ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES

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1 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 50, Number 11, November 1998, Pages S ) ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES PIOTR PAWLOWSKI Abstract If pz) is uivariate polyomial with complex coefficiets havig all its zeros iside the closed uit disk, the the Gauss-Lucas theorem states that all zeros of p z) lie i the same disk We study the followig questio: what is the maximum distace from the arithmetic mea of all zeros of pz) to a earest zero of p z)? We obtai bouds for this distace depedig o degree We also show that this distace is equal to 1 for polyomials of degree ad polyomials with real zeros 1 Itroductio Let P deote the family of all complex polyomials of exact degree, havig all zeros i the closed uit disk, ie 11) ad let 12) { P := pz) =a k=1 z z k ), z k 1 1 k, a 0) P := 1 P Let p P From the Gauss-Lucas theorem we kow that p P A well-kow attempt to further characterize the locatio of the zeros of p, which are also called the critical poits of p, is the Sedov Cojecture Cojecture 11 Sedov) If pz) P the each of the disks z C : z z k 1, k =1,,, cotais at least oe zero of p z) The Sedov Cojecture has bee verified i a umber of special cases [2] Let Hp) deote the covex hull of the zeros of a polyomial p Schmeisser [10] posed the followig geeralizatio of the Sedov Cojecture Cojecture 12 Schmeisser) If pz) P, the for every ζ Hp), the disk {z : z ζ 1},k=1,,, cotais at least oe zero of p z) Schmeisser [10] showed that this cojecture is true if Hp) is a triagular regio It is also show to be true if Hp) has all its vertices o the uit circle [2] We ivestigate a special case of the Schmeisser problem: we cocetrate o oe poit i Hp) - the average of zeros of p Although i this case the Schmeisser Cojecture follows immediately from the Gauss-Lucas Theorem, we show that the disk {z : Received by the editors Jue 27, Mathematics Subject Classificatio Primary 0C15 Key words ad phrases Polyomials, locatio of zeros } 4461 c 1998 America Mathematical Society Licese or copyright restrictios may apply to redistributio; see

2 4462 PIOTR PAWLOWSKI z ζ 1}i Cojecture 12) ca be replaced by a smaller cocetric disk with radius r We give bouds for r depedig o for ay p P We also obtai sharp estimates of r for polyomials of degree ad for polyomials havig oly real zeros Let 1) pz) = a k z k = a z z k ) P k=0 The the cetroid, A p, of the polyomial pz) is the arithmetic average of its zeros: A p := 1 z 1 + +z )= 1 a 14) a It is easy to check that the cetroid is differetiatio ivariat Namely, for >1 15) p z)= k +1)a k+1 z k = a z ζ k ), k=0 so that 16) A p = 1 1 ζ 1 + +ζ )= 1 )a = 1 a = A p, a a which gives 17) We also defie A p = A p k), 0 k 1 18) Jp) := mi A p ζ k ad γ := sup Jp) 1 k p P Remark 1 There exist results that characterize the locatio of zeros of p k) for k 1 for polyomials havig oly real zeros Milovaovic et al [6] give a accout of may results of this type For istace, if p is a polyomial with zeros x 1 x 2 x, the we defie spap) :=x x 1 Clearly, 19) spap k) ) spap), 0 k 1 by the Gauss-Lucas Theorem Robiso [8], [9] has show that the strict iequality holds i 19) if k is sufficietly large He also gave estimates o spap k) ), depedig o spap), ad k Sice the cetroid of a polyomial is ivariat uder differetiatio, the zeros of p k) must be gettig closer to A p as k grows big Although spap k) ) may ot decrease if k</2, we will show i sectio 4, that ot all zeros of p maystayawayfroma p k=1 k=1 2 A lower boud for the costat γ We first give simple upper ad lower bouds for γ Propositio 21 21) For ay >1, 1 γ 1 Proof The secod iequality follows from the fact that the cetroid of a polyomial is differetiatio ivariat, so the cetroid of pz) is also the cetroid of p z) Sice all of the zeros of p z) are iside the uit circle, at least oe of them lies withi distace 1 from the cetroid The first iequality follows from the special choice of Licese or copyright restrictios may apply to redistributio; see

