A refined realization theorem in the context of the Schur-Szegő composition
aa r X i v : . [ m a t h . C A ] A p r A REFINED REALIZATION THEOREM IN THE CONTEXT OFTHE SCHUR-SZEG ˝O COMPOSITION
VLADIMIR P. KOSTOV
Abstract.
Every polynomial of the form P = ( x + 1)( x n − + c x n − + · · · + c n − ) is representable as Schur-Szeg˝o composition of n − x + 1) n − ( x + a i ), where the numbers a i are unique up to permutation.We give necessary and sufficient conditions upon the possible values of the8-vector whose components are the number of positive, zero, negative andcomplex roots of a real polynomial P and the number of positive, zero, negativeand complex among the quantities a i corresponding to P . A similar result isproved about entire functions of the form e x R , where R is a polynomial. Introduction
Schur-Szeg˝o composition and the mapping Φ . In the present paperwe prove a realization theorem in the context of the
Schur-Szeg˝o composition of polynomials. For the two polynomials of degree n , A := P nj =0 (cid:0) nj (cid:1) a j x j and B := P nj =0 (cid:0) nj (cid:1) b j x j , their Schur-Szeg˝o composition is defined by the formula A ∗ B := n X j =0 (cid:18) nj (cid:19) a j b j x j . This formula is valid for any complex polynomials. In this paper we are interestedmainly in the case when A and B are real.Observe that when one considers the polynomials as degree n + k ones, with k leading zero coefficients, then the formula for their Schur-Szeg˝o composition willbe a different one. To avoid such ambiguity, we assume throughout the paper thatthe leading coefficient of at least one of the composed polynomials is nonzero.The polynomial ( x + 1) n plays the role of unity in the sense that(1.1) ( x + 1) n ∗ A = A for any polynomial A . Schur-Szeg˝o composition of polynomials is commutative andassociative. It can be defined for an arbitrary number of polynomials by the formula A ∗ · · · ∗ A s := n X j =0 (cid:18) nj (cid:19) a j · · · a sj x j , where A i := n X j =0 (cid:18) nj (cid:19) a ij x j . The reader can find a more detailed information about the Schur-Szeg˝o compositionin the monographies [11] and [12].
Mathematics Subject Classification.
Primary 12A10 Secondary 30D99.
Key words and phrases.
Schur-Szeg˝o composition; composition factor; entire function.Research supported by the French Foundation CNRS under Project 20682.
The Schur-Szeg˝o composition gives rise to a mapping Φ (defined below) from thespace of polynomials of degree n − n having one of their roots at ( − P := ( x + 1)( x n − + c x n − + · · · + c n − )Each such polynomial is representable as Schur-Szeg˝o composition of n − composition factors ) of the form K a i := ( x + 1) n − ( x + a i ), wherethe complex numbers a i are uniquely defined. In the case of real polynomials partof the numbers a i are real, while the rest form complex conjugate couples. Thisresult has been announced in [2] and proved in [1].Notice that the numbers ( − a i ) can be viewed as roots of another polynomial.In this sense we obtain a mapping from the space of degree n − σ j the elementary symmetric polynomialsof the quantities a i , i.e. σ j := P ≤ i
2] is the number of complex conjugate couples of roots of P/ ( x + 1) ([ . ] stands for the integer part of) and [( n − − r ) /
2] is the number ofcomplex conjugate couples of quantities a j . A priori, 0 ≤ ρ, r ≤ n and the parityof the numbers r and ρ must be the same. In paper [4] the following question isasked: When these natural restrictions are respected, what can be the values of the couple ( r, ρ ) ? The answer given there is:
All possible values are attained at some polynomials having all their roots distinctand for which the corresponding quantities a i are also distinct. In other words, all a priori admissible couples ( r, ρ ) are realizable. (This is arealization theorem.)A similar realization theorem has been proved in [7] for an analog of the Schur-Szeg˝o composition in the case of entire transcendental functions. Consider the twotranscendental functions f and g represented by the convergent everywhere in C series P ∞ j =0 γ j x j /j ! and P ∞ j =0 δ j x j /j !, respectively. Their Schur-Szeg˝o compositionis defined by the formula f ∗ g := ∞ X j =0 γ j δ j x j /j ! . As in the case of polynomials, Schur-Szeg˝o composition is commutative, associativeand can be defined for any finite number of entire transcendental functions by theformula f ∗ · · · ∗ f s := ∞ X j =0 γ j · · · γ sj x j /j ! , where f i := ∞ X j =0 γ ij x j /j ! . REFINED REALIZATION THEOREM IN THE CONTEXT OF THE SCHUR-SZEG ˝O COMPOSITION3
Consider transcendental functions of the form e x R , where R is a polynomial ofdegree n − R (0) = 1. Such a function is representable as Schur-Szeg˝o composition(1.3) e x R = κ a ∗ · · · ∗ κ a n − , where the composition factors κ a i are of the form e x (1 + x/a j ), the numbers a j being uniquely defined up to permutation. To extend the formula to the case when P (0) is not necessarily 1, one has to admit the presence of composition factors e x c , c = 0, and e x x (one can say that the latter correspond to the case a i = 0). Whenthe polynomial P is real, part of the numbers a j are real and the rest form complexconjugate couples.Set ˜ σ j := P ≤ i
When Schur-Szeg˝o composition of real polynomials is con-sidered, then it is important to distinguish the real positive, negative and zero rootsof the composed polynomials. (About the number of real positive or negative rootsof the Schur-Szeg˝o composition of two hyperbolic or real polynomials see respec-tively [2] and [6].) Propositions 1 and 2 show that the properties of the mappingΦ (defined for polynomials or for entire functions) are not the same with regard tothe real positive and the real negative roots. We use the following notation:
Notation 1.
