Positivity Conditions for Cubic, Quartic and Quintic Polynomials
PPositivity Conditions for Cubic, Quartic and Quintic Polynomials
Liqun Qi ∗ , Yisheng Song † , and Xinzhen Zhang ‡ ,September 21, 2020 Abstract
We present a necessary and sufficient condition for a cubic polynomial to bepositive for all positive reals. We identify the set where the cubic polynomialis nonnegative but not all positive for all positive reals, and explicitly give thepoints where the cubic polynomial attains zero. We then reformulate a necessaryand sufficient condition for a quartic polynomial to be nonnegative for all posi-tive reals. From this, we derive a necessary and sufficient condition for a quarticpolynomial to be nonnegative and positive for all reals. Our condition explic-itly exhibits the scope and role of some coefficients, and has strong geometricalmeaning. In the interior of the nonnegativity region for all reals, there is an ap-pendix curve. The discriminant is zero at the appendix, and positive in the otherpart of the interior of the nonnegativity region. By using the Sturm sequences,we present a necessary and sufficient condition for a quintic polynomial to bepositive and nonnegative for all positive reals. We show that for polynomials ofa fixed even degree higher than or equal to four, if they have no real roots, thentheir discriminants take the same sign, which depends upon that degree only,except on an appendix set of dimension lower by two, where the discriminantsattain zero.
Key words.
Cubic polynomials, quartic polynomials, quintic polynomials, theSturm theorem, discriminant, appendix.
AMS subject classifications. ∗ Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom,Kowloon, Hong Kong, China; ( [email protected] ). † School of Mathematical Sciences, Chongqing Normal University, Chongqing 401331 China;( [email protected] ). This author’s work was supported by NSFC (Grant No. 11571095,11601134). ‡ School of Mathematics, Tianjin University, Tianjin 300354 China; ( [email protected] ). Thisauthor’s work was supported by NSFC (Grant No. 11871369). a r X i v : . [ m a t h . G M ] S e p Introduction
In 1988, Schmidt and Heß[20] presented a necessary and sufficient condition for a cubicpolynomial to be nonnegative for all positive reals. In 1994, Ulrich and Watson [23]presented a necessary and sufficient condition for a quartic polynomial to be nonnega-tive for all positive reals. On the other hand, necessary and sufficient conditions for aquartic polynomial to be positive for all reals have been studied by Gadem and Li [7],Ku [13], Jury and Mansour [10], Wang and Qi [24], and Gao [8].In this paper, we first present a necessary and sufficient condition for a cubic polyno-mial to be positive for all positive reals. We identify the set where the cubic polynomialis nonnegative but not all positive for all positive reals, and explicitly give the pointswhere the cubic polynomial attains zero.We then reformulate the result of Ulrich and Watson [23]. Based upon this, wederive a necessary and sufficient condition for a quartic polynomial to be nonnegativeand positive for all reals. Comparing with the existing results in the literature, ourcondition has three merits. First, it explicitly states a necessary condition. Second, itexplicitly states a symmetric relation between two parameters. Third, its geometricalmeaning is clear. We also present a theorem on the geometrical properties of thenonnegativity region and the positivity region. In the interior of the nonnegativityregion for all reals, there is an appendix curve. The discriminant is zero at the appendix,and positive in the other part of the interior of the nonnegativity region.By using the Sturm sequences, we present a necessary and sufficient condition fora quintic polynomial to be positive and nonnegative for all positive reals. This is thefirst for quintic polynomials. It also indicates that such conditions are possible forpolynomials of even higher degrees.Finally, we show that such an appendix exists for all even degree polynomials withtheir degrees higher than or equal to 4.These results are useful in automatic control [7, 13, 10] and determining copositiv-ity and strict copositivity of symmetric tensors, and positive semidefiniteness of evenorder symmetric tensors [14, 15, 16, 21, 22]. These properties are further useful inoptimization [18, 19], hypergraphs [2, 3, 17] and physics [5, 9, 11, 12].Some preliminary knowledge on the Sturm theorem and quadratic polynomials aregiven in the next section. Then, four sections are about cubic polynomials, quarticpolynomials, quintic polynomials and higher even degree polynomials, respectively.All the polynomials studied in this paper are of real coefficients.2
Preliminaries
Suppose that we have an m th degree polynomial g ( t ). What is the analytically ex-pressed necessary and sufficient condition such that g ( t ) > t ≥ t > { g , g , · · · , g l } . Let g = g , g = g (cid:48) , and g k = − rem( g k − , g k ) for k ≥
1, where g (cid:48) is the derivative of g , rem( g k − , g k ) is theremainder of the division of g k − by g k . We have g l (cid:54) = 0 and g l +1 = 0. Then the length l + 1 of the Sturm sequence is not greater than m + 1. The number of variations of theSturm sequence S = { g , g , · · · , g l } at a real number ξ is the number of sign changes(ignore zero) of the real number sequence { g ( ξ ) , g ( ξ ) , · · · } . Denote it as V ( ξ ). By theSturm theorem, the number of real distinct roots of g ( t ) = 0, for t ≥
0, is V (0) − V ( ∞ ).If V (0) − V ( ∞ ) = 0, then g has no positive roots, i.e., g ( t ) > t ≥
0. Thisanswers the question.Note that g l is the greatest common divisor (GCD) of g and g (cid:48) . If g l is not aconstant term, then g has multiple roots. Any root of g l with multiple m is a m + 1multiple root of g .In the following, we simply say that g is strictly copositive if g ( t ) > t ≥ The following result should be known several centuries ago, and is easy to be derived.
Theorem 2.1
Suppose that we have a quadratic polynomial φ ( t ) = t + ut + v, where v (cid:54) = 0 . Then φ ( t ) > for all t ≥ if and only if either (i) u ≥ and v > ; or(ii) u < and v > u . If u < and v = u , then φ ( t ) ≥ for all t ≥ , and this isthe only case that φ ( t ) ≥ for all t ≥ , but not φ ( t ) > for all t ≥ .Furthermore, φ ( t ) > for all t if and only if v > u . If v = u , then φ ( t ) ≥ forall t , and this is the only case that φ ( t ) ≥ for all t , but not φ ( t ) > for all t . Cubic Polynomials
We call a cubic polynomial non-degenerate if its constant is nonzero. For a non-degenerate cubic polynomial, we may write it as h ( t ) = t + pt + qt + r, (3.1)where r (cid:54) = 0. If r <
0, then it cannot be always nonnegative for all positive reals.Thus, we may assume that r >
Theorem 3.1
Suppose that h ( t ) is defined by (3.1). Then h ( t ) ≥ for all t ≥ inand only in the following two cases: (A) p ≥ and q ≥ ; (B) (D) ∆( h ) ≤ . Here, ∆( h ) = p q + 18 pqr − r − p r − q (3.2) is the discriminant of h ( t ) . Suppose that we have a cubic polynomial h ( t ) defined by (3.1), with r > h ( t ) = φ ( t ) t + r, where φ ( t ) = t + pt + q . By Theorem 2.1, if either (I) p ≥ q ≥
0, or (II) q ≥ p ,then φ ( t ) ≥ t ≥
0. This implies that h ( t ) > t ≥
0, as r > q < p . We now construct the Sturm sequence for h ( t ). Let h ( t ) = h ( t ) = t + pt + qt + r,h ( t ) = h (cid:48) ( t ) = 3 t + 2 pt + q. Then h ( t ) − t h ( t ) = p t + 2 q t + r,h ( t ) − t h ( t ) − p h ( t ) = (cid:18) q − p (cid:19) t + r − pq .