3 ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES 446 pz) =z z Here, we have p z) =z 1 ad for every critical poit ζ k of pz), we have 22) ζ k = 1 1 so Jp) = 1 1 Remark ) log is asymptotically equal to, ie lim 1 1 log =1 We ca show it by takig Taylor series expasio of 1 1 : ) 1 1 = 1 exp = log 1 + o log 1 log 1 = log 1 1 2! ) Therefore, there exist costats c 1,c 2 such that 24) log 1 c 1 < 1 log <1 c2 A upper boud for costat γ ) 2 log + 1 ) log 1! 1 If p is moic ad A p = 0, the product of all critical poits is k=1 ζ k = a 1 /, so Jp) is bouded by a 1 / We ca use a good upper boud o a 1 give by the followig lemma: Lemma 1 1) If pz) =z + +a 1 z+a 0 P ad A p =0, the a 1 = p 0) 1 Proof If A p =0,the a 1 = z j k=1 j k = 2) = z k k=11 2 ) j k k=1 j k z j z j z k z j ) j=1 z k k=11 2 ) z j j k Let z k = r k, 0 r k 1, 1 k ad let ) We will show that B := rk) k=11 2 r j, j k 1 4) B 1 for every 1 First, we will show that B attais its maximum o the boudary of the set [0, 1] If = 1, the the maximum of B 1 =1 r 1 is attaied at r 1 =0 If 2adB Licese or copyright restrictios may apply to redistributio; see

4 4464 PIOTR PAWLOWSKI attais its maximum o the set [0, 1] at some poit r 1,,r ) 0, 1) The we have 5) The B = 2r k r j + 1 rj 2 ) r i =0, r k j k j k i k,j 1 k B 6) r k = 2rk 2 r j + rj r k j k j k1 2 ) r i i j = 1 rk 2 ) j + rj j kr j=11 2 ) r i i j = r k 1 ) r j + rj r k j=1 j=11 2 ) r i =0, 1 k i j For i j we get 7) Hece 8) r i B r i B r j = r j + 1 r i 1 ) r j r j r i r j + 1 r j = r i + 1 r i k=1 r k =0 Sice the fuctio fx) =x+ 1 x is oe to oe o 0, 1), r i = r j Thus the local maximum M of B ca be attaied at r 1,,r ) 0, 1) oly if all r k are equal Let r := r 1 = =r The 9) M sup r 1 r 2 ) ) 0<r<1 The above supremum is attaied at r = +1 ad 10) M ) 2, 2 It is well kow that the sequece e := ) 2 1 e as ad e is decreasig Thus, for, e e =1/2 Therefore, from 10) we have 11) M { 4 +1 < 1 if =2, < 1 if Sice B = 1 if r 1,,r )=0,1,,1), B must attai its maximum o the boudary of [0, 1] Now, we will prove 4) by iductio If =1,theB =1 r 1 1 Let N be fixed ad assume that 4) holds for =1,,N 1 If B N attais its maximum o the set [0, 1] at the poit r 1,,r ), the r k =0 or r k =1 for some k, 1 k N If r k =0,theB N = j k r j 1 Otherwise, if r k =1,the B N =B N 1 1 by iductio Remark 2 From Lemma 1 we get a estimate o Jp) i the special case whe A p =0 ThewehaveJp) p 0) /) 1 1/) 1 This estimate is sharp as show by pz) =z z Licese or copyright restrictios may apply to redistributio; see

