2 Basics of Polynomials

Transcription

1 Algebraic Geodesy and Geoinformatics PART I - METHODS 2 Basics of Polynomials 2-1 Representations of Polynomials List of monomials In Geodesy and Geoinformatics, most observations are related to the unknowns through equations of algebraic (polynomial) type. Let us consider the following univariate polynomial, poly Clear@"Global *"D poly = a + b - c x ^ 2; with variables, var = x; Test whether poly is a polynomial, PolynomialQ@poly, vard A polynomial is a multivariate product of monomials of the form x1 Α1 x2 Α2...xn Αn, where (Α1,...,Αn ) Î Zn in the variables x1,..., xn. Monomials of the polynomial poly, vard 9- c x2, a + b= The polynomial can be reconstructed from the monomials, Apply@Plus, %D a + b - c x2 Let us consider the following multivariable polynomial, polym = a0 + a1 x1 + a2 x2 + a3 x12 + a4 x22 + a5 x1 x2; with variables, varm = 8x1, x2<; Test it,

2 2 Basics_of_Polynomials_02.nb PolynomialQpolym, varm Monomials of this polynomial, MonomialListpolym, varm a3 x1 2, a5 x1 x2, a1 x1, a4 x2 2, a2 x2, a0 The polynomial can be reconstructed, ApplyPlus, a0 a1 x1 a3 x1 2 a2 x2 a5 x1 x2 a4 x List of coefficients of power of variable(s) A list of coefficients of powers of variables in polynomial, starting with power 0. coeffs CoefficientList poly, var a b, 0, c The polynomial can be reconstructed, FromDigitsReversecoeffs, var a b c x 2 In multivariable case, the list of coefficients of powers of variables (x1, x2) in polynomial, starting with power 0 (see reconstruction) coeffs CoefficientList polym, varm a0, a2, a4, a1, a5, 0, a3, 0, 0 Reconstruction a0, a2, a4.x1 0 x2 0, x1 0 x2 1, x1 0 x2 2 a1, a5, 0.x1 1 x2 0, x1 1 x2 1, x1 1 x2 2 a3, 0, 0.x1 2 x2 0, x1 2 x2 1, x1 2 x2 2 a0 a1 x1 a3 x1 2 a2 x2 a5 x1 x2 a4 x2 2 or employing the list of coefficients and the list of variables, FoldFromDigitsReverse1, 2 &, coeffs, varm Expand a0 a1 x1 a3 x1 2 a2 x2 a5 x1 x2 a4 x2 2 One may use different symbols for variables, FoldFromDigitsReverse1, 2 &, coeffs, x, y Expand a0 a1 x a3 x 2 a2 y a5 x y a4 y List of exponent vectors and coefficients of variable(s) The list of exponent vectors and coefficients for the monomial in polynomial with respect to the variable(s)

3 Basics_of_Polynomials_02.nb CoefficientRules poly, var 2 c, 0 a b FromCoefficientRules, x a b c x 2 In multivariable case, CoefficientRules polym, varm 2, 0 a3, 1, 1 a5, 1, 0 a1, 0, 2 a4, 0, 1 a2, 0, 0 a0 Reconstruction of the polynomial FromCoefficientRules, x1, x2 a0 a1 x1 a3 x1 2 a2 x2 a5 x1 x2 a4 x Polynomial Degree Let us consider the following polynomials p 2 x x ^ 3 y ^ 2 y ^ 2; It can be represented as a list of exponent vectors and coefficients of variables x and y, CoefficientRules p, x, y 3, 2 1, 1, 0 2, 0, 2 1 and Map1, 2 &, CoefficientRules p, x, y 3, 2, 1, 1, 0, 2, 0, 2, 1 The monomials are x 3 y 2, with coefficient 1, and x 1 y 0 with coefficient 2 and x 0, y 2 with coeffient 1. The degree of such a monomial is defined as the sum of the involved exponents of the variables. For example, the first monomial x 3 y 2, has degree = 5, etc. The degree of the polynomial is the maximum degree of its constituent monomials. In this case max( 5, 1, 2) = 5. In Mathematica one may define the following function to compute the degree of a polynomial, MultivariatePolynomDegree poly_, var_ : MaxMapApplyPlus, &, Map1 &, Map1, 2 &, CoefficientRules poly, var applying it to polynomial p MultivariatePolynomDegree p, x, y 5 Given m algebraic (polynomial) observational equations, where m is the dimension of the observation space Y of order L in n variables (unknowns) (where n is the dimension of the parameter space X, the application of algebraic least squares solution (ALESS) to the algebraic observation equations gives (2 L - 1) as the degree of the set of nonlinear algebraic normal equations. There exists n normal equations of the polynomial degree (2 L - 1) to be solved.

4 4 Basics_of_Polynomials_02.nb Example. (Pseudo-ranging problem). For pseudo-ranging or distance equations, the degree of the polynomials in the algebraic observational equations is L = 2. If we take the pseudo-ranges squared or distances squared, a necessary procedure in order to make the observational equations algebraic or polynomial, and implement ALESS, the objective function which is of degree L = 4 reduces by one to degree L = 3 upon differentiating once. The normal equations are of degree L = 3 as expected. Let us consider the following polynomials (now in this case n = m) f1 x1 2 2 a12 x1 x2 x2 2 a0; f2 x2 2 2 b23 x2 x3 x3 2 b0; f3 x3 2 2 c31 x3 x1 x1 2 c0; Let us consider f1, f2 and f3 as observational equations, then their degree L MapMultivariatePolynomDegree, x1, x2, x3 &, f1, f2, f3 2, 2, 2 The objective function and its degree g f1 2 f2 2 f3 2 ; MultivariatePolynomDegree g, x1, x2, x3 4 Differentiating to get the normal equations Eqs MapDg, &, x1, x2, x3; their degrees are indeed MapMultivariatePolynomDegree, x1, x2, x3 &, Eqs 3, 3, 3 2 L 1 3, 3, Operations "Addition" and "Multiplication" Addition is associative f1 f2 f3 f1 f2 f3 Zero element exists f1 0 f1 Inverse exists f1 f1 0

