Logarithm TheoremPythagorean TheoremCombinatoricsQuadratic EquationsSequence and SeriesLinear AlgebraDiophantine EquationElliptic Curve FactorMultiplication, DivisionIndicesHighest Common Factor, Lower Common MultipleEquationsQuadratic EquationsSimultaneous EquationsRatio and ProportionArithemetical ProgressionGeometrical ProgressionHarmonical ProgressionPermutations, CombinationsSurdsBinomial TheoremMultinomial TheoremLogarithmExponential TheoremContinued Fractions and ConvergentsIndeterminate EquationsSimultaneous Equations IIImaginary ExpressionsMethod of Indeterminate CoefficientsMethod of Proof by InductionPartial FractionsConvergency and Divergency of SeriesExpansion of a FractionRecurring SeriesSummation of SeriesPolygonal NumbersFigurate NumbersHypergeometrical SeriesInterest and AnnuitiesProbabilitiesInequalitiesScales of NotationTheory of Numbers Factors of EquationDescartes' Rule of SignsThe Derived Functions of π(π₯)Equal roots of an equationLimits of the RootsNewton's Method of Divisors Draft for Information Only
ContentTheory of Equation
Theory of EquationReciprocal Equations466 A reciprocal equation has its roots in pairs of the form π,1π; also the relation between the coefficients is ππ=ππβπ, or else ππ=βππβπ 467 A reciprocal equation of an even degree, with its last term positive, may be made to depend upon the solution of an equation of half the same degree. 468 Example4π₯6β24π₯5+57π₯4β73π₯3+57π₯2β24π₯+4=0 is a reciprocal equation of an even degree, with its last term positive.Any reciprocal equation which is not of this form may be reduced to it by dividing by π₯+1 if the last term be positive; and, if the last term be negative, by dividing by π₯β1 or π₯2β1, so as to bring the equation to an even degree. Then proceed in the following manner:- 469 First bring together equdistant terms, and divide the equation by π₯3; thus 4 π₯3+β24 π₯2++57 π₯+β73=0 By putting π₯+ π₯+2 the equation is reduced to a cubic in π¦, the degree being one-half that of the original equation. 470 Put π for π₯+ 1π₯, and ππ for π₯π+ 1π₯π. The relation between the successive factors of the form ππ may be expressed by the equation ππ=πππβ1βππβ2 471 The equation for ππ, in terms of π, is ππ=ππβπππβ2+ π(πβ3)1β 2ππβ4ββ―+(β1)π π(πβπβ1)β―(πβ2π+1)ππβ2π+β― By (54), putting π=1 Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdriveΒ©sideway ID: 210800010 Last Updated: 8/10/2021 Revision: 0 Ref: References
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