MSC 2010: 30C10The classical notion of apolarity is defined for two algebraic polynomials of equal degree. The main property of two apolar polynomials p and q is the classical Grace theorem: Every circular domain containing all zeros of p contains at least one zero of q and vice versa. In this paper, the definition of apolarity is extended to polynomials of different degree and an extension of the Grace theorem is proved. This leads to simplification of the conditions of several well-known results about apolarity

Sendov, Blagovest

Sendov, Hristo

Bulgarian Digital Mathematics Library at IMI-BAS

M a t h e m a t i c aB a l k a n i c a—————————NewSeries Vol. 27, 2013, Fasc. 1–2Extention of Apolarity and Grace TheoremBlagovest Sendov 1, Hristo Sendov 2Presented by Peta˘r KenderovThe classical notion of apolarity is defined for two algebraic polynomials of equaldegree. The main property of two apolar polynomials p and q is the classical Grace theorem:Every circular domain containing all zeros of p contains at least one zero of q and vice versa.In this paper, the definition of apolarity is extended to polynomials of different degree andan extension of the Grace theorem is proved. This leads to simplification of the conditions ofseveral well-known results about apolarity.MSC 2010: 30C10Key Words: zeros and critical points of polynomials, apolarity, polarderivative, Grace theorem1. PreliminariesLet C denote the complex plane and C∗ := C ∪ {∞} its one-point com-pactification. The set of all polynomialsp(z) = anzn + an−1zn−1 + · · ·+ a0 = an(z − α1)(z − α2) · · · (z − αn), (1.1)with an 6= 0 is denoted by Pn. The set of zeros of p ∈ Pn, each counted withits multiplicity, is denoted by Z(p) = {α1, α2, . . . , αn}. The closure of Pn withrespect to taking limits of the coefficients isP¯n = P0 ∪ P1 ∪ · · · ∪ Pn.The (real) degree of p ∈ P¯n is the highest power of z with non-zero coefficient.If p ∈ P¯n has degree n− s for some s ∈ {0, 1, . . . , n}, we say that p has a zero at∞ with multiplicity s. Thus, the set Z(p) of the zeros of a polynomial p ∈ P¯n ofdegree n− s contains the symbol ∞ repeated s times. Thus, every polynomial78 Bl. Sendov, Hr. Sendovp ∈ P¯n has n zeros, counting their multiplicity. We formally write p(∞) = 0 ifp ∈ Pn−s for some s = 1, 2, . . . , n and p(∞) 6= 0 if p ∈ Pn.A (non-degenerate) Mo¨bius transformationT (z) =az + bcz + d; a, b, c, d ∈ C, ad− bc 6= 0maps the extended complex plane C∗ one-to-one on itself, andT (−d/c) =∞, T (∞) = ac, T−1(z) =dz − b−cz + a, T−1(a/c) =∞, T−1(∞) = −dc.Every Mo¨bius transformation defines an operator on P¯n in the followingway:T [p](z) := (cz + d)np(az + bcz + d), p ∈ P¯n. (1.2)The operator T [p] maps the set P¯n onto itself. More precisely, we have thefollowing lemma.Lemma 1.1. Let p ∈ P¯n and {α1, α2, . . . , αn} be the zeros of p,counting multiplicities and zeros at ∞. Then, the zeros of T [p] are{T−1(α1), T−1(α2), . . . , T−1(αn)}.P r o o f. Let p∈Pn−s, for some s ∈ {0, 1, ..., n−1} and let {α1, α2, ..., αn−s}be the finite roots of p. We have the following cases:1) If p(a/c) 6= 0, then T [p] ∈ Pn andZ(T [p]) ={T−1(α1), T−1(α2), . . . , T−1(αn−s),−dc,−dc, . . . ,−dc}, (1.3)where the zero −d/c has multiplicity s. Informally, for a polynomial of degreen− s, with p(a/c) 6= 0, the operator T “brings” s zeros from ∞ to −d/c, whichconforms with the fact T−1(∞) = −d/c. In this case, if c = 0 then ∞ is not azero of p and we must have s = 0.2a) Suppose a/c is a zero of p of multiplicity l and c 