Showing papers in "The Mathematical Gazette in 2014"
TL;DR: The perimeter of the ellipse x 2/a 2 + y 2/b 2 = 1 is 4J (a, b), where J(a,b) is the "elliptic integral".
Abstract: The perimeter of the ellipse x 2/a 2 + y 2/b 2 = 1 is 4J (a, b), where J (a, b) is the ‘elliptic integral’ This integral is interesting in its own right, quite apart from its application to the ellipse. It is often considered together with the companion integral Of course, we may as well assume that a and b are non-negative.
17 citations
TL;DR: In this paper, the authors start from two simple identities: for any p > 0 and 0 ≥ b ≥ a, now let can we formulate statements about Gp(a, b) that generalise (1) and (2).
Abstract: We start from two simple identities: For any p > 0 and 0 ≥ b ≥ a, now let Can we formulate statements about Gp(a, b) that generalise (1) and (2)? We cannot hope for equalities, but perhaps we can establish inequalities which somehow reproduce (1) when p = 1 and (2) when p = 2. For (1), this might mean an inequality of the form Apap ≤ Gp (a, b) ≤ Bpap for certain constants Ap and Bp, and for (2) a similar statement with ap replaced by ap + bp. However, these are not the only possibilities, as we shall see.
12 citations
TL;DR: The reverse-comparison theorem of Sturm and Steiner as discussed by the authors states that if two angle bisectors of a triangle have equal length, then the triangle is necessarily isosceles.
Abstract: In 1840 C. L. Lehmus sent the following problem to Charles Sturm: ‘If two angle bisectors of a triangle have equal length, is the triangle necessarily isosceles?’ The answer is ‘yes’, and indeed we have the reverse-comparison theorem: Of two unequal angles, the larger has the shorter bisector (see [1, 2]). Sturm passed the problem on to other mathematicians, in particular to the great Swiss geometer Jakob Steiner, who provided a proof. In this paper we give several proofs and discuss the old query: ‘Is there a direct proof?’ before suggesting that this is no longer the right question to ask. We go on to discuss all cases when an angle bisector (internal orexternal) of some angle is equal to one of another.
6 citations
TL;DR: A method for classifying all integer triangles containing an angle whose cosine is known is given in this article. But this method assumes that the angle opposite c is known, which is not the case in this paper.
Abstract: 98.02 On integer triangles In this note, we derive a method for classifying all integer triangles containing an angle whose cosine is known. The term 'integer triangle' is taken to mean a triangle each of whose sides is an integer. We start off by considering such an integer triangle T with sides a, b, c and we shall use the convention that it is the angle opposite c whose cosine is known. Denoting the value of this cosine by k we see that a2+b-c2 k = 2ab which may be rewritten in the form I _k = (c + a b) (c + b a). 2ab 2 (1 k) we now have (c + a b) (c + b a) ab Since -1 < k < I, we see that 0 < K < 4. Furthermore, the fact that a, b, c are integers implies that both k and K are rational. Putting K
5 citations
5 citations
TL;DR: In this paper, the authors consider the problem of defining a continuous function f(x) which agrees with factorials at integers, and show that the gamma function is the only one satisfying a certain smoothness condition which we will specify below.
Abstract: Consider the problem of defining a continuous function f(x) which agrees with factorials at integers. There are many possible ways to do this. In fact, such a function can be constructed by taking any continuous definition of f(x) on [0,1] with f(0) = f(1) = 1 (such as f(x) = 1), and then extending the definition to all x > 1 by the formula f(x + 1) = (x + 1)f(x). This construction was discussed by David Fowler in [1] and [2]. For example, the choice f(x) = ½x(x − 1) + 1 results in a function that is differentiable everywhere, including at integers. However, this approach had already been overtaken in 1729, when Euler obtained the conclusive solution to the problem by defining what we now call the gamma function. Among all the possible functions that reproduce factorials, this is the ‘right’ one, in the sense that it is the only one satisfying a certain smoothness condition which we will specify below. Admittedly, Euler didn't know this. It is known as the Bohr-Mollerup theorem, and was only proved nearly two centuries later. First, a remark on notation: the notation Γ (x) for the gamma function, introduced by Legendre, is such that Γ (n) is actually (n − 1)! instead of n!. Though this might seem a little perverse, it does result in some formulae becoming slightly neater. Some writers, including Fowler, write x! for Γ (x + 1), and refer to this as the ‘factorial function’. However, the notation Γ (x) is very firmly entrenched, and I will adhere to it here.
