1 Introduction If p(x) is a polynomialof degree n 2 with n distinct realroots 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 ,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 0 and 2 1 2− 3 2m m 1 3 (B) m or 2 ≥ n n 2m m 1 3 (C) − − 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

Ratio vectors of polynomial-like functions

Similar Ratio vectors of polynomial-like functions