generally happen that the labour attendant upon it will be intolerably great. It may happen, too, that we may be in possession of no rule for actually performing the indicated operation, and in this case the problem could not be solved at all. As, for instance, to extract the fifth root of a +x, or more generally the nth root of a +x, to which no rule of extraction given in the earlier part of this work is applicable. A general method of expanding such expressions in a series of single terms, will be given a little farther on, by means of indeterminate coefficients, as well as solutions of one or two other problems which cannot be dispensed with in our future investigations. In this method, the development is assumed to be of a particular form, so far as indices are concerned, and equations by which the corresponding coefficients are obtained, are deduced from considerations presently to be explained. If these equations give real values of the coefficients, the development is effected by the solution of them: though, independently of an investigation specially directed to the decision of the point, we cannot be justified in affirming that the development so obtained is the only one that can possibly be made. The principle upon which the doctrine of indeterminate coefficients turns is this: That in all developments that are made according to successive powers of any quantity (whether integer or fractional), in whatever form the coefficients may appear, the one which belongs to any specified power of that quantity in one form must be equal in value to that which belongs to the same power in the other development. Thus, if the function fr could be written in two different ways, or so that the coefficients took different forms in the two developments, the following equation must be fulfilled independently of the specific value of x, viz.: A+ Bx+Cx2 + = A2+ B1x + С122 + For, by transposition, we have .... .... ....= 0. = (A — A1) + (B — B1) x + (C — C1) x2 + Now, if we have not simultaneously AA, = 0, BB, 0, CC, = 0, we should have an equation in x of some degree, n, and a would in this case have n values only, and not admit of all values indiscriminately. But since in development we have only to change the forms of the expressions, the quantities themselves must be capable of all possible values, and hence cannot be restricted to the special ones which constitute the roots of any given equations. It follows, therefore, that unrestricted values of x in the preceding equation are essential to the idea of a development, and hence that the coefficients of the several powers of x in the simplified equation are separately equal to zero. We may also make another remark which will hereafter be found useful. If the following equation be true for n + 1 values of x, it is true for all values of x, viz. : .... A+ Bx + Cx2 + For if it be true for m + 1 values of x, it is true for a value which is not a root of the equation of the nth degree; and hence it can only be true when A — A1 = 0, B — B1 = 0, .... . . N — N1 = 0. But in this case the equation is true whatever x may be, or for all values of x. It will at once appear, that to render this method effective, the development in symbolic coefficients must be of essentially different forms; and that in all cases where this cannot be effected, it will indicate that the powers of a, whose coefficients they are, do not enter into the true developement; or, in other words, that the assumed law of the indices of x is not possible. The same is true if any of the values of the coefficients become imaginary. A few examples will be necessary to illustrate the nature of the method. Ex. 1. Extract the square root of 1 + x2. Assume the root to be in positive powers of x; or √1 + x2 = A + Bx + Cx2+ Then squaring both sides, ...... 1 + x2 = = A2 + ABx + AСx2 + ADÃ3 + AEx1 + +ACx2+BCx3 + C2x2 + + ADX3 + BDÃ1 + .... ... ... The coefficients of the two sides of the equation being of different forms, and those of the first side given, we may proceed to equate those of the like powers of x. We have, therefore, as may be easily verified by actual extraction, according to the method pointed out at p. 148. Moreover, had we foreseen that only even powers entered into the expansion, we might have obtained twice as many terms of it as above with the same quantity of work, by assuming Assume it equal to A + Bx + Cx2 + ....; then multiply both sides by 1 + x + x2, and multiply the two factors of the numerator together. This Equating the coefficients of the like powers of x, we have A = 1, A + B = 0, A + B + C = − 1, B + C + D = 2, and so on. Resolving these equations, we have and so on. A = 1, B = − 1, C = 1. 