A Condition for Multiplicity Structure of Univariate Polynomials
aa r X i v : . [ c s . S C ] A ug A Condition for Multiplicity Structure of Univariate Polynomials
Hoon HongDepartment of Mathematics, North Carolina State UniversityBox 8205, Raleigh, NC 27695, [email protected] Yang ∗ SMS–KLSE–School of Mathemetics and Physics, Guangxi University for NationalitiesNanning 530006, [email protected]
Abstract
We consider the problem of finding a condition for a univariate polynomial having a given multiplicitystructure when the number of distinct roots is given. It is well known that such conditions can bewritten as conjunctions of several polynomial equations and one inequation in the coefficients, by usingrepeated parametric gcd’s. In this paper, we give a novel condition which is not based on repeated gcd’s.Furthermore, it is shown that the number of polynomials in the condition is optimal and the degree ofpolynomials is smaller than that in the previous condition based on repeated gcd’s.
In this paper, we consider the problem of finding a condition on the coefficients of a polynomial over thecomplex field C so that it has a given multiplicity structure. For example, consider a quartic polynomial F = a x + a x + a x + a x + a where a i ’s take values over C . We would like to find a conditionon a i ’s so that F has the multiplicity structure (3 , r and r , where the multiplicities of r and r are 3 and 1 respectively. The problem is important becausemany tasks in mathematics, science and engineering can be reduced to the problem. A prerequisite for theproblem is finding a condition on coefficients such that the polynomials has the given number of distinctroots. This is already well studied. For instance, the subdiscriminant theory provides a complete solutionto the sub-problem. More explicitly, a univariate polynomial of degree n has m distinct roots if and only ifits 0-th, . . . ,( n − m − n − m )-th psddoes not. For details, see standard textbooks on computational algebra (e.g., [1]).Thus from now on, we will assume that the number of distinct roots is fixed, say m . However, even withthis assumption, there can be several different multiplicity structures. For example, consider again a quarticunivariate polynomial F . Assume that it has two distinct roots. Then its multiplicity may be (3 ,
1) or (2 , Problem : Let µ = ( µ , . . . , µ m ) be such that µ , . . . , µ m ≥ µ + · · · + µ m = n . Find a condition onthe coefficients of a polynomial F over C of degree n with m distinct complex roots so that the multiplicitystructure of F is µ . (We will call the condition a µ - multiplicity-discriminating condition.) Due to its importance, the problem and several related problems have been already carefully studied. In[11], Yang, Hou and Zeng gave an algorithm to generate a multiplicity-discriminating condition (referred as ∗ Corresponding author. F ( x ) + I · G ( x ) where F, G ∈ R [ x ]and I is the imaginary unit. Further improvement and generalization can be found in [6, 7]. Multiplicitystructure is a particular root configuration of a univariate polynomial. In [10], another particular rootconfiguration is studied where there exists a symmetric triple of roots among which one root is the averageof the other two.It is known that a multiplicity-discriminating condition can be written as a conjunction of several poly-nomial equations and one inequation on the coefficients. For example, for a quartic polynomial with twodistinct roots, see two different conditions in Example 3. In general, there are infinitely many syntacticallydifferent conditions. Thus a challenge is to find a condition with “small” size. A natural way to measure the“size”of the condition is the number of polynomials appearing in the condition and their maximum degree.The main contribution in this paper is to provide a condition with only one polynomial with degree smaller than those in the previous method. The condition is novel in that it is based on a significantly differenttheory and techniques from the previous methods (which are essentially based on repeated parametric gcd orsubdiscriminant theory). In order to find the new condition we developed the following ideas and techniques.1. Convert the multiplicity condition in roots into an equivalent permanental equation in roots.2. Convert the permanent in roots into a sum of determinants in roots.3. Convert each determinant in roots into a determinant in coefficients.We found that the above ideas/techniques are interesting on their own. We hope that they could be usefulfor tackling other related problems.The paper is structured as follows. In Section 2, we give a precise statement of the main result of thepaper (Theorem 6). In Section 3, we give a proof of Theorem 6. The proof is long thus we divide theproof into three subsections which are interesting on their own. In Section 4, we compare the sizes of themultiplicity-discriminant condition in Theorem 6 and that given by a previous work. In this section, we give a precise statement of the main result of the paper. For this, we need a few notionsand notations.
Definition 1 (Multiplicity of a polynomial) . Let C be the complex field and F ∈ C [ x ] be with m distinctcomplex roots, say r , . . . , r m . The multiplicity of F , written as mult ( F ) , is defined by mult ( F ) = ( µ , . . . , µ m ) where µ i is the multiplicity of r i as a root of F . Without losing generality, we assume that µ ≥ · · · ≥ µ m . Assumption 1.
We assume that ≤ m ≤ n − . Remark 2.