4 citations
TL;DR: Euler as discussed by the authors gave the full details of his subsequent derivation of the general formula (1) where (Bn ) is a sequence of "strange constants" and showed that the sequence of constants occurring in the coefficients of the generalized ζ(2n) formula also occurs in the Euler-Maclaurin summation formula and in the Maclanar expansion of.
Abstract: The problem of finding a closed-form evaluation of baffled the pioneers of calculus such as Leibniz and James Bernoulli and, following the latter’s promulgation of the problem, it became known as the Basel problem after his home town (which was also Euler’s birthplace). Euler’s early sensational success in solving the Basel problem by identifying is extremely well-documented. In this paper, we give the full details of his subsequent derivation of the general formula (1) where (Bn ) is a sequence of ‘strange constants’. Euler’s polished account of his discovery, in which he popularised the designation of the strange constants as ‘Bernoulli numbers’, appears in Chapter 5 of Volume 2 of his great textbook Institutiones calculi differentialis [1; E212]: see [2] for an online English translation. Here, we will focus on his initial step-by-step account which appeared in his paper with Enestrom number E130, written c1739, carrying the rather nondescript title De seriebus quibusdam considerationes, ‘Considerations about certain series’. (For convenience, we will just use ‘Enestrom numbers’ when referencing Euler’s work: all are readily available on-line at [1].) Euler’s proof is notable for its early, sophisticated and incisive use of generating functions and for his brilliant insight that the sequence (Bn ) occurring in the coefficients of the general ζ(2n) formula (1) also occurs in the Euler-Maclaurin summation formula and in the Maclaurin expansion of . By retracing Euler’s original path, we shall not only be able to admire the master in full creative flow, but also appreciate the role played by recurrence relations such as (2) which, as our ample list of references (which will be reviewed later) suggests, have been rediscovered over and over again in the literature. Moreover, our historical approach makes it clear that, while deriving (2) is relatively straightforward (and may be used to calculate ζ(2n) recursively as a rational multiple of π2n ), it is establishing the connection between ζ(2n) and the Bernoulli numbers that was for Euler the more difficult step. Even today, this step presents pedagogical challenges depending on one’s starting definition for the Bernoulli numbers and what identities satisfied by them one is prepared to assume or derive.
4 citations
TL;DR: In this paper, it is shown that the set of all points (x, y, z) ∈ ℝ3 which satisfy the cubic equation is a surface of revolution.
Abstract: A well-known exercise in classical differential geometry [1, 2, 3] is to show that the set of all points (x, y, z) ∈ ℝ3 which satisfy the cubic equation is a surface of revolution. The standard proof ([2], [3, p. 11]), which, in principle, goes back to Lagrange [4] and Monge [5], is to verify that (1) satisfies the partial differential equation (here written as a determinant) which characterises any surface of revolution F (x, y, z) = 0 whose axis of revolution has direction numbers (l, m, n) and goes through the point (a, b, c). This PDE, for its part, expresses the geometric property that the normal line through any point of must intersect the axis of revolution (this is rather subtle; see [6]). All of this, though perfectly correct, seems complicated and rather sophisticated just to show that one can obtain by rotating a suitable curve around a certain fixed line. Moreover, to carry out this proof one needs to know a priori just what this axis is, something not immediately clear from the statement of the problem. Nor does the solution give much of a clue as to which curve one rotates. A search of the literature failed to turn up a treatment of the problem which differs significantly from that sketched above (although see [1]). The polynomial (1) is quite famous and has been the object of numerous algebraical and number theoretical investigations. See the delightful and informative paper [7].
4 citations
TL;DR: In this paper, the authors look at some common problem areas in A-level statistics and discuss the importance of the subtleties in the analysis of the results of A-Level statistics.
Abstract: Many students can get a lot of answers in A-level statistics right. But some of the ideas behind the calculations are somewhat subtle. Some of the subtleties are occasionally brought out in examination questions, usually those requiring verbal or discursive answers; unfortunately answers to these questions tend to be poor. Such questions are often dismissed as ‘waffle’ by strong students, while weaker ones generally try to avoid issues of understanding by memorising what they hope will be relevant slogans. Even respected textbooks are open to criticism. This is all very sad as I think it is here that much of the interest of the subject lies; ignoring the subtleties risks reducing the subject to a series of meaningless formulae. There should be time to focus on what is actually happening, as well as on mere technique. This article looks at some common problem areas.
3 citations
TL;DR: Tasks like the first can provide excellent experience of detecting and expressing generality as one route to algebra, but only if learners are first asked to describe a structural relationship which generates the succeeding items.
