Ming Antu's infinite series expansion of trigonometric functions

Fig 1Ming Antu model

Fig 3 Ming Antu discovered Catalan number

Ming Antu's infinite series expansion of trigonometric functions. Ming Antu, a court mathematician of the Qing dynasty did extensive work on infinite series expansion of trigonometric functions in his masterpiece Geyuan Milv Jifa(Quick Method of Dissecting the Circle and Determination of The Precise Ratio of the Circe). Ming Antu built geometrical models based on a major arc of a circle and nth dissection of the major arc. In Fig 1, AE is the major chord of arc ABCDE, and AB, BC, CD, DE are its nth equal segments. If chord AE = y, chord AB = BC = CD = DE = x, find chord y as infinite series expansion of chord x. He studied the cases of n = 2, 3, 4, 5, 10, 100, 1000 and 10000 in great detail in vol 3 and vol 4 of Geyuan Milv Jifa.

Historical background

In 1701, French Jesuit missionary(Pierre Jartoux 1668-1720) came to China, he brought along three infinite series expansion of trigonometry functions by Isaac Newton and J. Gregory:^[1]

$\pi=3\left(1+\frac{1}{4\cdot3!}+\frac{3^2}{4^2\cdot5!}+\frac{3^2\cdot5^2}{4^3\cdot7!}+\cdots\right)$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$

$\operatorname{vers} x=\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots.$

These infinite series stirred up great interest among Chinese mathematicians,as calculation of π with these "quick methods" involve only multiplication, addition or subtraction, much faster than classic Liu Hui's π algorithm which involves taking square roots. However, Jartoux did not bring along the method for deriving these infinite series.Ming Antu suspected that the westerner did not want to share their secret,hence he set to work on it, and spent on and off for thirty years and completed a manuscript Geyuan Milv Jifa, he created geometrical models for obtaining trigonometric infinite series, and not only found the method for deriving the above three infinite series, but also discovered six more infinite series. In the process, he discovered and applied Catalan number.

Two-segment chord

Fig 2 Ming Antu's geometric model of 2-segment chord

Figure 2 is Ming Antu's model of 2 segment chord. Arc BCD is a part of circle with unity radius. AD is the main chord, arc BCD is bissected at C, draw lines BC, CD, let BC = CD = x and let radius AC = 1.

Apparently, $BD = 2x-GH$ ^[2]

Let EJ = EF, FK = FJ; extend BE straight to L, and let EL = BE; make BF = BE, so F is inline with AE. Extended BF to M, let BF = MF; connect LM, LM apparently passes point C. The inverted triangle BLM along BM axis into triangle BMN, such that C coincident with G, and point L coincident with point N. The Invert triangle NGB along BN axis into triangle; apparently BI = BC.

$AB: BC: CI = 1: x: x^2$

BM bisects CG and let BM = BC; join GM, CM; draw CO = CM to intercept BM at O; make MP = MO; make NQ = NR, R is the intersection of BN and AC. ∠EBC = 1/2 ∠CAE = 1/2 ∠EAB; $\therefore$ ∠EBM = ∠EAB; thus we otain a series of similar triangles: ABE, BEF, FJK, BLM, CMO, MOP, CGH and triangle CMO = triangle EFJ;^[3]

$AB: BE: EF: FJ: JK = 1: p: p^2: p^3: p^4$

$1: BE = BE: EF;$ namely $EF = BE^2$

$1: BE ^ 2 = x: GH$

So $GH = x \cdot BE ^ 2 = x p^2$,

and $BD = 2x -x p^2$

Because kite-shaped ABEC and BLIN are similar,.^[3]

$EF = LC = CM = MG = NG = IN$

$LM + MN = CM + MN + IN = CI + OP = JK + CI$

$\therefore AB: (BE + EC) = BL: (LM + MN)$ and $AB: BL = BL: (CI + JK)$

Let $BL = q$

$AB: BL: (CI + JK) = 1: q: q ^ 2$

$JK = p ^ 4$

$CI = y ^ 2$

$CI + JK = q ^ 2 = BL ^ 2 = (2BE) ^ 2 = (2p) ^ 2 = 4p ^ 2$

Thus $q ^ 2 = 4p ^ 2$ or $p = \frac {q} {2}$

Further: $CI + JK = x ^ 2 + p ^ 4 = q ^ 2$.