5 Basics_of_Polynomials_02.nb It is commutative f1 f2 f2 f1 Multiplication is associative f1 f2 f3 f1 f2 f3 Identity element exists f1 1 f1 Multiplication is distributive for addition f1 f2 f3 f1 f2 f1 f3 Simplify 2-4 Polynomial Divison Definition (Polynomial division). Consider the polynomial ring k[x] whose elements are polynomials f (x) and g (x). There exists unique polynomials p(x) and r(x) also elements of polynomial ring k[x] such that f (x) = g (x) p (x) + r (x), with either r(x) = 0 or degree of r(x) is less than the degree of g(x). Let us consider the following two univariate polynomials, f x 2 ; The quotient g 1 2 x; q PolynomialQuotient f, g, x 1 4 x 2 the remainder r PolynomialRemainder f, g, x 1 4 indeed g q r f Simplify and MultivariatePolynomDegree g, x MultivariatePolynomDegree r, x

6 6 Basics_of_Polynomials_02.nb In multivariate case f x 2 y x y 2 3 x 2 y 2 3 x y 3 ; g x y x 2 ; the quotient in respect of variable x q PolynomialQuotient f, g, x y 3 y 2 the remainder r PolynomialRemainder f, g, x y 2 3 y 3 x y 2 y 2 3 y 3 indeed g q r f Simplify and again MultivariatePolynomDegree g, x MultivariatePolynomDegree r, x However, this relation is not true when we consider both variables (x, y) MultivariatePolynomDegree g, x, y MultivariatePolynomDegree r, x, y False Now, considering the other variable, y q PolynomialQuotient f, g, y x 3 3 x 4 3 x 5 x 3 x 3 y 3 x y 2 r PolynomialRemainder f, g, y x 4 2 x 5 6 x 6 3 x 7 indeed g q r f Simplify and MultivariatePolynomDegree g, y MultivariatePolynomDegree r, y but again MultivariatePolynomDegree g, x, y MultivariatePolynomDegree r, x, y False

7 Basics_of_Polynomials_02.nb 2-5 Factoring Polynomials In order to understand the factorization of polynomials, it is essential to revisit some of the properties of prime numbers of integers. This is due to the fact that polynomials behave like integers. Whereas for integers, any integer n > 1 is either prime (i.e. can only be factored by 1 and n itself) or a product of prime numbers, a polynomial f (x) k[x] is either irreducible in k[x] or factors as a product of irreducible polynomials in the field k[x]. Let us consider the following polynomial and test it whether irreducible over rationals, but IrreduciblePolynomialQ x 3 1 False Factorx x 1 x x 2 IrreduciblePolynomialQ 1 x x 2 However, this polynomial is not irreducible over algebraic number 3 1 IrreduciblePolynomialQ 1 x x 2, Extension 3 1 False because Solve1 x x 2 0, x x 1 13, x 1 23 In multivariable case f x 2 y x y 2 3 x 2 y 2 3 x y 3 ; fc Factorf x y x y 1 3 y the inverse operation Expandfc x 2 y x y 2 3 x 2 y 2 3 x y 3 These two multivariate polynomials are irreducibles over rationals IrreduciblePolynomialQ x ^ 2 2 y ^ 4, x ^ 4 3 y ^ 2, but over 3, the second polynomial is not irreducible, IrreduciblePolynomialQ x ^ 4 3 y ^ 2, Extension 3 False

8 8 Basics_of_Polynomials_02.nb indeed Factorx ^ 4 3 y ^ 2, Extension x2 3 y 3 x 2 3 y Let us factorize the following polynomials of three variables over the rational numbers p 9 y2 33 x y2 184 x2 y 2 96 x3 y x4 y x5 y 2 3 y3 8 x y3 192 x2 y x 3 y x4 y x 3 y z y 3 z 2 y4 12 x y4 48 x2 y 4 64 x3 y 4 9 y z x y z x2 y z 2 5 x4 y z x5 y z 2 3 y2 z x y3 z x2 y 3 z x3 y 3 z 2 9 z x y 2 z x2 y 2 z x3 y 2 z x4 y 2 z 2 33 x z x2 z 4 15 x3 z 4 9 x 4 z x 5 z 4 3 y z x y z4 3 x 2 y z 4 6 x 3 y z 4 4 x 4 y z 4 y2 z x y2 z x2 y 2 z 4 x 3 y 2 z 4 ; pf Factorp 1 4 x x 2 y 2 8 y 5 z One may try to factorize a polynomial over real numbers, too Factorx ^ 3 x ^ x x x x 2 However, the result can be incorrect! Considering the real representation of the polynomials of three variables pr Np 0.09 y x y x 2 y x 3 y x 4 y x 5 y y x y x 2 y x 3 y x 4 y y x y x 2 y x 3 y y z x y z x 2 y z x 3 y z x 4 y z x 5 y z y 2 z 2 2. x y 2 z x 2 y 2 z x 3 y 2 z x 4 y 2 z y 3 z x y 3 z x 2 y 3 z x 3 y 3 z z x z x 2 z x 3 z 4 9. x 4 z 4 4. x 5 z y z x y z 4 3. x 2 y z 4 6. x 3 y z 4 4. x 4 y z y 2 z x y 2 z x 2 y 2 z 4 1. x 3 y 2 z 4 Factorpr y x y x 2 y x 3 y 2 9. x 4 y 2 4. x 5 y y x y 3 3. x 2 y 3 6. x 3 y 3 4. x 4 y y x y x 2 y 4 1. x 3 y y z x y z x 2 y z x 3 y z x 4 y z 2 5. x 5 y z y 2 z x y 2 z x 2 y 2 z x 3 y 2 z 2 5. x 4 y 2 z y 3 z x y 3 z x 2 y 3 z x 3 y 3 z z x z x 2 z x 3 z x 4 z x 5 z y z x y z x 2 y z x 3 y z x 4 y z y 2 z x y 2 z x 2 y 2 z x 3 y 2 z 4 Length 2 Now, we have only two factors instead of three! Therefore it is better to rationalize the coefficients before factoring,

9 Basics_of_Polynomials_02.nb myfactorx_ : NFactorMapAllRationalize, 0 &, x, Precisionx myfactorpr x x 2. y 2 8. y 5. z 2 2 indeed Npf x x 2. y 2 8. y 5. z 2 2 As we have seen sometimes factoring over integer (rational, real) numbers is not possible, p 3 x ^ 3 7 x ^ 2 9; Factorp 9 7 x 2 3 x 3 IrreduciblePolynomialQ p In that case we can factor it approximately. Let us consider the solution Solve3 x ^ 3 7 x ^ 2 9 0, x x , x x , 13 Now we can use the algebraic number as extension Factorp, Extension x x x 2 in form of real numbers

