Jade Mirror of the Four Unknowns

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 hypothenuse 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：${\displaystyle (a+b-c)}$[4]

zhi ji：${\displaystyle ab}$

therefore ${\displaystyle (a+b-c)ab=24}$

Further, the sum of vertical and hypothenus is

${\displaystyle b+c=9}$

Set up the unknown unitary tian as the vertical

${\displaystyle x=a}$

We obtain the following equation

（${\displaystyle x^{5}-9x^{4}-81x^{3}+729x^{2}=3888}$）

太

Solve it and obtain x=3

太 Unitary

equation: ${\displaystyle -2y^{2}-xy^{2}+2xy+2x^{2}y+x^{3}=0}$;

from the given

太

equation: ${\displaystyle 2y^{2}-xy^{2}+2xy+x^{3}=0}$;

we get:

太

${\displaystyle 8x+4x^{2}=0}$

and

太

${\displaystyle 2x^{2}+x^{3}=0}$

by method of elimination, we obtain a quadratic equation

${\displaystyle x^{2}-2x-8=0}$

solution: ${\displaystyle x=4}$.

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

${\displaystyle -y-z-y^{2}x-x+xyz=0}$ .... I

${\displaystyle -y-z+x-x^{2}+xz=0}$.....II

太

${\displaystyle y^{2}-z^{2}+x^{2}=0;}$....III

Elimination of unknown between II and III

by manipulation of exchange of variables

We obtain

太

${\displaystyle -x-2x^{2}+y+y^{2}+xy-xy^{2}+x^{2}y}$ ...IV

and

太

${\displaystyle -2x-2x^{2}+2y-2y^{2}+y^{3}+4xy-2xy^{2}+xy^{2}}$.... V

Elimination of unknown between IV and V we obtain a 3rd order equation

${\displaystyle x^{4}-6x^{3}+4x^{2}+6x-5=0}$

Solve to this 3rd order equation to obtain ${\displaystyle x=5}$ ;

Change back the variables

We obtain the hypothenus =5 paces

This section deals with simultaneous equations of four unknowns.

Equations of four Elements

${\displaystyle {\begin{cases}-2y+x+z=0\\-y^{2}x+4y+2x-x^{2}+4z+xz=0\\x^{2}+y^{2}-z^{2}=0\\2y-w+2x=0\end{cases}}}$

Successive elimination of unknowns to get

${\displaystyle 4x^{2}-7x-686=0}$

Solve this and obtain 14 paces

There are 18 problems in this section.

Problem 18

Obtain a tenth order polynomial equation:

${\displaystyle 16x^{10}-64x^{9}+160x^{8}-384x^{7}+512x^{6}-544x^{5}+456x^{4}+126x^{3}+3x^{2}-4x-177162=0}$

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

Problem 1

Question: There is a rectangular town of unknown dimension which has one gate on each side. There is a pagoda located at 240 paces from the south gate. A man walking 180 paces from the west gate can see the pagoda, he then walks towards the south-east corner for 240 paces and reaches the pagoda; what is the length and width of the rectangular town? Answer: 120 paces in length and width one li

Let tian yuan unitary as half of the length, we obtain a 4th order equation

${\displaystyle x^{4}+480x^{3}-270000x^{2}+15552000x+1866240000=0}$[8]

solve it and obtain x=240 paces, hence length =2x= 480 paces=1 li and 120paces.

Similarity, let tian yuan unitary(x) equals to half of width

we get the equation:

${\displaystyle x^{4}+360x^{3}-270000x^{2}+20736000x+1866240000=0}$[9]

Solve it to obtain x=180 paces, length =360 paces =one li.

Problem 7

Identical to The depth of a ravine (using hence-forward cross-bars) in Haidao Suanjing.

Problem 8

Identical to The depth of a transparent pool in Haidao Suanjing.

Problem No 5 is the earliest 4th order interpolation formula in the world

men summoned :${\displaystyle na+{\tfrac {1}{2!}}n(n-1)b+{\tfrac {1}{3!}}n(n-1)(n-2)c+{\tfrac {1}{4!}}n(n-1)(n-2)(n-3)d}$[10]

In which

• a=1st order difference
• b=2nd order difference
• c=3rd order difference
• d=4th order difference

This section contains 20 problems dealing with triangular piles, rectangular piles

Problem 1

Find the sum of triangular pile

${\displaystyle 1+3+6+10+...+{\frac {1}{2}}n(n+1)}$

and value of the fruit pile is:

${\displaystyle v=2+9+24+50+90+147+224+\cdots +{\frac {1}{2}}n(n+1)^{2}}$

Zhu Shijie use Tian yuan shu to solve this problem by letting x=n

and obtained the formular

${\displaystyle v={\frac {1}{2\cdot 3\cdot 4}}(3x+5)x(x+1)(x+2)}$

From given condition ${\displaystyle v=1320}$, hence

${\displaystyle 3x^{4}+14x^{3}+21x^{2}+10x-31680=0}$[11]

Solve it to obtain ${\displaystyle x=n=9}$.

Therefore,

${\displaystyle v=2+9+24+50+90+147+224+324+450=1320}$。

Six problems of four unknowns.

Question 2

Yield a set of equations in four unknowns: .[12]

${\displaystyle {\begin{cases}-3y^{2}+8y-8x+8z=0\\4y^{2}-8xy+3x^{2}-8yz+6xz+3z^{2}=0\\y^{2}+x^{2}-z^{2}=0\\2y+4x+2z-w=0\end{cases}}}$