Cor. Should there be other equal roots, as p, roots each equal to r1, p2 roots 1, p1 — 2,... PROBLEM VI. To ascertain whether a given equation has equal roots, and if so to find Form the equation ƒ, (x) = 0, and perform the operation of finding the For, if the given equation ƒ (x) = 0 contain p roots equal to r, p, roots equal to As an example, let ƒ (x) = x + 3x5 6x4 Here f(x) = x6 + 3x5 .... 6x39x2+3x Of these two functions we find, as at p. 135, that the greatest common mea- EXAMPLES FOR PRACTICE. Find the equal roots of the following equations, if such exist: THEOREM V. If an equation whose coefficients are not imaginary have one Now, as the first member can never be equal to the second, except the bracketted coefficients themselves fulfil the condition, and as by hypothesis a+b=1 is a root of the equation, this equality must be fulfilled as a consequence of that hypothesis, we have simultaneously which is altogether independent of the sign of b-1. Hence if one sign + b√ — 1 fulfil the condition of the equation, the other sign — b √ also fulfil it: that is, if a + b N 1 be a root of the equation, a — also a root. 1 will - 1 is Cor. 1. Roots of such forms, (generally termed conjugate roots,) if they enter into an equation at all, enter it in pairs, and their number is always even. Cor. 2. If an equation be of an odd degree, there is at least one root free from the imaginary sign. Scholium. The same kind of reasoning will prove, that if one root be of the form a+b, there will be another of the form a b. For the same reason will exist for the separate bracketted coefficients being zero in this case as in the other. THEOREM VI. Change the signs of all the roots r1, r1⁄2, rз, equation of the form .... Tn, of an and combine the roots so changed by way of multiplication, in twos, threes, fours, and so on: then the sum of these changed roots will be equal to A; the sum of their products in twos will be equal to B; the sum of their products in threes will be equal to C; and so on, till the coefficient of the nth term is the sum of all the products of the roots (n-1) at a time, and their continued product will be equal to N *. 1 2 r2 = 0, which mul 203 (r1 + r2 + r3) x2 + (r1 r2 + r1 r3 + r2 r3) x − r1 r2 r3 = 0, in which, again, the statement is true. Proceeding thus to any extent, and observing the formation of the coefficients, we see that the theorem is generally true. Cor. 1. If any pair of roots had been conjugates, whether irrational or imaginary, we see that these several coefficients would have become rational or real, provided A, B, C, were so: for since {(a+b√ − 1) + (a − b √−1)} and {a+b√1} {a − b √ −1} are both real, the imaginary parts which would have come into the values of A, B, C... disappear from the result. * Some writers have considered it necessary to complete this proof, to show that, generally, if it be true for the pth coefficient, it will also be true for the (p+1)th. This is easily done by assuming the first, and proceeding by actual multiplication (literal) to the next; and the only reason for omitting it here is the space which the printing would occupy. Cor. 2. It may hence be inferred, that all the roots of an algebraic equation may be represented either by real quantities, r1ra ...., or imaginary ones, a+b√1, a, b, √1,... THEOREM VII. Every algebraic equation contains as many roots, either real or imaginary, as it has dimensions, and no more. For by Theorem VI. Cor. 2, every root may be represented by r,, r2 and a + b√ — 1, a, ± b, √ — 1; hence so many factors of the first degree may be formed of it as there are units in the degree and such equation admits of no other binomial factors but these, as then one of the binomial factors would be divisible by some other binomial factor, which is obviously absurd. THEOREM VIII. No equation can have a greater number of positive roots than there are changes of sign from + to, and from to +, in the terms of its first member; nor can it have a greater number of negative roots than of permanences, or successive repetitions of the same sign. To demonstrate this proposition, it will be necessary merely to show, that, if any polynomial, whatever be the signs of its terms, be multiplied by a factor x- a, corresponding to a positive root, the resulting polynomial will present at least one more change of sign than the original; and that if it be multiplied by xa corresponding to a negative root, the result will exhibit at least one more permanence of sign than the original. Suppose the signs of the proposed polynomial to succeed each other in any given order, as, for instance, then the multiplication of the polynomial, by x a, will give rise to two rows of terms, which, added vertically, furnish the product. The first row will, obviously, present the same lines of signs as the original; and the second, arising from the multiplication by the negative term a, will present the same lines of signs as we should get by changing every one of the signs of the first row. In fact, the two rows of signs would be ++ - ± + − +±±−±+ We have written the ambiguous sign in the product when the addition of unlike signs in the partial products occurs, and it is very plain that these ambiguities will, in this and in every other arrangement, be just as numerous as permanences in the proposed; thus, in the present arrangement, the proposed furnishes four permanences, viz. − − + + + +, − −; and there are, accordingly, in the product four ambiguities, the other signs remaining the same as in the proposed, with the exception of the final sign, which is superadded, and which is always contrary to the final sign in the proposed. It is an easy matter, therefore, when the signs of the terms of any polynomial are given to write down immediately the signs in the product of that polynomial, by x as far, at least, as these signs are determinable, without knowing the values of the quantities employed; for we shall merely have to change every permanency in the proposed into a sign of ambiguity, and to superadd the final sign changed. For instance, if the proposed arrangement were a, Again, if the signs of the proposed equation were in order As, therefore, in passing from the multiplicand to the product, it is the permanences only of the former can suffer any change, it is impossible that the variations can ever be