10 10 Basics_of_Polynomials_02.nb or N, x x x 2 Expand x x x 2 We have got the approximate factoring in form of real number, which can be written in rational form Rationalize, x x x Let us check this result, NExpand x 7. x 2 3. x 3 elinimating numerical error, indeed Chop x 2 3. x Polynomial Roots More often than not, the most encountered interaction with polynomials is the solution of a polynomial equations in terms of finding their roots. As an example, consider the simple planar ranging case where distances are measured from two known points to an unknown station. In such a case, the measured distances are normally related to the unknown station coordinates by multivariate quadratic equations whose intersection leads to the solution of two univariate quadratic equations. If for instance point P 1 whose coordinates are {x 1, y 1 } is occupied and distance s 1 measured to another point P 0 whose coordinates {x 0, y 0 } are unknown, the relationship between the measured distance and the coordinates would be the multivariate quadratic polynomial, or and similarly eq1 a x0 2 b x0 c 0; Clearf eq2 d y0 2 ey0 f 0; Solving this system, it is clear, that solution can exist over reals, but considering the physical problem, cannot exist over complexes.

11 Basics_of_Polynomials_02.nb Solveeq1, eq2, x0, y0 y0 ey0 f d, x0 b b2 4 a c 2 a, y0 ey0 f d, x0 b b2 4 a c 2 a, y0 ey0 f d, x0 b b2 4 a c 2 a, y0 ey0 f d, x0 b b2 4 a c 2 a Let us look for the solution of the following polynomial over integers eq y ; Reduceeq, y, Integers False which means, there is no solution over integers, Reduceeq, y, Rationals False even no solution over rationals. However it has solution over real numbers Reduceeq, y, Reals y 2 3 y 2 3 The following equation has solution over rational numbers, Reducey 2 9 0, y, Rationals y 3 y 3 This equation has solution over complex number, but has not over reals, Reducey 2 9 0, y, Reals False Reducey 2 9 0, y, Complexes y 3 y 3 Polynomials with degree higher than four, in general, have solution only over algebraic numbers represented by the roots of irreducible polynomials over rational. Remark: A polynomial p(x) is irreducible if there do not exist nonconstant polynomials p 1 (x), p 2 (x) such that p = p 1 p 2. sol Reduce 14 x 6 x 12 0, x, Complexes x Root &, 1 x Root &, 2 x Root &, 3 x Root &, 4 x Root &, 5 x Root &, 6 For example, the first polynomial Firstsol x Root &, 1

12 12 Basics_of_Polynomials_02.nb First2x 12 x 14 x 6 IrreduciblePolynomialQ 12 x 14 x 6 This can be solved approximately via numerical method with a prespecified precision NSolve12 x 14 x 6, x, WorkingPrecision 201 x Minimal Polynomials We presented the number ring concept and extended the sets from the sets of natural numbers to the set of complex number in order to cater for expanded operations. For polynomials, the roots may fail to exist in one set say but exist in another set. The polynomial y 2-2 for example, has no roots in but the roots exist in. Reducey 2 2 0, Rationals False Reducey 2 2 0, Reals y 2 y 2 The expansion of the set from to is also called field extension of k. It may occur however that in the polynomial ring k[x], the solution Ξ satisfies not only the polynomial p(x) but also another polynomial h(x), where p(x) and h(x) are both elements of k[x]. In case several polynomials in k[x] have Ξ as a root and the polynomials are multiples of the polynomial of least degree that also contains Ξ as root, the polynomial of least degree is termed the minimal polynomial. Let us construct a polynomial with a root 1 2, which polynomial has lowest degree minpoly MinimalPolynomial Sqrt1 Sqrt2, x 1 2 x 2 x 4 p1 Plotminpoly, x, 2, 2, Epilog Red, PointSizeLarge, PointSqrt1 Sqrt2, Fig. 2.1 The minimal polynomial for the root 1 2 Another polynomial, which has this root too, but having higher degree

13 Basics_of_Polynomials_02.nb poly minpoly x 1 x Expand x x 2 2 x 3 2 x 4 x 5 x 6 p2 Plotpoly, x, 2, 2, Epilog Blue, PointSizeLarge, Point1, 0, Point0, 0, Red, PointSizeLarge, PointSqrt1 Sqrt2, 0; Showp2, p Fig. 2.2 Another polynomial with the root 1 2, but it is not minimal polynomial for this root Solvepoly 0, x x 0, x 1, x 1 2, x 1 2, x 1 2, x 1 2 or Reducepoly 0, x, Rationals 1 2 Rationals && x Rationals && x 1 2 x 0 x 1 Minimal polynomials for different numbers b 3; Headb Integer MinimalPolynomial b, x 3 x b 5 2 ; Headb Rational MinimalPolynomial b, x 5 2 x b ; Headb Complex

14 14 Basics_of_Polynomials_02.nb MinimalPolynomial b, x x 16 x 2 b 2 5 ; AlgebraicIntegerQ b Representation of algebraic numbers as elements of a finite extension of rationals is the following, ToNumberFieldb AlgebraicNumber Root &, 4, 0, 1, 0, 0 Polynomial in x corresponding to the algebraic number object MinimalPolynomial b, x 9 14 x 2 x 4 Approximate solution NSolve 0, x x , x , x , x indeed, b in real form, b N and b as algebraic number is really root of the minimal polynomial, 9 14 x 2 x 4. x b Simplify 0 Let us consider {a 1,...,a n algebraic numbers. One has to define { f 1 (z),..., f n (z)}, f i [z] polynomials with rational coeffients and a b algebraic number like f i (b) = a i, i =1,...n. This is an algebraic field extention of [z] with b as a primitive element, Let a 1 and a 2 be two algebraic numbers, a ; a2 Sqrt3; Here is the function, which creates the polynomials f i and b for a i, PrimitiveElement z_, algs_list : Moduleans ToNumberFieldalgs, aa, IfListQans, aa Selectans, Head AlgebraicNumber &; ans Ifaa, 1, aa1, 1, AlgebraicNumberPolynomial ans, z; ans ; ListQans; b, f PrimitiveElement z, a1, a2; The algebraic number b b Root &, 4 The two polynomials with rational coefficients

15 Basics_of_Polynomials_02.nb f z 9 z3 z5 z z 63 z3 7 z5 z7, Substituting b into the first polynomials f1. z b Root &, Root &, Root &, Root &, 4 7 The reduced form RootReduce Root &, and we get a 1 ToRadicals Similarly f2. z b 4081 Root &, Root &, Root &, Root &, The reduced form RootReduce 3 now we get directly a 2 because it is real number.

