MST Topics in History of Mathematics

Transcription

1 MST Topics in History of Mathematics Paul Yiu Department of Mathematics Florida tlantic University Summer 017 8: rchimedes Measurement of a Circle

2 The works of rchimedes Recipient Propositions n the Sphere and Cylinder, ook I Dosithesus 44 n the Sphere and Cylinder, ook II Dosithesus 9 Measurement of a Circle 3 n Conoids and Spheroids Dosithesus 3 n Spirals Dosithesus 8 n the Equilibrium of Planes, ook I 15 n the Equilibrium of Planes, ook II 10 The Sand-Reckoner King Gelon Quadrature of the Parabola Dosithesus 4 n Floating odies, ook I 9 n Floating odies, ook II 10 ook of Lemmas 15 The Cattle Problem The Method Eratosthenes 15 1

3 rchimedes: Measurement of a Circle Proposition 1. The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference, of the circle. rea of circle= 1 radius circumference = π(radius). Proposition. The area of a circle is to the square on its diameter as 11 to 14. rea of circle= (diameter) = 7 (radius). Proposition 3. The circumference of any circle is less than3 1 7 but greater than of its diameter < π < 7.

4 rchimedes calculation ofπ: upper bound Consider a circle of unit diameter. Let Z n be the length of a side of a regular n gon circumscribing the circle. C If is a tangent to the circle, and if = 1 Z n, andc bisects, then C = 1 Z n. rchimedes showed that Z n = Z n Zn 3

5 rchimedes calculation ofπ: upper bound Consider a circle of unit diameter. Let Z n be the length of a side of a regular n gon circumscribing the circle. C If is a tangent to the circle, and if = 1 Z n, andc bisects, then C = 1 Z n. rchimedes showed that Z n = Z n Zn Since C bisects angle, C : C = : ; C : = : +; Z n : Z n = 1 : 1 + (1 ) + ( Zn ) = 1 : 1+ 1+Z n. 4

6 Upper bound ofπ (continued) eginning with the simple fact Z 6 = 3 1 3, and the inequality , rchimedes used the relation Z 6 Z n = Z n 1+ 1+Z n 30 4 times and obtained Thus, 1 Z 96 > circumference < 96Z 96 < = < =

7 rchimedes calculation ofπ: lower bound Consider again a circle of unit diameter. Let be a side of a regularn gon C inscribed in a circle, with lengthz n. IfK is a diameter andkc bisects K, then C is a side of a regularn gon inscribed in the circle. Suppose C has lengthz n respectively. K rchimedes showed that 1 = 1 (1+ 1 z z n z n), z 6 = 1 n. 6

8 rchimedes calculation ofπ: lower bound Let andkc intersect at H. Since KH bisects anglek, KC = HC, and the triangleskc, HC are similar. C H KC : C = C : CH = K : H = K : H = K+K : H +H = K+K :. K 7

9 rchimedes calculation ofπ: lower bound Let andkc intersect at H. Since KH bisects anglek, KC = HC, and the triangleskc, HC are similar. KC : C =... = K+K :. Therefore, KC : C = (K+K) : C H K d z n : z n = (d+ d z n) : z n d : z n = z n +(d+ d z n) : z n. From this, d zn +(d+ d zn) d(d+ d zn) = = z n z n z n With d = 1, this explains rchimedes 1 = 1 (1+ 1 z z n z n). n 8

10 Lower bound of π (continued) eginning with z 6 = 1, and an inequality , rchimedes iterated this relation four times, and obtained Consequently, 1 z 96 < D E F circumference > 96z 96 > C > K 9

11 ncient records ofπ (1) In his commentary of the Nine Chapters of Mathematical rt, LIU Hui (3rd century D) gave the approximation π by approximating a circle by circumscribed and inscribed regular polygons of96 sides. () ZU Chongzhi (49-500) gave the estimates forπ: He also gave the crude approximation 7 and the fine approximation forπ < π <

12 Reformulation of rchimedes method Leta n andb n respectively denote the perimeters of a circumscribing regular6 n gon and an inscribed regular6 n gon of a circle of diameter 1. Then a 0 = 3,b 0 = 3, and a n+1 = a nb n a n +b n, b n+1 = a n+1 b n. This was the basis of practically all calculations of π before the 16th century. Correct to the first 34 decimal places, π =