Abstract: Consider two related tasks: A) The diagram below displays a fragment of a pattern known to extend indefinitely and to be generated by an endlessly repeating pattern. What kind of shading will the 137th cell have? What number cell will the 137th white cell be in? B) What is the next term in the sequence 3, 6, 11, 18, 27, …? Tasks like the first can provide excellent experience of detecting and expressing generality as one route to algebra, but only if learners are first asked to describe a structural relationship which generates the succeeding items. Tasks like the second feature in intelligence tests, but it is well known that such tasks are mathematically incomplete.
3 citations
TL;DR: In this article, Mezo et al. showed that one can remove infinitely many numbers from a given set without losing the denseness of the remaining set, but the careful reader would have realised that the above construction does not work directly when p = O.
Abstract: The second and third examples show that we can remove infinitely many numbers from Q without losing the denseness of the remaining set. The careful reader would have realised that the above construction does not work directly when p = O. How is it possible to 'repair' the proof? (Or, why is it not necessary to repair it? That D is dense in IR\\{OJ does not imply that D is dense in the whole set IR?) Finally, we close with a small challenge: how should one generalise denseness and our above theorem to the Euclidean plane, three dimensional space or even higher dimensions? ISTVANMEZ6 Departamento de Matemdtica, Escuela Politecnica Nacional, Ladr6n de Guevara E11-253, Quito. Ecuador e-mail: istvan. mezo@epn.edu.ec
TL;DR: In this paper, the authors considered the arithmetic-geometric mean of positive numbers a > b, given a 0 = a, b 0 = b and at each stage the two new numbers are the arithmetic and geometric means of the previous two.
Abstract: Given positive numbers a > b, consider the ‘agm iteration’ given by a 0 = a, b 0 = b and At each stage, the two new numbers are the arithmetic and geometric means of the previous two. It is easily seen that bn < an, (an) is decreasing, (bn ) is increasing and an + 1 − bn + 1 < ½(an − bn), and hence that (an ) and (bn ) converge to a common limit, which is called the arithmetic-geometric mean of a and b. We will denote it by M (a, b).
TL;DR: In this article, a well-known particular value is ζ (2) = π2/6, and several alternative proofs of this fact have been presented in the Gazette, e.g., the recent notes [1], [2].
Abstract: Recall that for integers n ≥ 2, ζ (n) is defined by Of course, ζ (1) is not defined, since is divergent. A well-known particular value is ζ (2) = π2/6: numerous alternative proofs of this fact have been presented in the Gazette, e.g. the recent notes [1], [2].
TL;DR: In this paper, the authors tackle the problem of finding lx, Iy, I, about the analogous axes through 0 for a uniform elliptical wire of mass m. They will assume that 0 ~ b throughout.
Abstract: 98.08 The moment of inertia of an elliptical wire Introduction The standard moments of inertia of a uniform circular ring, mass m, radius a, are among the most straightforward to calculate: I, = Iy = !ma2 about perpendicular axes in its plane through its centre, 0, and l, = ma: about an axis perpendicular to its plane through O. In this note, we tackle the considerably harder problem of finding lx, Iy, I, about the analogous axes through 0 for a uniform elliptical wire of mass m: see Figure 1. We will assume that 0 ~ b throughout. Y
TL;DR: The incentre as mentioned in this paper is a metric-related function that has the property 0p2 = Ji2 'I' (P), where 0 is the circumcentre and R, r the circumradius and inradius, respectively.
Abstract: The incentre A key formula, which may be established by routine trigonometry, is 102 = Ji2 2Rr where 0 is the circumcentre and R, r the circumradius and inradius, respectively. This formula follows easily from a metric-related function 'I' (P). For a point P with actual or normalised areal coordinates P [x, y, zl (x + Y + z = 1), we have 'I' (P) = ayz + bzx + cxy which has the property 0p2 = Ji2 'I' (P). (See [1].)
TL;DR: In this paper, the authors derived Vieta's product of nested radicals for π from the Archimedean iterative algorithm for calculating π using a method involving the two equations: and in using this algorithm, they start with a circle of diameter 1.
