558 Y+Y-i+i- ^ m/ m 1.2 A L Gr E B R A [CONTINUED FRACTIONS. For n is evidently > 1. If then we suppose n>l<2, l.i x 2 1.2.3 a; 3 + &c. a? + &c. (/eN m x 2 1 H ) approaches the value 1 + x + - - + &c. m/ 1 . 2 or e*, as m becomes larger and larger. Ex. 2. (lH ) approaches to e* ( 1 - ) as7nincreases. V ml 2m J Ex. 3. *-^ (-!)" ( - 2)" - &c. = 1.2.. .71 when 7t is a whole number. 1.2 &c. JL + &c. Now I. (e" - But e*-l = /. II. TOCro-1) ,, 1.2 + &C, (e* -l} n = x n + &c. Equating coefficients of x* in I. and II., we get 71" -l) (TO -2)" -&c =1 1.2. ..TO 1.2 1.2... which is the required result. Cor. When r is less than n, n r - n(n -l) r + i~a^( n ~ %)" ~ &c - = Ex. 4. The logarithm of a number to the base a" is a mean proportional between its logarithms to the bases a and a" 2 . If .*, #, z, are the logarithms to the three bases in order, we have consequently Ex. 5. e x > 1 + x, whatever be x. If x be positive, or if it be negative and less than unity, the expansion may be thrown into the form -?> + &c. 1.2 every term of which after 1 + x is positive. en n >(l+n} n e.1^2 1 e.2 2 >3 2 Ex. G. e">- For Hence and, by multiplication, e"|TO>(l +n)*. Ex. 7. If n be a whole number > e, n* +l >(n + 1)". By the demonstration of Ex. G, ">(! +n) n But n>e .: n n+l >(l+n) Ex. 8. If n*+ l = (n + l )", then n > 1 < e. + &c. 1.2 3 = 2 + a series of positive terms by the hypo thesis n < 2, which is absurd, . . n> 2. Taking the Napierian logarithms of each side of the equation n = ( 1 + - ) , we get i i lo n = 1 - + 1/1 I | __ I --- n n<e. Ex. 9. Nap. log x>l-- <x-l. & 1 1 1 / 1 2 Because log=-log-=l hzll I + &c. x x 2 a;/ log a; > 1 x And because when x>, x< 1 + (x 1) + T t- (a; - I) 2 -f- 1. 4B log x<x 1 ; log a; = log (1-1 -a) &c. when #< 1, <#- 1. i ^Jc. 10. Nap. log x approaches to 2" (a; 2 "- 1) as n increases. SECT. XVII. CONTINUED FEACTIONS. 131. Every quantity which admits of being expressed by a common fraction may also be expressed in the form of what is called a continued fraction. The nature of such fractions will be easily understood by the following example : Let the fraction be - , or, which is the same, 3 + 100000 Since 100000 = 7 x 14159 + 887, therefore 14159 1 , 314159 14159 100000 14159 _ 100000 ~ 7x14159 + 887 = -^=, and 7 + 887 100000 = 3 + 14159 7 + 887 887 By treating the fraction n7g 9 in the same way, and con tinuing the process, we readily obtain 314159 1 1 100000 7 + 7T 1 By an operation in all respects the same as has been just now performed, may any proper fraction whatever be reduced to the form a i ^ "*"-+&<}. and it is then called a continued fraction.
132. When the root of any equation is found by the
