for a minimum. For other cases we proceed in a similar manner with the other terms. This demonstration will apply to any function of x from which we can find 'x, 'x, &c. If the student masters this demonstration, he will find it of great service to him when he takes up the Calculus, which at once furnishes the means of finding 'x, f''x, &c. Illustrative problem:- Divide a in to two such parts that the square of the first multiplied by the second shall be a maximum. Let be the first part, then ax is the second, and u = x2(a — x)· 3 x is to be a maximum. In this case 'x = 2ax · = = 2a 6x. 4a = If we put the value x = 0, in "x, we have 2a a positive quantity; and if we put the value x = a in "x we have 'x = 2a 2a, a negative quantity. Therefore, the first is a minimum, and the second a max imum. 2. The student will see that the difficulty consists mainly in finding 'x and 'x, and if we could readily find these we could solve many problems in maxima and minima without the Calculus. In many cases it is. unnecessary to find "x, since we can easily satisfy ourselves whether we have a maximum or a minimum. The following process, which I think the student of algebra can easily understand, will enable us to find both 'x and 'x, and more terms, if we choose, in algebraic functions, or quantities. If or be an algebraic function of x, then by writing a + h for x, and developing the resulting quantity according to the ascending or positive powers of h, as far as the first and the second power of h, the coefficient of h will be 'x (Eqs. (3) and (4)), and that of h2 will be 'x. This development can be effected by the Binomial Theorem, or by Indeterminate Coefficients. Suppose we have and we wish to find a value of x that will render u a maximum or a minimum. Put x = x +"h, then u' = (x + h)√(1 — x2 — 2xh — h2) + √(1 − x − h) (x + h)−1 'xh to the first power of h. Then We must now put the coefficient of h equal to 0, and we shall have The solution of this equation will make known the required values of x Take the equation u xe, then -x u' = (x + h)e−x+h = (x + h)e ̄*.e-" — e ̄* (x + h) (1 — h + }h2) = Since ''x = e* (-1), the value of x = 1, gives ''xe1(−1) e1, a negative quantity, so that this value of x makes ue1, a maximum. We could easily extend this demonstration to the case of two independent variables. ON THE SOLUTION OF CUBIC AND BIQUADRATIC BY G. W. HILL. In nearly all treatises on algebra the solution of these equations is presented as accomplished by the aid of analytical artifices, which one seems, by some happy hazard, to have stumbled upon. No doubt the processes were found in this manner by the original discoverers, Tartaglia, Carden and Ferrari. But, for many reasons, it would be better to treat the subject as one demanding invention rather than artifice. The equations can, as it were, be interrogated and compelled to yield up their secrets, if they have any. To say that an equation is solvable algebraically, is to say that an algebraic expression can be found equivalent to the general root, that is, one involving a finite number of the operations of addition, subtraction, multiplication, division and the extraction of roots of prime degree. If the expression does not involve the last mentioned operation, it is called rational, and if free from the two last, integral. However complex an algebraic expression involving radicals may be, it it is evident that there must be at least one radical which is involved in it rationally. Supposing this to be denoted by R", n being a prime integer, it is not difficult to convince one's self that, by the proper reductions, the expression can be exhibited thus where po, P1, &c., do not involve the radical". With no loss of generality we can suppose P1 = 1; for if p1 is not zero, we can multiply the quantity P1 under the radical sign by pi, and then take (p1R) 1 n as the radical; and in in the contrary case, if p is one of the quantities p which is not zero, the simplification can be accomplished by putting R' = p1⁄2R*. Then R2 2 Po + RTM* + p2RTM* + . . . + pm - R1 may be regarded as the most general form of an algebraic expression. Here may be enunciated a general proposition, which, although I am not aware that it has ever been proved, is doubtless true, and may be used for purposes of discovery. If an algebraic expression exists, equivalent to the general root of the equation x+ax+bxTM-2+...+g=0, it can be exhibited in the above form, n being equal to one of the prime factors of m. Thus the algebraic expression of the root of the general equation of the 5th degree, if it existed, could be presented in the form and that of the 6th degree in either of the two forms According to the foregoing proposition, the root of the general cubic equation 23 + ax2 + bx + c = 0, if it has an algebraic expression, must be presented in the form x = p + R 3 + p'R3 . But since we suppose that this is an irreducible expression involving radicals, it follows that it must satisfy the given equation, whichever of its three values is attributed to the radical R. Thus calling either of the imagin ary cube roots of unity a, the three roots of the cubic equation must be The first method that suggests itself for obtaining equations which shall give the values of p, p' and R, is to substitute these expressions in the sym metrical functions which are equivalent to the several coefficients a, b, c, viz., x1 + x2 + x 3 =—α, x1x2+x1x3 + x2x3 =b, X1X2X3=C. But a simpler proceeding is to employ the three symmetrical functions 2.x, 2.2 and 2.3. Since any cube root, as R, is a root of 2-R = 0, in which the coefficients denoted above by a and b are each zero, it follows that the sum of the three cube roots of any quantity, as well as the sum of their squares, is zero. Now it is plain that if the value of x is raised to the nth power, x"= A+ BR3 BR'S + CR3, where A, B and C are free from the radical same whichever of the three roots x denotes. 2 R, and are consequently the Thus, for computing the value of Σ.x", we need only the part A which is free from the radical R. In this way we obtain and equate to their known values in terms of the coefficients a, b, c, Σ.x2=3(p2+2p'R) = a2 — 2b 2.x2=3(p+R+ 6pp'R+p'3 R2)——a3 + 3ab—3c. These equations afford the values of p, p' and R; from the first two a quadratic equation in R; thus the general cubic admits of solution by radicals. and as we may take at our option either of the two roots, we have choice of the two expressions for æ, x = − }a + [B + √ ́ (B2 — A3)]* + A[B+ √(Ba— A3)] ̃ ̄3 · x = — }a + [B−√ (B2 — A3)]3 + A [B — √ (B2 — A2)] ̄ ̃3 — The three values of x are obtained by attributing in succession to the single cube root appearing in either of these expressions its three values. I do not know why almost all algebraists prefer to put the root iu the form x = − }a +v[B + √(B2 — A3)] + †[B— √(B2 —A3)]. It is certainly easier in practice to make a division than an extraction of a cube root; moreover we are troubled, in the last form, with the selection of the proper three values out of the nine of which it is susceptible, a difficulty which does not occur in the two former expressions. SOLUTION OF BIQUADRATIC EQUATIONS. An algebraic expression for the root of the general equation of the fourth degree, x2 + ax3 + bx2 + cx + d=0, if it exists, can be presented in the form P+Q. And if this denotes one of the roots, another will be P-√Q; but since x has four values, it is plain that P and Q must receive each two values. This condition will be fulfiled if we suppose that these quantities, in their turn, similarly to x, are rational functions of a second radical R. Thus we put Then we have P=p+vR, X= p+vR + v(q+q'√R). The four values of x are obtained by giving in succession to the radicals VQ and VR all the values they are, in combination, susceptible of. Thus x1 = p + √R+v (q + q'vR), |