SOMETHING YOU ALWAYS WANTED TO KNOW ABOUT REAL POLYNOMIALS (BUT WERE AFRAID TO ASK)

arXiv:1703.04436v1 [math.CA] 13 Mar 2017

VLADIMIR P. KOSTOV AND BORIS Z. SHAPIRO Abstract. The famous Descartes’ rule of signs from 1637 giving an upper bound on the number of positive roots of a real univariate polynomials in terms of the number of sign changes of its coefficients, has been an indispensable source of inspiration for generations of mathematicians. Trying to extend and sharpen this rule, we consider below the set of all real univariate polynomials of a given degree, a given collection of signs of their coefficients, and a given number of positive and negative roots. In spite of the elementary definition of the main object of our study, it is a non-trivial question for which sign patterns and numbers of positive and negative roots the corresponding set is non-empty. The main result of the present paper is a discovery of a new infinite family of non-realizable combinations of sign patterns and the numbers of positive and negative roots.

1. Introduction 1

This paper continues the line of study of Descartes’ rule of signs initiated in [4]. The basic set-up under consideration is as follows. Consider the affine space P old of all real monic univariate polynomials of degree d. Below we concentrate on polynomials from P old with all non-vanishing coefficients. An arbitrary ordered sequence σ = (σ0 , σ1 , . . . , σd ) of ±-signs is called a sign pattern. When working with monic polynomials we will use their shortened sign patterns σ b representing the signs of all coefficients except the leading term which equals 1. For the actual sign pattern σ, we write σ = (1, σ b) to emphasise that we consider monic polynomials. Given a shortened sign pattern σ b, we call by its Descartes’ pair (pσb , nσb ) the pair b). of non-negative integers counting sign changes and sign preservations of σ = (1, σ By Descartes’ rule of signs, pσb (resp. nσb ) gives the upper bound on the number of positive (resp. negative) roots of any monic polynomial from P old (b σ ). (Observe that, for any σ b , pσb + nσb = d.) To any monic polynomial q(x) with the sign pattern σ = (1, σ b), we associate the pair (posq , negq ) giving the numbers of its positive and negative roots counted with multiplicities. Obviously the pair (posq , negq ) satisfies the standard restrictions posq ≤ pσ , posq ≡ pσ (mod 2), negq ≤ nσ , negq ≡ nσ (mod 2).

(1)

We call pairs (pos, neg) satisfying (1) admissible for σ. Conversely, for a given pair (pos, neg), we call a sign pattern σ such that (1) is satisfied admitting the latter pair. It turns out that not for every pattern σ, all its admissible pairs (pos, neg) are realizable by polynomials with the sign pattern σ. Namely, D. J. Grabiner [5] found the first example of non-realizable combination for polynomials of degree 4. He has shown that the sign pattern (+, −, −, −, +) does not allow to realize the pair (0, 2) 2010 Mathematics Subject Classification. Primary 26C10, Secondary 30C15. Key words and phrases. standard discriminant, Descartes’ rule of signs, sign pattern. 1 The title of the present paper alludes to one of the funniest movies by Heywood “Woody” Allen, the favorite movie director of the second author. 1

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and the sign pattern (+, +, −, +, +) does not allow to realize (2, 0). Observe that their Descartes’ pairs equal (2, 2). His argument is very simple. (Due to symmetry induced by x 7→ −x it suffices to consider only the first case.) Observe that a fourth-degree polynomial with only two negative roots for which the sum of roots is positive could be factored as a(x2 + bx + c)(x2 − sx + t) with a, b, c, s, t > 0, s2 < 4t, and b2 ≥ 4c. The product of these factors equals a(x4 + (b − s)x3 + (t + c − bs)x2 + (bt − cs)x + ct). To get the correct sign pattern, we need b < s and bt < cs, which gives b2 t < s2 c and thus b2 /c < s2 /t. But we have b2 /c ≥ 4 > s2 /t. The following basic question and related conjecture were formulated in [4]. (Apparently for the first time Problem 1 was mentioned in [3].) Problem 1. For a given sign pattern σ, which admissible pairs (pos, neg) are realizable by polynomials whose signs of coefficients are given by σ? Observe that we have the natural Z2 × Z2 -action on the space of monic polynomials and on the set of all sign patterns respectively. The first generator acts by reverting the signs of all monomials in second, fourth etc. position (which for polynomials means P (x) → (−1)d P (−x)); the second generator acts by reading the pattern backwards (which for polynomials means P (x) → xd P (1/x)). If one wants to preserve the set of monic polynomials one has to divide xd P (1/x) by its leading term. We will refer to the latter action as the standard Z2 × Z2 -action. (Up to some trivialities) the properties we will study below are invariant under this action. The following initial results were partially proven in [3, 1] and in complete generality in [4]. Theorem 2. (i) Up to degree d ≤ 3, for any sign pattern σ, all admissible pairs (pos, neg) are realizable. (ii) For d = 4, (up to the standard Z2 × Z2 -action) the only non-realizable combination is (1, −, −, −, +) with the pair (0, 2); (iii) For d = 5, (up to the standard Z2 × Z2 -action) the only non-realizable combination is (1, −, −, −, −, +) with the pair (0, 3); (iv) For d = 6, (up to the standard Z2 × Z2 -action) the only non-realizable combinations are (1, −, −, −, −, −, +) with (0, 2) and (0, 4); (1, +, +, +, −, +, +) with (2, 0); (1, +, −, −, −, −, +) with (0, 4). The next two results can be found in [4] and [8]. Theorem 3. For d = 7, among the 1472 possible combinations of a sign pattern and a pair (up to the standard Z2 × Z2 -action), there exist exactly 6 which are non-realizable. They are: (1, +, −, −, −, −, −, +) with

(0, 5);

(1, +, −, −, −, −, +, +) with

(0, 5);

(1, +, −, +, −, −, −, −) with (3, 0); (1, +, +, −, −, −, −, +) with and, (1, −, −, −, −, −, −, +) with (0, 3) and (0, 5).

(0, 5);

Theorem 4. For d = 8, among the 3648 possible combinations of a sign pattern and a pair (up to the standard Z2 × Z2 -action), there exist exactly 13 which are non-realizable. They are: (1, +, −, −, −, −, −, +, +) with

(0, 6);

(1, −, −, −, −, −, −, +, +) with

(0, 6);

(1, +, +, +, −, −, −, −, +) with (0, 6); (1, +, +, −, −, −, −, −, +) with (0, 6); (1, +, +, +, −, +, +, +, +) with (2, 0); (1, +, +, +, +, +, −, +, +) with (2, 0); (1, +, +, +, −, +, −, +, +) with (2, 0) and (4, 0) ; (1, −, −, −, +, −, −, −, +) with (0, 2) and (0, 4); (1, −, −, −, −, −, −, −, +) with (0, 2), (0, 4), and (0, 6).

