Root Arrangements of Hyperbolic Polynomial-like Functions

Transcription

1 Root Arrangements o Hyperbolic Polynomial-like Functions Vladimir Petrov KOSTOV Université de Nice Laboratoire de Mathématiques Parc Valrose Nice Cedex France kostov@mathunicer Received: March, 005 Accepted: December 1, 005 ABSTRACT To Pro A Galligo A real polynomial P o degree n in one real variable is hyperbolic i its roots are all real A real-valued unction P is called a hyperbolic polynomial-like unction (HPLF) o degree n i it has n real zeros and P (n) vanishes nowhere Denote by x (i) k the roots o P (i), k = 1,, n i, i = 0,, n 1 Then in the absence o any equality o the orm one has i < j x (j) i x (i) k = x (l) k (1) < x(j) k < x (i) k+j i () (the Rolle theorem) For n 4 (resp or n 5) not all arrangements without equalities (1) o n(n + 1)/ real numbers x (i) k and compatible with () are realizable by the roots o hyperbolic polynomials (resp o HPLFs) o degree n and o their derivatives For n = 5 and when x (1) 1 < x (1) < x () 1 < x () < x (1) < x (1) 4 we show that rom the 40 arrangements without equalities (1) and compatible with () only 16 are realizable by HPLFs (rom which 6 by perturbations o hyperbolic polynomials and none by hyperbolic polynomials) Key words: hyperbolic polynomial, polynomial-like unction, root arrangement, coniguration vector 000 Mathematics Subject Classiication: 1D10 Rev Mat Complut 19 (006), no 1, ISSN:

2 1 Introduction 11 Hyperbolic polynomials, polynomial-like unctions, and their root arrangements Consider the amily o polynomials P (x, a) = x n + a 1 x n a n, x, a i R A polynomial o this amily is called (strictly) hyperbolic i all its roots are real (real and distinct) It is clear that i P is (strictly) hyperbolic, then such are P (1),, P (n 1) as well Hyperbolic are the polynomials o all known orthogonal amilies (eg, the Legendre, Laguerre, Hermite, Tchebyshev polynomials) A problem o interest in (real) algebraic geometry is the dependence o the roots o the irst, second, etc derivatives o a polynomial on the roots o the polynomial itsel For complex polynomials one has the Gauß-Lucas theorem which says that the roots o the derivative belong to the convex hull o the roots o the polynomial But in the case o hyperbolic polynomials there are also some properties speciic to this class For instance, one has the ollowing property Property 11 (see [1] or [6]) I one o the roots o a hyperbolic polynomial moves to the right (resp to the let) while the other roots remain ixed, then every root o every derivative o the polynomial moves to the right (resp to the let) or remains ixed The roots o the irst (resp o the second) derivative have a geometric interpretation because they deine the critical (resp the inlection) points I one is interested in these points in the graph o the irst (resp o the second) derivative, then one has to study the third (resp the ourth) derivative, thus one is led in a natural way to the study o the arrangements o the roots o a hyperbolic polynomial and o all its derivatives Notation 1 Denote by x 1 x n the roots o P and by x (k) 1 x (k) n k the ones o P (k) We set x (0) j = x j In the examples we never go beyond degree 5 and to avoid double indices we use also the notation j, s j, t j, l j or the roots respectively o P (1), P (), P (), P (4) The letters are chosen to match irst, second, third and last Deinition 1 The arrangement (or coniguration) deined by the roots o P, P (1),, P (n 1) is the complete system o strict inequalities and equalities that hold or these roots To explicit an arrangement one can write the roots in a chain in which any two consecutive roots are connected with a sign < or = An arrangement is called non-degenerate i there are no equalities between any two o the roots, ie, no equalities o the orm x (j) i = x (r) q or any indices i, j, q, r A partial arrangement is the arrangement o the roots o only part o the derivatives P (k), k = 0, 1,, n 1 The classical Rolle theorem implies that the roots o P and o its derivatives satisy 198

3 the ollowing inequalities: i < j, x (i) k x(j) k x (i) k+j i () One has also the sel-evident condition: ( (i) (x k = x(i+1) k ) or (x (i) k+1 = x(i+1) k ) ) (x (i) k = x(i+1) k = x (i) k+1 ) (4) Remark 14 In what ollows when speaking about root arrangements we always assume that they satisy conditions () and (4) The Rolle theorem provides only necessary conditions, ie, not all arrangements are realizable by the roots o hyperbolic polynomials and o their derivatives In previous papers (see [,, 7]) we ask the question: Which o the arrangements o the n(n + 1)/ real numbers x (k) j, k = 0,, n 1, j = 1,, n k, satisying conditions () and (4), can be realized by the roots o hyperbolic polynomials o degree n and o their derivatives? For n = 1,, this is the case o all arrangements (degenerate or not) For n 4 this is no longer like this Eg, or n = 4 only 10 out o 1 non-degenerate arrangements are realizable by hyperbolic polynomials (see [1] or []; this act is closely related to Property 11); or n = 5 these numbers equal respectively 116 and 86 (see []) Remark 15 For any n the number N(n) o non-degenerate arrangements compatible with () equals (see [9]) ( ) n + 1 1!! (n 1)! N(n) =! 1!! (n 1)! An obvious reason why this proportion o realizability is to drop urther when n increases is the lack o dimension a root arrangement o a hyperbolic polynomial and its derivatives is deined by n coeicients (By a shit o the origin o the x-axis one can transorm a 1 into 0; i ater this one has a 0, then a subsequent change o the scope o the x-axis transorms a into 1) while the total number o roots is n(n+1)/ (By similar transormations one can obtain the conditions x 1 = 0, x n = 1, ie, one can again kill two parameters) However, the Rolle theorem is ormulated not only or hyperbolic polynomials, but or smooth unctions Thereore one can introduce the ollowing generalization o a hyperbolic polynomial in the tentative to realize all non-degenerate arrangements Deinition 16 A polynomial-like unction (PLF) o degree n is a C -smooth unction whose n-th derivative vanishes nowhere (Hence, a PLF has at most n real roots counted with the multiplicities) A PLF o degree n is called (strictly) hyperbolic i it has exactly n real (and distinct) roots In what ollows all PLFs are presumed hyperbolic 199

