,n would be polynomial in n. Even if this conjecture could be resolved, we believe that the general case is much more difficult. We formally state this as an open decision problem. Conjecture 6.6.2. For a block set wn of integers and a bound B, determining whether wn can be partitioned to achieve an overhang greater than B is NPcomplete.
73
n
counterbalance
n
counterbalance
n
counterbalance
1
{}
26
{1, 2, 3, 4, 5, 6} \ {3}
51
{1, 2, 3, 4, 5, 6, 7}
2
{1}
27
{1, 2, 3, 4, 5, 6} \ {1}
52
{1, 2, 3, 4, 5, 6, 7, 8} \ {7}
3
{1, 2}
28
{1, 2, 3, 4, 5, 6}
53
{1, 2, 3, 4, 5, 6, 7, 8} \ {7}
4
{1, 2}
29
{1, 2, 3, 4, 5, 6}
54
{1, 2, 3, 4, 5, 6, 7, 8} \ {6}
5
{1, 2}
30
{1, 2, 3, 4, 5, 6}
55
{1, 2, 3, 4, 5, 6, 7, 8} \ {6}
6
{1, 2, 3}
31
{1, 2, 3, 4, 5, 6}
56
{1, 2, 3, 4, 5, 6, 7, 8} \ {5}
7
{1, 2, 3}
32
{1, 2, 3, 4, 5, 6}
57
{1, 2, 3, 4, 5, 6, 7, 8} \ {5}
8
{1, 2, 3}
33
{1, 2, 3, 4, 5, 6}
58
{1, 2, 3, 4, 5, 6, 7, 8} \ {4}
9
{1, 2, 3, 4}
34
{1, 2, 3, 4, 5, 6}
59
{1, 2, 3, 4, 5, 6, 7, 8} \ {4}
10
{1, 2, 3, 4}
35
{1, 2, 3, 4, 5, 6}
60
{1, 2, 3, 4, 5, 6, 7, 8} \ {3}
11
{1, 2, 3, 4}
36
{1, 2, 3, 4, 5, 6}
61
{1, 2, 3, 4, 5, 6, 7, 8} \ {3}
12
{1, 2, 3, 4}
37
{1, 2, 3, 4, 5, 6, 7} \ {6}
62
{1, 2, 3, 4, 5, 6, 7, 8} \ {2}
13
{1, 2, 3, 4}
38
{1, 2, 3, 4, 5, 6, 7} \ {5}
63
{1, 2, 3, 4, 5, 6, 7, 8} \ {2}
14
{1, 2, 3, 4}
39
{1, 2, 3, 4, 5, 6, 7} \ {4}
64
{1, 2, 3, 4, 5, 6, 7, 8} \ {1}
15
{1, 2, 3, 4, 5}
40
{1, 2, 3, 4, 5, 6, 7} \ {4}
65
{1, 2, 3, 4, 5, 6, 7, 8} \ {1}
16
{1, 2, 3, 4, 5}
41
{1, 2, 3, 4, 5, 6, 7} \ {3}
66
{1, 2, 3, 4, 5, 6, 7, 8}
17
{1, 2, 3, 4, 5}
42
{1, 2, 3, 4, 5, 6, 7} \ {2}
67
{1, 2, 3, 4, 5, 6, 7, 8}
18
{1, 2, 3, 4, 5}
43
{1, 2, 3, 4, 5, 6, 7} \ {2}
68
{1, 2, 3, 4, 5, 6, 7, 8}
19
{1, 2, 3, 4, 5}
44
{1, 2, 3, 4, 5, 6, 7} \ {1}
69
{1, 2, 3, 4, 5, 6, 7, 8}
20
{1, 2, 3, 4, 5}
45
{1, 2, 3, 4, 5, 6, 7} \ {1}
70
{1, 2, 3, 4, 5, 6, 7, 8}
21
{1, 2, 3, 4, 5}
46
{1, 2, 3, 4, 5, 6, 7}
71
{1, 2, 3, 4, 5, 6, 7, 8, 9} \ {8}
22
{1, 2, 3, 4, 5}
47
{1, 2, 3, 4, 5, 6, 7}
72
{1, 2, 3, 4, 5, 6, 7, 8, 9} \ {8}
23
{1, 2, 3, 4, 5}
48
{1, 2, 3, 4, 5, 6, 7}
73
{1, 2, 3, 4, 5, 6, 7, 8, 9} \ {7}
24
{1, 2, 3, 4, 5}
49
{1, 2, 3, 4, 5, 6, 7}
74
{1, 2, 3, 4, 5, 6, 7, 8, 9} \ {7}
25
{1, 2, 3, 4, 5, 6} \ {4}
50
{1, 2, 3, 4, 5, 6, 7}
75
{1, 2, 3, 4, 5, 6, 7, 8, 9} \ {6}
Table 6.4: The optimal counterbalance for c1,n = {1, . . . , n} given 1 n 75. 74
Chapter 7 Moments and centres of mass: preliminaries Our analysis of the block stacking problem depended fundamentally on the need to carefully consider the centres of mass of various configurations of blocks in the plane. And in doing so, the physical problem of maximising their overhang was rendered as a mathematical problem. We now invert this dynamic. By carefully arranging blocks and particles in the plane, we will then invoke physical principles to discover and prove various algebraic identities. In short, we will solve mathematical problems by way of physics. Before doing so, we first summarise the elementary physical principles that we will employ in the work to come. The reader may already be familiar with the details that follow, in which case this chapter can be safely skipped.
7.1
Informal definitions
Consider a collection of particles whose masses are m1 , . . . , mn . Let these be located at points x1 , . . . , xn along a horizontal axis, respectively. We think of this axis as being an inflexible and weightless rod suspended in a uniform gravitational field 75
whose gravitational constant is g. For convenience, we can assume that g = 1. If the system is supported below by a pivot located at point x, then each particle to the right of the pivot will impart a clockwise rotational force, while each particle to the left will impart a counterclockwise rotational force. The centre of mass of the system is denoted by x, and this is the location of the pivot for which the individual turning forces will cancel each other out. How might we locate this point?
m1
m2
x1
x2
x
m3 m4
m5
x3 x4
x5
Figure 7.1: This configuration of particles is in rotational equilibrium when balanced at x. The turning force that we allude to is often called torque, and this was first described by Archimedes in his book titled On the Equilibrium of Planes [7]. Consider a particle of mass mi located at position xi . The magnitude of the torque exerted by this particle about a pivot at point x is the product mi (xi and its displacement xi
x) of its mass mi
x from the pivot. The net torque ⌧x about the point x is
then ⌧x =
n X
mi (xi
x).
i=1
Although Archimedes’ language and notation was vastly di↵erent to what you see here, he was the first to theoretically show that such a system will be in rotational equilibrium about the point x if and only if the net torque about x is zero: n X
mi (xi
x) = 0.
i=1
This equation can easily be recast to deduce that the centre of mass is located at point
Pn mi xi x = Pi=1 . n i=1 mi 76
(7.1.1)
7.2
Formal definitions
Nowadays it is more fashionable to formally define centre of mass by (7.1.1) and then recognise that it corresponds to something physically meaningful. Although such formality serves to place these concepts on a firmer mathematical footing, we encourage the reader to keep in mind their physical interpretation. Definition 7.2.1 (systems of particles, signed systems of particles, subsystems). A system of particles (or, more simply, a system) in Rm is a finite multiset µ = {(mi , xi )}i2A such that xi 2 Rm and mi > 0 for every i 2 A. A signed system of particles is defined the same way, but without the requirement that mi > 0 for all i 2 A. We will say that (mi , xi ) is a particle belonging to µ. We call mi the mass of this particle, and xi the position of this particle. The total mass of µ is P M (µ) = i2A mi . Finally, if ✓ µ, then is called a subsystem of µ. We will primarily consider systems in R and, less often, R2 . Systems in R can
be depicted by placing particles along a horizontal axis. If two particles occupy the same position, then we can represent this by shifting one or more of these particles vertically. For instance, the signed system µ = {(1, 1), (1, 2), ( 2, 2), (3, 4)} can be depicted as shown in Figure 7.2 below.
2 +1 +1 1
+3
2
3
4
Figure 7.2: The signed system µ = {(1, 1), (1, 2), ( 2, 2), (3, 4)}. Definition 7.2.2 (The first moment of a signed system). If µ = {(mi , xi )}i2A is a signed system, then the first moment of µ is defined by m(µ) =
X i2A
77
mi x i .
(7.2.1)
If each mass in a system is positive, then we also have the following useful definition. Definition 7.2.3 (The centre of mass of a system). If µ = {(mi , xi )}i2A is a system, then the centre of mass of µ is defined by P mi x i m(µ) x(µ) = = Pi2A . M (µ) i2A mi
(7.2.2)
Note that if µ is a system in R2 , then Equation (7.2.2) can be decoupled. Indeed, if xi = (xi , yi ) for all i 2 A, then x(µ) = (x(µ), y(µ)) where P P mi yi i2A mi xi x(µ) = P and y(µ) = Pi2A . i2A mi i2A mi
7.2.1
(7.2.3)
Fundamental properties of the centre of mass
Centres of mass have many properties that we will find helpful in the work that follows. These properties are summarised below. If the proofs of these are obvious to the reader, then they can be safely ignored. We have nonetheless included them for the sake of completeness. Although many of the results in this section generalise in obvious directions, we find no reason to give, or prove, the more general statements as they serve no purpose for this thesis. Property 7.2.4 (Translation Property). If a system is translated by some vector, then its centre of mass is translated likewise. Proof. We translate the system µ = {(mi , xi )}i2A by the vector a, giving the trans= {(mi , xi + a)}i2A . It follows that P P P a i2A mi i2A mi (xi + a) i2A mi xi x( ) = = + = x(µ) + a. M( ) M (µ) M (µ)
lated system
Property 7.2.5 (Rotation Property). If a system in R2 is rotated through a given angle about a given point, then the centre of mass of the system is rotated likewise.
78
Proof. By applying the Translation Property 7.2.4, we can assume that the system µ = {(mj , xj )}j2A is rotated by angle ✓ about the origin. It is convenient to think of each position xj as a complex number, in which case the rotated distribution is = {(mj , xj ei✓ )}j2A . It follows that P P i✓ j2A mj xj e j2A mj xj i✓ x( ) = =e = ei✓ x(µ). M( ) M (µ) Properties 7.2.4 and 7.2.5 affirm that the centre of mass is an intrinsic property of a system of particles in that it does not depend on the choice of coordinates. Definition 7.2.6. A point c 2 Rm is a centre of symmetry of the signed system µ = {(mi , xi )}i2A if there is a bijection
: A ! A such that, for each i 2 A,
i 7! (i) = j where mi = mj and xi is the reflection of xj in the point c. That is, if
xi +xj 2
= c. If such a point c exits, then we say that µ is point symmetric about c.
Property 7.2.7 (Centre of mass of a point symmetric system). If a system is point symmetric, then its centre of mass is located at the centre of symmetry. Proof. Suppose that c is the centre of symmetry of µ = {(mi , xi )}i2A . Then there is some bijection and
xi +xj 2
: A ! A such that, for each i 2 A, i 7! (i) = j where mi = mj
= c. Therefore, mi xi + mj xj = 2mi c.
(7.2.4)
Summing each side of (7.2.4) over every i 2 A shows that x(µ) = c. Example 7.2.8. The system µ = {(1, 1), (2, 2), (3, 3), (2, 4), (1, 5)} shown below in Figure 7.3 is symmetric about the point 3. Therefore, x(µ) = 3.
+1 +2 +3 +2 +1 1
2
3
4
5
Figure 7.3: The system µ is symmetric about the point 3.