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Based on Theorems 2 – 4, we formulated in [4] the following guess. Conjecture 5. For an arbitrary sign pattern σ, the only type of pairs (pos, neg) which can be non-realizable has either pos or neg vanishing. In other words, for any sign pattern σ, each pair (pos, neg) satisfying (1) with positive pos and neg is realizable. At the moment Conjecture 5 has been verified by computer-aided methods up to d = 10. The main result of the present paper is a discovery of a new infinite series of non-realizable patterns which supports Conjecture 5. (Two other series can be found in [4].) Namely, for a fixed odd degree d ≥ 5 and 1 ≤ k ≤ (d − 3)/2, denote by σk the sign pattern beginning with two pluses followed by k pairs “−, +” and then by d − 2k − 1 minuses. Its Descartes’ pair equals (2k + 1, d − 2k − 1). Theorem 6. (i) The sign pattern σk is not realizable with any of the pairs (3, 0), (5, 0), . . ., (2k + 1, 0); (ii) the sign pattern σk is realizable with the pair (1, 0); (iii) the sign pattern σk is realizable with any of the pairs (2ℓ + 1, 2r), ℓ = 0, 1, . . ., k, r = 1, 2, . . ., (d − 2k − 1)/2. Notice that Cases (i), (ii) and (iii) exhaust all possible admissible pairs (pos, neg). It is also worth mentioning that the only non-realizable case for degree 5 (up to the Z2 × Z2 -action) and the third and the last two non-realizable cases for degree 7 mentioned above are covered by Theorem 6. The structure of the paper is as follows. In § 2 we present a proof of Theorem 6. In § 3 we present the detailed structure of the discriminant loci and (non)realizable patterns for polynomials of degrees 3 and 4. Acknowledgements. The first author is grateful to the Mathematics Department of Stockholm University for the hospitality. 2. Proofs P Proof of Theorem 6. Part (i): Suppose that a polynomial P := dj=0 aj xd−j has the sign pattern σk and realizes the pair (2s + 1, 0), 1 ≤ s ≤ k. Denote by (d−1)/2

Pe :=

X

ν=0

(d−1)/2

d−2ν−1

a2ν+1 x

and Po :=

X

a2ν xd−2ν

ν=0 (d−1)/2

its even and odd parts respectively. In each of the sequences {a2ν+1 }ν=0 and (d−1)/2 {a2ν }ν=0 there is exactly one sign change. Therefore each of the polynomials Pe and Po has exactly one real positive root (denoted by xe and xo respectively) which is simple. The polynomial Pe (resp. Po ) is positive and increasing on (xe , ∞) (resp. on (xo , ∞)) and negative on [0, xe ) (resp. on (0, xo )). The polynomial P has at least three distinct positive roots. Denote the smallest of them by 0 < ξ1 < ξ2 < ξ3 . Hence at any point ζ ∈ (ξ1 , ξ2 ) one has the P (ζ) > 0; clearly P is negative on (ξ2 , ξ3 ). One can choose ζ 6= xe and ζ 6= xo . Hence it is impossible to have Pe (ζ) < 0 and Po (ζ) < 0. It is also impossible to have Pe (ζ) > 0 and Po (ζ) > 0. Indeed, this would imply that xe < ζ and xo < ζ. Thus one would get Pe (x) > 0 and Po (x) > 0, i.e. P (x) > 0, for x ∈ (ξ2 , ξ3 ) – a contradiction. The two remaining possibilities are: a) Pe (ζ) > 0, Po (ζ) < 0; b) Pe (ζ) < 0, Po (ζ) > 0. The first one is impossible because it would imply that P (−ζ) = Pe (ζ) − Po (ζ) > 0 ,

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and since P (0) < 0 and P (x) → −∞ for x → −∞, the polynomial P would have at least one negative root in (−∞, −ζ) and at least one in (−ζ, 0) – a contradiction. So suppose that possibility b) takes place. In this case one must have xo < ζ < xe . Without loss of generality one can assume that ξ1 = 1; this can be achieved by a rescaling x 7→ αx with α > 0. Hence Po (1) = β > 0 and Pe (1) = −β. Considering the polynomial P/β instead of P, one can assume that β = 1. Lemma 7 below immediately implies that there are no real roots of P larger than 1 which is a contradiction finishing the proof of Part (i). Lemma 7. Under the above assumptions, P (m) (1) > 0, for any m = 1, 2, . . . , d. Proof of Lemma 7. For any m = 1, 2, . . ., d, it is true that if the sum of the coefficients δ := a2 + a4 + · · · + ad−1 is fixed (recall that all these coefficients are (m) negative), then Po (1) is minimal for a2 = δ, a4 = a6 = · · · = ad−1 = 0. Indeed, when taking derivatives and computing their values at x = 1, the monomial with the largest degree in x is multiplied by the largest factor (equal to this degree). Therefore in what follows we assume that a4 = a6 = · · · = ad−1 = 0, and hence a2 = 1 − a0 < 0. (m) Similarly, consider Pe (1). Recall that a1 > 0, a3 > 0, . . ., a2k+1 > 0, a2k+3 < 0, a2k+5 < 0, . . ., ad < 0. Hence for fixed sums δ∗ := a1 + a3 + · · · + a2k+1 and (m) δ∗∗ := a2k+3 + a2k+5 + · · · + ad , the value of Pe (1) is minimal if ( a1 = · · · = a2k−1 = 0 , a2k+1 = δ∗ (2) a2k+5 = · · · = ad = 0 , a2k+3 = δ∗∗ . Let us now assume that conditions (2) are valid. Thus Pe = a2k+1 xd−2k−1 + (m) a2k+3 xd−2k−3 and a2k+1 +a2k+3 = −1. One can further decrease Pe (1) by assumd d−2 d−2k−3 ing that a2k+1 = 0, a2k+3 = −1. Thus P (x) = a0 x +a2 x −x and a0 +a2 = 1. But then P (m) (x) = um a0 xd−m + vm a2 xd−2−m − wm xd−2k−3−m and P (m) (1) = um a0 + vm a2 − wm for some numbers 0 ≤ wm ≤ vm < um . Therefore P (m) (1) = =

wm (a0 + a2 − 1) + (vm − wm )(a0 + a2 ) + (um − vm )a0 (vm − wm )(a0 + a2 ) + (um − vm )a0 > 0 .

