Donald B. Wagner
8 June 2019
The mathematician Qin Jiushao 秦九韶 (ca. 1202–1261) in his Shushu jiuzhang 數書九章 gives an incorrect and very odd approximation formula for the area of ‘a field shaped like a banana leaf’. Hardly anyone, ancient or modern, seems to have attempted to explain the formula. The only attempt to deal with it that I know of is that of Qian Baocong (1966: 84–85), described further below. Libbrecht (1973: 108–109) gives a short account of Qian Baocong’s suggestion, but goes no further.
These scholars worked long before interactive mathematical software became widely available and made extensive experimentation possible. After a great deal of experimentation I propose below an explanation of Qin Jiushao’s formula.
The term jiaoyetian 蕉葉田, ‘banana leaf field’, does not occur elsewhere in extant classical Chinese mathematical texts. Judging from Qin Jiushao’s own illustration, Figure 8 below, it seems certain that the term refers to the intersection of two circles of equal radius, Figure 2.
Qin Jiushao’s text is translated further below. His approximation of the area of the ‘banana leaf field’ extracts the positive root of the quadratic equation (see Figure 2)
(1)
after which the area approximation is
(2)
The text gives the full numerical working for a particular case, and from this it is clear that the text of the formula is not corrupt; it is exactly as Qin Jiushao intended, and the text has been understood correctly.
The formula is not at all a good approximation, as we shall see further below, and (1) is dimensionally inconsistent. The first two terms are fourth powers, while the last is a cube.
An approximation for the area of a circle segment is given by the Jiuzhang suanshu 九章算術 (Arithmetic in nine chapters, perhaps 1st century CE). This is 
, in which k = the chord and s = the sagitta. This formula gives half of the area of the banana leaf with k = c and s = b/2, so that an approximation for the area of the banana leaf is
The particular case used in the text has b = 34 bu步 (paces) and c = 576 bu. The result is
AQin = 10,8715,213/63,070 bu2 ≈ 10,871.1 bu2
The approximation of the Jiuzhang suanshu gives
AJZSS = 10,081 bu2
So that in this particular case the two approximations are close to each other.
The exact value of the area is
= 13,065.1 bu2
and the error percentages of the two approximations are respectively 17% and 23%.
Plotting the values of A, AQin, and AJZSS for c = 576 and the full range of b gives the curves shown in Figure 3. It can be seen immediately that the moderate accuracy of AQin for this particular case is fortuitous. The formula does not in fact give a useful approximation for the area.
Going further, Figure 4 plots the error percentages of the two approximations for various values of c.
Qian Baocong (1966: 84–85) notes that if the constant term in (1) is changed to 
, a correct result would be obtained in the case b = c (a circle with diameter c) and 
. However, Figures 5 and 6 show that the resulting equation,
(3)
AQian = x/2
is moderately accurate for b > 0.4c, but is not in general a useful approximation.
I have found it useful, faced with this this type of philological challenge, to assume that at least two persons, Jia 甲 and Yi 乙, were involved in producing the text we have before us. Jia developed an algorithm which was useful in some way, but a scribal error on the way to Yi made it either incorrect or incomprehensible. Yi then, recognizing that there was something wrong with the text, modified it so that it made a kind of sense, but not the sense that Jia had intended.
The mathematical philologist’s task in such a case is to provide a credible hypothesis as to Jia’s original intention and how he arrived at it; how it was modified by a scribal error; and how and why Yi further modified the text to arrive at the text we have today. In the cases of two other Chinese mathematical texts I was able to arrive at what I consider to be such a ‘credible hypothesis’ (Wagner 2012a; 2012b).
Extensive experimentation with variations on Qin Jiushao’s formula has led me to this approximation:
(4)
(5)
which can be seen to be a variant of (1)–(2). This is a fair approximation, as can be seen in Figure 7.
Note the interesting similarity between Figure 5 and the curves for A and AJZSS in Figure 3. It turned out, to my amazement, that in fact Anew is equivalent to AJZSS ; it can be derived from AJZSS as follows.
Using 
,
Multiplying by 
,
Again using the Jiuzhang suanshu approximation, 
,
And this is equivalent to (4)–(5).
The algebraic manipulations shown here would not have been difficult for a mathematician of the Song period. Quite another question is why he would have developed this more complicated formula, which gives exactly the same result as the Jiuzhang suanshu formula. He may have believed it to be more accurate, or he may simply have wished to ‘show off’ with a more complicated calculation.
