ANCIENT APPROXIMATIONS

ANCIENT APPROXIMATIONS PR HEWITT 1. Square roots Let s look at a method for approximating square roots, which was discovered in ancient Iraq. We will not follow them in every detail, but close enough to illustrate how well positional systems work when approximating. And just for kicks we will work in sexagesimal. Suppose we want to approximate 185.0 or in sexagesimal, 3, 5; 0. We start with the fact that 13 is an underestimate of 3, 5; 0, since 13 2 < 3, 5; 0. Hence 3, 5; 0 13 is an overestimate. We might expect that the average of the underestimate and overestimate is a pretty good approximation: (1) 3, 5; 0 0; 5 (13; 0 + 3, 5; 0 13; 0). That is, 14; 13, 50, 46, 9 ---- 13; 0 3, 5; 0, 0, 0, 0 3, 2 3; 0 2; 49 11, 0 10, 50 10, 0 9, 58 2, 0 1, 57 ------ 3 (2) 13; 0 < 3, 5; 0 < 14; 13, 50, 46, 9. If we average these we get a much better estimate: 0; 5 (13; 0 + 14; 13, 50, 46, 9) = 13; 36, 55, 23, 4, 30. This is now just a little bit too big: (13; 36, 55, 23, 4, 30) 2 = 3, 5; 22,.... Date: 18 September 2008. 1

2 PR HEWITT (To be slightly more historically correct, I should say that the ancient Iraqis would not have used long division. Instead they would have first computed the reciprocal 1/13, using a table, and then multiplied this by 3, 5; 0.) We can get a lot closer by iterating this process. (We ignore everything after four sexagesimal places.) 13; 35, 15, 15, 28 ----- 13; 36, 55, 23, 5 3, 5; 0, 0, 0, 0 2, 57; 0, 0, 0, 5 -- 7; 59, 59, 59, 55 7, 56, 32, 18, 27 -------- 3, 27, 41, 28 3, 24, 13, 50 ---- 3, 30, 38 3, 24, 13 6, 25 6, 20 ------ 8 Conclusion: (3) 13; 36, 55, 23, 4 > 3, 5; 0 > 13; 35, 15, 15, 28. The average 0; 5 (13; 36, 55, 23, 4+13; 35, 15, 15, 28) = 13; 36, 5, 19, 16 is amazingly accurate: the error is less than one part in a million! This is good enough for most applications. But we are having fun, so let s continue until the process converges: 13; 36, 5, 15, 59 ----- 13; 36, 5, 19, 16 3, 5; 0, 0, 0, 0 2, 56; 49, 9, 10, 28 -- 8; 10, 50, 49, 32 8; 9, 39, 11, 33 -------- 1, 11, 37, 59 1, 8, 0, 26 ---- 3, 37, 33 3, 24, 1 13, 32 13, 22 ------- 10

ANCIENT APPROXIMATIONS 3 Conclusion: (4) 13; 36, 5, 19, 16 > 3, 5; 0 > 13; 36, 5, 15, 59. The average 0; 5 (13; 36, 5, 19, 16 + 13; 36, 5, 15, 59) = 13; 36, 5, 17, 37, 30 is now correct to four sexagesimal places (all but the last). Watch what happens when we now carry out our computations to eight places: 13; 36, 5, 17, 38, 5, 11, 7, 18-13; 36, 5, 17, 37, 30 3, 5; 0, 0, 0, 0, 0, 0, 0, 0 2, 56; 49, 8, 49, 7, 30 ------ 8; 10, 51, 10, 52, 30 8, 9, 39, 10, 34, 30 1, 12, 0, 18, 0, 0, 0 1, 8, 0, 26, 28, 7, 30 3, 59, 51, 31, 52, 30, 0 3, 51, 13, 29, 59, 37, 30 ------ 8, 38, 1, 52, 52, 30 8, 36, 51, 21, 9, 45 -- 1, 10, 31, 42, 45 1, 8, 0, 26, 28 -------- 2, 31, 16, 17 2, 29, 36, 58 ---- 1, 39, 19 1, 35, 12 4, 7 4, 4 ------ 3 The average of 13; 36, 5, 17, 37, 30 and 13; 36, 5, 17, 38, 5, 11, 7, 18 gives (5) 3, 5; 0 13; 36, 5, 17, 37, 47, 35, 33, 39. All eight sexagesimal places are correct! This is equivalent to fourteen correct decimal digits (since log 10 (60) = 1.77... and 8 1.77 = 14.22). This is well beyond almost any practical need. Roughly speaking the accuracy will continue double at each subsequent iteration (assuming we carry enough places in our computation). That is, at the next iteration we should expect around sixteen correct places; then around thirty-two; and so forth. This is called quadratic convergence, because at each step the new error is roughly the square of the previous one. For example, in the last iteration above we went from an error on the order of 60 4 = 7.7 10 8 to an error on the order of 60 8 = 5.95 10 15.

