П where is the angle nearest to (2r+1) π whose cosine and whose tangent = tan 0. This result includes those of the preceding two articles. For, if is odd, we may put 2r+1=q, so that cos 2r+1 = cos 0 рπ = ( − 1), and sin p=0. Hence the series for cos pe/q and sin pe/q reduce 9 2r+1 q to the forms given in Art. 620. And, if q is even and 9 odd, we may put 2r+1=1q, so that 2r+1 pr=(-1). Hence the series reduce 2r+1 COS PT=0 and sin to the forms given in Art. 621. Moreover, if 9 and are both even, we cannot make either cos p (2r+1)π/q or sin p (2r+1) π/q vanish, so that cos pe/q and sin pe/q cannot be expressed in a single series. 623. The result of the last article follows immediately from Art. 619. For, if (2r+ 1) is acute, the multipliers of cos (2r + 1) рπ/q and sin (2r + 1)рπ/q in the series of Art. 622 are, by Art. 619, respectively cos {0-(2r+1) π} p/q and sin {0 – (2r + 1) π} p/q. Hence the expressions given for cos pe/q and sin p0/q are where (sino 0) and 2 (− sin 0) are taken as real positive. where is the angle nearest to 1⁄2 (4r + 1) π whose sine = sin 0 and where is the angle nearest to 1⁄2 (4r + 3) π, whose sine = sin 0 and whose cot cot 0. Thus (A) If p is even, q of the form 4r + 1, and sin positive, Р p+2 C. (sin3 0) = (-1) cos20 and S. 2 (sino 0) = (-1) sin2 0. (B) If p is even, q of the form 4r + 3, and sin ◊ negative, In (A), (B), (C), (D), ✪ is the angle nearest to 19′′. 625. In the following article, an important application is made of the Binomial Theorem.. The proof given is equivalent to establishing an equation involving the first and second differential coefficients of the function to be expanded. 626. To expand {(1 + x2)*+x}" in powers of x, where x < 1. We may write 1+2+giờ + c +...+ c2 +...= {(1 + x) +} ...(I), the first two terms being immediately derivable from the binomial theorem. Now write x + h for x. The left-hand becomes 1 + n (x + h) + α2 (x + h)2 + ... + ar (x + h)" +... The right-hand becomes {(1 + x2 + 2hx + h2)3 +x+h}" = { (1+x2)1 + §. 2hx (1+x2) ̄3+x+h+... }~ ={(1+x2)1+x}"{1+h(1+x2) ̄*+...}"={(1+x2)*+x}"{1+nh(1+x2) ̄* + ...}. (I), Thus, equating the coefficients of h, after substituting from (1+x2) + (n+2αx+...+ra, x2-1+...)=n(1+nx+ax2+...+α, x2+...)...(II). Again in (II) write x+h for x. Then {(1 + x2)1 +hx (1 + x2) ̄* +.....} {n + 2α, (x + h) +...+ra, (x + h)r-1 +... } = n {1 + n (x + h) + a2 (x + h)2 +: (x + h)" +.....}. Equating the coefficients of h again, we have + ar (1+x2)3 {2a2+...+r(r−1)α, x2-2+...}+x(1+x2) ̄† (n+2a2+...+ra,x2-1+.....) = n (n + 2a,x+. +ra,.x2−1 +.....) = = n2 (1 + x2) ̄ * (1 + nx + a2x2 +...+ ax2+.....) by (II). Multiply by (1+x2). Thus (1 + x2) {2α +...+r (r− 1) α, x2+...} +x (n + 2α +...+ra,x-1+...) = n2 (1 + nx + α„x2 + •+α„x2+.....) In (III) equate the coefficients of x". Thus ......... .....(III) Putting in succession 0, 1, 2, 3.... we have Also substituting for a2, a........ in (II), we have nx2+.... 627. In these two expansions, the multiplier of the alternate 628. The above formulæ may be transformed in many ways. If x is wholly real and less than 1, the series are positive (as the student may show): hence the real positive values of the fractional powers must be taken. 629. A. Let (1+x2)2 + x = ≈ ; then (1 + x2)1 — x = z ̄1, 1 and + 1, ≈ lies between √2 - 1 and √√2 + 1; of z Hence any power of z may be expressed in ascending powers in a series, which is convergent if z lies between √2-1 and /2+1. It would thus appear that, by expanding the powers of z-z-1 by the binomial theorem, we could expand any power of ≈ in positive and negative integral powers of z. And this is in a sense true. But the resulting series would be divergent, unless we took the terms in groups corresponding to the integral powers of z-z-1. 630. B. Let x = cot 0; then √(1+x2)+x=cosec✪+cot0=cot10. If then x lies between 1 and +1, we may take 0 between and, so that cot 10 is positive. Thus 631. has any To expand cos n✪ and sin n✪ in powers of sin 0, when n real value. In the formulæ of Art. 626, put x = i sin 0. Then √(1 + x2) + x = cos + i sin 0. Hence the series represent some value of (cos + i sin 0)", i.e. of cos no + i sin no where is some angle whose cosine and sine are equal to those of respectively. If 0 = 0, x = 0; and the series become = 1 = cos no. Hence, by the principle of continuity, if 0 is acute, The first and last of these series terminate only if n is an even integer; the middle two terminate only if n is an odd integer. For these special cases, the above formulæ were proved in Art. 375. 632. To show that, if n is positive, the series, obtained by multiplying all the terms in the expansion of (1-1)" by any positive or negative finite quantities, is convergent. |