5 ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES 4465 Usig Lemma 1 we ca fid a upper boud o γ that is smaller tha 1 Theorem γ ) Proof If pz) P ad A p = r, the 1 1) qz)= r+1) pa p+r+1)z) P ad A q =0 The polyomial qz) is moic, so by Lemma 1, there exist w such that q w) =0 ad q 0) 14) w If ξ =r+1)w+a p,thep ξ)=0ad 15) ) Jp) ξ A p =r+1) w r+1 1 O the other had, there must be a zero of p z) i the half-plae Re za p ) r, so we have 16) Therefore, from 15) ad 16) 17) Jp) Jp) 1 r 2 sup mi 0 r 1 { r +1, } 1 r 2 1 As fuctios of r, the bouds i 15) ad 16) are respectively icreasig ad decreasig o the iterval [0, 1], so the supremum i 17) is attaied if r is the uique solutio of the equatio r +1 18) which gives 19) 1 r = = 1 r 2, 2 +1 ad Jp) Remark 4 Theorem 11 leads to the followig estimate ) ) γ = <1 c ) 2 log, where c is some costat, because as we have already oted i 2) 1 1 is asymptotically equal to log / However, the estimate 20) does t seem to be sharp Our cojecture is that a sharp iequality would have the form γ 1 c log, as i the example pz) =z z Licese or copyright restrictios may apply to redistributio; see

6 4466 PIOTR PAWLOWSKI 4 Extremal polyomials of low degrees I this sectio we compute γ exactly I order to do this we restrict our attetio to extremal polyomials, ie the subset of P cosistig of polyomials for which Jp) =γ Such polyomials exist sice Jp) is cotiuous i the roots of p,which are cotiuous i the coefficiets of p, which are cotiuous i the roots of p beig elemetary symmetric fuctios of those roots) The roots of p are cotaied i the compact set D, sojp) achieves its supremum o D For extremal problems of this type, Phelps ad Rodriguez [7]) have show that every extremal polyomial has at least 2 zeros o the uit circle ad that every sub-arc of the uit circle of legth at least π must cotai at least oe zero Otherwise, it would be possible to iscribe all zeros of the polyomial i the circle with radius smaller tha 1, which would cotradict extremality We ca coclude that uless the extremal polyomial has 2 zeros at the edpoits of diameter of the uit circle, it must have at least zeros o the uit circle This is also valid i our problem Propositio 41 γ = 2, ad all extremal polyomials are of the form pz) = qe it z)whereqz)=z z 2 z+1 Proof Let pz) P Without loss of geerality we ca assume that 1 is oe of the zeros of pz), ie 41) pz) =z 1)z z 1 )z z 2 ), ad A p = 1+z 1+z 2 42) Sice p z)/ is moic, the distace betwee A p ad the earest critical poit is bouded by p A p )/ Wehave 4) p A p ) = 1 9 z 2 1 z 2 2+z 1 +z 2 +z 1 z 2 1 By the maximum modulus priciple p A p )/ attais its maximum o D whe both z 1 ad z 2 are o the uit circle We ca also assume that pz) isextremal ad z 1 = e iφ1, 0 φ 1 π, adz 2 =e iφ2,π φ 2 2π,adφ 2 φ 1 πthe 44) p A p ) 1 z1 2 + z z 1 +z 2 +z 1 z 2 1 ) siφ1 si φ , where the last iequality follows from the fact that si φ 1 si φ 2 0 Thus we have 4 45) Jp) 9 = 2, with equality oly if si φ 1 si φ 2 = 0 I this case either z 1 = z 2 = 1 adpz)= z+1) 2 z 1) or z 1 =1,z 2 = 1adpz)=z 1) 2 z + 1) These are the oly extremal polyomials for which p1) = 0 I both cases Jp) = 2,soγ = 2 Remark 42 Numerical computatios usig oliear optimizatio algorithms suggest that γ 4 =2/ ad a example of extremal polyomial 46) p 4 z) =z 4 4 z + 2 z2 4 z+1 Licese or copyright restrictios may apply to redistributio; see

