Mathematical Recreations and Essays · W. W. Rouse Ball
The Three Problems Among the more interesting geometrical problems of antiquity are three questions which attracted the special attention of the early Greek mathematicians. Our knowledge of geometry is derived from Greek sources, and thus these questions have attained a classical position in the history of the subject. The three questions to which I refer are (i) the duplication of a cube, that is, the determination of the side of a cube whose volume is double that of a given cube; (ii) the trisection of an angle; and (iii) the squaring of a circle, that is, the determination of a square whose area is equal to that of a given circle--each problem to be solved by a geometrical construction involving the use of straight lines and circles only, that is, by Euclidean geometry.
With the restriction last mentioned all three problems are insoluble (F. Klein, Vorträge über ausgewählte Fragen der Elementargeometrie, Leipzig, 1895.). To duplicate a cube the length of whose side is a , we have to find a line of length x , such that x^3 = 2a^3. Again, to trisect a given angle, we may proceed to find the sine of the angle, say a , then, if x is the sine of an angle equal to one-third of the given angle, we have 4x^3=3x-a. Thus the first and second problems, when considered analytically, require the solution of a cubic equation; and since a construction by means of circles (whose equations are of the form x^2 + y^2 + ax + by + c = 0 ) and straight lines (whose equations are of the form α x + β y +γ = 0 )
cannot be equivalent to the solution of a cubic equation, it is inferred that the problems are insoluble if in our constructions we are restricted to the use of circles and right lines. If the use of the conic sections is permitted, both of these questions can be solved in many ways. The third problem is different in character, but under the same restrictions it also is insoluble.
I propose to give some of the constructions which have been proposed for solving the first two of these problems. To save space, I shall not draw the necessary diagrams, and in most cases I shall not add the proofs: the latter present but little difficulty. I shall conclude with some historical notes on approximate solutions of the quadrature of the circle.
[The Duplication of the Cube] The Duplication of the Cube (See Historia Problematis de Cubi Duplicatione by N.T. Reimer, Göttingen, 1798; and Historia Problematis Cubi Duplicandi by C.H. Biering, Copenhagen, 1844: also Das Delische Problem, by A. Sturm, Linz, 1895--7. Some notes on the subject are given in my History of Mathematics.)
tocsubsectionLegendary origin of the problem The problem of the duplication of the cube was known in ancient times as the Delian problem, in consequence of a legend that the Delians had consulted Plato on the subject. In one form of the story, which is related by Philoponus (Philoponus ad Aristotelis Analytica Posteriora, bk. i, chap. vii.), it is asserted that the Athenians in 430 b.c., when suffering from the plague of eruptive typhoid fever, consulted the oracle at Delos as to how they could stop it. Apollo replied that they must double the size of his altar which was in the form of a cube. To the unlearned suppliants nothing seemed more easy, and a new altar was constructed either having each of its edges double that of the old one (from which it followed that the volume was increased eight-fold) or by placing a similar cubic altar next to the old one. Whereupon, according to the legend, the indignant god made the pestilence worse than before, and informed a fresh deputation that it was useless to
trifle with him, as his new altar must be a cube and have a volume exactly double that of his old one. Suspecting a mystery the Athenians applied to Plato, who referred them to the geometricians. The insertion of Plato's name is an obvious anachronism. Eratosthenes (Archimedis Opera cum Eutocii Commentariis, ed. Torelli, Oxford, 1792, p. 144; ed. Heiberg, Leipzig, 1880--1, vol. iii, pp. 104--107.) relates a somewhat similar story, but with Minos as the propounder of the problem.
In an Arab work, the Greek legend was distorted into the following extraordinarily impossible piece of history, which I cite as a curiosity of its kind. "Now in the days of Plato," says the writer, "a plague broke out among the children of Israel. Then came a voice from heaven to one of their prophets, saying, 'Let the size of the cubic altar be doubled, and the plague will cease'; so the people made another altar like unto the former, and laid the same by its side. Nevertheless the pestilence continued to increase. And again the voice spake unto the prophet, saying, 'They have made a second altar like unto the former, and laid it by its side, but that does not produce the duplication of the cube.' Then applied they to Plato, the Grecian sage, who spake to them, saying, 'Ye have been neglectful of the science of geometry, and therefore hath God chastised you, since geometry is the most sublime of all the sciences.' Now, the duplication of a cube depends on a rare problem in geometry, namely". And then follows the solution of Apollonius, which is given later.
If a is the length of the side of the given cube and x that of the required cube, we have x^3= 2a^3 , that is, x: a = sqrt(2): 1. It is probable that the Greeks were aware that the latter ratio is incommensurable, in other words, that no two integers can be found whose ratio is the same as that of sqrt(2): 1 , but it did not therefore follow that they could not find the ratio by geometry: in fact, the side and diagonal of a square are instances of lines whose numerical measures are incommensurable.
I proceed now to give some of the geometrical constructions which have been proposed for the duplication of the cube (On the application to this problem of the traditional Greek methods of analysis by Hero and Philo (leading to the solution by the use of Apollonius's circle), by Nicomedes (leading to the solution by the use of the conchoid), and by Pappus (leading to the solution by the use of the cissoid), see Geometrical Analysis by J. Leslie, Edinburgh, second edition, 1811, pp. 247--250, 453.). With one exception, I confine myself to those which can be effected by the aid of the conic sections.
