472 
Mr . Hellins on the Rectification 
Example iii. 
25. Given a i== 1, and y = v /( 1000000 — 1) = 999*9995 
nearly, to find z. 
This arch, it is very obvious, may be computed by Theorem 
IHd, IVth, Vlth, and Vllth, the series in each of them con- 
verging, in this case, very swiftly. And it may be computed 
also by the IXth ; but the proper Theorem to be used in this 
case is the Vllth. 
Now, since ee is = 2, and y = y/( 1000000 — 1), we have 
(by Article 7,) u = 4/ (syy -f- 1) = y (2000000 — 1) = ^/2 x 
\/(ioooooo — i) = y 2 x (1000 — • k * co - | very nearly, = 
1000 v/2 — = 1414-2132088, which may be taken for the 
value of the whole series, since \u~ 3 , the second term of it, 
does not give a 1 in the tenth place of decimals. If, therefore, 
from u = 1414-2132088, we subtract d = 0-5990701, (by 
Article 17,) we shall have z == 1413-6141387,^ the length 
required. 
Example iv. 
2 6. Let a be given = 7, and y = 10, to find z. 
This Example may be computed by Theorem Hid, IVth, 
Vlth, and some others ; the Vlth is to be chosen rather than 
the Illd, and the IVth rather than the Vlth. 
* The computation of the value of z, in this example, is the problem alluded to in 
the Introduction to this Paper, which first turned my thoughts to the subject of it, in 
the year 1770. In the next year, two answers were given to it, by two persons of good 
reputation for their skill in mathematics, one of them making z'zz 1414-2132088, the 
other, z zz 1413-8921. These two are the only solutions of this problem that I know 
of; and, if my calculation be right, both are erroneous. 