4 Example 17 Prove that the unction (x) := e x x4 4 x 4 x x 1 is a hyperbolic PLF o degree 5 Indeed, one has (5) = e x which vanishes nowhere, hence, is a PLF o degree 5 Moreover, one has (0) = (1) (0) = () (0) = 0, () (0) = 1, lim x ± (x) = ±, ie, has a triple root at 0, a simple positive and a simple negative root Hence, is hyperbolic I ε > 0 is small enough, then the unction + εx is a strictly hyperbolic PLF o degree 5 (The triple root at 0 splits into three simple roots) In paper [4] we show that that or n = 4 PLFs realize all arrangements, and that one can choose these PLFs to be either hyperbolic polynomials o degree 4 or non-hyperbolic polynomials o degree 6 In particular, the two non-degenerate arrangements not realizable by hyperbolic polynomials are realizable by perturbations o such; these perturbations are polynomials o degree 6 Remark 18 Suppose that a strictly hyperbolic PLF and its derivatives realize a given non-degenerate arrangement One can approximate (n) by polynomials and keep the same constants o integration thus obtaining polynomials realizing the same arrangement Thereore all non-degenerate arrangements realizable by PLFs are realizable by polynomials as well (but not necessarily hyperbolic) In paper [5] we show that or n = 5 there are non-degenerate arrangements which are not realizable by PLFs As PLFs belong to an ininite-dimensional space, dimension is not the only obstacle towards realizability o arrangements by hyperbolic polynomials (or by PLFs) In the present paper we continue the study o the question or n = 5 which arrangements are realizable by PLFs The partial arrangements o the roots o the irst and third derivatives o a PLF deine our possible cases two o which are symmetric, see Subsection 1 The present paper oers the thorough study o one o the other two cases In paper [8] PLFs o degree are considered (The authors call them pseudopolynomials) and necessary and suicient conditions are given or the numbers x 1 < x < x, y 1 < y, and z 1 to be roots respectively o a PLF and o its irst and second derivatives In the present paper we use some o the ideas rom [8] 1 Coniguration vectors We use coniguration vectors (CVs) to deine arrangements On a CV the positions o the roots o P, P (1), P (), P (), P (4) are denoted by 0,, s, t, l We use two ways to present a CV on a line and by a partially illed matrix When the presentation is on a line, coinciding roots (i any) are put in square brackets Eg, or n = 5 the CV (compatible with () and (4)) indicates that one has ([00], s,, t, l, 0, s,, t, s, [00]) x 1 = 1 = x < s 1 < < t 1 < l 1 < x < s < < t < s < x 4 = 4 = x 5 00

5 We present by matrices only non-degenerate arrangements To present several arrangements at once we use the sign with the meaning that whenever two or three roots are surrounded by such signs, then any permutation o the surrounded roots is allowed Eg, the matrix l s t t s s (5) denotes each o the six CVs (0,, 0, s,, t, P, t,, s, 0,, 0) where P is any o the six permutations o 0, s and l We use also partial CVs when we need to denote the partial arrangement (presumed non-degenerate) deined by the relative position o the roots o only two o the derivatives Eg, the notation (tt) means that one has 1 < < t 1 < t < < 4 By {(tt)} we denote the subset o all non-degenerate arrangements or which the last chain o inequalities holds In what ollows we identiy or convenience arrangements with the CVs deining them (represented on a line or by a matrix) The ollowing lemma can be proved by ull analogy with Lemma 4 rom [7]: (The latter is ormulated or hyperbolic polynomials, not or PLFs) Lemma 19 A root o multiplicity m o a PLF o degree n, 0 m < i + 1 n, is at most a simple root o (i) Deinition 110 A degenerate arrangement (V ) is adjacent to the arrangement (W ) i (W ) is obtained rom (V ) by replacing one or several equalities between roots by strict inequalities Example 111 For n = 4 the arrangement ([00], s,, [t0], s,, 0) is adjacent to and only to the ollowing ive arrangements: (0,, 0, s,, [t0], s,, 0), ([00], s,, t, 0, s,, 0), ([00], s,, 0, t, s,, 0), (0,, 0, s,, t, 0, s,, 0), and (0,, 0, s,, 0, t, s,, 0) Only the last two o them are non-degenerate Proposition 11 I a degenerate arrangement (V ) is realizable by a PLF o degree n, then all arrangements to which (V ) is adjacent and with the same multiplicities o the roots o the PLF are realizable by PLFs which are perturbations o Proo 1 o It suices to prove the proposition or the arrangements obtained rom (V ) by replacing just one equality between roots by an inequality < or > Observe irst that by Lemma 19 there are cases in which this cannot be any equality Indeed, 01

6 is a root o o multiplicity m >, and i in this chain o equalities one wants to replace only one equality by an inequality, then this can be only the last one o Suppose that the arrangement (V ) contains the ollowing chain o equalities: i x (0) i 0 x (j1) i 1 = = x (m 1) i m 1 = = x (jq) i q (The indices j ν need not be consecutive integers) Suppose without loss o generality that x (j1) i 1 x (js) i s = x (js+1) i s+1 by the inequality x (js) i s = 0 and that one wants to change the equality > x (js+1) i s+1 or x (js) i s < x (js+1) i s+1 The new arrangement thus obtained is denoted by (W ) o Denote by χ a germ o a C -unction at 0 such that χ (r) (0) = 0 or r = 0, 1,, j s+1 1, j s+1 + 1, j s+1 +,, n, χ (js+1) (0) = 1 Denote by ψ a C -unction with compact support such that ψ 0, ψ(x) 1 or x η and ψ(x) 0 or x η where η > 0 Then or η small enough the support o ψ contains no zeros o, (1),, (n 1) other than 0 Hence, the perturbation realizing the arrangement (W ) can be chosen o the orm + εψχ where ε R is small enough and the choice o the sign o ε results in the choice o the sign o the inequality (< or >) 1 Aim, scope and basic results o the present paper In what ollows we ocus on non-degenerate arrangements The ollowing our partial arrangements are possible between the roots o P (1) and P () where P is a PLF: (tt), (tt), (tt), (tt) (6) The present paper is devoted to the irst o these cases Remarks 11 (i) It would be hard to imagine an entire study o the case n 6 due to N(6) = 59 (see Remark 15) unless some general rules and theorems about realizability o arrangements are proved On the other hand, the case n = 4 is thoroughly studied and there are only 1 non-degenerate arrangements there Thereore the case n = 5 is the irst truly interesting case to study To subdivide the study into several cases is reasonable because o the great number (namely, 86) o arrangements, see Remark 15 (ii) The cases (tt) and (tt) can be studied by analogy (the symmetry between these two cases is deined by the change o variable x x) This change o variable allows one to study in the cases (tt) and (tt) only the non-degenerate arrangements with l 1 < x or with x < l 1 This is what we oten do in the paper We say that two arrangements are symmetric (to one another) i the symmetry is induced by the change x x Example: or n = 4 such are the arrangements (0,, s, 0, t,, 0, s,, 0) and (0,, s, 0,, t, 0, s,, 0) (iii) For each o the our cases (see (6)) the number o all non-degenerate arrangements and the number o the ones realizable by hyperbolic polynomials are given in the ollowing table (see [, Observations 4, 5 and Lemmas 40 4]): 0