4e have h ( t ) = q t + r , where q = 2 p − q , r = pq − r. (3.3)Since q < p , we have q > h ( t ) ≡ r = − h (cid:18) − r q (cid:19) = − r q + 2 pr q − q = ∆( h )9 q , where ∆( h ) ≡ p q + 18 pqr − r − p r − q = − r + 54 pq r − qq . The expression of ∆( h ) can be found in [6].Suppose that ∆( h ) = 0. Then h ( t ) is the GCD of g and g (cid:48) , h ( t ) = ( t − α ) ( t + β ) , where α = − r q . We have q >
0. We now show that r <
0. Assume that r ≥
0. This means that pq ≥ r . Since we have already exclude the case that p ≥ q ≥
0, this means p < q < pq ≤ r . These further implies that ∆( h ) <
0, contradicting theassumption that ∆( h ) = 0.We have α β = r >
0. Thus, β > α (cid:54) = 0. Therefore, h ( t ) ≥ t ≥ t = α > h ( t ) = 0.Suppose now ∆( h ) (cid:54) = 0. We have S = { h , h , h , h } ,S ( ∞ ) = { , , q , ∆( h ) } ,S (0) = { , q, r , ∆( h ) } . Assume that ∆( h ) >
0. Then V ( ∞ ) = 0. Thus, h ( t ) > t ≥ V (0) = 0, i.e., q ≥ r ≥
0. However, this implies that p > q >
0, whichhas already been covered by Case (I).Assume that ∆( h ) <
0. Then V ( ∞ ) = 1. Thus, h ( t ) > t ≥ r ≤
0, or (ii) q ≥ r ≥
0. However, Case (ii) has already beencovered by Case (I) as discussed above. Thus, only Case (i) needs to be considered.As discussed above, the case that r >
0, i.e., pq > r , is covered by the condition (I)5 ≥ q ≥
0, and (III) ∆( h ) <
0. Thus, the condition that r ≤ h ( t ) > t ≥
0: (III) ∆( h ) < q ≥ p is coved by the union of Cases (I) and (III).Thus, we have the following theorem. Theorem 3.2
Suppose that h ( t ) is defined by (3.1). Then h ( t ) > for all t ≥ inand only in the following two cases: (A) p ≥ and q ≥ ; (B) ∆( h ) < . Here, ∆( h ) is the discriminant, defined by (3.2).If ∆( h ) = 0 and either p < or q < , then h ( t ) ≥ for all t ≥ , but there is α = 9 r − pq p − q > such that h ( α ) = 0 .In all the other cases, there is a t > such that h ( t ) < . Suppose that we have a quartic polynomial f ( t ) = t + αt + βt + γt + 1 . (4.4)By [23], its discriminant has the form:∆( f ) = 4[ β − αγ + 12] − [72 β + 9 αβγ − β − α − γ ] . (4.5)By [23], we have the following theorem. Theorem 4.1
Suppose that f is defined by (4.4). Let Λ = ( α − γ ) − α + β + γ + 2) , Λ = ( α − γ ) − β + 2) √ β − α + γ + 4 (cid:112) β − . Then f ( t ) ≥ for all t > if and only if either(1) β < − , ∆( f ) ≤ and α + γ > ; or(2) − ≤ β ≤ and either (i) ∆( f ) ≤ and α + γ > , or (ii) ∆( f ) ≥ and Λ ≤ ; or(3) β > and either (i) ∆( f ) ≤ and α + γ > , or (ii) α > and γ > ; or (iii) ∆( f ) ≥ and Λ ≤ .
6e may combine (1), (2i) and (3i) as(A) ∆( f ) ≤ α + γ > f ( t ) = ( t + βt + 1) + αt + γt, if α ≥ γ ≥ β ≥ −
2, then f ( t ) ≥ t ≥
0. Thus, we have(B) α ≥ γ ≥ β ≥ − ≤ ≤
0, by some constraints which arelinear with respect to α and γ .For the condition Λ ≤
0, from the discussion in Sections 2 and 3 of [23], we mayreplace it by | α − γ | ≤ r, where r is the value of | α − γ | at the two intersection pointsof α + γ = 0 and the curve Γ β in [23]. These are two points where α ( t ) + γ ( t ) = 0, for α ( t ) and γ ( t ) given by (7a) and (7b) of [23]. Solving this, we find that r = 4 √ β + 2.Thus, we may use | α − γ | ≤ √ β + 2 to replace Λ ≤
0. Note that we may rewrite theconstraint | α − γ | ≤ √ β + 2 such that α and γ are linear there.For the condition Λ ≤
0, by Figure 1 of [23], we may replace it by | α − γ | ≤ √ β + 2and α + γ ≥ − √ β − Theorem 4.2
Suppose that g ( z ) = az + bz + cz + dz + e be a quartic polynomialwith real coefficients and a > and e > . Let α = ba − e − , β = ca − e − , γ = da − e − ,f is defined by (4.4), and ∆( f ) is defined by (4.5). Then f ( t ) ≥ for all t > if andonly if either(A) ∆( f ) ≤ and α + γ > ; or(B) α ≥ , γ ≥ and β ≥ − ; or(C) ∆( f ) ≥ , | α − γ | ≤ √ β + 2 and either (i) − ≤ β ≤ , or (ii) β > and α + γ ≥ − √ β − . Proof
We may convert the general quartic polynomial g ( z ) to f ( t ), defined by (4.4),as in [23].We have already discussed to convert Conditions (1), (2) and (3) of Theorem 4.1to Conditions (A), (B) and (C). (cid:3) A merit of the format of Theorem 4.2 is that it is somewhat convenient to derivenonnegativity conditions for all t from conditions (A), (B) and (C).7 .2 Nonnegativity and Positivity Conditions for All Reals We now consider the conditions that f ( t ) ≥ f ( t ) >
0) for all t . This is equivalentto the condition that f ( t ) ≥ f ( t ) >
0) and g ( t ) ≥ g ( t ) >
0) for all t , where g ( t )is defined by g ( t ) = t − αt + βt − γt + 1 . Note that ∆( f ) = ∆( g ). By Theorem 4.2, we have the following theorem. Theorem 4.3