The assumption is natural and meaningful because otherwise there is nothing to discriminate:If m = 1 then the only possible multiplicity is ( n ) . If m = n then the only possible multiplicity is (1 , . . . , .If m = n − then the only possible multiplicity is (2 , , . . . , . Problem 1.
Let n ≥ m be fixed. Input: µ = ( µ , . . . , µ m ) such that µ ≥ · · · ≥ µ m ≥ and µ + · · · + µ m = n Output: a µ - multiplicity-discriminating condition, that is, a condition on the coefficients of a polynomial F of degree n with m distinct complex roots so that the multiplicity structure of F is µ .2 xample 3. Let F ( x ) = a x + a x + a x + a x + a be such that deg F = 4 and the number of distinctroots of F is . The followings two are (3 , -multiplicity discriminating conditions on F .1. C = 0 ∧ C = 0 where C = − a a a + 12 a a a + 576 a a a + 48 a a a a − a a a − a a a a +432 a a a + 128 a a a a + 4096 a a a − a a a C = 16 a a − a a . C ′ = 0 where C ′ = − a a + 64 a a a a − a a a + 8 a a a − a a a − a a . Remark 4.
As you see in the above example, in general, the µ -multiplicity discriminating condition of F isnot unique syntactically. In fact, there are infinitely many syntactically different µ -multiplicity discriminantconditions. Thus a challenge is to find a syntactically “small” condition. Definition 5 (Determinant of polynomials) . Let F , . . . , F k ∈ C [ x ] be such that deg F , . . . , deg F k ≤ k .Then the determinant of the polynomials dp is defined by dp F ... F k = det a ,k · · · a , ... ... a k,k · · · a k, where F i = P ≤ j ≤ k a i,j x j . We have introduced all the necessary notions and notations, and thus, now we give a precise statement ofthe main result of this paper.
Theorem 6 (Main result) . Let µ = ( µ , . . . , µ m ) be such that µ , . . . , µ m ≥ and µ + · · · + µ m = n , and F ∈ C [ x ] be of degree n . Let D µ ( F ) = X σ ∈ S p dp x n − µ m − F ... x Fx n − F ( σ ) /σ ! ... x F ( σ n ) /σ n ! where p = ( µ , . . . , µ | {z } µ , . . . , µ m , . . . , µ m | {z } µ m ) , S p is the set of all permutations of p and F ( i ) is the i -th derivativeof F in terms of x . Then D µ ( F ) = 0 is a µ -multiplicity-discriminanting condition. Remark 7.
Assume F = P ni =0 a i x i . Then a straightforward degree analysis of the expression of D µ inTheorem 6 shows that the degree of D µ ( F ) in a is n − µ m where a = ( a , a , . . . , a n ) . Example 8.
Let F ( x ) = a x + a x + a x + a x + a be such that deg F = 4 and the number of distinctroots of F is . We will construct a (3 , -multiplicity discriminating condition on F , using the main result(Theorem 6). Note1. µ = (3 , p = (3 , , , S p = { (3 , , , , (3 , , , , (3 , , , , (1 , , , } . D (3 , ( F ) = dp x Fx Fx Fx F (3) / x F (3) / x F (3) / x F (1) / + dp x Fx Fx Fx F (3) / x F (3) / x F (1) / x F (3) / + dp x Fx Fx Fx F (3) / x F (1) / x F (3) / x F (3) / + dp x Fx Fx Fx F (1) / x F (3) / x F (3) / x F (3) / = det a a a a a a a a a a a a a a a a a a a a a a a a a + det a a a a a a a a a a a a a a a a a a a a a a a a a +det a a a a a a a a a a a a a a a a a a a a a a a a a + det a a a a a a a a a a a a a a a a a a a a a a a a a = − a a + 64 a a a a − a a a + 8 a a a − a a a − a a Note that it is the polynomial C ′ in Example 3.5. The main result (Theorem 6) states that D (3 , ( F ) = 0 is a (3 , -multiplicity discriminating conditionon F . Remark 9.
Observe D (1 ,..., ( F ) is the Sylvester resultant of F and F ′ . Thus D µ can be viewed as a certaingeneralization of Sylvester resultant of F and F ′ (i.e., the traditional discriminant of F up to sign). Let F ∈ C [ x ] be of degree n and α , . . . , α n be the n roots of F . In this section, we give a proof of Theorem6. The proof is long thus we divide the proof into three steps (lemmas), which are interesting on their own.1. Lemma 10: We show that the multiplicity condition in roots can be converted into an equivalentpolynomial inequation which is a permanental expression in roots.2. Lemma 13: We show that the permanent in roots can be converted into a sum of determinants inroots.3. Lemma 15: We show that each determinant in roots can be converted into a determinant in coefficients.Finally, we combine the above three lemmas to prove the main result.4 .1 From a condition in roots to a permanental condition in roots Lemma 10.
