13 +23 +33 + ... n3 = (1+2+3+ + n)2. Ex. 10. To solve the equation x1 + 1 = 0. Assume 1 + 1 = (x2 m x + 1) (x2 − n x + 1), then x1 + 1 = x1 − (m + n) x3 + (mn + 2) x2 − (m + n) x + 1 ; and equating coefficients of like powers of a, m + n = 0, and mn + 2 = : 0, .. œ1 + 1 = (x2 + √2.x + 1) (x2 - √√2. x + 1) = 0, or x2 + √√2.x + 1 = 0, and a2 - √√√2.x + 1 = 0, from which two quadratics we obtain the four roots, Ex. 11. To expand a in a series of powers of x. and a*+= 1 + A (x + y) + B (x + y)2 + C (x + y)3 + ... Multiply the first two series together, and equate coefficients of y with the last, and the result will be A+ 2Bx + 3 Cx2 + 4 Dx3 + = A+ A2x + ABx2 + ACx3 + .... from which, by equating coefficients of like powers of x, we have CONTINUED FRACTIONS. a 343. To represent in a continued fraction*. b Let b be contained p times in a, with a re- b) a (p mainder c; again, let c be contained q times in b, with a remainder d, and so on; then we have c) b (q d) c (r 344. COR. 1. An approximation may thus be made to the value of a fraction whose numerator and denominator are in too high terms; and the farther the division is continued, the nearer will the approximation be to the true value. α 1 345. COR. 2. This approximation is alternately less and greater than the true value. Thus p is less than ; and p + b is greater, because a part of the denominator of the fraction is զ 1 is too great for the denominator, therefore p + 1 q+ α is less than ; and so on. b DEF. The quantities p, q, r, &c. are called the Partial Quotients ; α * Although a 'continued fraction' (see Def. Art. 8) may be of the form p+ a Converging Fractions, or Convergents, to . Ex. To find a fraction which shall be nearly equal to and in lower terms. 100000) 314159 (3 300000 14159) 100000 (7 314159 100000 887) 14159 (15 887 5289 4435 854) 887 (1 33 &c. Here p = 3, q = 7, r = 15, 8 = 1, &c. therefore The first approximation is 3, which is too little, the next is The proposed fraction expresses nearly the circumference of a circle whose diameter is 1; therefore the circumference is greater 346. To convert any continued fraction into a series of converging fractions. Let the continued fraction be (Art. 343) in which the law of formation is observed to be as follows: Write down in one line the quotients p, q, r, s, &c., and the first and second converging fractions at sight; then the other fractions may be obtained thus: For the 3rd, num'. = 3rd quot. num'. of 2nd fract. + num'. of 1st fract. denom'. = 3rd quot. × denom1. of 2nd fract. + denom1. of 1st fract. For the 4th, = S num❜. 4th quot. x num'. of 3rd fract. + num'. of 2nd fract. denom'. = 4th quot. × denom3. of 3rd fract. + denom'. of 2nd fract. And generally, for the nth fraction in the series, Multiply the nth quotient by the numerator of the n − 1]th fraction and add the product to the numerator of the n-2th fraction. This will give the numerator. h Multiply the nth quotient by the denominator of the n-1]" fraction and add the product to the denominator of the n 2th fraction. give the denominator. - This will 84 Ex. To find a series of converging fractions for 227 347. To express √a2 + 1 in the form of a "Continued Fraction." 1 2a +...... a result which is easily remembered and applied to any proposed case. Ex. √17 = √42 + 1 = 4 + The converging fractions will be 4 + 1 8+ &c. each of which is nearer to the true value of √17 than the one preceding. 348. To express ñ in a continued fraction; and to find the converging fractions. Let a be the greatest whole number less than √n; then and so on, until the quantity corresponding to r" is equal to 1, after which the quotients will recur. The quotients of the continued fraction being thus found, the converging fractions will be found as in Art. 346. |