79
4
Definition 7.2.9. A straight line L ⇢ R2 is an axis of symmetry of the signed system µ = {(mi , xi )}i2A if there is a bijection f : A ! A such that, for each i 2 A, i 7! (i) = j where mi = mj and xi is the mirror image of xj in the line L. If such a line L exits, then we say that µ is line symmetric about L. Property 7.2.10 (Centre of mass of a line symmetric system). If a system in R2 is line symmetric, then its centre of mass is on its line of symmetry. Proof. Suppose that L is a line of symmetry of µ = {(mi , xi )}i2A . According to Properties 7.2.4 and 7.2.5 we can assume that L is the vertical axis, x = 0. Therefore, if xi = (xi , yi ), then x = 0 is a centre of symmetry of the system {(mi , xi )}i2A in R. By Property 7.2.7, this implies that x(µ) = 0. Therefore, x(µ) = (x(µ), y(µ)) = (0, y(µ)), in which case the centre of mass lies on the axis of symmetry. Property 7.2.11 (Subsystem substitution). The centre of mass of a system does not change when one of its subsystems is replaced by a single particle located at the centre of mass of the subsystem with mass equal to the total mass of the subsystem. Proof. Any subsystem
of µ = {(mi , xi ))}x2A can be written in the form
=
{(mi , xi ))}x2B where B ✓ A. Therefore P P P P m x + m x M ( )x( ) + m x i i i i i i i2B i2A\B i2A\B mi xi x(µ) = i2A = = , M (µ) M (µ) M (µ) which gives the result. The following proposition is perhaps the most important property of the centre of mass. This is because it allows one to find the centre of mass of a system by strategically partitioning the system into subsystems whose centres of mass may be more easily found. Property 7.2.12. If a system is partitioned into disjoint subsystems, then the centre of mass of the system is equal to the mass-weighted mean (7.2.5) of the centres of mass of the subsystems, and is located in their convex hull. 80
Proof. Let µ = {(mi , xi )}i2A be a system and suppose that {Aj }j2B is a partition of A. For each subsystem µj = {(mi , xi )}i2Aj of µ, denote its total mass by mj = M (µj ) and its centre of mass by xj = x(µj ). We will prove that P j2B mj xj x(µ) = . M (µ)
(7.2.5)
To this end,
P P P P m(µ) j2B i2Aj mi xi j2B mj xj i2A mi xi x(µ) = = = = . M (µ) M (µ) M (µ) M (µ) P mj Finally, as j2B M (µ) = 1, Equation (7.2.5) implies that x(µ) lies in the convex hull of the centres of mass, {xj (µ)}j2B .
7.2.2
Fundamental properties of the first moment
If a signed system has a particle with negative mass, then the aforementioned properties do not apply. In this case, the centre of mass is not even defined. However, the first moment of a signed system is always defined, and this has various properties that will also prove useful. Property 7.2.13 (Point symmetric subsystems). If a signed system has a point symmetric subsystem, then the first moment of the system does not change when the subsystem is replaced by a single particle located at the centre of symmetry of the subsystem with mass equal to the total mass of the subsystem. Proof. Every subsystem
of µ = {(mi , xi )}i2A is of the form
= {(mi , xi )}i2B
for some B ✓ A. If c is a centre of symmetry of , then there is some bijection : B ! B such that, for each i 2 B, i 7! (i) = j where mi = mj and
xi +xj 2
= c.
Therefore, mi xi + mj xj = 2mi c.
(7.2.6)
Summing each side of (7.2.6) over each i 2 B shows that m( ) = M ( )c. Therefore, m(µ) =
X i2A
mi x i =
X
i2A\B
mi x i +
X i2B
mi xi = m(µ \ ) + m( ) = m(µ \ ) + M ( )c, 81
which proves the result. Example 7.2.14. To illustrate Property 7.2.13, in Figure 7.4 we have depicted the signed system µ = {(1, 1), ( 2, 2), (3, 3), ( 2, 4), (1, 5)}. As this system is symmetric about the point 3, we can find its first moment by concentrating its total mass at its centre of symmetry to give m(µ) = (1
2 + 1) ⇥ 3 = 1 ⇥ 3.
2+3
4
+1 2 +3 2 +1
2
+3
2
+1
1 2 3 4 5
+1
+1
1 2 3 4 5
1 2 3 4 5
Figure 7.4: The mass can be concentrated at the point of symmetry.
Finally, Property 7.2.15 below is obvious but helpful. It asserts that the first moment of a system is unchanged when a subsystem is deleted, provided that the first moment of that subsystem is zero. Any subsystem
of µ for which m( ) = 0
is called a zero-moment subsystem of µ. Property 7.2.15 (Subsystem deletion). If
is a zero-moment subsystem of µ,
then m(µ) = m(µ \ ). Proof. If m( ) = 0, then m(µ) = m(µ \ ) + m( ) = m(µ \ ).
7.2.3
The centre of mass of a planar region
If mass is distributed along a curve, or in some region of space, then total mass and centre of mass are defined using integrals rather than sums. These more general definitions can be found in many calculus textbooks (see, e.g., [67, pp. 329–346]). 82
Since such generalisations are required in Chapters 10 and 11, we now provide a cursory summary of the required theory. The centre of mass of a planar figure R ⇢ R2 of uniform density and area A is the mean position (x, y) of each point in R, defined by the integrals ˜ ˜ x dxdy y dxdy x= R and y = R . A A
(7.2.7)
Each of the properties of the centre of mass listed above carry over to the continuous setting, and their proofs are more or less identical. For instance, if {Ri }ni=1 is a partition of R, then its centre of mass can be found by first finding the centre of mass (xi , y i ) and area Ai of each part and then computing Pn Pn Ai y i i=1 Ai xi x= and y = i=1 . A A That is, the centre of mass of a region is well behaved with respect to decomposition, in the sense that it can be found as a mass-weighted mean of its constituent parts. The proof follows at once from the definition of the centre of mass and the calculation, x=
˜
R
x dxdy = A
Pn ˜ i=1
Ri
A
x dxdy
=
Pn
Ai x i . A
i=1
In much of the work that follows, we will only be interested in the x-component of the centre of mass. We call this the x-mean. Having now established the required physics, our new line of inquiry begins with an elaboration of work that we recently published in [77] and [78].
83
Chapter 8 Sum of powers using physical arguments Many mathematicians enjoy the demonstration of a mathematical identity by way of a simple diagram without accompanying text. These are called proofs without words and various books and countless articles are devoted to presenting the best of these, most notably [3], [54] and [56]. For instance, Figure 8.1 provides a well-known demonstration of the formula for the nth triangular number, Tn := 1 + 2 + · · · + n =
n(n + 1) . 2
(8.0.1)
Figure 8.1: A proof without words for a sum of consecutive integers. Readers may be surprised to know that the above configuration can be supplanted by a simpler figure if we allow a little physics. 84
8.1
On sums and alternating sums of integers
In this chapter we first provide a physical derivation of the formula for the the sum of consecutive integers, and then a variety of other integer summations. Many of the diagrams used in support of our arguments can be considered as alternatives to various well-known proofs without words of the same results.
8.1.1
Sums of consecutive integers
We consider the system µ = {(1, i)}ni=1 , which places a particle of unit mass at each point i 2 {1, . . . , n} on the number line. This is shown in Figure 8.2 below.
1
2 ··· n
Figure 8.2: Particles of unit mass arranged in a line.
Since this system has a centre of symmetry, its centre of mass x(µ) will be the midpoint
n+1 2
of the interval from 1 to n. The centre of mass can also be found
using Equation (7.2.2) so that 1 ⇥ 1 + 1 ⇥ 2 + ··· + 1 ⇥ n n+1 = , n 2 giving Equation (8.0.1). Note that the argument can also be easily adapted to find the sum of any arithmetic sequence; we simply place a particle of unit mass at each of the n points: a, a + d, . . . , a + (n
1)d.
We have not provided this demonstration because we regard it as an improvement of the proof implied by Figure 8.1. However, it does exemplify a fruitful approach to deducing various formulas, some familiar and others less-so. This approach is not new; Archimedes, most famously, determined the area and volume of various figures using similar arguments. This approach is fully expounded in The 85
Method of Archimedes, which can be found in [7]. However, the application of this approach in a discrete setting does not appear to have been adequately investigated.
8.1.2
Alternating sums of consecutive integers
We can use a variation of the above argument to physically demonstrate the formula for the alternating sum of integers,
1
2 + · · · + ( 1)n+1 n =
8 > : k, if n = 2k.
Of course, the result is algebraically trivial. Our intention is to simply demonstrate an approach that also works for more difficult alternating sums. The main idea is to utilise a signed system that allows for particles of negative mass.
Case 1 (n = 2k places 2k
1 1). We define a signed system µ = {(( 1)i+1 , i)}2k i=1 that
1 particles of alternating unit mass along the horizontal axis. This can
be seen below in Figure 8.3. The first moment of this system is the desired sum, P 1 i+1 m(µ) = 2k i. i=1 ( 1) +1 0
1 +1
1 ···
1 +1
k · · · 2k 1
+1 0 Figure 8.3: 2k
k
1 particles of alternating unit mass on a line.
The system has total mass M (µ) = k ⇥ 1 + (k
1) ⇥ ( 1) = 1 and is symmetric
about the point k. Therefore, by Property 7.2.13, we can find its first moment by 86
concentrating its total mass at its point of symmetry. This gives 2k X1
( 1)i+1 i = k.
i=1
Case 2 (n = 2k). We likewise define a signed system µ = {(( 1)i+1 , i)}2k i=1 , as shown below in Figure 8.4. The first moment of the system is the desired sum, P i+1 m(µ) = 2k i. i=1 ( 1) +1 0
1
1 +1 ···
1 +1
· · · 2k 1 2k
k
1
+1 0
1
2k
k
Figure 8.4: 2k particles of alternating unit mass on a line. The subsystem of µ comprising the first 2k
1 particles is symmetric about the
point k. Therefore, by Property 7.2.13, we can replace this subsystem with a single particle located at the point k with mass equal to the total mass of the subsystem, M (µ) = k ⇥ 1 + (k
1) ⇥ ( 1) = 1. It follows that 2k X i=1
8.2
( 1)i+1 i = 1 ⇥ k + ( 1) ⇥ 2k =
k.
On sums and alternating sums of squares
There are various well-known proofs without words for the sum of squares of consecutive natural numbers. For instance, Goldini [32] supplies a rather elaborate series of pictures to prove of the familiar identity, 12 + 2 2 + · · · + n2 = 87
n(n + 1)(2n + 1) . 6
(8.2.1)
Although more concise proofs are given by Alsina and Nelson [4] and Siu [71], each of these requires more than one picture to convey the idea. Can we provide a onepicture proof if we permit a little physics? A note by Lord [48] provides a clue to one such approach. His argument proceeds as follows.
n 3
(x, y)
2 1 0
1
2
3
n
Figure 8.5: Unit masses arranged in a triangular array
Arrange particles of unit mass in a triangular array as shown above in Figure 8.5, and let (x, y) 2 R2 be the centre of mass of this system. Now consider the rightangled triangle indicated on the same figure. Since each vertical and horizontal line of particles has a centre of mass on a median of this triangle, point (x, y) corresponds with intersection of the triangle’s medians. That is, at its centre of mass ✓ ◆ 2n + 1 n + 2 (x, y) = , . 3 3 On the other hand, for each i 2 {1, . . . , n}, there are i particles of unit mass located at point i. Therefore, the x-coordinate of the centre of mass can also be found by using Equation (7.2.5). It follows that 1 ⇥ 1 + 2 ⇥ 2 + ··· + n ⇥ n 2n + 1 = , 1 + 2 + ··· + n 3 from which we deduce Equation (8.2.1).
