Polynomials with non-negative coefficients

Transcription

1 Polynomials with non-negative coeicients R. W. Barnard W. Dayawansa K. Pearce D.Weinberg Department o Mathematics, Texas Tech University, Lubbock, TX Introduction Can a conjugate pair o zeros be actored rom a polynomial with nonnegative coeicients so that the resulting polynomial still has nonnegative coeicients? This question has attracted considerable attention during the last ew years because o its seemingly elementary nature and its potential or applications in number theory and control theory. A proposed answer to this question arose as a conjecture out o the work in [6], where explicit bounds were determined or the constants occurring in the Beauzamy-Enlo generalization [1], [2] o Jensen s Inequality. An airmative answer to this question was conjectured, independently, by B. Conrey in connection with some o his work in number theory. Conrey announced the conjecture at the annual West Coast Number Theory Conerence held in December The main theorem in this note gives a positive answer to the question. This will be proved in section 2. The principle ingredients o the proo are an idea rom index theory, classical properties o polynomials and a signiicant lemma Lemma 2.1) which we prove in section 3. This lemma states some strong consequences or the case o equality holding in Descartes Rule o Signs see [5]). We also prove a corollary o the main theorem which describes the region into which certain zeros can be moved while preserving the nonnegativity o the coeicients. Formerly at Texas Tech University, Currently in the Department o Electrical Engineering, University o Maryland,College Park, Md

2 Consider the polynomial deined by P N z) =1+z N.Wenotethat P N z) = N 1 l=0 1 z e i π N + 2πl N ) ) N 2 =1 2cosθ 0 z + z 2 ) b k z k 1) has b k > 0, 0 k N 2, i θ 0 = π/n, while any other choice or θ 0 produces a actor with some negative b k coeicients. Thus, one initial suggestion or the general question o actoring out a conjugate pair o zeros was to actor out a pair o zeros with greatest real part. The authors have used airly straightorward arguments to show that i the degree o the polynomial is less than or equal to 5, then a conjugate pair o zeros o greatest real part can be actored out and the resulting polynomial will still have nonnegative coeicients. However, i pz) = z + z z z 4 +5z 5 +20z 6 with z 0, z 0 approximately equal to i, i, resp., as the conjugate pair o zeros o greatest real part, then the resulting actors are z z 0 )z z 0 ) and z 1.53z z z 4 ). The other choice, or which zeros to actor out, suggested by the example P N in 1), is the pair determined by the zero in the upper hal-plane with smallest positive argument. Indeed, in this paper we prove the ollowing: Theorem 1.1 Let p be a polynomial o degree N, p0) = 1, with nonnegative coeicients and with zeros z 1,z 2,...,z N.Fort 0 write p t z) = ) 1 zzj 1 j N Argz j >t Then, i p t p, all o the coeicients o p t are positive. Remark 1 We note that a slightly sharper version o the Theorem 1.1 is true. Suppose that the given polynomial p has nonnegative coeicients and that we divide it by a single quadratic actor corresponding to any conjugate pair o zeros o smallest argument in magnitude. Then the quotient polynomial has nonnegative coeicients. Furthermore, i the conjugate pair 2

3 o zeros is unique and simple or i p has strictly positive coeicients then the quotient polynomial will have strictly positive coeicients. The example pz) =1+z 2 ) 2 shows the necessity o these conditions. The sharpened version o the theorem under the uniqueness hypothesis ollows rom the main theorem. Now let us consider the case when several complex conjugate pairs o zeros o p, say{z i, z i } i=1,...,k, have the smallest argument in magnitude and assume that all coeicients o p are strictly positive. A local perturbation can be made o z 1 and z 1 to w 1 and w 1 such that the argument o w 1 is less than that o z 1. Under the local perturbation the polynomial p changes to q which also has strictly positive coeicients. By Theorem 1.1, q/z w 1 )z w 1 ) has strictly positive coeicients. Since p/z z 1 )z z 1 )=q/z w 1 )z w 1 ) the desired conclusion ollows. Corollary 1.1 Let pz) = N 2 a k z 0 )z k =z z 0 )z z 0 ) b k z k be a real polynomial with a k z 0 ) 0. Let z 0 beazeroop in the upper hal plane with smallest argument. I z 1 is any complex number such that then has a k z 1 ) a k z 0 ). z 1 z 0 and Re{z 1 } Re{z 0 } N 2 a k z 1 )z k =z z 1 )z z 1 ) b k z k Proo: The proo ollows directly by noting, or z 0 = re iθ = x + iy, a k z 0 )z k = r 2 2rz cos θ + z 2 ) b k z k = x 2 + y 2 2xz + z 2 ) and comparing coeicients. 3 b k z k