Abstract: In this paper we show how to derive Vieta’s famous product of nested radicals for π [1] from the Archimedean iterative algorithm for π, [2, 3]. Only simple algebraic manipulations are needed. The Archimedean iterative algorithm for calculating π uses a method involving the two equations: and In using this algorithm, we start with a circle of diameter 1. Imagine two regular polygons each with the same number of sides, circumscribed and inscribed to this circle. The larger one has perimeter a 0, the smaller has perimeter b 0. Since the perimeter of the circle is π we have b 0 < π < a 0. Now consider regular polygons with twice the number of sides that circumscribe and inscribe the circle and call their perimeters a 1 and b 1 respectively. These can be calculated from (1) and (2). Continuing in this way we generate the perimeters of inscribed and circumscribed regular polygons, and in each case the number of sides is twice the sides of the previous polygon. Clearly the sequence {an } approaches π from above and {bn } approaches π from below.
TL;DR: In the half-circle centred at B with radius c = AB, draw the chord AA' with / and let C be the point where the chord meets the perpendicular bisector of A'B, so that ZA'BC = a as discussed by the authors.
Abstract: In the half-circle centred at B with radius c = AB, draw the chord AA' with /.BAA' = a and let C be the point where the chord meets the perpendicular bisector of A'B, so that ZA'BC = a. Simple angle chasing shows that y = 2a. The similar isosceles triangles BA'C and A'AB give § = ^-, that is, c = a(a + b) and AA' = ^. In the right-angled triangle BMC we have cos a = -%: the range of a is (c/2, c) since a < 60°. The bisector CD of y (with foot D on c) is parallel to A'B and of length "£• by similarity of ACD and BA'C.
TL;DR: In the July 2011 Gazette Paseau considers a population in which each birth is an independent event with probability g of being a girl and b ofBeing a boy (b + g = 1), and correctly states and shows that this will mean that the ratio Expected number of girl births : Ex expected number of boy births is g : b.
Abstract: In the July 2011 Gazette Paseau [1] considers a population in which each birth is an independent event with probability g of being a girl and b of being a boy (b + g = 1). He supposes that, due to a preference for girls, all couples stop after producing k girls, where k is an integer greater or equal to 1. Each new birth will, even after some couples have stopped producing, still have the probability g of being a girl and b of being a boy. Paseau correctly states and shows that this will mean that the ratio Expected number of girl births : Expected number of boy births is g : b. So it is the same as if all couples stopped randomly. Paseau was showing that this result was a consequence of the ratio g : b always being the same for all families. However in humans the ratio g : b is not always the same. The ratio g : b can vary significantly with many variables between couples, which influence either the ratio at conception or the subsequent implantation and survival of male or female foetuses to birth. For example, mothers whose menstrual cycles have a short follicular phase (normally days 1-14 of the 28-day cycle) tend to produce boys [2], resulting in a smaller g : b ratio. Also the timing of insemination within the menstrual cycle is associated with the gender outcome such that insemination early or late in the fertile period means that the offspring is more likely to be male; if insemination occurs in the middle it is more likely to be female [3]. It is therefore likely that changes in parental hormone concentrations during the fertile period can affect the sexes of the children [4].
TL;DR: A heptagonal triangle is defined by angles in the ratio 1 : 2 : 4 and is given by the first, second, and fourth vertices of a regular heptagon.
Abstract: For example, n = 2, m = 3 gives a = 4, b = 5, c = 6;« = 3, m = 4 gives 9, 7, 12; w = 3, m = 5 gives 9, 16, 15. A heptagonal triangle is defined by angles in the ratio 1 : 2 : 4 and is given by the first, second, and fourth vertices of a regular heptagon: it is the unique triangle shape that appears twice in our construction. As cos(-^p) = 0.90097, the triangle with sides 25, 56 and 45 (corresponding to n = 5,m = 9) is nearly heptagonal.
TL;DR: In this article, the authors studied the equilateral outscribed equilateral triangles of an arbitrary triangle and determined the area of the largest such triangles, and proved that the largest equilateral equilateral triangle of 4ABC can be constructed by ruler and compass and its area equals a +b+c 2 √ 3 +2S4ABC.
Abstract: An outscribed triangle of a triangle 4ABC is a triangle 4DEF such that each side of 4DEF contains a vertex of 4ABC In this article we study the equilateral outscribed triangles of an arbitrary triangle and determine the area of the largest such triangles We prove that the largest outscribed equilateral triangle of 4ABC can be constructed by ruler and compass and its area equals a +b+c 2 √ 3 +2S4ABC where S4ABC denotes the area of 4ABC Given two triangles 4ABC and 4DEF , if each side of 4DEF contains a vertex of 4ABC, then we call 4DEF an outscribed triangle of 4ABC Given 4ABC, let Φ4ABC be the set of all outscibed equilateral triangles of 4ABC Clearly Φ4ABC is non-empty In the following we will determine the area of the largest member of Φ4ABC and show that this largest member can be constructed by ruler and compass from 4ABC The corresponding problem on quadrilaterals has been considered in [1] 1 Area of the largest outscribed equilateral triangle Given a triangle 4ABC, let a and b denote the lengths of the sides BC and AC, respectively, and θ denote the angle ∠ACB Let 4DEF be any member in Φ4ABC as shown in Figure 1 and put t = ∠DCB
TL;DR: In this article, Dickson et al. presented a solution to the problem of approximating the construction of a radian for the Tonbridge School of Tonbridge, Kent, UK.