88
Perhaps the proof’s only flaw is that it might assume a little too much physics, and this makes the explanation subsequent to the diagram crucial. Specifically, Lord’s argument depends crucially on two facts. First, to locate (x, y), one must assume that the centre of mass of a triangle is located at the intersection of its medians. This is not hard to prove and, in the vein of Archimedes’ Method, we have given a calculus-free proof of this fact in [78]. Alternatively, we can recommend [6] for an argument that also avoids the use of calculus. Levi [47] also gives an informal argument that might be acceptable to a physicist. Secondly, Lord’s proof assumes that if a system of particles is divided into parts, and if the centre of mass of each of these parts lies on a given line, then the centre of mass of the entire system lies on the same line. This is a believable result, but it does require a little further thought and substantiation. Hint: use Property 7.2.12.
8.2.1
Sums of squares: An alternative configuration
We have found a new configuration of particles that obviates the need for explanation and which assumes a little less. We refer the reader to Figure 8.6 below. Before reading the subsequent explanation you might want to see for yourself how this configuration of particles provides a proof without words of Equation (8.2.1).
89
2 3
1
2
3
n
···
i
12 + 2 2 + · · · + n2 = 1 + 23 (n 1 + 2 + ··· + n
1)
Figure 8.6: A physical demonstration of the sum of squares of integers. The words. In case Figure 8.6 is insufficient, here’s the explanation. Recall that we use the term x-mean to refer to the x-coordinate of the centre of mass of a system of particles. This particular system of particles is line symmetric, so its x-mean coincides with the x-coordinate of its topmost particle, 2 x = 1 + (n 3
1) =
2n + 1 . 3
On the other hand, for each i 2 {1, . . . , n}, the ith diagonal comprises i particles of unit mass centred at i. Therefore, x can be found using Equation (7.2.5), giving the equation indicated in the figure. This is equivalent to the more familiar expression, 12 + 2 2 + · · · + n2 =
n(n + 1)(2n + 1) . 6
Figure 8.6 can be easily adapted to demonstrate the formula for the sum of triangular numbers.
8.2.2
Sums of triangular numbers
In Section 8.1 we physically demonstrated that the ith triangular number is Ti = i(i+1) . 2
The literature contains an assortment of proofs without words for the sum of 90
the first n triangular numbers,
Pn
i=1
Ti (see, e.g., [53], [63], [80], and [82]). We can
give a new physical derivation of this summation by first translating the horizontal axis in Figure 8.6 by one unit to the left, giving Figure 8.7 below. We then use precisely the same argument to obtain the equation indicated in Figure 8.7. This is equivalent to the more familiar expression, T1 + T2 + · · · + Tn =
n(n + 1)(n + 2) . 6
2 3
2
3
4
···
j
n+1
1 ⇥ 2 + 2 ⇥ 3 + · · · + n ⇥ (n + 1) = 2 + 23 (n 1 + 2 + ··· + n
1)
Figure 8.7: A physical demonstration of the sum of triangular numbers.
8.2.3
Alternating sums of squares
The formula for the alternating sum of squares is well-known to be 12
22 + · · · ± n2 = ±(1 + · · · + n) = ±Tn .
(8.2.2)
Of course, this has an obvious and elementary algebraic proof, and this corresponds to the proof without words demonstrated by Snover in [72]. Other authors have also found similar proofs for related summations (e.g., [62] and [25]). We will now provide a physical proof of (8.2.2). As before, the trick is to consider a signed 91
system of particles.
Case 1 (n = 2k
1). We first define a signed system µ that places i particles of
mass ( 1)i+1 at position i, for each i 2 {1, . . . , 2k
1}. This is depicted in the
left-hand side of Figure 8.8. The first moment of this system is an alternating sum P 1 i+1 2 of squares, m(µ) = 2k i. i=1 ( 1) +1
+1
1 +1 1 +1 1 +1
1 +1
1 +1
1
+1 +1 1
+1 1
+1
+1
+1 1
+1 1
+1
+1
1 +1
+1 1 +1
1
2
3
4
5 · · · 2k
+1 1
1
2
3
4
5 · · · 2k
1
Figure 8.8: An alternating sum of an odd number of squares. On the other hand, to evaluate m(µ), we note that with the exception of k particles of mass +1 located at 2k 1, each particle can be placed in a zero-moment subsystem of four particles. These zero-moment subsystems have been depicted in Figure 8.8 using red lines. Using Property 7.2.15, we may delete each of these zero-moment subsystems to show that 2k X1
( 1)i+1 i2 = k(2k
1) =
i=1
2k(2k 2
1)
= T2k 1 .
(8.2.3)
Case 2 (n = 2k). We likewise define a signed system µ by placing i particles of mass ( 1)i+1 at position i, for each i 2 {1, . . . , 2k}. This is depicted in the left-hand 92
side of Figure 8.9. The first moment of this system is also an alternating sum of P i+1 2 squares, m(µ) = 2k i. i=1 ( 1) 1
1
+1
+1 1 +1 1 +1
+1 1
+1
+1 1
+1
+1
+1
+1
+1 1
+1
+1
1
1
1
1
1
+1
1
1
1
1
1
1
1
1
1 +1
3
+1 1
1 2
+1 1
+1
+1
1
+1
4 · · · 2k
1
2
3
4 · · · 2k 2k + 1
Figure 8.9: An alternating sum of an even number of squares. To find m(µ), we now append k particles of mass +1 at position 2k + 1 to the system. These particles are depicted as orange circles. Now each particle in the enlarged system can be placed in a zero-moment subsystem of four particles. It follows that k(2k + 1) +
2k X
( 1)i+1 i2 = 0,
i=1
in which case 2k X
( 1)i+1 i2 =
k(2k + 1) =
i=1
2k(2k + 1) = 2
T2k .
(8.2.4)
Combining Equations (8.2.3) and (8.2.4) gives (8.2.2).
8.2.4
A demonstration with particles of positive mass
We note that the use of particles with negative mass can avoided altogether, if we’re willing to accept a less elegant argument. Suppose m > 0 and that the 93
particle ( m, x) belongs to some signed system µ. If we substitute this for the particle (m, x) of positive mass, then the first moment of µ is unchanged. With this in mind, the demonstration of the formula for the sum of alternating squares seen in the previous subsection can be recast using particles of positive mass as follows. We begin with the system shown in Figure 8.10 (a). For each i 2 {1, . . . , 2n
1}
we have placed i particles of positive unit mass at position ( 1)i+1 i. The first P 1 i+1 2 moment of this system is the desired sum, m(µ) = 2k i. i=1 ( 1)
On the left-hand side of the lever we shift equal sized groups of equal mass
particles in opposite directions so that the first moment of the system does not change. This process is seen in Figure 8.10 (a) and the result is seen in Figure 8.10 (b). This yields a zero-moment subsystem that is symmetric about the origin, depicted in blue. Deleting this zero-moment subsystem leaves k particles of unit mass at position 2k
1, giving the desired formula, 2k X1
( 1)i+1 i2 = k(2k
1).
i=1
The analogous result for the sum of an even number of alternating squares can be obtained likewise.
94
(a) (2k 1) 6 5 4 3 2 1 0 1 2 3 4 5 6 2k 1
(b) (2k 1) 6 5 4 3 2 1 0 1 2 3 4 5 6 2k 1
k (c) 0
Figure 8.10:
P2k
i=1
1
( 1)i+1 i2 = k(2k
95
2k 1
1)
8.2.5
Alternating sums of triangular numbers
Various authors have also given proofs without words for alternating sums of triangular numbers. For example, Nelson [55] uses a simple configuration of points to show that
2k X1
( 1)i+1 Ti = k 2 .
(8.2.5)
i=1
Plaza [61] subsequently used a similar argument to show that 2k X
( 1)i Ti = 2Tk .
(8.2.6)
i=1
We note that a demonstration of Equations (8.2.5) and (8.2.6) can be found by simply translating the axes in each of Figures 8.8 and 8.9 by one unit to the left.
8.3
On sums and alternating sums of cubes
There are also various well-known proofs without words that yield the formula for the sum of cubes of consecutive integers (see, e.g., [4], [30], [40] and [45]). We now give a physical proof of the same result. Once again, the reader is invited to establish the formula by inspection of Figure 8.11 below.
96
8.3.1
Sums of cubes 1
2
0
i
n
x = 12 (1 + 2 + · · · + n) 1 + 2 + · · · + n 1 2 i 2
1
2
0
i Ti
1 3 1 2
1
xi
n Ti
+ 12 23 + · · · + 12 n3 = 12 (1 + 2 + · · · + n) 1 + 2 + ··· + n
Figure 8.11: Physically determining the sum of cubes.
The words. Along the horizontal axis we place one particle of unit mass, then two particles of unit mass, and so on, concluding with n particles of unit mass, as shown in Figure 8.11 above. As the system of particles is symmetric, its centre of mass coincides with its midpoint, 1 x = (1 + 2 + · · · + n). 2 On the other hand, the ith subsystem has total mass i and is centred at xi = 1 (Ti 1 2
+ Ti ) = 12 i2 . Therefore, the centre of mass can also be found using Equation
(7.2.5), giving 1 ⇥ 12 12 + 2 ⇥ 12 22 + · · · + n ⇥ 12 n2 1 = (1 + 2 + · · · + n). 1 + 2 + ··· + n 2 This is equivalent to its more familiar form, 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2 . 97
In Chapter 11, precisely the same method will be further exploited to efficiently find various sums of products of Fibonacci and Lucas numbers.
8.3.2
Alternating sums of cubes
The formula for the alternating sum of cubes is best written in the form, 8 > : k 2 (4k + 3) if n = 2k.
Although this is algebraically easy to prove, we could not find a corresponding proof without words in the literature. We now o↵er two seperate physical demonstrations. Again, the reader is invited to establish the formula using Figure 8.12 alone. Alternating sums of cubes: Demonstration I. Case 1 (n = 2k
1). 1 2 i 2
1
2
+2
0
2
Ti
T1
+2
2
2
2
+2
+2
+2
2
2
2
Ti
1
+2
2k
···
i
+2
+2
2
T2k
2
2
2
2
+2
+2
1
+2
+2
+2
T2k
2
+2
+2
+2
+2
1
+2
0
T2k
1
k +2
0
T2k
+2
1
Figure 8.12: An alternating sum of an odd number of cubes. 98
+2
k 2
T2k
1
The words. Along the horizontal axis we have placed one particle of mass +2, and then two particles of mass
2, and so on, concluding with 2k
mass +2. The ith subsystem of particles is centred at xi = 12 (Ti
1
1 particles of + Ti ) = 12 i2 and
has total mass ( 1)i+1 2i. This is depicted in Figure 8.12. The desired sum is the P i+1 3 first moment of this system, m(µ) = 2k i. i=1 ( 1)
Aside from the final k particles, the system can be partitioned into zero-moment
subsystems. These are indicated using the red rectangles in Figure 8.12. We delete each of these zero-moment subsystems. The particles that remain have total mass 2k and are centred at the point T2k system is 2k X1
( 1)
✓
i+1 3
i = 2k T2k
i=1
Case 2 (n = 2k).
k . 2
1
1
Therefore, the first moment of this k 2
◆
= k 2 (4k
3).