Lemma of Hippocrates Hippocrates (Proclus, ed. Friedlein, pp. 212, 213.) (circ. 420 b.c.) was perhaps the earliest mathematician who made any progress towards solving the problem. He did not give a geometrical construction, but he reduced the question to that of finding two means between one straight line ( a ), and another twice as long ( 2a ). If these means are x and y , we have a: x = x: y = y: 2a , from which it follows that x^3 = 2a^3. It is in this form that the problem is always presented now. Formerly any process of solution by finding these means was called a mesolabum.
Solutions of Archytas, Plato, Menaechmus, Apollonius, and Sporus One of the first solutions of the problem was that given by Archytas (Archimedis Opera, ed. Torelli, p. 143; ed. Heiberg, vol. iii, pp. 98--103.) in or about the year 400 b.c. His construction is equivalent to the following. On the diameter OA of the base of a right circular cylinder describe a semicircle whose plane is perpendicular to the base of the cylinder. Let the plane containing this semicircle rotate round the generator through O , then the surface traced out by the semicircle will cut the cylinder in a tortuous curve. This curve will itself be cut by a right cone, whose axis is OA and semi-vertical angle is (say) 60^circ , in a point P , such that the projection of OP on the base of the cylinder will be to the radius of the cylinder in the ratio of the side of the required cube to that of the given cube. Of course the proof given by Archytas is geometrical; and it is interesting to note that in it he shows himself familiar with the results of the propositions Euc. iii, 18, iii, 35, and
xi, 19. To show analytically that the construction is correct, take OA as the axis of x , and the generator of the cylinder drawn through O as axis of z , then with the usual notation, in polar coordinates, if a is the radius of the cylinder, we have for the equation of the surface described by the semicircle r = 2a θ ; for that of the cylinder r θ = 2a φ ; and for that of the cone θ φ = 1/2. These three surfaces cut in a point such that ^3θ = 1/2 , and therefore (rθ)^3=2a^3. Hence the volume of the cube whose side is r θ is twice that of the cube whose side is a.
The construction attributed to Plato (ibid:5Archimedis Opera, ed. Torelli, p. 135; ed. Heiberg, vol. iii, pp. 66--71.) (circ. 360 b.c.) depends on the theorem that, if CAB and DAB are two right-angled triangles, having one side, AB , common, their other sides, AD and BC , parallel, and their hypothenuses, AC and BD , at right angles, then if these hypothenuses cut in P , we have PC : PB = PB : PA = PA : PD. Hence, if such a figure can be constructed having PD = 2PC , the problem will be solved. It is easy to make an instrument by which the figure can be drawn.
The next writer whose name is connected with the problem is Menaechmus (ibid:6Archimedis Opera, ed. Torelli, pp. 141--143; ed. Heiberg, vol. iii, pp. 92--99.), who in or about 340 b.c. gave two solutions of it.
In the first of these he pointed out that two parabolas having a common vertex, axes at right angles, and such that the latus rectum of the one is double that of the other will intersect in another point whose abscissa (or ordinate) will give a solution. If we use analysis this is obvious; for, if the equations of the parabolas are y^2 = 2ax and x^2=ay , they intersect in a point whose abscissa is given by x^3 = 2a^3. It is probable that this method was suggested by the form in which Hippocrates had cast the problem: namely, to find x and y so that a : x = x : y = y : 2a , whence we have x^2 = ay and y^2 = 2ax.
The second solution given by Menaechmus was as follows.
Describe a parabola of latus rectum l. Next describe a rectangular hyperbola, the length of whose real axis is 4l , and having for its asymptotes the tangent at the vertex of the parabola and the axis of the parabola. Then the ordinate and the abscissa of the point of intersection of these curves are the mean proportionals between l and 2l. This is at once obvious by analysis. The curves are x^2 = ly and xy = 2l^2. These cut in a point determined by x^3 = 2l^3 and y^3 = 4l^3. Hence l : x = x : y = y : 2l.
The solution of Apollonius (ibid:7Archimedis Opera, ed. Torelli, p. 137; ed. Heiberg, vol. iii, pp. 76--79. The solution is given in my History of Mathematics, London, 1901, p. 84.), which was given about 220 b.c., was as follows. The problem is to find two mean proportionals between two given lines. Construct a rectangle OADB , of which the adjacent sides OA and OB are respectively equal to the two given lines. Bisect AB in C. With C as centre describe a circle cutting OA produced in a and cutting OB produced in b , so that aDb shall be a straight line. If this circle can be so described, it will follow that OA : Bb = Bb : Aa = Aa : OB , that is, Bb and Aa are the two mean proportionals between OA and OB. It is impossible to construct the circle by Euclidean geometry, but Apollonius gave a mechanical way of describing it.
The only other construction of antiquity to which I will refer is that given by Diocles and Sporus (ibid:8Archimedis Opera, ed. Torelli, pp. 138, 139, 141; ed. Heiberg, vol. iii, pp. 78--84, 90--93.). It is as follows. Take two sides of a rectangle OA , OB , equal to the two lines between which the means are sought. Suppose OA to be the greater. With centre O and radius OA describe a circle. Let OB produced cut the circumference in C and let AO produced cut it in D. Find a point E on BC so that if DE cuts AB produced in F and cuts the circumference in G , then FE=EG. If E can be found, then OE is the first of the means between OA and OB. Diocles invented the cissoid in order to determine
E , but it can be found equally conveniently by the aid of conics.
Solutions of Vieta, Descartes, Gregory of St Vincent, and Newton In more modern times several other solutions have been suggested. I may allude in passing to three given by Huygens (Opera Varia, Leyden, 1724, pp. 393--396.), but I will enunciate only those proposed respectively by Vieta, Descartes, Gregory of St Vincent, and Newton.
