similarly may each of the other pairs of simple factors be compounded into a quadratic factor; and the last factor, if n is even, is +1. So that, if n is even, (x2 ...... n +1) (x2 - 2 x cos -2 (x2 - 2 x cos n T+1)(x+1). (52) r-1 = (x-1) (-2x cos2+1) (-2x cos+1) n n 65.] The solution of the equation "+1=0, and the resolu but if x+1=0, x=(−1)"; and therefore the several values of x, or roots of the equation "+1=0, are the values which the right-hand member of equation (54) admits of. Now k may be any integral number; let therefore after which the values recur, for the substitution of 1 gives the same values of (-1) as the substitution of n and the substitution of 2 +1 for k gives the same values of 1 n 2 (-1)" as the substitution of -2; and so on. And if n is odd, the substitutions for k must continue until after which the values recur; and thus in both cases we have n, 1 and only n, different values of (-1)"; and therefore x” + 1 = 0 has n, and only n, different roots. These roots may be expressed in the exponential form, in the same way as those considered in the preceding Article; but as the results are similar it is unnecessary to write them at length. Let us consider one or two examples. Ex. 1. Find the roots of x1+1= 0. In this case n = 4, and from (55) we have Thus also may 3п = COS + -1 sin 3п 5 "+1=0 be resolved into its factors; for π π (cos + √1 sin), and n may be compounded into a quadratic factor 2-2x cos +1. Similarly may each of n the other pairs of factors corresponding to k = 1, k = 2, ...... k = 2-1 be compounded into a quadratic factor; so that, if Similarly, if n is odd, there are quadratic factors correspond 3 ing to k = 0, k = 1, k = 2, ... k=”1⁄23; and when k = 2 appears by (56), there is a simple factor x + 1; so that T+1)(x+1). (58) 3п - 1 = (x2 – - 2 x cos + · 1 ) (x2 – 2 x cos +1) (x+1), п 5π 6 6 +1)(x2-2x cos+1)(x2-2rcos+1). 66.] If the problem is to resolve "-a", and "+a" into their factors, then, since so that, in both cases, there will be n different values of x, 1 arising from the n different values of (1)" and of (-1)", which have been found above. 67.] On the imaginary logarithms of positive and negative numbers. One of the most remarkable results of our admission into the subjects of the Calculus of such quantities and symbols as we are now considering, is the extension which they afford to the theory of logarithms. In ordinary treatises on logarithms it is shewn that the logarithm of O is ∞; of 1 is 0; and of +∞ is; so that the whole range of possible number from - ∞ to ∞ is exhausted in the logarithms of number between 0 and +∞; what then is the form of logarithm of negative numbers? The preceding theory enables us to give some answer to this question. PRICE, VOL. I. for let us write x + 2km, k being any integer; cos(2 +2k)+v−1sin(2+2km) = ex+2km)V-1 (59) ... (60) (61) in which remarkable results it must be remembered that e and are severally the symbols of the arithmetical numbers 2.7182818 and 3.14159. Therefore from (60) it follows, that 2k-1 is the general Napierian logarithm of 1, and (2k+1) π√1 of 1; and therefore 2 km√1 = log. 1, (2k+1) π √1 = loge (-1). (62) (63) In (62), if k = 0, log, 1= 0, which is the common arithmetical logarithm of 1; but as k may be any integral number, it follows that +1 has an infinite number of Napierian logarithms, all of which, except 0, are affected with follows, that every positive number has an -1; hence also it infinite number of logarithms to the same base. For suppose to be the arithmetical Napierian logarithm of y, so that y = e*; then y = e2 × 1 = ex xe2k π √ = 1 = @(x+2 kn√ = 1) (64) therefore x + 2kπ-1 is the general Napierian logarithm of y, which we will represent by Logy; and let us represent the arithmetical logarithm by loge y; hence the arithmetical logarithm is the particular value of the general logarithm corresponding to k = 0. and therefore, whatever is the base, every number has an infinite number of logarithms to that base. Again, in equation (63), since k is to be an integer, it follows, that loge (1) can never be a possible quantity, and therefore -1 has no arithmetical logarithm; yet, since k may be any integer, every negative number has an infinite number of Napierian logarithms, and therefore of logarithms to any other base, all of which are affected with 1. In equation (63), let k = 0, .'. and more generally, two of the most curious results in Analysis; in which however, and in all similar expressions, we must bear in mind that e and are the symbolical representations of certain series, and are therefore to be interpreted with respect to them; and in an algebraical system of course, which admits them amongst those quantities whose laws and combinations it takes cognizance of. SECTION 3.-Theory of the Equicrescent Variable, and 68.] In the two preceding Sections of this Chapter we have considered certain theorems which arise from an algebraical relation between f(x) and its several and successive derivedfunctions; but as these have been formed on an unexplained hypothesis of x increasing by equal increments, by virtue of which has been called the equicrescent variable, an adequate conception of the subject requires a discussion at greater length. We shall reserve other important algebraical relations to the succeeding Chapter. Let y = f(x) be an explicit function of a, finite and continuous for all values of x under which we consider it; whether any other values may make it infinite or discontinuous is what we are not concerned with. Let a receive an increment dx, so that, in accordance with Art. 17, y becomes y+dy, and we have y+ dy = f(x+dx). (69) And let receive another increment, not necessarily equal to dr; and let it be dx+d.dx = dx+ d2x; so that we have y+dy + d (y+dy) = f {x + dx + d (x+dx)}, y + 2 dy + d2y = f (x + 2 dx+d2x). (70) |