Commensurable Roots of any Equation. 3. Find the commensurable roots of the equation 24 x3-26 x29x-1=0. Ans.,, and . 4. Find the commensurable roots of the equation 3 x3-14x2+21 x — 10 = 0. Ans. 1, §, and 2. 5. Find the commensurable roots of the equation 8x438x3 49 x2-22x+3=0. Ans. 1, 1, 1, and 3. 6. Find all the roots of the equation 6 x3 +7 x2+39 x + 63 = 0 which has a commensurable root. Ans., and -251. 7. Find the commensurable roots of the equation 9x6+30x5+22x4 +10x3 +17 x2-20x+4= 0. Ans. and 2. Value of Continued Fractions. CHAPTER IX. CONTINUED FRACTIONS. 316. A continued fraction is one whose numerator is unity, and its denominator an integer increased by a fraction, whose numerator is likewise unity, and which may be a continued fraction. 317. Problem. To find the value of a continued fraction which is composed of a finite number of fractions. Solution. Let the given fraction be An approximate value of this fraction is obviously obtained by omitting all its terms beyond any assumed fraction, and obtaining the value of the resulting fraction, as in the previous article. and each of these values is easily shown to be more accurate than the preceding; for the second value is what the first becomes by substituting, for the denominator a, the 1 Ъ more accurate denominator a+ ; the third is what the second becomes by substituting, for the denominator b, the more accurate denominator b + ; and so on. 1 C Approximate Values of Continued Fractions. 320. Theorem. The numerator of any approximate value, as the nth, is obtained from the nume rators of the two preceding approximate values, the 1)st, and the (n — 2)nd, by multiplying the (n (n 1)st numerator by the nth denominator contained in the given continued fraction, and adding to the result the numerator of the (n proximate value. 2)nd ap The denominator of the nth approximate value is obtained in the same way from the two preceding denominators. Demonstration. Let the (n-3)rd, (n—2)nd, (n-1)st, and nth approximate values be, respectively, and let the (n-1)st and the nth denominators, contained in the given continued fraction, be p and q. We shall suppose the proposition demonstrated for the (n-1)st approximate value, and shall prove that it can thence be continued to the nth value; that is, we shall suppose it proved that M = PL K Now it is plain, from the remarks at the end of the preceding article, that the nth value is deduced from the (n—1)st, 1 by changing p into p+; which change, being made in the preceding value, gives |