7 ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES 4467 Zeros of p 4 are z 1 = z 2 =1,z = 1/+ 2 2 iad z 4 = 1/ ) i Zeros of p 4 are 48) w 1 =1,w 2 = iad w = i γ 5 =/4 ad a example of extremal polyomial is 49) p 5 z) =z z4 5 4 z z z+1 Zeros of p 5 are z 1 = z 2 = z =1,z 4 = ) i ad z 5 = 7 8 Zeros of p 5 are 411) w 1 = w 2 =1adw =w 4 = 1/ i Both the Sedov Cojecture ad our problem are special cases of extremal problems cosidered by Miller [4],[5]) He defied S, β) to be the set of complex polyomials of degree that have all their zeros i the uit disk ad at least oe zero at β ad p α to be the distace betwee α ad the closest root of p Polyomial p S, β) issaidtobemaximal with respect to α if for every q S, β) we have p α q α He also defied critical circle to be the circle with ceter α ad radius p α Our problem ca be formulated as follows: fid maximum of p Ap for p S, β) Miller [4]) made the followig cojecture: Cojecture 4 Miller) Let p S, β) be maximal with respect to α, letsbe the umber of zeros of p o the uit circle, ad let r be the umber of zeros of p o the critical circle The 412) r = 1, ad 41) give that all zeros of p lie o a circle, s is as large as possible He also proved that uder the hypothesis of Cojecture 1, 2r + s +1 I our problem, if =,4 or 5, it seems that all zeros of a maximal polyomial p lie o the uit circle ad all zeros of p lie o the critical circle p S, β) issaid to be maximal with respect to α if for every q S, β) wehave p α q α He also defied critical circle to be the circle with ceter α ad radius p α Our problem ca be formulated as follows: 5 Polyomials with real zeros The polyomial pz) =z zwith high Jp) has all its zeros, except oe, spread evely aroud the uit circle If this situatio caot happe, for istace for polyomials with real zeros, we may expect Jp) to be lower Let s cosider the subclass of P cosistig of polyomials havig oly real zeros I this case we get the followig theorem: Theorem 51 If px) P is a polyomial havig oly real zeros, the 51) Jp) 2 Licese or copyright restrictios may apply to redistributio; see

8 4468 PIOTR PAWLOWSKI Equality i 51) is attaied if ad oly if is a multiplicity of ad px)=x 1) 2 x +1) or px) =x 1) x +1) 2 Proof It s eough to show that 51) is true if A p is o-egative; otherwise we ca cosider p x) istead of px) If A p > 1/, the obviously Jp) < 2/, so let us assume that A p [0, 1/] Let I =A p 2/,A p +2/) We wat to show that the iterval I [ 1, 1] cotais at least oe zero of p x) By Rolle s Theorem, this is true if I cotais at least two zeros of px), coutig multiplicities Let us cosider the remaiig two cases Case 1: I does t cotai ay zeros of px) Let A p [0, 1/] be fixed ad let 1 ad 2 = 1 be the umbers of zeros of px) i the itervals I 1 =[ 1,A p 2/] ad I 2 =[A p +2/,1] respectively The coditio x 1 + +x =A p imposes some costraits o 1 ad 2 Namely, 1 is maximal ad 2 is miimal if we move all zeros to the right as far as possible I the extreme situatio we would have 1 zeros at A p 2/ ad 2 zeros at 1 We the have a liear system { 1 A p 2/) + 2 = A p, 52) = Although the solutio of 52) is ot a pair of itegers i geeral, they give a upper boud for 1 ad a lower boud for 2 Similarly, if we put 1 zeros at 1 ad 2 zeros at A p +2/ we get a liear system { A p +2/) = A p, 5) = that gives a lower boud for 1 ad a upper boud for 2 Combiig 52) ad 5) we obtai rage of 1 ad 2 depedig o A p A p) 54), 5+A p 5 A p A p) 55) 5 A p 5+A p Sice all zeros of px) lie i 2 disjoit itervals, we ca apply Walsh s two circle theorem [2, p 89]) Theorem 52 Walsh) If all the zeros of a 1 -degree polyomial f 1 z) are cotaied i the closed iterior of the circle C 1 with ceter c 1 ad radius r 1 ad all the zeros of a 2 -degree polyomial f 2 z) are cotaied i the closed iterior of the circle C 2 with ceter c 2 ad radius r 2, the all zeros of the derivative of the product fz) =f 1 z)f 2 z)are cotaied i the uio of C 1,if 1 >1,adC 2,if 2 >1,adathirdcircleCwith ceter c ad radius r where c = 1c c 1 ad r = 1r r 1 56) Moreover, if the closed iteriors of the circles C 1,C 2 ad C have o poit i commo, the umber of zeros of f z) which they cotai is respectively 1 1, 2 1 ad 1 By 56), all zeros of p x) are cotaied i the uio of the itervals I 1,I 2 ad I, where 57) [ 1 A p +2/) 2 I =, ] A p 2/) Licese or copyright restrictios may apply to redistributio; see

