History of Mathematics

Transcription

1 History of Mathematics Paul Yiu Department of Mathematics Florida tlantic University Spring : rchimedes

2 Quadrature of the Parabola onsider the parabola y = x 2. Let 1, 1 be two points on the x axis, 1 the midpoint of 1 1, D 1 the midpoint of 1 1, and E 1 the midpoint of 1 1. Let,,, D, E be the points on the parabola above 1, 1, 1,D 1,E 1 respectively. (1) rchimedes first showed that D + E = 1 4. Exercise: Suppose the vertical line through D 1 cuts and atx andy respectively. y calculating coordinates, show thatxy = 2DY. From this, it follows that E Z Y X D 1 E 1 1 D 1 1 D = DY + DY = 1 2 DY 1D DY 1D 1 = 1 2 DY 1 1 = 1 4 XY 1 1 = Z 1 1 = 1 4 Z. 1

3 Quadrature of the Parabola D = DY + DY = = 1 4 Z. Similarly, E = 1 4 Z, and Z Y X D + E = 1 ( Z + Z) 4 D = 1 4 Z. (2) Now, repeated bisections yield E 1 E 1 1 D 1 1 rea of parabolic segment = +( D + E)+ 1 ( D + E)+ 4 = = =

4 rchimedes: Measurement of a ircle 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 2 radius circumference = π(radius)2. Proposition 2. The area of a circle is to the square on its diameter as 11 to 14. rea of circle= (diameter)2 = 22 7 (radius)2. Proposition 3. The circumference of any circle is less than3 1 7 but greater than of its diameter < π <

5 rchimedes calculation ofπ: upper bound onsider a circle of unit diameter. Let Z n be the length of a side of a regular n gon circumscribing the circle. If is a tangent to the circle, and if = 1 2 Z n, ando bisects O, then = 1 2 Z 2n. O rchimedes showed that Z 2n = Z n Zn 2 4

6 rchimedes calculation ofπ: upper bound onsider a circle of unit diameter. Let Z n be the length of a side of a regular n gon circumscribing the circle. If is a tangent to the circle, and if = 1 2 Z n, ando bisects O, then = 1 2 Z 2n. O rchimedes showed that Z 2n = Z n Zn 2 Since O bisects angleo, : = O : O; : = O : O+O; Z 2n : Z n = 1 2 : (1 2) 2 + ( Zn 2 ) 2 = 1 : 1+ 1+Z 2 n. 5

7 Upper bound ofπ (continued) eginning with the simple fact Z 6 = 3 1 3, and the inequality , rchimedes used the relation Z 6 2 Z 2n = Z n 1+ 1+Z 2 n 30 O 4 times and obtained Thus, 1 Z 96 > circumference < 96Z 96 < = < =

8 rchimedes calculation ofπ: lower bound onsider again a circle of unit diameter. Let be a side of a regularn gon inscribed in a circle, with lengthz n. IfK is a diameter andk bisects O K, then is a side of a regular2n gon inscribed in the circle. Suppose has lengthz 2n respectively. K rchimedes showed that 1 = 1 2(1+ 1 z z 2n z n), 2 z 6 = 1 n 2. 7

9 rchimedes calculation ofπ: lower bound Let andk intersect at H. Since KH bisects anglek, K = H, and the trianglesk, H are similar. H K : = : H = K : H = K : H = K+K : H +H = K+K :. O K 8

10 rchimedes calculation ofπ: lower bound Let andk intersect at H. Since KH bisects anglek, K = H, and the trianglesk, H are similar. K : =... = K+K :. Therefore, H O K From this, d z 2n = K 2 : 2 = (K+K) 2 : 2 d 2 z 2 2n : z 2 2n = (d+ d 2 z 2 n) 2 : z 2 n d 2 : z 2 2n = z 2 n +(d+ d 2 z 2 n) 2 : z 2 n. z 2 n +(d+ d 2 z 2 n) 2 z n = With d = 1, this explains rchimedes 1 = 1 2(1+ 1 z z 2n z n). 2 n 9 2d(d+ d 2 z 2 n) z n

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12 Lower bound ofπ (continued) eginning with z 6 = 1 2, and an inequality , rchimedes iterated this relation four times, and obtained onsequently, 1 z 96 < D E F circumference > 96z 96 > O > K 11

13 ncient records ofpi (1) In his commentary of the Nine hapters of Mathematical rt, LIU Hui (3rd century D) gave the approximation π by approximating a circle by circumscribed and inscribed regular polygons of96 sides. (2) ZU hongzhi ( ) gave the estimates forπ: He also gave the crude approximation 22 7 and the fine approximation forπ < π <

14 Reformulation of rchimedes method Leta n andb n respectively denote the perimeters of a circumscribing regular6 2 n gon and an inscribed regular6 2 n gon of a circle of diameter 1. Then a 0 = 2 3,b 0 = 3, and a n+1 = 2a 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. orrect to the first 34 decimal places, π =