Proof of Part (ii): The polynomial xd − 1 has the necessary signs of the leading coefficient and of the constant term. It has a single real simple root at 1. One can Pd−1 construct a polynomial of the form S := xd − 1 + ε j=1 cj xj , where cj = 1 (resp. cj = −1) if the sign at the corresponding position of σk is + (resp. −). For a small enough ε > 0, the polynomial S has a single simple real root close to 1, and its coefficients have the sign pattern σ. Finally, our approach how to settle Part (iii) is based on the following lemma borrowed from [4]. For a monic polynomial we might write 1 instead of the first + sign in its sign pattern. Recall that the shortened sign pattern of a monic polynomials is what remains from its sign pattern when this initial 1 is deleted. Lemma 8 (See Lemma 14 in [4]). Suppose that the monic polynomials P1 and P2 of degrees d1 and d2 with sign patterns σ ¯1 = (1, σ ˆ1 ) and σ ¯2 = (1, σ ˆ2 ), respectively, realize the pairs (pos1 , neg1 ) and (pos2 , neg2 ). Then (i) if the last position of σ ˆ1 is +, then for any small enough ε > 0, the polynomial ˆ1 , σ ˆ2 ) and the pair (pos1 +pos2 , neg1 + εd2 P1 (x)P2 (x/ε) realizes the sign pattern (1, σ neg2 ).

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(ii) if the last position of σ ˆ1 is −, then for any ε > 0 small enough, the polynomial ˆ1 , σ ˆ2 ) and the pair (pos1 +pos2 , neg1 + εd2 P1 (x)P2 (x/ε) realizes the sign pattern (1, σ neg2 ). (Here −ˆ σ is the sign pattern obtained from σ ˆ by changing each + by − and vice versa.) Remark 9. Example 15 in [4] explains some of the possible applications of Lemma 8. We present and extend this example below. If P2 = x − 1 , x + 1 , x2 + 2x + 2 , x2 + 2x + 0.5 , x2 − 2x + 2 or x2 − 2x + 0.5 , then (pos2 , neg2 ) = (1, 0), (0, 1), (0, 0), (0, 2), (0, 0) and (2, 0) respectively. Denote by τ the last entry of σ ˆ1 . When τ = +, then one has respectively σ ˆ2 = (−), (+), (+, +), (+, +), (−, +) and (−, +) and the sign pattern of εd2 P1 (x)P2 (x/ε) equals (1, σ ˆ1 , −) , (1, σ ˆ1 , +) , (1, σ ˆ1 , +, +) , (1, σ ˆ1 , +, +) , (1, σ ˆ1 , −, +) or (1, σ ˆ1 , −, +) . If τ = −, then σ ˆ2 = (+), (−), (−, −), (−, −), (+, −) and (+, −) and the sign pattern of εd2 P1 (x)P2 (x/ε) equals (1, σ ˆ1 , +) , (1, σ ˆ1 , −) , (1, σ ˆ1 , −, −) , (1, σ ˆ1 , −, −) , (1, σ ˆ1 , +, −) or (1, σ ˆ1 , +, −) . Proof of Part (iii): Recall that the sign pattern σk ends with d − 2k − 1 minuses. Set σk = (+, +, σ ∗ , σ † ), where the sign patterns σ ∗ (resp. σ † ) consist of a minus followed by k pairs (+, −) (resp. of d − 2k − 2 minuses). The sign pattern (+, +) is realizable by the polynomial x + 1 (hence with the pair (0, 1)). To obtain a polynomial realizing the sign pattern (+, +, σ ∗ ) with the pair (2ℓ + 1, 1) one applies Lemma 8, first k − ℓ times with P2 = x2 − 2x + 2, and then 2ℓ + 1 times with P2 = x − 1. After this one applies Lemma 8, first 2r − 1 times with P2 = x + 1, and then (d − 2k − 1)/2 − r times with P2 = x2 + 2x + 2 to realize the sign pattern σk with the pair (2ℓ + 1, 2r). 3. Discriminant loci of cubic and quartic polynomials under a microscope The goal of this section is mainly pedagogical. For the convenience of our readers, we present below detailed descriptions and illustrations of cases of (non)realizability of sign patterns and admissible pairs for polynomials of degree up to 4. Define the standard real discriminant locus Dd ⊂ P old as the subset of all polynomials having a real multiple root. (Detailed information about a natural stratification of Dd can be found in e.g., [6].) It is a well-known and simple fact that P old \ Dd consists of d2 + 1 components distinguished by the number of real simple roots. Moreover, each such component is contractible in P old . Obviously, the number of real roots in a family of monic polynomials changes if and only if this family crosses the discriminant locus Dd . 3.1. Degrees 1 and 2. Clearly, a polynomial x+u has a single real root −u whose sign is opposite to the sign of the constant term. For degrees 2, 3 and 4 we will use the invariance of the zero set of the family of polynomials xn + a1 xn−1 + · · · + an with respect to the group of quasi-homogeneous dilatations x 7→ tx, aj 7→ tj aj , to set the subdominant coefficient to 1. Thus for n = 2, we consider the family P2 := x2 + x + a. For a ≤ 1/4, it has two real roots; for a < 1/4, these are distinct. For a ∈ (0, 1/4), they are both negative while for a < 0, they are of opposite signs. 3.2. Degree 3. For n = 3, we consider the family P3 := x3 + x2 + ax + b. Its discriminant locus Σ is defined by the equation 4a3 − a2 + 4b − 18ab + 27b2 = 0. This is a curve shown in Fig. 1. It has an ordinary cusp for (a, b) = (1/3, 1/27) and an ordinary tangency to the a-axis at the origin. In the eight regions of the complement to its union with the coordinate axes, the polynomial has roots as

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V. KOSTOV AND B. SHAPIRO b

(0,1)

(0,1) (0,3) (2,1) a

(1,2) (1,0)

(1,2)

(1,0)

Figure 1. The discriminant locus of the family x3 + x2 + ax + b. indicated in Fig. 1. (Here (0, 1) means 0 positive and 1 negative real roots hence there exists a complex conjugate pair as well.) The point of the cusp corresponds to a triple root at −1/3, the upper arc corresponds to the case of one double real root to the right and a simple one to the left (and vice versa for the lower arc).

Figure 2. The projection of the discriminant locus of x4 + x3 + ax2 + bx + c to the plane of parameters (a, b). (Picture on the right shows the enlarged portion of the projection near the cusp point.) 3.3. Degree 4. For n = 4, we consider the family P4 := x4 + x3 + ax2 + bx + c. ˜ of its discriminant locus Φ in the (a, b)-plane. In Fig. 2 we show the projection Φ (For the other sets their projections in (a, b) are denoted by the same letters with tilde.) By the dashed line we show the set Σ for the family P3 . One has Φ ∩ {c = 0} = Σ ∪ {b = c = 0}. By the solid line we represent the projection ˜ : 64a3 − 18a2 + 54b − 216ab + 216b2 = 0 Λ of the subset Λ ⊂ Φ for which the polynomial P4 has a real root of multiplicity at ˜ is the projection of the point (3/8, 1/16, 1/256) least 3. The ordinary cusp point of Λ which defines the polynomial x4 + x3 + 3x2 /8 + x/16 + 1/256 = (x + 1/4)4 to the plane (a, b). At this point the set Φ has a swallowtail singularity, see e.g. [2]. On the upper arc of Λ the polynomial P4 has one triple root to the right and a simple one to the ˜ has an ordinary tangency left (and vice versa for the lower arc). The upper arc of Λ to the a-axis at the origin. Along the curve Λ the intersections of the hypersurface Φ with planes transversal to Λ have cusp points.