1 2 i 2
1 +2
2 2
3 2
0 T1
+2
+2
T2
i +2
Ti
2
2
2
2
+2
+2
··· +2
2k +2
Ti
1
+2
2
T2k
1
2
2
2
2
2
T2k
k +2
2
2
+2
+2
+2
2
2
2
2
+2
+2
0
+2
+2
+2
2
2
2
2
2
2
+2
+2
+2
T2k T2k + k 2
Figure 8.13: An alternating sum of an even number of cubes
The words. Once again, along the horizontal axis we place one particle of mass +2, and then two particles of mass particles of mass
2, and so on, concluding with a placement of 2k
2. The ith subsystem of particles is centred at xi = 12 (Ti 1 +Ti ) = 99
1 2 i 2
and has total mass ( 1)i+1 2i. This is shown in Figure 8.13. Again, the desired P i+1 3 sum is the first moment of this system, m(µ) = 2k i. i=1 ( 1)
We now append k particles of mass +2 to the end of this system. These are
shown above in orange. They have total mass of 2k and are centred at T2k + k2 . The augmented system can now be partitioned into zero-moment subsystems. These are indicated using red rectangles. Therefore, 2k X
( 1)
✓
k i + 2k T2k + 2
i+1 3
i=1
◆
= 0.
From this, we readily obtain 2k X
( 1)i+1 i3 =
k 2 (4k + 3).
i=1
Alternating sums of cubes: Demonstration II We now give an alternative demonstration that yields the same result. Pleasingly, this method generalises to yield a closed formula for more general alternating sums. For this demonstration, we will require the identity 1 + 2 + · · · + (n
1) + n + (n
1) + · · · + 2 + 1 = n2 .
(8.3.1)
The usual demonstration of this identity is given in [54, p. 74], and is reproduced below in Figure 8.14. One simply observes that the total number of dots is equal to the sum of dots in each column.
Figure 8.14: 1 + 2 + · · · + (n
1) + n + (n 100
1) + · · · + 2 + 1 = n2
It is worth noting that we can physically demonstrate the same identity using moments. Define a system µ that places one particle of unit mass at each of the points 0 and n, and two particles of unit mass each each of the points i 2 {1, . . . , n 1}. This can be seen below in Figure 8.15.
n
1
n+1
0
1 ··· n
1n
Figure 8.15: 1 + 2 + · · · + (n
0 1) + n + (n
n 2
n
1) + · · · + 2 + 1 = n2 .
Notice that the first moment of µ is the desired sum, m(µ) = 1 · 0 + 2 · 1 + · · · + 2 · (n
1) + 1 · n.
On the other hand, the system is symmetric about the point
n . 2
Therefore, using
Property 7.2.13, we can find the first moment of µ by concentrating its total mass n + 1 + (n
1) = 2n at the point n2 . Therefore, 1 · 0 + 2 · 1 + · · · + 2 · (n
1) + 1 · n = 2n ⇥
n = n2 . 2
With Equation (8.3.1) in mind, we now begin our second physical demonstration that yields the formula for the alternating sum of cubes.
101
Case 1 (n = 2k). 2k 5
(a)
+5
5
+4
4
4
+4
4
3
+3
3
+3
3
+3
3
+2
2
+2
2
2
+2
2
+2
2
1
+1
1
+1
1
+1
1
+1
1
+1
1
1
2
3
4
2k 2k 1
4 +3
2 +1
(b)
· · · 2k 2k+1
1
2
3
4
· · · 2k 2k+1
Figure 8.16: An alternating sum of an even number of cubes. The words. We define a signed system µ as follows. At each point i 2 {1, 2, . . . , 2k} we place 2i
1 particles whose masses have magnitudes 1, 1, 2, 2, . . . , i
1, i
1, i
and sign ( 1)i+1 . This is depicted in Figure 8.16 (a). Using Equation (8.3.1), we note that the total mass of the subsystem located at point i is ( 1)i+1 i2 . That is, at each consecutive position the total mass is a square of alternating sign. Therefore, the first moment of µ is the desired alternating sum of cubes, P i+1 3 m(µ) = 2k i. i=1 ( 1)
To this system, we now append k particles at the point 2k + 1, whose masses are
1, 3, . . . , 2k
1. These are depicted as orange circles. We have used red connect-
ing lines to indicate the symmetric and zero-moment subsystems of four particles. 102
Deleting each of these subsystems leaves k particles with masses
2, 4, . . . ,
2k,
each located at position 2k. It follows that (2k + 1)[1 + 3 + · · · + (2k
1)] +
2k X
( 1)i+1 i3 = 2k( 2
4
i=1
···
2k), (8.3.2)
from which it easily follows that 2k X
( 1)i+1 i3 =
k 2 (4k + 3).
i=1
Case 2 (n = 2k
1). 2k 1
+4
(a)
4
+4
3
+3
+3
3
+3
+2
2
+2
2
+2
2
+2
1
+1
1
+1
+1
1
+1
1
+1
1
2
3
2k 1
2k 2
+3
2
+1
(b)
· · · 2k 1 2k
1
2
3
· · · 2k 1 2k
Figure 8.17: An alternating sum of an odd number of cubes.
The words. We likewise define a signed system µ as follows. At each point i 2 {1, 2, . . . , 2k
1} we place 2i
1 particles. These particles have masses of
magnitude 1, 1, 2, 2, . . . , i
103
1, i
1, i
and sign ( 1)i+1 . This is shown in Figure 8.17 (a). Once again, the total mass of the subsystem located at point i is ( 1)i+1 i2 . Therefore, the first moment of µ is P 1 i+1 3 the required sum, m(µ) = 2k i. i=1 ( 1)
To this system we append k 1 particles at point 2k, whose masses are 2, 4, . . . ,
2(k
1). These are depicted as orange circles. In Figure 8.17 (b) we have indi-
cated the symmetric and zero-moment subsystems of four particles using red lines. Deleting each of these subsystems leaves k particles with masses 1, 3, . . . , 2k each located at position 2k 2k[2 + 4 + · · · + 2(k
1,
1. It follows that
1)] +
2k X1
( 1)i+1 i3 = (2k
i=1
1)[1 + 3 + · · · + (2k
1)], (8.3.3)
from which we readily obtain 2k X1
( 1)i+1 i3 = k 2 (4k
3).
i=1
8.3.3
A demonstration with positive masses
The demonstration shown above can also be recast using a system of particles whose masses are all positive. We refer the reader to Figure 8.18 (a). We have depicted a system µ for which particles of mass 1, 1, 2, 2, . . . , i
1, i
1, i are placed at point
( 1)i+1 i, for each i 2 {1, 2, . . . , 2k}. The total mass at each point ( 1)i+1 i is equal to i2 . Therefore, the first moment of this system is equal to the desired sum, P i+1 3 m(µ) = 2k i. i=1 ( 1) As indicated in Figure 8.18 (a) using arrows, we then shift pairs of particles with
equal mass in opposite directions by one unit. Note that none of these shifts will a↵ect the first moment of the system. This yields the (almost symmetric) system depicted in Figure 8.18 (b). We delete each pair of particles with equal mass that are symmetric about 0. This yields the system depicted in Figure 8.18 (c). All that remains are k particles with masses 2, 4, . . . , 2k at point
104
2k, and k particles with
masses 1, 3, . . . , 2k 2k X i=1
( 1)i+1 i3 =
1 at point
(2k + 1). It follows that
2k(2 + 4 + · · · + 2k)
105
(2k + 1)(1 + 3 + · · · + 2k
1).
2k 2k 1
(a) (2k+1)
2k 1
2k 1
4
4
4
4
4
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
2k
5
4
3
1
2
1 0
1
2
3
4
2k 2k 1
(b)
5 2k
2k 1
4
3
2k 1
4
3
4
3
3
4
3
3
2
1
2
1
1
2
1
4
1
2
1
2
1
2
(2k+1)
2k
5
3
4
3
2
2
1
1 0
1
2
1
2
1
1
2
3
4
5 2k
2k 1 2k
(c)
3
4
1
2
(2k+1)
Figure 8.18:
P2k
i=1 (
2k
1)i+1 i3 =
0
2k(2 + 4 + · · · + 2k)
106
(2k + 1)(1 + 3 + · · · + 2k
1)
8.3.4
A generalisation
The argument given in Section 8.3.2 can be easily generalised. For any given sequence (ai )ni=1 we will define the sequence of partial sums (Ai )ni=0 by A0 = 0 and Ai = a1 + · · · + ai for i > 0. We then define a signed system µ = {(mi , i)}ni=1 such that mi = ( 1)i+1 (Ai + Ai 1 ). Note that Ai + Ai
1
= (a1 + a1 ) + (a2 + a2 ) + · · · + (ai
1
+ ai 1 ) + ai .
Therefore, at each point i 2 {1, 2, . . . , n}, we can think of this system as comprising 2i
1 separate particles whose masses have magnitudes a1 , a1 , a2 , a2 , . . . , ai 1 , ai 1 , ai
and sign ( 1)i+1 . This system is shown below in Figure 8.19, assuming that n is odd. an a4 a4 a3 a3 a2
a3
.. .
a2
a2
a2
a2
a1
a1
a1
a1
a1
a1
a1
1
2
3 ··· n
Figure 8.19: A signed system of particles of alternating mass.
For this system we can repeat the same argument outlined in Section 8.3.2. That is, we append an appropriate subsystem of particles before deleting each zeromoment subsystem of four particles. Equations (8.3.2) and (8.3.3) can then be 107
generalised to give 2k X
i=1 2k X1
( 1) ( 1)
i+1
i+1
i(Ai + Ai 1 ) =
2k
i(Ai + Ai 1 ) = 2k
i=1
k X
a2i
(2k + 1)
i=1 k X1
k X
a2i 1 ;
(8.3.4)
i=1
a2i + (2k
i=1
1)
k X
a2i 1 .
(8.3.5)
i=1
Example 8.3.1. We give just one application of (8.3.4) and (8.3.5) to the Fibonacci numbers, which we study at length in Chapters 10 and 11. Recall that these are defined for every integer i by the recurrence relation Fi = Fi
1
and F2 = 1. We let ai = Fi 3 . Using the familiar formula and the fact that F
2
=
1 and F
1
+ Fi Pi
j=1
2
where F1 = 1
Fj = Fi+2
= 1 we can show that Ai + Ai
1
= Fi
1 2.
We substitute these into Equations (8.3.4) and (8.3.5) to show, after a routine calculation, that 2k X
i=1 2k X1
( 1)i+1 iFi = 2
2kF2k
1
F2k 3 ;
( 1)i+1 iFi = 2 + 2kF2k
2
F2k 3 .
i=1
A result such as this would ordinarily be found with the use of generating functions, Binet’s Formula [38, p. 11] or Abel’s Lemma 3.5.1.
8.4
4
The sum and alternating sum of powers of four
In this section we will physically establish formulas for the sum and alternating sum of powers of four. For the first of these, our argument requires a regrettable dose of algebra. By contrast, demonstrating the formula for the alternating sum of powers of four requires almost no algebra at all.
108
8.4.1
The sum of powers of four
We begin with the system µ depicted in Figure 8.20 (a). Each of the particles shown has unit mass. Along the horizontal axis we have placed a 1 ⇥ 1 array of particles, then a 2 ⇥ 2 array, and so on, concluding with an n ⇥ n array. For each i 2 {1, . . . , n}, the ith array has total mass i2 and is centred at 12 (Ti
1
+ Ti ) = 12 i2 .
Therefore, the desired sum is twice the first moment of this system, 2m(µ) =
n X
i4 .
(8.4.1)
i=1
(a)
(b)
0
Ti
1
0
Ti
1
1 2 i 2
Ti
Ti
Tn
Tn
1 +Tn
2
Figure 8.20: The sum of powers of four.