Let F be of degree n and α , . . . , α n be its n complex roots. Then mult ( F ) = µ ⇐⇒ D µ ( F ) = 0 where • D µ ( F ) = per F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p ! ... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! / c ; • per stands for the permanent operation; • p = ( µ , . . . , µ | {z } µ , . . . , µ m , . . . , µ m | {z } µ m ) ; • c is an integer determined by p ; specifically if p consists of ℓ distinct numbers occurring q , . . . , q ℓ times,then c = Q ℓi =1 q i ! .Proof. We will prove the lemma by “deriving” the condition D µ ( F ) = 0 instead of merely “verifying” thecorrectness of the lemma, since it will be much more interesting to read, bringing out the underlying ideasand intuitions. The derivation will be driven by two wishes:1. Wish to find a condition that involves a single polynomial on the all the (not necessarily distinct) roots α , . . . , α n .2. Wish to find the polynomial which is symmetric in the roots, so that later we can turn it into anexpression in the coefficients.The strategy is to repeatedly rewrite the condition mult ( F ) = µ with the above two wishes in mind.1. Note that the condition mult ( F ) = µ is written in terms of distinct roots. We rewrite the conditioninto a symmetric condition on all the (not necessarily distinct) roots α , . . . , α n .mult ( F ) = µ ⇐⇒ _ σ ∈ S p n ^ i =1 mult ( α i ) = σ i Proof: Obvious from the definition of multiplicity of a root.2. We rewrite the symmetric condition into a symmetric polynomial condition _ σ ∈ S p n ^ i =1 mult ( α i ) = σ i ⇐⇒ _ σ ∈ S p n ^ i =1 F ( σ i ) ( α i ) = 0Proof: We will show each direction of ⇐⇒ one by one.(a) W σ ∈ S p n V i =1 mult ( α i ) = σ i = ⇒ W σ ∈ S p n V i =1 F ( σ i ) ( α i ) = 0From the elementary calculus, we havemult ( α i ) = σ i = ⇒ F ( σ i ) ( α i ) = 0 . Thus the direction follows immediately. 5b) W σ ∈ S p n V i =1 F ( σ i ) ( α i ) = 0 = ⇒ W σ ∈ S p n V i =1 mult ( α i ) = σ i It is immediate from the following two sub-claims.i. n V i =1 F ( σ i ) ( α i ) = 0 = ⇒ n V i =1 mult ( α i ) ≤ σ i It is immediate from elementary calculus.ii. n V i =1 mult ( α i ) ≤ σ i = ⇒ n V i =1 mult α i = σ i It is immediate from ∀ σ ∈ Sp n P i =1 mult ( α i ) = n P i =1 σ i , which is again obvious from the definitionsof mult and S p .3. We rewrite the condition so that a fewer polynomials are involved. _ σ ∈ S p n ^ i =1 F ( σ i ) ( α i ) = 0 ⇐⇒ _ σ ∈ S p n Y i =1 F ( σ i ) ( α i ) = 0Proof: Obvious.4. We rewrite the condition so that only one polynomial is involved. _ σ ∈ S p n Y i =1 F ( σ i ) ( α i ) = 0 ⇐⇒ X σ ∈ S p n Y i =1 F ( σ i ) ( α i ) = 0Proof: We prove the implication for both directions.(a) W σ ∈ S p n Q i =1 F ( σ i ) ( α i ) = 0 = ⇒ P σ ∈ S p n Q i =1 F ( σ i ) ( α i ) = 0Immediate from the fact that if n V i =1 mult ( α i ) = σ i then ∀ π ∈ S p ,π = σ Q ni =1 F ( π i ) ( α i ) = 0 . (b) P σ ∈ S p n Q i =1 F ( σ i ) ( α i ) = 0 = ⇒ W σ ∈ S p n Q i =1 F ( σ i ) ( α i ) = 0.Obvious.Now we have arrived at our goal by deriving a single symmetric polynomial condition from the multi-plicity condition given at the beginning.We will carry out a few “cosmetic” rewritings: (1) remove some redundancies and (2) write the conditionmore compactly by recalling permanent.5. We remove some redundancies in the coefficients of F ( σ i ) ( α i ). X σ ∈ S p n Y i =1 F ( σ i ) ( α i ) = 0 ⇐⇒ X σ ∈ S p n Y i =1 F ( σ i ) ( α i ) /σ i ! = 0Proof: Obvious.6. We rewrite the condition more compactly by recalling permanent.6onsider the following permanent: P := per F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! Expanding P , we get P = X π ∈ S n n Y i =1 F ( p π ( i ) ) ( α i ) (cid:14) p π ( i ) !Since, for σ ∈ S p , there are c = Q ℓi =1 q i ! distinct permutations π ’s in S n such that π ( p ) = σ where q , . . . , q ℓ are the occurrences of distinct numbers in p , we have X σ ∈ S p n Y i =1 F ( σ i ) ( α i ) (cid:14) σ i ! = per F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! /c
7. By denoting per F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! /c by D µ ( F ), we finally havemult ( F ) = µ ⇐⇒ D µ ( F ) = 0 Example 11.