6= 0. Let α1, α2,. . . , αn−s−l be the finite zeros of p different from a/c. Then T [p] ∈ Pn−l andZ(T [p])={T−1(α1), T−1(α2), ..., T−1(αn−s−l),−dc,−dc, ...,−dc,∞,∞, ...,∞},(1.4)where the zero −d/c has multiplicity s and the zero ∞ has multiplicity l. In-formally, the operator T “brings” s zeros from ∞ to −d/c and “sends” a/cExtention of Apolarity and Grace Theorem 79to ∞ with multiplicity l. This conforms with the fact T−1(∞) = −d/c andT−1(a/c) =∞.2b) If a/c is a zero of p of multiplicity l and c = 0, then we must havel = s andZ(T [p]) ={T−1(α1), T−1(α2), . . . , T−1(αn−s),∞,∞, . . . ,∞}, (1.5)where the zero ∞ has multiplicity s. That is, T [p] ∈ Pn−s. Informally, theoperator T does not “bring” or “send” zeros to ∞.For any a ∈ C and p ∈ Pn, n ≥ 1, define the polynomial Da(p; z) ∈ Pn−1byDa(p; z) = np(z)− (z − a)p′(z). (1.6)The linear operator (1.6) is called polar derivative with pole a, see [1, p. 97]. Itis clear thatlima→∞Da(p; z)a= p′(z),and we defineD∞(p; z) := p′(z).It is important to see how the operator T transforms the zeros of the derivative ofp (i.e. the critical points), when p ∈ Pn. Naturally, we consider the polynomialsDa(p; z) and D∞(p; z) to be members of P¯n−1.Lemma 1.2. Let p ∈ P¯n. The critical points of T [p] are{T−1(β1), T−1(β2), . . . , T−1(βn−1)},where {β1, β2, . . . , βn−1} are the zeros of Da/c(p; z), counting multiplicities andzeros at infinity.P r o o f. Let p ∈ Pn−s, for some s ∈ {0, 1, . . . , n − 1}, implying thatDa/c(p; z) ∈ Pn−s−1. Let β1, β2, . . . , βn−s−1 be the finite roots of Da/c(p; z). Weconsider two cases:1) If c 6= 0, then from (1.2), we calculateT [p]′(z) = nc(cz + d)n−1p(T (z)) + (cz + d)np′(T (z))T ′(z)= c(cz + d)n−1(np(T (z)) + 1cad−bccz+d p′(T (z)))= (cz + d)n−1(np(T (z))− (T (z)− a/c)p′(T (z)))= (cz + d)n−1Da/c(p;T (z))= (cz + d)s∏n−s−1k=1 (z − T−1(βk)).Thus, the critical points of T [p] are80 Bl. Sendov, Hr. SendovZ(T [p]′) ={T−1(β1), T−1(β2), . . . , T−1(βn−s−1),−dc,−dc, . . . ,−dc},where the zero −d/c has multiplicity s. Since T−1(∞) = −d/c, we are done.2) If c = 0, then both a and d are non-zero and observing thatT [p]′(z) = adn−1p′(T (z)) = adn−1D∞(p;T (z)),we are done.We may consider the ordinary derivative as a polar derivative with pole∞. When we are interested in the critical points of the polynomial T [p], wehave to get the polar derivative of p with pole T (∞) = a/c.2. ApolarityThe notion of apolarity, see [1, p. 102], is defined as a symmetric relationbetween two polynomials of the same degree.Definition 2.1. The polynomialsp(z) =n∑k=0akzk ∈ Pn and q(z) =n∑k=0bkzk ∈ Pnare called apolar, ifn∑k=0(−1)k(nk) an−kbk = 1n!n∑k=0(−1)kp(n−k)(0)q(k)(0) = 0. (2.1)The basic relation between the zeros of two apolar polynomials is theclassical theorem of Grace, see [1, p. 107]. A circular domain, open or closed,is a proper subset of C, bounded by a circle, or a line.Theorem 2.2. (Grace) Let p and q be apolar. Then every circulardomain containing all the zeros of one of them contains at least one zero of theother.It is natural to extend the notion apolarity to polynomials of arbitrarydegree.Definition 2.3. We say that p, q ∈ P¯n are n-apolar, or just apolar, ifn∑k=0(−1)kp(n−k)(0)q(k)(0) = 0. (2.2)Such a definition of apolarity does not require the degree of the polyno-mials to be fixed or the same.Extention of Apolarity and Grace Theorem 81For example, let p(z) = a1z + a0 and q(z) = b1z + b0. If we consider pand q to be in P1, then they are 1-apolar if a1b0 − a0b1 = 0. If we consider pand q to