7 T b S D F C a O A B Figure 1: The hyperbolicity domain in degree 5 Case All arrangements Realizable by hyperbolic polynomials (tt) 40 0 (tt) or (tt) 7 5 (tt) (iv) Our interest in the irst o the cases is motivated by the absence o non-degenerate arrangements realizable by hyperbolic polynomials, see the table (so one expects to ind many examples o non-degenerate arrangements not realizable by PLFs o degree 5) Geometrically this can be explained by ig 1 On the igure we show the hyperbolicity domain Π o the amily o polynomials P = x 5 x +ax +bx+c, ie, the set o values o (a, b, c) or which the polynomial is hyperbolic (The reader will ind more details about Π in []) The axis Oc is vertical, ie, perpendicular to the plane o the sheet The hyperbolicity domain is the curvilinear tetrahedron ABCD (The concavity o the aces is everywhere towards its interior) The set D(1, ) o values o (a, b, c) or which the derivatives P (1) and P () have a common root (we call such sets discriminant sets) is the union o two vertical planes (ie, parallel to Oc) represented on the igure by the lines BT and AS They intersect along a vertical line which has a single point in common with Π the point F Such a situation might appear to be quite non-generic This impression is alse Indeed, discriminant sets are deined by algebraic equations whose coeicients are relatively small integers, so it makes no sense to speak about genericity The two planes rom the set D(1, ) divide the space into our sectors These are the our sets deined by the partial arrangements (6) The case (tt) is the 0

8 sector above (which intersects with Π only at the point F ), the case (tt) is the sector below The basic result o the present paper is the ollowing Theorem 114 (i) In the case (tt) exactly 16 out o the 40 non-degenerate arrangements are realizable by PLFs o degree 5 From them exactly 6 arrangements (represented by matrix (5)) are realizable by perturbations o hyperbolic polynomials The other 10 are the, 1, arrangements represented respectively by matrices (7), (11), (1), and their symmetric ones (ii) The remaining 4 arrangements (which are not realizable by PLFs o degree 5) are the 8,, and arrangements represented respectively by matrices (15), (16), and (17), and their symmetric ones Proo o Theorem 114 The proo o Theorem 114 occupies almost the whole o the rest o the paper In subsection 1 we prove the realizability by the roots o PLFs and their derivatives o the 16 arrangements mentioned in part 1) o the theorem Lemma 1 claims that exactly 6 o them are realizable by perturbations o hyperbolic polynomials Lemmas, 6 and 8 claim the realizability respectively o 4,, and 4 o the remaining 10 arrangements by PLFs (which, by Lemma 1, are not perturbations o hyperbolic polynomials) In subsection we prove that the 4 arrangements rom part (ii) o Theorem 114 are not realizable by the roots o PLFs and their derivatives Matrices (15) and (16) at the beginning o subsection represent respectively 8 and non-degenerate arrangements These 10 arrangements and their symmetric ones (hence, 0 arrangements altogether) are not realizable by the roots o PLFs and their derivatives; this ollows rom the results in [5] The non-realizability o the remaining 4 arrangements is claimed by Lemma 10 The proo o that lemma is long Thereore it is subdivided into several parts and is preceded by a plan 1 Proo o the existence part Lemma 1 (i) There are exactly 4 non-degenerate arrangements to which the ollowing arrangement is adjacent: (A) : ([00], s, [t], [0sl], [t], s, [00]) They are all realizable by perturbations o hyperbolic polynomials (ii) Exactly six out o these 4 non-degenerate arrangements belong to the set {(tt)} These six arrangements are deined by matrix (5) 04

9 (iii) Out o the remaining 18 arrangements, exactly 8 are not realizable by hyperbolic polynomials, see [, Observation ] Proo Part (ii) o the lemma is to be checked straightorwardly, and part (iii) needs no proo Prove part (i) 1 o The non-degenerate arrangements to which (A) is adjacent are obtained by replacing each o the two groups [00] by 0,, 0, o each o the two groups [t] either by, t or by t,, (This gives our possibilities) and o [0sl] by one o the six permutations o 0, s and l This gives 4 non-degenerate arrangements o Consider the polynomial P := x 5 x + x 4 = x(x 1 ) It realizes arrangement (A) and deines the point F on ig 1 For the roots o P and its derivatives one has x 1, = 1 = 1, x = s = l 1 = 0,, = t 1, = ± 1 10 x 4,5 = 4 = 1, s 1, = ± 10 It ollows rom Proposition 11 that all arrangements satisying the ollowing three conditions are realizable: (a) arrangement (A) is adjacent to them; (b) the groups [t] and [0sl] in arrangement (A) are replaced respectively by (, t) or (t, ) (independently or each o the two groups [t]) and by any o the 6 permutations o 0, s, l; (c) the groups [0 0] rom arrangement (A) remain the same To prove the realizability o the 4 arrangements rom the lemma one has ater this to make the double roots x 1 = x and x 4 = x 5 biurcate This can be done by adding an aine unction ε(x x ), where ε > 0 is small enough Remark In what ollows we oten draw the graphs o a unction (in most cases this is a PLF) and o its derivatives one upon the other On the igures we use the notation x j (and similarly j, s j, t j ) both in the sense the roots o the unction g and in the sense the point with coordinate x j on the x-axis in the picture representing the graph o g This might sometimes oblige the reader when looking at a igure to search dierent roots, say, and s, on dierent x-axes the ones o g (1) and o g () 05

10 t 1 t s 1 U s s 1 R V 4 Q T x 1 x x x 4 x 5 Figure : Construction o the PLF realizing arrangement (7) Lemma The ollowing two non-degenerate arrangements and their symmetric ones are realizable by PLFs o degree 5: l s t s t s (7) Proo 1 o To construct PLFs which realize the two arrangements rom matrix (7) we irst construct an odd C -smooth unction g which realizes the arrangement (0,, [0s], [t], [0sl], [t], [0s],, 0) Properly speaking, the presence o the letter l is not justiied here given that the unction is only C -smooth We set l 1 = 0, the sense o this equality will become at least partially clear in o and completely clear in 6 o The graphs o g, g (1), g (), and g () are drawn one above the other on ig The x-axis or each o the our graphs are the solid lines o One has g (4) (x) 1 or x < 0 and g(4) (x) 1 or x > 0, (This explains why we set l 1 = 0) and one can think o g (5) as o δ(x) (the delta-unction) The graph o g () is piecewise-linear and symmetric wrt the line s R, the one o g () 06