Suppose that g ( z ) = az + bz + cz + dz + e is a quartic polynomialwith real coefficients and a > and e > . Let α = ba − e − , β = ca − e − , γ = da − e − ,f be defined by (4.4), and ∆( f ) be defined by (4.5). Then f ( t ) ≥ for all t if andonly if ∆( f ) ≥ , | α − γ | ≤ √ β + 2 and either (i) − ≤ β ≤ , or (ii) β > and | α + γ | ≤ √ β − .Furthermore, f ( t ) > for all t if and only if either(A) ∆( f ) = 0 , α = γ , α + 8 = 4 β < ; or(B) ∆( f ) > , | α − γ | ≤ √ β + 2 and either (i) − ≤ β ≤ , or (ii) β > and | α + γ | ≤ √ β − . Proof
Let ¯ f ( t ) = f ( − t ). Then¯ f ( t ) = t − αt + βt − γt + 1 . Then f ( t ) ≥ t if and only if f ( t ) ≥ f ( t ) ≥ t ≥
0. From Theorem4.2, we have conditions that f ( t ) ≥ f ( t ) ≥ t ≥
0. Combining theseconditions, i.e., taking the intersection of the conditions that f ( t ) ≥ t ≥ f ( t ) ≥ t ≥
0, we have the conditions for the first partof this theorem.We now prove the second part of this theorem. If ∆( f ) >
0, it is in the interiorof the nonnegativity region, we have f ( t ) > t . We have condition (B). Nowassume that ∆( f ) = 0. Then f ( t ) has a multiple root. To make f ( t ) > t , thismultiple toot has to be complex. Thus, f ( t ) = ( t + ut + 1) with u <
4. Comparingthe coefficients of f ( t ) = ( t + ut + 1) = t + αt + βt + γt + 1 , We have α = γ = 2 u and u + 2 = β . Then, we have condition (A). (cid:3) β , the nonnegativity region of ( α, γ ) for g ( t ) is the convexhull of the central convex compact part of the positive sign region in Figure 5b and 5cof [23], while the positivity region is the interior of the nonnegativity region. In Figures5b of [23], for 2 ≤ β <
6, there are two points missing. These two points ( α, β, γ ) aredefined by α = γ and α + 8 = 4 β . At these two points, ∆( f ) = 0. These two pointsare missing in Figure 5b of [23], but do not affect the final results of [23].Note that β + 2 ≥ f ( t ) ≥ t . Thisexplicitly stated in Theorem 4.3. Actually, if f ( t ) ≥ t , then β + 2 = f (1) + f ( − ≥ . The following theorem presents the geometrical features of the nonnegativity regionand the positivity region.
Theorem 4.4
Suppose that f ( t ) is defined by (4.4). Let S = { ( α, β, γ ) : f ( t ) > t } and ¯ S = { ( α, β, γ ) : f ( t ) ≥ t } . Then ¯ S is a closed convex cone with a recession direction (0 , , and an apex (0 , − , .It is symmetric with respect to α and γ , i.e., if ( α, β, γ ) ∈ ¯ S , then ( γ, β, α ) , ( − α, β, − γ ) ∈ ¯ S . For any ( α, β, γ ) in the interior of ¯ S , if ≤ β < , α = β and α + 8 = 4 β , wehave ∆( f ) = 0 . Otherwise, we have ∆( f ) > . For any ( α, β, γ ) at the boundary of ¯ S ,we have ∆( f ) = 0 . The set S is the interior of ¯ S . Proof
Consider ¯ S . Suppose that f ( t ) and g ( t ) are two quartic polynomials with theform of (4.4), f ( t ) ≥ g ( t ) ≥ t . Then ( f ( t ) + g ( t )) is still a quarticpolynomial with the form of (4.4), and is nonnegative for all t . This shows that ¯ S isconvex. Similarly, we may show that it is closed. Let f ( t ) be defined by (4.4), and δ >
0. Let g ( t ) = t + αt + ( β + δ ) t + γt + 1 . Then g ( t ) = f ( t ) + δt ≥ f ( t ) ≥ t . This shows that (0 , ,
0) is a recessiondirection of ¯ S and ¯ S is a cone. From Theorem 4.3, the point ( α, β, γ ) = (0 , − ,
0) isthe unique point of ¯ S with the smallest value of β . Hence, it is an apex of ¯ S . Theother properties of ¯ S and S can be derived from Theorem 4.3 accordingly. (cid:3)
9n Figure 1, the α = γ plane is exhibited. The set ¯ S is the area above β = 2 | α | − | α | ≤
4, and the area above the parabola β = α + 2 for | α | ≥
4. The discriminant∆( f ) vanishes at the line segments β = 2 | α | − | α | ≤ β = α + 2for all α . The truncated parabola β = α + 2 for | α | < S . InFigure 2, the α = − γ plane is exhibited. The set ¯ S is the area above the parabola β = α −
2. We may see that ( α, β, γ ) = (0 , − ,
0) is the apex of the cone ¯ S .The truncated parabola β = α + 2 for | α | < f ) = 0. The other part of the surface ∆( f ) = 0 is of dimension 2, butthe truncated parabola β = α + 2 for | α | < appendix of the surface ∆( f ) = 0. Such an appendix exists for polynomials with theirdegrees higher than or equal to 4. 10his problem has been studied by Gadem and Li [7], Ku [13], Jury and Mansour[10], Wang and Qi [24], and Gao [8].We may compare Theorem 4.3 with the result of [24], which is a correction of theresult of [13]. The following is the result of [24].The polynomial considered in [24] has the following form: f ( t ) = a t + 4 a t + 6 a t + 4 a t + a . Thus, a = a = 1 , a = α , a = β , a = γ . Let G = a a − a a a + 2 a = γ − αβ α
32 = 8 γ − αβ + α .H = a a − a = β − α
16 = 8 β − α .I = a a − a a + 3 a = 1 − αγ β
12 = 12 − αγ + β .J = a a a +2 a a a − a a − a a − a = β αβγ − α + γ − β
