Clearing the equation of fractions, we have 6- 2x2-4 (1-x) (1 + x) + B (1 + x) + C(1 − x)2 = = ( − A + C) x2 + (B − 2 C) x + A + B + C. = Since this equation must be true for all values of x, we have The forms of the partial fractions are assumed the same as in Ex. 1. We have 3+x2=(- A+ C) x2+(B-2 C) x + A+B+C, and then - A+C=1,) B-2C 0, Whence A=0, B2, C=1. = When the factors of the denominator of the given fraction are of the first degree, as in Exx. 1 and 2, the work may be shortened. Begin with the equation 6 − 2 x2 = A (1 − x) (1 + x) + B (1 + x) + C (1 − x)2, of Ex. 1. Since this equation is true for all values of x, we may substitute in it for x any value we please. Let us take such a value as will make one of the prime factors zero. Substituting 1 for x, we obtain 4 2 B, whence B=2. Next, letting x = — -1, we have 44 C, whence C=1. There is no other value of x which will make a prime factor zero, but any other value, the smaller the better, will give an equation in which we may substitute the values of B and C already obtained. Letting x=0, we obtain 6=A+B+C, whence A= 3. The same method can be applied to Ex. 2. In this example the one prime factor being of the second degree we assume the corresponding numerator to be a complete linear expression. Clearing of fractions, we have x2−x+3= A (x2 + x + 1) + (Bx + C) (x − 1) = (A + B) x2 + (A − B + C) x + A— C. Equating coefficients of like powers of x, we obtain whence, A=1, B=0, C=-2. Or, we might have used the second method, beginning with x2-x+3=4(x2 + x + 1) + (Bx + C) (x − 1). Letting x1, we obtain 3 = 3 A, whence A= 1. Since no other value of x will make a factor vanish, we take any simple values. When x = 0, we have 3A-C, whence C-2. Finally, letting x=-1, we have The prime factors in the denominators of the first two partial fractions being of the second degree, expressions of the first degree are assumed as numerators. Clearing of fractions, we have 2-2x+4x2 = (Ax + B) (1 + x2) (1 − x) + (Сx + D) (1 − x) + E (1 + x2)2 Equating coefficients of like powers of x, we obtain whence · A-B+C-D=−2, B+D+E=2; A=1, B=1, C2, D=0, E = 1. EXERCISES IV. CHAPTER XXXIV. CONTINUED FRACTIONS. 1. If the numerator and denominator of be divided by the numerator, we have Reducing, and subsequent fractions, in a similar way, we The complex fraction thus obtained is usually written more compactly thus: 1 1 1 1 1+2+3+4 Observe that in the last form the signs + are written on a line with the denominators to distinguish the complex fraction from the sum of common fractions. It is important to keep in mind that in both forms the numerator at any stage is the numerator of a fraction whose denominator is the entire complex fraction which is written below and to the right of that particular numerator. 2. A Continued Fraction is a fraction whose numerator is an integer, and whose denominator is an integer plus a fraction whose numerator is an integer, and whose denominator is an integer plus a fraction, etc. A continued fraction frequently occurs in connection with an integral term. In such cases it is customary to call the entire mixea number the continued fraction. The general form of a continued fraction, therefore, is: 3. We shall confine ourselves in this chapter to continued fractions in which the numerators are all 1, and the denominators all positive integers; of the general form, therefore, in which the d's are all positive integers, and n is a positive integer or 0. Then and the d's are called Partial Quotients. 4. A Terminating, or Finite Continued Fraction, is one in which the number of partial quotients is limited, as in the example given above. A Non-terminating, or Infinite Continued Fraction, is one in which the number of partial quotients is unlimited or infinite. |