We set b j := − j/ ( n − j ) for j = 0 , , . . . , n − b n = −∞ . Proposition 1. (1)
If the polynomial P from (1.2) has m positive roots countedwith multiplicity ( m ≥ ) and a k -fold root at ( k ≥ ), then there are at least m +max(0 , k − negative and distinct among the numbers a i out of which max(0 , k − equal b , . . . , b k − ; if k ≥ , then one of the numbers a i equals . (2) If there are q numbers a i which equal and q which are positive, then thepolynomial P has at least q + max(0 , q − negative roots counted with multiplicity;for q ≥ it has a root at . Proposition 2. (1)
If the polynomial R has m positive roots counted with multi-plicity and a k -fold root at , then there are at least m + max(0 , k − negative anddistinct among the numbers a i out of which max(0 , k − equal ( − j ) , j = 1 , . . . , k − .For k ≥ one composition factor equals e x x . (2) If there are q numbers a i which equal and q which are positive, then thepolynomial R has at least q + max(0 , q − negative roots counted with multiplicity;for q ≥ it has a root at . The propositions are proved in Section 3. In their proofs we use some facts aboutthe Schur-Szeg˝o composition, see Section 2.
VLADIMIR P. KOSTOV
Theorems 1 and 2 below show that the necessary conditions expressed by thepropositions are in fact sufficient as well. In this sense they are realization theorems.Before formulating the theorems we analyse in detail the possible number of realpositive, negative and zero among the roots of the polynomial P or R on the one-hand side and the numbers ( − a i ) on the other. We use the same notation as in thepropositions.Suppose that k ≥ q ≥
1) and that among the numbers ( − a i ) corre-sponding to the polynomial P or R there are q negative ones and q C / − a i ) is k − m + r, for some r ≥ . There are q − q + s negative among the roots of the polynomial (where s ≥ k C / k = 0,then q = 0 and the above two numbers equal m + r and q + s .We distinguish four cases. In Cases 1) and 2) (respectively 3) and 4)) we supposethat k C ≤ r (respectively k C > r ). One has k = 1 in Cases 1) and 3) and k = 0 inCases 2) and 4). Case 1)
Set r = k C + k . Hence s = q C + k . We present the situation schemat-ically like this: k C roots of P/ ( x + 1) or R q − q + s z }| { q C + k k m − − − − − − − − a i ) q q k − m + k C + k | {z } r q C The sequences of minus and plus signs to the left and right in the middle linesymbolize the negative and positive half-axes; the numbers k C and q C are put awayfrom that line, i.e. “away from the real axis” because these are the quantities ofcomplex (not real) roots. Case 2) is defined by the conditions k = 0 (hence q = 0), k C ≤ r . The case canbe presented like this: k C roots of P/ ( x + 1) or R q + s z }| { q C + k m − − − − − − − − a i ) q m + k C + k | {z } r q C In Case 3) , i.e. when k ≥ q ≥
1) and k C > r , one can set k C := r + δ , q C := s + δ and the situation admits the following presentation: REFINED REALIZATION THEOREM IN THE CONTEXT OF THE SCHUR-SZEG ˝O COMPOSITION5 r + δ roots of P/ ( x + 1) or R q − q + s k m − − − − − − − − a i ) q q k − m + rs + δ Finally, in
Case 4) , i.e. when k = 0 = q and k C > r , the presentation looks likethis: r + δ roots of P/ ( x + 1) or R q + s m − − − − − − − − a i ) q m + rs + δ We say that a polynomial P or R realizes Case 1), 2), 3) or 4) if the numberof its positive, zero, negative and complex roots and these numbers defined for thequantities ( − a i ) are as shown on the above figures, and if the non-zero roots aredistinct and the non-zero quantities a i are also distinct. Theorem 1.
Cases 1) – 4) are realizable by some polynomials P . Theorem 2.
Cases 1) – 4) are realizable by some polynomials R . The theorems are proved in Sections 4 and 5, respectively.2.