Let F = ( x − α )( x − α )( x − α )( x − α ) and µ = (3 , . Then p = (3 , , , and c = 3! · .Thus D µ ( F ) = per F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (3) ( α )3! F (1) ( α )1! F (1) ( α )1! F (1) ( α )1! F (1) ( α )1! / (3! · P ≤ i ≤ i =1 ( α − α i ) P ≤ i ≤ i =2 ( α − α i ) P ≤ i ≤ i =3 ( α − α i ) P ≤ i ≤ i =4 ( α − α i ) P ≤ i ≤ i =1 ( α − α i ) P ≤ i ≤ i =2 ( α − α i ) P ≤ i ≤ i =3 ( α − α i ) P ≤ i ≤ i =4 ( α − α i ) P ≤ i ≤ i =1 ( α − α i ) P ≤ i ≤ i =2 ( α − α i ) P ≤ i ≤ i =3 ( α − α i ) P ≤ i ≤ i =4 ( α − α i ) Q ≤ i ≤ i =1 ( α − α i ) Q ≤ i ≤ i =2 ( α − α i ) Q ≤ i ≤ i =3 ( α − α i ) Q ≤ i ≤ i =4 ( α − α i ) / (3! · − ( α + α − α − α ) ( α + α − α − α ) ( α + α − α − α ) . If we know that F has two distinct roots, then D µ ( F ) = 0 if and only if µ = (3 , . Remark 12.
Suppose F = a n Q ni =1 ( x − α i ) . Let µ = (1 , . . . , . Then D µ ( F ) = n Y i =1 F ′ ( α i ) = a n − n Y i = j ( α i − α j ) which is the well known discriminant up to sign. Thus D µ ( F ) in Lemma 10 can be viewed as a certaingeneralization of discriminant. The results presented in this subsection and the next subsection are more general than what are needed forproving the main result (Theorem 6). We present the more general results in the hope that they would beuseful for some other related problems. The following lemma shows that one can rewrite a permanent interms of determinants.
Lemma 13.
Let A and B be square matrices of size n . We have per ( A ) = 1det ( B ) X τ ∈ S n det (( P τ A ) ◦ B ) where • The notation ◦ stands for the entry-wise (Hadamard) product; in other words, the ( i, j ) -th entry of A ◦ B is the product of the ( i, j ) -th entries of A and B ; • The notation P τ B stands for the matrix obtained by permuting the rows of B as indicated by τ .Proof. We will rewrite a permanent in terms of determinants as follows.1. Recalling the definition of permanent, we haveper ( A ) = X τ ∈ S n n Y j =1 a τ ( j ) ,j
2. Now we make a simple, but crucial rewriting of the above expression into the followingper ( A ) = X τ ∈ S n n Y j =1 a ( τ ◦ π )( j ) ,j for arbitrary π ∈ S n . Note that τ is replaced with τ ◦ π . This is correct because τ ◦ π also rangesover S n . Why we make the above rewriting will be made clear in the following steps.8. Recalling the definition of determinant, we haveper ( A ) det ( B ) = X τ ∈ S n n Y j =1 a τ ( π ( j )) ,j X π ∈ S n sgn ( π ) n Y j =1 b π ( j ) ,j
4. Rearranging the sums and the products, we haveper ( A ) det ( B ) = X τ ∈ S n X π ∈ S n sgn ( π ) n Y j =1 (cid:0) a τ ( π ( j )) ,j b π ( j ) ,j (cid:1)
5. Writing in terms of determinants and Hadamard product, we haveper ( A ) det ( B ) = X τ ∈ S n det (( P τ A ) ◦ B )6. Finally we have per ( A ) = 1det ( B ) X τ ∈ S n det (( P τ A ) ◦ B ) Example 14.
When n = 2 , we have S = { (1) , (12) } . Let A = (cid:20) a a a a (cid:21) and B = (cid:20) b b b b (cid:21) . Thenwe have per ( A ) = a a + a a , det ( B ) = b b − b b X τ ∈ S det (( P τ A ) ◦ B )= det (cid:0) P (1) A ◦ B (cid:1) + det (cid:0) P (12) A ◦ B (cid:1) = det (cid:20) a b a b a b a b (cid:21) + det (cid:20) a b a b a b a b (cid:21) =( a a b b − a a b b ) + ( a a b b − a a b b )=( a a + a a )( b b − b b )= per ( A ) det ( B ) Thus, per ( A ) = 1det ( B ) X τ ∈ S det (( P τ A ) ◦ B ) Let C [ x ] k stand for the set of all polynomials in C [ x ] with degree at most k .9 emma 15. Let ω , . . . , ω k be a canonical basis of C [ x ] k for every k ≥ . Let F = a n ( x − α ) · · · ( x − α n ) and G , . . . , G n ∈ C [ x ] n − . Then we have dp ω n − F ... ω FG ... G n = a n − n · det G ( α ) · · · G ( α n ) ... ... G n ( α ) · · · G n ( α n ) det ω n − ( α ) · · · ω n − ( α n ) ... ... ω ( α ) · · · ω ( α n ) Proof.
