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1 Introduction If p(x) is a polynomialof degree n 2 with n distinct realroots r <r < 1 2 ≥ 7 0 <r and critical points x <x < <x , let n 1 2 n 1 0 ··· ··· − 2 n x r k k a σk = − ,k =1,2,...,n 1. J rk+1 rk − − 5 ] (σ ,...,σ ) is called the ratio vector of p, and σ is called the kth ra- A 1 n 1 k − C tio. Ratio vectors were first discussed in [4] and in [1], where the inequality . h 1 k t <σ < ,k =1,2,...,n 1 was derived. For n=3 itwas shown a n k+1 k k+1 − m − in [1] that σ and σ satisfy the polynomial equation 3(1 σ )σ 1 = 0. In [ 1 2 − 1 2− 3 addition,necessaryandsufficientconditionsweregivenin[5]for(σ ,σ )tobea 1 2 v 8 ratiovector. Forn=4,apolynomial,Q,inthreevariableswasgivenin[5]with 8 4 9 thepropertythatQ(σ1,σ2,σ3)=0foranyratiovector(σ1,σ2,σ3). Itwasalso 0 5 shown that the ratios are monotonic–that is, σ <σ <σ for any ratio vector 1 2 3 0 / 1 1 1 2 h (σ ,σ ,σ ). For n=3, <σ < and <σ < , and thus it follows imme- 1 2 3 1 2 t 3 2 2 3 a m diately that σ < σ . The monotonicity of the ratios does not hold in general 1 2 : v for n 5(see [5]). Further results on ratio vectors for n = 4 were proved by i ≥ X r the author in [6]. In particular, necessary and sufficient conditions were given a for (σ ,σ ,σ ) to be a ratio vector. We now want to extend the notion of ratio 1 2 3 vectortopolynomiallikefunctionsoftheformp(x)=(x r1)m1 (x rN)mN, − ··· − where m ,...,m are given positive real numbers and r < r < < r . We 1 N 1 2 N ··· extend some of the results and simplify some of the proofs in [5] and in [6], and we prove some new results as well. In particular, we derive more general 1 bounds on the σ (Theorem 3). Even for N = 3 or N =4, the monotonicity of k the ratiosdoes nothold in generalfor allpositive realnumbers m ,...,m . We 1 N provide examples below and we also derive necessary and sufficient conditions on m ,m ,m which imply that σ < σ (Theorem 5).Finally, we prove some 1 2 3 1 2 general results for any N using Projective Elimination Theory(see Proposition 12). Proposition 12 can be used to provide necessary and sufficient conditions for(σ ,...,σ )tobearatiovector. Inparticular,weshow(Corollary13)that 1 N 1 − there is a polynomial Q=0 in N 1 variables with real coefficients and which 6 − does not depend on r ,...,r ,σ ,...,σ , such that Q(σ ,...,σ ) = 0 for 1 N 1 N 1 1 N 1 − − every ratio vector (σ ,...,σ ). 1 N 1 − 2 Main Results Throughout, p(x)=(x r )m1 (x r )mN, 1 N − ··· − N where m ,...,m are given positive real numbers with m = n and r < 1 N k 1 k=1 P r < <r . We need the following lemmas. 