9 ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES 4469 Applyig iequalities 54) ad 55) we obtai 58) 1 A p +2/) 2 A p 2 ) 1 A p 0 ad A p + 2 ) 59) A p 2/) A p + 1 > 0 If 1 A p > 0 i 58), the the itervals I 1,I 2 ad I are disjoit ad Walsh s theorem guaratees that there is a uique zero of p i I I ad therefore Jp) < 2 The equality i 51) holds oly if A p = 1 i 58) The, from 52) ad 5) we must have 1 = ad 2 = 2 Therefore, the oly extremal polyomial for A p > 0ispx)=x 1) 2 x +1) Case 2: I cotais oe zero of px) Let t be the uique zero of px) i I Let 1 ad 2 = 1 1 be the umbers of zeros of px) i the itervals I 1 =[ 1,A p 2/] ad I 2 =[A p +2/,1] respectively As i the first case, whe we fix t ad A p, we ca get bouds for 1 ad 2 If we move all other zeros to the right as far as possible, ie put 1 zeros at A p 2/ ad 2 zeros at 1, the the solutio of the liear system { 1 A p 2/) t = A p, 510) = gives a upper boud for 1 ad a lower boud for 2 Similarly, if we move all other zeros to the left as far as possible, ie put 1 zeros at 1 ad 2 zeros at A p +2/ we get a liear system { A p +2/) + t = A p, 511) = that gives a lower boud for 1 ad a upper boud for 2 Combiig these two we obtai that the rage of 1 ad 2 depedig o A p ad t is give by 2 +t+a p A p)+t 1 512) 5+A p 5 A p ad 2 t +A p A p) t 1 51) 5 A p 5+A p We will see that by keepig t ad A p fixed ad movig the other zeros of px) we ca get ot oly the rage of 1 ad 2, but also estimate the distace from A p to a earest zero of p x) I order to do that we apply so called Root Draggig Theorem [1]) which states that if we move ay of the roots of the real-root polyomial to the right left) the all of the critical poits other tha multiple roots move to the right left) Sice t is the oly zero of px) ii,ica cotai at most two zeros of p x) ad they must be those that are closest to t from the left ad right Let s call them w ad v, w<t<v We will show that at least oe of w ad v is i I A p w will attai maximum whe we move all zeros as far as possible to the left ad so A p w A p w where w is the zero of derivative of 514) x +1) 1 x A p +2/)) 2 x t) Licese or copyright restrictios may apply to redistributio; see

10 4470 PIOTR PAWLOWSKI that is closest to t from the left, ie w is the leftmost zero of the quadratic equatio x 2 1)t + 1 A p + 2 ) 2 + A p 1 ) x + t 1 A p + 2 ) ) 2 A p 2 =0 From this ad the bouds o 1 ad 2, 515) A p w F t) := 1 2A p t t+ 1 ) 2 + 4t +1)A p t+ 2 ) We eed to show that F t) < 2/ for every t I F t) 2/ has at most 2 zeros, which are the zeros of the quadratic fuctio F t) := 4t +1)A p t+ 2 ) + t+ 1 ) 2 t+1 2A p ) 2 = 4 t A p 1 ) t 4 A 2 p A p A ) p +2 However, F t) 2/ has at most oe zero i the iterval I, sice F A p 2 ) 2 = 1 A p A p 1 ) ) 1 + A p > 1 A p A p 1 ) =0 ad 516) F A p + 2 ) 2 = 1 A p + A p +1 5 ) =A p Also the leadig coefficiet i F t) is egative ad 517) F A p 2 ) = 16 ) 1 + A p > 0adF A p + 2 ) = 16 A p 1 ) 0, so the oly zero of F t) 2/iIca be η, the rightmost zero of F t)adft) 2 i I if t η Also v A p will attai maximum whe we move all zeros as far as possible to the right ad v A p v A p where v is the zero of derivative of 518) x A p 2/)) 1 x 1) 2 x t) that is closest to t from the right, ie v is the rightmost zero of the quadratic equatio x 2 1)t t 2 A p 2 ) A p + 1 ) x A p 2 ) ) A p 2 =0 Licese or copyright restrictios may apply to redistributio; see