216 = 72 β + 9 αβγ − α − γ − β . ∆ = I − J = (cid:18) − αγ + β (cid:19) − (72 β + 9 αβγ − α − γ − β ) × = ∆( f )4 × . Hence, ∆ in [24] has the same sign as ∆( f ) and plays the same role as ∆( f ). Theresult of [24] can be formulated with the language of this paper as follows.11 heorem 4.5 Let f ( t ) be defined by (4.4). Then f ( t ) > for all t if and only if either(1) ∆( f ) = 0 , G = 0 , H − I = 0 and H > ; or(2) ∆( f ) > and either (i) H ≥ , or (ii) H < and H − I < . We have12 H − I = 12 (cid:18) β − α (cid:19) − − αγ + β
12 = 64 β − α β + 9 α −
192 + 48 αγ − β β − α β + 3 α −
64 + 16 αγ . Corollary 4.6
Let f ( t ) be defined by (4.4). Then f ( t ) ≥ for all t if and only if ∆( f ) ≥ and either (i) H ≥ , or (ii) H < and H − I ≤ . The conclusions of Theorem 4.3 should be the same with the conditions of Theorem4.5 and Corollary 4.6. However, Theorem 4.3 has three merits. First, it explicitly statedthat a necessary condition is that β ≥ −
2. Second, it treats α and γ in a symmetric way.Third, its geometrical meaning is clear, as shown by Theorem 4.4. The set indicatedby Theorem 4.3 (A) and Theorem 4.5 (1) is the appendix of the surface ∆( f ) = 0. Itis actually in the interior of the nonnegativity region. Suppose that we have a quintic polynomial g ( t ) = t + at + bt + ct + dt + e, (5.6)where e > g , we may findthe GCD of g and g (cid:48) if g has a multiple root.(A). Suppose that the GCD of g and g (cid:48) is a linear polynomial. We may assumethat it is φ ( t ) = t − α . Then g has a double root α . Since e > α (cid:54) = 0. We may writethat g ( t ) = h ( t )( t − α ) , (5.7)where h ( t ) = t + pt + qt + r . Then p = a + 2 α, q = b + 2 αp + α , r = eα . (5.8)Since e > r >
0, i.e., h has the form (3.1). By Theorem 3.2, we have the followingproposition. 12 roposition 5.1 Suppose that g ( t ) is defined by (5.6), g and g (cid:48) has a GCD φ ( t ) = t − α . Then g has the form (5.7), where h ( t ) has the form (3.1), with p, q and r ,given by (5.8). Then g ( t ) ≥ for all t ≥ , if and only if h ( t ) ≥ for all t ≥ . Iffurthermore α < , then g ( t ) > for all t ≥ , if and only if h ( t ) > for all t ≥ . (B). Suppose that the GCD of g and g (cid:48) is a quadratic polynomial. We may assumethat it is φ ( t ) = t + ut + v . There are three subcases.(B1). u (cid:54) = 4 v . In this subcase, φ ( t ) has two distinct roots α and β . Then g has adouble root α and a double root β . Since e > α (cid:54) = 0 and β (cid:54) = 0. We have g ( t ) = φ ( t ) ( t − γ ) = ( t − α ) ( t − β ) ( t − γ ) , (5.9)where γ is the fifth root of g . Since e > γ <
0. Then we have the followingconclusion.
Proposition 5.2
Suppose that g ( t ) is defined by (5.6), g and g (cid:48) has a GCD φ ( t ) = t + ut + v . If u (cid:54) = 4 v , then g ( t ) ≥ for all t ≥ . If furthermore φ ( t ) > for all t ≥ , then g ( t ) > for all t ≥ . We may use Theorem 2.1 to determine if φ ( t ) > t > u = 4 v . Then φ has a double root α = − u , and α is a triple root of g . Since e > α (cid:54) = 0. We have g ( t ) = (cid:16) t + u (cid:17) ψ ( t ) , (5.10)where ψ ( t ) = t + ˆ ut + ˆ v. (5.11)Then ˆ v = 8 eu , ˆ u = a − u . (5.12)We have the following conclusion. Proposition 5.3
Suppose that g ( t ) is defined by (5.6), g and g (cid:48) has a GCD φ ( t ) = t + ut + v . Suppose that u = 4 v . Then u (cid:54) = 0 . Let ψ ( t ) be calculated by (5.11) and(5.12). Then g ( t ) ≥ for all t ≥ if and only if u > and ψ ( t ) ≥ for all t ≥ , and g ( t ) > for all t ≥ if and only if u > and ψ ( t ) > for all t ≥ . We may use Theorem 2.1 to determine the situation of ψ ( t ).(C). Suppose that the GCD of g and g (cid:48) is a cubic polynomial. We may assume thatit is h ( t ) = t + pt + qt + r = ( t − α ) ( t + β ). There are two subcases.(C1) q = p and r = p . This implies α = − β = − p , and g ( t ) = (cid:16) t + p (cid:17) ( t − γ ) . e > γ < p (cid:54) = 0. Thus, g ( t ) ≥ t ≥
0. If furthermore p >
0, then g ( t ) > t ≥ α (cid:54) = − β , α (cid:54) = 0 and β (cid:54) = 0. We have α from ( ?? ) and (3.3). Then β = rα . We have g ( t ) = ( t − α ) ( t + β ) . Then − α β = e . Since e > α <
0. Thus, we always have g ( t ) ≥ t ≥
0. Iffurthermore r >
0, then g ( t ) > t ≥ Proposition 5.4
Suppose that g ( t ) is defined by (5.6), g and g (cid:48) has a GCD h ( t ) = t + pt + qt + r . Then g ( t ) ≥ for t ≥ . We have g ( t ) > for all t ≥ if and onlyif either (i) q = p , r = p and p > ; or (ii) r > and either q (cid:54) = p or r (cid:54) = p . The case that the GCD of g and g (cid:48) is a quartic polynomial will be analyzed in thenext section. In that case, g ( t ) > t ≥ For constructing the Sturm sequence of a quintic polynomial g ( t ), we keep to use a, b, c, d and e to denote the coefficients of the polynomials g i . We have g ( t ) = g ( t ) = t + at + bt + ct + dt + e, and g ( t ) = g (cid:48) ( t ) = 5 t + 4 at + 3 bt + 2 ct + d. For i ≥