Abstract: References I. J. Dickson. Problem 93.H, Math. Gaz. 93 (July 2009) p. 367 & Solution to Problem 93.H, Math. Gaz. 94 (March 2010) pp. 163-165. 2. I. Niven, Irrational numbers, Math. Assoc. of America (1956) p. 131. 3. J. M. Child, Approximate construction for a radian, Math. Gaz. 5 (October 1909) pp. 142-143. JIM DICKSON Flat 2, 1 Park Crescent, Leeds LS8 1DH NICK LORD Tonbridge School, Kent TN9 11P
TL;DR: In this article, the authors show that unless C is a square, it is impossible to find a convex quadrilateral that encloses a larger area than C. The argument is in three steps.
Abstract: IQ < is unless C is a square, when equality holds. Standard proofs of this statement use trigonometry and several pages of argument, [1]. Here we offer a short mostly geometric argument. The argument is in three steps. We fix P and think of maximising A. 1) If C is a quadrilateral that is non-convex, then there is a convex quadrilateral with the same P that encloses a larger area. 2) If C is a convex quadrilateral that is not a rhombus, then there is a rhombus with the same P and strictly larger enclosed area. 3) If C is a rhombus but not a square then the square with the same P has strictly larger area. Let ABCD be a quadrilateral with vertices named counterclockwise. On 1: If one diagonal, say AC, is external, then let ABCD\" be the quadrilateral with Lf the mirror image of D through AC. This satisfies the claim. On 2: Let ABCD be a convex quadrilateral with \\AD\\ * \\DC\\. Consider the ellipse with foci A and C and passing through D. If we take ZX on this ellipse near D, then ABCD' has the same perimeter as ABCD. The enclosed area depends only on the distance of D\" from AC. It is maximised when D\" is at the end of the minor axis. Since D was not there, ABCD\" has strictly larger area. Now do the same argument by passing an ellipse through B and replacing B by B\" at the minor axis. The quadrilateral AB\"CD\" has \\AEf\\ = \\CLf\\, \\AB\"\\ = \\CB\"\\, and area larger than ABCD. Next do the ellipse argument a third time with the foci at B\" and D'. The ellipse through A will also pass through C so we can simultaneously replace A and C by A' and C\" at the minor axis. We have a rhombus as required. On 3: In the rhombus ABCD draw a diagonal, say AC. The enclosed area is the sum of two triangles with parallel bases of fixed length. Their heights are simultaneously maximised when the rhombus is a square.
TL;DR: A strategy is described that improves on the value of 22 weighings, and is structured as follows, which proves the main result, that 20 weighings is always sufficient to sort 15 marbles.
Abstract: Many mathematical puzzles involve using a scale to compare the weights of two objects, or two groups of objects. The present work studies a similar but different type of question involving a scale. We are given a set of marbles, whose weights are all different. The only way to distinguish them is to use a special kind of scale. The scale has three trays and each can accept exactly one marble. The scale then indicates which is the heaviest, the lightest and consequently the middle one of the three marbles. The paper studies the question of ordering 15 different marbles with as few weighings as possible. The problem is taken from the website Enigmes, casse-têtes, curiosités et autres bizarreries [1] and Toppuzzle [2] where the best reported strategy requires 23 weighings. The problem can also be found at the website Trick of Mind [3] where a strategy requiring 22 weighings is proposed. These websites contain many challenging mathematical puzzles, some are classical problems, others are original creations. The author is not aware of any other work on this problem. The present paper describes a strategy that improves on the value of 22 weighings, and is structured as follows. Section 2 introduces some useful definitions and provides strategies to sort up to 9 marbles. Section 3 proves our main result, that 20 weighings is always sufficient to sort 15 marbles.
TL;DR: The theorem of this paper was discovered by the computer program 'Discoverer' created by the authors as discussed by the authors, which gave a proof which is rewritten in a traditional manner, and the list was partially produced by using the theorem.
