As soon as becomes greater than n, the terms in the expansion of (1 − 1)" after the rth have all the same sign. If n is positive, the series is convergent by Art. 461, and is in fact equal to zero. Hence the sum of the terms after the rth is finite, being numerically equal but opposite in sign to the sum of the terms up to the rth. Hence, if all the terms after the rth are multiplied by any positive finite quantities, the resulting series will be convergent by Art. 453. And if any of the signs are then changed, the resulting series will still be convergent by Art. 445. Hence, the series obtained by multiplying all the terms in the expansion of (1-1)", where n is positive, by any positive or negative finite quantities is convergent. Q. E. D. 633. To expand (2 cos 0)" in ratios of multiples of 0, where n has any (integral or fractional) positive value. The general value of (2 cos 0)" is given by the expression Again (2 cos 0)" {cos + i sin + (cos + i sin 0)-1. = Expanding by the binomial theorem, the general value of the above is 0 (cos +isine)"+n (cos + i sin 0)"-2+ {n(n−1) (cos +isinė)"→1+... where, though each of the expressions (cos + i sin 0)n-2r is manyvalued, yet each is obtained from the preceding by dividing by the single-valued expression (cos + i sin 0)2, i.e. cos 20 + i sin 20. Hence, by De Moivre's theorem, the above expansion = cos no + n cos (n − 2) 0 + 1⁄2 n (n − 1) cos (n − 4) 0 + ... where must have the same value throughout. Since n is positive, this series is convergent by the last article. Hence also, by making a sufficiently small change in the value of 0, we may make the change in the value of the series as small as we please. Now suppose 0=2r, where r has any specified integral value. Then the series becomes (cos 2r. n + i sin 2rπ . n) {1 + n + ‡n (n − 1) + ... } = (cos 2rπ. n + i sin 2rπ. n) (positive value of 2") by the binomial theorem, n being greater than 1. (Art. 462.) Thus, if cos = 1, we have (cos 2r. n + i sin 2r. n) {positive value of (2 cos 0)"}, = cos no + n cos (n − 2) 0 + 1n (n − 1) cos (n − 4) 0 + ... + i sin no + in sin (n-2)0+ in (n - 1) sin (n-4) 0+... where 02гπ. If then changes continuously from 2rπ to (2r±1) π, cos 0 remains positive, the positive value of (2 cos 0)" also changes continuously, and the above series changes continuously. Hence the multiplier cos 2rπ. n + i sin 2r. n must not continuously into cos 2r'π . n + i sin 2r'π . n. change dis Thus, if cos has any positive value, the above equation still holds if, in the series, we take to be the angle nearest to 2rπ whose cosine = cos 0. Next suppose = (2r+1) π, where r has any specified value. Then the series becomes {cos (2r + 1) π . n + i sin (2r + 1) π . n} (positive value of 2′′). Thus, if cos -= 1, we have {cos (2r + 1) π . n + i sin (2r + 1) π . n} {positive value of (− 2 cos 6)"}, = cos n✪ + n cos (n − 2) 0 + žn (n − 1) cos (n − 4) 0 + - ... + i sin n✪ + in sin (n − 2) 0 + žin (n − 1) sin (n − 4) 0 + ... where 0 = (2r + 1) π. As before then, if cos 0 has any negative value, this equation still holds, if, in the series, we take ◊ to be the angle nearest to (2r+1) π whose cosine = cos 0. 634. In the first result of the last article suppose 0 is acute. Then, r=0; and, equating real and imaginary terms, we have cos n✪ + n cos (n − 2) 0 + 1n (n − 1) cos (n − 4) 0 +... and Ꮎ sin no + n sin (n − 2) 0 + in (n − 1) sin (n - 4) 0+... equal, respectively, to the real positive value of (2 cos 0)" and to 0 A. If however is not acute, but differs from 2гя bу an acute angle, the above series are equal to the real positive value of (2 cos 0)" multiplied, respectively, by cos 2rπ. n and sin 2ræ. n. Thus, if n=p/q, where q is even, we may put 2r=q, and the above series become equal respectively to the real value of — (2 cos 0)" and to 0. If further is even, we may put 2r=19, and the above series become equal, respectively, to 0 and to the real value of p-1 (-1) 2 (2 cos 0)". B. If 0 differs from (2r+ 1)π by an acute angle, the above series are equal to the real positive value of (- 2 cos 0)" multiplied, respectively, by cos (2r+ 1) . n and sin (2r + 1) π. n. π. Thus, if n=p/q, where q is odd, we may put 2r+1=q; and the above series become equal, respectively, to (-1) (-2 cos 0)" and to 0. If is even and 19 is odd, we may put 2r+ 1 = 19; and they become equal, respectively, to 0 and (-1) (-2 cos 0)". p-1 635. To extend the above results to the case, when n is negative. Let (n), represent the coefficient of a" in (1+x)”. Then, as shown in Art. 464, (n)r_1 + (n), = (n + 1),. This formula follows at once from the fact that (1 + x)+1 must be obtainable from (1 + x)" by multiplying by 1+x. Now let + (n), cos (n − 2r) 0 2) 0 + ... + i (n), sin (n − 2r) ✪ (1 + x)" and the series termiresolve each product and cos (p − 1) 0, and Thus, by the above ƒ (n) = cos n + (n), cos (n − 2) 0 + 2 cos 0. f(n) = f(n + 1) + (n), cos (n − 2r − 1) 0 + i (n), sin (n − 2r − 1) 0. If then, when r is indefinitely increased, (1)f(n + 1) remains finite, (2) (n), remains finite, (3) cos is not zero, then ƒ(n) will be finite. Now it was shown in Art. 519 (and also in Art. 462), that, when is infinite, (n), is zero or infinite according as n is greater or less than - 1. 1, (n), becomes ± 1. If n - If then n lies between 0 and 1, the series f(n) is convergent and = (2 cos 0)" provided cos 0 is not zero. If n == − 1, the series ƒ(n) is finite, but indeterminate. If n < -1, the series f(n) is infinite for all but certain special forms of r, when r is infinite. 636. In the expansions of the ratios of ne in terms of the ratios of 0, we may, by putting n=∞, obtain expansions of the ratios of in terms of 0. This has been done in Art. 501. 637. But, by putting n = 0, we obtain expansions of 0 in terms of the ratios of 0. The results are true, in each case, only if represents the circular measure of an acute angle. Thus, from Art. 619, we have, if tan <1, Putting n = 0, and observing that the expansion of cos” ◊ in powers of n gives 1 + n log cos 0 + 1⁄2 (n log cos 0)2 + we have = log sec 0 [The latter result follows at once from the equation sec = (1 + tan3 0)*.] Other results may be obtained on the same principle. § 3. COMPLEX OR IMAGINARY INDICES. 638. The use of complex expressions of the form a+bi, where a and b are real, is explained for symbolical purposes by the understanding that the equation a + bi = c + di shall be interpreted to mean that a = c and b = d. It remains then to examine what use can be made of imaginary or complex indices. |