into a series, arranged according to 2 4 8 16 32 5. To convert 1-2 into a series, arranged according to the ascending powers of 2. We may clearly assume - 22 1-2=1 + A2 + B2 + C26 etc., or 1-22= (1 + A+ B+ etc.) = 1 + 2A22 + (A2 + 2B)≈1 + 2(AB + C)26 + etc.; since it is clear that the odd powers of z can not enter into the series. Comparing the coefficients of like powers of 2, we get A= 1 B 2' 1 8 2.4' X 3 2.4.6' etc. Hence, by substitution, we shall have 1 -- 2 2.4 2.4.6 6. To find two fractions whose sum equals the fraction. fractions, and suppose A and B to be independent of x. Hence, B A + B − (4A + 3B), or by rejecting the common 1-4.c (13x). (1-4x) = denominators we shall have 1= A + B (4A + 3B)x, which must clearly be an identical equation; consequently, we must have the equations A + B = 1 and 4A +3B=0, which give A = -3 and B = 4, and hence 7. To find two fractions, whose sum equals the fraction and C for 2m, we may assume A = + (1 − 3x). (1 + x2) 1-3x B+ Cx 1+22 equation 1= A + B + (C - 3B)x + (A — 3C). Hence, we have the equations A + B =1, C=3B, and A= 3C, ; which, freed from fractions, gives the identical 3 10 ; consequently, we 1+ 3x 10(1 — 3x) ' 10(1 + x2) = + (13x). (1+x) 10(13x) (the most simple), whose sum shall equal the given fraction. 9. To find two fractions, whose sum equals the fraction 2x 5 1 3 x - 2)3. (x 1 + 5) 3(x-2)3 1 2 + + 3(x-2)3 9(x-2)2 11. To find three fractions whose sum equals the fraction 12. To find three fractions whose sum equals the fraction 13. To convert a + bx c + mx + nx2 into a series, arranged accord ing to the ascending powers of x. Dividing each term of the numerator and denominator quently, if we convert this into a series, we shall readily obtain the required development. It is easy to perceive that we may assume a' + b'x 1 + m2x + n'x2 =a+Ax+ Ba2+ Ca+ etc.; which, freed from the fraction, gives the identical equation a' + b'aa' + (A + a'm') x + (B+ Am' + a'n')x2 + (C + Bm' + An')x3 + (D+ Cm' + Bn') + etc. Hence we get the equations a' a', A = — a' × (— n') + A × (— m'), C = A × (— n') D=Bx (− n') + Cx (m'), and so on; which give Ax (— a'm' + b')x, Bx2 = a' × (— n'x2) + Ax × (— m'x), Ca3 = Ax × (— n'x2) + Bæ2 × (— m'x), Da1 = Bx2 × (− n'x2) + Сx3 × (− m'x), and so on. a'm' + b', B. + B × (m'), = Hence, the development is easily found; since any term after the second is found from the two preceding terms by multiplying the first of the two by — n'x2, and the second - m'x, and then adding the products according to their by signs. Remark. It is clear that fractions of the forms a' + b'x + c'a2 a' + b'x + c'x2 + d'as etc., can 1 + m'x + n'x2 + p'x32 1 + m'x + n'x2 + p'x3 + q'x22 be developed in a similar way, when their first three, four, etc., terms are known. 1 14. To develop into a series, arranged accord 1 x x2 ing to the ascending powers of x. By actual division, we find 1 + x for the first two terms of the development; consequently, since n'a and -m'x are here represented by a and x, by proceeding as directed. in the preceding example, we have 1 1 x x2 = − 1 + x + into a series. =1 By division we get 1 + 5x + 13a2 for the first three terms of the development; and since p'a3, n'x2, m'x are here expressed by a3, 2, and 3x, we easily get 1+5x+13x2 + 30x2 + 69x + 160 + 4 - = 4+16x+64x2 + 256x3 + etc., it follows that etc., 1 4.0 by adding the last two series, according to their signs, the sum must equal the first series. If we use n to stand for the number of any term in either of the two last series, it is clear that 3-1 will represent the nth term in the first of them, and that 4""-1 will represent the nth term of the second. Hence, if we add these terms, we shall get (4" — 3")11 for their sum, which will equal the nth term (or what is called the general term) of the development of 1 1-7x+12x2 ; thus, if n = 5, (4"-3")-1 becomes (1024-243)=781, which is the fifth term of the series 1 + 7x + 37x2 + 175x3 + 781+ etc. Remark. From this example we perceive how we may proceed in order to find the general term of any series of the kind given above. REVERSION OF SERIES. 1. To revert a series is to express the unknown letter contained in it by another series, which may be regarded as the development of the value of the unknown letter. 2. We shall suppose that we have y = ax + bx2 + cx3 + dx1 + etc., (1), and that it is proposed to find x. Then it is clear that we may assume x = Ay + By2+ Cy3 + Dy1 + etc., (2), and regard A, B, C, etc., as being independent of y. If we substitute the values of y, ya, y3, etc., from (1) in (2), it will become a A(ax + bx2 + etc.) + B(ax + bx2 + etc.)2 + C(ax + bx2 + etc.)3 + etc., (3), which must be an identical equation. = It is easy to perceive that the developments required in the right member of (3) can be obtained from the expression Aax + Ba2x2 + Ca3a + etc., by changing a into a + bx, b into b+cx, e into c + dx, and so on successively, and developing by the rule given in Involution for the expansion of a function of a polynomial. Thus, by using the vertical bar to express the coefficients |