Jade Mirror of the Four Unknowns, Siyuan yujian (simplified Chinese: 四元玉鉴; traditional Chinese: 四元玉鑒), also referred to as Jade Mirror of the Four Origins, is a 1303 mathematical monograph by Yuan dynasty mathematician Zhu Shijie. The book consists of an introduction and three books, with a total of 288 problems. The first four problems in the introduction illustrate his method of the four unknowns. He showed how to convert a problem stated verbally into a system of polynomial equations (up to the 14th order), by using up to four unknowns: 天 Heaven, 地 Earth, 人 Man, 物 Matter, and then how to reduce the system to a single polynomial equation in one unknown by successive elimination of unknowns. He then solved the high-order equation by Southern Song dynasty mathematician Qin Jiushao's "Ling long kai fang" method published in Shùshū Jiǔzhāng (“Mathematical Treatise in Nine Sections”) in 1247 (more than 570 years before English mathematician William Horner's method using synthetic division). To do this, he makes use of the Pascal triangle, which he labels as the diagram of an ancient method first discovered by Jia Xian before 1050. Zhu also solved square and cube roots problems by solving quadratic and cubic equations, and added to the understanding of series and progressions, classifying them according to the coefficients of the Pascal triangle. He also showed how to solve systems of linear equations by reducing the matrix of their coefficients to diagonal form. Jade Mirror of the Four Unknowns consists of four books, with 24 classes and 288 problems, in which 232 problems deal with Tian yuan shu, 36 problems deal with variable of two variables, 13 problems of three variables, and 7 problems of four variables.
Introduction
The four quantities are x, y, z, w can be presented with the following diagram
x y 太w z The square of which is:
The Unitary Nebuls This section deals with Tian yuan shu or problems of one unknown.
Question: Given the product of huangfan and zhi ji equals to 24 paces, and the sum of vertical and hypotenuse equals to 9 paces, what is the value of the base? Answer: 3 paces Set up unitary tian as the base (that is let the base be the unknown quantity x) Since the product of huangfang and zhi ji = 24 in which
huangfan is defined as： ( a + b − c ) {\displaystyle (a+b-c)}
zhi ji： a b {\displaystyle ab}
therefore ( a + b − c ) a b = 24 {\displaystyle (a+b-c)ab=24}
Further, the sum of vertical and hypotenuse is
b + c = 9 {\displaystyle b+c=9}
Set up the unknown unitary tian as the vertical
x = a {\displaystyle x=a}
Then use Pythagoras to isolate c and b: a 2 + b 2 = c 2 ⟺ c 2 − b 2 = a 2 ⟺ ( c − b ) ( c + b ) = a 2 ⟺ c − b = a 2 c + b = x 2 9 {\displaystyle a^{2}+b^{2}=c^{2}\Longleftrightarrow c^{2}-b^{2}=a^{2}\Longleftrightarrow (c-b)(c+b)=a^{2}\Longleftrightarrow c-b={\frac {a^{2}}{c+b}}={\frac {x^{2}}{9}}}
such that we obtain:
2 c = ( c + b ) + ( c − b ) = 9 + x 2 9 {\displaystyle 2c=(c+b)+(c-b)=9+{\frac {x^{2}}{9}}}
2 b = ( c + b ) − ( c − b ) = 9 − x 2 9 {\displaystyle 2b=(c+b)-(c-b)=9-{\frac {x^{2}}{9}}}
Combining everything, we obtain the following equation:
（ x 5 − 9 x 4 − 81 x 3 + 729 x 2 = 3888 {\displaystyle x^{5}-9x^{4}-81x^{3}+729x^{2}=3888} ） 太
Solve it and obtain x=3
The Mystery of Two Natures 太 Unitary
equation: − 2 y 2 − x y 2 + 2 x y + 2 x 2 y + x 3 = 0 {\displaystyle -2y^{2}-xy^{2}+2xy+2x^{2}y+x^{3}=0} ; from the given
太
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