diminished, however they may be increased; consequently, the most unfavourable supposition for our purpose is, that the permanences (omitting the superadded sign) remain the same in number; and, in this case, if the proposed terminate with a variation, the superadded sign in the product will introduce another variation; but if it terminate with a permanency, then the corresponding ambiguity in the result will, obviously, substitute for it what sign we will, form a variation, either with the preceding, or with the superadded sign. It follows, therefore, that no equation can have a greater number of positive roots than variations of signs. To demonstrate the second part of the proposition, it will suffice to remark, that, if we change all the signs in an equation, we change the roots from positive to negative, and vice versa (Theor. IV.) The equation thus changed would have its permanences replaced by variations, and its variations by permanences; and since by the foregoing the changed equation cannot have a greater number of positive roots than variations, the proposed cannot have a greater number of negative roots than permanences. This proposition constitutes the rule of Harriot*, and serves to point out limits which the number of the positive and negative roots of an equation can never exceed. It does not, however, furnish us with the means of ascertaining how many real roots, of either kind, any proposed equation may involve; nor, indeed, does it enable us to affirm that even one positive or negative root actually exists in any equation; it merely shows, that if real roots exist, those which are positive, or those which are negative, cannot exceed a certain number; they may, however, fall greatly short of its number, and, indeed, all be imaginary. But the rule is not without its use, even in detecting imaginary roots, as it sometimes discovers discrepancies incompatible with the existence of real roots, in those equations which are incomplete, or have terms wanting. This will be *By the foreign writers this rule is always attributed to Descartes, and most English writers follow their example. There is, however, undeniable evidence that the rule was obtained indirectly by Descartes from Harriot; and it may be mentioned in support of this view, that Harriot gives a reason for the rule, while Descartes gives none. On the other hand, it has been alleged that the failure of its generality in consequence of the existence of imaginary roots was not perceived by Harriot, and that there is no evidence that he was even acquainted with the existence of imaginary roots. It must, however, be replied, that the Ars praxis Analytica was a posthumous work, edited by Warner, who does not appear to have fully understood Harriot's views, and who, therefore, thought he exercised a sound and kind discretion towards his friend in suppressing certain parts of the work; a suppression which we know did take place. We cannot, therefore, say more as to the views which Harriot entertained on this subject, till some of his papers, still in existence, are more completely examined than they have been. With respect, however, to his knowledge of imaginary roots, we have sufficient proof that he understood their forms and their meaning too. In the Supplement to the Works of Bradley, published by my estimable friend, the late Professor Rigaud, plate 5, will be seen a solution of the equation 1 aα = 2a+34, and the solutions are separately put down; viz. a = = 1 +√—32, and a = 1-√-32. Even this, were this all, would remove the imputation of his ignorance of the existence of imaginary roots.-EDITOR. made apparent in the proof, of De Gua's Criterion of imaginary roots, a little further on (see p. 224). .... THEOREM IX. If r1, T2, T3, T-2 be the real roots of the equation f(x) = 0 of the nth degree, in the descending order of their magnitudes, and quantities ❤, P1, P2, Pn-2p taken so that p, 71, P1, T2, (2,....Tn-2p, Pn-2p be also in descending order of magnitude: then we shall have f(p) = +k, ƒ (91) - k1, ƒ (2) .... = + k2, Let the p pairs of imaginary roots of ƒ (x)=0 be a,b,√√—1, a2+b2√—1,... abp√1: then the portion of ƒ (x) depending upon these is F(x) = {(a,—x)2 + b,2} {(α, — x)2 + b22} .... {(a, - x)2 + b,2 }; in which, since every factor is essentially positive, their product will be +, whatever be the value given to x. .... (x — r1) (x — r2) ........ (x — r2n−p) . F(x) = 0 is the general form of ƒ (x) = 0; and of this F(x) being always +, it will not affect the signs depending on the values of x in the other factors. Substituting then p, 1, successively (pr) (pr2) (p — r2) .... ... Pa-2p in the other part of the function, we have (9 — Ta−2p) = +k, since all the factors are +. - k1, since only one factor is (91 — Sn−2p) = (92— T1) (92 — r2) (P2 — r3) •••• (P2 — ̃n−2p) +k,, since only two factors are Cor. If in the results of any two substitutions p' and p′′ in ƒ (x) we find different signs, there are 1, 3, 5, or some odd number of roots in the interval p' and p"; and if the signs of those results be alike, then 0, 2, 4, or some even number of roots are situated in that interval. THEOREM X. In any function f (x) proceeding by decreasing powers of x, a value may be found for a which shall render the sign of the result the same with that of the first term. .... Let f (x) = x + Ax”¬1 + +La+M; and suppose the most unfavourable case, where all the coefficients after the first, are different from the first, and K that which is numerically greatest. Then we shall have K{x1−1 + x2−2+ x+1} greater than {Ax11 + Bæ”¬2 + all the terms of the function after the first. It will, therefore, be sufficient to show that such a value can be found for x as shall render a greater than or (x — 1)x" greater than K(x-1). Now the value K + 1 given but K(K + 1)" is greater than K{(K + 1)" — 1}, and hence such a value of x as was asserted possible has been found. .... -1 THEOREM XI. In any function, as N + Mx + Lx2 + +Bæ11 + Ax", values of a may be found which will render the result of the entire function of the same sign as the first term N. Take, as before, the most unfavourable case, where all the terms after N have their signs different from that of N. Then if K be greater than either of these coefficients, we shall have -} greater than x {M + Lx + ... Bæ"~2 + Ax"−1}. + Now the greater of these is |