16 16 Basics_of_Polynomials_02.nb 2-8 Univariate Polynomials with Real Coefficients Quadratic Polynomials We shall see later, that the solution of a system of polynomial equations can be reduced to the solution of a univariate polynomial. Therefore it is useful to study the solution of polynomials with a single variable. We revisit the various types of univariate polynomials with the coefficients in the field, which we often use to manipulate our measurements. We recapture the basic high school mathematics of inferring the roots of the polynomials from the coeffiients. Polynomials of degree 2 are known as quadratic polynomials. For univariate cases, they take the form ax 2 + bx + c = 0. The general solution of quadratic equations given by the quadratic formula or This is a solution indeed, substituting it into the equation Simplifya x 2 b x c. x 0 b 2 c b 2 4 a c If a and c are small numbers, the possibility of the numerical error is increasing, since we should substract numbers which are nearly the same. In that case, it is better to use the following formula, 2 Remark: Sign(x) gives 1, 0 or 1 depending on whether x is negative, zero, or positive. then Let us consider a quadratic polynomial with the following coefficients a 1 The exact solution is ; b 12; c ; 10 7

17 Basics_of_Polynomials_02.nb Solve x2 12 x 2 0, x 7 10 x , x The approximated value of the smallest root with 25 digits 1 N, Employing the standard formula with 5 digits, b SetPrecision b or using 25 digits 4 a c 2 a, 5 b SetPrecision b a c 2 a, 25 Using the special formula with 5 digits N 1 2 c b Signb b 2 4 a c, In general, every quadratic polynomial has exactly two real or two complex roots. From the coefficients, if b 2-4 ac > 0, the roots are real but if b 2-4 ac < 0 the roots are a pair of non real complex numbers.the case where b 2-4 ac = 0 gives real and identical roots and is also known as the bifurcation point upon which the roots change sign. Let us consider the pairs of roots of the equation : as function of the parameter c on the complex plain!

18 18 Basics_of_Polynomials_02.nb ListPlotAppendRe1, Im1 & FlattenTablex. NSolve2 x 2 5 x c 0, x, c, 2.5, 3.5, 0.02, 5 4, 0, Axes False, Frame, FrameLabel "Real", "Imaginery" where 2.5 c 3.5 with increment The critical value of parameter c is Fig. 2.3 The roots of 2 x 2 5 x c = 0 as function of the parameter c, in range 2.5 c 3.5 c cirt then Solve2 x 2 5 x c cirt 0, x x 5 4, x 5 4 If c c cirt then the roots are reals, while c > c cirt the roots are complex. Let us consider, 0.1; Solve2 x 2 5 x c cirt 0, x x , x Solve2 x 2 5 x c cirt 0, x x , x In stability analysis of a dynamical system this critical parameter has vital importance Cubic Polynomials The polynomials of degree 3 take the form Considering a = 1, the solution is,

19 Basics_of_Polynomials_02.nb 13 2 b 2 13 Clearb, c, d Solvex 3 b x 2 c x d 0, x x b 2 13 b 2 3 c b 3 9 b c 27 d 3 3 b 2 c 2 4 c 3 4 b 3 d 18 b c d 27 d b c 27 d 3 3 b 2 c 4 c 3 4 b 3 d 18 b c d 27 d , x b 1 3 b 2 3 c b 3 9 b c 27 d 3 3 b 2 c 2 4 c 3 4 b 3 d 18 b c d 27 d , x b 1 3 b 2 3 c b 3 9 b c 27 d 3 3 b 2 c 2 4 c 3 4 b 3 d 18 b c d 27 d b 3 9 b c 27 d 3 3 b 2 c 2 4 c 3 4 b 3 d 18 b c d 27 d 2 13 the discriminant D from the coefficients b, c, d as If D > 0 then the roots of the cubic polynomial are real and distinct. If D < 0, then one of the roots is real and the remaining two non real complex conjugate. In a case where D = 0, multiple roots all which are real are given. In case the coefficients b, c, d are all positive, then all of the three roots will be negative while if b, d are negative and c positive, all of the roots will be positive Quartic Polynomials Quartic polynomials are those of degree 4. For example, one of the solution in case of a = 1 sol Solvex 4 b x 3 c x 2 d x e 0, x;

20 20 Basics_of_Polynomials_02.nb sol1 Simplify x b 3 3 b 2 8 c c 2 3 b d 12 e 2 c 3 9 b c d 27 d 2 27 b 2 e 72 c e 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e c 3 9 b c d 27 d 2 27 b 2 e 72 c e 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e b 2 8 c c 2 3 b d 12 e 2 c 3 9 b c d 27 d 2 27 b 2 e 72 c e 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e c 3 9 b c d 27 d 2 27 b 2 e 72 c e 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e b 3 4 b c 8 d 3 b 2 8 c c 2 3 b d 12 e 2 c 3 9 b c d 27 d 2 27 b 2 e 72 c e 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e c 3 9 b c d 27 d 2 27 b 2 e 72 c e 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e 2 13 The discriminant term, 4 c 2 3 b d 12 e 3 2 c 3 9 c b d 8 e 27 d 2 b 2 e 2 Expand 27 b 2 c 2 d c 3 d b 3 d b c d d b 2 c 3 e 432 c 4 e 486 b 3 c d e 2160 b c 2 d e 162 b 2 d 2 e 3888 c d 2 e 729 b 4 e b 2 c e c 2 e b d e e 3 Usually we consider its normalized form 27 Expand b 2 c 2 d 2 4 c 3 d 2 4 b 3 d 3 18 b c d 3 27 d 4 4 b 2 c 3 e 16 c 4 e 18 b 3 c d e 80 b c 2 d e 6 b 2 d 2 e 144 c d 2 e 27 b 4 e b 2 c e c 2 e b d e e 3 - > 0: four different real solutions or two complex cunjugate solutions - < 0: two different real solutions and one complex cunjugate solution - = 0: at least two of solutions are the same