11 consists o parts o two parabolas The axes o symmetry o the entire parabolas are the lines t 1 U and t V, the point s is the center o symmetry o the graph o g () The graph o g (1) consists o the graphs o two cubic unctions deined respectively or x 0 and x 0 When these two unctions are deined on R, then their graphs have respectively the points and as centers o symmetry The graph o g (1) has the line s Rx as axis o symmetry o One can think o g as o the result o a ive-old integration o δ(x) with suitable constants c 1,, c 5 o integration Decrease c 5 so that the graph o g still cut the x-axis in ive distinct points (The graphs o g (4), g (), g (), and g (1) do not change) The new position o the x-axis is the lowest dash-dotted line on ig Ater this change o constant one has s 1 > x, x > s, s > x 4 (8) and s = l 1 = 0 4 o Ater this increase c 4 (without changing c 5 ) This increasing must be much smaller than the decreasing o c 5 because it changes the graph o g; i it is small enough, then conditions (8) are preserved and one has < t 1, t < (9) 5 o Ater this increase (resp decrease) c without changing c 4 and c 5 I this change o c is small enough, then conditions (8) and (9) are preserved and one has l 1 < s < x (resp s < l 1 < x ) (10) 6 o Change g (5) rom δ(x) to an even C -smooth unction h with compact support [ η, η] where η is so small that no root o g, g (1), g (), g () belongs to [ η, η], and one has h > 0 or x ( η, η), η h(x)dx = 1 I η is small enough, then conditions η (8), (9), and (10) still hold and g is a C -smooth unction Moreover, l 1 = 0 is the only zero o g (4) The unction g is still not a PLF because g (5) vanishes outside ( η, η) To make it a PLF one has to change g (5) rom h to h + b where b > 0 is so small that conditions (8), (9), and (10) still hold This means that the unction g thus constructed is a PLF and realizes one o the arrangements rom matrix (7); which one exactly depends on whether one increases or decreases c in 5 o Their symmetric arrangements are realized by the unction g( x) Remark 4 I one decreases c 4 in 4 0 instead o increasing it, then in the same way one can construct a PLF realizing one o the arrangements (0,, 0, s, t,, l, s, 0,, t, 0, s,, 0) and (0,, 0, s, t,, s, l, 0,, t, 0, s,, 0) and their symmetric ones; these our arrangements belong not to {(tt))}, but to {(tt)} 07

12 t 1 t s 1 U K s B L s D M 1 S V H A P I 4 N G W C E J x 1 x x x x 4 5 Z Y Figure : Construction o the PLF realizing arrangement (11) Notation 5 We use the notation S( ) or the area o a curvilinear igure and S( ) or the area o a (rectilinear) polygon We use dierent notation or the same thing in order to distinguish rectilinear rom curvilinear polygons We denote the igures whose area is expressed by closed contours Example: S(ABCA) denotes the area o the curvilinear triangle ABC Lemma 6 The non-degenerate arrangement represented by the ollowing matrix and its symmetric one are realizable by PLFs o degree 5: l t t s s s Proo 1 o The proo ollows the same ideas as the ones o Lemma, this is why ig which illustrates the construction is so much like ig Not completely, though We start by deining the unction g and the constants c j in the same way as in 1 o o o the proo o Lemma So the initial graphs o the unctions g, g (1), g (), g () are given on ig o Decrease the constant c The change is presumed to be small The new position o the x-axis is the dash-dotted line on ig The point D (and not the root s ) is the center o symmetry o the graph o g () (11) 08

13 o Choose the new value o the constant c 4 such that = t This means that it is close to the old one Hence, the graph o the new unction g (1) looks like on ig Notice that one has indeed < t 1 This is so because when a point ollows the graph o g (1) moving to the let or right, its initial position being the point H, then it descends to 0 which is due to the integration o g () When moving to the right, the point descends to 0 at, so one has HG = S(s DV LBs ) One has also S(s DV LBs ) > S(s UKs ) (remember that the point D is the center o symmetry o the graph o g () ) Hence, when the point is moving to the let, it will go beyond the vertical line t 1 UKP NY beore descending to 0 4 o Choose the new constant c 5 such that x = 0 Hence, when c is changed little enough, the new constant c 5 is close to the old one and the graph o g looks like on ig Notice that one has x 4 < s and x < s 1 The irst o these inequalities ollows rom S(V Ms LV ) > S(V DBLV ) (remember that the line t LV is the axis o symmetry o the parabola MV D; hence, is the center o symmetry o the curve A J); this in turn implies S( JI ) > S( AW ) Hence, the area S( AW ) = x g (1) (x) dx is suicient to make the point E rom the graph o g go down to 0 at x when moving to the let, and the area S( JI ) = s g (1) (x) dx makes this point descend below 0 when moving to the right The inequality x < s 1 ollows rom S(s 1 KUs 1 ) = S(s KUs ) (remember that the line t 1 UK is the axis o symmetry o the parabola DUs 1 ), hence, S( SC ) < S(NW AP N) and the point Y rom the graph o g will go aster to 0 when moving to the right than when moving to the let In act, when moving to the let it will irst descend (till ) and only then go up 5 o Ater this increase the constant c 5 so that x become < 0 The increasing can be chosen so small that all strict inequalities between roots o g and its derivatives be preserved Then increase c 4 so that become greater than t and all strict inequalities between roots o g and its derivatives be preserved Finally, repeat the reasoning rom 6 o o the proo o Lemma (ie, change g (5) rom δ(x) to h(x) and then to h(x) + b) The unction g(x) thus obtained is a PLF and realizes the arrangement rom matrix (11) Its symmetric arrangement is realized by the unction g( x) Remark 7 I in 5 0 o the proo one decreases c 5 instead o increasing it, and i ater this one decreases c 4, then one similarly realizes the arrangement (0,, 0, s,, t, s, l, 0,, t, 0, s,, 0) {(tt)} by a PLF g(x), and by g( x) its symmetric one Lemma 8 (i) The two non-degenerate arrangements deined by matrix (1) 09

