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ContentTheory of Equation
Theory of EquationSymmetrical Functions of the Roots of an EquationNotation: Let π, π, π, β― be the roots of the equation π(π₯)=0.Let π π denote ππ+ππ+β―, the sum of the πth powers of the roots. Let π π,π denote ππππ+ππππ+ππππ+β―, through all the permutations of the roots, two at a time. Similarly, let π π,π,π denote ππππππ+ππππππ+β―, taking all the permutations of the roots three at a time; and on. 534 Sums of the powers of the rootsπ π+π1π πβ1+π2π πβ2+β―+ππβ1π 1+πππ=0 where π is less than π, the degree of π(π₯).Obtained by expanding by division each term in the value of πβ²(π₯) given at (432), arranging the whole in powers of π₯, and equating coefficients in the result and in the value of πβ²(π₯), found by differentiation us in (424). 535 where π is less than π, the formula will be π π+π1π πβ1+π2π πβ2+β―+πππ πβπ=0 Obtained by multiplying π(π₯)=0 by π₯πβπ, substituting for π₯ the roots π, π, π, β― in succession and adding the results. By these formula π 1, π 2, π 3, β― may be calculated successively. 536 To find the sum of the negative powers of the roots, put π equal to πβ1, πβ2, πβ3, β―, successively in (535), in order to obtain π β1, π β2, π β3, β― 537 To calculate π π independently, Rule: π π=βπΓcoefficient of π₯βπ in the expansion of >π(π₯)π₯π Proved by taking π(π₯)=(π₯βπ)(π₯βπ)(π₯βπ)β―, dividing by π₯π, and expanding the logarithm of the right side of the equation by (456). 538 Symmetrical Functions which are not powers of the rootsThese are expressed in terms of the sums of powers of the roots as under, and thence, by (534), in terms of the roots explicitly, π π,π=π ππ πβπ π+π 539 π π,π,π=π ππ ππ πβπ π+ππ πβπ π+ππ πβπ π+ππ π+2π π+π+π The last equation may be proved by multiplying π π,π by π π; and expansions of other symmetrical functions may be obtained in a similar way. 540 If π(π₯) be a rational integral function of π₯, then the symmetrical function of the roots of π(π₯), denoted by π(π)+π(π)+π(π)+β― is equal to the coefficient of π₯πβ1 in the remainder obtained by dividing π(π₯)πβ²(π₯) by π(π₯).Proved by multiplying the equation (432) by π(π₯)π(π₯), and by theorem (401). 541 To find the equation whose roots are the squares of the differences of the roots of a given equation. Let πΉ(π₯) be the given equation, and ππ the sum of the πth powers of its roots. Let π(π₯) and π π have the same meaning with regard to the required equation. The coefficients of the required equation can be calculated from those of the given one as follows: The coefficients of each equation may be connected with the sums of the powers of its roots by (534); and the sums of the powers of the roots of the two equations are connected by the formula. 542 2π π=ππ2πβ2ππ1π2πβ1+ 2π(2πβ1)1β 2π2π2πβ2ββ―+ππ2π Rule: 2π π is equal to the formal expansion of (πβπ)2π by the Binomial Theorem, with the first and last terms each multiplied by π, and the indices all changed to suffixes. As the equi-distant terms are equal we can divide by 2, and take half the series. Demonstration: Let π, π, π, β― be the roots of πΉ(π₯) Let π(π₯)=(π₯βπ)2π+(π₯βπ)2π+β―i. Expand each term on the right by the Bin. Theor., and add, substituting π1, π2, β―. In the result change π₯ into π, π, π, β― successively, and add the π equations to obtain the formula, observing that, by [i.]. π(π)+π(π)+β―=2π π If π be the degree of πΉ(π₯), then 12π(πβ1) is the degree of π(π₯). 543 The last term of the equation π(π₯)=0 is equal to πππΉ(πΌ)πΉ(π½)πΉ(πΎ)β― where πΌ, π½, πΎ, β―, are the roots of πΉ(π₯). Proved by shewing that πΉβ²(πΌ)πΉβ²(π½)β―=πππΉ(πΌ)πΉ(π½)β― 544 If πΉ(π₯) has negative or imaginary roots, π(π₯) must have imaginary roots. 545 The sum of the πth powers of the roots of the quadratic equation π₯2+ππ₯+π=0 π π=ππβπππβ2π+ π(πβ3)|2ππβ4π2ββ―+(β1)π π(πβπβ1)β―(πβ2π+1)|πππβ2πππ+β― By (537) expanding the logarithm by (156) 546 The sum of the πth powers of the roots of π₯πβ1=0 is π if π be a multiple of π, and zero if it be not. By (537); expanding the logarithm by (156) 547 If π(π₯)=π0+π1π₯+π2π₯2+β―i. then the sum of the selected terms πππ₯π+ππ+ππ₯π+π+ππ+2ππ₯π+2π+β― will be π = 1π{πΌπβππ(πΌπ₯)+π½πβππ(π½π₯)+πΎπβππ(πΎπ₯)+β―} where πΌ, π½, πΎ, β―, are the πth roots of unity. For proof, multiply (i.) by πΌπβπ, and change π₯ into πΌπ₯; so with π½, πΎ, β―, and add the resulting equations. 548 To approximate to the root of an equation by means of the sums of the powers of the roots. By taking π large enough, the fraction π π+1π πwill approximate to the value of the numberically greatest root, unless there be a modulus of imaginary roots greater than any real root, in which case the fraction has no limiting value. 549 Similarly the fraction π ππ π+2βπ 2ππ πβ1π π+1βπ 2πapproximates, as π increases, to the greatest product of any pair of roots, real or imaginary; excepting in the case in which the product of the pair of imaginary roots, though less than the product of the two real roots, is greater than the square of the least of them, for then the fraction has no limiting value. 550 Similarly the fraction π ππ π+3βπ π+1π π+2π ππ π+2βπ 2π+1approximates, as π increases, to the sum of the two numerically greatest roots, or to the sum of the two imaginary roots with the greatest modulus. Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdriveΒ©sideway ID: 210800019 Last Updated: 8/19/2021 Revision: 0 Ref: References
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