21 Basics_of_Polynomials_02.nb 21 This can be computed with Mathematica, Discriminant Ax4 + b x3 + c x2 + d x + e, xe b2 c2 d2-4 c3 d2-4 b3 d b c d3-27 d4-4 b2 c3 e + 16 c4 e + 18 b3 c d e 80 b c2 d e - 6 b2 d2 e c d2 e - 27 b4 e b2 c e2-128 c2 e2-192 b d e e3 2-9 Methods for Investigating Roots Logaritmic and Contour Plots on Complex Plane Here we introduce two graphical and an algebraic - numeric method to find roots of polynomials. Let us consider the following polynomial f = z z z z z5 ; Ploting the logaritmic of the absolute value of f (z) + 1 on the complex plain. We use this additive constant to avoid singularity at abs( f(z)) = 0. Plot3D@Log@Abs@f. 8z x + ä y<d + 1D, 8x, - 0.5, 1<, 8y, - 3, 2<, PlotPoints 61, Mesh False, AxesLabel 8"ReHzL", "ImHzL", None<, PlotRange All, Ticks 8None, None, None<, ViewPoint , , <D ReHzL ImHzL Fig. 2.4 The surface of the function log(abs (f (z)) + 1) on the complex plain. The "holes" on the surface show the approximate positions of the roots. To get better approximation, we can display Re(f(z)) = 0 and Im(f(z)) = 0 contours on the complex plain. The cross points of these contours represent locations of the roots. sol = NSolve@f 0, zd; zz = z. sol , ä, ä, ä, ä<

22 22 Basics_of_Polynomials_02.nb ShowContourPlotRef. z x I y, x, 0.5, 1, y, 3, 2, Contours 0, ContourShading False, DisplayFunction Identity, ContourStyle RGBColor1, 0, 0, Thickness0.001, PlotPoints 61, ContourPlotImf. z x I y, x, 0.5, 1, y, 3, 2, Contours 0, ContourShading False, DisplayFunction Identity, PlotPoints 61, ContourStyle RGBColor0, 0, 1, Thickness0.001, GraphicsPointSize0.02, PointRe1, Im1 & zz, DisplayFunction $DisplayFunction, FrameTicks None, FrameLabel "Rez", "Imz" Imz Fig. 2.5 The contour plot of f(z) on the complex plain. Along the lines the real and imaginary parts are zero Isograph Simulator Let us consider the exponential form of the complex numbers z = r i Φ, and display f(z) as function of Φ at constant r. Changing r value, find the proper r value at which the curve of f(z) crosses the origin of the coordinate system of the complex plane (0, 0). The function IsoGraphSimulator simulates this process. The numerical values of the roots are computed via NSolve. IsoGraphSimulator eqn_equal, var_ : Modulep, Show GraphicsGridAppendPartition1, 3, Complement1, FlattenPartition1, 3 & ParametricPlotEvaluateRe1, Im1 &Subtract eqn. var 1 Exp p, p, 0, 2 Π, AspectRatio Automatic, Axes, FrameTicks False, Frame, AxesOrigin 0, 0, PlotPoints 20, PlotStyle RGBColor1, 0, 0, PlotLabel "" ToStringvar " " ToStringN1, 3, DisplayFunction Identity & Sortx. NSolveeqn, var, Abs1 Abs2 &

23 Basics_of_Polynomials_02.nb IsoGraphSimulator[(f/.z->x)==0, x] Fig. 2.6 The roots of f(z) displayed by isograph simulator Application of Inverse Series We consider the x = r(y), as the inverse mapping of y = f (x). Then the roots of f(x) can be computed as r(0), while f(r(0)) = 0. In order to illustrate the idea let us consider the power series expansion for following second order polynomial, about the point x = 0 to order four, Cleara Seriesa x ^ 2 b x c, x, 0, 4 c b x a x 2 Ox 5 The inverse series provides the approximation of the inverse function r(y), r InverseSeries, y y c a y c2 b b 3 2 a2 y c3 b 5 The r (0) gives the root of the polynomial f (x) Normalr. y 0 c b a c2 b 3 2 a2 c3 b 5 5 a3 c4 b 7 5 a3 y c4 b 7 Oy c 5 One may recognize that the k 1 th term can be expressed as

24 24 Basics_of_Polynomials_02.nb -c^(k + 1) / b^(2k + 1) a^k Binomial[2 k, k]/(k + 1) ak b 12 k c 1k Binomial2 k, k 1 k Then, for example, considering the first four terms, we get Sumc^k 1 b ^ 2 k 1 a ^ k Binomial2 k, k k 1, k, 0, 3 c b a c2 b 3 2 a2 c3 b 5 5 a3 c4 However, with infinity terms we get the analytical solution b 7 Sumc^k 1 b ^ 2 k 1 a ^ k Binomial2 k, k k 1, k, 0, Infinity b 1 b 2 4 a c b 2 2 a Now we can apply this method to the polynomial considered as an example, the inverse SeriesNf, z, 0, z z z z z 5 Oz 21 r InverseSeries, y y y y y y y y y y y y y y y y y y y y y Oy One of the real root is Normalr. y Using this root, we can reduce the order of our polynomial to a quartic ff PolynomialQuotient f, z, z z z z z 4 The remainder close to zero PolynomialRemainder f, z, z Then we can use factoring to reduce this quartic. Do not forget that quintic can not be factored in general! Factorf z z z z z 5 However

25 Basics_of_Polynomials_02.nb Factorff z z z z 2 The roots of the second order polynomials NSolve2 0, z z , z NSolve3 0, z z , z Now, we have all of the roots. In special case, canonical form (quintic) x 5 - x - Ρ inverse series method can provide even analytical solution. Let us apply the method r = InverseSeries[Series[x^5 - x - Ρ, {x, 0, 15}], y]; Normal[r] /. y -> 0 Ρ Ρ 5 5 Ρ 9 35 Ρ 13 Now the general term of the approximating series then - Ρ^(4k + 1) Binomial[5k, k] / (4k + 1) Ρ14 k Binomial5 k, k 1 4 k Sum Ρ^4 k 1 Binomial5 k, k 4 k 1, k, 0, 3 Ρ Ρ 5 5 Ρ 9 35 Ρ 13 Considering infinite terms sol=sum[- Ρ^(4k + 1) Binomial[5k, k] / (4k + 1), {k, 0, Infinity}] Ρ HypergeometricPFQ 1 5, 2 5, 3 5, 4 5, 1 2, 3 4, Ρ4, Now, for example, let us to solve fff 3 z z 5 3 z z 5 Employing our result we get a complex root, m sol. Ρ Then we can reduce the quintic to a third order polynomial ffff PolynomialQuotient fff, z m z Conjugatem, z z z 2 z 3 PolynomialRemainder fff, z m z Conjugatem, z z

