On Descartes' rule of signs
Hassen Cheriha, Yousra Gati, Vladimir Petrov Kostov

TL;DR
This paper investigates the realizability of specific sign patterns and root sign distributions in real polynomials, providing new conditions for cases with exactly two sign changes, extending Descartes' rule of signs.
Contribution
It offers new sufficient conditions for the realizability of root sign pairs in polynomials with exactly two sign changes, advancing understanding of Descartes' rule of signs.
Findings
Identifies conditions for the realizability of certain root sign pairs
Extends Descartes' rule of signs to specific cases with two sign changes
Provides criteria for polynomial sign pattern realizability
Abstract
A sequence of signs and beginning with a is called a {\em sign pattern (SP)}. We say that the real polynomial , , defines the SP ,sgn, , sgn. By Descartes' rule of signs, for the quantity of positive (resp. of negative) roots of , one has (resp. ), where and are the numbers of sign changes and sign preservations in ; the numbers and are even. We say that realizes the SP with the pair . For SPs with , we give some sufficient conditions for the (non)realizability of pairs of the form , , , .
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Taxonomy
TopicsMathematics and Applications · History and Theory of Mathematics · Advanced Mathematical Theories and Applications
On Descartes’ rule of signs
Hassen Cheriha, Yousra Gati and Vladimir Petrov Kostov
Université Côte d’Azur, LJAD, France and University of Carthage, EPT - LIM, Tunisia
[email protected], [email protected]
University of Carthage, EPT - LIM, Tunisia
Université Côte d’Azur, LJAD, France
Abstract.
A sequence of signs and beginning with a is called a sign pattern (SP). We say that the real polynomial , , defines the SP ,sgn, , sgn. By Descartes’ rule of signs, for the quantity of positive (resp. of negative) roots of , one has (resp. ), where and are the numbers of sign changes and sign preservations in ; the numbers and are even. We say that realizes the SP with the pair . For SPs with , we give some sufficient conditions for the (non)realizability of pairs of the form , , , .
Key words: real polynomial in one variable; Descartes’ rule of signs; sign pattern
AMS classification: 26C10, 30C15
1. Introduction
In the present paper we consider a problem which is a natural continuation of Descartes’ rule of signs. The latter states that the number of positive roots of a real univariate polynomial (counted with multiplicity) is majorized by the number of sign changes in the sequence of its coefficients. We focus on polynomials without zero coefficients. Such a polynomial (say, of degree ) is representable in the form , . Denoting by and the numbers of sign changes and sign preservations in the sequence , , , , and by and the number of positive and negative roots of (hence ) one obtains the conditions
[TABLE]
(the condition results from Descartes’ rule applied to the polynomial ).
We call sign pattern (SP) a sequence of or signs of length beginning with a . We say that the polynomial defines the SP , sgn, , sgn, sgn. A pair satisfying conditions (1.1) is called admissible. An admissible pair (AP) is called realizable if there exists a polynomial with exactly positive distinct and exactly negative distinct roots.
Example 1**.**
For , the all-pluses SP is realizable with any AP (which is of the form , , , , , where denotes the integer part of ). Indeed, one can construct a polynomial with distinct negative roots and distinct critical levels. Then in the family of polynomials , , one encounters polynomials with exactly , , , negative distinct roots and with no positive roots (as increases, the polynomial loses two-by-two its real roots; each time two coalescing real roots give birth to a complex conjugate pair).**
Proposition 1**.**
(1) Any SP with is realizable with any AP of the form , .
(2) Any SP is realizable with the AP .
Proposition 1 is proved in Section 2. Part (1) of it shows that in terms of the value of , the first truly nontrivial case is . Its study is the object of the present paper. We should point out that due to the possibility to consider instead of the polynomial the polynomial (this change exchanges the quantities and and the quantities and ), it suffices to consider (for a given degree ) the cases with .
Notation 1**.**
We denote by the SP consisting of pluses followed by minuses followed by pluses, where . For a given polynomial , we denote by the corresponding reverted polynomial, i.e. . If the polynomial defines the SP , then defines the SP . The roots of are the reciprocals of the roots of .**
For small values of or , we have the following result:
Theorem 1**.**
(1) For , , and for , , any SP is realizable with the AP .
(2) For and , any SP is realizable with the AP . For , the SP is not realizable with the AP .
(3) For , the SP is realizable with the AP if , and , or if and (hence ).
(4) For and , the SP is not realizable with the AP .
Theorem 1 is proved in Section 3.
