659. To expand in powers of x, when sin 0 = x sin (0 + a). = expansion of {log (1 − xe ̄at) — log (1 − xa1)} ; ..0x sin a + x2 sin 2a + 3 sin 3a + ... = Bernoulli's Numbers. 660. The following is an application of the exponential theorem, which is extensively used in the expansion of functions. 661. In the expansion of (x + 1)" − x", for integral values of n greater than 1, substitute A, for x" and equate the result to zero. The substitutive function (x + 1)" — " may be simplified by subtracting its values for different values of n; since, after substituting, the result has to be equated to zero. Thus take (x+1)" − x2 from (x + 1)n+1 − x2+1 ; take the remainder i.e. x (x + 1)” — x2 (x − 1) from x(x+1)+1 − xn+1 (x − 1); take the remainder i.e. x2 (x + 1)” − x2 (x − 1 )2 and so on. from x2 (x+1)+1 − x2+1 (x − 1 )2; Thus the substitutive function becomes xm (x + 1)1 − x2 (x − 1)m. (1) Here put m = n; then we have x2 (x + 1)” – x2 (x − 1)”. Here all the terms involving even powers of x cancel; hence the odd terms between A, and Am alone remain. Put n 2; thus A1 = 0. And since every increase of n by 1 brings in one new odd term, we see that After A, all the coefficients with odd suffixes vanish. (2) Now put n = m + 1; then we have xm (x + 1)m+1 − xm+1 (x − 1)m. In this expansion we may omit the odd powers of x since the substitutes for them are zero. The term involving am-2 in x3 (x + 1)m+1 is (m + 1) m ... (m − 2n + 1) 12n + 1 2m-2n is i.e. Hence in the substitutive function the term involving ä2m-2n Let us write then C2 = (2r + 1) A1⁄2r. Then the values of Car may be found by substituting Car for x2 in the expansion of xm (x+1)m+1 equated to zero, the odd coefficients being dropped. Put m = 2; then 3C, + C2 = 0; . C1 = -3. 4 Put m = 3; then 4C + 4C4=0; .. C1 = 8 4 3 Put m = 4; then 5Cg+ 10C + C1 = 0; . Cg = -10. Put m = 12 6 6; then 7C12+ 35C10 + 21Cg+ Cε = 0; .. C'12 And so on. In the above, write B, = (-1)+1A; then B1, B2, B... are called Bernoulli's numbers. The relations between the three sets of coefficients here introduced should be carefully noted. Thus Br=(-1)+ Agr; Car (2r+1) Ar; . Br= = (-1)+1 Car 2r+ 1 The introduction of the A-coefficients leads to the simplest definition; while that of the C-coefficients simplifies the calculations. [It should be observed that if e*=1+y; this expression becomes log (1+ y) which (if y<1) can be expanded in powers of y.] y Hence multiplying and equating the coefficient of a (where That is, A, may be found by substituting A, for ✰ in the expansions of (x + 1)" — x" and equating the results to zero. Hence the coefficients A1, A2, A.... are those found in the last article. Since A1 - ; and after A, the odd coefficients vanish, while B1 =(-1)+1 Agr, we may write y = cos ky and sinh } = i sin hy, B2 – B3 6 cosec y = } + B, + (2o − 1) B, + ( − 1) B, C + .. ..(4), + .....(5), y (22 B1 y 6 уб } tan =(2*− 1) Bỉ +(2 - 1) B+ (2 - 1) B (6) 2 664. Now we have (Art. 526) 1 2 6 cot y If y < 2π, all the terms can be expanded in ascending powers of y2 by division. Hence y y Equating coefficients of yr in the two expansions for cot 2' Hence the limit of the ratio of each term to the preceding in y the expansion of cotis (2)* so that the series is convergent if y < 2π. 1. series. EXAMPLES XXI. The quantities 1, i, 1, -i, 1, i, &c. form a geometric 2. Show that the addition, subtraction, multiplication, and division of complex quantities yields always a complex quantity. 3. If (a+bi) (c + di) is wholly real, a, b, c, and -d must form&proportion; if it is wholly imaginary, a, b, d, c must form a proportion. 4. If the ratio of two complex quantities is wholly imaginary, then the norm of their sum is equal to the sum of their norms. 5. The modulus of the sum of any number of complex numbers is never greater than the sum of their moduli. J. T. 30 |