(III) Let x be positive, and m + 1 positive. Then all the terms in f (m, 2) up to x", where r is not > m + 1, are positive, after which they are alternate in sign but numerically decreasing; :. f (m, x) is positive. (IV) Let x be positive, and m +1 negative. f (m, x) = f (m + n, x) = f(n,). Since then f (m, 2) is always positive, when it is endless but convergent, (m and being real), When m is integral f (m, 2) = (1 + 3)". When m is fractional, f (m, x) = thė one positive value of (1 + )". = § 3. EXPONENTIAL AND LOGARITHMIC SERIES. 468. To find the limit of the product of an infinite number of factors, the limit of each of which is unity. A. Let each factor reach its limit unity, independently of the indefinite increase in the number of factors. Then, call the first n factors 1+pi, 1 + P2, ...l + Pn; where P1, P2, ... Pn are ultimately zero. Assume the theorem that for n factors (1 + p) (1 + p) ... (1 + Pn) = 1 + R, where R is ultimately zero. Introduce another factor 1 Pn+1; then R=1+R', where R' is ultimately zero. Hence, if the theorem is true for n factors, it is true for n + 1. Hence it is true, however large the number of factors may be. + = Observe here particularly that we have allowed each factor to reach its limiting value, before introducing the next factor. This was legitimate because of the special hypothesis (A). B. Let each factor only reach its limiting value, when the number of factors is indefinitely increased. In other words, let each factor be a function of the number of factors. Then the above argument will not hold. An important example is investigated in Art. 469. Similar propositions hold with respect to the sum of an infinite number of terms, the limit of each of which is zero. If each term reaches its limit zero independently of the indefinite increase in the number of terms, the limit of the sum is zero. But if each term is a function of the number of terms, this is not necessarily the case. These cases of multiplication and summation are particular cases of the evaluation of expressions of the form 1" and 0 xoc. Now it is clear that however many times we multiply 1 by itself the result is 1; and however many times we add 0 to itself, the result is 0. Hence, when the symbols represent mere independent numbers, we have 1"=1 and 0 Xco=0. But, when the symbols represent the limiting values of co-varying quantities, these results are not necessarily true : and the values cannot be determined without assigning what function the one quantity is of the other. The student may be warned against confounding the products or series here discussed, with those in which the limit of the wth factor or term, not of each factor or term, is unity or zero. (1 + )" 1, when 469. To find an expression for the limit of n is indefinitely increased. n [Here 1+1/n is a factor whose limit is unity and which has to be multiplied by itself an infinite number of times. But each factor only reaches its limit, as the number of factors is indefinitely increased. Hence we cannot employ the argument of Art. 468 (A).] For all finite values of n greater than 1, the positive value of 1 nac 1+ n 2 =l+na. 1 nx (n2-1) + + ... (9) ==+1)(A)+. =1+2+ (1 - .) + .+ (1 -) ---- (1 -->) +.. .. (1 -... 1-*z , nx In this series, the multiplier This decreases as r increases, and is less than a quantity less than 1, if r is large enough. This is true, even if n is infinite. Hence the series is convergent for all values of n however great. 1 n; since the term decreases independently of the increase in r. nx Hence, the limit of each of the factors (1-1-1-1) is unity. And they reach their limit independently of the increase in their number r; :their product is unity. When r becomes commensurate with n, the sum of the terms beginning with ach is indefinitely small (because the series is convergent): hence no error arises from falsely evaluating this sum. Thus the limit when n is indefinitely increased of the positive n value of for all real finite values of x. 470. In the above result, put a = 1. Thus the limit of 1 1 1 ...(1). 2 3 r (1+2)" - } (1 + a), we have n { 1 1+1+ + 2 1 + n + = 1 + 2 + This important formula is called the Exponential Theorem. Example. Prove (2) directly by means of the index theorem. If f(x) represents the right hand of (2); the terms of the oth dimension in f(x) x f(y), are clearly 1 r(r – 1) r(n-1)(r—2) 21–2 y2 + acra +r.xn-ly + X=-343 +. 2 13 (x+y)" by the Binomial Theorem for a positive integer. 1 .: f(x) xf (y)=f(x+y). :: f(x)= {f(1)} Q. E. D. 3 [+ ...] n 471. The limit, when n is infinite, of (1 + )" is usually de n , n , (1 noted by the symbol e (or e). Thus (1) and (2) of the last article become 1 1 1 e=1 +1+ 2 13 = Example. Find the value of }(ex te-x) and 3 (ex - e-x). 22 23 e-s=1- x + 2 13 472. To find the limits within which the value of e lies. Take each of the terms in (3) expressed decimally and found by dividing the preceding by 2, 3, 4, &c.; and then add up the terms as far as calculated. Thus e> 2; > 2:5; > 2:6; > 2.7083; > 2.716. 1 1 1 l < 3 2 22 2 1 1 1 1 e+ 2 3 32 1 2 . x2 + + + + Logarithms to base e. 473. The expansion ca eit = 1 + 2 + :(4), 1 2 IT enables us to find logarithms to base e as follows. Let x = y loge a. Then ex = (elog a)ý = a”, ay = 1 + y loge a + (y loge a) (y loge a)" ...(5). 2 The series (4) holds for all finite values of x. Hence the series (5) holds for all finite values of y and a, provided that a is positive, so that logea has a single real value positive or negative. (See Art. 243.) Now let a =%+1, where z is any quantity numerically not greater than 1: so that a is positive as required. Also, by the binomial theorem, if y is positive, [See Arts. 461, 462.] y(y – 1)...(y-p+1) (1 +2)4 = 1 + y2 + 22+ +... (6), 1.2 17 for any value of z numerically not greater than 1. y(y-1) + + |