On the other hand, the ith row of particles shown in Figure 8.20 (b) has total mass Tn
Ti
1
and is balanced at its midpoint, 12 (Ti
1
+ Tn ). Therefore, the first
moment of this system can also be found by evaluating n
1X m(µ) = (Tn 2 i=1
n
1X 2 Ti 1 )(Tn + Ti 1 ) = (T 2 i=1 n
109
Ti2 1 ).
(8.4.2)
Comparing (8.4.1) with (8.4.2) gives, n X
i4 =
i=1
n X
(Tn2
Ti2 1 )
i=1
=
nTn2
n X
Ti2
1
i=1
n
=
nTn2
1X 2 i (i 4 i=1
1)2
n
= nTn2
1X 4 (i 4 i=1 n
=
nTn2
2i3 + i2 ) n
1X 4 1X 3 i + i 4 i=1 2 i=1
n
1X 2 i. 4 i=1
The desired sum can be seen on either side of the above equation. Each of the other sums on the righthand side have already been physically established. Therefore, by using these prior results, we can obtain n X i=1
8.4.2
i4 =
1 n(n + 1)(2n + 1)(3n2 + 3n 30
1).
The alternating sum of powers of four
The formula for the alternating sum of powers of four is best written in the form, 8 > n : k(2k + 1)(4k 2 + 2k 1), if n = 2k. i=1
We now indicate how one might physically demonstrate this formula. As we have done earlier, we consider a signed system of particles, considering separately the cases when n is odd and even. Although the system we study is not globally symmetric, look out for symmetric zero-moment subsystems that can be exploited to establish the above result. Again, the relevant figures are intentionally shown before the corresponding explanation.
110
Case 1 (n = 2k
1). +1 +1 +1 +1 +1
(a)
+1
1
1
1
1 +1 +1 +1 +1 +1
+1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
1
1 +1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
1
1 +1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
Ti
1
0
Ti
1 2 i 2
T2k
1
+1 +1 +1 +1 +1
(b)
+1
1
1
1
1 +1 +1 +1 +1 +1 +1
+1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
1
1 +1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1 +1
1
1 +1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
0
1 +1
T2k
1 +1
1 +1
1 +1
1
1
1
+1 +1 +1 +1 +1
(c)
+1
1
1
1
1 +1 +1 +1 +1 +1 +1
+1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
1
1 +1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1 +1
1
1 +1 +1 +1
1
1
1
1 +1 +1 +1 +1 +1
0
T2k
+1
+1 +1
1 +1
1 +1
1 +1
1
1
1
+1 +1
+1
+1 +1
+1 +1
+1
(d)
1 +1
+1
0
T2k
+1 +1
+1
1
k +1 +1 +1 +1 +1 +1
2k
1
+1 +1 +1 +1 +1 +1
(e)
+1 +1 +1
0 Figure 8.21:
T2k P2k
i=1
1
( 1)i+1 i4 = 2k(2k 111
1
1)(T2k
1 2
1
1 ) 2
The words. Along the horizontal axis we place a 1 ⇥ 1 array of particles of mass +1, then a 2 ⇥ 2 array of particles of mass (2k
1) ⇥ (2k
1) array of particles of mass +1. This is depicted in Figure 8.21
(a). For each i 2 {1, 2, . . . , 2k centred at 12 (Ti
1, and so on, concluding with a
1
1}, the ith array has total mass ( 1)i+1 i2 and is
+ Ti ) = 12 i2 . Therefore, the desired sum is twice the first moment
of this system, 2m(µ) =
2k X1
( 1)i+1 i4 .
(8.4.3)
i=1
To this system we append the subsystem shown using orange circles in Figure 8.21 (b). This is a zero-moment subsystem since the total mass in each column is zero. In Figure 8.21 (c), we observe that all except k diagonals can be placed in a zero-moment subsystem. Each of these subsystems are depicted using a di↵erent colour. If we delete each of these subsystems, then all that remains is a subsystem of total mass k(2k + 1) centred at T2k 2k X1
( 1)i+1 i4 = 2m(µ) = 2k(2k
1
1 . 2
1)(T2k
i=1
112
It follows that
1
1 ) 2
= k(2k
1)(4k 2
2k
1).
Case 2 (n = 2k).
(a)
(b)
(c)
+1
1
1
1
1
+1 +1 +1
1
1
1
1
1
1 +1 +1 +1
1
1
1
1
1
1 +1 +1 +1
1
1
1
1
0
Ti
+1
1 1 2 i 2
Ti
T2k
1
1
1
1
+1 +1 +1
1
1
1
1 +1
1
1 +1 +1 +1
1
1
1
1
1
1 +1 +1 +1
1
1
1
1 +1
1 +1
1
1
1 +1
0
+1
1
1
1
1
+1 +1 +1
1
1
1
1 +1
1
1 +1 +1 +1
1
1
1
1
1
1 +1 +1 +1
1
1
1
1 +1
1 +1
1
1
1 +1
0
1
1 1
1 1
(d)
1 1
1
0
1
1
k
2k + 1 (e)
0
Figure 8.22:
1
1
1
1
1
1
1
1
1
1
T2k P2k
i=1 (
1)i+1 i4 = 113
1 2
2k(2k + 1)(T2k
1 ) 2
The words. Along the horizontal axis we place a 1 ⇥ 1 array of particles of mass +1, then a 2 ⇥ 2 array of particles of mass 2k ⇥ 2k array of particles of mass
1, and so on, concluding with a
1. This is depicted in Figure 8.22 (a). For
each i 2 {1, 2, . . . , 2k}, the ith array has total mass ( 1)i+1 i2 and is centred at 1 (Ti 1 2
+ Ti ) = 12 i2 . Therefore, the desired sum is twice the first moment of this
system, 2m(µ) =
n X
( 1)i+1 i4 .
(8.4.4)
i=1
To this system we append the subsystem depicted using orange circles in Figure 8.22 (b). Once again, this is a zero-moment subsystem as the total mass in each column is zero. In Figure 8.22 (c) we now observe that all except k diagonals can be placed in a zero-moment subsystem. Each of these zero-moment subsystems are depicted using a di↵erent colour. If we delete each of these subsystems, then all that remains is a subsystem of total mass
k(2k + 1) centred at T2k
1 . 2
It follows
that 2k X
( 1)i+1 i4 = 2m(µ) =
2k(2k + 1)(T2k
1 ) 2
=
k(2k + 1)(4k 2 + 2k
1).
i=1
8.5
The sum of powers of five
The argument given to find the formula for the sum of powers of four can be adapted to find the formula for the sum of powers of five. We begin with the system µ depicted in Figure 8.23 (a) below. Along the horizontal axis we place a 1 ⇥ 1 array of particles, then a 2 ⇥ 2 array, and so on, concluding with an n ⇥ n array. For each i 2 {1, . . . , n}, the ith array has total mass i(1 + 3 + · · · + (2i centred at 12 (Ti
1
1)) = i3 and is
+ Ti ) = 12 i2 . Therefore, the desired sum is twice the first moment
of this system, 2m(µ) =
n X i=1
114
i5 .
(8.5.1)
2n 1 2n 1 2n 1 2n 1 2n 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1
1
(a)
5
5
5
5
5
5
5
5
5
5
5
5
3
3
3
3
3
3
3
3
3
3
3
3
3
3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
Ti
1
Ti
1 2 i 2
Tn
2n 1 2n 1 2n 1 2n 1 2n 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1 2i 1
1
(b)
5
5
5
5
5
5
5
5
5
5
5
5
3
3
3
3
3
3
3
3
3
3
3
3
3
3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
Ti
1
Ti
Tn
1 +Tn
2
Figure 8.23: Physically demonstrating the sum of powers of five. On the other hand, the ith row shown in Figure 8.23 (b) has total mass equal to (2i
1)(Tn
Ti 1 ) and is balanced at its midpoint, 12 (Ti
1
+ Tn ). Therefore, the
first moment of this system is also given by n
1X m(µ) = (2i 2 i=1
n
1X Ti 1 )(Tn + Ti 1 ) = (2i 2 i=1
1)(Tn
1)(Tn2
Ti2 1 ). (8.5.2)
By comparing (8.5.1) with (8.5.2) we see that n X
i5 =
i=1
n X
(2i
1)(Tn2
Ti2 1 ).
i=1
Expanding the right-hand side then solving for the desired sum gives n X i=1
i5 =
1 2 n (n + 1)2 (2n2 + 2n 12
1).
We have given an alternative physical derivation of this result in [78], by considering the centre of mass of a right-angled triangle dissected into trapezia. So far as we can tell, neither of these arguments can be easily adapted to find a corresponding 115
formula for the alternating sum of powers of five. For this, we require an approach outlined in the following section.
8.6
Sums of odd powers of consecutive integers
For p 2 N, we define Sp (n) =
Pn
i=1
ip . We now present our e↵orts to physically
determine a formula for Sp (n). We have found a clean approach to this task when p is odd but which unfortunately fails when p is even. Let p
3 be an odd natural number. We now define a symmetric system µ as
follows. At each point i 2 {1, 2, . . . , n}, we place two particles: one whose mass is ip and another whose mass is ((n + 1)
0
i)p . This is depicted below in Figure 8.24.
np
(n 1)p
···
2p
1p
1p
2p
···
(n 1)p
np
1
2
··· n
1 n
Figure 8.24: A symmetric system of particles.
The first moment of µ can be calculated two ways. First, we can find it directly using Equation (7.2.1) followed by a binomial expansion. At each point i the total mass is ip + ((n + 1)
i)p . Therefore,
116
m(µ) = =
n X i=1 n X
i(ip + ((n + 1) p+1
i
+
i=1
i)p )
p ✓ ◆ n X X p
j
i=1 j=0
= Sp+1 (n) +
p ✓ ◆ X p
j
j=0
(n + 1)p j ( 1)j ij+1
(n + 1)p j ( 1)j Sj+1 (n)
= Sp+1 (n) + ( 1)p Sp+1 (n) + ( 1)p 1 p(n + 1)Sp (n) p 2✓ ◆ X p + (n + 1)p j ( 1)j Sj+1 (n). j j=0
(8.6.1)
Since p is odd, we know that Sp+1 (n) + ( 1)p Sp+1 (n) = 0. Therefore, Equation (8.6.1) becomes m(µ) = p(n + 1)Sp (n) +
p 2✓ ◆ X p j=0
j
(n + 1)p j ( 1)j Sj+1 (n)
On the other hand, note that µ is symmetric about the point
n+1 . 2
(8.6.2)
Therefore, we
can use Property (7.2.13), which states that the first moment of µ can be found by concentrating the total mass 2Sp (n) of the system at the point
n+1 . 2
m(µ) = (n + 1)Sp (n).
It follows that (8.6.3)
Finally, by comparing (8.6.2) with (8.6.3), we obtain the following proposition. Proposition 8.6.1. If p Sp (n) =
3 is odd, then 1
p
1
p 2✓ ◆ X p j=0
j
(n + 1)p
1 j
( 1)j+1 Sj+1 (n).
(8.6.4)
For odd values of p, this result is no less appealing than Pascal’s Formula for Sp (n), ! p p 1 (n + 1)p+1 n+1 X j Sp (n) = + Sp j (n) . (8.6.5) p+1 p + 1 j=1 j + 1 117
A clear rendition of the method that Pascal employed to find this result is given by Laosinchai and Panijpan [46], who also give a geometric interpretation of Equation (8.6.5). Likewise, for the alternating sum of powers, we define Ap (n) = By an identical argument we can prove a corresponding result. Proposition 8.6.2. If p Ap (2k
1) =
i=1 (
1)i+1 ip .