4 PR HEWITT 2. Cube roots The Iraqi method for square roots is part of a more general procedure, now called Newton s method. Newton s method can be applied to approximate the roots of arbitrary polynomials. For example, to approximate 3 A you start with a guess r and then iteratively improve it by replacing r by the weighted average 1 3 (2r+A/r2 ). The Chinese had a different method, centuries before Newton. Their counting board and rod numerals a decimal positional system were ideal for accurate computation with polynomials of arbitrary degree. Although they did not have a symbolic algebra they regularly stated and solved polynomials of degree 4 and above. We illustrate their method by approximating 3 9745. The idea is to locate digits one at a time, starting with the most significant and working down. Before we look at the Chinese method let s look at how the Iraqi method for square roots might have been modified by the Chinese to handle cube roots. The Chinese had discovered the binomial theorem, which says that if n is a positive integer then n ( ) n (a + n) n = a n k b k, k k=0 where the binomial coefficient ( n k) is the k-th entry in the n-th row of Pascal s triangle : 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 (Both row and column indices start counting at 0.) For example,. (a + b) 6 = a 6 + 6a 5 b + 15a 4 b 2 + 20a 3 b 3 + 15a 2 b 4 + 6ab 5 + b 6. The power of this in approximations arises when b is small compared to a: (a + b) n a n + na n 1 b = a n (1 + nb/a), when a/b < 1. Although it is called Pascal s triangle, it was known in medieval China, India, and the Arab world, centuries before Pascal. Let s use the binomial theorem to approximate r = 3 9745. Since 9745 is a bit bigger than 9 10 3, its cube root is a bit bigger than 2 10 1. That is, 3 9745 has the form 2x.xxx.... If we let r = 20 + b then Hence 9745 = r 3 = 20 3 (1 + b/20) 3 = 20 3 (1 + 3b/20 + 3(b/20) 2 + (b/20) 3 ) 20 3 (1 + 3b/20). b 1 3 (9745/202 20) = 1.....

ANCIENT APPROXIMATIONS 5 Of course, the computation above would have to be done on the counting board, but note that in the hardest part of the computation the division we are only after one digit. At any rate, this computation tells us that this next digit is 1. That is, 3 9745 = 21..... Similarly if r = 21 + b then b 1 3 (9745/212 21) = 0.3.... That is, 3 9745 = 21.3.... We continue this way for a few more steps: 1 3 (9745/21.32 21.3) = 0.05... 1 3 (9745/21.352 21.35) = 0.009... 1 3 (9745/21.3592 21.359) = 0.0006... This is all very efficient, and it is close to the Iraqi method, since in effect what we are doing is improving on an approximation ˆr r by replacing it with the weighted average ˆr + 1 3 (9745/ˆr2 ˆr) = 1 3 (2ˆr + 9745/ˆr2 ). However, this is not how the Chinese did it. Instead they developed general techniques for working with polynomials, and applied this to special cases such as x 3 9745. First, they had an efficient method for evaluation polynomials, which is now called Horner s method. It is based on the identity x n + a n 1 x n 1 + a n 2 x n 2 + + a 0 = x (x (x (x + a n 1 ) + a n 2 ) + a 1 ) + a 0. For example, to evaluate 7 4 + 3 7 3 5 7 2 + 2 7 9 they would compute 7 (7 (7 (7 + 3) 5) + 2) 9. Note that doing it this way uses exactly n 1 multiplications for a polynomial of degree n, whereas doing it the other way would require 2(n 1). [Check this!] Many compilers for modern computer languages use this transformation to minimize computational time. Here is how the Chinese would do this computation on their counting board: 7 1 3-5 2-9 ----- 7 70 455 3199 ----- 1 10 65 457 3190 ------ (Actually, they set up the coefficients in a column, not a row.) This tells us not only that the value of the polynomial at 7 is 3190 but also what are the coefficients of the quotient of the polynomial when we divide by x 7: x 4 + 3x 3 5x 2 + 2x 9 = (x 7)(x 3 + 10x 2 + 65x + 457) 3190. Hence this technique is now called synthetic division.