Remarks 1**.**
(1) If a SP is realizable with the AP , then it is realizable with any AP of the form , , , . Indeed, if a polynomial with distinct nonzero roots realizes the SP , then one can perturb to make all its critical levels distinct. In the family one encounters (for suitable positive values of ) polynomials with exactly distinct negative roots and no positive ones, for , , . As , the constant term of the polynomial is positive hence defines the SP .
(2) The exhaustive answer to the question which couples (SP, AP) are realizable for is given in [3], [1], [2] and [4]. From the results in these papers one deduces that for , the SP is not realizable with the AP . For , and , the only cases when the AP is and which are not covered by Theorem 1 are the ones of and for and of for . These cases are settled by Proposition 2.
(3) The following result is proved in [2] (see Proposition 6 therein): If , then the SP is not realizable with the AP . This seems to be the only result concerning nonrealizability of the couple (, ) known up to now. Part (4) of Theorem 1 implies nonrealizability of cases which are not covered by the cited result. These are for (with which is for ). **
Proposition 2**.**
(1) For , the SPs and are not realizable with the AP .
(2) For , the SP is not realizable with the AP .
Proposition 2 is proved in Section 4. Our next result contains sufficient conditions for realizability of a SP with the AP :
Theorem 2**.**
The SP is realizable with the AP if
[TABLE]
Theorem 2 is proved in Section 5.
Remarks 2**.**
(1) Condition (1.2) is sharp in the following sense: in the two nonrealizable cases (, ) and (, ) (see Proposition 2) one has .
(2) The condition of realizability (1.2) can be compared with the condition of nonrealizability (see part (3) of Remarks 1). To this end the latter can be given the following equivalent form:**
[TABLE]
(3) As the SPs and are simultaneously (non)realizable with the AP (see the definition of in Notation 1), one can assume that hence . Condition (1.2) can be presented in the form
[TABLE]
which allows, for given , to find such that for , one has . Then for and , condition (1.2) is fulfilled and the corresponding SP is realizable with the AP .**
2. Proof of Proposition 1
Part (1). We remind the formulation of a concatenation lemma (see [2]):
Lemma 1**.**
Suppose that the monic polynomials and of degrees and with SPs and respectively realize the pairs and . Here denote what remains of the SPs when the initial sign is deleted. Then
(1) if the last position of is , then for any small enough, the polynomial realizes the SP and the pair ;
(2) if the last position of is , then for any small enough, the polynomial realizes the SP and the pair . Here is obtained from by changing each by and vice versa.
For , the SP (resp. ) is realizable with the AP (resp. ) by the polynomial (resp. ). Applying Lemma 1 with and of the form one realizes for all the three SPs with or with the APs of the form or .
Suppose that for all SPs with or are realizable by monic polynomials (denoted by ). Then to realize for a SP with or with the pair or it suffices to apply Lemma 1 with and with (resp. ) if and the last two signs of the SP defined by are (resp. if and these last two signs are ).
To realize for a SP with any AP , , it suffices to perturb a polynomial realizing this SP with the pair so that all critical levels become distinct and then choose suitable values of in the family of polynomials .
To prove part (2) one has to apply times Lemma 1. When it is applied for the first time one sets (resp. ) if the second entry of the SP is (resp. ). Each time the polynomial equals or .
3. Proof of Theorem 1
Part (1). For and , the polynomials
[TABLE]
realize the APs and with the SPs and respectively. If a polynomial realizes a SP (with or ) with the AP , then the concatenation of with realizes the SP with the AP , and the polynomial (the reverted of ) realizes the SP with the AP . Thus by means of concatenation and reversion one can realize all SPs and with the AP .
Part (2). For , the nonrealizability of the SP with the AP is proved in [3]. For , the SP is realizable with the AP , see [1]. To prove the first claim of part (2) one has to combine concatenation and reversion as in the proof of part (1) (applied to ).
Part (3). For , the polynomial
[TABLE]
defines the SP . (The quadratic factor is without real roots.) One can perturb its -fold root at so that the latter splits into negative simple roots. Thus the perturbation realizes this SP with the AP . Similarly, for , a suitable perturbation of the polynomial
[TABLE]
realizes the SP with the AP . As in the proof of parts (1) and (2), one deduces the realizability of all SPs as claimed by part (3) by applying concatenation and reversion.