We will derive the expression step by step.1. Let M F = ω n − F ... ω F M G = G ... G n
2. Let M ∈ C (2 n − × (2 n − be such that (cid:20) M F M G (cid:21) = M ω n − ... ω Then by the definition of dp, we have dp (cid:20) M F M G (cid:21) = det ( M ).3. Let us partition M naturally as M = (cid:20) A BC D (cid:21) where A ∈ C ( n − × ( n − B ∈ C ( n − × n C ∈ C n × ( n − D ∈ C n × n
4. Now we introduce a crucial object in the derivation. W = (cid:20) I n − UV (cid:21) where U = ω n − ( α ) · · · ω n − ( α n )... ... ω n ( α ) · · · ω n ( α n ) and V = ω n − ( α ) · · · ω n − ( α n )... ... ω ( α ) · · · ω ( α n ) Note that V is the generalized Vandermonde matrix of F (up to ordering of rows). A similar objectwas also used in [4] for studying Sylvester double sum.5. Note M W = (cid:20) A BC D (cid:21) (cid:20) I n − U V (cid:21) = (cid:20) A AU + BVC CU + DV (cid:21) (cid:20) A M F ( α ) · · · M F ( α n ) C M G ( α ) · · · M G ( α n ) (cid:21) = (cid:20) AC M G ( α ) · · · M G ( α n ) (cid:21) since F ( α i ) = 06. Thus det ( M ) · det ( W ) = det ( A ) · det (cid:2) M G ( α ) · · · M G ( α n ) (cid:3) = det ( A ) · det (cid:2) G ( α ) · · · G ( α n ) (cid:3) where G ( α i ) = (cid:2) G ( α i ) · · · G n ( α i ) (cid:3) T .7. Note that det ( W ) = det ( I n − ) · det ( V ) = det ( V ) since W is block-triangular.8. Note that det ( A ) = a n − n since A is triangular and the diagonal elements are a n .9. By putting together we have dp (cid:20) M F M G (cid:21) = det ( M ) = a n − n · det (cid:2) G ( α ) · · · G ( α n ) (cid:3) det ( V ) Example 16.
Let F = a x + a x + a x + a = a ( x − α )( x − α )( x − α ) and G i = x F ( i ) /i ! . Let ω i = x i .Then dp ω Fω FG G G = det a a a a a a a a a a a a a a = 9 a a det G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) = det a α ( α − α ) ( α − α ) a α ( α − α ) ( α − α ) a α ( α − α ) ( α − α ) a α (2 α − α − α ) a α (2 α − α − α ) a α (2 α − α − α ) a α a α a α =9 a α α α ( α − α ) ( α − α ) ( α − α )=9 a a ( α − α ) ( α − α ) ( α − α )det ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) = det α α α α α α α α α = ( α − α )( α − α )( α − α ) Thus we have dp ω Fω FG G G = a · det G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) G ( α ) det ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) ω ( α ) .4 Proof of Main Result (Theorem 6) Finally we will prove the main result by combining the above three lemmas (Lemma 10, 13 and 15).1. From Lemma 10, we havemult ( F ) = µ ⇐⇒ per F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! = 02. Applying Lemma 13 to A = α n − · · · α n − n ... ... α · · · α n , B = F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! and dividing det ( A ) on both sides, we haveper F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! = P τ ∈ S n det α n − · · · α n − n ... ... α · · · α n ◦ P τ F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! det α n − · · · α n − n ... ... α · · · α n Since for σ ∈ S p , there are c = Q ℓi =1 q i ! τ ’s such that τ ( p ) = σ where q , . . . , q ℓ are the occurrences ofdistinct numbers in p , we haveper F ( p ) ( α ) p ! · · · F ( p ) ( α n ) p !... ... F ( p n ) ( α ) p n ! · · · F ( p n ) ( α n ) p n ! = c · P σ ∈ S p det α n − · · · α n − n ... ... α · · · α n ◦ F ( σ ) ( α ) σ ! · · · F ( σ ) ( α n ) σ !... ... F ( σ n ) ( α ) σ n ! · · · F ( σ n ) ( α n ) σ n ! det α n − · · · α n − n ... ... α · · · α n c · X σ ∈ S p det α n − F ( σ )1 ( α ) σ ! · · · α n − n F ( σ ) n ( α n ) σ !... ... α F ( σ n )1 ( α ) σ n ! · · · α n F ( σ n ) n ( α n ) σ n ! det α n − · · · α n − n ... ... α · · · α n
3. By applying Lemma 15 to G i ( x ) = x n − i F ( σ i ) ( x ) σ i ! and ω i = x i , we have a n − n · det α n − F ( σ ) ( α ) σ ! · · · α n − n F ( σ ) ( α n ) σ !... ... α F ( σ n ) ( α ) σ n ! · · · α n F ( σ n ) ( α n ) σ n ! det α n − · · · α n − n ... ... α · · · α n = dp x n − F ... x Fx n − F ( σ ) /σ !... x F ( σ n ) /σ n ! . Finally, combining the above three steps, we havemult ( F ) = µ ⇐⇒ D µ ( F ) = 0 . We have proved the main result (Theorem 6).