4 It has been recently shown by R. Evans and P. Montgomery[4] that in the special case when p is the polynomial P N deined in 1) that the corresponding reduced p t polynomials are all strictly unimodal in their coeicients). Also, R. Evans and J. Greene [3] have shown or polynomials p o the orm pz) = z Mk 1)/z k 1) that the corresponding reduced p t polynomials all have positive coeicients. 2 Proo o theorem 1.1 We note to prove Theorem 1.1 it suices, by a successive reduction argument, to show that i we remove rom p a conjugate pair o zeros o smallest argument in the absolute value, then the resulting polynomial still has nonnegative coeicients. Proo. Let pz) = a k z k =1 2cosθ 0z r 0 + z2 N 2 ) b r 2 k z k 2) 0 with a k 0, 0 k N. We may assume a N > 0. Let {r l e iθ l } be the zero set o p with 0 <θ 0 θ l or each l. We may assume r 0 =1in2)by using pr 0 z). Since the coeicients o p depend continuously on the zeros o p, we may assume that p has only one zero, say z 0 = e iθ 0,ontheray {re iθ 0 :0<r< } and that this zero is simple. Otherwise, a sequence o polynomials with this property can be chosen to converge to the desired polynomial with the appropriate inequalities holding. We will prove that b k > 0 or 0 k N 2. Now N 2 b k z k = = pz) 1 2cosθ 0 z + z 2 sink +1)θ 0 z k sin θ 0 a k z k = N 2 k ) sinl +1)θ 0 a k l z k. sin θ 0 l=0 4

5 Figure 1: This gives Noting that b k = k l=0 sinl +1)θ 0 sin θ 0 a k l. 3) z N p1/z) = a N k z k = z 2 2cosθ 0 z +1 ) N 2 b N 2 k z k it will suice to show that b k > 0 or 0 k [ N 2 ]. From 3) i 2 sinl +1)θ 0 > 0 or l =0, 1,,k, i.e., when 0 θ 0 <π/k +1)it ollows that b k > 0. Also, i all the zeros o p are in the closure o the let hal plane then the result is clear since p can be put into the orm: pz) = 1 2cosθ l r l z + z2 ) 1 + z rl 2 r m ). Thus, we may assume [ ] ) [ ]) N 2 N π/ +1 = π/ θ 0 <π/2. 4) 2 2 We will assume that there is a k with 5

6 b k 0 5) to reach a contradiction. Consider the unction deined by F z) =z k 1 pz) = a l z l k 1 l=0 having the same zeros as p. We observe rom the hypothesis that F has no zeros between the rays deined by {te iθ 0 :0<t< } and {te iθ 0 :0<t< }. Motivated by this, letγ=γ + Γ + be the curve shown in Figure 1, which is symmetric about the reals with Γ + = 5 l=1 Γ l where Γ 1 = {Re iθ :0 θ θ 0 }, Γ 2 = {te iθ 0 : 1+s t R}, Γ 3 = {z 0 +se iθ : θ 0 π θ θ 0 }, Γ 4 = {te iθ 0 : r t 1 s} and Γ 5 = {re iθ :0 θ θ 0 } and r, s, and1/r will be chosen suiciently small. We will show that under the assumption 5) the index o F with respect to Γ about any interior point is positive, implying the existence o a zero o F inside Γ. This would contradict that θ 0 is the smallest positive argument o the zeros o p in the upper hal plane. The unction F has the orm F z) =a 0 z k a k+1 + a k+2 z + a N z N k 1. 6) From symmetry we can examine arg F z), the change in the argument o F, as z traverses Γ +.Let l equal arg F z)asz traverses Γ l.forr suiciently large the term a N z N k 1 in 6) dominates so that 1 =N k 1)θ 0 +O 1 ). R For r suiciently small and positive the term a 0 z k 1 in 6) dominates so that 5 = k 1)θ 0 +Or) =k+1)θ 0 +Or), noting the clockwise transversal o Γ 5. Hence, ) = Nθ 0 + O + Or) 7) R or 1/R and r suiciently small. Let Γ 6 = {te iθ 0 : r t R}, notingthat Γ 2 Γ 3 Γ 4 approaches Γ 6 as s O. For notational convenience let ˆ 6 denote lim 3 as s 0. We observe that as t crosses 1 the change in argument o π accounts or the change in argument determined by traversing Γ 3 or s suiciently small, i.e., 3 = π + Os). 6