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Figure 3. Intersections of the discriminant locus of x4 + x3 + ax2 + bx + c with the planes a = −0.1 (the first three pictures); a = 0.15 (the fourth and the fifth pictures); and a = 0.26 (the last picture). The cusp point of Σ belongs to Λ. At this point Λ intersects the (a, b)-plane. ˜ : b = a/2 − 1/8 to Λ ˜ at its cusp at (3/8, 1/16) is tangent to The tangent line L ˜ is shown by the dotted line.) The set L corresponds the curve Σ at (1/4, 0). (L to polynomials having two double roots. For a < 3/8, these roots are real, and for a > 3/8, they are complex conjugate. The curve L is tangent to the (a, b)-plane at the point (1/4, 0, 0). It belongs to the half-space {c ≥ 0}. Now we consider the intersections of Φ with the planes parallel to the (b, c)plane. For a < 3/8, they have two ordinary cusps (which are the points of Λ) and a transversal self-intersection point (which belongs to L). The first three pictures in Fig. 3 show this intersection with the plane a = −0.1 in different scales. The curves are tangent to the a-axis. Inside the curvilinear triangle (denoted by H4 )

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V. KOSTOV AND B. SHAPIRO

the polynomial has four distinct real roots. In the domain H2 which surrounds H4 , the polynomial P4 has two distinct real roots and a complex conjugate pair. In the domain H0 above the self-intersection point it has two complex conjugate pairs. These domains are defined in the same way for all a < 3/8. For a > 3/8, the domain H4 does not exist.

Figure 4. The intersection of the discriminant locus of x4 + x3 + ax2 + bx + c with the planes a = 0.29; 0.31; 0.335; 0.4. The set Φ ∩ {a < 0, b < 0, c > 0} divides the set {a < 0, b < 0, c > 0} into four sectors, see the first picture in Fig. 3. The intersection {a < 0, b < 0, c > 0} ∩ H2 consists of two contractible components. They correspond to the two cases (0, 2) (the right sector, bordering {a < 0, b > 0, c > 0}) and (2, 0) (the left sector) realizable with the sign pattern (+, +, −, −, +). The other two cases realizable in {a < 0, b < 0, c > 0} are (2, 2) (the sector below) and (0, 0) (the sector above). For a < 0, b > 0, c > 0, and when the polynomial P4 belongs respectively to H4 , H2 or H0 , it realizes the cases (2, 2), (0, 2) and (0, 0). The set {a < 0, b > 0, c > 0} ∩ H2 is contractible, so only one of the cases (0, 2) and (2, 0) (namely, (0, 2)) is realizable with the sign pattern (+, +, −, +, +) (see the first picture in Fig. 3). In {a < 0, b < 0, c < 0} one can realize the cases (1, 3) and (1, 1). They correspond to the domains {a < 0, b < 0, c < 0} ∩ H4 (the curvilinear triangle) and {a < 0, b < 0, c < 0} ∩ H2 (its complement). In {a < 0, b > 0, c < 0} one can similarly realize the cases (3, 1) (the curvilinear triangle) and (1, 1) (its complement). On the fourth and fifth pictures in Fig. 3 we present the intersection of Φ with the plane {a = 0.15}. The figures are quite similar to the first three pictures in Fig. 3, and the realizable pairs are the same with one exception. Namely, for a > 0, b > 0, c > 0 in the domain H4 it is the pair (0, 4) which is realized. And, clearly, the third component of the sign patterns changes from − to +.

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The intersections of Φ with the planes {a = 0.26}, {a = 0.29}, {a = 0.31} and {a = 0.335} are shown on the last picture in Fig. 3 and in Fig. 4. For a0 > 0.375, the intersections of Φ with the planes {a = a0 } resemble the lower right picture in Fig. 4. 4. Final Remarks The following important and closely related to the main topic of the present paper questions remained unaddressed above. Problem 10. Is the set of all polynomials realizing a given pair (pos, neg) and having a sign pattern σ path-connected (if non-empty)? Given a real polynomial p of degree d with all non-vanishing coefficients, consider the sequence of pairs {(pos0 (p), neg0 (p)), (pos1 (p), neg1 (p)), (pos2 (p), neg2 (p)), . . . , (posd−1 (p), negd−1 (p))}, where (posj (p), negj (p)) is the numbers of positive and negative roots of p(j) respectively. Observe that if one knows the above sequence of pairs, then one knows the sign pattern of a polynomial p which is assumed to be monic. Additionally it is easy to construct examples when the converse fails. Problem 11. Which sequences of pairs are realizable by real polynomials of degree d with all non-vanishing coefficients? Notice that a similar problem for the sequence of pairs of real roots (without division into positive and negative) was considered in [7]. One can find easily examples of non-realizable sequences {(posj (p), negj (p))}d−1 j=0 . E. g. for d = 4 this is the sequence (2, 0), (2, 1), (1, 1), (0, 1). Indeed, the sign pattern must be (+, +, −, +, +) about which we know that it is not realizable with the pair (2, 0). However it is not self-evident that all non-realizable sequences are obtained in this way. Our final question is as follows. Problem 12. Is the set of all polynomials realizing a given sequence of pairs as above path-connected (if non-empty)? References [1] A. Albouy, Y. Fu, Some remarks about Descartes’ rule of signs, Elemente der Mathematik, 69 (2014), pp. 186–194. [2] V. I. Arnold, The Theory of Singularities and Its Applications, Cambridge, UK: Cambridge University Press, 1993. [3] B. Anderson, J. Jackson, M. Sitharam, Descartes’ rule of signs revisited, The American Mathematical Monthly 105 (1998) pp. 447– 451. [4] J. Forsg˚ ard, Vl. Kostov, B. Shapiro, Could Ren´ e Descartes have known this? Experimental Mathematics, vol 24, issue 4, (2015) 438–448. [5] D. J.Grabiner, Descartes Rule of Signs: Another Construction, The American Mathematical Monthly 106 (1999) pp. 854–856. [6] B. Khesin and B. Shapiro, Swallowtails and Whitney umbrellas are homeomorphic, J. Algebraic Geom. 1(4), (1992), 549–560. [7] V. P. Kostov, Topics on hyperbolic polynomials in one variable. Panoramas et Synth` eses 33 (2011), vi + 141 p. SMF. [8] V. P. Kostov, On realizability of sign patterns by real polynomials, arXiv:1703.03313. ˆ te dAzur, CNRS, LJAD, France Universit´ e Co E-mail address: [email protected] Department of Mathematics, Stockholm University, SE-106 91 Stockholm, Sweden E-mail address: [email protected]

arXiv:1703.04436v1 [math.CA] 13 Mar 2017

VLADIMIR P. KOSTOV AND BORIS Z. SHAPIRO Abstract. The famous Descartes’ rule of signs from 1637 giving an upper bound on the number of positive roots of a real univariate polynomials in terms of the number of sign changes of its coefficients, has been an indispensable source of inspiration for generations of mathematicians. Trying to extend and sharpen this rule, we consider below the set of all real univariate polynomials of a given degree, a given collection of signs of their coefficients, and a given number of positive and negative roots. In spite of the elementary definition of the main object of our study, it is a non-trivial question for which sign patterns and numbers of positive and negative roots the corresponding set is non-empty. The main result of the present paper is a discovery of a new infinite family of non-realizable combinations of sign patterns and the numbers of positive and negative roots.

