To illustrate 809, transform the equation 2 x 13 x16x9x+20=0, 811. Fourth Transformation. To transform an equation into another which lacks any assigned term. (2) Po(y+k)"+p ̧(y+k)n¬1+P2(y+k)"¬2+ . . . +Pn-1(y+k)+P2=0. The development of the binomial in the several terms of (2) gives P1(y+k)n−1=p1 (yn−1 + (n − 1)y"−2k + ...) P2 (y+k)2=P2 (y"-2 + ..) Adding the second members of these equations and arranging the terms with respect to the descending powers of y, we have for the transformed equation, (3) _P ̧y"+(»pk+p})y"-'+["("271p&2+(n−1)p;k+P3]y"−2+......=0, The second term of equation (3) will be wanting if we put and the corresponding substitution is a = y - . EXAMPLE.. -Thus the solution of the equation 192 x3 +144x2 + 132 x + 91 = 0 npo can be made to depend upon the solution of the simpler cubic, 6 y3 + 3 x + 2 = 0, by substituting a = 1 - 1 in the first equation. y If one desires to remove the third term in equation (3) put from which can be found by solving a quadratic equation. Similarly other terms may be removed. 812. Fifth Transformation. -Reciprocal roots and reciprocal equations. Let the equation be then, to transform equation (1) into another equation whose roots are the reciprocal of those of equation (1), put a = l and obtain 1 + P2y2+ P1 Y + 1 = 0. Definition. If equation (3) takes the same form as equation (1). then equation (1) is called a reciprocal equation. and sufficient condition that this may be so is that The necessary Thus, in the first class of reciprocal equations the coefficients reckoned from the beginning and the end are equal in magnitude and have the same signs, Thus, in the second kind of reciprocal equations the coefficients reckoned from the beginning and the end are equal in magnitude and have contrary signs. In case n = 2m; then one of the conditional equations becomes Pm = Pm, i. e., Pm 0; thus, the middle term is absent in a reciprocal equation of the second class and of an even degree. 1 1 If is a root of a reciprocal equation, then is also a root, for is a root of the transformed equation and by definition it is identical with the given equation; hence, the roots of a reciprocal equation. enter in pairs r,;,; t, ; etc. 813. The Standard Form of Reciprocal Equations. Let f(x) = 0 be a reciprocal equation. If f(x) =0 is of the first kind and of an odd degree, then it is evident from the form of the equation that 1 is a root. Hence, the equation is divisible by x+1 and the depressed equation is of the first kind and of an even degree. If f(x) = 0 is of the second kind and of an odd degree then it is evident from the form of the equation that +1 is a root. Hence the equation is divisible by x-1 and the depressed equation is of the first kind and of an even degree. If f(x) = 0 is of the second kind and of an even degree it is divisible by x2-1 since the equation can be written in the form By dividing by x2 - 1 the depressed equation is of the first kind and of an even degree. Therefore all reciprocal equations may be reduced to those of the first kind whose degree is even, and hence it is regarded as the standard form of reciprocal equations. 814. The Reciprocal Equation of the Standard Form can have its Degree Diminished One-Half.-Let the equation be 2m-1 (1) P。 x2m+ P ̧x2m−1+ P2 x2m−2+ · · · · Pmxm + ... + P2x2+P1x+P2=0. Divide this equation by am, and regrouping the terms we have x2 + 1 = (x + 1)(x + 1)(1+1)=-2, x2 + 1/3 = (x2 + 1 ) ( x + 1) − (x + 1) = y3 — 3y. -- x2 * + 1 = — 815. Definition.-If both the signs of two succeeding terms are + or a permanence is said to occur. ceeding terms are respectively + and is said to occur. If the signs of two sucand +, a variation - or 816. Descartes's Rule of Signs for Positive Roots. This rule enables one, by merely inspecting the signs of the terms of an equation, to assign a superior limit to the number of positive roots of the given equation. It may be stated as follows: The number of positive roots of a given equation can not be greater than the number of variations of the signs of its terms. This rule is but a particular case of the more general theorem by Budan and Fourier. The usual proof of this celebrated theorem of Descartes is given, though it amounts to but little more than a verification instead of a full demonstration. Suppose that the given equation is f(x) = 0, and that the signs of the polynomial f(x) succeed each other in the following order: If a positive root a was introduced into the equation f(x)=0, we would multiply the equation by x-a. There are six variations in the signs of the given sequence, and it is proposed to show that if the positive root a is introduced into the equation f(x) = 0, there will be at least seven variations in the signs of the resulting equation, ( a) f(x) = 0. Writing down only the signs which occur in the operation, we have The double sign occurs whenever there are two terms with different signs to be added. On examining the product it follows that: I. The ambiguous sign occurs whenever + follows + and follows in the original sequence of signs. II. The signs before and after an ambiguity, or set of ambiguities, are unlike. III. A change of sign is introduced at the end. Take the most unfavorable case, that in which all the ambiguities in signs are taken as continuations; then it follows from II that the number of changes in signs is the same whether the upper or lower sign is taken; e. g., take the upper sign, then the number of changes of sign can not be less than the number in which has the same arrangement of signs as the original polynomial excepting a change of signs at the end. Suppose now that a polynomial is formed of the factors corresponding to the negative and imaginary roots of an equation; the result of multiplying this product by each of the factors x-a, x-b, x-c, etc., corresponding to the positive roots a, b, c, etc., is that at least one variation in sign for each root is introduced; therefore an equation can not have more positive roots than variations in sign. 817. Descartes's Rule of Signs for Negative Roots.—If — x is substituted for x in the equation f(x) = 0, the resulting equation f(x)=0 has the same roots as the equation f(x)=0, excepting that their signs are changed. This follows from the identical equation f(x) = (x − a) (x − a).... (x — a„) from which we deduce ƒ(— x) = (— 1)" (x + a ̧) (x + a2) . (x + an). |