Jade Mirror of the Four Unknowns,[1] Siyuan yujian (Chinese: 四元玉鉴), also referred to as Jade Mirror of the Four Origins,[2] is a 1303 mathematical monograph by Yuan dynasty mathematician Zhu Shijie.[3] Zhu advanced Chinese algebra with this Magnum opus.
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. His methods predate Blaise Pascal, William Horner, and modern matrix methods by many centuries. The preface of the book describes how Zhu travelled around China for 20 years as a teacher of mathematics.
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.
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:
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)
zhi ji：
ab
therefore
(a+b-c)ab=24
Further, the sum of vertical and hypotenuse is
b+c=9
Set up the unknown unitary tian as the vertical
x=a
We obtain the following equation
（
x5-9x4-81x3+729x2=3888
太
Solve it and obtain x=3
太 Unitary
equation:
-2y2-xy2+2xy+2x2y+x3=0
from the given
太
equation:
2y2-xy2+2xy+x3=0
we get:
太
8x+4x2=0
and
太
2x2+x3=0
by method of elimination, we obtain a quadratic equation
x2-2x-8=0
solution:
x=4
Template for solution of problem of three unknowns
Zhu Shijie explained the method of elimination in detail. His example has been quoted frequently in scientific literature.[5] [6] [7]
Set up three equations as follows
太
-y-z-y2x-x+xyz=0
-y-z+x-x2+xz=0
太
y2-z2+x2=0;
by manipulation of exchange of variables
We obtain
太
-x-2x2+y+y2+xy-xy2+x2y
太
-2x-2x2+2y-2y2+y3+4xy-2xy2+xy2
Elimination of unknown between IV and V we obtain a 3rd order equation
x4-6x3+4x2+6x-5=0
Solve to this 3rd order equation to obtain
x=5
Change back the variables
We obtain the hypothenus =5 paces
This section deals with simultaneous equations of four unknowns.
\begin{cases} -2y+x+z=0\\ -y2x+4y+2x-x2+4z+xz=0\\ x2+y2-z2=0\\ 2y-w+2x=0 \end{cases}
4x2-7x-686=0
Solve this and obtain 14 paces
There are 18 problems in this section.
Problem 18
Obtain a tenth order polynomial equation:
16x10-64x9+160x8-384x7+512x6-544x5+456x4+126x3+3x2-4x-177162=0
The root of which is x = 3, multiply by 4, getting 12. That is the final answer.
There are 18 problems in this section
There are 9 problems in this section
There are 6 problems in this section
There are 7 problems in this section
There are 13 problems in this section
There are eight problems in this section
Let tian yuan unitary as half of the length, we obtain a 4th order equation
x4+480x3-270000x2+15552000x+1866240000=0
solve it and obtain =240 paces, hence length =2x= 480 paces=1 li and 120 paces.
Similarity, let tian yuan unitary(x) equals to half of width
we get the equation:
x4+360x3-270000x2+20736000x+1866240000=0
Solve it to obtain =180 paces, length =360 paces =one li.
Problem No 5 is the earliest 4th order interpolation formula in the world
men summoned :
na+\tfrac{1}{2!}n(n-1)b+\tfrac{1}{3!}n(n-1)(n-2)c+\tfrac{1}{4!}n(n-1)(n-2)(n-3)d
In which
This section contains 20 problems dealing with triangular piles, rectangular piles
Problem 1
Find the sum of triangular pile
| 1+3+6+10+...+ | 1 |
| 2 |
n(n+1)
and value of the fruit pile is:
v=2+9+24+50+90+147+224+ … +
| 1 | |
| 2 |
n(n+1)2
Zhu Shijie use Tian yuan shu to solve this problem by letting x=n
and obtained the formular
v=
| 1 | |
| 2 ⋅ 3 ⋅ 4 |
(3x+5)x(x+1)(x+2)
From given condition
v=1320
3x4+14x3+21x2+10x-31680=0
Solve it to obtain
x=n=9
Therefore,
v=2+9+24+50+90+147+224+324+450=1320
Six problems of four unknowns.
Question 2
Yield a set of equations in four unknowns: .[12]
\begin{cases} -3y2+8y-8x+8z=0\\ 4y2-8xy+3x2-8yz+6xz+3z2=0\\ y2+x2-z2=0\\ 2y+4x+2z-w=0 \end{cases}
Sources