1. Introduction 1

This paper continues the line of study of Descartes’ rule of signs initiated in [4]. The basic set-up under consideration is as follows. Consider the affine space P old of all real monic univariate polynomials of degree d. Below we concentrate on polynomials from P old with all non-vanishing coefficients. An arbitrary ordered sequence σ = (σ0 , σ1 , . . . , σd ) of ±-signs is called a sign pattern. When working with monic polynomials we will use their shortened sign patterns σ b representing the signs of all coefficients except the leading term which equals 1. For the actual sign pattern σ, we write σ = (1, σ b) to emphasise that we consider monic polynomials. Given a shortened sign pattern σ b, we call by its Descartes’ pair (pσb , nσb ) the pair b). of non-negative integers counting sign changes and sign preservations of σ = (1, σ By Descartes’ rule of signs, pσb (resp. nσb ) gives the upper bound on the number of positive (resp. negative) roots of any monic polynomial from P old (b σ ). (Observe that, for any σ b , pσb + nσb = d.) To any monic polynomial q(x) with the sign pattern σ = (1, σ b), we associate the pair (posq , negq ) giving the numbers of its positive and negative roots counted with multiplicities. Obviously the pair (posq , negq ) satisfies the standard restrictions posq ≤ pσ , posq ≡ pσ (mod 2), negq ≤ nσ , negq ≡ nσ (mod 2).

(1)

We call pairs (pos, neg) satisfying (1) admissible for σ. Conversely, for a given pair (pos, neg), we call a sign pattern σ such that (1) is satisfied admitting the latter pair. It turns out that not for every pattern σ, all its admissible pairs (pos, neg) are realizable by polynomials with the sign pattern σ. Namely, D. J. Grabiner [5] found the first example of non-realizable combination for polynomials of degree 4. He has shown that the sign pattern (+, −, −, −, +) does not allow to realize the pair (0, 2) 2010 Mathematics Subject Classification. Primary 26C10, Secondary 30C15. Key words and phrases. standard discriminant, Descartes’ rule of signs, sign pattern. 1 The title of the present paper alludes to one of the funniest movies by Heywood “Woody” Allen, the favorite movie director of the second author. 1

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and the sign pattern (+, +, −, +, +) does not allow to realize (2, 0). Observe that their Descartes’ pairs equal (2, 2). His argument is very simple. (Due to symmetry induced by x 7→ −x it suffices to consider only the first case.) Observe that a fourth-degree polynomial with only two negative roots for which the sum of roots is positive could be factored as a(x2 + bx + c)(x2 − sx + t) with a, b, c, s, t > 0, s2 < 4t, and b2 ≥ 4c. The product of these factors equals a(x4 + (b − s)x3 + (t + c − bs)x2 + (bt − cs)x + ct). To get the correct sign pattern, we need b < s and bt < cs, which gives b2 t < s2 c and thus b2 /c < s2 /t. But we have b2 /c ≥ 4 > s2 /t. The following basic question and related conjecture were formulated in [4]. (Apparently for the first time Problem 1 was mentioned in [3].) Problem 1. For a given sign pattern σ, which admissible pairs (pos, neg) are realizable by polynomials whose signs of coefficients are given by σ? Observe that we have the natural Z2 × Z2 -action on the space of monic polynomials and on the set of all sign patterns respectively. The first generator acts by reverting the signs of all monomials in second, fourth etc. position (which for polynomials means P (x) → (−1)d P (−x)); the second generator acts by reading the pattern backwards (which for polynomials means P (x) → xd P (1/x)). If one wants to preserve the set of monic polynomials one has to divide xd P (1/x) by its leading term. We will refer to the latter action as the standard Z2 × Z2 -action. (Up to some trivialities) the properties we will study below are invariant under this action. The following initial results were partially proven in [3, 1] and in complete generality in [4]. Theorem 2. (i) Up to degree d ≤ 3, for any sign pattern σ, all admissible pairs (pos, neg) are realizable. (ii) For d = 4, (up to the standard Z2 × Z2 -action) the only non-realizable combination is (1, −, −, −, +) with the pair (0, 2); (iii) For d = 5, (up to the standard Z2 × Z2 -action) the only non-realizable combination is (1, −, −, −, −, +) with the pair (0, 3); (iv) For d = 6, (up to the standard Z2 × Z2 -action) the only non-realizable combinations are (1, −, −, −, −, −, +) with (0, 2) and (0, 4); (1, +, +, +, −, +, +) with (2, 0); (1, +, −, −, −, −, +) with (0, 4). The next two results can be found in [4] and [8]. Theorem 3. For d = 7, among the 1472 possible combinations of a sign pattern and a pair (up to the standard Z2 × Z2 -action), there exist exactly 6 which are non-realizable. They are: (1, +, −, −, −, −, −, +) with

(0, 5);

(1, +, −, −, −, −, +, +) with

(0, 5);

(1, +, −, +, −, −, −, −) with (3, 0); (1, +, +, −, −, −, −, +) with and, (1, −, −, −, −, −, −, +) with (0, 3) and (0, 5).

(0, 5);

Theorem 4. For d = 8, among the 3648 possible combinations of a sign pattern and a pair (up to the standard Z2 × Z2 -action), there exist exactly 13 which are non-realizable. They are: (1, +, −, −, −, −, −, +, +) with

(0, 6);

(1, −, −, −, −, −, −, +, +) with

(0, 6);

(1, +, +, +, −, −, −, −, +) with (0, 6); (1, +, +, −, −, −, −, −, +) with (0, 6); (1, +, +, +, −, +, +, +, +) with (2, 0); (1, +, +, +, +, +, −, +, +) with (2, 0); (1, +, +, +, −, +, −, +, +) with (2, 0) and (4, 0) ; (1, −, −, −, +, −, −, −, +) with (0, 2) and (0, 4); (1, −, −, −, −, −, −, −, +) with (0, 2), (0, 4), and (0, 6).