Vieta's construction is as follows (Opera Mathematica, ed. Schooten, Leyden, 1646, prop, v, pp. 242--243.). Describe a circle, centre O , whose radius is equal to half the length of the larger of the two given lines. In it draw a chord AB equal to the smaller of the two given lines. Produce AB to E so that BE = AB. Through A draw a line AF parallel to OE. Through O draw a line DOCFG , cutting the circumference in D and C , cutting AF in F , and cutting BA produced in G , so that GF = OA. If this line can be drawn then AB : GC = GC : GA = GA : CD.
Descartes pointed out (Geometria, bk. iii, ed. Schooten, Amsterdam, 1659, p. 91.) that the curves x^2 = ay and x^2 + y^2 = ay + bx cut in a point (x, y) such that a : x = x : y = y : b. Of course this is equivalent to the first solution given by Menaechmus, but Descartes preferred to use a circle rather than a second conic.
Gregory's construction was given in the form of the following theorem (Gregory of St Vincent, Opus Geometricum Quadraturae Circuli, Antwerp, 1647, bk. vi, prop. 138, p. 602.). The hyperbola drawn through the point of intersection of two sides of a rectangle so as to have the two other sides for its asymptotes meets the circle circumscribing the rectangle in a point whose distances from the asymptotes are the mean proportionals between two adjacent sides of the rectangle. This is the geometrical expression of the proposition that the curves xy = ab and x^2 + y^2 = ay + bx cut in a point (x, y) such that a : x = x : y = y : b.
One of the constructions proposed by Newton is as follows (Arithmetica Universalis, Ralphson's (second) edition, 1728, p. 242; see also pp. 243, 245.).
Let OA be the greater of two given lines. Bisect OA in B. With centre O and radius OB describe a circle. Take a point C on the circumference so that BC is equal to the other of the two given lines. From O draw ODE cutting AC produced in D , and BC produced in E , so that the intercept DE=OB. Then BC : OD = OD : CE = CE : OA. Hence OD and CE are two mean proportionals between any two lines BC and OA
.
The trisection of an angle is the second of these classical problems, but tradition has not enshrined its origin in romance. The following two constructions are among the oldest and best known of those which have been suggested; they are quoted by Pappus (Pappus, Mathematicae Collectiones, bk. iv, props. 32, 33 (ed. Commandino, Bonn, 1670, pp. 97--99). On the application to this problem of the traditional Greek methods of analysis see Geometrical Analysis, by J. Leslie, Edinburgh, second edition, 1811, pp. 245--247.), but I do not know to whom they were due originally.
Solutions quoted by Pappus (three) The first of them is as follows. Let AOB be the given angle. From any point P in OB draw PM perpendicular to OA. Through P draw PR parallel to OA. On MP take a point Q so that if OQ is produced to cut PR in R then QR = 2 OP. If this construction can be made, then AOR=1/3AOB. The solution depends on determining the position of R. This was effected by a construction which may be expressed analytically thus. Let the given angle be ^-1 (b/a). Construct the hyperbola xy=ab , and the circle (x-a)^2 + (y-b)^2 = 4(a^2 + b^2). Of the points where they cut, let x be the abscissa which is greatest, then PR = x-a , and ^-1 (b/x)=1/3^-1(b/a).
The second construction is as follows. Let AOB be the given angle. Take OB=OA , and with centre O and radius OA describe a circle. Produce AO indefinitely and take a point C on it external to the circle so that if CB cuts the circumference
in D then CD shall be equal to OA. Draw OE parallel to CDB. Then, if this construction can be made, AOE = 1/3 AOB. The ancients determined the position of the point C by the aid of the conchoid: it could be also found by the use of the conic sections.
I proceed to give a few other solutions; confining myself to those effected by the aid of conics.
Among the other constructions given by Pappus (Pappus, bk. iv, prop. 34, pp. 99--104.) I may quote the following. Describe a hyperbola whose eccentricity is two. Let its centre be C and its vertices A and A'. Produce CA' to S so that A'S = CA'. On AS describe a segment of a circle to contain the given angle. Let the orthogonal bisector of AS cut this segment in O. With centre O and radius OA or OS describe a circle. Let this circle cut the branch of the hyperbola through A' in P. Then SOP = 1/3SOA.
Solutions of Descartes, Newton, Clairaut, and Chasles In modern times one of the earliest of the solutions by a direct use of conics was suggested by Descartes, who effected it by the intersection of a circle and a parabola. His construction (Geometria, bk. iii, ed. Schooten, Amsterdam, 1659, p. 91.) is equivalent to finding the points of intersection other than the origin, of the parabola y^2 = 1/4x and the circle x^2 + y^2 - 13/4x + 4ay = 0. The ordinates of these points are given by the equation 4y^3 = 3y - a. The smaller positive root is the sine of one-third of the angle whose sine is a. The demonstration is ingenious.
One of the solutions proposed by Newton is practically equivalent to the third one which is quoted above from Pappus. It is as follows (Arithmetica Universalis, problem xlii, Ralphson's (second) edition, London, 1728, p. 148; see also pp. 243--245.). Let A be the vertex of one branch of a hyperbola whose eccentricity is two, and let S be the focus of the other branch. On AS describe the segment of a circle containing an angle equal to the supplement of the given angle. Let this circle cut the S branch of the hyperbola in P. Then PAS will be equal to one-third of the given angle.