Part (4). Suppose that the SP with is realizable by the polynomial
[TABLE]
where , and is the th elementary symmetric function of the moduli of the negative roots of . (As , the coefficient of of the quadratic factor must be negative, otherwise all coefficients of will be positive, so ). Thus one obtains the conditions
[TABLE]
Keeping in mind that , and , one gets
[TABLE]
The quadratic polynomial has a positive discriminant (this follows from Newton’s inequalities). Hence it has two real roots, so the last inequality implies
[TABLE]
and as , one deduces the condition
[TABLE]
One has . Indeed, every product is encountered exactly two times in and three times in (and there are also the products in ). Hence one can take squares of both hand-sides of inequality and then divide by to obtain the condition
[TABLE]
We are going to show that for ,
[TABLE]
which contradiction proves part (4). For , one has (see Proposition 2 on page 2 of [6]). Suppose that holds true up to degree . We proceed by induction on . Recall that we denote by the negative roots of . For degree , we have to show that
[TABLE]
where are the elementary symmetric functions of the quantities , which is simplified to
[TABLE]
Newton’s inequality implies the following ones:
[TABLE]
From inequalities (3.7) we conclude that
[TABLE]
which proves (3.6) and hence as well.
4. Proof of Proposition 2
We give in detail the proof of part (1). For part (2), we point out only the differences w.r.t. the proof of part (1). These differences are only technical in character. In order to give easily references to the different parts of the proof, the latter are marked by , , , .
Proof of part (1) of Proposition 2.
. Suppose that there exists a polynomial , where
[TABLE]
which realizes one of the two SPs or with the AP . We set and , , . We show that for , there exists no polynomial satisfying the conditions
[TABLE]
Hence this holds true also for because , and the polynomial has all coefficients positive. This in turn implies that for , there exists no polynomial realizing the SP or . So from now on we concentrate on the case .
. Suppose that a polynomial with and satisfying the left or right couple of inequalities (4.8) exists. We make the change of variables and after this we multiply by (these changes preserve the signs of the coefficients), so now we are in the case . Denote by the set on which one has conditions (4.8). The closure of this set is compact. Indeed, one has hence
[TABLE]
The set can be stratified according to the multiplicity vector of the variables and the possible equalities , and/or . Suppose that the set contains a polynomial satisfying the inequalities (4.8).
Remarks 3**.**
(1) For this polynomial one has , otherwise all its coefficients are nonnegative. One has also , , , . Indeed, in the case of (resp. ), if three or more (resp. if four or more) of the variables are [math], then the polynomial has less than two sign changes in the sequence of its coefficients and by the Descartes rule of signs cannot have two positive roots counted with multiplicity. For , if exactly one or two of the variables equal [math], then the polynomial is the product of with a polynomial defining the SP or of with a polynomial defining the SP . However these SPs are not realizable with the APs or respectively, see [4] and [2]. For , if exactly one, two or three of the variables equal [math], then is the product of , or with a polynomial defining respectively the SP , or which is not realizable with the AP , or , see [4], [2] and [1].
(2) The set being compact the quantity , resp. , attains its minimum on it (). Consider the set on which one has , resp. . On this set one has . Indeed, , so any coefficient of is not less than , where is the sum of all coefficients of (they are all nonnegative); clearly (follows from ).
(3) There exists such that on the set , one has also . This follows from part (1) of the present remarks.**
. We need some technical lemmas:
Lemma 2**.**
The minimum of the quantity , resp. , is not attained at a point of the set with three or more distinct and distinct from among the quantities , .
The lemmas used in the proof of part (1) of Proposition 2 are proved after the proof of part (1).
Lemma 3**.**
Conditions (4.8) fail for and any .
Thus to prove Proposition 2 we have to consider only the case when exactly one or two of the quantities are distinct from . We use the following result:
Lemma 4**.**
For , set , where , , . Then the coefficients of , , , , are quadratic polynomials in with positive leading coefficients and with two distinct positive roots.
. Further we consider several different cases according to the multiplicity of , the smallest of the variables . In the proofs we use linear changes , , followed by . These changes preserve the signs of the coefficients; the condition is lost and the condition , , is obtained. The aim of this is to have more explicit computations. In all the cases the polynomial is of the form , , and one has , , but and are not necessarily distinct and we do not suppose that or (which permits us to assume that ). Allowing the equality means treating together cases of exactly two or exactly three distinct quantities (counting also ). We list the triples defining the cases:
[TABLE]
The cases when there are exactly two different quantities one of which is can be coded in a similar way. E.g. means that , . The nonrealizability of these cases follows automatically from the one of the above ones (when and coalesce), with the only exception of (the case ).
Lemma 5**.**
Conditions (4.8) fail in case .
. We consider the SP first. We compute using MAPLE the resultant Res as a function of and . Then we set , , , . In all cases this resultant is a polynomial in and with all coefficients positive. Hence for no value of and do the coefficients and vanish together.