Abstract: The theorem of this paper was discovered by the computer program 'Discoverer', created by the authors. The proofs of 'Discoverer' are nonstandard. In this paper we gave a proof which is rewritten in a traditional manner. In [6] the authors present a list of more than 2000 remarkable points which lie on the Kiepert hyperbola. The list was partially produced by using the theorem of this paper. The complete list was produced by the computer program 'Discoverer'.
TL;DR: The well-known Fibonacci and Lucas numbers continue to faxcinate the mathematical community with their beauty, elegance, ubiquity, and applicability as discussed by the authors, and they are still a fertile ground for additional activities, for fibonacci enthusiasts and amateurs alike.
Abstract: The well-known Fibonacci and Lucas numbers continue to faxcinate the mathematical community with their beauty, elegance, ubiquity, and applicability. After several centuries of exploration, they are still a fertile ground for additional activities, for Fibonacci enthusiasts and amateurs alike. Fibonacci numbers F n and Lucas numbers L n belong to a large integer family { x n }, often defined by the recurrence x n = x n −1 + x n −2 , where x 1 = a , x 2 = b , and n ≥ 3. When a = b = 1, x n = F n ; and when a = 1 and b = 3, x n = L n . Clearly, F 0 = 0 and L 0 = 2. They satisfy a myriad of elegant properties [1,2,3]. Some of them are: In this article, we will give a brief introduction to the Q -matrix, employ it in the construction of graph-theoretic models [4, 5], and then explore some of these identities using them. In 1960 C.H. King studied the Q -matrix
TL;DR: In this paper, a family has two children. One of them is a girl, and the other is a boy, and both of them were born on a Tuesday, and one of them has green hair.
Abstract: We came across the ‘two girls’ version of the children's gender problem nearly 35 years ago. How we came to it we cannot remember, but Martin Gardner had published a variant of it in the Scientific American in 1959. It re-emerged for us in the summer of 2010, following the publication of an article in Science News [1]. Subsequently Keith Devlin wrote about how this re-emergence impacted on him, and noting that ‘Probability Can Bite“ [2]. The mathematics herein reflects and extends that in Devlin's article. In case the reader has not encountered the problem before, we first pose four problems. 1. A family has two children. One of them is a girl. What is the probability that they are both girls? 2. A family has two children. The younger is a girl. What is the probability that they are both girls? 3. A family has two children. One of them is a girl, and she was born on a Tuesday. What is the probability that they are both girls? 4. A family has two children. One of them is a girl, and she has green hair. What is the probability that they are both girls?
TL;DR: In this article, it was shown that each of the first and last terms of the continued fraction algorithm is centrally symmetric (a palindrome) and that the process continues indefinitely unless the latter occurs.
Abstract: Each of them, excluding the first and the last terms, is centrally symmetric (a palindrome). The goal of this note is to prove that this is true in general. Alan Baker's beautifully written book, A Concise Introduction to the Theory of Numbers, [1], is an excellent reference for our purpose. It is an ideal place to start. This is how he describes the continued fraction algorithm [1, p. 44]. To describe the algorithm, let z (our notation) be any real number. We put a0 = [zj. If a0 * z we write z = a0 + 1/zi, so that Z\\ > 1, and we put a! = [z\\]. Ifai * Z\\ we write z\\ = a\\ + llzi, so that zi > 1, and we put a2 = [zi]The process continues indefinitely unless an = z„ for some n. It is clear that if the latter occurs then z is rational; in fact we have
TL;DR: In the course of a coffee-table conversation with my friends regarding the nature of static equilibrium of different solid objects, the situation involving a uniform hemisphere came up as mentioned in this paper, and they decided to probe into the matter quantitatively, and found that if the prolate hemispheroid is made indefinitely taller, keeping its equatorial radius fixed, then the equilibrium should eventually become unstable.
Abstract: In the course of a coffee-table conversation with my friends regarding the nature of static equilibrium of different solid objects the situation involving a uniform hemisphere came up. Intuition (and perhaps experience) tells that a uniform hemisphere as shown in Figure 1 resting on a flat surface will be at stable equilibrium, and so will an oblate hemispheroid as shown in Figure 2. Things get complicated when we move to a prolate hemispheroid like the one shown in Figure 3, for the nature of its equilibrium is less obvious. The intuition does come to mind though that if the prolate hemispheroid is made indefinitely taller, keeping its equatorial radius fixed, then the equilibrium should eventually become unstable. Intrigued, we decided to probe into the matter quantitatively.