14 (hence, their symmetric ones as well) are realizable by PLFs o degree 5: l t t s s s (1) (ii) The two arrangements rom matrix (1) dier rom the two arrangements rom matrix (17) (not realizable by PLFs) only by the sign o the inequality between l 1 and s which is an inequality between roots o derivatives o highest possible orders Proo 1 o Part (ii) o the lemma is sel-evident, so we prove only part (i) o it For each o the arrangements ([00], s, [t], l, s, [t0],, P,, 0) where P = (0, s) or P = (s, 0), we construct C -smooth unctions g which realize them (In principle, one should not put the letter l in the CV given that the unctions are only C -smooth; however, in the subsequent approximation o g by a PLF the root l 1 will be in the same place as indicated in the CV) The unction g (4) is piecewise-constant, it equals 4 on some interval (, τ) and 4a, a > 0 on (τ, ) (In the PLF approximating g the root l 1 will be close to τ) On (, τ) one has g(x) = P (x) := x 4 + x 1 4 Notice that the polynomial P (x) realizes the arrangement ([00], s, [t], s, [00]) We choose τ to be bigger than s = 1 6 On (τ, ) one has g(x) = a(x τ) 4 + P () (τ) 6 (x τ) + P () (τ) (x τ) + P (1) (τ)(x τ) + P (τ) (1) Observe irst that the unction g possesses the ollowing properties: (a) I one sets τ = 1 and g(x) = P 1(x) := x 4 8x + 7x x or x τ, then such a unction g realizes the arrangement ([00], s, [t], s, [0t],, s, [00]) This can be checked directly the roots o P 1 (resp P () 1 ) equal 0,, 1, 1 (resp ); one has P (i) ( 1 ) = P (i) 1 ( 1 ), i = 0, 1,, (b) The unctions g which we construct below are obtained by increasing τ (τ ( 1, )) and by keeping the same root t = The restriction o g to (τ, ) is a polynomial P k, k = or, o degree 4, whose roots we denote by x, x, x 4, x 5 (and not by x 1, x, x, x 4 ) in order to have the same notation or the roots o g or o P k ; we make a similar shit by 1 o the indices o the roots o the derivatives o P k We list all the roots o P k and its derivatives, but the ones smaller than τ are o no importance or the arrangement realized by the roots o g and its derivatives because or x τ one has g(x) = P (x) 10

15 (c) The condition t = implies that one has a = P () (τ) 4( τ) = τ τ (14) When τ is close to 1 one has s < x 4 (see the polynomial P ), when it becomes big enough, then one has s > x 4 (see the polynomial P ) o For τ = The roots o P equal and or x one has a = and g(x) = P (x) := 15 ( x 51 ) 4 51 ( x ( x 51 ) 149 ( + x ) ) = x x x x x = , x = , x 4 = , x 5 = , the ones o P () equal s = , s = (and one has t = < x ) Thus the unction g deined or τ = realizes the arrangement o For τ = 6 10 ([00], s, [t], s, l, t, 0,, s, 0,, 0) 6 and or x 10 ( x 6 g(x) = P (x) := 9 The roots o P equal ( x 6 10 one has a = 9 and ) 4 1 ( x ) = 9x 4 4x x 16 5 x ) 9 ( x 6 ) 5 10 x = , x = , x 4 = , x 5 = , the ones o P () equal s = , s = One has again t = < x The unction g deined or τ = 6 10 realizes the arrangement ([00], s, [t], s, l, t, 0,, 0, s,, 0) 11

16 4 o Add small positive constants to the unctions g constructed or τ = and τ = 6 10 so that the double root x 1 = x split into two simple roots while preserving all strict inequalities between roots Ater this add to the thus modiied unctions g aine unctions ε(x ) where ε > 0 is small enough (to preserve all existing strict inequalities between roots) Ater this last modiication the unction g (denoted rom now on by g) with τ = (resp τ = 6 10 ) realizes the arrangement (Ω) : (0,, 0, s,, t, s, l, t, 0,, s, 0,, 0) (resp (Ξ) : (0,, 0, s,, t, s, l, t, 0,, 0, s,, 0)) 5 o One can smoothen the unctions g () in a neighborhood o τ to make them C -smooth and so that g (4) be increasing The new unctions are denoted by g The smoothening can make the unctions g as close to the unctions g as to preserve all strict inequalities between roots Hence, the unctions g realize the same arrangements as the unctions g and the root l 1 o g (4) appears in the CV indeed between s and t There remains to change g into a PLF by adding a small positive constant to g (5) (Remember that g (5) is with compact support; it is greater than 0 because g (4) is increasing) When this constant is small enough the unctions g are PLFs and realize arrangements (Ω) and (Ξ) Proo o the non-existence part It ollows rom the results in [5] that all non-degenerate arrangements with s 1 < x, < t 1, s < x, or with x 4 < s, t <, x < s, are not realizable by PLFs o degree 5 This makes 46 arrangements Out o them exactly 0 belong to the case (tt) (the reader will easily check this onesel) Up to the symmetry induced by the change x x (see Remarks 11) these are the arrangements deined by the ollowing two matrices: l s t s t s (15) This matrix deines eight arrangements (there are three couples where one can choose each o the two permutations) The ollowing matrix deines two more ar- 1

17 rangements: t l t s s s (16) Remark 9 Matrix (15) implies the non-realizability o the ollowing partial arrangements o the roots o a PLF o degree 5 and its irst three (and not our) derivatives: (0,, s, 0,, t, s, t, 0,, P,, 0) where P = (s, 0) or (0, s) Indeed, in the absence o equalities between roots these partial arrangements allow only two possibilities or l 1 : s < l 1 < t and t 1 < l 1 < s For each o them the corresponding (complete) arrangements are deined by matrix (15), hence, are not realizable by PLFs On the other hand, all non-degenerate partial arrangements o the roots o a PLF (even better o a hyperbolic polynomial) and its irst two derivatives are realizable This ollows rom [7, Theorem ] it suices to realize the partial arrangement o the roots o the polynomial and its second derivative which deines the relative positions o the roots o its irst derivative Lemma 10 The two non-degenerate arrangements represented by matrix (17) and their symmetric ones are not realizable by PLFs o degree 5 t l t s s s 0 0 (17) Proo I) Plan o the proo Deinition 11 We deine the set W as the one consisting o the two non-degenerate arrangements deined by matrix (17) and o the degenerate arrangement obtained rom anyone o them by replacing the inequality s < x 4 or s > x 4 by s = x 4 We irst change a PLF g supposed to realize an arrangement rom the set W to another unction g with simpler graph o g () which also realizes such an arrangement, see II) The unction g is not a PLF but can be approximated by PLFs which also realize arrangements rom the set W (In act, one can perturb them so that they 1