2, we use b i , c i , d i and e i to denote the coefficients of g i ( t ), with b i for thecoefficient of t , c i for t , d i for t and e i for the constant term. If there are differentcoefficients used, we use an additional index to distinguish them. For example, inthe following, e was used in Case (2), and in Case (3), we use e , to denote a newcoefficient. We use g to denote different polynomials in different parts of this section,while e and e , are uniquely used in this section, as g will not appear in the theoremsin the next sections, but e and e , play a role in establishing those theorems.However, e and e , are fractional functions of the coefficients a, b, c, d and e . Thisis not convenient. Then, we use the bar symbol to denote some polynomial functionsto replace them. For example, we write e = ¯ e d , where ¯ e is a polynomial functionof the coefficients a, b, c, d and e , and we use ¯ e instead of e in our theorems. Totally,eleven such fractional functions are replaced.14n the following, S (0) and S ( ∞ ) are the Sturm sequence at 0 and ∞ . Since onlysigns are important, we may replace their entries by other numbers as long as the signsare not changed.Note that when there are no multiple roots, either g ( t ) has no positive root, or has atleast two positive roots. This means that either V (0) − V ( ∞ ) = 0 or V (0) − V ( ∞ ) ≥ g . We have g ( t ) − t g ( t ) = a t + 2 b t + 3 c t + 4 d t + e,g ( t ) − t g ( t ) − a g ( t ) = (cid:18) b − a (cid:19) t + (cid:18) c − ab (cid:19) t + (cid:18) d − ac (cid:19) t + (cid:18) e − ad (cid:19) ,g ( t ) = 4 a − b t + 3 ab − c t + 2 ac − d t + ad − e b t + c t + d t + e . There are four possibilities.(1) b = c = d = 0. We further divide this case to two subcases.(1A) e = 0. This implies that b = a , c = ab = a , d = ac = a and e = ad = a . Thus, g ( t ) = (cid:16) t + a (cid:17) . Since e >
0, we have a >
0. Then, g is strictly copositive in this subcase.(1B) e (cid:54) = 0. Then S = { g , g , g } , and b = a , c = ab and d = ac , and g ( t ) = e .We have d = ac
10 = a b
50 = a ≥ . Then V ( ∞ ) = 0 if e >
0. Otherwise V ( ∞ ) = 1. On the other hand, V (0) = 1 if e < V (0) = 0 if e >
0. Thus, V (0) − V ( ∞ ) ≡
0, and g is strictly copositive in thissubcase.Thus, g is strictly copositive in Case (1).(2) b = c = 0 and d (cid:54) = 0. Then b = a , c = ab = a , g ( t ) = d t + e ,g ( t ) = 5 t + 4 at + 6 a t + 4 a t + d, rem { g , g } = g (cid:18) − e d (cid:19) ,g ( t ) ≡ e = − g (cid:18) − e d (cid:19) = ¯ e d , e = − e + 100 ae d − a e d + 4 a e d − dd . We need to divide this case to two subcases.(2A) e = 0, i.e., ¯ e = 0. In this subcase, α = − e d is a double root of g . We mayapply Proposition 5.1 and Theorem 3.2 to determine the situation of g ( t ).(2B) e (cid:54) = 0, i.e., ¯ e (cid:54) = 0. Then S = { g , g , g , g } ,S ( ∞ ) = { + ∞ , + ∞ , d , e } ,S (0) = { e, d, e , e } . Clearly, we may replace e by ¯ e here. Then V (0) = V ( ∞ ) if and only if (i) V (0) = V ( ∞ ) = 0; or (ii) V (0) = V ( ∞ ) = 1; or (iii) V ( ∞ ) = 2. As we discussed early, V (0) − V ( ∞ ) ≥
2. Hence, in (iii), we have V (0) = 2 too as V (0) cannot be 4 in thissubcase.Thus, in this subcase, g is strictly copositive if and only if either (i) ¯ e > d ≥ d ≥ e ≥
0; or (ii) ¯ e <
0, and d ≥ e >
0; or (iii) ¯ e > d < b = 0 and c (cid:54) = 0. Then b = a , c = 3 ab − c
25 = 6 a − c ,g ( t ) = c t + d t + e ,g ( t ) = 5 t + 4 at + 6 a t + 2 ct + d.g ( t ) − t c g ( t ) = (cid:18) a − d c (cid:19) t + (cid:18) a − e c (cid:19) t + 2 ct + d,g ( t ) − t c g ( t ) − (cid:18) a − d c (cid:19) tc g ( t )= (cid:20)(cid:18) a − e c (cid:19) − (cid:18) a − d c (cid:19) d c (cid:21) t + (cid:20) c − (cid:18) a − d c (cid:19) e c (cid:21) t + d = (cid:20) a − e + 4 ad c + 5 d c (cid:21) t + (cid:20) c − ae c + 5 d e c (cid:21) t + d = 6 a c − e + 4 ad ) c + 25 d c t + 2 cc − ae c + 5 d e c t + d ( t ) − t c g ( t ) − (cid:18) a − d c (cid:19) tc g ( t ) − a c − e + 4 ad ) c + 25 d c g ( t )= (cid:20) cc − ae c + 5 d e c − a c d − e + 4 ad ) c d + 25 d c (cid:21) t + d − a c e − e + 4 ad ) c e + 25 d e c = 10 cc − ae c + 25 c d e − a c d + 5(5 e + 4 ad ) c d − d c t + 5 c d − a c e + 5(5 e + 4 ad ) c e − d e c . Then g ( t ) = d t + e , , where d = ¯ d c , e , = ¯ e , c , ¯ d = − cc + 20 ae c − c d e + 6 a c d − e + 4 ad ) c d + 25 c d , ¯ e , = − c d + 6 a c e − e + 4 ad ) c e + 25 c d e . There are three subcases:(3A) d = e , = 0, i.e., ¯ d = ¯ e , = 0. Then g is the GCD of g and g (cid:48) . Let u = d c and v = e c . We may use Propositions 5.2, 5.3 and Theorem 3.2 to determinethe situation of g ( t ).(3B) d = 0 and e , (cid:54) = 0, i.e., ¯ d = 0 and ¯ e , (cid:54) = 0. Then g ( t ) = e , , S = { g , g , g , g } , S ( ∞ ) = { + ∞ , + ∞ , c , e , } ,S (0) = { e, d, e , e , } . We may replace e , by ¯ e , here. Then V (0) = V ( ∞ ) if and only if (i) V (0) = V ( ∞ ) =0; or (ii) V (0) = V ( ∞ ) = 1; or (iii) V ( ∞ ) = 2, which implies V (0) = 2 by the earlydiscussion.We may replace e , by ¯ e , here. Thus, in this case, g is strictly copositive if andonly if either (i) ¯ e , > c ≥ d ≥ e ≥