11 ON THE ZEROS OF A POLYNOMIAL AND ITS DERIVATIVES 4471 From this ad the bouds o 1 ad 2, 519) v A p Gt) := 1 2A p +t t+ 1 ) t +1) A p+t+ 2 ) We eed to show that Gt) < 2/ for every t I Gt) 2/ has at most 2 zeros, which are the zeros of the quadratic fuctio Gt) := 4 t +1) A p+t+ 2 ) + t+ 1 ) 2 t 1 2A p ) 2 = 4 t A p + 1 ) t 4 A 2 p+a p + A ) p +2 However, Gt) 2/ has at most oe zero i the iterval I, because G A p 2 ) 2 = 1 A p + A p 1 5 ) 520) = A p 1 2 <0 ad G A p + 2 ) 2 = 1 A p A p + 1 ) ) 1 A p 1 A p A p + 1 ) =0 Also the leadig coefficiet i Gt) is egative ad 521) G A p 2 ) = 16 ) 1 + A p < 0adG A p + 2 ) = 16 ) 1 A p 0, so the oly zero of Gt) 2/ iica be ξ, the leftmost zero of Gt) adgt) 2 i I if t ξ We wat to show that ξ η The equatio F t) =Gt) has the uique solutio t =A p so it s eough to show that F A p )=GA p ) < 0 But F A p )=8 A p 1 ) A p + 1 ) 522) 0for0 A p 1 ad equality is attaied oly if A p = 1 ad t = 1, but this is case I, adwemust have px) =x 1) 2 x +1) with Jp) = 2 Otherwise ξ>ηad either ξ or η lie i I ad Jp) < 2 IfA p<0 the above argumet applied to polyomial p x) shows that Jp) = 2 oly if px) =x+1)2 x 1) Refereces [1] B Aderso, Polyomial Root Draggig, Amer Math Mothly ), MR 94i:26006 [2] M Marde, Cojectures o the critical poits of a polyomial, Amer Math Mothly ), MR 84e:0007 [] M Marde, Geometry of Polyomials, Math Surveys, Amer Math Soc, Providece, RI 1966 MR 7:1562 [4] MJ Miller, Maximal polyomials ad the Ilieff-Sedov cojecture, TrasAmerMathSoc ), MR 90m:0007 Licese or copyright restrictios may apply to redistributio; see

12 4472 PIOTR PAWLOWSKI [5] MJ Miller, Cotiuous idepedece ad the Ilieff-Sedov cojecture, Proc Amer Math Soc ), 79-8 MR 92h:0012 [6] GV Milovaovic, DS Mitriovic, ad ThM Rassias, Topics i polyomials : extremal problems, iequalities, zeros, World Scietific, River Edge, NJ, 1994 MR 95a:0009 [7] D Phelps ad RS Rodriguez, Some properties of extremal polyomials for the Ilieff cojecture, Kodai Math Sem Report ), MR 46:75 [8] R M Robiso, O the spa of derivatives of polyomials, Amer Math Mothly ), MR 28:515 [9] R M Robiso, Algebraic equatios with spa less tha 4, Math Comp ), MR 29:6624 [10] G Schmeisser, O Ilieff s Cojecture, Math Z ), MR 58:6182 Departmet of Mathematics ad Computer Sciece, Ket State Uiversity, Ket, Ohio Curret address: Summit Systems, Ic, 22 Cortlad St, New York, New York address: piotr-pawlowski@summithqcom Licese or copyright restrictios may apply to redistributio; see