3 is odd, then
1 p
Pn
1
Ap (2k) = Ap (2k
p 2✓ ◆ X p j=0
1)
j
(2k)p
1 j
( 1)j+1 Aj+1 (2k
1),
(8.6.6)
(2k)p .
(8.6.7)
It is straight-forward to verify that Equations (8.6.4), (8.6.6) and (8.6.7) yield results that are consistent with all those that we found earlier in this chapter. We can also find the formula for the alternating sum of powers of five, which was missing in Section 8.5. Example 8.6.3. If p = 5, then 2k X1
( 1)
( 1)
i+1 5
i=1
2k X i=1
8.7
3 ✓ ◆ 1X 5 i = (2k)4 j ( 1)j+1 Aj+1 (2k 4 j=0 j
i+1 5
i =
2k X1
( 1)i+1 i5
(2k)5 =
1) = k 2 (16k 3
k 2 (16k 3 + 20k 2
20k 2 + 5),
5).
i=1
4
A continuous counterpart
We conclude our work in this chapter by showing that the argument given in the previous section has a continuous counterpart. For p 2 N we define a region R = {(x, y) : 0 x 1, 0 y xp + (1 in R2 , depicted below in Figure 8.25.
118
x)p }
y
x 0
1
1 2
Figure 8.25: The symmetric region R is balanced at x = 12 . Let (x, y) be the centre of mass of R. The region R is symmetric about 1 x= . 2
(8.7.1)
On the other hand, x can also be computed directly using Equation (7.2.7). This gives x=
˜
R
´1 ´ 1 ´ xp +(1 x)p x(1 x dxdy x dy dx 0 0 0 = ´ 1 ´ xp +(1 x)p = A dy dx 0
0
By equating (8.7.1) and (8.7.2) we obtain ˆ 1 x(1 x)p dx = 0
1 p+1
x)p dx + 2 p+1
1 p+2
.
1 . p+2
(8.7.2)
(8.7.3)
If we expand the brackets on the left-hand side of (8.7.3) and then interchange the order of integration and summation we obtain ˆ 1 ˆ 1X p ✓ ◆ p ✓ ◆ X p p 1 p i i+1 x(1 x) dx = ( 1) x dx = ( 1)i . i i i + 2 0 0 i=0 i=0 This, by comparison with (8.7.3), yields p ✓ ◆ X p 1 1 ( 1)i = i i+2 p+1 i=0
1 . p+2
This calculation is not particularly deliberate in that it yielded a combinatorial identity that we could not have possibly anticipated in advance. The next chapter focusses on further combinatorial identities that can be obtained by way of physical arguments. 119
Chapter 9 Combinatorial Identities We have seen how to demonstrate certain well-known formulas by applying physical arguments to systems of particles that possess either global or internal symmetry. How might this idea be extended? In this chapter, we focus on summations involving binomial coefficients.
9.1
Moments of a symmetric system
We now take any sequence {mi }ni=0 for which mi = mn
i
for all i 2 {0, . . . , n}. We
call such a sequence a symmetric n-sequence. For each symmetric n-sequence we define a corresponding system µ = {(mi , i)}ni=0 , which is depicted in Figure 9.1.
m0
m1
m2
···
mn
0
1
2
···
n
Figure 9.1: A symmetric system.
The first moment of µ is m(µ) =
Pn
i=0
imi . However, the system is symmetric
about n2 . Therefore, by Property 7.2.13, the first moment can also be found by con120
centrating the entire mass
Pn
i=0
mi at the point of symmetry, n2 . This observation
yields the following simple proposition. Proposition 9.1.1. If {mi }ni=0 is a symmetric n-sequence, then n X i=0
n
nX imi = mi . 2 i=0
(9.1.1)
To demonstrate the fruitfulness of Proposition 9.1.1 we now give a range of examples, all of which pertain to the binomial coefficients. n i
}ni=0 , the symmetric n-sequence
✓ ◆ n X n i = n2n 1 . i i=0
4
Example 9.1.2. Applying equation (9.1.1) to { of binomial coefficients, gives
This result is ordinarily obtained by either a combinatorial argument or by di↵erentiating the generating function for the binomial coefficients, (1 + x)n . It can also be obtained by an application of the simple but useful identity, ✓ ◆ ✓ ◆ n n 1 i =n . i i 1
(9.1.2)
Here, the physical argument makes the result almost self-evidently true. As seen in the following example, this technique obviously generalises to any symmetric and evenly-spaced sequence of binomial coefficients, {
mn mk
}nk=0 , the sums of which are
considered by Benjamin, Chen and Kindred [14] as well as Gould [33]. 2n 2i
}ni=0 we find that
= n22n 2 .
4
Example 9.1.3. By considering the symmetric n-sequence { ✓ ◆ ◆ n n ✓ X 2n n X 2n n i = = 22n 2i 2 i=0 2i 2 i=0
1
Example 9.1.4. Likewise, by considering the symmetric n-sequence { have
✓ ◆2 ✓ ◆ n n ✓ ◆2 X n nX n n 2n i = = . i 2 i 2 n i=0 i=0 121
n 2 n }i=0 , i
we
4
Example 9.1.5. With the aid of (9.1.2) we can show that {i ni }n+1 i=0 is a symmetric P n (n + 1)-sequence. Moreover, since n+1 = 0 and ni=0 i ni = n2n 1 , we can employ (9.1.1) to show that
9.2
✓ ◆ n X 2 n i = n(n + 1)2n 2 . i i=0
4
Pairs of symmetric sequences
Now consider a symmetric n-sequence {mi }ni=0 and a symmetric (n
1)-sequence
{ai }ni=01 . The first of these will define the masses of a system of particles while the second will define the distances separating these particles. Specifically, we define a system µ as follows. We first place a particle of mass m0 at the origin. Subsequent to this, we place a particle of mass mi on the x-axis so that its distance from the previous particle is ai 1 , as shown in Figure 9.2. That is, for each i 2 {0, . . . , n}, we place a particle of mass mi at position xi = with the convention that
P
a0
i 1 X
ak ,
(9.2.1)
k=0
1 k=0
ak = 0. an
a1
m0
m1
m2
0
a0
a0 + a1
an
2
mn
Pn
2 i=0
1
ai
1
mn
Pn
1 i=0
ai
Figure 9.2: A symmetric configuration of a symmetric n-sequence. This system is symmetric about the point x =
1 2
Pn
1 i=0
ai . Therefore, its first moPn ment can be found by supposing that its total mass i=0 mi is concentrated at its point of symmetry. That is,
m(µ) =
n X i=0
n 1
mi ·
122
1X ai . 2 i=0
(9.2.2)
The first moment can also be found directly using Equation (7.2.1). So, with the added use of (9.2.1), we obtain m(µ) =
n X
n X i 1 X
mi xi =
i=0
m i ak .
(9.2.3)
i=0 k=0
Comparing (9.2.2) with (9.2.3) yields the following proposition. Proposition 9.2.1. If {mi }ni=0 is a symmetric n-sequence and {ai }ni=01 is a symmetric (n
1)-sequence, then n X i 1 X i=0
n
n 1
X 1X m i ak = mi · ai . 2 i=0 i=0 k=0
(9.2.4)
Note that we can obtain Propsition 9.1.1 as a special case of Proposition 9.2.1 by letting ai = 1 for all i 2 {0, 1, . . . , n
1}. To demonstrate the usefulness of the
more general result we prove two identities. Others can be discovered with equal ease. Example 9.2.2. We can apply (9.2.4) to the symmetric n-sequence { the symmetric (n
1)-sequence {
n X i 1 ✓ ◆✓ X n n i=0 k=0
i
1 k
◆
n 1 i
}ni=01 . We obtain,
◆ n ✓ ◆ n 1✓ 1X n X n 1 1 = · = · 2n · 2n 2 i=0 i i 2 i=0
1
n i
}ni=0 and
= 22n 2 .
4
Example 9.2.3. For our final application Proposition (9.2.1) we begin with a special case of Vandermonde’s Identity, ◆ ✓ ◆ n ✓ ◆✓ X m m 2m = i n i n i=0 for m, n
0, a combinatorial proof of which appears in [15, p. 66]. We consider the
symmetric n-sequence { {
p i
p n 1 i
(9.2.5)
m i
m n i
}ni=0 , and an additional symmetric (n 1)-sequence
}ni=01 . Then with (9.2.4) and (9.2.5) we obtain ◆✓ ◆✓ n X i 1 ✓ ◆✓ X m m p i n i k n i=0 k=0 123
p 1
k
◆
✓ ◆✓ ◆ 1 2m 2p = . 2 n n 1
4
Having now found physical derivations of formulas for sums of powers and other combinatorial identities, we turn our attention to identities involving Fibonacci and Lucas numbers. First, we apply a similar approach to determine a new formula for the sum of cubes of Fibonacci numbers. We then systematically study various sums of products of Fibonacci and Lucas numbers. We will find simple proofs of various known identities, and make fortuitous discoveries of others.
124
Chapter 10 Summations of Fibonacci numbers The Fibonacci and Lucas numbers are defined for all integers i by the recurrence relations Fi = Fi
1
+ Fi
2
where F1 = 1 and F2 = 1;
Li = Li
1
+ Li
2
where L1 = 1 and L2 = 3.
The method that we have outlined in the previous two chapters can also be applied to summations involving these two sequences. The contents of this chapter first appeared in [75], and are primarily concerned with Fibonacci numbers.
10.1
Hidden identities in geometric tilings
In 1972, Brousseau, a founding editor of the Fibonacci Quarterly, published an entertaining account of how geometric tilings can be used to generate identities involving Fibonacci numbers [17]. For instance, one now famous example is seen in Figure 10.1. If we compare the total area to the sum of its parts, then we see that F12 + F22 + · · · + Fn2 = Fn Fn+1 .
125
F6
F4 F2 F1
F1 F3
F5
F7
Figure 10.1: The sum of squares of Fibonacci numbers.
In this chapter, we explain how there is a hidden formula in each of Brousseau’s tilings. These can be discovered by considering each tiling’s centre of mass. By applying this approach to various geometric tilings we find a range of formulas for summations involving Fibonacci numbers. For instance, we provide a simple derivation of a formula for the sum of cubes of Fibonacci numbers.
10.2
The method
The elementary physics required for this chapter can be found in Section 7.2.3. We remind the reader that the x-mean refers to the x-component of the centre of mass of a planar region. We have already seen how geometric diagrams can illustrate algebraic identities. Perhaps the simplest such example can be seen in Figure 10.2, which demonstrates the fact that for non-negative real numbers a, b and c, c(a + b) = ca + cb.
126
(10.2.1)
a
b
c
1
a
b
1
2
O
a+b 2
a+b
Figure 10.2: This figure can demonstrate identities (10.2.1) and (10.2.2).
It may be surprising that lurking in that same simple diagram one can also find that (a + b)2 = a2 + 2ab + b2 .
(10.2.2)
To see this, we let c = 1 and then construct perpendicular axes along the left and lower edges of the rectangle. As the rectangle is symmetric, its x-mean is located at x=
a+b . 2
(10.2.3)
Now decompose the figure into two parts, as indicated in Figure 10.2. Here, and later, we let xi and Ai be the x-mean and area of the rectangle denoted by the symbol i . We can also locate the x-mean of the entire rectangle by taking a mass-weighted average of its two parts. Since c = 1, x=
a a + (a + 2b )b x1 A1 + x2 A2 = 2 . A1 + A2 (a + b)
(10.2.4)
Equating (10.2.3) and (10.2.4) gives (10.2.2). Likewise, Figure 10.3 below is ordinarily used to demonstrate the truth of (10.2.2). However, it can also be leveraged to demonstrate that (a + b)3 = a3 + 3a2 b + 3ab2 + b3 .