0; or (ii) ¯ e , <
0, and d ≥ e >
0; or(iii) ¯ e , > c < d (cid:54) = 0, i.e., ¯ d (cid:54) = 0. Then g ( t ) − c d tg ( t ) = (cid:18) d − c d e , (cid:19) t + e = d d − c e , d t + e .g ( t ) − c d tg ( t ) − d d − c e , d g ( t ) = e − d d − c e , d e , , ( t ) = e = d d − c e , d e , − e = d d e , − c e , − e d d = ¯ e ¯ d , where ¯ e = d ¯ d ¯ e , − c ¯ e , − ¯ e ¯ d . There are two further subcases.(3Ca) e = 0, i.e., ¯ e = 0. Then g ( t ) is the GCD of g and g (cid:48) , and α = − ¯ e , ¯ d isa double root of g . We may use Proposition 5.1 and Theorem 3.2 to determine thesituation of g ( t ).(3Cb) e (cid:54) = 0, i.e., ¯ e (cid:54) = 0. We have S = { g , g , g , g , g } ,S ( ∞ ) = { + ∞ , + ∞ , c , d , e } ,S (0) = { e, d, e , e , , e } . We may replace d , e , and e by ¯ d , ¯ e , and ¯ e here. Then V (0) = V ( ∞ ) if and onlyif (i) V (0) = V ( ∞ ) = 0; or (ii) V (0) = V ( ∞ ) = 1; or (iii) V (0) = V ( ∞ ) = 2; or (iv) V ( ∞ ) = 3, which implies V (0) = 3 by the early note.Thus, in this case, g is strictly copositive if and only if either (i) ¯ e > c ≥ d ≥ d ≥ e ≥
0, ¯ e , ≥
0; or (ii) ¯ e < c ≥ d > d ≥ { e , ¯ e , } > e ≥ e , >
0; or (iii) ¯ e >
0, min { c , ¯ d } <
0, min { d, e , ¯ e , } < d ≥ e > e <
0, ¯ d > c < b (cid:54) = 0. Then g ( t ) = b t + c t + d t + e .g ( t ) − b g ( t ) = (cid:18) a − c b (cid:19) t + (cid:18) b − d b (cid:19) t + (cid:18) c − e b (cid:19) t + d,g ( t ) − b g ( t ) − b (cid:18) a − c b (cid:19) g ( t )= (cid:20)(cid:18) b − d b (cid:19) − c b (cid:18) a − c b (cid:19)(cid:21) t + (cid:20)(cid:18) c − e b (cid:19) − d b (cid:18) a − c b (cid:19)(cid:21) t + d − e b (cid:18) a − c b (cid:19) = 3 bb − (5 d + 4 ac ) b + 5 c e b t + 2 cb − (5 e + 4 ad ) b + 5 c d b t + db − ae b + 5 c e b . g ( t ) = c t + d , t + e , , where c = ¯ c b , d , = ¯ d , b , e , = ¯ e , b , ¯ c = − bb + (5 d + 4 ac ) b − c e , ¯ d , = − cb + (5 e + 4 ad ) b − c d , ¯ e , = − db + 4 ae b − c e . We have four subcases.(4A) c = d , = e , = 0, i.e., ¯ c = ¯ d , = ¯ e , = 0. This implies that g is theGCD of g and g (cid:48) . Let p = c b , q = d b and r = e b . Then Proposition 5.4 determines thesituation of g ( t ).(4B) c = d , = 0 and e , (cid:54) = 0, i.e., ¯ c = ¯ d , = 0 and ¯ e , (cid:54) = 0. Then g ( t ) = e , ,S = { g , g , g , g } ,S ( ∞ ) = { + ∞ , + ∞ , b , e , } ,S (0) = { e, d, e , e , } . We may replace e , by ¯ e , here. Then, g is strictly copositive if and only if either (i)¯ e , > b ≥ d ≥ e ≥
0; or (ii) ¯ e , < d ≥ e >
0; or (iii) ¯ e , > b < c = 0 and d , (cid:54) = 0. Then g ( t ) = d , t + e , ,g ( t ) ≡ e , = − g (cid:18) − e , d , (cid:19) = ¯ e , ¯ d , , where ¯ e , = b ¯ d , ¯ e , − c ¯ d , ¯ e , + d ¯ d , ¯ e , − e ¯ d , . Again, there are two further subcases.(4Ca) e , = 0, i.e., ¯ e , = 0. Then g ( t ) is the GCD of g and g (cid:48) , and g has a doublenonzero root α = − ¯ e , ¯ d , . We may use Proposition 5.1 and Theorem 3.2 to determine the situation of g ( t ).(4Cb) e , (cid:54) = 0, i.e., ¯ e , (cid:54) = 0. We have S = { g , g , g , g , g } , ( ∞ ) = { + ∞ , + ∞ , b , d , , e , } ,S (0) = { e, d, e , e , , e , } . Then V (0) = V ( ∞ ) if and only if (i) V (0) = V ( ∞ ) = 0; or (ii) V (0) = V ( ∞ ) = 1; or(iii) V (0) = V ( ∞ ) = 2; or (iv) V ( ∞ ) = 3, which implies that V (0) = 3 by the earlynote.We may replace d , , e , and e , by ¯ d , , ¯ e , and ¯ e , Thus, in this subcase, g isstrictly copositive if and only if either (i) ¯ e , > b ≥
0, ¯ d , ≥ d ≥ e ≥ e , ≥
0; or (ii) ¯ e , < b ≥ d , > d ≥ { e , ¯ e , } > e ≥ e , > e , >
0, min { b , ¯ d , } <
0, min { d, e , ¯ e , } < d ≥ e >
0; or (iv) ¯ e , < d , > b < c (cid:54) = 0, i.e., ¯ c (cid:54) = 0. Then g ( t ) = c t + d , t + e , ,g ( t ) = b t + c t + d t + e ,g ( t ) − b c tg ( t ) = (cid:18) c − b d , c (cid:19) t + (cid:18) d − b e , c (cid:19) t + e ,g ( t ) − b c tg ( t ) − c (cid:18) c − b d , c (cid:19) g ( t )= (cid:20)(cid:18) d − b e , c (cid:19) − d , c (cid:18) c − b d , c (cid:19)(cid:21) t + e − e , c (cid:18) c − b d , c (cid:19) = (cid:18) d − b e , c − c d , c + b d , c (cid:19) t + e − c e , c + b d , e , c . Then g ( t ) = d t + e , , where d = ¯ d ¯ c , e , = ¯ e , ¯ c , ¯ d = − ¯ c d + b ¯ c ¯ e , + c ¯ c ¯ d , − b ¯ d , , ¯ e , = − ¯ c e + c ¯ c ¯ e , − b ¯ d , ¯ e , . There are three further subcases.(4Da) d = e , = 0, i.e., ¯ d = ¯ e , = 0. Then g is the GCD of g and g (cid:48) . Let u = ¯ d , ¯ c and v = ¯ e , ¯ c . We may use Propositions 5.2, 5.3 and Theorem 3.2 to determinethe situation of g ( t ).(4Db) d = 0 and e , (cid:54) = 0, i.e., ¯ d = 0 and ¯ e , (cid:54) = 0. Then g ( t ) = e , , S = { g , g , g , g , g } , ( ∞ ) = { + ∞ , + ∞ , b , c , e , } ,S (0) = { e, d, e , e , , e , } . Then V (0) = V ( ∞ ) if and only if (i) V (0) = V ( ∞ ) = 0; or (ii) V (0) = V ( ∞ ) = 1; or(iii) V (0) = V ( ∞ ) = 2; or (iv) V ( ∞ ) = 3, which implies V (0) = 3 by the early note.We may replace c , e , and e , here by ¯ c , ¯ e , and ¯ e , . Thus, in this case, g isstrictly copositive if and only if either (i) ¯ e , > b ≥
