.. log e2 = log(1 + i tan 6) - log(1 — i tan 6). .. 2i02 i(tan ✪ — } tan3 0 + tan3 0 — etc.) (Art. 130) .. 0 = tan 0 – 1 tan3 0 + } tan30 — etc. which is Gregory's Series. (1) This series is convergent if tan 0 or <1, i.e., if lies Sch. This series may also be obtained by reverting (5) in Cor. 2, Art. 156. Cor. 1. If tan = x, we have from (1) a series which is very slowly convergent, so that a large number of terms would have to be taken to calculate to a close approximation. We shall therefore show how series, which are more rapidly convergent, may be obtained from Gregory's series. a series which converges much more rapidly than (3) of In this way it is found that = 3.141592653589793 ........ 70 1 -1 99 are much more convenient for pur poses of numerical calculation than the series for tan-1 1. 239 Example. Find the numerical value of π to 6 figures by Machin's series, 165. Given sin = x sin (0 + «); expand 0 in a Series of Ascending Powers of x. 0 sin 20 sin 40+ sin 60 sin 80+.... = 166. Given tan x = n tan 0; expand x in Powers of n. .. 2 ix2i0+ log(1+me-2i0)-log (1+ me2i0) = 2i0 — m (e2io − e−2i0) + m2 (etio — e-4io) — 2 (1) When n is even. If r = 0, we obtain from (1) a real responding factors are x —- 1 and x + 1. 1, If we put - 2 additional roots, since 1 roots are found by (2) When n is odd. The only real root is 1, found by putting r = 0 in (1); the other n putting r=1, 2, 3,... n- - 1 in (1) or (2) in succession. 168. Resolve " + 1 into Factors. Since cos (2r+1)π ±√ −1 sin (2r+1)π = − 1, where r is any integer, and .. ï”= cos (2r+1)π ± √−1 sin (2r+1)π. (1) When n is even. There is no real root; the n roots are all imaginary, and are found by putting |