*MISCELLANEOUS EXAMPLES. V. (1) Since a*={1+(a−1)}*, prove by expanding the righthand side that where α=1+Д1x+А2x2 + A3x3 + etc., A1 =(a-1)-(a−1)2 + (a−1)3. (2) Since a*+y=a* × a", expand a*+y and a* by the theorem of Ex. 1 and by equating coefficients of x, prove that Expand a", and by equating the coefficients of the various powers of y find A2, A3, etc. in terms of A1. that (3) Show that a41 in the last example is e. Hence by Ex. 1 prove that log, a (a-1)-(a− 1)2+(a− 1)3- etc. (4) Prove that 1 log, n=m {(n′′ – 1) – ↓ (n′′ – 1)2 + } (n′′ – 1)3 — etc.}. 1 Hence, having given that 10232 − 1 = '000000000536112, prove log, 10=2.30258. 2n2 + 3n3 - etc.}. 10000 and d a number d Hence if n be a number greater than less than 1, prove that proximation for all practical purposes. == to a sufficient ap log sin 2a + log cot a = cos 2a - cos2 2a + cos3 2a - etc. 11. DEF. CHAPTER II. DE MOIVRE'S THEOREM. 1 is a symbolical expression, whose square is −1, which is capable of obeying the ordinary laws of Algebra. Since 1 obeys the laws of Algebra √-a2 = √1x a3 = a√ — 1. The student must observe that such an equation as A+B√ −1 =a+b√-1 can only be true when A = a and B=b. We shall often use the letter i as an abbreviation for √-1. 12. De Moivre's Theorem. Whatever be the value of n positive or negative, integral or fractional, cos na+√-1 sin na is one of the values of (cos a +/-1 sin a)". I. When n is a positive integer. Consider the product (cos a +/- 1 sin a) × (cos ẞ+ √-1 sin ß). It is equal to cos a. cos ß - sin a. sin ẞ+ √-1 (cos a. sin ß + sin a . cos B). That is to cos (a + B) + √-1 sin (a + ẞ). Similarly the product {cos (a +ẞ) + √1 sin (a + B)} x {cos y + √-1 sin y} is equal to cos (a + B + y) + √ − 1 sin (a + B + y). - 1 sin a Proceeding in this way we obtain that the product of any number n of factors, each of the form cos a + is equal to cos (a +ẞ+y+... n terms) + √-1 sin (a+B+y+... n terms). In this result let ẞ = y = etc. = α, and we have that (cos a+√ - 1 sin a)” = cos na + √−1 sin na. Thus, when ʼn is a positive integer, De Moivre's Theorem is true. II. When n is a negative integer. Let n=- m. Then m is a positive integer. And (cos a + √− 1 sin a)" = (cos a + √ — 1 sin a)-m 1 1 cos ma + √1 sin ma - 1 sin ma cos ma [By I.] cos ma-√-1 sin ma √1 sin ma cos ma + sin2 ma Therefore (cos a +/-1 sin a)" = cos ma 1 sin ma = cos (− m) a + √ − 1 sin (− m)a = cos na + √ −1 sin na. Thus De Moivre's Theorem is true when n is a negative integer. III. When n is a fraction, positive or negative. Now (cos +/- 1 sin ẞ)2 = cos qẞ + √ − 1 sin qß. [By I. and II.] Therefore taking the qth root of both sides is one of the values of (cos a +/- 1 sin a)2. p is one of the values of (cos a+/-1 sin a)". Thus the theorem is completely established. EXAMPLES. VI. (1) If A stand for cos 2a+isin 2a, and B, C, D for similar expressions in terms of ß, y, d, prove that AB+CD =2 cos (a + ẞ − y − d) {cos (a+B+y+8)+i sin (a+B+y+8)}. (2) With the notation of Ex. 1, prove that 1 AB-CD sin (a+B+y+8) − i cos (a+B+y+d) (3) With the same notation, prove that (A-B) (C – D) - 4 sin (a-B) sin (y-d) {cos (a+B+y+8) - i sin (a+B+y+8)}. (4) With the same notation prove that 1 (A+B) (C+D) cos (a+B+y+8) - i sin (a+B+y+8) 4 cos (a - ẞ) cos (y – d) (5) Prove that cos (a+B+y...)+isin (a+B+y+...) where 81 stands for the sum of tan a +tan ẞ+tan y + etc., 83 stands for the sum of the products of these tangents three at a time, and so on. (7) Prove that tan (a+ẞ+y+...): 82,... are defined in Ex. 6. = 81-83+8,- etc. where $1, (8) Write down the last term of the numerator of the fraction in Ex. 7, (i) when n is even, (ii) when n is odd. 13. It is known from the Theory of Equations that there are q different values of x, and no more, which satisfy the equation = a, where a is real or of the form A + √(−1) B. We can prove that we may obtain q different values by de Moivre's theorem, and no more. 14. The expression cos 0+/1 sin 0 is unaltered if for we put (0+2), where r is an integer. which is one of the values of {cos (0 + 2rπ) + √=1 sin (0 + 2rπ)}", is one of the values of (0+ {cos + √ = 1 sin |