26 26 Basics_of_Polynomials_02.nb Chop 0 Further reduction can be achieved with factoring Factorffff Chop z z z 2-10 The zeros of Polynomial Systems Bezout s Theorem In one variable case, the polynomial, has d roots, counting multiplicities, in the field of complex numbers according to the fundamental theorem of algebra. However in multivariable case, the situation is more complicated. To keep the discussion easy and transparent, we consider a system of two polynomials. Let us consider the following system, G a 1 a 2 x a 3 x y a 4 y; H b 1 b 2 x 2 y b 3 x y 2 ; BØzout s Theorem: Consider two polynomial equations in two unknowns: g(x,y) = h(x,y) = 0. If this system has only finitely many zeros (x, y) 2, then the number of zeros is at most deg(g)* deg(h). Here deg(g) and deg(h) are the total degree of g(x,y) and h(x,y), respectively, see Section 2.2 Bezout Theorem is the best possible in the sense that almost all polynomial systems have deg(g)* deg(h) distinct solutions. degg MultivariatePolynomDegree G, x, y 2 degh MultivariatePolynomDegree H, x, y 3 These two polynomials have precisely four distinct zeros (x, y) 2 for generic choices of coefficients a i and b j. It means that a certain polynomial in the coefficients a i, b j, called the discriminant, should be non- zero. The discriminant can be computed as, R ResultantG, H, x; S FactorResultantR, DR, y, y;

27 Basics_of_Polynomials_02.nb discriminantgh LastS 4 a 2 a 3 3 a 4 4 b 1 3 b 2 3 a 1 2 a 3 2 a 4 4 b 1 2 b a 1 a 2 a 3 a 4 5 b 1 2 b a 2 2 a 4 6 b 1 2 b a 1 3 a 4 5 b 1 b a 2 a 3 6 a 4 b 1 4 b 2 b 3 4 a 1 2 a 3 5 a 4 b 1 3 b 2 2 b 3 96 a 1 a 2 a 3 4 a 4 2 b 1 3 b 2 2 b a 2 2 a 3 3 a 4 3 b 1 3 b 2 2 b 3 22 a 1 3 a 3 3 a 4 2 b 1 2 b 2 3 b 3 94 a 1 2 a 2 a 3 2 a 4 3 b 1 2 b 2 3 b a 1 a 2 2 a 3 a 4 4 b 1 2 b 2 3 b a 2 3 a 4 5 b 1 2 b 2 3 b 3 22 a 1 4 a 3 a 4 3 b 1 b 2 4 b 3 14 a 1 3 a 2 a 4 4 b 1 b 2 4 b 3 4 a 1 2 a 2 a 3 5 b 1 3 b 2 b a 1 a 2 2 a 3 4 a 4 b 1 3 b 2 b a 2 3 a 3 3 a 4 2 b 1 3 b 2 b 3 2 a 1 4 a 3 4 b 1 2 b 2 2 b a 1 3 a 2 a 3 3 a 4 b 1 2 b 2 2 b a 1 2 a 2 2 a 3 2 a 4 2 b 1 2 b 2 2 b a 1 a 2 3 a 3 a 4 3 b 1 2 b 2 2 b a 2 4 a 4 4 b 1 2 b 2 2 b a 1 5 a 3 2 a 4 b 1 b 2 3 b a 1 4 a 2 a 3 a 4 2 b 1 b 2 3 b a 1 3 a 2 2 a 4 3 b 1 b 2 3 b 3 2 a 1 6 a 4 2 b 2 4 b a 2 4 a 3 3 a 4 b 1 3 b a 1 3 a 2 2 a 3 3 b 1 2 b 2 b a 1 2 a 2 3 a 3 2 a 4 b 1 2 b 2 b a 1 a 2 4 a 3 a 4 2 b 1 2 b 2 b a 2 5 a 4 3 b 1 2 b 2 b a 1 5 a 2 a 3 2 b 1 b 2 2 b a 1 4 a 2 2 a 3 a 4 b 1 b 2 2 b a 1 3 a 2 3 a 4 2 b 1 b 2 2 b a 1 7 a 3 b 2 3 b a 1 6 a 2 a 4 b 2 3 b 3 3 a 1 2 a 2 4 a 3 2 b 1 2 b a 1 a 2 5 a 3 a 4 b 1 2 b a 2 6 a 4 2 b 1 2 b a 1 4 a 2 3 a 3 b 1 b 2 b a 1 3 a 2 4 a 4 b 1 b 2 b 3 4 a 1 6 a 2 2 b 2 2 b a 1 3 a 2 5 b 1 b 3 5 MultivariatePolynomDegree discriminantgh, a 1, a 2, a 3, a 4, b 1, b 2, b 3, b 4 14 If this polynomial of degree 14 is non- zero, then the system {G(x, y) = 0, H(x, y) = 0} has four distinct complex zeros. Here the function Resultant computes the resultant of the two polynomials, see Section and more details in Chapter Bernstein s Theorem Bezout s theorem would predict 2*3 = 6 common complex zeros for our system. Indeed, in projective geometry we would expect the cubic curve ( H = 0) and the quadratic curve (G = 0) to intersect in six points. To understand why is four and not six let us consider convex polygons associated with our system. A polytope is a subset of n which is the convex hull of a finite set of points. A 2-dimensional polytope is called polygon. Considering a polynomial f(x,y), Each term x u i y v i can be regarded as a lattice point u i, v i in the plane 2. The convex hull of all these points is called the Newton polygon of f (x, y). Let us express it as, This is a polygon in 2 having at most m vertices. In general, every polynomial in n unknowns can be represented by a Newton polytope in n. Let us consider the Newton polygons of our systems. The Newton polygon of the polynomial G is quadrangel with points (0,0), (1,0),(1,1) and (0,1), while that of the polynomial H is a triangle (0,0), (2,1), (1,2), see Fig. 2.7.