0, ¯ c ≥ d ≥ e ≥ e , ≥
0; or (ii) ¯ e , < b ≥ c > d ≥ { e , ¯ e , } > e ≥ e , > e , >
0, min { b , ¯ c } <
0, min { d, e , ¯ e , } < d ≥ e >
0; or (iv) ¯ e , < c > b < d (cid:54) = 0, i.e., ¯ d (cid:54) = 0. Then g ( t ) = c t + d , t + e , ,g ( t ) = d t + e , ,g ( t ) ≡ e = − g (cid:18) − e , d (cid:19) = ¯ e b ¯ d , where ¯ e = − ¯ c ¯ e , + ¯ d , ¯ e , ¯ d − ¯ e , ¯ d . There are two further subcases.(4Dc1) e = 0, i.e., ¯ e = 0. Then g ( t ) is the GCD of g and g (cid:48) , and g has a doublenonzero root α = − ¯ e , ¯ d . We may use Proposition 5.5 and Theorem 3.2 to determine the situation of g ( t ).(4Dc2) e (cid:54) = 0, i.e., ¯ e (cid:54) = 0. We have S = { g , g , g , g , g , g } ,S ( ∞ ) = { + ∞ , + ∞ , b , c , d , e } ,S (0) = { e, d, e , e , , e , , e } . Then V (0) = V ( ∞ ) if and only if (i) V (0) = V ( ∞ ) = 0; or (ii) V (0) = V ( ∞ ) = 1; or(iii) V (0) = V ( ∞ ) = 2; or (iv) V (0) = V ( ∞ ) = 3, or (v) V ( ∞ ) = 4, which impliesthat V (0) = 4 by the early note.We may replace c , d , e , , e , and e here by ¯ c , ¯ d , ¯ e , , ¯ e , and ¯ e . Thus, in thiscase, g is strictly copositive if and only if either (i) ¯ e > b ≥
0, ¯ c ≥
0, ¯ d ≥ e , ≥ d ≥ e ≥
0, ¯ e , ≥
0; or (ii) ¯ e < b ≥ { ¯ c , ¯ d } >
0, ¯ c ≥ d > d ≥ { e , ¯ e , , ¯ e , } > e ≥ { ¯ e , , ¯ e , } >
0, ¯ e , ≥ e , > e >
0, min { b , ¯ c , ¯ d } < b ≥ c >
0, min { d, e , ¯ e , , ¯ e , } < d ≥ { e , e , } > e ≥ e , >
0; or (iv) e <
0, min { b , ¯ c } <
0, max { ¯ c , ¯ d } > { d, e , ¯ e , } <
0, max { e , ¯ e , , ¯ e , } > d ≥ e >
0; or (v) ¯ e >
0, ¯ d < c > b <
0. 21 .3 Positivity Conditions for Quintic Polynomials
Summarizing the coefficients, we have b = 4 a − b , c = 3 ab − c , d = 2 ac − d , e = ad − e , (5.13)¯ e = − e + 100 ae d − a e d + 4 a e d − dd , (5.14)¯ d = − cc + 20 ae c − c d e + 6 a c d − e + 4 ad ) c d + 25 c d , (5.15)¯ e , = − c d + 6 a c e − e + 4 ad ) c e + 25 c d e , (5.16)¯ e = d ¯ d ¯ e , − c ¯ e , − ¯ e ¯ d , (5.17)¯ c = − bb + (5 d + 4 ac ) b − c e , (5.18)¯ d , = − cb + (5 e + 4 ad ) b − c d , (5.19)¯ e , = − db + 4 ae b − c e , (5.20)¯ e , = b ¯ d , ¯ e , − c ¯ d , ¯ e , + d ¯ d , ¯ e , − e ¯ d , , (5.21)¯ d = − ¯ c d + b ¯ c ¯ e , + c ¯ c ¯ d , − b ¯ d , , (5.22)¯ e , = − ¯ c e + c ¯ c ¯ e , − b ¯ d , ¯ e , , (5.23)¯ e = − ¯ c ¯ e , + ¯ d , ¯ e , ¯ d − ¯ e , ¯ d . (5.24)We now can state our theorem. Theorem 5.5
Let g ( t ) be defined by (5.6), with e > . Let the additional coefficientsbe defined by (5.13-5.24). Then we have cases.(1) b = c = d = 0 . In this case, g ( t ) > for all t ≥ .(2) b = c = 0 , d (cid:54) = 0 and ¯ e (cid:54) = 0 . In this case, g ( t ) > for all t ≥ if and onlyif either (i) ¯ e > , d > , d ≥ , e ≥ ; or (ii) ¯ e < , and d ≥ if e > ; or (iii) ¯ e > , d < .(3) b = 0 , c (cid:54) = 0 , ¯ d = 0 and ¯ e , (cid:54) = 0 . In this case, g ( t ) > for all t ≥ if andonly if either (i) ¯ e , > , c ≥ , d ≥ , e ≥ ; or (ii) ¯ e , < , and d ≥ if e > ;or (iii) ¯ e , > , c < .(4) b = 0 , c (cid:54) = 0 , ¯ d (cid:54) = 0 and ¯ e (cid:54) = 0 . In this case, g ( t ) > for all t ≥ if andonly if either (i) ¯ e > , c ≥ , ¯ d ≥ , d ≥ , e ≥ , ¯ e , ≥ ; or (ii) ¯ e < , c ≥ if ¯ d > , d ≥ if max { e , ¯ e , } > , e ≥ if ¯ e , > ; or (iii) ¯ e > , min { c , ¯ d } < , min { d, e , ¯ e , } < , d ≥ if e > ; or (iv) ¯ e < , ¯ d > , c < .(5) b (cid:54) = 0 , ¯ c = ¯ d , = 0 , and ¯ e , (cid:54) = 0 . In this case, g ( t ) > for all t ≥ if andonly if either (i) ¯ e , > , b ≥ , d ≥ , e ≥ ; or (ii) ¯ e , < , d ≥ if e > ; or(iii) ¯ e , > , b < . b (cid:54) = 0 , ¯ c = 0 , ¯ d , (cid:54) = 0 and ¯ e , (cid:54) = 0 . In this case, g ( t ) > for all t ≥ if andonly if either (i) ¯ e , > , b ≥ , ¯ d , ≥ , d ≥ , e ≥ , e , ≥ ; or (ii) ¯ e , < , b ≥ if ¯ d , > , d ≥ if max { e , ¯ e , } > , e ≥ if ¯ e , > ; or (iii) ¯ e , > , min { b , ¯ d , } < , min { d, e , ¯ e , } < , d ≥ if e > ; or (iv) ¯ e , < , ¯ d , > , b < .(7) b (cid:54) = 0 , ¯ c (cid:54) = 0 and ¯ d = 0 . In this case, g ( t ) > for all t ≥ if and only ifeither (i) ¯ e , > , b ≥ , ¯ c ≥ , d ≥ , e ≥ , ¯ e , ≥ ; or (ii) ¯ e , < , b ≥ if ¯ c > , d ≥ if max { e , ¯ e , } > , e ≥ if ¯ e , > ; or (iii) ¯ e , > , min { b , ¯ c } < , min { d, e , ¯ e , } < , d ≥ if e > ; or (iv) ¯ e , < , ¯ c > , b < .