The following elegant solution is due to Clairaut (I believe that this was first given by Clairaut, but I have mislaid my reference. The construction occurs as an example in the Geometry of Conics, by C. Taylor, Cambridge, 1881, No. 308, p. 126.). Let AOB be the given angle. Take OA = OB , and with centre O and radius OA describe a circle. Join AB , and trisect it in H , K , so that AH= HK= KB. Bisect the angle AOB by OC cutting AB in L. Then AH = 2 HL. With focus A , vertex H , and directrix OC , describe a hyperbola. Let the branch of this hyperbola which passes through H cut the circle in P. Draw PM perpendicular to OC and produce it to cut the circle in Q. Then by the focus and directrix property we have AP : PM = AH : HL = 2 : 1 , Therefore AP = 2 PM = PQ. Hence, by symmetry, AP = PQ = QR. Therefore AOP = POQ = QOR.
I may conclude by giving the solution which Chasles (Traité des sections coniques, Paris, 1865, art. 37, p. 36.) regards as the most fundamental. It is equivalent to the following proposition. If OA and OB are the bounding radii of a circular arc AB , then a rectangular hyperbola having OA for a diameter and passing through the point of intersection of OB with the tangent to the circle at A will pass through one of the two points of trisection of the arc.
The object of the third of the classical problems was the
determination of a side of a square whose area should be equal to that of a given circle.
The investigation, previous to the last two hundred years, of this question was fruitful in discoveries of allied theorems, but in more recent times it has been abandoned by those who are able to realize what is required. The history of this subject has been treated by competent writers in such detail that I shall content myself with a very brief allusion to it.
Archimedes showed (Archimedis Opera, , prop. i, ed. Torelli, pp. 203--205; ed. Heiberg, vol. i, pp. 258--261, vol. iii, pp. 269--277.) (what possibly was known before) that the problem is equivalent to finding the area of a right-angled triangle whose sides are equal respectively to the perimeter of the circle and the radius of the circle. Half the ratio of these lines is a number, usually denoted by π.
Incommensurability of pi That this number is incommensurable had been long suspected, and has been now demonstrated. The earliest analytical proof of it was given by Lambert (Mémoires de l'Académie de Berlin for 1761, Berlin, 1768, pp. 265--322.) in 1761; in 1803 Legendre (Legendre's Geometry, Brewster's translation, Edinburgh, 1824, pp. 239--245.) extended the proof to show that π^2 was also incommensurable; and recently Lindemann (Ueber die Zahl π , Mathematische Annalen, Leipzig, 1882, vol. xx, pp. 213--225. The proof leads to the conclusion that, if x is a root of a rational integral algebraical equation, then e^x cannot be rational: hence, if π i was the root of such an equation, e^(π i) could not be rational; but e^(π i) is equal to -1 , and therefore is rational; hence π i cannot be the root of such an algebraical equation, and therefore neither can π .) has shown that π cannot be the root of a rational algebraical equation.
An earlier attempt by James Gregory to give a geometrical demonstration of this is worthy of notice. Gregory proved (Vera Circuli et Hyperbolae Quadratura, Padua, 1668: this is reprinted in Huygens's Opera Varia, Leyden, 1724, pp. 405--462.) that the ratio of the area of any arbitrary sector to that of the inscribed or circumscribed polygons is not expressible by a finite number of algebraical terms. Hence he inferred that the quadrature was impossible. This was accepted by Montucla,
but it is not conclusive, for it is conceivable that some particular sector might be squared, and this particular sector might be the whole circle.
In connection with Gregory's proposition above cited, I may add that Newton (Principia, bk. i, section vi, lemma xxviii.) proved that in any closed oval an arbitrary sector bounded by the curve and two radii cannot be expressed in terms of the co-ordinates of the extremities of the arc by a finite number of algebraical terms. The argument is condensed and difficult to follow: the same reasoning would show that a closed oval curve cannot be represented by an algebraical equation in polar co-ordinates. From this proposition no conclusion as to the quadrature of the circle is to be drawn, nor did Newton draw any. In the earlier editions of this work I expressed an opinion that the result presupposed a particular definition of the word oval, but on more careful reflection I think that the conclusion is valid without restriction.
With the aid of the quadratrix, or the conchoid, or the cissoid, the quadrature of the circle is easy, but the construction of those curves assumes a knowledge of the value of π , and thus the question is begged.
Approximations to the value of π.
Definitions of pi I need hardly add that, if π represented merely the ratio of the circumference of a circle to its diameter, the determination of its numerical value would have but slight interest. It is however a mere accident that π is defined usually in that way, and it really represents a certain number which would enter into analysis from whatever side the subject was approached.
I recollect a distinguished professor explaining how different would be the ordinary life of a race of beings born, as easily they might be, so that the fundamental processes of arithmetic, algebra and geometry were different to those which seem to us so evident, but, he added, it is impossible to conceive of a universe in which e and π should not exist.
I have quoted elsewhere an anecdote, which perhaps will bear repetition, that illustrates how little the usual definition
of π suggests its properties. De Morgan was explaining to an actuary what was the chance that a certain proportion of some group of people would at the end of a given time be alive; and quoted the actuarial formula, involving π , which, in answer to a question, he explained stood for the ratio of the circumference of a circle to its diameter. His acquaintance, who had so far listened to the explanation with interest, interrupted him and exclaimed, "My dear friend, that must be a delusion, what can a circle have to do with the number of people alive at the end of a given time?" In reality the fact that the ratio of the length of the circumference of a circle to its diameter is the number denoted by π does not afford the best analytical definition of π , and is only one of its properties.
Origin of symbol pi The use of a single symbol to denote this number 3.14159... seems to have been introduced about the beginning of the eighteenth century. W. Jones (Synopsis Palmariorum Matheseos, London, 1706, pp. 243, 263) in 1706 represented it by π ; a few years later (See notes by G. Eneström in the Bibliotheca Mathematica, Stockholm, 1889, vol. iii, p. 28; , 1890, vol. iv, p. 22.) John Bernoulli denoted it by c ; Euler in 1734 used p , and in 1736 used c ; Chr. Goldback in 1742 used π ; and after the publication of Euler's Analysis the symbol π was generally employed.