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Based on Theorems 2 – 4, we formulated in [4] the following guess. Conjecture 5. For an arbitrary sign pattern σ, the only type of pairs (pos, neg) which can be non-realizable has either pos or neg vanishing. In other words, for any sign pattern σ, each pair (pos, neg) satisfying (1) with positive pos and neg is realizable. At the moment Conjecture 5 has been verified by computer-aided methods up to d = 10. The main result of the present paper is a discovery of a new infinite series of non-realizable patterns which supports Conjecture 5. (Two other series can be found in [4].) Namely, for a fixed odd degree d ≥ 5 and 1 ≤ k ≤ (d − 3)/2, denote by σk the sign pattern beginning with two pluses followed by k pairs “−, +” and then by d − 2k − 1 minuses. Its Descartes’ pair equals (2k + 1, d − 2k − 1). Theorem 6. (i) The sign pattern σk is not realizable with any of the pairs (3, 0), (5, 0), . . ., (2k + 1, 0); (ii) the sign pattern σk is realizable with the pair (1, 0); (iii) the sign pattern σk is realizable with any of the pairs (2ℓ + 1, 2r), ℓ = 0, 1, . . ., k, r = 1, 2, . . ., (d − 2k − 1)/2. Notice that Cases (i), (ii) and (iii) exhaust all possible admissible pairs (pos, neg). It is also worth mentioning that the only non-realizable case for degree 5 (up to the Z2 × Z2 -action) and the third and the last two non-realizable cases for degree 7 mentioned above are covered by Theorem 6. The structure of the paper is as follows. In § 2 we present a proof of Theorem 6. In § 3 we present the detailed structure of the discriminant loci and (non)realizable patterns for polynomials of degrees 3 and 4. Acknowledgements. The first author is grateful to the Mathematics Department of Stockholm University for the hospitality. 2. Proofs P Proof of Theorem 6. Part (i): Suppose that a polynomial P := dj=0 aj xd−j has the sign pattern σk and realizes the pair (2s + 1, 0), 1 ≤ s ≤ k. Denote by (d−1)/2

Pe :=

X

ν=0

(d−1)/2

d−2ν−1

a2ν+1 x

and Po :=

X

a2ν xd−2ν

ν=0 (d−1)/2

its even and odd parts respectively. In each of the sequences {a2ν+1 }ν=0 and (d−1)/2 {a2ν }ν=0 there is exactly one sign change. Therefore each of the polynomials Pe and Po has exactly one real positive root (denoted by xe and xo respectively) which is simple. The polynomial Pe (resp. Po ) is positive and increasing on (xe , ∞) (resp. on (xo , ∞)) and negative on [0, xe ) (resp. on (0, xo )). The polynomial P has at least three distinct positive roots. Denote the smallest of them by 0 < ξ1 < ξ2 < ξ3 . Hence at any point ζ ∈ (ξ1 , ξ2 ) one has the P (ζ) > 0; clearly P is negative on (ξ2 , ξ3 ). One can choose ζ 6= xe and ζ 6= xo . Hence it is impossible to have Pe (ζ) < 0 and Po (ζ) < 0. It is also impossible to have Pe (ζ) > 0 and Po (ζ) > 0. Indeed, this would imply that xe < ζ and xo < ζ. Thus one would get Pe (x) > 0 and Po (x) > 0, i.e. P (x) > 0, for x ∈ (ξ2 , ξ3 ) – a contradiction. The two remaining possibilities are: a) Pe (ζ) > 0, Po (ζ) < 0; b) Pe (ζ) < 0, Po (ζ) > 0. The first one is impossible because it would imply that P (−ζ) = Pe (ζ) − Po (ζ) > 0 ,

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and since P (0) < 0 and P (x) → −∞ for x → −∞, the polynomial P would have at least one negative root in (−∞, −ζ) and at least one in (−ζ, 0) – a contradiction. So suppose that possibility b) takes place. In this case one must have xo < ζ < xe . Without loss of generality one can assume that ξ1 = 1; this can be achieved by a rescaling x 7→ αx with α > 0. Hence Po (1) = β > 0 and Pe (1) = −β. Considering the polynomial P/β instead of P, one can assume that β = 1. Lemma 7 below immediately implies that there are no real roots of P larger than 1 which is a contradiction finishing the proof of Part (i). Lemma 7. Under the above assumptions, P (m) (1) > 0, for any m = 1, 2, . . . , d. Proof of Lemma 7. For any m = 1, 2, . . ., d, it is true that if the sum of the coefficients δ := a2 + a4 + · · · + ad−1 is fixed (recall that all these coefficients are (m) negative), then Po (1) is minimal for a2 = δ, a4 = a6 = · · · = ad−1 = 0. Indeed, when taking derivatives and computing their values at x = 1, the monomial with the largest degree in x is multiplied by the largest factor (equal to this degree). Therefore in what follows we assume that a4 = a6 = · · · = ad−1 = 0, and hence a2 = 1 − a0 < 0. (m) Similarly, consider Pe (1). Recall that a1 > 0, a3 > 0, . . ., a2k+1 > 0, a2k+3 < 0, a2k+5 < 0, . . ., ad < 0. Hence for fixed sums δ∗ := a1 + a3 + · · · + a2k+1 and (m) δ∗∗ := a2k+3 + a2k+5 + · · · + ad , the value of Pe (1) is minimal if ( a1 = · · · = a2k−1 = 0 , a2k+1 = δ∗ (2) a2k+5 = · · · = ad = 0 , a2k+3 = δ∗∗ . Let us now assume that conditions (2) are valid. Thus Pe = a2k+1 xd−2k−1 + (m) a2k+3 xd−2k−3 and a2k+1 +a2k+3 = −1. One can further decrease Pe (1) by assumd d−2 d−2k−3 ing that a2k+1 = 0, a2k+3 = −1. Thus P (x) = a0 x +a2 x −x and a0 +a2 = 1. But then P (m) (x) = um a0 xd−m + vm a2 xd−2−m − wm xd−2k−3−m and P (m) (1) = um a0 + vm a2 − wm for some numbers 0 ≤ wm ≤ vm < um . Therefore P (m) (1) = =

wm (a0 + a2 − 1) + (vm − wm )(a0 + a2 ) + (um − vm )a0 (vm − wm )(a0 + a2 ) + (um − vm )a0 > 0 .