7 Since a bound or ˆ 6 can be determined by the maximum number o times the image o Γ 6 crosses the real axis we consider α 0 ), the number o real positive zeros o 0 deined by 0 t) = Im [ t k+1 F te iθ 0 ) ] 8) = k a l [sink +1 l)θ 0 ] t l + l=0 l=k+2 a l [sinl k 1)θ 0 ] t l. Descartes Rule o Signs says that α 0 ) is bounded above by γ 0 ), the number o sign changes in the coeicients o 0. Since, a l 0 or each l, γ 0 ) is determined by the number o sign changes in sin lθ 0 as l goes rom k 1toN k 1, i.e., over a range o N. It is clear that there exists an m such that mπ N <θ m +1)π 0. 9) N Since the number o sign changes as lθ 0 varies over an Nθ 0 range is determined by Nθ 0 /π, it ollows that γ 0 ) m +1. In counting α 0 )andγ 0 ) i a 0 sink +1)θ 0 = 0, then we actor out the leading power o t l in 0. So, we are only considering positive zeros o 0. In the case when α 0 ) is strictly less than m, saym 1 with 0 m 1 <m, it ollows that ˆ 6 <m 1 π + π mπ. Thus, by using 7), 8) and 9) we obtain 5 argf z) 2 > 2Nθ 0 m 1 +1)π) 0. l=1 This would imply that Γ encloses a zero o F, contradicting the construction o Γ. Now we consider the remaining cases, i.e., when γ 0 )=m or m +1 and α 0 ) m. I γ 0 ) α 0 ) = 1, then one o the positive zeros o 0 occurs with even multiplicity. This can be easily seen by observing that the sign o 0 t) or large t can be determined by multiplying the sign o the constant term o 0 by 1) γ0) and, alternatively, by multiplying the sign o the constant term o 0 by 1) α0) i all o the positive zeros o 0 have odd multiplicity. But, a positive zero o 0 with even multiplicity does not correspond to a proper crossing o the real axis by te iθ 0 ) and, hence, ˆ 6 mπ in this case, which is impossible. 7

8 The only remaining case is when α 0 )=γ 0 )=m or m +1. Weshall need an involved technical lemma, which we state and prove aslemmas 3.3 and 3.4 in the next section. To complete the proo we state Lemma 3.3 as Lemma 2.1 in the ollowing orm: Lemma 2.1 Let g and h be two real polynomials with hz) =a l z l + +a n z n where a l > 0 and the degree o g is less than l with the last coeicient o g being negative. I p is the polynomial deined by p = g + h and the number o sign charges in the coeicients o p equals the number o real positive zeros o p then hz j ) is positive at each o the zeros, z j, o p. We observe that the unction 0 deined in 8) satisies the hypothesis o k Lemma 2.1 with 0 = g + h where gt) = a l [sink +1 l)θ 0 ] t l and ht) = a l [sinl k 1)θ 0 ] t l. Since α 0 )=γ 0 )=m or m +1, it l=k+2 ollows, rom a partitioning argument, that the last nonzero term o gt) is negative and that the irst term o ht) is positive, or otherwise we would have γ 0 ) m 1, which would reduce to the previous case. Now, since 0 1) = 0, Lemma 2.1 gives that h1) > 0, which implies that g1) < 0. But, using 3) it ollows that g1) = b k sin θ 0, which contradicts the assumption in 5). Our main theorem ollows. 3 Lemmas on Zeros, Critical Points, and Sign Changes Let x) denote a polynomial with real coeicients. Write where l=1 x) =± p 0 x) p 1 x)+p 2 x) p 3 x)+ ) p i x) = c i0 x n i 1 + c i1 x n i c i,mi x n i 1+m i n 1 = 0,n i = n i 1 + m i +1, c ij 0,i 0,j 0. 8