4x4 +0x5+ 4x6 as may be easily verified by Synthetic Division. .... These questions, and others of similar kinds, admit, however, of easier solution by other methods, and they are only instanced here for showing the nature of the operations to be performed, whilst their results admit of verification by those easier methods. The following is one of a class to which this method is the only one that can be applied without great labour; and it is one of perpetual occurrence in integration. solvable into factors of the first, second, third, &c. degrees: to transform it into a series of partial fractions, each of which shall be one of those factors in succession. As the method will be quite as well seen by a numerical example, it will be sufficient to so transform 1 - 2x Here the denominator is x(x + 1) (x2) we may assume its partial fractions to be Whence, equating the homologous coefficients of the numerator, we get which may be easily verified by reducing to a common denominator. It is necessary to remark, that the power of x in the assumed numerators must in all cases be one degree lower than in the denominator. 8. Expand 1 1+x a+b'1 a' a2 62, square root of 4 in a series according to positive, and and a2 a -X 6x+5x2 9x3. √1 + 2x + 3x2, √√2—3x + 5x2 and √1+x+x2 + x3. 1 + x2 9. Resolve into partial fractions the following expressions: the denominators being the same as those in Examples 2 and 4, p. 236. The preceding simple applications of the method will suffice to show its character; and we pass on to some of its applications in the investigation of general theorems for summation and expansion. PILING OF BALLS. The usual forms of piles of balls are the triangular, the square, and the rectangular, and they take their names from the figure of their lowest courses. In all cases the successive courses have one ball less in each of their sides than the one upon which they respectively rest; the highest course being in the triangular and square piles a single ball, and in the rectangular, a line of balls. A pile is said to be incomplete or broken when it has either not been finished, or when some of the balls have been removed from a finished pile. The following figures represent the three kinds, as named below them respectively. The number of courses, therefore, is the same as the number of balls in the shortest side of the lowest course. 1. To find the number of balls in a triangular pile of n courses. Hence, as there are n courses expressible in the same manner, we may infer that the highest power of n that enters the expression for the sum to be n3. Let, then, S1 = C2+ Cn−1 + Cn−2 + .. . C2 + C1 = An3 + Bn2 + Cn + D. Then, substituting in this the first four values of n, we shall have S1 = A+ B+C+D= 1, for the first course; which gives the sum of the n courses. 2. To find the number of balls in a square pile of n courses. In this we have for the n courses: and for the same reason as before the form of the function will be S, = An2 + Bn2 + Cn + D. Substitute for n the successive values 1, 2, 3, 4, as in the last case: then 3. The rectangular pile. Let n be the number of balls in the shorter side, and n+m the number in the longer side of the lowest rectangular course. Then it will be obvious, by reference to the structure of the pile, that it is composed of a square pile of n courses, with m oblique triangular courses added successively to one of the oblique faces. Hence the entire pile is which gives the balls in the rectangular pile. It may be well to recollect that the ridge of the rectangular pile has m + 1 balls. 4. The incomplete pile. This will require the whole pile to be computed, and then the partial pile removed, both by the appropriate formula for the kind of pile. The difference is the number in the incomplete pile. Formulæ, indeed, may be given for all the cases in terms either of the number of courses taken off or the courses left, and the original number of courses: but these would be much more complex, and the implied numerical operations more laborious, than the unreduced formula and the work which it involves. Such formulæ, therefore, would be without utility, and from having little mathematical elegance, would be destitute of sufficient interest to justify their introduction here. Ex. 1. Find the number of balls in a triangular and in a square pile, each composed of twelve courses. Ex. 2. In a rectangular pile are 18 courses, and the number in the ridge is 45. How many balls are there in the entire pile? Here m + 1 = 45, or m = 44, and n = 18. Hence by the formula, 18. 19. 169 =9633. 6 Ex. 3. Of the preceding rectangular pile there are to be taken away 1031 balls; how many complete courses must be removed? Let a be the number: then x(x + 1) (3.44 + 2x + 1) 6 =1031, or by reduction, 2x3+135x2 +133x=6186. The real root of this is between 6 and 7; and hence six complete courses, together with 96 balls more, must be removed; for |