6 PR HEWITT Let s return to approximating 3 9745 that is, finding a root of the polynomial x 3 9745. We have already determined that x has the form 2x.xxx... that is, that 20 < x < 30. We use synthetic division to shift the axes to the right 20 units. That is, we set x 0 = x 20 and rewrite the polynomial in terms of x 0. For this we divide the polynomial by x 20; then divide the quotient by x 20; and so forth: 20 0 0-9745 ----- 20 400 8000 ----- 1 20 400-1745 20 800 ------- ------- 1 40 1200 20 ----- 1 60 ---- This says that if x 0 = x 20 then x 3 9745 = (x 20)((x 20)((x 20) + 60) + 1200) 1745 = x 0 (x 0 (x 0 + 60) + 1200) 1745 = x 3 0 + 60x 2 0 + 1200x 0 1745. Now we are looking for a root of this new polynomial in the range 0 < x 0 < 10. As with the use of the binomial theorem above, the higher-degree terms are least significant. Hence 1200x 0 1745 0, which tells us that x 0 = 1..... We now shift the axes 1 more unit to the right: 1 1 60 1200-1745 ----- 1 61 1261 ----- 1 61 1261-484 1 62 ------- ------- 1 62 1323 1 ------ 1 63

ANCIENT APPROXIMATIONS 7 ---- This says several things. First of all, 21 3 9745 = 484 < 0, so the root is a bit to the right of 21. This will always serve as a check that we have not overshot. Second, if x 1 = x 0 1 then we are looking for a root of x 3 0 +60x 2 0 +1200x 0 1745 = x 1 (x 1 (x 1 +63)+1323) 484 = x 3 1 +63x 2 1 +1323x 1 484 in the range 0 < x 1 < 1. Again we ignore higher-degree terms and estimate that x 1 484/1323 = 0.3.... Let s continue this process a few more times: 0.3 1 63 1323-484 ----- 0.3 18.99 402.597 --- 1 63.3 1341.99-81.403 0.3 19.08 -- 1 63.6 1361.07 0.3 --------- -- 1 63.9 ------ 81.403/1361.07 = 0.05... 0.05 1 63.9 1361.07-81.403 ------ 0.05 3.1975 68.213375 1 63.95 1364.2675-13.189625 0.05 3.2 -- ----- 1 64.0 1367.4675 0.05 - --- 1 64.05 ------- 13.189625/1367.4675 = 0.009... 0.009 1 64.05 1367.4675-13.189625 ------- 0.009 0.576531 12.312396279 ---- 1 64.059 1368.044031-0.877228721 0.009 0.576612 ----- -------- 1 64.068 1368.620643 0.009 --- ---- 1 64.077 --------