$x ^ 2 + \frac {q ^ 4} {16} = q ^ 2,$

then

$x ^ 2 = q ^ 2- \frac {q ^ 4} {16}$

Square up the above equation on both sides and divide by 16:^[4]

$\frac {(x ^ 2)^2} {16} = \frac {(q^2- \frac {q^4} {16})^2} {16} = \sum_{j=0}^2 (-1)^j {2 \choose j} \frac {q ^ {2 (2 + j)}} {16 ^ j}$

$\frac {x^4} {16} = \frac {q^4} {16} - \frac {q^6} {128} + \frac {q^8} {4096} {16 }$

And so on

$\frac {x ^ {2n}} {16^{n-1}} = \sum_ {j = 0}^n (-1)^j {n \choose j} \frac {q ^ {2 (n + j)}} {16 ^ {n + j-1}}$.^[5]

Add up the following two equations to eliminate $q ^ 4$ items:

$x ^ 2 = q ^ 2- \frac {q ^ 4} {16}$

$\frac {x ^ 4} {16} = {\frac {q ^ 4} {16} - \frac {2 q ^ 6} {16 ^ 2} + \frac {q ^ 8} {4096}} {16}$

$x ^ 2 + \frac {x ^ 4} {16} = q ^ 2- \frac {q ^ 6} {128} + \frac {q ^ 8} {4096}$

$x ^ 2 + \frac {x ^ 4} {16} + \frac {2x ^ 6} {16 ^ 2} = q ^ 2- \frac {5q ^ 8} {4096} + \frac {3q ^ {10}} {32768} - \frac {q ^ {12}} {524288},$(after eliminated $q ^ 6$ item).

......................................

$\begin{align} & x ^ 2 + \frac{x^4}{16} + \frac{2 x^6}{16^2} + \frac{5x^8}{16^3} + \frac{14 x^{10}}{16^4} + \frac{42x^{12}}{16^5} \\[10pt] {} & +\frac{132 x^{14}}{16^6} + \frac{429x^{16}}{16^7} + \frac{1430x^{18}}{16^8} + \frac{4862 x^{20}}{16^9} \\[10pt] & {} + \frac{16796 x^{22}}{16^{10}} + \frac{58786 x^{24}}{16^{11}} + \frac{208012 x^{26}}{16^{12}} \\[10pt] & {} + \frac{742900 x^{28}}{16^{13}} + \frac{2674440 x^{30}}{16^{14}} + \frac{9694845 x^{32}}{16^{15}} \\[10pt] & {} + \frac{35357670 x^{34}}{16^{16}} + \frac{129644790 x^{36}}{16^{17}} \\[10pt] & {} + \frac{477638700 x^{38}}{16^{18}} + \frac{1767263190 x^{40}}{16^{19}} + \frac{6564120420 x^{42}}{16^{20}} \\[10pt] & = q^2 + \frac{62985}{8796093022208} q^{24} \end{align}$

Expansion coefficients of the numerators: 1,1,2,5,14,42,132 ...... (see Figure II Ming Antu original figure bottom line, read from right to left) are none other than the Catalan numbers , Ming Antu is the first person in history to discover the Catalan number.^[6][7]

Thus :

$q ^ 2 = \sum_ {n = 1} ^ \infty C_n \frac {x ^ {2n}} {4 ^ {2n-2}}$ ^[8][9]

in which $C_n = \frac {1} {n + 1} {2n \choose n}$ is Catalan number. Ming Antu pioneered the use of recursion relations in Chinese mathematics^[10]

$C_n = \sum_k (-1) ^ k {nk \choose k + 1} C_ {nk}$

$\because BC: CG: GH = AB: BE: EF = 1: p: p ^ 2 = x: px: p ^ 2x$

$\therefore GH: = p ^ 2 x = (\frac {q} {2}) ^ 2 x = \frac {q ^ 2 x} {4}$

substituted into $BD = 2x-GH$

Finally he obtained^[11]

