Multiply numerator and denominator of this fraction by cos (0+2j't)/q-isin (0+2u'r)/q. +2' θ + 2μπ θ+ 2μπ q 9 2u – 2n' 2μ -- 2μ' i.e. h, a qth root of unity. 9 9 COS (+ 8+247) a ti sin 610. To express x" - 1 in real factors, n being integral. Solving the equation 3c" = 1, i.e. = cos 220 + i sin 21tr, we have 2λπ 2λπ + i sin = COS This expression has n values obtained by putting successively equal to 0, 1, 2...n-1. Putting, here I = 0, we have for the first factor x – 1. n If n is even, the value .= in is unpaired, and gives X - COST - i sin = 3 + 1. 611. To express x" + 1 in real factors, n being integral. Solving the equation ac" =-1, i.e. cos (21 + 1) + + isin (21 + 1)T, + T 21 + 1 22+1 we have a = cos a + i sin T. n n This expression has n values obtained by putting i successively equal to 0, 1, 2...n-1. 22+1 21 + 1 i sin .. by Art. 598, 2*+1=Pd=-(* + COS T 1-). λ=0 T If n is odd, the value 2 = 1 (n − 1) is unpaired, and gives 2 - COST - i sin a = + 1. Hence, if n is even, 2c" + 1 = Зп 2* – ac . 612. To resolve x2 – 2x* cos n0 + 1 into factors, where n is integral and cos no is not + 1 nor 1. Solving the equation 22n – 20cn cos no 1, 2021 – 2” cos no + cosa no (1 - cos no)=- sino no. . . c = cos no + i sin no = cos (n0 + 217) + i sin (n0 + 217) 2λπ 2λπ ... X = COS O + + i sin ( 0 + If to i we give different values j and r', then the two values of x will be different unless 604. Whatever positive onu'min) or (0 + 2p:/n) – (+ 2m'r/n) cos no + i sin no is one o hultiple of t, or u~ ' is a multiple We have cos (0 ++4+. = (cos 6 + i sin 8) (cos 's no=1: which has been excluded. Now let 0=$=y=... ar; Thus, if n is a positive intege,' values 0, 1, 2...n-1 we have 2n ? 0 cos no + i sin a Next let n be a negative 2λπ 2λπ cos no + i sin no = cos mb x cos (A + A cosam? 2 i sin (0+297)} ( n n (@+ 2001 ) ) + 1}, = COS m n. = co rém, n being integral = (c and cos no – i sin no = 2-n. ..cos no + i sin no is one of t : subtraction cos no + i sin no is one 2i sin e = 3 – 2-1. 2i sin no = 2c" — 20-. 605. To find q values prime to one another. .1 transformations. We have cos 0 + i sin 0 = 3, 418, that has any integral value. (2014 – x-) = (x - 2-1) .. any one of the value inctions of x + 2-1 as 2 cos no and 19. But we did not there equate a value of (cos 0 + i sin 6). To do this we must implicitly Giving to d any two v: ? cos must be < 2 for any real angle, ny real value of x (: 0 – 2 +is + cos? (0 614. In Art. 610, where n is even, write 2m for n and divide both sides by xn. Thus om - 30 1 2C + -1 – 2 cos 8C + 2-1 – 2 cos 20 + x-1 -2 cos -m m m = Now put x+x-1= 2 cos 0; then x - x-1= 2i sin 0; and 2c— x-= 2i sin mo. Thus sin morsin o x m 1 2 cos 0 - 2 cos m (2.00 ) (2 1-). T Again, in Art. 611, where n is even, write 2m for n, and divide both sides by xn. Thus xm + x-m Again in Art. 612, divide both sides by a". Now put x + +1 = 2 cosa; then ac" + xcN = 2 cos na. Thus 27 2 cos na — 2 cos no = (2 cos a – 2 cos 0) {2 cos a – 2 cos (0 + These 6 results are the same as those obtained in Arts. 419 420, 421. § 2. EXPANSIONS BY DE MOIVRE'S THEOREM. 615. In the following propositions we shall require to make use of the Principle of Continuity. DEF. A function of a variable, whose value may be made to change by as small a quantity as we please, by making the change in the variable sufficiently small, is said to be continuous. PROP. An endless series in ascending integral powers of a is a continuous function of x, for all values for which it is convergent. For let, f (x) = co + C72C + Cy@c2 + + Crack + ... f (+ d) = 0,+ c (x+d) + C2 (20+d)2 + ... + Cn (x + d)" + :: f (w+d)-f(a) = d.S where. S is finite, since f (c + d) is convergent. Now, if a is any assignable quantity not zero, we can make d less than a - S, since S is not infinite. Hence f (x + d)-f(a) may be made less than assignable quantity a by sufficiently diminishing d. That is, f (x) is continuous. The above proposition is required, when we know the value of a series for some particular value of x, but cannot otherwise determine which of a number of different forms must be assigned to its general value. The proposition shows that we must assign such a form to the general value of the series as will make it change continuously from its known value when a changes continuously. 616. In deducing expansions involving complex quantities, our results are based on the Binomial Theorem. We have, then, first to examine whether the expansions are convergent. This will require that both the series of real terms and the series of imaginary terms are convergent. |