13. To expand (90° – x) when x is small. 5 (m - 2) = flog cos x d = 5 } + x log cos 2 + f tan x . d. Let tan x = ≈, and assume + A ̧1⁄23 – A‚≈5 + A ̧z1 − &c.. .§ whence A, 1, 4, = 1 + }, 4, = 1 +} + }, &c... = 2 + } {}z2 − }zo + &c...} } &c. &c..... Let us henceforth use x for Stan”-1 x . dx, n so that ò̟1 x = x; p2 = log sec x; and •„+2x = and we finally obtain 5 + Ò ̧ x − } Þ¿x + }p, x − ¦p,x + &c.. .. .(24), When x is < 10°, 4, x will not affect the sixth decimal. 7 1.3 -2 and we get B1 = 1, B2= B1+1⁄2.¦2 ̄1; B ̧=B2- + .}22; &c... 2.4 -3 To increase therefore the convergence, put C = π 2√2 B.; . 23+ &c... {1 - 1} = .. C ̧ = C, − 1 ; C = C, − §. } 2`; C = C − 1.3. C-1; .122; &c. ... 1 2.4 and C, C, C.. may be easily tabulated. Finally, observing that √(20) = 2 sin x, and modifying the fraction under log tanπ+x) ↓ (1⁄2π − x) = ↳ 1⁄2 π + x log cos x + π log tanπ x) 2 3 4 sin x {C + C1 ⋅ }v + C2 · }v2 + C ̧ . {v3 + &c.. . }. .(25) which converges well even when x is as large as 60°. 14. In constructing a table of x we need to find Ax, or ↳ (x + h) - ↳x. Taylor's theorem may of course be used, but the law of the terms is cumbrous. As a substitute for it, let us recal equation (2), which gave ..(w + a) - Sa = ↳w + w log x - 1-2x sin a 2-2x2 sin 2a 3 dx ω. ; 0 X sin 3a - &c. To conform to the usual notation, write x, h for a, w, and then we must put y for x. We hereby find that - 1-2y sin x 22y sin 2x-3-2y sin 3x - &c...(26). If x is large compared with h, and h is known, y will be small enough to give a good convergence, and the law of the series is simple and convenient. Nevertheless, methods of interpolation, such as the following, will be often better. First, let m1, m,, m... satisfy the equation x = log (1 + x) = 1 + m1x + m ̧x2 + M ̧x3 + •••• then AfFx dx = h {Fx + m,^Fx + m2▲2Fx + &c.. ...} and by slightly modifying the process, it is easy to shew that we have also Take the sum: Al sin (xh) - ▲l sin (x – 2h)} .(28). x = − h [l sin (x + h) + 1 sin x − {▲ sin x - + ▲2l sin. (x − h)}]. .(29). But perhaps certain series of Legendre's are better still. Let M, M2, M.....be determined by the equation ≈ ÷ sin ̈1(x) = 1 + M1x2 + M ̧x* + M ̧1⁄2o +.... ▲ [Fx dx=h{ F(x+}h)+M1▲2F(x− }h) +M‚▲*F(x− }3h)+ &c...} also = 2 h{F(x+}h)+M ̧▲3F(x+}3h)+M ̧▲*F(x+{5h)+&c...} 17 + 1⁄2▲23⁄4 sin (x − ¦h) – 31⁄4‰▲1l sin (x − {3h) + &c.} + 1⁄2▲2l sin (x + 13h) – 5 ▲l sin (x + 15h) + &c.} which are easy to us, because we have tables of log sin. (30), Again, let N1, N2, N3. 1 − N ̧22 + N ̧1⁄2“ – No + &c... = (1 + M1x2 + M2x + &c...)2 ; :. ▲2ƒFxdx = h2 {F"(x + h) + N ̧▲2F"x + N2A*F′(x − h) + &c.} also = h2 {F"(x + h + N ̧Ã3F"(x + 2h) + N2▲*F"(x + 3h) +&c.}, Θ - sin2 a = sin (0 + a). sin (0 − a). ℗ = ↳ (0 + a) + ↳ (0 − a), or = ·5 (a + 0)-5(a). flog (cos - cos a) do. COS α = 2 (cos2 10 - cos2 ja), 0 log 2 +27 = log (1 + sec μ cos 0) d0, α 2 since log (1 + sec μ cos 0) = log {cos μ - cos (π-0)}– log cos μ, (4) If=log (1 + sin μ cos 0) do, put tan ŋ = and we had by equation (6) Θ = ·5 (0 - n) + ↳n - 50 + 0 log sin 1⁄2μ 50+ log sin μ-n log tanμ. Thus flog (a + b cos 0) do can always be found by ↳. (5) If = log (tan 0 + tan a) do ; -2 = (6) To find S = 12x - 3 ̃2x23 + 5 ̃2x3 − 72x2 + &c.. ...In the process which gave rise to equation (26), put a x= tan w; and observing that (+) T = T - ↳ (π-w), we get w S = ∞ log x + w + ↳ (π – w) – 14π. חל (7) If=Лlog (1 − 2n cos a tan 0 + n2 tan2 0) d0, we reduce this to No. (4), as in the process for finding X. See equation (5). (8) If = log (tan2 0 + tan2 ß) do, make n = cot ß; = 20 log tan ẞ+ flog (1 + n2 tan' 0) do. The last falls under Ex. (7), as a particular case, when α = 1⁄2π. - (9) If = log (1 2r cos a sin @ + r2 sin2 w) dw, this also may be reduced to, by the following process. Supposer positive, x = tan w, or sin w = 1 + x2= sec2 w; 2x 1 + x : and .. Q = ƒ log {1 – 4rx cos a + (4r2 + 2) x2 − 4rx3 cos a + x1} dw :. (n + n`1) cos y = 2r cos a, and 4r2 + 2 = W 4 cos2 γ + n2 + n22. Let tan v = n, tan p = r ; .. 4r2 + 4 = 4 cos2 y + (n + n−1)2 ; or cos y = r cos a sin 2v, and 2r cosec 2p = cos2 γ + cosec2 2v. Eliminate Y, and solve for cosec 2v. The result is, that if we take sin 25 = cos a sin 2p, and select that root of which makes sin least, we have where we pass from ' to " by changing v to (Tv); then |