In all cases, the leading coefficients of and considered as quadratic polynomials in are positive. In fact, they are polynomials in and with all coefficients positive. For , we compute the two roots of and the two roots of . In all cases, one has . By continuity, these inequalities hold true for all values of and . Hence the intervals and on which and are negative do not intersect for any , . This proves the proposition in the case of .
. Consider now the SP . Recall that the polynomials and have one and the same numbers of positive and negative roots. Their roots are mutually reciprocal and they define the same SP. Hence the non-realizability of the case (resp. , or , or ) implies the one of (resp. and , or , or ).
As in the case of , we express Res as a polynomial of and , and then of and . In cases , and , this resultant has a single monomial with negative coefficient, this is . We give the monomials , and for these three cases:
[TABLE]
The discriminants of these quadratic homogeneous polynomials are negative hence they are nonnegative (and positive for , ). In the case of , there are exactly two monomials with negative coefficients, namely and . The resultant equals
[TABLE]
(we skip all other monomials; their coefficients are positive). The two quadratic homogeneous polynomials have negative discriminants, so they are positive for , .
The rest of the reasoning goes by exact analogy with the case of .
∎
Proof of Lemma 2.
Denote by , and three distinct and distinct from of the variables . We prove that one can choose , , , such that the infinitesimal change , , , , , , results in , , where or and , . Hence locally the quantity is not minimal.
Set , where , , and are not roots of and are the multiplicities of throots of . Set , , , etc. Then the above infinitesimal change transforms into
[TABLE]
We show that one can choose and such that the coefficients of and of the polynomial (where or ) are both negative from which the lemma follows. To this end we observe that each of the polynomials and is a linear combination of , , and .
We consider first the case of , i.e . The -vectors of coefficients of and of the polynomials , , and equal , , and respectively. For , the first two of them are not collinear. As (see parts (2) and (3) of Remarks 3), for , the second and fourth of these vectors are not collinear and the choice of and is possible.
If , then the -vectors of coefficients of and equal , , and . One has either or . Indeed, the polynomial has all roots real and by Rolle’s theorem this is the case of and as well. If (resp. or ), then (resp. or ) has not all roots real. Thus either , or , are not collinear and the choice of and is possible.
∎
Proof of Lemma 3.
For the polynomial , we list its coefficients , , and and their roots:
[TABLE]
Hence for no value of does one have the left or the right two of conditions (4.8) together.
∎
Proof of Lemma 4.
Set , , . The polynomial has negative roots. Hence Newton’s inequalities hold true:
[TABLE]
The coefficient equals , , , , . This quadratic polynomial has two distinct positive roots if and only if . These inequalities result from (4.9) because (the latter inequality is equivalent to which is true).
∎
Proof of Lemma 5.
In case (6,1), with , one has
[TABLE]
For , the roots of (resp. of ) equal and (resp. and ). As Res has no positive roots, for any fixed, the two intervals of values of , for which or , do not intersect. Hence the couple of conditions , fails.
One has Res , so only for do the polynomials and have a root in common. For , and , the roots of and equal respectively
[TABLE]
Hence again the intervals of values of for which or do not intersect and the couple of conditions , fails.
∎
Proof of part (2) of Proposition 2.
. In the analog of part of the proof of part (1), we set , , and the analog of inequalities (4.8) reads , .
. In the analog of part we make the change of variables and then we multiply by . We denote by the set on which one has the conditions , . On the closure of this set one has hence
[TABLE]
The analog of Remarks 3 reads:
Remarks 4**.**
(1) One has , , , . Indeed, if exactly one of the quantities is [math], then , where the polynomial defines the SP which by part (1) of Proposition 2 is impossible. If more than one of the quantities is [math], then see part (1) of Remarks 3 about .
(2) In the proof of part (2) of Proposition 2 we define the set as the one on which one has . On this set one has . Indeed, as , any coefficient of is not less than , where is the sum of all coefficients of (they are all nonnegative); clearly (follows from ). **
. The analog of Lemma 2 reads: The minimum of the quantity is not attained at a point of the set with three or more distinct and distinct from among the quantities , .
The proof is much the same as the one of Lemma 2. One sets . Each of the polynomials and is a linear combination of , , and . The -vectors of coefficients of and of the polynomials , , and equal , , and respectively. If or , there are two noncollinear among the first three of these vectors and the choice of and is possible. If , then, as , either the polynomial or one of its derivatives is not with all roots real which is a contradiction.
The analog of Lemma 3 reads: Conditions , fail for and any .
Here’s the proof of this. For the polynomial , we list its coefficients , and their roots:
[TABLE]
Hence for no value of does one have , .
We remind that Lemma 4 is formulated for any .