Properties of the Schur-Szeg˝o composition
The following formulae concerning the composition of polynomials can be checkedstraighforwardly (see [2]). In the second of them S denotes a polynomial of degree n −
1. Notice that in their left-hand sides (respectively, in their right-hand sides)the polynomials are composed as degree n (respectively degree n −
1) ones:(2.4) ( A ∗ B ) ′ = 1 n ( A ′ ∗ B ′ ) , ( xS ∗ B ) = xn ( S ∗ B ′ )The analogs of these formulae in the case of entire transcendental functions read:(2.5) ( f ∗ g ) ′ = f ′ ∗ g ′ , xf ∗ g = x ( f ∗ g ′ )For the composition factors of the form K a i the following formula holds:(2.6) K a i = ( x + 1) n − ( x + a i ) = n X j =0 (cid:18) nj (cid:19) (cid:18) n − jn a i + jn (cid:19) x j Its analog in the case of composition factors κ a i looks like this:(2.7) κ a i := e x (cid:18) xa i (cid:19) = ∞ X j =0 j ! (cid:18) ja i (cid:19) x j The numbers ( n − j ) a i /n + j/n (see formula (2.6)) form an arithmetic progression.Present the numbers 1 + j/a i from formula (2.7) in the form ( a i + j ) /a i . Hence thenumerators form also an arithmetic progression. This implies the following result: VLADIMIR P. KOSTOV
Corollary 1.
When a i is real, there is at most one sign change in the sequence ofcoefficients of a composition factor K a i or κ a i . The following proposition is Proposition 1.4 in [9].
Proposition 3.
If the degree n polynomials A and B have roots x A = 0 and x B = 0 of multiplicities m A and m B respectively, where m A + m B ≥ n , then − x A x B is aroot of A ∗ B of multiplicity m A + m B − n . The conditions x A = 0, x B = 0 are omitted in [9] which is an error.Set K ∞ := ( x + 1) n − . For a degree n polynomial P denote by P R its reverted polynomial x n P (1 /x ). The following facts are straightforward: Proposition 4. (1) If P = K a ∗ · · · ∗ K a n − , then P R = K /a ∗ · · · ∗ K /a n − . (2) For any two polynomials A and B one has ( A ∗ B ) R = A R ∗ B R . (3) For any degree n polynomial A one has ( x + 1) n − ∗ A = A − xA ′ /n and ( x + 1) n − ∗ A = A − xA ′ /n + x A ′′ / ( n ( n − . Proofs of Propositions 1 and 2 . We begin the proof with the following Observation. (1) The polynomial P has a k -fold root at 0 ( k >
0) if and onlyif there are k composition factors K a i such that in K a i the coefficient of x i is 0.According to formula (2.6), one must have a i = b i , i = 1 , . . . , k −
1, and a k = 0(after a suitable permutation of the indices if necessary).(2) In the same way, the polynomial R has a k -fold root at 0 ( k >
0) if and onlyif there are k composition factors κ a i with a i = 0 , − , . . . , − ( k − . For k = 0, Remark 6 in [4] states that in the case of a polynomial P thereare at least m different negative among the numbers a i . In the case of a polynomial R the same statement is contained in Corollary 2 in [8].3 . Suppose that k >
0. Set P := x ( x + 1) S , where deg S = n −
2. Then one ofthe numbers a i defined after the polynomial P equals 0. Consider the presentation x ( x + 1) S = x ( x + 1) n − ∗ ( x + 1) n − ( x + a ) ∗ · · · ∗ ( x + 1) n − ( x + a n − ) . Using both formulae (2.4) and formula (1.1) we present the right-hand side in theform x [( x + 1) n − ( x + d ) ∗ · · · ∗ ( x + 1) n − ( x + d n − )] , where d i = ( n − a i + 1 n . (The polynomials are composed as degree n − d i arethe numbers a i computed for the degree n − x + 1) S . One has a i = ( nd i − / ( n −
1) which implies that if d i <
0, then a i < b i depend on i and n . In this proof we denote themfurther by b i,n because we need to compare them for different values of n .When passing from the numbers a i to the numbers d i , the number 0 correspond-ing to the factor x in the last displayed formula is lost, and these of the numbers a i which equal b j,n change as follows: b j,n b j − ,n − (to be checked directly). And inthe same way, if a number a i is different from b j,n for all j , then d i is different from REFINED REALIZATION THEOREM IN THE CONTEXT OF THE SCHUR-SZEG ˝O COMPOSITION7 all b j − ,n − . Thus the rest of part (1) of Proposition 1 follows by finite inductionon k . Part (1) of Proposition 2 is proved by analogy.4 . Prove part (2) of Proposition 1. The composition U of all composition fac-tors K a i , where a i is either complex or negative (we denote their quantity by ν ), isa polynomial having an ( n − ν )-fold root at ( − ν − U consecutively with the com-position factors K a i with 0 < a i <
