1+2x =A+Вx+Сx2+Dx3+Ex2+, etc., where the coefficients A, B, C, D are supposed to be independ ent of x, but dependent on the known terms of the fraction. In order to obtain the values of these coefficients, let us clear this equation of fractions, and we shall have 1+2x=A+(B-3A)x+(C−3B)x2+(D−3C)x3+ Now, since this is supposed to be an identical equation, the coefficients of the like powers of x in the two members are equal each to each. E-3D=0, E=135, etc. Substituting these values of the coefficients in the assumed series, we obtain 1+2x 1-3x =1+5x+15x2+45x3+135x+, etc., where the coefficient of each term after the second is three times the coefficient of the preceding term. 380. The method thus exemplified is expressed in the fol lowing RULE. Assume the proposed expression equal to a series of the form A+Bж+Сx2+, etc.; clear the equation of fractions, or raise it to its proper power, and place the coefficients of the like powers of x in the two members equal each to each. Then find from these equations the values of A, B, C, etc., and substitute these values in the assumed development. 1=A+(B-2A)x+(C-2B+A) x2+(D−2C+B)x3 Ans. 1+3x+4x2+7x3+11x2+18x+29x+, etc., where the coefficient of each term is equal to the sum of the coefficients of the two preceding terms. Ex. 4. Expand 1-x 1-2x-3x2 into an infinite series. Ans. 1+x+5x2+13x3+41x+121x+, etc. What is the law of the coefficients in this series? Ex. 5. Expand √1-x into an infinite series. Ans. 1. x x2 ნეი 7.5 etc. into an infinite series. 1+x+x2 Ans. 1-2x+x2+x3-2x2+x2+x6—, etc. Ex. 7. Expand Va2x2 into an infinite series. 381. Proper Form of the assumed Series.-In applying the method of undetermined coefficients to develop algebraic ex-. pressions into series, we should determine what power of the variable will be contained in the first term of the development, and assume a corresponding series of terms. Generally the first term of the development is constant, or contains xo; but the first term of the series may contain x with any exponent either positive or negative. If the assumed development commences with a power of x lower than is necessary, no error will result, for the coefficients of the redundant terms will reduce to zero. But if the assumed development commences with a power of a higher than it should, the fact will be indicated by an absurdity in one of the resulting equations. The form of the series which should be adopted in each case may be determined by putting x=0, and observing the nature of the result. If in this case the proposed expression becomes equal to a finite quantity, the first term of the series will not contain x. If the expression reduces to zero, the first term will contain x; and if the expression reduces to the form A, then the first term of the development must contain x with a negative exponent. Clearing of fractions, we have 1=3Ax+(3B—A)x2+, etc., whence, according to Art. 378, we obtain 1=0, which is absurd, and shows that the assumed form is not applicable in the present case. Let us, however, assume 1 3x-x2 =Ax¬1+B+Cx+Dx2+, etc. Clearing of fractions, we have 1=3A+(3B—A)x+(3C−B)x2+(3D−C)x3+, etc. To Resolve a Fraction into Simpler Fractions. 382. When the denominator of a fraction can be resolved into factors, the principles now developed enable us to resolve the fraction itself into two or more simpler fractions, having these factors for denominators. In such a case, the given fraction is the sum of the partial fractions. Ex. 1. Resolve the fraction 5x12 x_5x+6 into partial fractions. A B We perceive that x2-5x+6=(x—2)(x—3). = + in which the values of A and B are to be determined. Clearing of fractions, we have 5x-12=(A+B)x−(3A+2B). By the principle of Art. 378, A+B=5, and 3A+2B=12. From which we obtain A=2 and B=3. Substituting in the assumed equation, we have 383. The reversion of a series is the finding the value of the unknown quantity contained in an infinite series by means of another series involving the powers of some other quantity. This may be accomplished by the method of undetermined coefficients in a mode similar to that employed in Art. 379. Ex. 1. Given the series y=x+x2+x3+, etc., to find the value of x in terms of y. х Assume x=Ay+By2+Cy3+Dy*+, etc. Find, by involution, the values of x2, x3, x1, and x3, carrying each result only to the term containing y. Then, substituting these values for x, x2, x3, etc., in the given equation, we shall have y=Ay+B | y2+C 3+E +A2 y3+D +2AB yo+, etc. Since this is an identical equation, we place the coefficients of the like powers of y in the two members equal to each other, and we obtain A=+1, B=-1, C=+1, D=-1, E=+1, etc. IIence we have x=y—y2+y3—y1+y3—, etc., Ans. |