Proof of Part (ii): The polynomial xd − 1 has the necessary signs of the leading coefficient and of the constant term. It has a single real simple root at 1. One can Pd−1 construct a polynomial of the form S := xd − 1 + ε j=1 cj xj , where cj = 1 (resp. cj = −1) if the sign at the corresponding position of σk is + (resp. −). For a small enough ε > 0, the polynomial S has a single simple real root close to 1, and its coefficients have the sign pattern σ. Finally, our approach how to settle Part (iii) is based on the following lemma borrowed from [4]. For a monic polynomial we might write 1 instead of the first + sign in its sign pattern. Recall that the shortened sign pattern of a monic polynomials is what remains from its sign pattern when this initial 1 is deleted. Lemma 8 (See Lemma 14 in [4]). Suppose that the monic polynomials P1 and P2 of degrees d1 and d2 with sign patterns σ ¯1 = (1, σ ˆ1 ) and σ ¯2 = (1, σ ˆ2 ), respectively, realize the pairs (pos1 , neg1 ) and (pos2 , neg2 ). Then (i) if the last position of σ ˆ1 is +, then for any small enough ε > 0, the polynomial ˆ1 , σ ˆ2 ) and the pair (pos1 +pos2 , neg1 + εd2 P1 (x)P2 (x/ε) realizes the sign pattern (1, σ neg2 ).

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(ii) if the last position of σ ˆ1 is −, then for any ε > 0 small enough, the polynomial ˆ1 , σ ˆ2 ) and the pair (pos1 +pos2 , neg1 + εd2 P1 (x)P2 (x/ε) realizes the sign pattern (1, σ neg2 ). (Here −ˆ σ is the sign pattern obtained from σ ˆ by changing each + by − and vice versa.) Remark 9. Example 15 in [4] explains some of the possible applications of Lemma 8. We present and extend this example below. If P2 = x − 1 , x + 1 , x2 + 2x + 2 , x2 + 2x + 0.5 , x2 − 2x + 2 or x2 − 2x + 0.5 , then (pos2 , neg2 ) = (1, 0), (0, 1), (0, 0), (0, 2), (0, 0) and (2, 0) respectively. Denote by τ the last entry of σ ˆ1 . When τ = +, then one has respectively σ ˆ2 = (−), (+), (+, +), (+, +), (−, +) and (−, +) and the sign pattern of εd2 P1 (x)P2 (x/ε) equals (1, σ ˆ1 , −) , (1, σ ˆ1 , +) , (1, σ ˆ1 , +, +) , (1, σ ˆ1 , +, +) , (1, σ ˆ1 , −, +) or (1, σ ˆ1 , −, +) . If τ = −, then σ ˆ2 = (+), (−), (−, −), (−, −), (+, −) and (+, −) and the sign pattern of εd2 P1 (x)P2 (x/ε) equals (1, σ ˆ1 , +) , (1, σ ˆ1 , −) , (1, σ ˆ1 , −, −) , (1, σ ˆ1 , −, −) , (1, σ ˆ1 , +, −) or (1, σ ˆ1 , +, −) . Proof of Part (iii): Recall that the sign pattern σk ends with d − 2k − 1 minuses. Set σk = (+, +, σ ∗ , σ † ), where the sign patterns σ ∗ (resp. σ † ) consist of a minus followed by k pairs (+, −) (resp. of d − 2k − 2 minuses). The sign pattern (+, +) is realizable by the polynomial x + 1 (hence with the pair (0, 1)). To obtain a polynomial realizing the sign pattern (+, +, σ ∗ ) with the pair (2ℓ + 1, 1) one applies Lemma 8, first k − ℓ times with P2 = x2 − 2x + 2, and then 2ℓ + 1 times with P2 = x − 1. After this one applies Lemma 8, first 2r − 1 times with P2 = x + 1, and then (d − 2k − 1)/2 − r times with P2 = x2 + 2x + 2 to realize the sign pattern σk with the pair (2ℓ + 1, 2r). 3. Discriminant loci of cubic and quartic polynomials under a microscope The goal of this section is mainly pedagogical. For the convenience of our readers, we present below detailed descriptions and illustrations of cases of (non)realizability of sign patterns and admissible pairs for polynomials of degree up to 4. Define the standard real discriminant locus Dd ⊂ P old as the subset of all polynomials having a real multiple root. (Detailed information about a natural stratification of Dd can be found in e.g., [6].) It is a well-known and simple fact that P old \ Dd consists of d2 + 1 components distinguished by the number of real simple roots. Moreover, each such component is contractible in P old . Obviously, the number of real roots in a family of monic polynomials changes if and only if this family crosses the discriminant locus Dd . 3.1. Degrees 1 and 2. Clearly, a polynomial x+u has a single real root −u whose sign is opposite to the sign of the constant term. For degrees 2, 3 and 4 we will use the invariance of the zero set of the family of polynomials xn + a1 xn−1 + · · · + an with respect to the group of quasi-homogeneous dilatations x 7→ tx, aj 7→ tj aj , to set the subdominant coefficient to 1. Thus for n = 2, we consider the family P2 := x2 + x + a. For a ≤ 1/4, it has two real roots; for a < 1/4, these are distinct. For a ∈ (0, 1/4), they are both negative while for a < 0, they are of opposite signs. 3.2. Degree 3. For n = 3, we consider the family P3 := x3 + x2 + ax + b. Its discriminant locus Σ is defined by the equation 4a3 − a2 + 4b − 18ab + 27b2 = 0. This is a curve shown in Fig. 1. It has an ordinary cusp for (a, b) = (1/3, 1/27) and an ordinary tangency to the a-axis at the origin. In the eight regions of the complement to its union with the coordinate axes, the polynomial has roots as

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V. KOSTOV AND B. SHAPIRO b

(0,1)

(0,1) (0,3) (2,1) a

(1,2) (1,0)

(1,2)

(1,0)

Figure 1. The discriminant locus of the family x3 + x2 + ax + b. indicated in Fig. 1. (Here (0, 1) means 0 positive and 1 negative real roots hence there exists a complex conjugate pair as well.) The point of the cusp corresponds to a triple root at −1/3, the upper arc corresponds to the case of one double real root to the right and a simple one to the left (and vice versa for the lower arc).

Figure 2. The projection of the discriminant locus of x4 + x3 + ax2 + bx + c to the plane of parameters (a, b). (Picture on the right shows the enlarged portion of the projection near the cusp point.) 3.3. Degree 4. For n = 4, we consider the family P4 := x4 + x3 + ax2 + bx + c. ˜ of its discriminant locus Φ in the (a, b)-plane. In Fig. 2 we show the projection Φ (For the other sets their projections in (a, b) are denoted by the same letters with tilde.) By the dashed line we show the set Σ for the family P3 . One has Φ ∩ {c = 0} = Σ ∪ {b = c = 0}. By the solid line we represent the projection ˜ : 64a3 − 18a2 + 54b − 216ab + 216b2 = 0 Λ of the subset Λ ⊂ Φ for which the polynomial P4 has a real root of multiplicity at ˜ is the projection of the point (3/8, 1/16, 1/256) least 3. The ordinary cusp point of Λ which defines the polynomial x4 + x3 + 3x2 /8 + x/16 + 1/256 = (x + 1/4)4 to the plane (a, b). At this point the set Φ has a swallowtail singularity, see e.g. [2]. On the upper arc of Λ the polynomial P4 has one triple root to the right and a simple one to the ˜ has an ordinary tangency left (and vice versa for the lower arc). The upper arc of Λ to the a-axis at the origin. Along the curve Λ the intersections of the hypersurface Φ with planes transversal to Λ have cusp points.