1, the number of negative roots (counted withmultiplicity) of the given polynomial does not decrease. It suffices to consider thecase when all negative roots are distinct, in the general case the result follows bycontinuity.Indeed, one has V := ( x + 1) n − ( x + a i ) ∗ U = a i U + (1 − a i ) xU ′ (use formulae(2.4) and (1.1) and part (3) of Proposition 4). The signs of the polynomials U and V are the same at the negative roots of U ′ and at 0. These signs alternate. Hence V has at most one negative root less than U .Notice that sgn V = − sgn U ′ =sgn((1 − a i ) xU ′ ) at the smallest real root α of U (which is negative). Hence there is a root of V between α and the smallest realroot of U ′ which is > α , i.e. V has at least as many negative roots as U .5 . To prove that composition with a i > U , one can consider instead of U and K a i the reverted polynomials x n U (1 /x ) and K /a i using Proposition 4. One can skip the composition factors with a i = 1 due to (1.1).For a i = 0 one has ( x + 1) n − x ∗ U = xU ′ . It is easy to show that only thefirst such composition can decrease by 1 the number of negative roots, while thesubsequent ones preserve this number. Part (2) of Proposition 1 is proved.6 . To prove part (2) of Proposition 2 (by analogy with 4 – 5 ) one can use thefollowing formula (derived from formulae (2.5)): V := e x (cid:18) xa i (cid:19) ∗ e x U = e x (cid:18)(cid:18) xa i (cid:19) U + xa i U ′ (cid:19) . We prove the statement in the case when all negative roots of U are distinct. Inthe presence of multiple roots the proof follows by continuity.The polynomial xU ′ changes sign at the consecutive roots of U . Hence thereis a root of V between any two consecutive negative roots of U . The signs of V are different at 0 and at the smallest in absolute value root of U . They are alsodifferent at −∞ and at the largest in absolute value root of U (the details are leftfor the reader). Hence V has one negative root more than U . (cid:3) Proof of Theorem 1
We prove first a proposition from which the theorem is deduced below.
Proposition 5.
The following composition (with l ≥ composition factors ( x +1) n − x and with l + µ ≤ n ) (4.8) U := ( x +1) n − x ∗· · ·∗ ( x +1) n − x ∗ ( x +1) n − ( x + b ) ∗· · ·∗ ( x +1) n − ( x + b µ ) , is a polynomial with a ( µ + 1) -fold root at , with an ( n − µ − l ) -fold root at ( − and with l − simple roots belonging to the interval ( − , .Proof: VLADIMIR P. KOSTOV
Set T := ( x + 1) n − x ∗ ( x + 1) n − ( x + b ) ∗ · · · ∗ ( x + 1) n − ( x + b µ ). It follows fromProposition 3 and from the Observation from 1 of the proof of Propositions 1 and2 (see Section 3) that T is a polynomial of degree n , with a ( µ + 1)-fold root at 0and with an ( n − µ − − T k the result of composing k times T with ( x + 1) n − x . Apply thesecond of formulae (2.4) and then formula (1.1):( x + 1) n − x ∗ T k = x (( x + 1) n − ∗ T ′ k ) = xT ′ k . If T k has a ( µ + 1)-fold root at 0, then this is the case of xT ′ k as well. The multi-plicity of ( −
1) decreases by 1. If − < ζ < · · · < ζ k < T k inthe interval (0 , − , ζ ), ( ζ , ζ ), . . . , ( ζ k ,
0) thereis exactly one root of xT ′ k . These roots are simple (because deg( xT ′ k ) = n ) and theproposition is thus proved by finite induction on k . (cid:3) Proof of the theorem:Case 1) .1 . Use the proposition with l = q + q + q C , µ = k − m + k + k C . Hence thepolynomial U (see (4.8)) has q − q + q C distinct roots belonging to ( − , k + m + k + k C )-fold root at 0 and a simple root at ( − x + 1) n − x as follows:– q of them do not change;– q of them are replaced by composition factors ( x + 1) n − ( x + εg j ), where g j are distinct positive numbers;– q C of them are replaced by factors ( x + 1) n − ( x + εh j ), where the numbers h j form q C / ε > U which belong to( − ,
0) are perturbed and its other roots do not change. The perturbed rootsremain negative and distinct.2 . Change the ˜ m := m + k + k C of the numbers b i with largest absolute valuesto b i + λ i , where λ i are small real parameters. Before the change the polynomial U was of the form x k + ˜ m U , U (0) = 0. After the change it becomes V := U + x k ( w λ x ˜ m − + w λ x ˜ m − + · · · + w ˜ m λ ˜ m + P ) , where w i are non-zero real numbers and P is a polynomial in x , λ , . . . , λ ˜ m of totaldegree ˜ m and only with monomials whose total degree w.r.t. the variables λ i is ≥ V /x k is a versal deformation of the germ of a function U at 0which has a root at 0 of multiplicity ˜ m . Hence one can choose the values of theparameters λ i such that this ˜ m -fold zero splits into m positive, k negative rootsand k C / Case 2) .Use Case 1) of the theorem with k = q = 1. Exactly one of the compositionfactors equals ( x + 1) n − x . Perturb it into ( x + 1) n − ( x − ε ) ( ε > REFINED REALIZATION THEOREM IN THE CONTEXT OF THE SCHUR-SZEG ˝O COMPOSITION9 which is the sign of the product of all numbers a i ). The existing roots were simple,therefore they remain simple after the perturbation. The numbers of negative andpositive (perturbed existing) roots remain the same. Thus one obtains instead ofthe figure describing Case 1) the one describing Case 2) with m replaced by m + 1.Hence the possibility to have m = 0 in Case 2) has to be considered separately.In Case 2) with m = 0 one can again use Case 1) with k = 1, but this time oneperturbs the composition factor ( x + 1) n − x into ( x + 1) n − ( x + ε ), ε >