We can imagine that the original text, by our hypothetical mathematician Jia, may have been something like this:
術曰：以長併半廣，再自乘，又
Somehow it ended up in Qin Jiushao’s book as:
術曰：以長併 廣，再自乘，又
As to the sequence of events by which the first was transformed to the second, numerous scenarios can be imagined. Here is one. There might well have been an expectation that the breadth and height, b and c, would be treated symmetrically, leading Yi to a text that amounts to
AYi = x/2
But when Qin Jiushao received this text and applied the formula to the case b = 34 bu, c = 576 bu, he obtained the result AYi = 27,878 bu2, which is far from AJZSS = 10,081 bu2. In dealing with this problem he focused, for whatever reason, on the multiplication by the breadth in the linear term of the equation. Experimenting, he found that substituting a constant 10 for the breadth, b, gave the result AQin = 10,871 bu, which is close to AJZSS . He therefore emended the text to what we see in Qin Jiushao’s book,
but did not test the formula for other values of the breadth and length.
The original text is in Figure 8. My comments are indented in the text of the translation.
A field shaped like a banana leaf [see Figure 2)] has central length [c =] 576 bu 步 and central breadth [b =] 34 bu. The circumference is not known. What is the area in mu?
Answer: The area of the field is 45 mu 1 jiao 11 5,213/63,070 [square] bu.
One mu is equal to 240 square bu, and one jiao is 60 square bu.
45 mu × 240 bu2/mu + 1 jiao × 60 bu2/jiao + 11 5,213/63,070 bu2 ≈ 10,871.1 bu2
Method: Multiply the sum of the breadth [b] and the length [c] twice by itself. Further multiply this by 10 to make the shi 實 [the constant term of the quadratic equation to be solved].
shi = 10(b+c)3
Halve the breadth [b]; halve the length [c]; multiply each by itself. Subtract the one from the other to make the zongfang 從方 [the linear coefficient].
zongfang = (c/2)2 – (b/2)2
Let the zongyu 從隅 [the quadratic coefficient] be 1.
zongyu = 1
Extract the square root and halve it to obtain the area.
Working: Adding the length, [c =] 576 bu, and the breadth, [b =] 34 bu, gives 610. Multiplying this twice by itself gives 226,981,000 [cubic] bu. Shifting up one position, that is, multiplying by 10, gives 2,269,810,000 [cubic] bu, obtaining this number to be the shi.
shi = (576+34)3 × 10 = 2,269,810,000 bu3
Setting up the length, [c =] 576 bu, and halving it gives 288. Multiplying this by itself gives 82,944, at the top [of the counting board]. Further setting up the breadth, [b =] 34 bu, and halving it gives 17. Multiplying this by itself gives 289. Subtracting this from the top, the difference is 82,655, and this is the zongfang.
zongfang = (576/2)2 – (34/2)2 = 82,655 bu2
Letting the zongyu be 1,
The equation to be solved numerically is then
x2 + 82,655x = 2,269,810,000
and extracting the square root gives 21,742 bu with a remainder of 10,426 [bu2].
The numbers on the counting board are now
result = 21,742
remainder (shi) = 10,426
zongfang = 126,139
zongyu = 1
representing the equation
y2 + 126,139 y = 10,426
in which y = x + 21,742.
Entering the shang sheng yu into the fang, and further adding the yusuan [??] gives 126,140, to be the denominator.
I do not understand the terminology here, but clearly the calculation is
denominator = zongfang + zongyu = 126,139 + 1 = 126,140
and the numerator is the remainder of the shi, 10,426.
Halving both the remainder and the area result of the root extraction gives the final result,
10,871 5,213/63,070 . Dividing by the mu factor, 240 [bu2/mu] and simplifying gives 45 mu, 1 jiao,
11 5,213/63,070 [square] bu.
Chemla, Karine, and Guo Shuchun. 2004. Les neuf chapitres: Le classique mathématique de la Chine ancienne et ses commentaires. Paris: Dunod.
Guo Shuchun 郭书春, Joseph W. Dauben, and Xu Yibao 徐义保. 2013. Nine chapters on the art of mathematics. 3 vols. Shenyang: Liaoning Education Press. ‘A critical edition and English translation based upon a new collation of the ancient text and modern Chinese translation by Guo Shuchun; English critical edition and translation, with notes by Joseph W. Dauben and Xu Yibao.’
Libbrecht, Ulrich. 1973. Chinese mathematics in the thirteenth century: The Shu-shu chiu-chang of Ch’in Chiu-shao. (MIT East Asian science series 1). Facs. repr. Mineola, NY: Dover, 2005.
Qian Baocong 錢寶琮, ed. 1963. Suanjing shi shu (Ten mathematical classics). Beijing: Zhonghua Shuju.
———. 1966. ‘Qin Jiushao “Shushu jiuzhang” yanjiu 秦九韶《数书九章》研究’ (A study of Qin Jiushao’s Shushu jiuzhang). In 宋元数学史论文集 (Studies of the history of mathematics in the Song and Yuan periods), edited by Qian Baocong. Bejing: Kexue Chubanshe, pp. 60–103.
Wagner, Donald B. 2012a. ‘Shen Gua and an ignorant editor on the length of an arc’.donwagner.dk/Shen-Gua-arc.htm.
———. 2012b. ‘Another ignorant editor? Note on a mathematical problem in a 14th-century Chinese treatise on river conservancy’. donwagner.dk/another.htm.