Multiply numerator and denominator of this fraction by cos (0+2μm)/q-isin (0+2μ')/q. The new denominator becomes unity. And the new numerator becomes 610. To express x1-1 in real factors, n being integral. This expression has n values obtained by putting λ successively equal to 0, 1, 2...n-1. Putting, here λ=0, we have for the first factor x-1. If n is even, the value λ = n is unpaired, and gives 1. n 611. To express x2 + 1 in real factors, n being integral. This expression has n values obtained by putting λ successively equal to 0, 1, 2...n - 1. If n is odd, the value λ = (n − 1) is unpaired, and gives 612. To resolve x2 - 2x2 cos no + 1 into factors, where n is integral and cos ne is not + 1 nor 1. .. xn = cos no i sin no '= cos (n0 + 2λπ) ± i sin (n✪ + 2λπ) 2λπ 2λπ 2~). x= cos (6+) i sin (+2) If to A we give different values μ and μ', then the two values of x will be different unless х either (θ + 2μπ/n) + (0 + μ'π|n) or (θ + 2μπ/η) - (θ +2μπ/n) is a multiple of 2′′; i.e. unless either ne is a multiple of π, or μ~μ' is a multiple of n. If no is a multiple of π, cos no: =1: which has been excluded. Hence, giving to λ the values 0, 1, 2...n − 1 we have 2n values of x. where the two factors, obtained by taking the upper and lower signs before i, are multiplied together. Also, by De Moivre's theorem, n being integral cos no + i sin no = xn and cos no- i sin n0 = x-n. Moreover by addition and subtraction -1 2 cos 0 These are extremely useful transformations. We showed, in Arts. 413, 418, that x2+x-" and (x2 - x-") ÷ (x-x-1) are respectively the same functions of x+x-1 as 2 cos no and 2 sin n0/2 sin are of 2 cos 0. But we did not there equate the corresponding expressions. To do this we must implicitly introduce imaginaries. For 2 cos 0 must be <2 for any real angle, and x+x-1 must be > 2 for a ny real value of x (··· x 2+x1 is a square quantity). 614. In Art. 610, where n is even, write 2m for n and divide both sides by xm. Thus 0 1 (2 cos 0-2 cos) (2 cos 0-2 cos)... (2 cos - 2 cos). Again, in Art. 611, where n is even, write 2m for n, and Again in Art. 612, divide both sides by ". Thus x” + x ̄” — 2 cos no = (x + x ̄1 − 2 cos 0) {x+x ̄ {ဆ -1 2 cos(+ 2π n n Now put x + x1= 2 cos a; then x2 + x ̄ˆ = 2 cos na. Thus 2 cos na - 2 cos no = (2 ̊cos a − 2 cos 0) 2 cos a {2 {2 cos a (2 008 - 2008 (6+12)}. ... 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 x is a continuous function of x, for all values for which it is convergent. ·· ƒ (x + d) − ƒ (x) = d. S where. S is finite, since ƒ(x+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(x) 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 x 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. |