Methods of approximating to the numerical value of pi The numerical value of π can be determined by either of two methods with as close an approximation to the truth as is desired.
The first of these methods is geometrical. It consists in calculating the perimeters of polygons inscribed in and circumscribed about a circle, and assuming that the circumference of the circle is intermediate between these perimeters (The history of this method has been written by K.E.I. Selander, Historik öfver Ludolphska Talet, Upsala, 1868.). The approximation would be closer if the areas and not the perimeters were employed. The second and modern method rests on the determination of converging infinite series for π.
We may say that the π -calculators who used the first
method regarded π as equivalent to a geometrical ratio, but those who adopted the modern method treated it as the symbol for a certain number which enters into numerous branches of mathematical analysis.
Geometrical methods of approximation It may be interesting if I add here a list of some of the approximations to the value of π given by various writers (For the methods used in classical times and the results obtained, see the notices of their authors in M. Cantor's Geschichte der Mathematik, Leipzig, vol. i, 1880. For medieval and modern approximations, see the article by A. De Morgan on the Quadrature of the Circle in vol. xix of the Penny Cyclopaedia, London, 1841; with the additions given by B. de Haan in the Verhandelingen of Amsterdam, 1858, vol. iv, p. 22: the conclusions were tabulated, corrected, and extended by Dr J.W.L. Glaisher in the Messenger of Mathematics, Cambridge, 1873, vol. ii, pp. 119--128; and , 1874, vol. iii, pp. 27--46.). This will indicate incidentally those who have studied the subject to the best advantage.
tocsubsectionResults of Egyptians, Babylonians, Jews The ancient Egyptians (Ein mathematisches Handbuch der alten Aegypter ( the Rhind papyrus), by A. Eisenlohr, Leipzig, 1877, arts. 100--109, 117, 124.) took 256/81 as the value of π , this is equal to 3.1605... ; but the rougher approximation of 3 was used by the Babylonians (Oppert, Journal Asiatique, August, 1872, and October, 1874.) and by the Jews (1 Kings, ch. 7, ver. 23; 2 Chronicles, ch. 4, ver. 2.). It is not unlikely that these numbers were obtained empirically.
We come next to a long roll of Greek mathematicians who attacked the problem. Whether the researches of the members of the Ionian School, the Pythagoreans, Anaxagoras, Hippias, Antipho, and Bryso led to numerical approximations for the value of π is doubtful, and their investigations need not detain us. The quadrature of certain lunes by Hippocrates of Chios is ingenious and correct, but a value of π cannot be thence deduced; and it seems likely that the later members of the Athenian School concentrated their efforts on other questions.
It is probable that Euclid (These results can be deduced from Euc. iv, 15, and iv, 8: see also book xii, prop. 16.), the illustrious founder of the
Alex-and-rian School, was aware that π was greater than 3 and less than 4 , but he did not state the result explicitly.
tocsubsectionResults of Archimedes and other Greek writers The mathematical treatment of the subject began with Ar-chi-medes, who proved that π is less than 31/7 and greater than 310/71 , that is, it lies between 3.1428... and 3.1408.... He established (Archimedis Opera, , prop. iii, ed. Torelli, Oxford, 1792, pp. 205--216; ed. Heiberg, Leipzig, 1880, vol. i, pp. 263-271.) this by inscribing in a circle and circumscribing about it regular polygons of 96 sides, then determining by geometry the perimeters of these polygons, and finally assuming that the circumference of the circle was intermediate between these perimeters: this leads to the result 6336/2017frac14 <π<14688/4673frac12 ,Inserted 14688/
from which he deduced the limits given above. This method is equivalent to using the proposition θ < θ < θ , where θ= π/96 : the values of θ and θ were deduced by Archimedes from those of frac13 π and frac13π by repeated bisections of the angle. With a polygon of n sides this process gives a value of π correct to at least the integral part of (2 n - 1.19) places of decimals. The result given by Archimedes is correct to 2 places of decimals. His analysis leads to the conclusion that the perimeters of these polygons for a circle whose diameter is 4970 feet would lie between 15610 feet and 15620 feet--actually it is about 15613 feet 9 inches.
Apollonius discussed these results, but his criticisms have been lost.
Hero of Alexandria gave (Mensurae, ed. Hultsch, Berlin, 1864, p. 188.) the value 3 , but he quoted (Geometria, ed. Hultsch, Berlin, 1864, pp. 115, 136.) the result 22/7 : possibly the former number was intended only for rough approximations.
The only other Greek approximation that I need mention is that given by Ptolemy (Almagest, bk. vi, chap. 7; ed. Halma, vol. i, p. 421.), who asserted that π = 3^circ 8' 30". This is equivalent to taking π = 3 + frac860 + 30/3600 = 317/120 = 3.1416.
tocsubsectionResults of Roman surveyors and Gerbert The Roman surveyors seem to have used 3 , or sometimes 4 , for rough calculations. For closer approximations they often employed 31/8 instead of 31/7 , since the fractions then introduced are more convenient in duodecimal arithmetic. On the other hand Gerbert (uvres de Gerbert, ed. Olleris, Clermont, 1867, p. 453.) recommended the use of 22/7.
Before coming to the medieval and modern European mathematicians it may be convenient to note the results arrived at in India and the East.
tocsubsectionResults of Indian and Eastern writers Baudhayana (The Sulvasutras by G. Thibaut, Asiatic Society of Bengal, 1875, arts. 26--28.) took 49/16 as the value of π.