9 Notation α) denotes the number o strictly positive real zeros, γ) denotes the number o sign changes, β) denotes the number o strictly positive critical points, 0 i c 00 β) = β)+1 i p 0 x) c 00 β) otherwise. Let 0 <x 1 ) <x 2 ) <x 3 ) < denote the distinct positive real zeros o. Lety i ) denote the critical point between x i ) and x i+1 ) in the case where α) =γ)). Lemma 3.1 i) α) β) γ). ii) I α) =γ), then α k) )=γ k) ) or all k such that k) 0. Proo. i) This is an immediate consequence o Rolle s Theorem and Descartes Rule o Signs. ii) It needs only to be shown that α )=γ ) ¾ Case 1: p 0 x) c 00.Thenα )= β) =β) =γ) =γ ). Case 2: p 0 x) =c 00. Then, α )= β) =β) 1=γ) 1=γ ) Remark 2 The remaining lemmas will each contain the hypothesis that α) = γ). It ollows rom Lemma 3.1i) that α) =β). Thereore, the positive real zeros and critical points o must strictly interlace. We emphasize again that all coeicients o each o the polynomials p i x) appearing below are nonnegative. It will be useul to the reader at this stage to note that the essentially dierent cases are, i) p 0 c 00 > 0 9

10 ii) p 0 c 00 > 0 iii) c 00 =0 It also may be beneicial to draw rough sketches or each o the cases. Lemma 3.2 Let x) =p 0 x) p 1 x)+hx), where hx) =p 2 x) p 3 x)+. Suppose that α) =γ), then hx i )) > 0 or all i. Proo. Let p 1 x) =c n x n + c n+1 x n c r x r,wherec r > 0andn =0is allowed. I n =0,thenp 0 x) 0.) Since the lemma is trivial i αh) = 0, let us assume that αh) > 0. Observe that x 1 h j) ) is deined or 0 j r +1, since h j) 0) = 0 or 0 j r. Now r) x) = r!c r +h r) has a critical point at x 1 h r+1) ). Since α r) )=γ r) ) Lemma 3.1ii)), it ollows that x 1 r) ) <x 1 h r+1) )) < x 1 h r) ), and r) x 1 h r+1) )) > 0. Furthermore, since r) <h r) it ollows that x 1 r) ) <x 1 h r+1) ) < x 2 r) ) < x 1 h r) ). Now r 1) has critical points at x 1 r) )andx 2 r) ) and since α r 1) )=γ r 1) ) it ollows that x 1 r) ) <x 1 r 1) ) <x 2 r) ). Thereore, r 1) x 2 r) )) > 0, and since r 1) <h r 1), it now ollows that x 2 r) ) <x 2 r 1) ) <x 1 h r 1) ). Thus, x 1 r 1) ) <x 2 r) ) <x 2 r 1) ) <x 1 h r 1) ). An easy induction then shows that x 1 j) ) <x 2 j) ) <x 1 h j) ),n j r 1. Now let p 0 x) =d 0 + d 1 x + + d n 1 x n 1. Since it suices to prove the lemma in the generic case, we assume that d i > 0; 0 i n 1. Since n 1) has critical points at x 1 n) )andx 2 n) )andα n 1) )=γ n 1) ), it ollows that n 1) x 1 n) )) < 0and n 1) x 2 n) )) > 0. Since n) x) < h n) x) or x>0, it now ollows that or all x x 1 n) ), n 1) x) = n 1) x 1 n) )) + x x 1 n) ) n) t)dt < x x 1 n) ) hn) t)dt < h n 1) x). By a previous argument, we have that x 1 n) ) <x 2 n) ) <x 1 h n) ) < x 1 h n 1) ). Then, since n 1) x) <h n 1) x) or x x 1 n) ) and since 10