127
(10.2.5)
a
b
a
b
a
a
1
2
b
b
3
4
O
a+b
a+b 2
Figure 10.3: The above picture demonstrates both identities (10.2.2) and (10.2.5). Once again, the x-mean is located at x=
a+b 2
(10.2.6)
and alternatively, x=
a 2 a + (a + 2b )ab + a2 ab + (a + 2b )b2 x1 A1 + x2 A2 + x3 A3 + x4 A4 = 2 . (10.2.7) A1 + A2 + A3 + A4 (a + b)2
Equating (10.2.6) and (10.2.7) clearly gives (10.2.5). Of course, we don’t give these examples because they are new or profound. We o↵er them because they indicate a novel way of generating identities from rectangular regions possessing vertical symmetry. The procedure is simple. Take any such figure and then: 1. locate the x-mean along the line of symmetry, then 2. establish the x-mean by dividing the figure into parts, then 3. equate the two results. This procedure is particularly fruitful when applied to rectangular figures whose side lengths are Fibonacci numbers. With scarce e↵ort we will physically show that n X i=1
Fi Fi+1 Fi+2 =
3 3 Fn3 + Fn+1 + Fn+2 + Fn 1 Fn Fn+1 4
128
2
.
(10.2.8)
We also obtain a formula for the sum of cubes of Fibonacci numbers. Notably, we make no appeal to the Binet formula, which features prominently in much of the Fibonacci literature. Nor do we make any combinatorial arguments. In the work that follows, we will state results true for Fibonacci numbers. Some of our results generalise in obvious directions to other sequences of numbers generated by the same recurrence relation, such as the Lucas numbers.
10.3
A Simple Fibonacci tiling
To illustrate the ease and utility of this technique we begin with the simple Fibonacci tiling depicted in Figure 10.4 below.
Fn+1
Fn
Fn+1
Fn+1
Fn+1
Fn
1
2
O
Fn+2 2
Fn+2
Figure 10.4: A simple Fibonacci tiling. 2 Superficially, the tiling demonstrates that Fn+2 Fn+1 = Fn+1 + Fn Fn+1 . Con-
sideration of the rectangle’s x-mean reveals a further identity. First, its x-mean is located at x=
Fn+1 + Fn Fn+2 = . 2 2
129
(10.3.1)
Now by considering a mass-weighted average of its parts we also see that x 1 A1 + x 2 A2 A1 + A2 Fn+1 2 · Fn+1 + (Fn+1 + F2n ) · Fn+1 Fn = 2 Fn+1 (Fn+1 + Fn ) 2 F + (2Fn+1 + Fn ) · Fn = n+1 2Fn+2 2 F + Fn+3 Fn = n+1 . 2Fn+2
x=
(10.3.2)
Equating (10.3.1) and (10.3.2) gives the identity, 2 Fn+2
2 Fn+1 = Fn Fn+3 .
(10.3.3)
No doubt Equation (10.3.3) is a trivial result, and algebraically obvious from the 2 factorisation of Fn+2
2 Fn+1 . Nonetheless, this example serves to demonstrate the
technique that we will deploy in our subsequent work, where more complicated configurations of rectangles lead to more interesting results.
10.4
A preliminary result
However, before considering further tilings, we remind the reader of Equation (10.4.1) below, which we will call upon in the following section. This appears to have first been proved by Block [16]. Clary and Hemenway [23, p.135] have subsequently given a particularly clever proof of this identity. Here, we have provided a proof by induction, as this warrants a few subsequent remarks. Proposition 10.4.1. For n n X i=1
1, 1 Fi2 Fi+1 = Fn Fn+1 Fn+2 . 2
130
(10.4.1)
Proof. The base case is obvious. For the inductive step, 2
k+1 X
Fi2 Fi+1
=2
i=1
k X
2 Fi2 Fi+1 + 2Fk+1 Fk+2
i=1
2 = Fk Fk+1 Fk+2 + 2Fk+1 Fk+2
= Fk+1 Fk+2 (Fk + 2Fk+1 ) = Fk+1 Fk+2 Fk+3 . We note that Proposition 10.4.1 is as geometrically impressive as it algebraically impressive. To see this, if we take two copies of each of the boxes with dimensions (F12 ⇥ F2 ), (F22 ⇥ F3 ), . . . , (Fn2 ⇥ Fn+1 ), then these 2n boxes can be packed into one box of dimensions Fn ⇥ Fn+1 ⇥ Fn+2 . This fact is not implied by formula (10.4.1), but by the preceding calculation that demonstrates its truth. Indeed, assume that the first 2k boxes can be packed into a box of dimensions Fk ⇥ Fk+1 ⇥ Fk+2 . We take this box along with two boxes with di2 mensions Fk+1 ⇥Fk+2 and pack these into a box with dimensions Fk+1 ⇥Fk+2 ⇥Fk+3 .
This is shown in Figure 10.5 below.
Fk+1
+
+
=
Fk+1
Fk+1
Fk+1
Fk+1
Fk
Fk+2 Fk+1 Fk+3
Figure 10.5: Packing the first 2(k + 1) boxes inductively.
131
10.5
The sum of products of three consecutive Fibonacci numbers
Our second example considers Figure 10.6 below, depicted by Brousseau in [17]. It shows four rectangles with sides Fn and Fn+1 arranged into the corners of a square of side length Fn+2 . At the square’s centre is a smaller square of side length Fn+1 Fn = Fn 1 . Brousseau notes that the diagram provides a one-glance demonstration of the identity 2 Fn+2 = 4Fn Fn+1 + Fn2 1 .
Fn+1
Fn
(10.5.1)
Fn+1
Fn
3 4 Fn
Fn
1
1
5 2
1 O
Fn+2
Fn+2 2
Figure 10.6: This spiral of tiles yields two formulas.
Before considering the x-mean, we first note that Equation (10.5.1) and a telescoping sum can be used to find a formula for the sum of products of two consecutive Fibonacci numbers, n X i=1
n
Fi Fi+1 =
1X 2 F 4 i=1 i+2
Fi2
1
=
2 2 Fn+2 + Fn+1 + Fn2 4
2
.
Surprisingly, the same diagram can be used to find a formula for the sum of products of three consecutive Fibonacci numbers. To this end, we first note that the x-mean 132
is located at x=
Fn+2 . 2
(10.5.2)
As before, this can also be found by considering a mass-weighted average of its parts. Forgoing the algebraic details, we can show that x1 A1 + x2 A2 + x3 A3 + x4 A4 + x5 A5 A1 + A2 + A3 + A4 + A5 3 F + 2Fn2 1 Fn + 4Fn Fn+1 Fn+2 = n 1 . 2 2Fn+2
x=
(10.5.3)
As before, we equate (10.5.2) and (10.5.3) to discover the identity, 3 2Fn2 1 Fn + 4Fn Fn+1 Fn+2 = Fn+2
Fn3 1 .
(10.5.4)
Now with the aid of a telescoping sum we obtain, 2
n X i=1
Fi2 1 Fi
+4
n X
3 3 Fi Fi+1 Fi+2 = Fn3 + Fn+1 + Fn+2
2.
(10.5.5)
i=1
And finally, Equations (10.4.1) and (10.5.5) together yield an appealing formula for the sum of products of three consecutive Fibonacci numbers. Proposition 10.5.1. For n n X
Fi Fi+1 Fi+2 =
1,
3 3 Fn3 + Fn+1 + Fn+2
Fn 1 Fn Fn+1
2
4
i=1
.
(10.5.6)
Armed with this result, we now find a formula for the sum of cubes of Fibonacci numbers.
10.6
The sum of cubes of Fibonacci numbers
For this we now consider the tiling shown in Figure 10.7 below.
133
Fn+1
Fn
Fn
Fn+1
Fn
Fn
1
2
Fn+1
Fn+1
3
4
O
Fn+2
Fn+2 2
Figure 10.7: Summing the cubes of Fibonacci numbers.
The x-mean is located at x=
Fn+2 . 2
(10.6.1)
Considering a mass-weighted average of its parts, a minor calculation shows that x=
3 F 3 + Fn+1 + 3Fn Fn+1 Fn+2 x 1 A1 + x 2 A2 + x 3 A3 + x 4 A4 = n . 2 A1 + A2 + A3 + A4 2Fn+2
(10.6.2)
By equating (10.6.1) and (10.6.2) we can establish that 3 Fn3 + 3Fn Fn+1 Fn+2 = Fn+2
3 Fn+1 .
(10.6.3)
Using Equation (10.6.3), it follows in the now customary fashion that n X
Fj3
+3
i=1
n X
Fi Fi+1 Fi+2 =
i=1
n X
3 Fi+2
3 3 Fi+1 = Fn+2
1.
(10.6.4)
i=1
Finally, Equations (10.6.4) and (10.5.6) together yield a closed formula for the sum of cubes of Fibonacci numbers, n X i=1
Fi3 =
3 Fn+2
3 3Fn+1
3Fn3 + 3Fn 1 Fn Fn+1 + 2 . 4
(10.6.5)
A cleaner expression can be obtained by substituting Fn+2 = Fn+1 + Fn and Fn Fn+1
Fn into Equation (10.6.5). We obtain the following result.
134
1
=
Proposition 10.6.1. For n n X
1,
Fi3 =
2 3Fn+1 Fn
i=1
3 Fn+1 2
Fn3 + 1
.
(10.6.6)
Certainly, Equation (10.6.6) isn’t as impressive as the closed and factored expression found by Clary and Hemenway [23], However, the grace of the physical argument perhaps o↵sets its ungainly answer. Interestingly, the formula is comparable to that obtained by Benjamin and Quinn [13] using a combinatorial argument, which is equivalent to the identity n X
Fi3 =
i=1
2 Fn+1 Fn+2 + ( 1)n Fn 2
2Fn3
.
The reader is invited to discover further identities using this approach, perhaps beginning with Figure 10.1.
135
Chapter 11 Balancing finite sums of Fibonacci numbers In the previous chapter we showed that essentially physical arguments can be used to find various summations involving Fibonacci numbers, including a formula for the sum of cubes of Fibonacci numbers. In this chapter we detail a more systematic treatment of this approach, the details of which were first published in [76]. We will now show how various summation formulas involving Fibonacci and Lucas numbers can be interpreted physically, and that an appealing secondary summation can often be deduced as a natural consequence. For instance, we will demonstrate how the first summation in (11.0.1) implies the second summation. That is,
n X
Fi2
= Fn Fn+1 =)
i=1
n X
2 Fi2 F2i = Fn2 Fn+1 .
(11.0.1)
i=1
While both of these are well-known, we believe that the derivation of the second identity as a physical consequence of the first identity to be new. We will also establish some identities that we believe are original. Most notably, in Theorem 11.2.5 we prove that
n X i=1
3 Fi3 Fi+1
=
✓X n i=1
136
Fi2 Fi+1
◆2
.
(11.0.2)
The reader will note the resemblance between (11.0.2) and the iconic formula for the sum of cubes,
n X i=1
✓X ◆2 n i = i , 3
(11.0.3)
i=1
which we established in Section 8.3. The similarity is not superficial. We can deduce each of these formulas using essentially the same general method that we outline in Section 11.1. In this respect, the work in this chapter can be regarded as a generalisation of (11.0.3). By looking at the proofs of these identities we will come to see each as a physical imperative rather than a neat algebraic coincidence.