Arya-Bhata (Leçons de calcul d'Aryabhata, by L. Rodet in the Journal Asiatique, 1879, series 7, vol. xiii, pp. 10, 21.), circ. 530, gave 62832/20000 , which is equal to 3.1416. He showed that, if a is the side of a regular polygon of n sides inscribed in a circle of unit diameter, and if b is the side of a regular inscribed polygon of 2n sides, then b^2=1/2-1/2(1-a^2)^(1/2). From the side of an inscribed hexagon, he found successively the sides of polygons of 12 , 24 , 48 , 96 , 192 , and 384 sides. The perimeter of the last is given as equal to sqrt(9.8694) , from which his result was obtained by approximation.
Brahmagupta (Algebra from Brahmegupta and Bhascara, trans. by H.T. Colebrooke, London, 1817, chap. xii, art. 40, p. 308.), circ. 650, gave sqrt(10) , which is equal to 3.1622.... He is said to have obtained this value by inscribing in a circle of unit diameter regular polygons of 12 , 24 , 48 , and 96 sides, and calculating successively their perimeters, which he found to be sqrt(9.65) , sqrt(9.81) , sqrt(9.86) , sqrt(9.87) respectively; and to have assumed that as the number of sides is increased indefinitely the perimeter would approximate to sqrt(10).
Bhaskara, circ. 1150, gave two approximations. One (ibid:9Algebra Bhascara, p. 87.)--possibly copied from Arya-Bhata, but said to have been calculated afresh by Archimedes's method from the perimeters of regular polygons of 384 sides--is 3927/1250 , which is equal to 3.1416 :
the other (ibid:10Algebra Bhascara, p. 95.) is 754/240 , which is equal to 3.1416 , but it is uncertain whether this was not given only as an approximate value.
Among the Arabs the values 22/7 , sqrt(10) , and 62832/20000 were given by Alkarisimi (The Algebra of Mohammed ben Musa, ed. by F. Rosen, London, 1831, pp. 71--72.), circ. 830; and no doubt were derived from Indian sources. He described the first as an approximate value, the second as used by geometricians, and the third as used by astronomers.
In Chinese works the values 3 , 22/7 , 157/50 are said to occur: probably the last two results were copied from the Arabs.
tocsubsectionResults of European writers, 1200--1630 Returning to European mathematicians, we have the following successive approximations to the value of π : many of those prior to the eighteenth century having been calculated originally with the view of demonstrating the incorrectness of some alleged quadrature.
Leonardo of Pisa (Boncompagni's Scritti di Leonardo, vol. ii (Practica Geometriae), Rome, 1862, p. 90.), in the thirteenth century, gave for π the value 1440/4581/3 which is equal to 3.1418.... In the fifteenth century, Purbach (Appendix to the De Triangulis of Regiomontanus, Basle, 1541, p. 131.) gave or quoted the value 62832/20000 , which is equal to 3.1416 ; Cusa believed that the accurate value was 3/4(sqrt(3) + sqrt(6)) which is equal to 3.1423... ; and, in 1464, Regiomontanus (In his correspondence with Cardinal Cusa, De Quadratura Circuli, Nuremberg, 1533, wherein he proved that Cusa's result was wrong. I cannot quote the exact reference, but the figures are given by competent writers and I have no doubt are correct.) is said to have given a value equal to 3.14243.
Vieta (Canon Mathematicus seu ad Triangula, Paris, 1579, pp. 56, 66: probably this work was printed for private circulation only, it is very rare.), in 1579, showed that π was greater than
31415926535/10^10 , and less than 31415926537/10^10. This was deduced from the perimeters of the inscribed and circumscribed polygons of 6 × 2^16 sides, obtained by repeated use of the formula 2 ^21/2θ = 1- θ. He also gave (Vietae Opera, ed. Schooten, Leyden, 1646, p. 400.) a result equivalent to the formula 2/π = 2/2 (2+ 2)/2 {2+ (2+ 2)}/2 ;.
The father of Adrian Metius (Arithmeticae libri duo et Geometriae, by A. Metius, Leyden, 1626, pp. 88, 89. [Probably issued originally in 1611.]), in 1585, gave 355/113 , which is equal to 3.14159292... , and is correct to 6 places of decimals. This was a curious and lucky guess, for all that he proved was that π was intermediate between 377/120 and 333/106 , whereon he jumped to the conclusion that he should obtain the true fractional value by taking the mean of the numerators and the mean of the denominators of these fractions.
In 1593 Adrian Romanus (Ideae Mathematicae, Antwerp, 1593: a rare work, which I have never been able to consult.) calculated the perimeter of the inscribed regular polygon of 1073,741824 ( 2^30 ) sides, from which he determined the value of π correct to 15 places of decimals.
L. van Ceulen devoted no inconsiderable part of his life to the subject. In 1596 (Vanden Circkel, Delf, 1596, fol. 14, p. 1; or De Circulo, Leyden, 1619, p. 3.) he gave the result to 20 places of decimals: this was calculated by finding the perimeters of the inscribed and circumscribed regular polygons of 60 × 2^33 sides, obtained by the repeated use of a theorem of his discovery equivalent to the formula 1- A = 2 ^21/2A. I possess a finely executed engraving of him of this date, with the result printed round a circle which is below his portrait. He died in 1610, and by his directions the result to 35 places of decimals
(which was as far as he had then calculated it) was engraved on his tombstone (The inscription is quoted by Prof. de Haan in the Messenger of Mathematics, 1874, vol. iii, p. 25.) in St Peter's Church, Leyden. His posthumous arithmetic (De Arithmetische en Geometrische Fondamenten, Leyden, 1615, p. 163; or p. 144 of the Latin translation by W. Snell, published at Leyden in 1615 under the title Fundamenta Arithmetica et Geometrica. This was reissued, together with a Latin translation of the Vanden Circkel, in 1619, under the title De Circulo; in which see pp. 3, 29--32, 92.) contains the result to 32 places; this was obtained by calculating the perimeter of a polygon, the number of whose sides is 2^62 , 4,611686,018427,387904. Van Ceulen also compiled a table of the perimeters of various regular polygons.
