William Oughtred: A Great Seventeenth-Century Teacher of Mathematics - Florian Cajori

and

C_qL=420000(A+E).

Subtract from 247651713 the sum of the known terms A³ (his A_c) and 420000 A (his C_qA). This sum is 232000000 the remainder is 15651713.

“Exemplum II

1c+42̣00̣00̣l=247̇651̇7̣1̣3̣̇

Hoc est, L_c+C_qL=D_c

2 4 7̇ | 6 5 1̇ | 7̣ 1̣ 3̣̇ | ( 4 1 7
------+-------+-------+------------
4 2 | 0 0 0 | 0 | C_q ------+-------+-------+------------
6 4 | | | A_c 1 6 8 | 0 0 0 | 0 | C_q A ------+-------+-------+------------
2 3 2 | 0 0 0 | 0 | Ablatit.
===================================
R 1 5 | 6 5 1̇ | 7 1 3̣ |
------+-------+-------+------------
4 | 8 | | 3 A_q | 1 2 | | 3 A 4 | 2 0 0 | 0 0 | C_q ------+-------+-------+------------
9 | 1 2 0 | 0 0 | Divisor.
------+-------+-------+------------
4 | 8 | | 3 A_q E | 1 2 | | 3 A E_q | 1 | | E_c 4 | 2 0 0 | 0 0 | C_q E ------+-------+-------+------------
9 | 1 2 1 | 0 0 | Ablatit.
===================================
R 6 | 5 3 0 | 7 1 3̣̇ | 4 | 1 |
------+-------+-------+------------ ----+-----+---
| 5 0 4 | 3 | 3 A_q | |
| 1 | 2 3 | 3 A 1 6 | 8 |
| 4 2 0 | 0 0 0 | C_q | 1 |
------+-------+-------+------------ ----+-----+---
| 9 2 5 | 5 3 0 | Divisor. 1 6 8 1
------+-------+-------+------------
3 | 5 3 0 | 1 | 3 A_q E | 6 0 | 2 7 | 3 A E_q | | 3 4 3 | E_c 2 | 9 4 0 | 0 0 0 | C_q E ------+-------+-------+------------
6 | 5 3 0 | 7 1 3 | Ablatit.”

Next, he evaluates the coefficients of E in 3A²E and 420000E, also 3A, the coefficient of E². He obtains 3A²=480000, 3A=1200, C_q=420000. He interprets 3A² and C_q as tens, 3A as hundreds. Accordingly, he obtains as their sum 9120000, which is the divisor for finding the second digit in the approximation. Observe that this divisor is the value of |f(a+s₁)-f(a)|-s₁ⁿ in our general expression, where a=400, s₁=10, n=3, f(x)=x³+420000x.

Dividing the remainder 15651713 by 9120000, he obtains the integer 1 in ten’s place; thus E=10, approximately. He now computes the terms 3A²E, 3AE² and E³ to be, respectively, 4800000, 120000, 1000. Their sum is 9121000. Subtracting it from the previous remainder, 15651713, leaves the new remainder, 6530713.

From here on each step is a repetition of the preceding step. The new A is 410, the new E is to be determined. We have now in closer approximation, L=A+E. This time we do not subtract A³ and C_qA, because this subtraction is already affected by the preceding work.