be in P2, then they are 2-apolar if a1b1 = 0. Finally, they are n-apolarfor every n ≥ 3.As another example, consider p(z) = a2z2 +a1z+a0 and q(z) = b1z+ b0.If we consider p and q to be in P2, then they are 2-apolar if a2b0 − a1b1/2 = 0.If we consider p and q to be in P3, then they are 3-apolar if a2b1 = 0. Finally,they are n-apolar for every n ≥ 4.Statement 2.1. Let s ∈ {0, 1, . . . , n − 1}. If p ∈ Pn and q ∈ Pn−s aren-apolar, then p(l) and q are (n− l)-apolar for l = 0, 1, . . . , s.P r o o f. As p and q are n-apolar and q ∈ Pn−s, we have0 =n∑k=0(−1)kp(n−k)(0)q(k)(0) =n−l∑k=0(−1)k dn−l−kdzn−l−kp(l)(z)∣∣∣z=0q(k)(0)for any l = 1, 2, . . . , s.Lemma 2.4. If p, q ∈ P¯n are apolar, then so are T [p] and T [q].The statement of Lemma 2.4 is analogous to the one given in [1, Re-mark 3.3.4, p. 103]. Only there, it is required that both p and q be of degree n,as well as that both T [p] and T [p] be of degree n. The justification of Lemma 2.4is similar: Every Mo¨bius transformation T , with ad − bc 6= 0, is a compositionof transformations of the type 1/z and az+ b for a 6= 0. It is not difficult to seethat Lemma 2.4 holds for these two types of transformations.We formulate an extension of the Grace theorem for two arbitrary poly-nomials.Theorem 2.5. Let p and q in P¯n be apolar. Then every circulardomain containing all the zeros of one of them contains at least one zero of theother.P r o o f I. If p and q are both of degree n, then the theorem coincideswith the classical theorem of Grace. If p and q are both of degree strictly lessthan n, then both polynomials have ∞ as a zero. Hence, the theorem is true.The more interesting case is when the degree of p is m < n, while the degree ofq is n. Let p(z) =∑mk=0 akzk and consider the polynomialpε(z) = εzn +m∑k=0ak(ε)zk = ε(z − ζ1(ε))(z − ζ2(ε)) · · · (z − ζn(ε)).It is possible to choose the coefficients {ak(ε) : k = 0, . . . ,m} so that ak(ε)approaches ak as ε approaches 0 and so that pε is apolar to q. Then, ε approaches82 Bl. Sendov, Hr. Sendov0, m of the zeros {ζ1(ε), ζ2(ε), . . . , ζn(ε)} approach the finite zeros of p, and therest n−m escape to ∞. The classical Grace theorem is valid for the pair pε, qand by continuity it remains valid for the pair p, q.P r o o f II. Similarly to the first proof, the interesting case is when thedegree of p is m < n, while the degree of q is n. The rest follows immediatelyfrom the original Grace theorem and Lemma 2.4 after choosing the Mo¨biustransformation T to be such that T−1 sends the roots of p and ∞ into C andreferring to Lemma 1.1.Observe that if the degree of q is less than n, then every circular domain,containing all the zeros of q is a half plane or the exterior of a open disk. Asimple corollary, which is a priori obvious from the Gauss-Lucas theorem, is thefollowing. Let p(z) = (z−z1)(z−z2) · · · (z−zn) = zn−(z1+z2+ · · ·+zn)zn−1+· · ·+ a0; n ≥ 2 and q(z) = z− b be apolar to p. According to Definition 2.3, wehave b = (z1 + z2 + · · · + zn)/n. The extended Grace theorem says that everyhalf plane having b on its boundary contains at least one of the zeros of p. Thisis a priori trivial, as b is the center of gravity of the zeros of p.2.1. Properties of extended apolarity. The apolarity relation, see[1, p. 102-104], has several important properties. In some of them, it is necessaryto state that the two polynomials are of the same degree. We will rephrasethese properties for the extended definition of apolarity and see that in someinstanced the underlying