In this section, we compare the “sizes” of the multiplicity-discriminanting condition in Theorem 6 and thatgiven by a complex root version of Yang-Hou-Zeng (YHZ) [11]. Specifically we compare the number and themaximum degrees of polynomials appearing in the conditions. In Table 1, we show a comparison for n = 8.They were determined through brute-force computations. In the table, we used the following short-hands: • NEW denotes the number of polynomials appearing in the new condition (Theorem 6) • YHZ denotes the number of polynomials appearing in the YHZ’s condition • d NEW denotes the degree of the polynomial D µ appearing in the new condition (Theorem 6) • d YHZ denotes the maximum of the degrees of the polynomials appearing in the YHZ’s condition.We make a few observations on the table. They introduced the key concept of complete discriminant system for polynomials which is a set of explicit expressions ofthe coefficients to determine the numbers and multiplicities of real and non-real roots. The conditions in the solution to thecomplete root classification problem consists of equations, inequations and inequalities. When restricted to discriminating onlymultiplicities of roots (without discriminating between real and non-real roots), computing a complete discriminant system isequivalent to computing greatest common divisors iteratively. n m µ
NEW
YHZ d NEW d YHZ ,
4] 1 7 12 81[5 ,
3] 1 8 13 81[6 ,
2] 1 11 14 63[7 ,
1] 1 16 15 333 [3 , ,
2] 1 3 14 75[4 , ,
2] 1 4 14 75[4 , ,
1] 1 5 15 75[5 , ,
1] 1 7 15 75[6 , ,
1] 1 11 15 454 [2 , , ,
2] 1 1 14 49[3 , , ,
1] 1 2 15 49[3 , , ,
1] 1 3 15 63[4 , , ,
1] 1 4 15 635 [2 , , , ,
1] 1 1 15 45[3 , , , ,
1] 1 2 15 45[4 , , , ,
1] 1 4 15 456 [2 , , , , ,
1] 1 1 15 33[3 , , , , ,
1] 1 2 15 331. Concerning the number of polynomials:(a) Observe that
NEW = 1 always. It is obvious from Theorem 6.(b) Observe that
YHZ = 1 when the entries of µ are at most 2 and that YHZ is large when someentries of µ are large. In fact, the observations hold in general, since straightforward book-keepingof YHZ’s algorithm immediately shows that YHZ = 1 + m X i =1 (cid:18) µ i − (cid:19) For a proof, see Lemma 17 in Appendix.(c) Hence
NEW ≤ YHZ always and = holds only when the entries of µ are at most 2.2. Concerning the maximum degree of polynomials:(a) Observe that d NEW ≤ n − d NEW = 2 n − µ m in Remark 7.(b) Observe that d YHZ ≥ n − d YHZ is large when some entries of µ are large. Infact, the observations is conjectured to hold in general, since it can be shown, under some minorand reasonable assumption, that d YHZ ≥ n + 3 µ − µ For a proof, see Lemma 17 in Appendix.(c) Hence most likely
NEW ≤ YHZ always.
Acknowledgements.