Willebrord Snell (Cyclometricus, Leyden, 1621, p. 55.), in 1621, obtained from a polygon of 2^30 sides an approximation to 34 places of decimals. This is less than the numbers given by van Ceulen, but Snell's method was so superior that he obtained his 34 places by the use of a polygon from which van Ceulen had obtained only 14 (or perhaps 16 ) places. Similarly, Snell obtained from a hexagon an approximation as correct as that for which Archimedes had required a polygon of 96 sides, while from a polygon of 96 sides he determined the value of π correct to seven decimal places instead of the two places obtained by Archimedes. The reason is that Archimedes, having calculated the lengths of the sides of inscribed and circumscribed regular polygons of n sides, assumed that the length of 1/n th of the perimeter of the circle was intermediate between them; whereas Snell constructed from the sides of these polygons two other lines which gave closer limits for the corresponding arc. His method depends on the theorem 3 θ /(2 + θ) < θ < (2 1/3θ + 1/3θ) , by the aid of which a polygon of n sides gives a value of π correct to at least the integral part of (4 n -.2305) places of decimals, which is more than twice the number given by the older rule. Snell's proof of his theorem is incorrect, though the result is true.
Snell also added a table (It is quoted by Montucla, ed. 1831, p. 70.) of the perimeters of all regular inscribed and circumscribed polygons, the number of whose sides is 10× 2^n where n is not greater than 19 and not less than 3. Most of these were quoted from van Ceulen, but some were recalculated. This list has proved useful in refuting circle-squarers. A similar list was given by James Gregory (Vera Circuli et Hyperbolae Quadratura, prop. 29, quoted by Huygens, Opera Varia, Leyden, 1724, p. 447.).
In 1630 Grienberger (Elementa Trigonometrica, Rome, 1630, end of preface.), by the aid of Snell's theorem, carried the approximation to 39 places of decimals. He was the last mathematician who adopted the classical method of finding the perimeters of inscribed and circumscribed polygons. Closer approximations serve no useful purpose. Proofs of the theorems used by Snell and other calculators in applying this method were given by Huygens in a work (De Circula Magnitudine Inventa, 1654; Opera Varia, pp. 351--387. The proofs are given in G. Pirie's Geometrical Methods of Approximating the Value of π , London, 1877, pp. 21--23.) which may be taken as closing the history of this method.
Theorems of Wallis and Brouncker In 1656 Wallis (Arithmetica Infinitorum, Oxford, 1656, prop. 191. An analysis of the investigation by Wallis was given by Cayley, Quarterly Journal of mathematics, 1889, vol. xxiii, pp. 165--169.) proved that π/2=2 2 4 4 6 6/1 3 3 5 5 7 7;, and quoted a proposition given a few years earlier by Viscount Brouncker to the effect that π/4=1+1^2/2 0pt+ 3^2/2 0pt+ 5^2/2 0pt+...;, but neither of these theorems was used to any large extent for calculation.
Subsequent calculators have relied on converging infinite series, a method that was hardly practicable prior to the
invention of the calculus, though Descartes (See Euler's paper in the Novi Commentarii Academiae Scientiarum, St Petersburg, 1763, vol. viii, pp. 157--168.) had indicated a geometrical process which was equivalent to the use of such a series. The employment of infinite series was proposed by James Gregory (See the letter to Collins, dated Feb. 15, 1671, printed in the Commercium Epistolicum, London, 1712, p. 25, and in the Macclesfield Collection, Correspondence of Scientific Men of the Seventeenth Century, Oxford, 1841, vol. ii, p. 216.), who established the theorem that θ = θ - 1/3^3θ + 1/5^5θ -... , the result being true only if θ lies between -1/4π and 1/4π.
Analytical methods of approximation. Gregory's series
tocsubsectionResults of European writers, 1699--1873 The first mathematician to make use of Gregory's series for obtaining an approximation to the value of π was Abraham Sharp (See Life of A. Sharp by W. Cudworth, London, 1889, p. 170. Sharp's work is given in one of the preliminary discourses (p. 53 ) prefixed to H. Sherwin's Mathematical Tables. The tables were issued at London in 1705: probably the discourses were issued at the same time, though the earliest copies I have seen were printed in 1717.), who, in 1699, on the suggestion of Halley, determined it to 72 places of decimals ( 71 correct). He obtained this value by putting θ = 1/6π in Gregory's series.
Machin (W. Jones's Synopsis Palmariorum, London, 1706, p. 243; and Maseres, Scriptores Logarithmici, London, 1796, vol. iii, pp. vii--ix, 155--164.), earlier than 1706, gave the result to 100 places (all correct). He calculated it by the formula 1/4π = 4 ^-11/5 -^-11/239;.
De Lagny (Histoire de l'Académie for 1719, Paris, 1721, p. 144.), in 1719, gave the result to 127 places of decimals ( 112 correct), calculating it by putting θ = 1/6π in Gregory's series.
