whence
| loge | 1 + x | = 2 (x + 1⁄3x3 + 1⁄5x5 + &c.), |
| 1 − x |
and, therefore, replacing x by (p − q)/(p + q),
| loge | p | = 2 { | p − q | + 1⁄3 ( | p − q | ) 3 + 1⁄5 ( | p − q | ) 5 + &c. }, |
| q | p + q | p + q | p + q |
in which the series is always convergent, so that the formula affords a method of deducing the logarithm of one number from that of another.
As particular cases we have, by putting q = 1,
| loge p = 2 { | p − 1 | + 1⁄3 ( | p − 1 | ) 3 + 1⁄5 ( | p − 1 | ) 5 + &c. }, |
| p + 1 | p + 1 | p + 1 |
and by putting q = p + 1,
| loge(p + 1) − loge p = 2 { | 1 | + 1⁄3 | 1 | + 1⁄5 | 1 | + &c. }; |
| 2p + 1 | (2p + 1)3 | (2p + 1)5 |
the former of these equations gives a convergent series for logep, and the latter a very convergent series by means of which the logarithm of any number may be deduced from the logarithm of the preceding number.