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Figure 3. Intersections of the discriminant locus of x4 + x3 + ax2 + bx + c with the planes a = −0.1 (the first three pictures); a = 0.15 (the fourth and the fifth pictures); and a = 0.26 (the last picture). The cusp point of Σ belongs to Λ. At this point Λ intersects the (a, b)-plane. ˜ : b = a/2 − 1/8 to Λ ˜ at its cusp at (3/8, 1/16) is tangent to The tangent line L ˜ is shown by the dotted line.) The set L corresponds the curve Σ at (1/4, 0). (L to polynomials having two double roots. For a < 3/8, these roots are real, and for a > 3/8, they are complex conjugate. The curve L is tangent to the (a, b)-plane at the point (1/4, 0, 0). It belongs to the half-space {c ≥ 0}. Now we consider the intersections of Φ with the planes parallel to the (b, c)plane. For a < 3/8, they have two ordinary cusps (which are the points of Λ) and a transversal self-intersection point (which belongs to L). The first three pictures in Fig. 3 show this intersection with the plane a = −0.1 in different scales. The curves are tangent to the a-axis. Inside the curvilinear triangle (denoted by H4 )

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V. KOSTOV AND B. SHAPIRO

the polynomial has four distinct real roots. In the domain H2 which surrounds H4 , the polynomial P4 has two distinct real roots and a complex conjugate pair. In the domain H0 above the self-intersection point it has two complex conjugate pairs. These domains are defined in the same way for all a < 3/8. For a > 3/8, the domain H4 does not exist.

Figure 4. The intersection of the discriminant locus of x4 + x3 + ax2 + bx + c with the planes a = 0.29; 0.31; 0.335; 0.4. The set Φ ∩ {a < 0, b < 0, c > 0} divides the set {a < 0, b < 0, c > 0} into four sectors, see the first picture in Fig. 3. The intersection {a < 0, b < 0, c > 0} ∩ H2 consists of two contractible components. They correspond to the two cases (0, 2) (the right sector, bordering {a < 0, b > 0, c > 0}) and (2, 0) (the left sector) realizable with the sign pattern (+, +, −, −, +). The other two cases realizable in {a < 0, b < 0, c > 0} are (2, 2) (the sector below) and (0, 0) (the sector above). For a < 0, b > 0, c > 0, and when the polynomial P4 belongs respectively to H4 , H2 or H0 , it realizes the cases (2, 2), (0, 2) and (0, 0). The set {a < 0, b > 0, c > 0} ∩ H2 is contractible, so only one of the cases (0, 2) and (2, 0) (namely, (0, 2)) is realizable with the sign pattern (+, +, −, +, +) (see the first picture in Fig. 3). In {a < 0, b < 0, c < 0} one can realize the cases (1, 3) and (1, 1). They correspond to the domains {a < 0, b < 0, c < 0} ∩ H4 (the curvilinear triangle) and {a < 0, b < 0, c < 0} ∩ H2 (its complement). In {a < 0, b > 0, c < 0} one can similarly realize the cases (3, 1) (the curvilinear triangle) and (1, 1) (its complement). On the fourth and fifth pictures in Fig. 3 we present the intersection of Φ with the plane {a = 0.15}. The figures are quite similar to the first three pictures in Fig. 3, and the realizable pairs are the same with one exception. Namely, for a > 0, b > 0, c > 0 in the domain H4 it is the pair (0, 4) which is realized. And, clearly, the third component of the sign patterns changes from − to +.

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The intersections of Φ with the planes {a = 0.26}, {a = 0.29}, {a = 0.31} and {a = 0.335} are shown on the last picture in Fig. 3 and in Fig. 4. For a0 > 0.375, the intersections of Φ with the planes {a = a0 } resemble the lower right picture in Fig. 4. 4. Final Remarks The following important and closely related to the main topic of the present paper questions remained unaddressed above. Problem 10. Is the set of all polynomials realizing a given pair (pos, neg) and having a sign pattern σ path-connected (if non-empty)? Given a real polynomial p of degree d with all non-vanishing coefficients, consider the sequence of pairs {(pos0 (p), neg0 (p)), (pos1 (p), neg1 (p)), (pos2 (p), neg2 (p)), . . . , (posd−1 (p), negd−1 (p))}, where (posj (p), negj (p)) is the numbers of positive and negative roots of p(j) respectively. Observe that if one knows the above sequence of pairs, then one knows the sign pattern of a polynomial p which is assumed to be monic. Additionally it is easy to construct examples when the converse fails. Problem 11. Which sequences of pairs are realizable by real polynomials of degree d with all non-vanishing coefficients? Notice that a similar problem for the sequence of pairs of real roots (without division into positive and negative) was considered in [7]. One can find easily examples of non-realizable sequences {(posj (p), negj (p))}d−1 j=0 . E. g. for d = 4 this is the sequence (2, 0), (2, 1), (1, 1), (0, 1). Indeed, the sign pattern must be (+, +, −, +, +) about which we know that it is not realizable with the pair (2, 0). However it is not self-evident that all non-realizable sequences are obtained in this way. Our final question is as follows. Problem 12. Is the set of all polynomials realizing a given sequence of pairs as above path-connected (if non-empty)? References [1] A. Albouy, Y. Fu, Some remarks about Descartes’ rule of signs, Elemente der Mathematik, 69 (2014), pp. 186–194. [2] V. I. Arnold, The Theory of Singularities and Its Applications, Cambridge, UK: Cambridge University Press, 1993. [3] B. Anderson, J. Jackson, M. Sitharam, Descartes’ rule of signs revisited, The American Mathematical Monthly 105 (1998) pp. 447– 451. [4] J. Forsg˚ ard, Vl. Kostov, B. Shapiro, Could Ren´ e Descartes have known this? Experimental Mathematics, vol 24, issue 4, (2015) 438–448. [5] D. J.Grabiner, Descartes Rule of Signs: Another Construction, The American Mathematical Monthly 106 (1999) pp. 854–856. [6] B. Khesin and B. Shapiro, Swallowtails and Whitney umbrellas are homeomorphic, J. Algebraic Geom. 1(4), (1992), 549–560. [7] V. P. Kostov, Topics on hyperbolic polynomials in one variable. Panoramas et Synth` eses 33 (2011), vi + 141 p. SMF. [8] V. P. Kostov, On realizability of sign patterns by real polynomials, arXiv:1703.03313. ˆ te dAzur, CNRS, LJAD, France Universit´ e Co E-mail address: [email protected] Department of Mathematics, Stockholm University, SE-106 91 Stockholm, Sweden E-mail address: [email protected]