11 n 1) x 2 n) )) > 0, we have that x 3 n 1) ) <x 1 h n 1) ). Thereore, x 1 n 1) ) <x 1 n) ) <x 2 n 1) ) <x 2 n) ) <x 3 n 1) ) <x 1 h n 1) ). Now suppose that or some j with 0 <j n 1, x 1 j) ) <x 2 j) ) <x 3 j) ) <x 1 h j) ) and j) x) <h j) x) or all x>y 1 j) ). Then, exactly the same reasoning as above applies and it ollows that x 1 j 1) ) <x 2 j 1) ) <x 3 j 1) ) <x 1 h j 1) ) j ) and j 1) x) <h j 1) x) or all x>y 1 j 1) ). Thereore, j ) is true or 0 j n 1. In particular, 0 )showsthatx 1 ) <x 1 h). Thus, p 0 x 1 ))+p 1 x 1 )) = hx 1 )) > 0. By Lemma 3.1 p 1 p 0 has only one zero. Thereore, p 1 p 0 )x) is monotone increasing or all x>x 1 p 1 p 0 ). Since p 1 p 0 )x 1 )) > 0, we have x 1 ) >x 1 p 1 p 0 ) and it ollows that hx j )) > 0 or all j. ¾ Lemma 3.3 Let =p 0 p 1 )+p 2 p 3 )+ +p 2k+2 p 2k+3 )+p 2k+4 p 2k+5 )+p 2k+6 ), where g 0 = p 0 p 1, g 1 = p 2 p 3,,g k+1 = p 2k+2 p 2k+3, h = p 2k+4 p 2k+5 + p 2k+6. Suppose that α) =γ), then hx i )) > 0 or all i. Proo. We may assume that γh) 0, or otherwise the result is trivial. Hence, γh j) ) 0 or 0 j degp 2k+3 ). Let n j =degg j )andm j = n j +1 or 0 j k +1. Letm 1 =0. We now regard k as ixed and proceed by induction on l, where 1 l k. From the previous lemma it ollows that x 1 m k) ) <x 2 m k) ) <x 3 m k) ) <x 1 h m k) ) and h m k) m k) )x) is positive and monotone increasing or x y 1 m k) ) We now make the ollowing induction hypothesis: For some l with 0 l k, x 1 m l) ) < <x 2k l)+3 m l) ) <x 1 h m l) ) 11

12 and h ml) ml) )x) is positive and monotone increasing or x y 2k l)+1 ml) ). Let p 2l+1 x) =a 0 x s + a 1 x s a r x r,wherer = n l = m l 1. Now r) x) = x ml) t)dt c, wherec is a nonnegative constant. Observe that 0 r) x) is positive or Thereore, or x satisying 10) x 2k l)+3 r) ) <x<x 2k l)+4 r) ). 10) 0 r) x) = x x 2k l)+3 m l) t)dt x x 2k l)+3 h m l) t)dt by the induction hypothesis since x 2k l)+3 r) ) >y 2k l)+1 m l) )) <h r) x 2k l)+3 r) )) + x x 2k l)+3 h m l) t)dt. Thereore, x 2k l)+4 r) ) <x 1 h r) ) and since α r) )=γ r) ), it ollows that x 1 r) ) <x 1 m l) ) <x 2 r) ) < <x 2k l)+3 m l) ) <x 2k l)+4 r) ) <x 1 h r) ) 11) and h r) r) )x) is positive and monotone increasing or x>y 2k l)+2 r) ). A repetition o the above argument shows that 11) remains true when r is replaced by j and m l by j + 1) or s j r and also shows that x 2k l)+5 ml 1) ) <x 1 h ml 1) ). This veriies the induction hypothesis or l 1). In particular, when l = 1 wehavex 1 ) <x 2 ) < <x 2k+5 ) < x 1 h) andh )x) > 0 or x y 2k+3 ). Since hx) > 0 or 0 <x<x 1 h) and since k+1 j=0 g jx) < 0 or x y 2k+3 ) andy 2k+3 ) <x 1 h), we inally have hx i )) > 0 or all i. ¾ The same argument establishes 12

13 Lemma 3.4 Let = p 0 +p 1 p 2 )+ +p 2k 1 p 2k )+p 2k+1 p 2k+2 )+ Let h = p 2k+1 p 2k+2 +. Suppose that α) =γ), then hx i )) > 0 or all i. Acknowledgements We wish to thank the mathematics department at the University o Caliornia at San Diego and the Math Sciences Research Institute at Berkeley or supporting the irst and last authors respectively as visiting scholars during some o the preparation time o this paper. We also wish to thank the number theorist R. Evans o the University o San Diego or many helpul discussions. Reerences [1] Beauzamy, B. Jensen s inequality or polynomials with concentration at low degrees. Numer. Math. No ) pp [2] Beauzamy, B. and Enlo, P. Estimations de produits de polynômes. J. Number Theory. No ) pp [3] Evans, R. and Greene, J. Polynomials with positive coeicients whose zeros have modulus one. preprint. [4] Evans, R. and Montgomery, P. Elementary Problem Proposal or American Mathematical Monthly. Amer.Math.Mon.toappear. [5] Polya, G. and Szego, G. Problems and theorems in analysis II. Springer- Verlag, [6] Rigler, A.K., Trimble, S.Y., and Varga, R.S. Sharp lower bounds or a generalized Jensen inequality. preprint. 13