0. Thusone obtains Case 2) with q replaced by q + 1. So one has to consider separatelythe possibility m = q = 0.For m = q = 0 apply Proposition 5 with l = q C +1, µ = k + k C −
1. Hence U has q C negative simple roots and a ( µ + 1)-fold root at 0. Perturb the k C compositionfactors with largest | b j | into ( x + 1) n − ( x + b j + λ j ) . (For k = 0 one perturbs all k C − x + 1) n − x into ( x + 1) n − ( x + λ ).) Theperturbation can be carried out so that the root of U at 0 split into k C / k -fold root at 0 (by analogy with Case 1), see2 ).When k >
0, the k − x + 1) n − ( x + b j ) and one factor( x + 1) n − x are perturbed so that U have k negative roots close to 0 (the previ-ously existing q C negative roots remain such). Its k C complex couples remain such.Finally, perturb the remaining factors ( x + 1) n − x into ( x + 1) n − ( x + h i ), wherethe numbers h i form q C / Case 3)
Apply Proposition 5 with l = q + q + s , µ = k + r + m . Notice that for δ = 0one has l + µ < n . The polynomial U has a ( µ + 1)-fold root at 0 and l − r + m composition factors with largest | b j | so that the root of U at 0 split into r/ k -fold root at 0 and m positiveroots. Then perturb the factors ( x + 1) n − x into ( x + 1) n − ( x + η i ) as follows:– s of the numbers η i form distinct complex conjugate couples;– q of them are negative and distinct;– q of them are 0.The last perturbation does not change the number of positive, negative, zero andcomplex roots of U . For δ = 0 the case is completely solved.Denote by S the sector { u + iv ∈ C | < u < v } . When δ >
0, we need thefollowing lemma (proved after the proof of the theorem).
Lemma 1. (1)
For all ε ∈ S sufficiently close to the polynomial V := ( x + 1) n − ( x + 1 + ε ) ∗ ( x + 1) n − ( x + 1 + ¯ ε ) has two complex conjugate roots close to ( − and an ( n − -fold root at ( − . (2) Suppose that the polynomial U has s ∗ negative and t ∗ positive simple roots, r ∗ / distinct conjugate couples and an ( n − s ∗ − t ∗ − r ∗ ) -fold root at ( − , n − s ∗ − t ∗ − r ∗ ≥ . Then one can choose ε ∈ S so close to that the polynomial U ∗ V have s ∗ negative (different from − ) and t ∗ positive simple roots, r ∗ / distinctcomplex conjugate couples and an ( n − s ∗ − t ∗ − r ∗ − -fold root at ( − . (3) The multiplicity of as a root of U and U ∗ V is the same for ε small enough. To complete the proof in Case 3) one sets s ∗ := q − q + s , t ∗ := m , r ∗ := r and then applies the lemma δ/ Case 4)
Use Case 3) with k = q = 1. Perturb the factor ( x +1) n − x into ( x +1) n − ( x − ζ ), ζ >
0. This changes m to m + 1. Thus Case 4) is deduced from Case 3) except for m = 0. For m = 0 use again Case 3) with k = q = 1 changing this time ( x + 1) n − x into ( x + 1) n − ( x + ζ ). This changes q into q + 1 and there remains to consideronly the possibility m = q = 0.In this particular case r must be even. Hence such are δ and s as well. ApplyProposition 5 with l = s + 1, µ = r −
1. Hence the polynomial U has s negativedistinct roots and an r -fold root at 0. Perturb one factor ( x +1) n − x and the factors( x + 1) n − ( x + b j ) to make the r -fold root of U split into r/ x + 1) n − x becomes ( x + 1) n − ( x − ε ) with ε > r being even). After this perturb the remaining s factors ( x + 1) n − x into ( x + 1) n − ( x + ζ i ), where the numbers ζ i form s/ δ = 0. For δ > δ/ r ∗ = r , t ∗ = 0 and s ∗ = q + s . (cid:3) Proof of Lemma 1:
Using equality (1.1) and part (3) of Proposition 4 one finds that V = ( x + 1) n − (( x + 1) + ( ε + ¯ ε + ε ¯ ε/n )( x + 1) + ( n − ε ¯ ε/n ) . Set ε = u + iv . The discriminant ∆ of the quadratic factor (considered as apolynomial in x + 1) equals(4 /n )( u − ( n − v ) + o ( u + v ) . When ε ∈ S is close to 0, one has ∆ <
0. The coefficient of ( x + 1) and the constantterm of the quadratic factor tend to 0 as ε →
0. Hence its roots also tend to 0.This proves part (1) of the lemma.Set p := n − s ∗ − t ∗ − r ∗ . Present U as a polynomial in x + 1: U = U ( x + 1) p + · · · + U n − p ( x + 1) n , U = 0 . One has U ∗ V = U ( x + 1) p ∗ V + · · · + U n − p ( x + 1) n ∗ V .
