Jiuzhang Suanshu Shaoguang 24

## **Problem**

又有積一萬六千四百四十八億六千六百四十三萬七千五百尺。問：為立圓徑幾何？

There is another volume measuring 1,644,866,437,500 cubic chi. What is the diameter of the sphere?

## **Solution**

答曰：一萬四千三百尺。

開立圓術曰：置積尺數，以十六乘之，九而一，所得開立方除之，即丸徑。

Answer: 14300 chi.

Spherical extraction method: Take the volume, multiply by 16, and divide by 9. Extract the cube root to obtain the diameter of the ball.

This solution found in the *Jiuzhang Suanshu* is not accurate. The formula used here is

# V

9
16

d
3

{\displaystyle V = \frac{9}{16} d^3 }

.
This meant the diameter was taken as the cube root of 2,924,207,000,000. This value turns out to be 14300.

However the Chinese did discover the precise formula of the volume of the sphere

# V

π
6

d
3

{\displaystyle V = \frac{\pi}{6} d^3 }

.
Using this formula the correct solution (rounded to the nearest whole number) is

# d

(

6
⋅
1644866437500

π

)

1

/

3

{\displaystyle d = {\left(\frac{6 \cdot 1644866437500}{\pi}\right)}^{1/3} }

d
≈
14646
c
h
i

{\displaystyle d \approx 14646 chi }

.
## **Additional Remark**

Yet one question remains. How did one calculate the cube root of a large number like 2,924,207,000,000? The answer lies in scaling and an iterative algorithm for extracting cube roots. Here I present the algorithm with algebraic equations. The algorithm employed in ancient China two millennia ago was reduced to clever computations that replicate these equations (more specifically the coefficients of these equations).

**Initialization**

We need to solve

x
3

=
2
,
924
,
207
,
000
,
000

{\displaystyle x^3 = 2,924,207,000,000 }

. Notice the six zeroes at the end?

Let

# x
100
y

{\displaystyle x = 100y }

. This will help reduce the number to cube root.

10
6

y
3

=
(
2924207
)
(

10
6

)

{\displaystyle 10^6 y^3 = (2924207)(10^6) }

y
3

=
2924207

{\displaystyle y^3 = 2924207 }

**Iteration 1**

Make the first estimate for determining the error bound

100
3

<
2924207
<

200
3

{\displaystyle 100^3 < 2924207 < 200^3 }

.

Let

# y
z
100

{\displaystyle y = z + 100 }

, then

(
z
100

)
3

=
2924207

{\displaystyle (z+100)^3 = 2924207 }

z
3

300

z
2

30000
z
# 1000000
2924207

{\displaystyle z^3 + 300z^2 + 30000z + 1000000 = 2924207 }

z
3

300

z
2

30000
# z
1924207

{\displaystyle z^3 + 300z^2 + 30000z = 1924207 }

z
3

300

z
2

30000
# z
1924207

{\displaystyle z^3 + 300z^2 + 30000z = 1924207 }

300
(
40

)
2

30000
(
40
)
<
1924207
<
300
(
50

)
2

30000
(
50
)

{\displaystyle 300(40)^2 + 30000(40) < 1924207 < 300(50)^2 + 30000(50) }

**Iteration 2**

Since

140
3

=
2744000
<
292407

{\displaystyle 140^3 = 2744000 < 292407 }

, update the substitution

# y
z
140

{\displaystyle y = z + 140 }

.

(
z
140

)
3

=
2924207

{\displaystyle (z+140)^3 = 2924207 }

z
3

420

z
2

58800
z
# 2744000
2924207

{\displaystyle z^3 + 420z^2 + 58800z + 2744000 = 2924207 }

z
3

420

z
2

58800
# z
180207

{\displaystyle z^3 + 420z^2 + 58800z = 180207 }

420
(
3

)
2

58800
(
3
)
<
180207
<
420
(
4

)
2

58800
(
4
)

{\displaystyle 420(3)^2 + 58800(3) < 180207 < 420(4)^2 + 58800(4) }

**Iteration 3**

Since

143
3

=
292407

{\displaystyle 143^3 = 292407 }

, we may end the procedure.

Therefore, if

(
z
143

)
3

=
2924207

{\displaystyle (z+143)^3 = 2924207 }

, then

# z
0

{\displaystyle z=0}

# y
0
# 143
143

{\displaystyle y = 0 + 143 = 143 }

# x
100
(
143
# )
14300

{\displaystyle x = 100(143) = 14300 }

.