$BD = 2x- \frac {x} {4} q ^ 2$

$= 2x- \sum_ {n = 1} ^ \infty C_n \frac {x ^ {2n + 1}} {4 ^ {2n-1}}$

In Figure 1 BAE angle = α, BAC angle = 2α

• x = BC = sinα
• q = BL = 2BE = 4sin (α /2)
• BD = 2sin (2α)

Ming Antu obtained $BD = 2x -x\cdot BE ^ 2$

That is

$\sin (2 \alpha) = 2 \sin \alpha- \sum_ {n = 1} ^ \infty C_n \frac {(\sin \alpha) ^ {2n + 1}} {4 ^ {n-1}}$

$= 2 \sin (\alpha) - \frac {2 \sin (\alpha) ^ 3} {1+ \cos (\alpha)}$

$q ^ 2 = BL ^ 2 = \sum_ {n = 1} ^ \infty C_n \frac {x ^ {2n}} {4 ^ {2n-2}}$

Ie $\sin (\frac {\alpha} {2}) ^ 2 = \sum_ {n = 1} ^ \infty C_n \frac {(sin \alpha) ^ {2n}} {4 ^ {2n}}$

Three-segment chord

Fig 3. Ming Antu's geometrical model for three segment chord

As shown in Fig 3,BE is a whole arc chord, BC = CE = DE = a are three arcs of equal portions. Radii AB = AC = AD = AE = 1. Draw lines BC, CD, DE, BD, EC; let BG=EH = BC, Bδ = Eα = BD, then triangle Cαβ = Dδγ; while triagnle Cαβ is similar to triangle BδD.

As such:

$AB: BC = BC: CG = CG: GF$, $BC: FG = BD: \delta \gamma$

$2 BD = BE + \delta \alpha$

$2 BD- \delta \gamma = BE + BC$

$\therefore 2 * BD- \delta \gamma-BC = BE$

Eventually, he obtained

$BE=3*a-a^3$^[12][13]

Four-segment chord

Ming Antu 4 segment chord model

Let $y_4$ denotes the length of the main chord, and let the length of four equal segment chord =x,

$y_4=4*a-\frac{10*a^3}{4}+\frac{14*a^5}{4^3}-\frac{12*a^7}{4^5}$+……

$4a-10*a^3/4+\sum_{n=1}^\infty (16C_n-2C_{n+1})*\frac{a^{2n+1}}{4^{2n-1}}$。^[14]。

Trigonometry meaning:

$\sin(4*\alpha)=4*sin(\alpha)-10*sin^3\alpha$ $+\sum_{n=1}^\infty(16*C_n-2C_{n+1})*\frac{\sin^{2n+3}(\alpha)}{4^n}$ ^[14]。

Five-segment chord

Ming Antu 5 segment chord model

$y_5=5a-5a^3+a^5$

that is

$\sin(5\alpha)=5\sin(\alpha)-20\sin^3(\alpha)+16\sin^5(\alpha)$^[15]。

Ten-segment chord

Ming Antu 10 segment chord diagram

From here on, Ming Antu stop building geometrical model, he carried out his computation by pure algebraic manipulation of infinite series.

Apparently ten segments can be considered as a composite 5 segment, with each segment in turn consist of two subsegments.

$\therefore y_{10}=y_5(y_2)$

$y_{10}(a)=5*y_2-5*(y_2)^3+(y_2)^5$,

He computed the third and fifth power of infinite series $y_2$ in the above equation, and obtained:

$y_{10}(a)=10*a-\frac{165*a^3}{4}+\frac{3003*a^5}{4^3}-\frac{21450*a^7}{4^5}$+……^[16][17]

Hundred-segment chord

Ming Antu 100 segment chord diagram

Facsimile of Ming Antu's calculation of 100 segment chord

A hundred segment arc's chord can be considered as composite 10 segment-10 subsegments, thus sustutde $a=y_{10}$ into $y_{10}$, after manipulation with infinite series he obtained:

$y_{100}=y_{10}(a=y_{10})$

$y_{100}(a)=100*a-166650*\frac{a^3}{4}[phone]*\frac{a^5}{4*16}-316350028500*\frac{a^7}{4*16^2}[phone]*\frac{a^9}{4*16^3}+$……^[17][18]

Thousand-segment chord

$y_{1000}=y_{100}(y_{10})$

$y_{1000}(a)=1000*a-1666666500*\frac{a^3}{4}[phone]*\frac{a^5}{4*16}-3174492064314285000\frac{a^7}{4*16^2}+$……^[17][19]

Ten-thousand-segment chord

$y_{10000}=10000*a-\frac{166666665000*a^3}{4}+\frac{33333330000000300*a^5}{4^3}+$…………^[20]。

When number of segments approaches infinity

After obtained the infinite series for n=2,3,5,10,100,1000,10000 segments, Ming Antu went on to handle the case when n approaches infinity.

y100,y1000 and y10000 can be rewritten as:

$y100=100a-\frac{(100a)^3}{24.002400240024002400}+\frac{(100a)^5}{24.024021859697730358*80}+$..........

$y1000:=1000a-\frac{(1000a)^3}{24.000024000024000024}+\frac{(1000a)^5}{24.000240002184019680*80}+$..............

$y10000:=10000a-\frac{(10000a)^3}{24.000000240000002400}+\frac{(10000a)^5}{24.000002400000218400*80}+$..................

He noted that obviously, when n approaches infinity, the denominators 24.000000240000002400，24.000002400000218400*80 approach 24 and 24*80 respectively, and when n -> infinity, na (100a,1000a,1000a)becomes the length of the arc; hence^[21]

$chord= arc -\frac{arc^3}{4*3!}+\frac{arc^5}{4^2*5!}-\frac{arc^7}{4^3*7!}+$.....

$=\sum_{n=1}^\infty \frac{(-1)^{n-1}*arc^{2*n-1}}{(4^{n-1}*(2*n-1)!)}$

Ming Antu then performed an infinite series reversion and expressed the arc in terms of its chord

^[21]

$arc := chord+\frac{chord^3}{24}+\frac{3*chord^5}{640}+\frac{5*chord^7}{7168}+$............

References

1. He Shaodong, "A Key Problem in the Study of Infinite Series", in The Qing Dynasty, Studies in the History of Natural Sciences vol 6 No3 1989 pp 205–214
2. Li Yan "Selected Papers in History of Chinese Mathematics",book III, "Li Yan Qian Baocong History of Science Collection" Volume 7, 300
3. J.Luo p96
4. Luo Jianjin p100
5. Luo p106
6. J.Luo, "Ming Antu and his power series expansion" Mathematical Journal 34 volume 1, pp. 65-73
7. P Larcombe, The 18th Century Chinese Discovery of Catalan Numbers, Mathematical Spectrum, Vol 32, No 1, pp5-7, 1999/2000
8. Luo 113
9. Yan Xue-min Luo Jian-jin, Catalan Numbers, A Geometric Model J.of Zhengzhou Univ Vol 38 No2, Jun 2006, p22
10. Luo 114
11. Luo p114
12. Yoshio Mikami, p147
13. Luo p148
14. Luo p153
15. Luo p156
16. Luo p164
17. Yoshio Mikami p147
18. Li Yan p320
19. Li Yan p320页
20. Yoshio Mikami, p147
21. Yoshio Mikami, p148

• Luo A Modern Chinese Translation of Ming Antu's Geyuan Milv Jifa, translated and annotated by Luo Jianjin, Inner Mongolia Education Press 1998(明安图原著 罗见今译注 《割圆密率捷法》 内蒙古教育出版社 This is the only modern Chinese translation of Ming Antu's book, with detailed annotation with modern mathematical symbols). ISBN 7-5311-3584-1
• Yoshio Mikami The Development of Mathematics in China and Japan, Leipzig, 1912