By Proposition 3 all terms have a root at ( −
1) of multiplicity at least p −
2. As V is a perturbation of ( x + 1) n , for ε small enough the polynomial U ∗ V has r ∗ / s ∗ negative and t ∗ positive roots close to the ones of U . Weshow that when ε ∈ S is small enough, then the first of the terms to the right hastwo complex conjugate roots close to −
1. One has( x + 1) p ∗ V = ( x + 1) p − ((1 + o (1))( x + 1) + (2 pu/n + o ( | u | + | v | ))( x + 1)+(( u + v ) p ( p − /n )) . The discriminant of the quadratic factor equals(4 /n )( p u − p ( p − u + v )) + o ( u + v ) . REFINED REALIZATION THEOREM IN THE CONTEXT OF THE SCHUR-SZEG ˝O COMPOSITION11
For ε ∈ S small enough it is <
0. The coefficient of x + 1 and the constant termof the quadratic factor tend to 0 as ε → x + 1) remainsclose to 1, so its two roots also tend to − ε = τ η , where τ = | ε | . Set x + 1 = τ y . Set L := ( V / ( x + 1) n − ) | x = τy − and B := (( x + 1) p ∗ V / ( x + 1) p − ) | x = τy − . As τ →
0, the two complex roots of thepolynomial L (respectively B ) are of the form α i τ + o ( τ ), (respectively β i τ + o ( τ )), α i = 0 = β i , i = 1 ,
2. The numbers α i and β i are roots respectively of thepolynomials y + ( η + ¯ η ) y + (( n − /n ) η ¯ η and y + ( p ( η + ¯ η ) /n ) y + ( p ( p − /n ) η ¯ η . Consider two circles of radius τ min( | β | , | β | ) / B . Allterms ( U j ( x + 1) p + j ∗ V ) | x = τy − , j = 1 , . . . , n − p when restricted to these circleshave their module tending to 0 (as τ →
0) faster than the module of the term U B .(This is due to their higher power of ( x + 1), i.e. of τ y .) By Rouch´e’s theoreminside each of the circles there is exactly one root of the polynomial U ∗ V . Thisproves part (2) of the lemma.Part (3) is evident – the multiplicity of 0 as a root of U and U ∗ V is defined bythe number of the first consecutive coefficients of these polynomials which are 0.As V is a perturbation of ( x + 1) n , these numbers are the same. (cid:3) Proof of Theorem 2
The theorem is proved with the help of the following proposition:
Proposition 6.