conditions are simplified. Definition 2.3 immediatelyimplies that: (i) Every polynomial of odd degree is apolar to it self. (ii) If twopolynomials q1 and q2 are both apolar to p and λ1, λ2 ∈ C, then the polynomialq = λ1q1 + λ2q2 is also apolar to p.For polynomials p, q∈P¯n given by p(z)=∑nk=0 akzk and q(z)=∑nk=0 bkzk,introduce the functionalΛp[q] := n!n∑k=0(−1)k(nk) an−kbk = n∑k=0(−1)kp(n−k)(0)q(k)(0)defined in P¯n for every fixed p ∈ P¯n. The following lemma is proved in [1, p.103] for p, q ∈ Pn.Lemma 2.6. Let p, q ∈ P¯n and let deg(p) = λ, deg(q) = µ. DefineP (z) = p(az + b), G(z) = g(az + b), a 6= 0.Then, Λp[q] = a−mΛP [Q], where m := max{λ, µ}.P r o o f. It is easy to see thatΛP [Q] = amn∑k=0(−1)kp(n−k)(b)q(k)(b) (2.3)Extention of Apolarity and Grace Theorem 83and∂ΛP [Q]∂b= am(p(n+1)(b)q(b) + (−1)np(b)q(n+1)(b))= 0.The last equality implies that the sum on the right-hand side of (2.3) does notdepend on b. Letting b = 0 completes the proof.The proof of Lemma 2.6 shows that for any b ∈ C we haveΛp[q] =n∑k=0(−1)kp(n−k)(b)q(k)(b) (2.4)The following lemma, see [1, p. 104], suggests that apolarity is connected withcommon zeros.Lemma 2.7. Let p, q ∈ P¯n have a common zero z1. Suppose the sumof the multiplicities of z1 as a zero of p and q, respectively, is larger than n.Then, Λp[q] = 0 and so q is apolar to p.P r o o f. Let l and m be the multiplicities of z1 as a zero of p and q,respectively. By assumption, we have l +m > n.If z1 =∞, then p(k)(0) = 0 for all k = n− l+ 1, . . . , n and q(n−k)(0) = 0for all k = 0, . . . ,m− 1. Since n− l + 1 ≤ (m− 1) + 1, we get Λp[q] = 0.If z1 ∈ C, then using (2.4), we getΛp[q] =n∑k=0(−1)kp(k)(z1)q(n−k)(z1),and the proof concludes as in the first case.The following theorem describes the set of all polynomials q(z) ∈ P¯nwhich are apolar to a given polynomial p(z) ∈ Pn in terms of its zeros. Theproof follows the proof of the analogous theorem in [1, p. 104], keeping inmind that q(z) ∈ P¯n. Note that while the result in [1, p. 104] requires that theconstants cj,l be non-zero, this restriction is now relaxed, thanks to the extendednotion of apolarity.Theorem 2.8. Let p(z) = an∏kj=1(z− zj)mj ∈ Pn, where z1, z2, . . . , zkbe distinct, and {mj}kj=1 be natural numbers with∑kj=1mj = n. Then, thepolynomialsq(z) =k∑j=1ml−1∑l=0cj,l(z − zj)n−l, (2.5)where cj,l ∈ C for j = 1, 2, . . . , k, l = 0, 1, . . . ,mj − 1, constitute the set of allpolynomials q ∈ P¯n apolar to p.84 Bl. Sendov, Hr. SendovP r o o f. Each polynomialqj(z) =ml−1∑l=0cj,l(z − zj)n−l, j = 1, 2, . . . , k, (2.6)has zj as a zero of multiplicity at least n −mj + 1. Hence, Lemma 2.7 impliesthat Λp[qj ] = 0. Thus, Λp[q] =∑kj=1 Λp[qj ] = 0 showing that every polynomialof the form (2.5) is apolar to p. It remains to show that there are no otherpolynomials apolar to p.The functional Λp[q] is not identically zero, since for the polynomialh(z) :=n∑k=0(−1)kk!p(n−k)(0)zk,we haveh(k)(0) = (−1)kp(n−k)(0) and Λp[h] =n∑k=0|p(n−k)(0)|2 > 0.By a standard result from linear algebra, the kernel of Λp[q], that is, the set of allpolynomials q ∈ P¯n satisfying Λp[q] = 0, is a linear subspace Vn of dimension n.A polynomial q, apolar to p, must belong to Vn. By Lemma 2.7, the polynomialsqj,l(z) := (z − zj)n−l, j = 1, 2, . . . , k, l = 0, 1, . . . ,mj − 1do all belong to Vn. Since there is total of n polynomials qj,l(z), it is enough toshow that they are linearly independent. Assume the contrary thatk∑j=1mj−1∑l=0cj,lqj,l(z) ≡ 0with some of the coefficients cj,l being non-zero. Writing the left-hand side as∑nν=0 Lνzν , we must have Ln = Ln−1 = · · · = L0 = 0. SinceLν =k∑j=1mj−1∑l=0(n− lν)(−zj)n−l−νcj,l, ν = 0, 1, . . . , n,where(n−lν)= 0 when ν > n− l, it follows that the systemk∑j=1mj−1∑l=0(n− lν)(−zj)n−l−νuj,l = 0, ν = n, n− 1, . . . , 1, 0Extention of Apolarity and Grace Theorem 85of n equation in the n unknowns uj,l has a non-trivial solution. Hence, thedeterminant ∆ of this system must vanish. Its entries are1 0 . . . 