. In the analog of part of the proof, one has , , and one has to consider the following cases of exactly three different quantities :
[TABLE]
The cases with exactly two different quantities are treated in the same way. The exceptional case is the one with (the case ).
Lemma 6**.**
The conditions , fail in case .
Proof.
Set . Then
[TABLE]
One has Res . We list the roots of and for , and :
[TABLE]
As in the proof of Lemma 5, we conclude that the conditions , fail for . ∎
. We compute Res as a function of and and then set , . Our aim is to show that in all cases, the leading coefficients of and considered as quadratic polynomials in are positive. The rest of the reasoning is done by analogy with part of the proof of part (1) of Proposition 2.
. It is in the analog of that there is much more technical work to be done. Of the twelve cases listed in , in three there is a single monomial with a negative coefficient, and this is . We list the coefficients of the monomials , and of the cases , and respectively:
[TABLE]
Everywhere in quadratic and biquadratic polynomials have negative discriminants. There are four cases in which exactly two monomials have negative signs, namely , , and in which we give only the monomials forming quadratic homogeneous polynomials with negative discriminants (multiplied by or ); we skip all other monomials (their coefficients are positive):
[TABLE]
In the cases and there are four and five negative monomials respectively. These cases are treated in a similar way:
[TABLE]
and
[TABLE]
In the case , there are four negative monomials which we include in polynomials as follows:
[TABLE]
For the third polynomial in brackets its corresponding inhomogeneous polynomial
[TABLE]
has one negative and two complex conjugate roots. For a univariate real polynomial with positive leading coefficient and having only negative and complex conjugate roots we say that it is of type P. It is clear that the homogeneous polynomial corresponding to a type P univariate polynomial (we say that it is also of type P) is nonnegative.
In the case , there are seven negative monomials:
[TABLE]
The third and fourth of the polynomials in brackets are of type P hence nonnegative.
Finally, in the case we have also seven negative monomials:
[TABLE]
The third and the last two polynomials in brackets are of type P. ∎
5. Proof of Theorem 2
Consider the polynomial , where the quadratic factor has no real roots, i.e. . Hence and (otherwise all coefficients of must be positive). If the polynomial defines the SP , then any perturbation of with distinct negative roots close to defines also the SP . We expand in powers of :
[TABLE]
where with if . The coefficients of define the SP , so
[TABLE]
The latter inequalities (combined with ) yield:
[TABLE]
This means that :
[TABLE]
where . Indeed, the polynomial (quadratic in ) has a positive discriminant hence its value is negative precisely when is between its roots. Set
[TABLE]
Lemma 7**.**
One has , where
[TABLE]
Lemma 8**.**
The quantities are decreasing functions in (for ).
Lemmas 8 and 7 are proved after the proof of Theorem 2. It follows from Lemma 8 that one can find a value of satisfying conditions (5.10) if
[TABLE]
or equivalently
[TABLE]
where
[TABLE]
One has
[TABLE]
which permits to take squares in (5.13) to obtain the condition :
[TABLE]
which is equivalent to (5.12). If , then (5.14) is trivially true. If , then (5.14) is equivalent to , i.e. to (1.2) (the latter equivalence can be proved using MAPLE). Theorem 2 is proved.
Proof of Lemma 7.
[TABLE]
We substitute this expression of in (5.11) to obtain
[TABLE]
∎
Proof of Lemma 8.
Both factors and of are decreasing in (for ) hence is also decreasing. We represent the quantity in the form
[TABLE]
The inequality is equivalent to
[TABLE]
This follows from , and
. ∎
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] A. Albouy, Y. Fu: Some remarks about Descartes’ rule of signs. Elem. Math., 69 (2014), 186–194. Zbl 1342.12002, MR 3272179
- 2[2] J. Forsgård, B. Shapiro and V. P. Kostov: Could René Descartes have known this? Exp. Math. 24 (4) (2015), 438-448. Zbl 1326.26027, MR 3383475
- 3[3] D. J. Grabiner: Descartes’ Rule of Signs: Another Construction. Am. Math. Mon. 106 (1999), 854–856. Zbl 0980.12001, MR 1732666
- 4[4] V. P. Kostov, On realizability of sign patterns by real polynomials, Czechoslovak Math. J. 68 (143) (2018), no. 3, 853–874.
- 5[5] V. P. Kostov, Polynomials, sign patterns and Descartes’ rule of signs, Mathematica Bohemica 144 (2019), No. 1, 39-67.
- 6[6] Mitev New inequalities between elementary symmetric polynomials, J. of Inequalities in Pure and Appl. Math. 4 (2) (2003), 11 p.