The second author’s work was supported by National Natural Science Foundation ofChina (Grant No. 11801101) and Guangxi Science and Technology Program (Grant No. 2017AD23056).14 eferences [1] Basu, S., Pollack, R., Roy, M.-F.: Algorithms in Real Algebraic Geometry. Springer-Verlag, Berlin-Heidelberg (2006)[2] Brown, W. S.: On Euclid’s Algorithm and the Computation of Polynomial Greatest Common Divisors.JACM 18, 476–504 (1971)[3] Collins, G.E.: Subresultants and Reduced Polynomial Remainder Sequences. Journal of the Associationfor Computing Machinery, 14, 128–142 (1967)[4] D’Andrea, C., Hong, H., Krick, T., Szanto, A.: Elementary proof for Sylvester’s double sum for subre-sultants. Journal Symbolic Computation 42(3), 290–297 (2007)[5] Gonz´alez-Vega, L., Recio, T., Lombardi, H., Roy, M.-F.: Sturm-Habicht Sequences, Determinants andReal Roots of Univariate Polynomials. In: Quantifier Elimination and Cylindrical Algebraic Decompo-sition. Texts and Monographs in Symbolic Computation (A Series of the Research Institute for Sym-bolic Computation, Johannes-Kepler-University, Linz, Austria) (Caviness B.F. and Johnson J.R. eds.).Springer, Vienna (1998)[6] Liang, S., Jeffrey, D.: An Algorithm for Computing the Complete Root Classification of a ParametricPolynomial. In: Proceedings of AISC 2006 (Calmet, J., Ida, T., Wang, D. eds). LNAI 4120, pp. 116-130.Springer-Verlag, Berlin Heidelberg (2006)[7] Liang, S., Jeffrey, D., Moreno Maza, M.: The Complete Root Classification of a Parametric Polynomialon an Interval. In: Proceedings of ISSAC 2008, pp. 189-196, ACM Press, New York, NY, USA (2008)[8] Liang, S., Zhang, J.: A Complete Discrimination System for Polynomials with Complex Coefficientsand Its Automatic Generation. Science in China (Series E), 42, 113–128 (1999)[9] Loos, R.: Generalized Polynomial Remainder Sequences. In: Computer Algebra. Computing Supple-mentum (Buchberger, B., Collins, G.E., Loos, R. eds.), vol. 4, pp. 115–137. Springer, Vienna (1982)[10] Wang, D., Yang, J.: The Second Discriminant of a Univariate Polynomial. Science China Mathematics,DOI: https://doi.org/10.1007/s11425-018-1594-2 (2019)[11] Yang, L., Hou, X., Zeng, Z.: A Complete Discrimination System for Polynomials. Science in China(Series E), 39(6), 628–646 (1996)
Appendix: Analysis of size of YHZ’s condition
We reproduce the result for the complex root case of the YHZ’s method for readers’ convenience. Assume F is of degree n with m distinct roots. Let µ be an m -partition of n . Then we havemult ( F ) = µ ⇐⇒ µ − ^ i =1 s i +1 − ^ j =0 S j ( G i ) = 0 ∧ S ( G µ − ) = 0where • S k ( G ) = the k -th subresultant of G and G ′ . • S k = the coefficient of x k in S k ( G ) • s i = P mj =1 max( µ j − i,
0) 15 G i = (cid:26) F if i = 0 S s i ( G i − ) if i > Assumption 2.
We will assume that S ( G j ) in the above YHZ’s condition is not identically as a poly-nomial on the coefficients of F . We make this assumption because1. It simplifies the analysis of the size of the condition produced by the YHZ’s method.2. Numerous direct computations support its truth.3. However, so far, we were not able to prove it.
Lemma 17 (Size of YHZ’s condition) . Under Assumption 2, we have1.
YHZ = 1 + m P i =1 (cid:0) µ i − (cid:1) .2. d YHZ = µ − Q j =0 (2 m j − if µ = µ m µ − − if µ = µ + 1(2 ( µ − µ ) − if µ > µ + 1 where m i is the largest k such that µ k > i .3. d YHZ ≥ n + 3 µ − µ .Proof. We will prove each one by one.1. Immediate from
YHZ = 1+ µ − X j =2 s j = 1+ µ − X j =2 m X i =1 max( µ i − j,
0) = 1+ m X i =1 µ − X j =2 max( µ i − j,
0) = 1+ m X i =1 µ i − X j =2 ( µ i − j ) = 1+ m X i =1 (cid:18) µ i − (cid:19) .