18 realize non-degenerate arrangements rom W ) Ater this in III) we prove that it is impossible to realize an arrangement rom W by the unction g The change o g into g is done in two steps, see IIA), IIB) At these steps we modiy the graph o g respectively on (, t 1 ] and [t, ) The unctions thus obtained are denoted by g 1 and g We include them into a homotopy g τ, τ [0, ]; we set g 0 = g (When g j and g j 1 are deined, one sets g τ = (τ j + 1)g j + (j τ)g j 1 or τ [j 1, j]) II) Replacing the PLF realizing a given arrangement by another unction IIA) Modiication o the graph on (, t 1 ] 1 o Suppose that at least one o the arrangements o the set W is realizable by a PLF g The graphs o g, g (1), g () and g () are shown on ig 4, one above the other Deinition 1 An almost PLF (APLF) o degree n is a unction g : R R or which g (n) 0 and g (n) is piecewise-smooth, with a inite number o discontinuities at which there exist the let and right limits I an APLF g has n real zeros (counted with the multiplicities), then it is called hyperbolic In what ollows all APLFs are presumed hyperbolic Remark 1 It is clear that i a non-degenerate arrangement is realizable by an APLF, then it is realizable by a PLF as well (one has to approximate the n-th derivative by a smooth positive-valued unction; i the approximation is good enough, then all strict inequalities between roots are preserved) We use APLFs in situations when they yield easier estimations o Consider a PLF g realizing an arrangement rom the set W as a ive-old integral o g (5) Integration is perormed rom a ixed point rom the interval (t 1, t ) Replace g by an APLF g 1 (with the same constants o integration) or which g (5) 1 g (5) or x t 1 and g (5) 1 0 or x < t 1 This means that g (4) 1 is constant or x < t 1 ; to obtain the graph o g () 1 or x < t 1 one has to replace the one o g () by the tangent to this graph at (t 1, 0), see ig 4 Statement 14 The APLF g 1 and the PLF g realize the same arrangement Proo o Statement 14 Consider the homotopy g τ := (1 τ)g + τg 1 deined ater (1 τ)g (5) + τg (5) 1 with the same constants o integration or all τ [0, 1] For each τ this is an APLF (and a PLF or τ = 0) For all τ [0, 1] the unction g τ () has exactly one zero in (, t 1 ) because one is integrating the positive-valued unction g τ () rom t 1 to x, x < t 1, and g τ () (t 1 ) > 0 is ixed When τ increases rom 0 to 1, then the root s 1 moves to the let, and or each a (, t 1 ) ixed the value o g τ () increases with τ Hence, the root moves to 14

19 A B E J I t 1 X Z N M t s 1 I J L X K Z s F U s V R W 1 A B N M Q Q S P P 4 T G H x 4 x 1 x Y x x 5 C Figure 4: The graphs o g, g (1), g (), and g () 15

20 the right (without reaching t 1 because g τ (1) (t 1 ) > 0 does not depend on τ) For each a (, t 1 ) ixed the value o g τ (1) (a) decreases For each τ ixed one has g τ (1) (x) when x Hence, g τ (1) has a root 1 <, and it is clear that it can have no roots other than 1,,, 4 In the same way one shows that the root x o g τ moves to the right when τ increases (without reaching t 1 because g τ (t 1 ) < 0 is ixed) As g τ (x) when x, the unction g τ has a root x 1 < x Hence, it is a hyperbolic APLF The arrangement realized by g τ may change only i or some τ one has x = s 1 (because t 1 and all roots to the right o t 1 do not change their positions) The ollowing statement implies that this doesn t happen Statement 15 For each τ [0, 1] one has (x) dx = S( AB ) < S( MN ) = and g τ (1) s 1 s g τ (1) (x) dx (18) AB < MN (19) The statement implies that one cannot have g τ (s 1 ) > g τ (t ) because g τ (1) s 1 s (x) dx < g τ (1) t (x) dx < g τ (1) (x) dx, hence, g τ (s 1 ) < g τ (s ) < g τ (t ) (0) However, i one has x = s 1 or some τ, then one must have g τ (s 1 ) = 0 and g τ (t ) < 0 a contradiction with (0) Hence, the arrangement realized by g τ does not change throughout the homotopy Proo o Statement 15 Consider two points I, Z such that I (s 1, K), Z (K, s ), IK = KZ (see the graph o g () on ig 4) One has IJ < ZX Indeed, consider g () as a primitive o g () The graph o g () is convex and lies above its tangent at (t 1, 0) Hence, i a point ollows the graph o g () starting at L, then it descends aster when it is moving to the let than to the right because one has t1 KL IJ = g () (x) dx = S(t 1J I t 1 ) > S(t 1 X Z t 1 ) I X = g () (x) dx = KL ZX t 1 This implies that s 1 K < Ks In the same way one shows that when a point ollows the graph o g (1) starting at, then it climbs aster to the right than it descends to the let Moreover, one has N > A This proves condition (18) and (having in mind that s 1 < s 1 t 1 = s 1 K < Ks = t 1 s < s ) also condition (19) 16

21 Remark 16 One can prove in a similar way (using the concavity o g () on (, t 1 ]) that one has s 1 1 < s 1, and, hence, s 1 1 < t 1 s 1 < s t 1 IIB) Modiication o the graph on [t, ) o Replace the APLF g 1 by another APLF g this time modiying the graph o g 1 to the right o t Namely, we set g (5) (x) = 0 or x > t, g (5) (x) = g(5) 1 (x) or x t, and we keep the same constants o integration when deining g (j), j = 0,, 4 Hence, g (4) is constant or x t and to obtain the graph o g () rom the one o g () 1 one has to replace this graph to the right o t by the tangent line at (t, 0) to the graph o g () 1, see ig 4 Statement 17 The unction g is a hyperbolic APLF Proo o Statement 17 We ollow the same ideas as in the proo o Statement 14 First o all, include g in the homotopy g τ (see I)), ie, set g τ = ( τ)g 1 + (τ 1)g, τ [1, ] For τ close to 1 the APLFs g 1 and g τ deine the same arrangement The ollowing property is evident: Property 18 When τ increases rom 1 to, then or each a > t ixed, the values o g τ (j) (a) (j = 0,, 4) decrease Hence, the roots s, 4, and x move to the right while the root moves to the let (without attaining t because g τ (1) (t ) > 0 remains ixed) For x > t, g τ () (resp g τ (1) ) is a polynomial o x o degree (resp ) with positive leading coeicient, and one has g τ () (x) > 0 or x > t This implies that g τ () has exactly one root (namely, s ) or x > t, and g τ (1) has exactly two roots (namely, and 4 ) there To prove that g is a hyperbolic APLF it suices to prove the ollowing Statement 19 For no τ [1, ] does one have g τ ( ) g τ ( 1 ) Indeed, it ollows rom Property 18 that g τ ( ) decreases when τ increases to obtain g τ ( ) one integrates the unction g τ (1) (x) (positive-valued on (t, ) and whose value as a unction o τ is decreasing or each x > t ixed) along an interval whose length decreases with τ On the other hand, the dierence g τ ( ) g τ ( 4 ) increases with τ due to Property 18 Thereore one also has g τ ( 4 ) < 0 which means that g τ has a root x 4 (, 4 ) As g τ has at least our roots x 1,, x 4, then it has also a ith one x 5 > 4, (For x it behaves like bx 4, b > 0) hence, g τ is a hyperbolic APLF Remarks 0 (i) In Statement 17 we do not claim that the APLFs g and g 1 (hence, the PLF g as well) realize the same arrangement This is so because we do not take into account the relative position o s and x 4 However, i this 17