0−(n1)zj 1 . . . 0(n2)z2j −(n−11)zj . . . 0....... . ....(−1)n−1( nn−1)zn−1j (−1)n−2(n−1n−2)zn−2j . . . (−1)n−mj(n−mj+1n−mj )zn−mjj ,where j = 1, 2, . . . , k. From ∆ = 0 follows that a non-trivial linear combinationof the rows is zero. Hence, there exists a polynomialf(z) =n−1∑ν=0(−1)ν(nν)λνzν ,which is not identically zero and such that(−1)l n!(n− l)!f(l)(zj) = 0, j = 1, 2, . . . , k, l = 0, 1, . . . ,ml − 1.Thus, f(z) would have a total of at least∑kj=1mj = n zeros, which is impossiblesince the degree of f(z) is at most n− 1.The next lemma gives a connection between apolarity and polar deriva-tives.Lemma 2.9. Let p ∈ Pn, q ∈ P¯n, and let z∗ be a zero of p(z) =∑nk=0 akzk. If p and q are apolar, thenp1(z) ={p(z)/(z − z∗) if z∗ ∈ C,∑n−1k=0 akzk if z∗ =∞,is apolar to the polar derivative Dz∗(q; z).P r o o f. We have p1 ∈ Pn−1 and Dz∗(q; z) ∈ P¯n−1. If z∗ = ∞, then wehave bn = 0 and0 =n∑k=0(−1)k(nk) an−kbk = 1nn∑k=1(−1)k(n−1k−1) an−k(kbk).This shows that Λq′ [p1] = 0, or that p1 and D∞(q; z) are apolar.Suppose now, that z∗ ∈ C. Letp(z) = ank∏j=1(z − zj)mj ,k∑j=1mj = n, an 6= 0,86 Bl. Sendov, Hr. Sendovwhere z1, z2, . . . , zk are distinct numbers. Then, by Theorem 2.8, q(z) must beof the form q(z) =∑kj=1 qj(z), whereqj(z) :=mj−1∑l=0cj,l(z − zj)n−l, j = 1, 2, . . . , k.Hence, we may writeDz∗(q; z) = nk∑j=1qj(z)− (z − z∗)k∑j=1q′j(z) =k∑j=1Dz∗(qj ; z).Clearly, Dz∗(qj ; z) is a polynomial of degree at most n − 1 in z, with zero zjhaving multiplicity at least n−mj + 1 when zj = z∗ and at least n−mj whenzj 6= z∗. Therefore, Dz∗(qj ; z) is of the formDz∗(qj ; z) =nj−1∑ν=0γj,ν(z − zj)n−1−ν , where nj :={mj if zj 6= z∗,mj − 1 if zj = z∗.This means, by Theorem 2.8, that Dz∗(q; z) =∑kj=1Dz∗(qj ; z) is amongst thepolynomials apolar toank∏j=1(z − zj)nj = p1(z),completing the proof.Combining Lemma 2.9 and Statement 2.1, we obtain the following corol-lary.Corollary 2.1. Let s ∈ {0, 1, . . . , n− 1}. Let p ∈ Pn, q ∈ Pn−s, and z∗be a zero of p =∑nk=0 akzk. If p and q are apolar, thenp1(z) ={p(z)/(z − z∗) if z∗ ∈ C,∑n−1k=0 akzk if z∗ =∞,is apolar to the polar derivative Dz∗(q(s); z).Acknowledgements. The first author (Bl. S.) was partially supportedby Bulgarian National Sci. Fund under Grant # DTK 02/44, and the secondauthor (Hr. S.) was partially supported by the Natural Sciences and EngineeringResearch Council (NSERC) of Canada.Extention of Apolarity and Grace Theorem 87References[1] Q. I. R a h m a n, G. S c h m e i s s e r. Analytic Theory of Polynomials, OxfordUniv. Press, New York, 2002.Received: July 2, 20121 Institute of Information and Communication TechnologiesBulgarian Academy of SciencesAcad. G. Bonchev Str., Block 25A,1113 Sofia, BULGARIAe-mail: sendov2003@yahoo.com2 Department of Statistical and Actuarial SciencesWestern University1151 Richmond Str.,London, ON, N6A 5B7 CANADAe-mail: hssendov@stats.uwo.ca 

Extention of Apolarity and Grace Theorem

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Extention of Apolarity and Grace Theorem

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