(8) b (cid:54) = 0 , ¯ c (cid:54) = 0 , ¯ d (cid:54) = 0 and ¯ e (cid:54) = 0 . In this case, g ( t ) > for all t ≥ if andonly if either (i) ¯ e > , b ≥ , ¯ c ≥ , ¯ d ≥ , ¯ e , ≥ , d ≥ , e ≥ , ¯ e , ≥ ; or(ii) ¯ e < , b ≥ if max { ¯ c , ¯ d } > , ¯ c ≥ if ¯ d > , d ≥ if max { e , ¯ e , , ¯ e , } > , e ≥ if max { ¯ e , , ¯ e , } > , ¯ e , ≥ if ¯ e , > ; or (iii) ¯ e > , min { b , ¯ c , ¯ d } < , b ≥ if ¯ c > , min { d, e , ¯ e , , ¯ e , } < , d ≥ if max { e , ¯ e , } > , e ≥ if ¯ e , > ; or (iv) ¯ e < , min { b , ¯ c } < , max { ¯ c , ¯ d } > , min { d, e , ¯ e , } < , max { e , ¯ e , , ¯ e , } > , d ≥ if e > ; or (v) ¯ e > , ¯ d < , ¯ c > , b < .In these eight cases, g ( t ) > for all t ≥ if and only if g ( t ) ≥ for all t ≥ .(9) (i) b = c = 0 , d (cid:54) = 0 and ¯ e = 0 . Let α = − e d .(ii) b = 0 , c (cid:54) = 0 , ¯ d (cid:54) = 0 and ¯ e = 0 . Let α = − ¯ e , ¯ d .(iii) b (cid:54) = 0 , ¯ c = 0 , ¯ d , (cid:54) = 0 and ¯ e , = 0 . Let α = − ¯ e , ¯ d , .(iv) b (cid:54) = 0 , ¯ c (cid:54) = 0 , ¯ d (cid:54) = 0 and ¯ e = 0 . Let α = − ¯ e , ¯ d .Then we may apply Proposition 5.1 and Theorem 3.2 to determine the situation of g ( t ) .(10) (i) b = 0 , c (cid:54) = 0 and ¯ d = ¯ e , = 0 . Let u = d c and v = e c .(ii) b (cid:54) = 0 , ¯ c (cid:54) = 0 and ¯ d = ¯ e , = 0 . Let u = ¯ d , ¯ c and v = ¯ e , ¯ c .Then we may use Propositions 5.2, 5.3 and Theorem 3.2 to determine the situationof g ( t ) .(11) b (cid:54) = 0 and ¯ c = ¯ d , = ¯ e , = 0 . Let p = c b , q = d b and r = e b . ThenProposition 5.4 determines the situation of g ( t ) . Proof
The 11 cases are summarized from the discussion in the last subsection. Cases(1-8) are corresponding to Cases (1), (2B), (3B), (3Cb), (4B), (4Cb), (4Db) and (4Dc2)of the last subsection, respectively. Case (9) is corresponding Cases (2A), (3Ca), (4Ca)and (4Dc1) of the last subsection. Case (10) is corresponding Cases (3A) and (4Da) ofthe last subsection. Case (11) is corresponding to Case (4A) of the last subsection. (cid:3) Higher Even Degree Polynomials and Appendices
We consider non-degenerate even degree polynomials with their degrees higher than orequal to 4. Suppose that g ( t ) = m (cid:88) i =1 a i t m − i , where m ≥ a = a m = 1, a , · · · , a m − are real numbers. Denote S = { ( a , · · · , a m − ) ∈ (cid:60) m − : g ( t ) > t } , ¯ S = { ( a , · · · , a m − ) ∈ (cid:60) m − : g ( t ) ≥ t } and A = { ( a , · · · , a m − ) ∈ (cid:60) m − : g ( t ) = ( t + ut + v ) φ ( t ) , u < v, φ ( t ) ≥ t } . We call A the appendix of ¯ S . We have the following theorem. Theorem 6.1
In the above setting, ¯ S is a closed convex set, S is its interior, and A ⊂ S . The dimension of ¯ S and S is m − , while the dimension of A is m − . Thediscriminant ∆( g ) is equal to zero on A and the boundary of ¯ S , and has the same signat the other part of ¯ S . Proof
Suppose that g ( t ) = m (cid:88) i =1 a i t m − i , ˆ g ( t ) = m (cid:88) i =1 ˆ a i t m − i ,g ( t ) , ˆ g ( t ) ≥ t , a = a m = ˆ a = ˆ a m = 1. Then ( g ( t ) + ˆ g ( t )) ≥ t . Thisshows that ¯ S is convex. Taking limiting points, we see that ¯ S is closed. Similarly, S isalso convex. Consider min { g ( t ) } . By the continuity property, we see that S is an openset, and ¯ S is the closure of S . Then S is the interior of ¯ S . For any ( a , · · · , a m − ) ∈ A , g ( t ) > t . Thus, A ⊂ S . We also see that there is an (cid:15) > a , · · · , a m − ) ∈ S for | a i | ≤ (cid:15), i = 1 , · · · , m −
1. This shows that the dimension of¯ S and S is m −
1. Consider the number of independent parameters of g ( t ) in A , weconclude that the dimension of A is m −
3. For g ( t ) in A , g ( t ) has multiple roots. Thus∆( g ) = 0. On a boundary point of ¯ S , as it neighbors some parts not in ¯ S , we alsohave ∆( g ) = 0. On the other part of ¯ S , as it is in the interior of ¯ S , and g ( t ) has nomultiple roots there, we have ∆( g ) (cid:54) = 0. As A is of dimension m −
3, the other partsis connected. Thus, ∆( g ) has the same sign there. (cid:3) g ( t ) without real roots,∆( g ) takes the same sign, which depends upon m only, except at an appendix set ofdimension lower by two, where ∆( g ) = 0.Note that such an appendix also exists for odd degree polynomials with their degreeshigher than 4. In Subsection 5.1, for the subcase (B1), if u < v , then there existsalso an appendix similarly. However, as the property that g ( t ) ≥ t only worksfor even degree polynomials, only for even degree polynomials, the properties of theappendix are distinguished. Acknowledgment
We are thankful to Chen Ouyang and Jinjie Liu for drawingFigures 1 and 2.
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