2 N ··· Lemma 1 p has exactly one root, x I =(r ,r ),k =1,2,...,N 1. ′ k k k k+1 ∈ − Proof. By Rolle’s Theorem, p has at least one root in I for each k = ′ k p N m ′ k 1,2,...,N 1. Now = , which has at most N 1 real roots − p x r − k=1 − k 1 P since is a Chebyshev system. x r (cid:26) − k(cid:27)k=1,...,N 2 Now we define the N 1 ratios − x r k k σ = − ,k =1,2,...,N 1. (1) k r r − k+1 k − (σ ,...,σ ) is called the ratio vector of p. We shall derive a system of 1 N 1 − nonlinear equations in the r and σ . By the product rule, p(x) = (x k k ′ { } { } − r1)m1−1 (x rN)mN−1 ··· − × N N mj (x ri) . Since p′(x)=n(x r1)m1−1 (x rN)mN−1 j=1 i=1,i=j − ! − ··· − × 6 N P1 Q − (x x ) as well, we have k − k=1 Q N 1 N N − n (x x )= m (x r ) . (2) k j i −  −  k=1 j=1 i=1,i=j Y X Y6   Let e e (r ,...,r ) denote the kth elementary symmetric function of the k k 1 n ≡ r , j = 1,2,...,n, starting with e (r ,...,r ) = 1,e (r ,...,r ) = r + +r , j 0 1 n 1 1 n 1 n ··· and so on. Let e (r ,...,r )=e (r ,...,r ,r ,...,r ), k,j 1 N k 1 j 1 j+1 N − that is, e (r ,...,r ) equals e (r ,...,r ) with r removed, j =1,...,n. Since k,j 1 N k 1 N j p(x+c)andp(x)havethesameratiovectorsforanyconstantc,wemayassume that r =0. 2 3 Equatingcoefficients in(2) usingthe elementarysymmetric functions yields N ne (x ,...,x )= m e (r ,0,r ,...,r ),k =1,2,...,N 1. k 1 N 1 j k,j 1 3 N − − j=1 X e (r ,r ,...,r ) if j =2 and k N 2 k,j 1 3 N 6 ≤ −  Since ek,j(r1,0,r3,...,rN)=  ek(r1,r3,...,rN) if j =2 we 0 if j =2 and k =N 1 have  6 − ne (x ,...,x )=m e (r ,r ,...,r )+ k 1 N 1 2 k 1 3 N − N m e (r ,r ,...,r ),k =1,...,N 2 (3) j k,j 1 3 N − j=1,j=2 X6 nx x =m r r r 1 N 1 2 1 3 N ··· − ··· Solving (1) for x yields k x =∆ σ +r ,k =1,2,...,N 1, (4) k k k k − 4 where∆ =r r . Substituting(4)into(3)givesthefollowingequivalent k k+1 k − system of equations involving the roots and the ratios. ne ((1 σ )r ,r σ ,∆ σ +r ,...,∆ σ +r ) = (5) k 1 1 3 2 3 3 3 N 1 N 1 N 1 − − − − N m e (r ,r ,...,r )+ m e (r ,r ,...,r ),k = 1,...,N 2 2 k 1 3 N j k,j 1 3 N − j=1,j=2 X6 n(1 σ )r (r σ )(∆ σ +r ) (∆ σ +r ) = m r r r 1 1 3 2 3 3 3 N 1 N 1 N 1 2 1 3 N − ··· − − − ··· Note that this system is homogeneous in the r since ∆ σ +r is a linear k k k k function of r for each k. This will be crucial later in our use of projective k elimination theory. 2.1 Bounds 1 k The inequality < σ < ,k = 1,2,...,n 1 was first derived in k n k+1 k+1 − − [4] and later in [1] for polynomials of degree n 2 with n distinct real roots. ≥ Critical in proving the inequality was the root–dragging theorem(see [2]). We now extend this inequality to the ratios defined in (1) for functions of the form p(x)=(x r1)m1 (x rN)mN. Firstwegeneralizetheroot–draggingtheorem. − ··· − The proof is very similar to the proof in [2] where m = = m = 1. For 1 N ··· completeness, we provide the details here. Lemma 2 Let x <x < <x be the N 1 critical points of p lying in 1 2 N 1 ··· − − Ik =(rk,rk+1),k =1,2,...,N −1. Let q(x)=(x−r1′)m1···(x−rN′ )mN, where r > r ,k = 1,2,...,N 1 and let x < x < < x be the N 1 critical k′ k − ′1 ′2 ··· ′N−1 − points of q lying in J = (r ,r ),k = 1,2,...,N 1. Then x > x ,k = k k′ k′+1 − ′k k 5 1,2,...,N 1. − N m Proof. Suppose that for some i,x < x . Now p(x ) = 0 k = 0 ′i i ′ i ⇒ x r k=1 i− k N m P k and q (x )=0 =0. r >r and x <x implies that ′ ′i ⇒ x r k′ k ′i i k=1 ′i− k′ P x r <x r ,k =1,2,...,N 1. (6) ′i− k′ i− k − m m k k Sincebothsidesof(6)havethesamesign, > ,k =1,2,...,N x r x