28 28 Basics_of_Polynomials_02.nb Fig. 2.7 Mixed area of Newton-polygons of the two polynomials The Minkowski sum of the two polygons, P and Q in the plane is, P + Q = {p + q : p P, q Q} The Newton polygon of the product of two polynomials is the Minkowski sum of the Newton polyon of these two polynomials, namely The product of the two polynomials, G H Expand New (G*H) = New (G) + New (H) a 1 b 1 x a 2 b 1 x y a 3 b 1 y a 4 b 1 x 2 y a 1 b 2 x 3 y a 2 b 2 x 3 y 2 a 3 b 2 x 2 y 2 a 4 b 2 x y 2 a 1 b 3 x 2 y 2 a 2 b 3 x 2 y 3 a 3 b 3 x y 3 a 4 b 3 The corresponding points can be seen on Fig So the Minkowski sum of the polygons mirrors the algebraic operation of multiplying polynomials. If P and Q are any two polygons then their mixed area can be defined as it follows, In our case, (P, Q) = area (P + Q) - area (P) - area (Q) 2 2 This number coincides with the number of common zeros of G and H. This is not an accident, but is an instance of the general theorem of Bernstein. Bernstein s Theorem: If G and H are two generic bivariate polynomials, then the number of non-zero solutions of G(x,y) = H(x,y) =0 in 2 equals the mixed area (New (G), New (H)). It is remarkable that Bezout s theorem follows as a special case from Bernstein s theorem. Namely, if g and h a general polynomials of degree d and e respectively,

29 Basics_of_Polynomials_02.nb then their Newton polygons are triangles, P = New (g) = conv ((0, 0), (0, d), (d, 0)) Q = New (h) = conv ((0, 0), (0, e), (e, 0)) P + Q = New (g* h) = conv ((0, 0), (0, d + e), (d + e, 0)) The areas of these triangles are, d 2 /2, e 2 /2, d e 2 /2, and hence Hence two general plane curves of degree d and e meet in d*c points Computation zeros of Polynomial Systems Resultants Elimination theory deals with the problem of eliminating one or more variables from a system of polynomial equations, thus reducing the problem to a smaller problem in fewer variables. For instance, if we wish to solve, U a 0 a 1 x a 2 x 2 ; V b 0 b 1 x b 2 x 2 ; U = V = 0 with a 2 0 and b 2 0 then we can eliminate the variable x to get, EliminateU 0, V 0, x Expand a 0 a 2 b a 0 a 2 b 0 b 2 a 0 a 1 b 1 b 2 a 0 2 b 2 2 a 2 2 b 0 2 a 1 a 2 b 0 b 1 a 1 2 b 0 b 2 This polynomial of of degree 4 is the resultant. It vanishes if and only if the given quadratic polynomials have common complex root x. ResultantU, V, x a 2 2 b 0 2 a 1 a 2 b 0 b 1 a 0 a 2 b 1 2 a 1 2 b 0 b 2 2 a 0 a 2 b 0 b 2 a 0 a 1 b 1 b 2 a 0 2 b 2 2 Consider two polynomials in two variables, G(x, y) and H(x, y). Let us eliminate y ResGH y ResultantG, H, y x 2 a 3 2 b 1 2 x a 3 a 4 b 1 a 4 2 b 1 x 3 a 1 a 3 b 2 x 4 a 2 a 3 b 2 x 2 a 1 a 4 b 2 x 3 a 2 a 4 b 2 x a 1 2 b 3 2 x 2 a 1 a 2 b 3 x 3 a 2 2 b 3 We get an univariate polynomial in x of degree 4, which should be zero, Similarly, MapExponentResGH y, &, x, y 4, 0

30 30 Basics_of_Polynomials_02.nb ResGH x ResultantG, H, x a 2 2 b 1 2 y a 2 a 3 b 1 y 2 a 3 2 b 1 y a 1 2 b 2 2 y 2 a 1 a 4 b 2 y 3 a 4 2 b 2 y 2 a 1 a 2 b 3 y 3 a 1 a 3 b 3 y 3 a 2 a 4 b 3 y 4 a 3 a 4 b 3 So the two multivariate polynomials can be separated into two univariate polynomials. This resultant method can be extended for case n > 2 and various determinantal formulas are known for multivariate resultant, see Chapter Groebner Basis Let K be a field, K be its algebraic closure, R = K[x,y] be the algebra of polynomials in the variables {x, y}. Let G (x, y) and H (x, y) Ε R. Our objective is to solve the system f = G= H= 0. Let I be the ideal generated by G and H. Then exist polynomials g and h, that and similarly G (x, y)= Α g(x, y) + Β h (x, y) H (x, y) = Γ g (x, y) + h (x, y) If the basis {G, H} then the trivial coefficients are Α = = 1 and Γ = Β = 0. However, we can find other set of {g, h} polynomials as basis for the ideal. Groebner basis can provide a special basis, which is triangular, namely e.g., G (x, y)= Α gr 1 (x) + Β gr 2 (x, y) = {Α, Β}.{gr 1 (x), gr 2 (x, y)} where {Α, Β} is a not vanishing coefficient vector. Therefore {G, H} = 0 implies that {gr 1 (x), gr 2 (x, y)}= 0, and vica versa. This triangular system {gr 1 (x), gr 2 (x, y)}= 0 can be solved by successive elimination. In our case the Groebner basis of the ideal generated by G(x, y) and H(x, y) is, gr GroebnerBasisG, H, x, y; TableFormgr a 2 2 b 1 2 y a 2 a 3 b 1 y 2 a 3 2 b 1 y a 1 2 b 2 2 y 2 a 1 a 4 b 2 y 3 a 4 2 b 2 y 2 a 1 a 2 b 3 y 3 a 1 a 3 b 3 y 3 a 2 a 4 b 3 y 4 a 3 a 4 b 3 a 2 a 2 3 b 1 y a 3 3 b 1 a 2 1 a 3 b 2 x a 1 a 2 a 3 b 2 a 1 a 2 a 4 b 2 x a 2 2 a 4 b 2 2 y a 1 a 3 a 4 b 2 y a 2 a 2 4 b 2 y 2 a 3 a 2 4 b 2 y 2 a 1 a 2 3 b 3 y 3 a 2 3 a 4 b 3 a 1 x a 2 x y a 3 y a 4 a 2 a 3 b 1 y a 3 2 b 1 a 1 2 b 2 x a 1 a 2 b 2 2 y a 1 a 4 b 2 x y a 2 a 4 b 2 y 2 a 4 2 b 2 y 2 a 1 a 3 b 3 y 3 a 3 a 4 b 3 a 2 b 1 y a 3 b 1 x y a 1 b 2 x y 2 a 4 b 2 y 2 a 1 b 3 y 3 a 4 b 3 a 3 b 1 x a 1 b 2 x 2 a 2 b 2 x y a 4 b 2 y a 1 b 3 x y a 2 b 3 y 2 a 4 b 3 b 1 x 2 y b 2 x y 2 b 3 This is a triangular system for x and y, TransposeMapExponentgr, &, x, y MatrixForm G can be expressed in this basis. The corresponding coefficient vector is c and the r is the remainder, which should be zero, since G belonging to the ideal,