11.1
The method and the main result
The required theory is elementary. Consider a rectangle R ⇢ R2 of uniform density and area A, placed lengthwise along the x-axis. The balancing point x of this rectangular figure is its midpoint. If R is partitioned into sub-rectangles {Ri }ni=1 by a series of vertical lines, then the balancing point of R can also be found by finding the midpoint xi and area Ai of each part and then computing a mass-weighted average, x=
Pn
Ai x i . A
i=1
(11.1.1)
This is is illustrated below in Figure 11.1.
0
x1
x2
x3
x4 x
xn
Figure 11.1: The balancing point of a rectangle is a mass-weighted average of its parts. We now show how (11.1.1) can be used to deduce each of (11.0.1), (11.0.2) and (11.0.3), amongst other results. To this end, we take any finite sequence of positive 137
numbers (ai )ni=1 and let (Si )ni=0 be the sequence of partial sums, defined so that Pi S0 = 0 and Si = k=1 ak , for i 2 {1, . . . , n}. We then construct rectangles of dimensions
a1 ⇥ 1, a2 ⇥ 1, . . . , an ⇥ 1, arranging these along the x-axis as shown below in Figure 11.2. a1
a2
a3
an
···
1 0
Sn
Sn 2
Figure 11.2: Rectangles placed side by side along the x-axis. For each i 2 {1, . . . , n}, the midpoint xi of rectangle i is 1 xi = (ai + 2Si 1 ). 2
(11.1.2)
The balancing point x of the combined rectangle will be its midpoint, 1 x = Sn , 2
(11.1.3)
which can also be found using Equations (11.1.1) and (11.1.2) giving Pn 1 2 Pn (a + 2ai Si 1 ) i=1 ai xi x= = i=1 2 i . Sn Sn
(11.1.4)
By equating the righthand sides of (11.1.3) and (11.1.4) we prove Theorem 11.1.1. Theorem 11.1.1. If (ai )ni=1 is a finite sequence of positive numbers whose partial sums are (Si )ni=0 , then
n X
a2i + 2ai Si
1
= Sn2 .
(11.1.5)
i=1
This is the central result of this chapter, and will be deployed extensively throughout. A given sequence (ai )ni=1 must satisfy two informal conditions for Theorem 11.1.1 to yield an interesting result: 138
1. the closed form of Sn must be known, and 2. (a2i + 2ai Si 1 )ni=1 is a sequence that we would like to sum. As we will now see, such examples are not hard to come by.
11.2
Applications
To verify that Equation (11.1.5) works, we will first check that we obtain the expected formula for the sum of cubes of consecutive integers. Example 11.2.1. Let ai = i so that Sn = a2i + 2ai Si
Pn
i=1
i = 12 n(n + 1). It follows that
1 = i2 + 2i · i(i 2
1
1) = i3 ,
in which case (11.1.5) yields n X i=1
3
i =
✓
n(n + 1) 2
◆2
✓X ◆2 n = i .
4
i=1
We will now see that Equation (11.1.5), while valid for any finite sequence of positive numbers, is often productive when each term of the sequence is some product of Fibonacci or Lucas numbers, as the following two simple examples serve to illustrate. Example 11.2.2. We first let ai = Fi2 . In Figure 10.1 we see that Sn = Fn Fn+1 . It follows that a2i + 2ai Si
1
= Fi4 + 2Fi3 Fi
1
= Fi3 (Fi + 2Fi 1 ) = Fi3 Li
(as Li = Fi + 2Fi 1 )
= Fi2 F2i . Therefore, (11.1.5) yields
n X
(as F2i = Fi Li )
2 Fi2 F2i = Fn2 Fn+1 .
i=1
139
Pn
i=1
Fi2 =
This identity was also deduced by Melham [50] using an altogether di↵erent ap4
proach.
Example 11.2.3. For our second example we let ai = F2i . We will employ the P well-known result that Sn = ni=1 F2i = F2n+1 1. Subsequently, a2i + 2ai Si
1
= F2i2 + 2F2i (F2(i
= F2i (F2i + 2F2i 1 ) = F2i L2i = F4i
1)
1)+1
2F2i
2F2i
(as L2i = F2i + 2F2i 1 )
2F2i .
(as F4i = F2i L2i )
Yet another application of (11.1.5) yields n X
(F4i
2F2i ) = (F2n+1
1)2
i=1
from which we readily obtain
n X
2 F4i = F2n+1
4
1.
i=1
Of course, the same procedure works for other summations and eight such results are summarised in Table 11.1 below. Although each of these will be familiar, the approach taken to establish them is original. (1) (2)
Pn
ai = F i
ai = F2i
(4)
ai = L2i ai = F2i
1
(6)
ai = L2i
1
ai = Fi2
(8)
ai = L2i
1
i=1
Li = Ln+2
3
i=1
F2i = F2n+1
1
=)
i=1
L2i = L2n+1
1
=)
i=1
F2i
1
= F2n
i=1
L2i
1
= L2n
i=1
Fi2 = Fn Fn+1
i=1
L2i = Ln Ln+1
Pn
(5)
(7)
Fi = Fn+2
Pn
ai = Li
(3)
i=1
Pn Pn Pn Pn Pn
=) =)
=) 2
=) =) 2
=)
Pn
i=1
2 Fi Fi+3 = Fn+2
1
i=1
Li Li+3 = L2n+2
9
i=1
2 F4i = F2n+1
i=1
F4i =
i=1
F4i
2
2 = F2n
i=1
F4i
2
= 15 (L22n
i=1
2 Fi2 F2i = Fn2 Fn+1
i=1
L2i F2i =
Pn Pn Pn Pn Pn Pn Pn
1 5
1
L22n+1
1 5
1
4)
L2n L2n+1
Table 11.1: Each of the first summations implies the second summation. 140
4
Interestingly, by comparing row (3) with row (4), and row (5) with row (6), we obtain the well-known and so-called fundamental identity, L2n
5Fn2 = 4( 1)n ,
(11.2.1)
the importance of which is detailed by Rabinowitz [68]. By looking at individual rows we can make further observations. For instance, by considering row (5) we see that
n X
F4i
i=1
2
=
✓X n i=1
F2i
1
◆2
.
Example 11.2.4. Suppose we now begin with the remarkable (and remarkably imposing) result deduced by Melham [50, Equation (1.3)], n X i=1
1 2 Fi Fi+1 Fi+2 Fi+3 Fi+4 = Fn Fn+1 Fn+2 Fn+3 Fn+4 Fn+5 . 4
(11.2.2)
2 As before, we let ai = Fi Fi+1 Fi+2 Fi+3 Fi+4 so that by (11.2.2) we have
1 Sn = Fn Fn+1 Fn+2 Fn+3 Fn+4 Fn+5 . 4 It follows that a2i + 2ai Si
1
1 2 4 2 2 2 3 2 2 = Fi2 Fi+1 Fi+2 Fi+3 Fi+4 + Fi 1 Fi2 Fi+1 Fi+2 Fi+3 Fi+4 2 1 2 2 3 2 2 = Fi Fi+1 Fi+2 Fi+3 Fi+4 (2Fi+2 + Fi 1 ) 2 1 2 3 2 2 = Fi2 Fi+1 Fi+2 Fi+3 Fi+4 Li+2 (as Li+2 = 2Fi+2 + Fi 1 ) 2 1 2 2 2 2 = Fi2 Fi+1 Fi+2 Fi+3 Fi+4 F2i+4 . (as F2i+4 = Fi+2 Li+2 ) 2 Therefore, Equation (11.1.5) yields n X i=1
1 2 2 2 2 2 2 2 2 2 Fi2 Fi+1 Fi+2 Fi+3 Fi+4 F2i+4 = Fn2 Fn+1 Fn+2 Fn+3 Fn+4 Fn+5 . 8
(11.2.3)
Equation (11.2.3) is also deduced by Melham [51, Equation (2.7)], independently of (11.2.2) rather than as a consequence of it. 141
4
The method that we have discovered can also be applied sequentially. For instance, suppose we begin with n X i=1
1 Fi2 Fi+1 = Fn Fn+1 Fn+2 , 2
(11.2.4)
which we first encountered in Section 10.4. We can use (11.2.4) to establish Equations (11.2.5) and (11.2.6) below. We believe that these are original identities. Theorem 11.2.5. If n
1, then Equation (11.2.4) implies (11.2.5) which in turn
implies (11.2.6) where n X
1 3 2 2 Fi3 Fi+1 = Fn2 Fn+1 Fn+2 ; 4
(11.2.5)
1 5 4 4 Fi5 Fi+1 F2i+1 = Fn4 Fn+1 Fn+2 . 8
(11.2.6)
i=1
n X i=1
3 Proof. We prove that Equation (11.2.5) implies (11.2.6). Let ai = Fi3 Fi+1 so that
by Equation (11.2.5) we have 1 2 2 Sn = Fn2 Fn+1 Fn+2 . 4 Therefore a2i + 2ai Si
1
1 6 5 = Fi6 Fi+1 + Fi2 1 Fi5 Fi+1 2 1 5 5 = Fi Fi+1 (2Fi Fi+1 + Fi2 1 ) 2 1 5 = Fi5 Fi+1 F2i+1 2
(11.2.7)
where (11.2.7) follows from the fact that 2Fi Fi+1 + Fi2 1 = 2(Fi + Fi 1 )Fi + Fi2 = Fi2 + Fi2 + 2Fi Fi = Fi2 + (Fi + Fi 1 )2 2 = Fi2 + Fi+1
= F2i+1 . 142
1
1
+ Fi2
1
Finally, we obtain Equation (11.2.6) by making use of Equation (11.2.7) in (11.1.5). The proof that Equation (11.2.4) implies (11.2.5) is both similar and simpler, and is therefore omitted. A couple remarks. First, note that Equations (11.2.4) and (11.2.5) imply (11.0.2), which is given in the introduction to this chapter. Additionally, note that we do not obtain a formula of any particular interest if we apply the method once more to (11.2.6). The method of discovery of Equations (11.2.5) and (11.2.6) does not work quite as gracefully for Lucas numbers. Nonetheless, each does have a corresponding counterpart. In this case, we begin with the Lucas counterpart to Equation (11.2.4) [50, Equation (3.2)],
n X
L2i Li+1 =
i=1
[Li Li+1 Li+2 ]n0 . 2
(11.2.8)
Here, and later, we follow the convention that i serves as a dummy variable on the right-hand side of (11.2.8) so that [Li Li+1 Li+2 ]n0 = Ln Ln+1 Ln+2
L 0 L 1 L2 .
Assuming (11.2.8) we easily obtain Theorem 11.2.6, which we also believe to be original. We have omitted the proof as it is no more illuminating than the proof of Theorem 11.2.5. Theorem 11.2.6. If n
0, then n X
L3i L3i+1 =
[L2i L2i+1 L2i+2 ]n0 , 4
(11.2.9)
L5i L5i+1 F2i+1 =
[L4i L4i+1 L4i+2 ]n0 . 40
(11.2.10)
i=1
n X i=1
We conclude with one final result, the proof of which depends on an identity that has recently been considered by Sury [74], and which has received further attention in [29] and elsewhere (see, e.g., [43] and [49]), n X 2i Li = 2n+1 Fn+1 . i=0
143
(11.2.11)
Using the method outlined in this paper, the reader might like to show that Equation (11.2.11) yields what we believe is a further original identity, n X
2 22i Li Fi+3 = 22n+2 Fn+1 .
(11.2.12)
i=0
11.3
Further comments
We could not possibly provide a complete account of the Fibonacci and Lucas summations discoverable by the method that we have detailed. Therefore, the reader has ample opportunity to discover additional results using this method. Moreover, combinatorists might also like to ponder if the new identities presented here have combinatorial proofs, similar to those detailed by Benjamin and Quinn in [11], [12] and [15].
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