r − ′i− k′ i− k N m N m k k 1,which contradictsthe factthat and areboth zero. x r x r k=1 i− k k=1 ′i− k′ P P Theorem 3 If σ ,...,σ are defined by (1), then 1 N 1 − m m + +m k 1 k <σ < ··· (7) k m + +m m + +m k N 1 k+1 ··· ··· Proof. To obtain an upper bound for σ we use Lemma 2. Arguing as in [1], k we can move the critical point x (r ,r ) as far to the right as possible k k k+1 ∈ by letting r ,...,r r and r ,...,r . Let s = m + +m ,t = 1 k 1 k k+2 N 1 k − → → ∞ ··· mk+2 + + mN, and let qb(x) = (x rk)s(x rk+1)mk+1(x b)t. Then ··· − − − q (x)=(x r )s b′ − k × (x rk+1)mk+1t(x b)t−1+mk+1(x rk+1)mk+1−1(x b)t + − − − − (cid:0) (cid:1) s(x rk)s−1(x rk+1)mk+1(x b)t = − − − (x rk+1)mk+1−1(x rk)s−1(x b)t−1 − − − × (t(x r )(x r )+m (x r )(x b)+s(x r )(x b)). k+1 k k+1 k k+1 − − − − − − x is the smallest root of the quadratic t(x r )(x r )+ k k+1 k − − m (x r )(x b)+s(x r )(x b)=(m +t+s)x2+ k+1 k k+1 k+1 − − − − 6 ( tr tr m r m b sr sb)x+tr r +sr b+m r b. k+1 k k+1 k k+1 k+1 k+1 k k+1 k+1 k − − − − − − As b ,x increases and approaches the root of ( m s)x+sr + k k+1 k+1 → ∞ − − sr +m r k+1 k+1 k m r . Thus x k+1 k k ↑ m +s ⇒ k+1 sr +m r k+1 k+1 k σ r /(r r )= k k k+1 k ↑ m +s − − (cid:18) k+1 (cid:19) sr +m r r (m +s) s m + +m k+1 k+1 k k k+1 1 k − = = ··· . Simi- (m +s)(r r ) m +s m + +m k+1 k+1 k k+1 1 k+1 − ··· larly, to obtain a lower bound for σ , move the critical point x (r ,r ) as k k k k+1 ∈ far to the left as possible by letting r ,...,r r and r ,...,r . k+2 N k+1 1 k 1 → − →−∞ Byconsideringqb(x)=(x rk)mk(x rk+1)s(x+b)t,wheres=mk+1+ +mN − − ··· m k and t=m + +m , one obtains σ . 1 k 1 k ··· − ↓ mk+ +mN ··· 2.2 N = 3 The following Theorem generalizes ([5],Theorem 1). Throughout, n=m +m +m 1 2 3 Theorem 4 Let p(x) = (x r1)m1(x r2)m2(x r3)m3. Then (σ1,σ2) is a − − − m m m m +m 1 1 2 1 2 ratio vector if and only if < σ < , < σ < , 1 2 n m +m m +m n 1 2 2 3 m 2 and σ = 2 n(1 σ ) 1 − Proof. To prove the necessity part, m m m m +m 1 1 2 1 2 <σ < , <σ < (8) 1 2 n m +m m +m n 1 2 2 3 7 follows from Theorem 3 with N =3. With N =3 (5) becomes n(r σ +(r r )σ +r ) = m (r +r )+m r +m r 2 1 3 2 2 2 1 2 3 2 3 3 2 − nr σ ((r r )σ +r ) = m (r r ) 2 1 3 2 2 2 1 2 3 − Since p(cx) and p(x) have the same ratios when c>0, in addition to r =0 we 1 may also assume that r =1. Let r =1 and r =r to obtain 2 2 3 (nσ m m )r+n(σ σ )+m = 0 2 1 2 1 2 2 − − − (nσ σ m )r+nσ (1 σ ) = 0 (9) 1 2 1 1 2 − − m +m 1 2 Notethatnσ m m =0sinceσ < by(8)andnσ σ m = 2 1 2 2 1 2 1 − − 6 n − 6 m m +m m 1 1 2 1 0 since σ σ < = by (8). Hence we can solve each 1 2 m +m n n 1 2 n(σ σ ) m σ 1 2 1 2 2 equation in (9) for r to obtain r = − − and r =nσ − . 