Hutton (Philosophical Transactions, 1776, vol. lxvi, pp. 476--492.), in 1776, and Euler (Nova Acta Academiae Scientiarum Petropolitanae for 1793, St Petersburg, 1798, vol. xi, pp. 133--149: the memoir was read in 1779.), in 1779, suggested the use of the formulae 1/4π = ^-11/2 + ^-11/3 or 1/4π = 5 ^-11/7 + 2^-13/79 ,
but neither carried the approximation as far as had been done previously.
Vega, in 1789 (Nova Acta Academiae Scientiarum Petropolitanae for 1790, St Petersburg, 1795, vol. ix, p. 41.), gave the value of π to 143 places of decimals ( 126 correct); and, in 1794 (Thesaurus Logarithmorum (logarithmisch-trigonometrischer Tafeln), Leipzig, 1794, p. 633.), to 140 places ( 136 correct).
Towards the end of the last century Baron Zach saw in the Radcliffe Library, Oxford, a manuscript by an unknown author which gives the value of π to 154 places of decimals ( 152 correct).
In 1841 Rutherford (Philosophical Transactions, 1841, p. 283.) calculated it to 208 places of decimals ( 152 correct), using the formula 1/4π=4^-11/5 - ^-11/70 + ^-11/99.
In 1844 Dase (Crelle's Journal, 1844, vol. xxvii, p. 198.) calculated it to 205 places of decimals ( 200 correct), using the formula 1/4π= ^-11/2 +^-11/5 + ^-11/8.
In 1847 Clausen (Schumacher, Astronomische Nachrichten, vol. xxv, col. 207.) carried the approximation to 250 places of decimals ( 248 correct), calculating it independently by the formulae 1/4π = 2^-11/3 + ^-11/7 and 1/4π = 4^-11/5-^-11/239.
In 1853 Rutherford (Proceedings of the Royal Society, Jan. 20, 1853, vol. vi, pp. 273-275.) carried his former approximation to 440 places of decimals (all correct), and William Shanks prolonged the approximation to 530 places. In the same year Shanks published an approximation to 607 places (Contributions to Mathematics, W. Shanks, London, 1853, pp. 86, 87.): and in 1873 he carried the approximation to 707 places of decimals (Proceedings of the Royal Society, 1872--3, vol. xxi, p. 318; 1873--4, vol. xxii, p. 45.). These were calculated from Machin's formula.
In 1853 Richter, presumably in ignorance of what had been done in England, found the value of π to 333 places (Grunert's Archiv, vol. xxi, p. 119.) of
decimals ( 330 correct); in 1854 he carried the approximation to 400 places (ibid:11Grunert's Archiv, vol. xxiii, p. 476: the approximation given in vol. xxii, p. 473, is correct only to 330 places.); and in 1855 carried it to 500 places (ibid:12Grunert's Archiv, vol. xxv, p. 472; and Elbingen Anzeigen, No. 85.).
Of the series and formulae by which these approximations have been calculated, those used by Machin and Dase are perhaps the easiest to employ. Other series which converge rapidly are the following, π/6 = 1/2 + 1/2 1/3 2^3 + 1 3/2 4 1/5 2^5 +...;, and π/4 = 2 + 22^-11/28 + ^-11/443 - 5^-11/1393 - 10^-11/11018;, the latter of these is due to Mr Escott (L'Intermédiaire des mathématiciens, Paris, Dec. 1896, vol. iii, p. 276.).
As to those writers who believe that they have squared the circle their number is legion and, in most cases, their ignorance profound, but their attempts are not worth discussing here. "Only prove to me that it is impossible," said one of them, "and I will set about it immediately"; and doubtless the statement that the problem is insoluble has attracted much attention to it.
Geometrical approximations Among the geometrical ways of approximating to the truth the following is one of the simplest. Inscribe in the given circle a square, and to three times the diameter of the circle add a fifth of a side of the square, the result will differ from the circumference of the circle by less than one-seventeen-thousandth part of it.
Approximations by the theory of probability An approximate value of π has been obtained experimentally by the theory of probability. On a plane a number of equidistant parallel straight lines, distance apart a , are ruled; and a stick of length l , which is less than a , is dropped on to the plane. The probability that it will fall so as to lie across one of the lines is 2l/π a. If the experiment is repeated many
hundreds of times, the ratio of the number of favourable cases to the whole number of experiments will be very nearly equal to this fraction: hence the value of π can be found. In 1855 Mr A. Smith (A. De Morgan, Budget of Paradoxes, London, 1872, pp. 171, 172 [quoted from an article by De Morgan published in 1861].) of Aberdeen made 3204 trials, and deduced π= 3.1553. A pupil of Prof. De MorganSmith:footnote:278DeMorgan:footnote:278A. De Morgan, Budget of Paradoxes, London, 1872, pp. 171, 172 [quoted from an article by De Morgan published in 1861]., from 600 trials, deduced π = 3.137. In 1864 Captain Fox (Messenger of Mathematics, Cambridge, 1873, vol. ii, pp. 113, 114.) made 1120 trials with some additional precautions, and obtained as the mean value π = 3.1419.
Other similar methods of approximating to the value of π have been indicated. For instance, it is known that if two numbers are written down at random, the probability that they will be prime to each other is 6/π^2. Thus, in one case (Note on π by R. Chartres. Philosophical Magazine, London, series 6, vol. xxxix, March, 1904, p. 315.) where each of 50 students wrote down 5 pairs of numbers at random, 154 of the pairs were found to consist of numbers prime to each other. This gives 6/π^2 = 154/250 , from which we get π=3.12
.
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End of part 10 of 16
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