The following composition (with l ≥ composition factors xe x ) (5.9) U := e x x ∗ · · · ∗ e x x ∗ e x ( x − ∗ · · · ∗ e x ( x − µ ) , is of the form e x Y , where Y is a degree l + µ polynomial with a ( µ + 1) -fold root at and with l − simple negative roots.Proof: With the help of formulae (2.5) and using finite induction on µ one shows thatthe composition of the last µ + 1 composition factors is exactly e x x µ +1 .Suppose that the proposition is true for l = l . Then for l = l + 1 one has U = e x x ∗ e x Y = e x x ( Y + Y ′ ). The sign of x ( Y + Y ′ ) changes alternatively atthe consecutive negative roots of Y ′ . Hence there is a root of U between any twonegative roots of Y ′ . Denote the latter roots by β < · · · < β l − and by α thegreatest (i.e. smallest in absolute value) negative root of Y .One has sgn U ( −∞ ) = − sgn Y ( −∞ ), sgn U ( β ) = − sgn Y ( β ) and sgn Y ( −∞ ) = − sgn Y ( β ). Hence there is a root of U in ( −∞ , β ). In the same way, sgn U ( β l − ) = − sgn Y ( β l − ), sgn U ( α ) = − sgn Y ′ ( α ) =sgn Y ( β l − ). Hence U has a root in( β l − , α ) as well.Thus the product x ( Y + Y ′ ) has l distinct negative roots and a ( µ + 1)-fold rootat 0. As deg( Y + Y ′ ) = l + µ , all negative roots are simple. (cid:3) Proof of the theorem:
Case 1) .Use the proposition with l = q + q + q C , µ = k − m + k + k C . Hence thepolynomial Y has q − q + q C distinct negative roots and a ( k + m + k + k C )-foldroot at 0. Perturb the composition factors e x x as follows:– q of them do not change;– q of them are replaced by composition factors e x ( x + εg j ), where g j are distinctpositive numbers;– q C of them are replaced by factors e x ( x + εh j ), where the numbers h j form q C / ε > Y are per-turbed and its other roots do not change. The perturbed roots remain negativeand distinct.Change the ˜ m := m + k + k C of the composition factors e x ( x − j ) with largestabsolute values of j to e x ( x − j + λ j ), where λ j are small real parameters chosensuch that the root of Y at 0 split into a k -fold root at 0, k C / k negative roots. The proof of Case 1) is finished by complete analogywith the proof of this case in Theorem 1.The rest of the proof of Theorem 2 is done also by analogy with the rest of theproof of Theorem 1 (modulo some technical details) – the role of the compositionfactor ( x + 1) n − x in the latter is played by e x x , the one of ( x + 1) n − ( x + b j ) isplayed by e x ( x − j ). Case 2)
Use Case 1) of the theorem with k = q = 1. Exactly one of the factors equals e x x . Perturb it into e x ( x − ε ) ( ε > m replaced by m + 1. The possibilityto have m = 0 in Case 2) has to be considered separately.For m = 0 one can again use Case 1) with k = 1, this time perturbing the factor e x x into e x ( x + ε ), ε >
0. Thus one obtains Case 2) with q replaced by q + 1.There remains to consider the possibility m = q = 0.For m = q = 0 apply Proposition 6 with l = q C + 1, µ = k + k C −
1. Hence Y has q C negative roots and a ( µ + 1)-fold root at 0. Perturb the k C compositionfactors e x ( x − j ) with largest j into e x ( x − j + λ j ). (When k = 0 one perturbs all k C − e x ( x − j ) as indicated and one factor e x x into e x ( x + λ ).) The rootof Y at 0 splits into k C / k -fold root at0. When k >
0, the k − e x ( x − j ) and one factor e x x areperturbed so that Y have k negative roots close to 0 (the previously existing q C negative roots remain such). Its k C complex couples remain such. Finally, perturbthe remaining factors e x x into e x ( x + h i ), where the numbers h i form q C / Case 3)
Apply Proposition 6 with l = q + q + s , µ = k − r + m . (For δ = 0 one has l + µ = n , otherwise l + µ < n .) The polynomial Y has a ( µ + 1)-fold root at 0 and l − r + m composition factors e x ( x − j ) with largest REFINED REALIZATION THEOREM IN THE CONTEXT OF THE SCHUR-SZEG ˝O COMPOSITION13 j so that the root of Y at 0 split into r/ k -fold rootat 0 and m positive roots. Then perturb the factors e x x into e x ( x + η i ) as follows:– s of the numbers η i form distinct complex conjugate couples;– q of them are negative and distinct;– q of them are 0.The last perturbation does not change the number of positive, negative, zero andcomplex roots of Y . For δ = 0 this finishes the proof of Case 3).Denote by S some sector centered at 0 and avoiding (except 0) the real axis. For δ > Lemma 2. (1)