1 nσ m m nσ σ m 2 1 2 1 2 1 − − − n(σ σ ) m σ 1 2 1 2 2 Equating these two expressions yields − − = nσ − 1 nσ m m nσ σ m ⇒ 2 1 2 1 2 1 − − − m 2 (nσ σ nσ +m )(m nσ ) = 0 σ = since m nσ = 1 2 2 2 1 1 2 1 1 − − ⇒ n(1 σ ) − 6 1 − m 1 0. To prove sufficiency, suppose that u is any real number with < u < n m m 1 2 . We want to show that (u,v) is a ratio vector, where v = . m +m n(1 u) 1 2 − (1 u)n m m 2 1 Let r = u − − . u < m (m +m )u > 0 and 1 1 2 m (m +m )u m +m ⇒ − 1 1 2 1 2 − m n m m m m 2 1 2 2 3 (1 u)n m > n m = m − − = > 0. Also, 2 2 2 − − m +m − m +m m +m 1 2 1 2 1 2 (1 u)n m 2 u − − > 1 u((1 u)n m ) > m (m +m )u 2 1 1 2 m (m +m )u ⇐⇒ − − − ⇐⇒ 1 1 2 − u((1 u)n m ) (m (m +m )u) = (1 u)(un m ) > 0, which holds 2 1 1 2 1 − − − − − − 8 m m since 1 < u and u < 1 < 1. Thus r > 1. Now let p(x) = xm1(x n m +m − 1 2 1)m2(x r)m3. Then the ratios of p,σ1 and σ2, must satisfy (9) by (5) with − (1 σ )n m 1 2 N =3. Thusr =σ − − . Foranyr,(9)withσ =uandσ =v 1 1 2 m (m +m )σ 1 1 2 1 − isequivalenttof(u)=0,wheref(u)= nu2+(rm +rm +n m )u rm . 1 2 2 1 − − − m m m m m m 1 3 1 1 2 3 Then f = m (r 1)− < 0 and f = > 0. (cid:16) n (cid:17) 1 − n (cid:18)m1+m2(cid:19) (m1+m2)2 (1 u)n m 2 If r = u − − , then f(u) = 0. Since lim f(u) = , f has m1−(m1+m2)u u→±∞ −∞ m m m m 1 1 1 1 exactly one solution < u < . Now < σ < and 1 n m +m n m +m 1 2 1 2 f(σ )=0 as well. Thus u=σ . That finishes the proof of Theorem 4. 1 1 2.2.1 Monotonicity 1 1 1 2 For m = m = m = 1, Theorem 3 yields < σ < and < σ < , and 1 2 3 1 2 3 2 2 3 thus it follows immediately that σ < σ . σ < σ does not hold in general 1 2 1 2 for all positive real numbers(or even positive integers) m ,m , and m . For 1 2 3 example, if m = 6, m = 1, and m = 2, then it is not hard to show that 1 2 3 σ < σ for all r < r < r . Also, if m = 4, m = 3, and m = 6, then 2 1 1 2 3 1 2 3 σ < σ for certain r < r < r , while σ < σ for other r < r < r . For 1 2 1 2 3 2 1 1 2 3 6 1 1 1 1 p(x) = x4(x 1)3 x+ √13 ,σ = σ = √13. One can easily 1 2 − 2 − 2 2 − 26 (cid:18) (cid:19) derive sufficient conditions on m ,m ,m which imply that σ < σ for all 1 2 3 1 2 m m r < r < r . For example, if m m < m2, then 1 < 2 , which 1 2 3 1 3 2 m +m m +m 1 2 2 3 implies that σ < σ by (8). Also, if m +m < 3m , then n < 4m , which 1 2 1 3 2 2 m 1 2 implies that σ = > σ since 4x(1 x) 1 for all real 2 1 n(1 σ ) 4(1 σ ) ≥ − ≤ 1 1 − − x. We shallnowderivenecessaryandsufficientconditionsonm ,m ,m which 1 2 3 imply that σ <σ . 1 2 9 Theorem 5 σ <σ for all r <r <r if and only if 1 2 1 2 3 (A) m2+m (m m )>0 and 2 1 2− 3 2m m 1 3 (B) m or 2 ≥ n n 2m m 1 3 (C) <m < 2 4 n Proof. Asnotedabove,wemayassumethatp(x)=xm1(x 1)m2(x r)m3,r> − − 1. A simple computation shows that 1 σ = (n m )r n m √A +1 1 3 2 2n − − − − (cid:16) (cid:17) 1 (n m )r n m +√A 3 2 2n − − − σ = 2 (cid:16) r 1 (cid:17) − where A=(m +m )2r2+2(m m m n)r+(m +m )2 1 2 2 3 1 1 3 − 1 (n m )r n m +√A 3 2 2n − − − Thus σ σ = 2− 1 (cid:16) r 1 (cid:17)− − 1 (n m )r n m √A 1= 3 2 2n − − − − − 1 (cid:16)(n m )r2 ( n+2m (cid:17)m )r 2m +√Ar 3 3 2 2 − − − − − − > 0 when r > 1 2 n(r 1) ⇐⇒ − √Ar >(n m )r2+( n+2m m )r+2m 3 3 2 2 − − − ⇐⇒ Ar2 > (n m )r2+( n+2m m )r+2m 2 3 3 2 2 − − − ⇐⇒ (cid:0) (cid:1) 4(r 1) m2+m m m m r2+ m m m m m2 r+m2 >0 − 2 1 2− 1 3 2 3− 1 2− 2 2 ⇐⇒ (cid:0)(cid:0) (cid:1) (cid:0) (cid:1) (cid:1) h(r)>0, where h(r)= m2+m (m m ) r2+m (m m m )r+m2. 2 1 2− 3 2 3− 2− 1 2 (cid:0) (cid:1) 10