8 PR HEWITT 0.877228721/1368.620643 = 0.0006... Hence 3 9745 21.3596, and 21.3596 3 9745 = 0.877228721. 3. π In Euclid s Elements you will find proofs that the ratio of the circumference of a circle to its diameter is constant, independent of the particular circle. We (but not Euclid!) call π. You will also find a proof that the ratio of the area to the square of the diameter is another constant. What is this second constant? Euclid is silent on this matter. We now know that it is π/4, but this was not proven until a generation or so after Euclid, by one of the greatest mathematicians of all time, Archimedes of Syracuse. The basic idea is this. Since the area of a triangle is 1 2hb, where h is the height and b is the base, the area of a regular polygon is 1 2hP, where h is the height of one of its constituent triangles and P is the perimeter that is, the sum of the bases of the constituent triangles. h b Now consider the inscribed and a circumscribed regular polygons above. In the limit we have that h r = 1 2D and P C = πd, for each of the polygons. Hence the area of the circle is 1 2 CD = 1 4 πd2. Archimedes did not have a notion of limit. In fact he did not even have a formal notion of real number system, and would not have expressed his results as an equation between real numbers. The ancient Greeks has been spooked by the Pythagorean discovery of incommensurables, and so followed Plato s stricture to separate the concept of whole number from that of geometric quantities such as length and area. The first steps towards a single unifying number system were taken over 1500 years later, in the late Arabic empire. This journey was completed in the European industrial age, another 500 years later. But the ancient Greeks needed some way to manipulate and compare geometric quantities. To address this need Eudoxus of Cnidos developed an elaborate theory of proportion and a method of exhaustion which presaged concepts such as the ɛ-δ definition of limit. He used his method to provide a rigorous proof for the Egyptian formula that the volume of a pyramid is one third the product of its height and the area of its base. Archimedes used Eudoxus method to prove his formula that the area of a circle is one fourth the product of its diameter and its circumference. He then used the ideas he introduced in this proof to compute upper and lower bounds for π.

ANCIENT APPROXIMATIONS 9 For the lower bound he took a circle of radius 1 and computed the area of an inscribed regular polygon of n sides. The area of one constituent triangle from such a polygon is sin(π/n) cos(π/n): sin π/n π/n π/n cos π/n Hence the area of the polygon if n sin(π/n) cos(π/n). Let me be clear: we are phrasing Archimedes formulas in modern notation. Archimedes would not have used sine and cosine to phrase his formula. These were not introduced until medieval India, 1000 years later. But these formula does capture the main ideas, in a way which is easily read by us. For the upper bound he took the same circle but computed the area of an circumscribed regular polygon of n sides. The area of one constituent triangle from such a polygon is tan(π/n) = sin(π/n) cos(π/n): tan π/n π/n π/n Hence the area of the polygon if n sin(π/n)/ cos(π/n). Let c n = cos(π/n), so that sin 2 (π/n) = 1 c 2 n. When we put the above estimates together we get the double inequality A n = n 1 c 2 n c n < π < B n = n 1 c 2 n/c n Archimedes now began with a hexagon, and repeatedly doubled the number of sides, until he got two estimates that were quite close together. When n is doubled the central angled is halved. We have a nice formula for cosine of the half angle: c 2 2n = 1 2 (1 + c n) We summarize this calculation in the table below: start with c 6 = 3/2 and repeatedly apply the half-angle formula to compute c 12, C 24, etc. For each of these use the formulas above to compute A n and B n : n c n A n B n 6 0.866025 2.598076 3.464102 12 0.965926 3.000000 3.215390 24 0.991445 3.105829 3.159660 48 0.997859 3.132629 3.146086 96 0.999465 3.139350 3.142715 192 0.999866 3.141032 3.141873 Of course, Archimedes did not have available a decent numeration system, and so his calculation was much more complicated, involving rational approximations to each of the multitude of square roots.

10 PR HEWITT Exercises (1) Use the Iraqi method to find a 6-place sexagesimal approximation to 3. (2) Use the Chinese synthetic division method to find a 10-place decimal approximation to 3 2. (3) Use the Chinese binomial theorem in the Iraqi-style method (the weighted average, not synthetic division ) to find a 10-place decimal approximation to 3 2. Which method is better? Explain! (4) We see from the chart above that it requires 192-sided polygons to guarantee 3 accurate decimal places with Archimedes method. The best approximation before the introduction of infinite series was given by Zu Chongzhi, 700 years after Archimedes. (See The story of π in our text.) How many sides would be required to achieve Zu s accuracy with Archimedes method? Did Zu use this method, or another? (You may want to consult MacTutor.) Univ of Toledo