.. 2 + 3x = A + (B+ A)x + (C + B+ A ) x2 + (D + C + B) x3 + .... A = 2, B+ A = 3, C + B + A = 0, D + C + B = 0 ; B = 1, C = −3, D = 2; and so on. By § 396, whence, The series is of course equal to the fraction for only such values of x as make the series convergent. NOTE. In employing the method of Indeterminate Coefficients, the form of the given expression must determine what powers of the variable x must be assumed. It is necessary and sufficient that the assumed equation, when simplified, shall have in the right member all the powers of x that are found in the left member. If any powers of x occur in the right member that are not in the left member, the coefficients of these powers in the right member will vanish, so that in this case the method still applies; but if any powers of x occur in the left member that are not in the right member, then the coefficients of these powers of x must be put equal to 0 in equating the coefficients of like powers of x; and this leads to absurd results. Thus, if it were assumed that there would be in the simplified equation no term on the right corresponding to 2 on the left; so that, in equating the coefficients of like powers of x, 2, which is 2 xo, would have to be put equal to 0xo; that is, 20, an absurdity. 398. To resolve a fraction into partial fractions is to express it as the sum of a number of fractions of which the respective denominators are the factors of the denominator of the given fraction. This is the reverse of the process of adding fractions that have different denominators. Resolution into partial fractions may be easily accomplished by the use of indeterminate coefficients and the theorem of § 396. In decomposing a given fraction into its simplest partial fractions, it is important to determine what form the assumed fractions must have. Since the given fraction is the sum of the required partial fractions, each assumed denominator must be a factor of the given denominator; moreover, all the factors of the given denominator must be taken as denominators of the assumed fractions. Since the required partial fractions are to be in their simplest form, incapable of further decomposition, the numerator of each required fraction must be assumed with reference to this condition. Thus, if the denominator is x2 or (x± a)", the assumed fraction must be of the form A хп or A for, if it had the form Ax+B Ax+B or хп (x ± a)n3 it could be decomposed into two fractions, and the partial fractions would not be in the simplest form possible. When all the monomial factors, and all the binomial factors, of the form xa, have been removed from the denominator of the given expression, there may remain quadratic factors that cannot be further resolved; and the numerators corresponding to these quadratic factors may each contain the first power of x, so that the assumed Ax+ B fractions must have either the form x2± ax + b or the Since x2 + 1 = (x + 1) (x2 − x + 1), the denominators will be x + 1 and x2x + 1. The denominators may be x, x2, x + 1, (x + 1)2. .. 4x3-x23x whence, 2 = Ax (x + 1)2 +B (x + 1)2 + Cx2 (x+1)+Dx2 =(A+C)x+(24+B+C+D) x2 + (A+2 B) x+B; and CHAPTER XXV. BINOMIAL THEOREM. 399. Binomial Theorem, Positive Integral Exponent. By successive multiplication we obtain the following identities: (a + b)2 = a2 + 2 ab + b2; (a + b)3 = a3 + 3 a2b + 3 ab2 + b3 ; (a + b) = a* + 4 a3b + 6 a2b2 + 4 ab3 + ba. The expressions on the right may be written in a form better adapted to show the law of their formation: NOTE. The dot between the Arabic figures means the same as the sign X. 400. Let n represent the exponent of (a + b) in any one of these identities; then, in the expressions on the right, we observe that the following laws hold true : 1. The number of terms is n + 1. 2. The first term is a", and the exponent of a is one less in each succeeding term. 3. The first power of b occurs in the second term, the second power in the third term, and the exponent of bis one greater in each succeeding term. 4. The sum of the exponents of a and b in any term is n. |