One has e x (1 + εx ) ∗ e x (1 + ¯ εx ) = e x V , where V = 1 + ( ε + ¯ ε + ε ¯ ε ) x + ε ¯ εx . (2) Suppose that the polynomial Y is of degree p . Then the function ( e x Y ) ∗ ( e x V ) is of the form e x Y , where Y is a degree p + 2 polynomial. (3) Suppose that the degree p polynomial Y has s ∗ negative and t ∗ positive simpleroots, r ∗ / distinct conjugate couples and a ( p − s ∗ − t ∗ − r ∗ ) -fold root at . One canchoose ε ∈ S so close to that the polynomial Y have s ∗ negative and t ∗ positivesimple roots, r ∗ / distinct conjugate couples and a ( p − s ∗ − t ∗ − r ∗ ) -fold rootat . Set r ∗ := r , t ∗ := m and s ∗ := q − q + s . Applying the lemma δ/ Case 4)
Use Case 3) with k = q = 1. Perturb the factor e x x into e x ( x − ζ ), ζ >
0. Thischanges m to m + 1. Thus Case 4) is deduced from Case 3) except for m = 0. For m = 0 use again Case 3) with k = q = 1 changing this time e x x into e x ( x + ζ ). Thischanges q into q +1 and there remains to consider only the possibility m = q = 0.In this particular case r , δ and s are even. Apply Proposition 6 with l = s + 1, µ = r −
1. Hence the polynomial Y has s negative roots and an r -fold root at 0.Perturb one factor e x x and the factors e x ( x − j ) to make the r -fold root of Y splitinto r/ e x x becomes e x ( x − ε ) with ε > r being even). After this perturb the remaining s factors e x x into e x ( x + δ i ), where the numbers δ i form s/ δ = 0.For δ > δ/ (cid:3) Proof of Lemma 2:
Parts (1) and (2) of the lemma follow from the second of formulae (2.5) with f = e x x or f = e x x and g = e x Y . Prove part (3). The polynomial Y is aperturbation of the polynomial Y , therefore for ε close to 0 it has p roots close tothe respective roots of Y and two roots (called distant ) whose moduli tend to ∞ as ε → Y and Y have the same multiplicity of the root at 0. Indeed,for ε nonreal all coefficients of the function e x V are nonzero and this multiplicityis defined by the number of first consecutive coefficients of e x Y which are 0. As both Y and Y are real polynomials, Y has the same number of distinct nega-tive and distinct positive roots and the same number of distinct complex conjugatecouples as Y (excluding the two distant roots).Suppose that Y is monic (this is not restrictive). For p ∗ ∈ N one has e x V ∗ e x x p ∗ = x p ∗ ( ε ¯ εx + ( ε + ¯ ε + (2 p ∗ + 1) ε ¯ ε ) x + 1 + p ∗ ( ε + ¯ ε ) + p ∗ ε ¯ ε ) . Set W := ε ¯ εx + ( ε + ¯ ε ) x + 1. Hence the function e x V ∗ e x Y is of the form e x x p ( W + T ), where T is a Laurent series in x whose coefficients are polynomialsin ε and ¯ ε . It contains only monomials x α ε β ¯ ε γ with α − β − γ < W are 1 /ε and 1 / ¯ ε . Consider two circles C and C centered atthem and of radius 1. When ε ∈ S is small, the values of | T | at each point of eachof the two circles are much smaller than the respective values of W . By Rouch´e’stheorem each of the circles contains exactly one root of x p ( W + T ). (cid:3) References [1] S. Alkhatib and V.P. Kostov, The Schur-Szeg¨o composition of real polynomials of degree2, Rev. Mat. Complutense 21 (2008) no. 1, 191–206.[2] V. P. Kostov, The Schur-Szeg¨o composition for hyperbolic polynomials, C.R.A.S. S´er.I 345/9 (2007), 483-488, doi:10.1016/j.crma.2007.10.003.[3] V. P. Kostov, Eigenvectors in the context of the Schur-Szeg¨o composition of polynomials,Math. Balkanica 22(2008) Fasc. 1-2, 155–173.[4] V. P. Kostov, Teorema realizatsii v kontekste kompozitsii Shura-Sege, Funkcional’nyyAnaliz i ego Prilozheniya 43 (2009) no. 2, 79-83. (A realization theorem in the contextof the Schur-Szeg¨o composition, Funct. Anal. Appl. 43 (2009) no. 2, 147-150.)[5] V. P. Kostov, A mapping connected with the Schur-Szeg˝o composition, C.R.A.S. S´er. I347 (2009) 1355-1350.[6] V.P. Kostov, The Schur-Szeg¨o composition for real polynomials, C.R.A.S. S´er. I, 346(2008), 271-276.[7] V.P. Kostov, A realization theorem about the Schur-Szeg¨o composition for entire func-tions, Comptes Rendus Acad. Sci. Bulgare 62, No. 1 (2009), 17-22.[8] V. P. Kostov, A mapping defined by the Schur-Szeg˝o composition, Comptes RendusAcad. Sci. Bulgare Vol. 63 (2010) No. 7 943-952.[9] V. P. Kostov and B. Z. Shapiro, On the Schur-Szeg¨o composition of polynomials,C.R.A.S. S´er. I 343 (2006) 81–86.[10] V. P. Kostov, B. Z. Shapiro and A. Martinez-Finkelstein, Narayana numbers and Schur-Szeg¨o composition, J. Approx. Theory, 161 (2) (2009) 464-476.[11] V. Prasolov, Polynomials, Translated from the 2001 Russian second edition by DimitryLeites. Algorithms and Computation in Mathematics, 11. Springer-Verlag, Berlin, 2004.[12] Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, London Math. Soc.Monogr. (N.S.), vol. 26, Oxford Univ. Press, New York, NY, 2002.
Universit´e de Nice, Laboratoire de Math´ematiques, Parc Valrose, 06108 Nice, Cedex2, France
E-mail address ::