2. The proof is a bit long and so we divide it into several steps.(a) Note d YHZ = max µ − [ i =1 n deg S ( G i ) , . . . , deg S s i +1 − ( G i ) o! [ n deg S ( G µ − ) o! = max ≤ i ≤ µ − deg S ( G i )= max ≤ i ≤ µ − deg a ( G i ) (2 s i − ≤ i ≤ µ − i − Y j =0 (2 m j − (2 s i −
1) since the size of matrices for the coefficients of G j +1 is 2 m j −
1= max ≤ i ≤ µ − i − Y j =0 (2 m j − µ − X j = i m j − (b) The above motivates the following notations. d i = i − Y j =0 (2 m j − µ − X j = i m j − A = i − Y j =0 (2 m j −
1) and B = µ − X j = i m j and C = m i − i such that d i is the maximum. Note d i ≤ d i − ⇐⇒ A (2 C −
1) (2 B − − A (2 ( B + C ) − ≤ ⇐⇒ A (cid:18) ( B −
1) ( C − − (cid:19) ≤ ⇐⇒ B = 1 ∨ C = 1 since A, B, C ≥ ⇐⇒ µ − X j = i m j = 1 ∨ m i − = 1 ⇐⇒ ( m i = 1 ∧ i = µ − ∨ m i − = 1 ⇐⇒ µ > µ ∧ ( i = µ − ∨ i > µ ) ⇐⇒ ( µ = µ + 1 ∧ ( i = µ − ∨ i > µ )) ∨ ( µ > µ + 1 ∧ ( i = µ − ∨ i > µ )) ⇐⇒ ( µ = µ + 1 ∧ i ≥ µ ) ∨ ( µ > µ + 1 ∧ i > µ )(d) Thus d YHZ = max ≤ i ≤ µ − d i = d µ − if µ = µ d µ − if µ = µ + 1 d µ if µ > µ + 1= (cid:16)Q µ − − j =0 (2 m j − (cid:17) (cid:16) P µ − j = µ − m j − (cid:17) if µ = µ (cid:16)Q µ − − j =0 (2 m j − (cid:17) (cid:16) P µ − j = µ − m j − (cid:17) if µ = µ + 1 (cid:16)Q µ − j =0 (2 m j − (cid:17) (cid:16) P µ − j = µ m j − (cid:17) if µ > µ + 1= (cid:16)Q µ − j =0 (2 m j − (cid:17) (2 m µ − −
1) if µ = µ (cid:16)Q µ − j =0 (2 m j − (cid:17) (2 (1 + m µ − ) −
1) if µ = µ + 1 (cid:16)Q µ − j =0 (2 m j − (cid:17) (2 ( µ − µ ) −
1) if µ > µ + 1= (cid:16)Q µ − j =0 (2 m j − (cid:17) if µ = µ (cid:16)Q µ − j =0 (2 m j − (cid:17) (2 m µ − + 1) if µ = µ + 1 (cid:16)Q µ − j =0 (2 m j − (cid:17) (2 ( µ − µ ) −
1) if µ > µ + 1= µ − Y j =0 (2 m j − µ = µ m µ − − if µ = µ + 1(2 ( µ − µ ) −
1) if µ > µ + 13. We will divide the proof into three cases. 17a) µ = µ .i. We rewrite d YHZ = µ − Y j =0 (2 m j −
1) = µ − Y j =0 (2 m j − − µ − X j =0 m j + 2 n. ii. Let G ( x , . . . , x µ − ) = µ − Y j =0 ( x j − − µ − X j =0 x j + 2 n over R = { ( x , . . . , x µ − ) : x j ≥ · } . Then we have d YHZ ≥ min x ∈ R G ( x )iii. Note ∂G/∂x i = Y ≤ j ≤ µ − j = i ( x j − − > R Hence min x ∈ R G ( x ) = µ − Y j =0 (4 − − µ − X j =0 n = 3 µ − µ + 2 n iv. Hence d YHZ ≥ µ − µ + 2 n (b) µ = µ + 1.i. We rewrite d YHZ = µ − Y j =0 (2 m j − · (cid:18) m µ − − (cid:19) = µ − Y j =0 (2 m j − (2 m µ − + 1) − µ − X j =0 m j + 2 n. ii. Let G ( x , . . . , x µ − ) = ( x µ − + 1) µ − Y j =0 ( x j − − µ − X j =0 x j + 2 n over R = { ( x , . . . , x µ − ) : x j ≥ · } . Then we have d YHZ ≥ min x ∈ R G ( x )iii. Note ∂G/∂x i = ( ( x µ − + 1) Q ≤ j ≤ µ − j = i ( x j − − , if i < µ − Q ≤ j ≤ µ − ( x j − − , if i = µ − ∂G/∂x i > R . Hencemin x ∈ R G ( x ) = (4 + 1) µ − Y j =0 (4 − − µ − X j =0 + 2 n
18 5 · µ − − µ − n = 53 · µ − µ − n = (3 µ − µ ) + (cid:18) · µ − (cid:19) + 2 n ≥ µ − µ + 2 n iv. Hence d YHZ ≥ µ − µ + 2 n (c) µ > µ + 1.i. We rewrite d YHZ = µ − Y j =0 (2 m j − (2 ( µ − µ ) − µ − Y j =0 (2 m j − (2 ( µ − µ ) − − µ − X j =0 m j + 2( µ − µ ) + 2 n ii. Let G ( x , . . . , x µ ) = µ Y j =0 ( x j − − µ X j =0 x j + 2 n over R = { ( x , . . . , x µ ) : x j ≥ · } . Then we have d YHZ ≥ min x ∈ R G ( x )iii. Note ∂G/∂x i = Y ≤ j ≤ µ j = i ( x j − − > R Hence min x ∈ R G ( x ) = µ Y j =0 (4 − − µ X j =0 n = 3 µ +1 − µ + 1) + 2 n = 3 · µ − µ + 1) + 2 n = (3 µ − µ ) + (2 · µ −
4) + 2 n ≥ µ − µ + 2 n iv. Hence d YHZ ≥ µ − µ + 2 nn