31 Basics_of_Polynomials_02.nb indeed Similarly and c 1, r PolynomialReduce G, gr, x, y a 3 2, a 1 a 2 a 3 b 2 a 2 2 a 4 b 2 c 1.gr r G Simplify a 2 a 1 a 2 a 3 b 2 a 2 2 a 4 b 2 c 2, r PolynomialReduce H, gr, x, y y a 3 x a 3 y a 3 a 2 4 b 2 a 1 a 2 a 3 a 2 2 a 4 a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 a 1 a 2 a 3 b 2 a 2 2 a 4 b 2, 0, 0, 0, 0, 0, 0 a 3 3 b 1 a 1 a 3 a 4 b 2 a 2 a 2 4 b 2 y 2 a 2 3 a 4 b 3, a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 x y a 1 a 2 a 3 a 2 2 a 4 a 2 a 3 b 1 y a 2 3 b 1 a 1 a 4 b 2 a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 y 3 a 3 a 4 b 3 y 2 a 2 4 b 2 a 2 a 4 b 3, 0, 0, 0, 0, 0, 0 a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 a 1 a 2 a 3 a 2 2 a 4 a 1 a 3 b 2 a 2 a 4 b 2 c 2.gr r H Simplify We can carry out the solution of the triangular Groebner basis directly, (computing Groebner basis followed by elimination) e.g. eliminating y from the Groebner basis to get x, and similarly gry GroebnerBasisG, H, x, y, y x 2 a 3 2 b 1 2 x a 3 a 4 b 1 a 4 2 b 1 x 3 a 1 a 3 b 2 x 4 a 2 a 3 b 2 x 2 a 1 a 4 b 2 x 3 a 2 a 4 b 2 x a 1 2 b 3 2 x 2 a 1 a 2 b 3 x 3 a 2 2 b 3 grx GroebnerBasisG, H, x, y, x a 2 2 b 1 2 y a 2 a 3 b 1 y 2 a 3 2 b 1 y a 1 2 b 2 2 y 2 a 1 a 4 b 2 y 3 a 4 2 b 2 y 2 a 1 a 2 b 3 y 3 a 1 a 3 b 3 y 3 a 2 a 4 b 3 y 4 a 3 a 4 b 3 which equal with the first polynomial in the Groebner basis, grx gr1 Simplify This is called reduced Groebner basis corresponding to variables x and y, respetively. Let us compare the result of the resultants, ResGH y and ResGH x., and ResGH y gry Simplify

32 32 Basics_of_Polynomials_02.nb ResGH x grx Simplify Further details about Groebner basis and its computation can be found in Chapter Homotopy Method This method is originated from numerical algebraic geometry and belongs to the general class of numerical continuation method. The method operates in two stages. Firtsly, it exploits the structure of the system f(x) = 0 to find a root count (see Bezout s and Bernstein Theorem) and construct a start system g(x) = 0 that has exactly as many regular solutions as the root count. This start system is embedded in the homotopy, with Γ a random number. Let the system to be solved, f 1 x 2 4 y 2 4; f 2 2 y 2 x; h (x, t) = Γ (1 - t) g (x) + t f (x) = 0, t [0, 1] an obvious the start system having the same polynomial degree, g 1 x 2 1; g 2 y 2 1; Forming homotopy function, Γ RandomReal ; H 1 Γ g 1 1 t f 1 t; H 2 Γ g 2 1 t f 2 t; Let us solve this system from t = 0 (x =1, y = 1) up to t =1 employing Newton- Raphson method. Tracing the first path from x 0 1, y 0 1 : t 0.2; x 0 1; y 0 1; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 0.4; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 0.6; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y

33 Basics_of_Polynomials_02.nb t 0.8; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 1.; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y Eliminating small imaginary parts, Chop x , y Checking the result f 1, f 2. 0., Tracing the second path from x 0 1, y 0 1 : t 0.2; x 0 1; y 0 1; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 0.4; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 0.6; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 0.8; x 0 x. sol1; y 0 y. sol2; sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y t 1.; x 0 x. sol1; y 0 y. sol2;

34 34 Basics_of_Polynomials_02.nb sol FindRootH 1 0, H 2 0, x, x 0, y, y 0 x , y Eliminating small imaginary parts, Chop x , y Checking the result f 1, f 2. 0., The other two paths belong to the x 0 1, y 0 1 and x 0 1, y 0 1 initial conditions. We can employ GeoAlgebra package to solve the problem. First let us to load the package, GeoAlgebra LinearHomotopy The function we shall use LinearHomotopyFR,? LinearHomotopyFR Computes the homotopy paths with direct path tracing. Input parameters: F list of functions of the target system, G list of functions of the start system, X list of variables, X0 list of initial values, Γ list of complex weigths, n number of subintervals, Λ dummy variable. Output variables: sol1 list of the solutions, sol2 list of homotopy paths Therefore F f 1, f 2 ; G g 1, g 2 ; X x, y; X0 1, 1, 1, 1, 1, 1, 1, 1; Γ 1, 1 ; sol LinearHomotopyFR F, G, X, X0, Γ, 100, Λ; sol , , , , , , , These pairs are solution of the system, Mapf 1, f 2. x 1, y 2 &, sol1 0., , 0., 0., 0., , 0., 0. The visualization of the paths can be achieved by the function Paths,

35 Basics_of_Polynomials_02.nb 35? Paths Display homotopy paths. Input parameters: X list of variables, sol list of homotopy paths, X0 list of initial values PathsX, sol2, X0, Λ Im Im Im Im Im xλ Re Im yλ Re Fig. 2.8 Homotopy paths belonging to the different solutions of the start system Employing Global Numerical Solver (GNS), built - in Mathematica, we get, NSolvef 1, f 2, x, y x , y , x , y , x , y , x , y A computational advantage of the homotopy method that the tracing of the different paths can be carried out simultaneously and independently, consequently the homotopy is a parallel algorithm by nature. See further details in Chapter 5.

36 36 Basics_of_Polynomials_02.nb A computational advantage of the homotopy method that the tracing of the different paths can be carried out simultaneously and independently, consequently the homotopy is a parallel algorithm by nature. See further details in Chapter 5.