As soon as r becomes greater than n, the terms in the expansion of (1 - 1)" after the pth 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 path is finite, being numerically equal but opposite in sign to the sum of the terms up to the goth. Hence, if all the terms after the goth 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@)in ratios of multiples of 0, where n has any (integral or fractional) positive value. The general value of (2 cos 6)" is given by the expression (2 cos 6)" (cos 217 . n + i sin 22. n). Expanding by the binomial theorem, the general value of the above is (cos 8 + i sin 6)" +n (cos8 + i sin 6)n-2 + 1 n(n-1) (cos @+isin ()n-4+... where, though each of the expressions (cos 0 + i sin )"-2 is manyvalued, yet each is obtained from the preceding by dividing by the single-valued expression (cos 0 + i sin 0)*, i.e. cos 20 + i sin 20. A 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. + n-2 + . n. Now suppose 0 = 2rm, where r has any specified integral value. Then the series becomes (cos Art. n + i sin 2rt. n){1+n + n(n − 1) + ...} =(cos 2rt.n+ i sin 2rt. n) (positive value of 2") by the binomial theorem, n being greater than – 1. (Art. 462.) Thus, if cos 0 = 1, we have (cos 2rt.n+ i sin 2r7. n) {positive value of (2 cos 0)"}, = cos no + n cos (n − 2) 0 + {n (n − 1) cos (n – 4) 8+ +i sin no + in sin (n − 2) 0 + žin (n − 1) sin (n – 4) 0 + ... where A = 2ra. If then a changes continuously from 2rt to (2r+1) , cos e remains positive, the positive value of (2 cos 6)" also changes continuously, and the above series changes continuously. Hence the multiplier cos 2rt . n + i sin 2ri.n must not change discontinuously into cos 2r'n. n + i sin 2r'.. Thus, if cos 0 has any positive value, the above equation still holds if, in the series, we take 0 to be the angle nearest to 2rt whose cosine Next suppose 0 = (2r + 1) +, where r has any specified value. Then the series becomes {cos (2r + 1) .N+ i sin (2r + 1) 7. n} (positive value of 2"). Thus, if cos 0 -1, we have {cos (2r + 1). n + i sin (2r +1) 7. n} {positive value of (– 2 cos 6)"}, = cos no +n cos (n − 2) 0 + in (n − 1) cos (n – 4) 0 + ... + i sin n8 + in sin (n-2) 0 + }in (n − 1) sin (n – 4) 0 + ... where 0 = (2r + As before then, if cos 0 has any negative value, this equation still holds, if, in the series, we take o to be the angle nearest to (2x + 1) + whose cosine 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 no + n cos (n − 2) 0 + {n (n − 1) cos (n – 4) 0 +... and sin no + n sin (n - 2) + n(n-1) sin (n - 4) 6+... 0 equal, respectively, to the real positive value of (2 cos 6)" and to 0 = cos 0. . 1) T = cos 0. . p-1 p-1 A. If however 0 is not acute, but differs from 2rt by an acute angle, the above series are equal to the real positive value of (2 cos 6)" multiplied, respectively, by cos 2ra . n and sin 2rt. n. Thus, if n= =P/9, where 9 even, we may put 2r=9, and the above series become equal respectively to the real value of - (2 cos 6)” and to 0. If further žq is even, we may put 2r = £9, and the above series become equal, respectively, to 0 and to the real value of (-1) (2 cos 6)". B. If 0 differs from (2r + 1) + 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) . n. Thus, if n= =p/9, where 9 is odd, we may put 2r + 1 =9; and the above series become equal, respectively, to (-1) (- 2 cos 6) and to 0. If q is even and 9 is odd, we may put 2r + 1 = £9; and they become equal, respectively, to 0 and (-1) (-2 cos 6)". 635. To extend the above results to the case, when n is negative. Let (n), represent the coefficient of " in (1 + x)". (n)r-, + (n) = (n + 1)r. 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 f(n) = cos no + (n), cos (n − 2) 0 + + (n), cos (n – 2r) 0 + i sin n6 + i(n), sin (n − 2) @ + ... +i(n), sin (n – 2r) e ñ0 0 where the coefficients are those of (1 + )" and the series terminates at (n), Multiply by 2 cos 0, and resolve each product 2 cos A cos po into the sum of cos (p+1) 6 and cos (P-1)0, and 2 2 cos 6 sin po into sin (p+1) 8+ sin(P-1)0. Thus, by the above relation between the coefficients of the binomial expansions, 2 cos 6. f(n) == f(n+1) +(n), cos (n - 2r - 1) @+i (n), sin (n – 2r - 1) 0. -1 If then, when r is indefinitely increased, (1)f (n + 1) remains finite, (2) (n), remains finite, (3) cos 0 is not zero, then f(n) will be finite. r Now it was shown in Art. 519 (and also in Art. 462), that, when r is infinite, (n), is zero or infinite according as n is greater or less than – 1. If n=-1, (n), becomes +1. If then n lies between 0 and -1, the series f (n) is convergent and = (2 cos () provided cos 0 is not zero. - 1, the series f(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. If n= < 636. In the expansions of the ratios of no in terms of the ratios of 0, we may, by putting n=c, obtain expansions of the ratios of A in terms of 0. This has been done in Art. 501. 637. But, by putting n= 0, we obtain expansions of 6 in terms of the ratios of 0. The results are true, in each case, only if o represents the circular measure of an acute angle. Thus, from Art. 619, we have, if tan 0<1, (n-1)(n-2) (n-1)(n-2)(n-3)(n-4) tan tanA + tan" - ... 1.2.3 1.2.3.4.5 sin no n cos” A' (n-1)(n-2)(n-3) cos no - cos” tano 0 - n cos” Putting n=0, and observing that the expansion of cos" in powers of n gives 1 + n log cos 0 + ] (n log cos 6)2 + we have tan 6 – } tano 0 + } tan 0 0 :(1), 3 tan” 0 – 1 tan* 0 + tano 0 - = log sec ...(2). [The latter result follows at once from the equation sec 0 = (1 + tan’6)}.] n-1 + 3 n-1 (n-1)(n-2) cos (2 – n) 0+ cos (4 – n) 0 + cos (6 – n) 0 + ... 1.2 1.2.3 (2 cos O)" – cos no n Putting n=0, we have sin 20 – sin 40 + f sin 68 - ... = 6..... (5), cos 20 – } cos 40 + { cos 68 = log 2 cos 6 .... (6). Other results may be obtained on the same principle. § 3. COMPLEX OR IMAGINARY INDICES. 638. The use of complex expressions of the form a + big 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. |