22 position changes (rom x 4 < s to x 4 > s or vice versa), then the arrangement changes to another arrangement rom the set W (ii) One need not consider separately the particular situation when x 4 = s (all other strict inequalities between roots being preserved) because one can approximate the APLF g by a PLF so that this equality changes to x 4 < s or to x 4 > s (iii) To prove that g realizes one o the arrangements rom the set W it is suicient to show that g is a hyperbolic APLF because the relative position o t and o the roots to the let o t are the same or g 1 and g Proo o Statement 19 One has g τ ( 1 ) g τ ( ) = (g τ ( 1 ) g τ ( )) (g τ ( ) g τ ( )) = (x) dx τ (x) dx g τ (1) 1 g (1) = S( 1 B A 1 ) S( NP QM ) We show that S( 1 B A 1 ) < S( NP QM ) which implies the statement The last inequality ollows rom inequality (18) and rom S( 1 BA 1 ) S(NP QMN) = which we prove now One has because s1 g τ (1) 1 (x) dx g τ (1) s (x) dx < 0 (1) g () τ (s ) = N M B A = g () τ (s 1 ) () S(A B t 1 A ) = B A s 1 K S(t 1 N M t 1 ) N M Ks and s 1 K Ks (see Remark 16) The absolute value o g τ () grows to the let o s 1 not slowlier than to the right o s on (s, t ) (This ollows rom the convexity o the graph o g τ () ) Thereore the same statement holds or the absolute value o g τ () This in turn means that when a point is ollowing the graph o g τ (1), then it climbs aster on (s 1 1 ) than it descends on (s, ) As one has also (19), this implies inequality (1) III) End o the proo 4 o Denote by s 1 V the tangent line at (s 1, 0) to the graph o g () and by s S the one at (s, 0) (in dash-dotted lines on ig 4) Denote by BR and MQ P the graphs 18

23 o g (1) when its derivative g () is replaced by the aine unction whose graph is one o these tangent lines We keep the constant o integration the same; integration starts at s 1 or s Remark 1 We use the act that l 1 s as ollows: the curve s U, see ig 4, lies above the tangent line s S or x (s, t ] (For l 1 > s this is not true) This implies that the curve MQ P (which is a parabola) lies below the curve MQ Statement I there exists an APLF g realizing an arrangement rom the set W, then there exists an APLF realizing an arrangement obtained rom one o the set W in which the inequalities < t 1, t < x, x 4 < x 5 are replaced by the corresponding equalities Proo o Statement Consider the one-parameter deormation g (x) + s(x t 1 ), s 0 By abuse o notation we denote it again by g It deines (or all s 0 such that t 1 ) an APLF Indeed, g ( 1 ) increases and g ( ) decreases when s decreases (The positions o 1 and also change) Hence, g has two real roots which are t 1 (It behaves like bx 4, b > 0, or x ) For all s 0 one has g ( ) > g ( 1 ) > 0 (This can be proved by analogy with Statement 19) As g ( 4 ) decreases, g must have three real roots which are t As or x g behaves like bx 4, b > 0, g has exactly ive real roots As g (5) does not depend on s, g is an APLF or all s 0 until one has = t 1 So assume that = t 1 Case 1) Suppose that one has g (t ) g ( 4 ) Set g (x) g (x) g (t ) Hence, the new unction g (x) is also an APLF and one has g (t ) = 0, t = x I one has x 4 = x 5, then there is nothing to do I x 4 < x 5, then observe irst that one has t t 1 < t 1 1 Indeed, deine r R by the condition r t 1 = t 1 s 1 One proves by analogy with Statement 15 that one has r t 1 g (1) (x) dx t1 (x) dx and g (1) (r) g(1) (s 1) = AB () s 1 g (1) The value o the unction g (1) (x) decreases aster when x decreases rom s 1 to 1 than when x grows rom r to t (In act, g (1) (x) increases or x [r, s ], and then decreases or x [s, t ]) This can be proved by analogy with the proo o Statement 15 Hence, i one has t t 1 t 1 1, then one must have g (t ) g (t 1 ) = t t 1 g (1) (x) dx > t1 1 g (1) (x) dx = g ( 1 ) g (t 1 ), ie, g (t ) > g ( 1 ) which is impossible; the inequality in the middle ollows rom () and rom the lines above Thus one has t t 1 < t 1 1 Set g (x) g (x) + u((x t 1 ) (t t 1 ) ), u 0 Hence, g (t 1 ) decreases when u increases, g (t ) does not change while g ( 1 ), g ( ), 19

24 and g ( 4 ) increase This means that there exists u > 0 or which one has x 4 = x 5 (and one has already = t 1, x = t ) Case ) Suppose that one has g (t ) < g ( 4 ) Set g (x) g (x) vq(x), q(x) := (x t 1 ) ( t 1 ), v 0 Notice that q (1) (t 1 ) = 0 When v increases, then and g ( ) remain the same, x and x 4 decrease, x 5, g(t ), and g ( 4 ) increase Prove that g ( 1 ) increases One has s t 1 > t 1 s 1 (see Remark 16) and s > P N > AR A 1 = s 1 1 (This ollows rom the act that the parabolas MP and BR have horizontal tangents at M and B, rom (19) and rom ()) Hence, t 1 = ( s ) + (s t 1 ) > (s 1 1 ) + (t 1 s 1 ) = t 1 1 which means that q( 1 ) < 0 and that g ( 1 ) increases with v When increasing v, i one has g (t ) < g ( 4 ) < 0, then one cannot have x = t 1 or any v > 0 Indeed, as one has g (1) (t 1) = 0 or all v, the equality x = t 1 would imply that x is at least a double root o g which together with x 1, x, x 4 and x 5 makes at least 6 real roots (counted with the multiplicities) a contradiction Hence, one can choose v such that g (t ) = g ( 4 ) Ater this set g (x) g (x) g (t ) Thus one has all three equalities x 4 = x 5, = t 1, x = t Assumption We assume till the end o the proo o Lemma 10 that the APLF g realizes an arrangement satisying the conclusion o Statement Statement 4 One has S(BR 1 AB) = 4 S( 1BA 1 ) Proo o Statement 4 Recall that the graph o g () restricted to (, s 1 ] is an arc o a parabola To ease the computation assume (ater a change o the scopes o the axes) that its equation is y = 1 x Hence, t 1 = 0, s 1 = 1 and 1 = The last equality ollows rom the condition s 1 1 g () (x) dx = s 1 g () (x) dx Ater this one inds that S( AB ) = 5 1, S( 1 BA 1 ) = 1 The equation o the tangent line s 1V is y = x +, the one o the parabola RB is y = x + x + 1, the x-coordinate o the point R equals 1 which implies the equality rom the state- and one inds that S(BR 1 AB) = 4 9 ment Notation 5 On ig 5 we show parts o the graphs o g (), g(), g(1), and g under Assumption The curve UU is the continuation o the arc o parabola s U As the arc N t lies above the tangent line to the graph o g () at (t, 0), the arc UU lies, 0

25 N t N s U U F s U S S M Q M M N Q 4 P P Q Q 0 T x x 4 = x 5 Figure 5: A detail o the graphs o g (), g(), g(1), and g 1

26 above the arc Us ; these two arcs are (horizontally) tangent at U Denote by S U a tangent line to the arc o parabola UU Statement 6 One has S(S U F S ) S(UU F U) Proo o Statement 6 By changing the scopes o the axes one can assume that the equations o the parabola U Us and o its tangent line S U are respectively y = x 1 and y = x 0 x x 0 1 where (x 0, x 0 1) is the point o tangency, x 0 < 0 Hence, S F = x 0 + 1, U F = (x 0 + 1)/ x 0, and S(S U F S ) = φ(s) := (s +1) 4s, s = x 0 When s > 0, the unction φ attains its minimum or s = 1 and this 4 minimum equals One has S(UU F U) = 0 1 (x 1) dx = rom where the statement ollows 5 o Consider g (1) as a primitive o g () when integration starts at t Construct the arc QM when one deines the graph o g () to be not the arc s U but the arc U U Hence, the arc QM lies below the arc QM and is tangent to it at Q One has QQ QM = Q 0 Q The last equality ollows rom the act that the arc (o parabola) U Us is symmetric wrt the vertical line F U (which implies that the arc M QT is symmetric wrt the point Q) Suppose that t = 0 and that the equation o the parabola U Us is y = x 1 Hence, the one o the arc M QT is o the orm y = x x + a, a R Statement 7 One has S(QP Q) = S( T 4 ), a = g (1) (0) = P Q = and P Q 0 = Proo o Statement 7 One has S(QP Q) = 0 g (1) (x) dx = 4 g (1) (x) dx = S( T 4 ), and i g = x4 1 x + ax + b, b R, then one has the ollowing system o equalities: ie, ( ) + a = ( 4) 4 + a = 0, g ( ) g (0) = g ( ) g ( 4 ), ( 4 ) 4 1 ( 4) + a 4 = 0 with unknown variables, 4 and a The last equation combined with the second o the irst two yields 4 = Hence, a = and = 6 One has P Q 0 = g (1) (s ) = g (1) (1) =

27 Remark 8 One has Q Q 4 Indeed, suppose that on ig 5 the tangent line U S to the parabola U Us is parallel to the tangent line s S to the curve s U at s Then one has Q Q = 0 g () (x) dx = S(s SF s ) S(S U F S ) S(UU F U) = 4 s The last inequality ollows rom Statement 6 Statement 9 One has S(MNP QM) > S( 1 BA 1 ) ie, The statement and condition (18) together imply that one has g (t ) g (t 1 ) = S( NP QM ) > S( B 1 A ) = g ( 1 ) g (t 1 ) g (t ) > g ( 1 ) which is a contradiction This contradiction proves the lemma Proo o Statement 9 One has S(MNP QM) S(MNP Q M) and we show that S(MNP Q M) S( 1 BA 1 ) which implies the statement One has S(MNP Q M) S(BR 1 AB); this ollows rom BR and MP being parabolas with horizontal tangents at B and M, and rom conditions (19) and () On the other hand, one has τ := S(MNP Q M) S(MNP Q M) = > 4 = S( 1BA 1 ) S(RBAR) Indeed, the last equality ollows rom Statement 4 The inequality is to be checked directly So there remains to prove only the second equality One has P Q P Q = P Q P Q + Q Q < see Statement 7 and Remark 8 Hence, P Q P Q + Q Q σ := + 4, τ = b 0 A numeric computation yields / 1 (1 x ) dx (1 x b b ) dx = 0 where 1 b = σ σ = , b = , τ =

28 Conclusions The present paper gives the answer to the question which non-degenerate arrangements are realizable by the roots o a PLF o degree 5 and o its derivatives in the case (tt) The author intends to publish two more papers (one treating the cases (tt) and (tt), and one treating the case (tt)), ater which the answer to this question will be known or all non-degenerate arrangements o degree 5 Computations made by the author up to now show that in the other three cases (not covered by the present paper) there appear no non-realizable non-degenerate arrangements dierent rom the ones mentioned in [5] In particular, in the case (tt) all non-degenerate arrangements are realizable This means that in the case o degree 5 exactly 50 out o 86 non-degenerate arrangements are not realizable by the roots o PLFs and o their derivatives Out o these 50 arrangements, 46 are the ones described in paper [5] and 4 are the ones described by Lemma 10 o the present paper The author does not intend to consider the case o degree 6, at least not in detail, because o the enormous number o non-degenerate arrangements (namely, 59; see Remark 15) The non-realizability o some o them ollows rom the results o [5] and o the present paper For instance, i a non-degenerate arrangement (U) o degree 6 is such that the partial arrangement deined by the roots o the irst ive derivatives o the PLF is non-realizable (say, by the results o the present paper), then arrangement (U) is also non-realizable Instead o considering the case o degree 6 in detail it would be more interesting to see whether the ratio non-degenerate arrangements o degree n realizable by the roots o PLFs and o their derivatives /N(n) (see Remark 15 or N(n)) tends to a inite limit or not when n, and whether this limit is 0 or not This has been suggested to the author by V I Arnold Acknowledgements The idea to study root arrangements o hyperbolic PLFs and their derivatives (as well as the deinition o a PLF) was suggested to the author by B Z Shapiro rom the University o Stockholm during the author s visit to his university The author is deeply grateul both to him and to his home institution Reerences [1] B Anderson, Polynomial root dragging, Amer Math Monthly 100 (199), no 9, [] V P Kostov, Discriminant sets o amilies o hyperbolic polynomials o degree 4 and 5, Serdica Math J 8 (00), no, [], Root conigurations or hyperbolic polynomials o degree, 4, and 5, Funktsional Anal i Prilozhen 6 (00), no 4, (Russian); English transl, Funct Anal Appl 6 (00), no 4, [4], On root arrangements o polynomial-like unctions and their derivatives, Serdica Math J 1 (005), no,