On Hyperbolic Polynomials and Four-term Recurrence with Linear Coefficients
aa r X i v : . [ m a t h . C V ] A p r a, b , and c , we form the sequence of polynomials { P n ( z ) } ∞ n =0 satisfying the four-term recurrence P n ( z ) + azP n − ( z ) + bP n − ( z ) + czP n − ( z ) = 0 , n ∈ N , with the initial conditions P ( z ) = 1 and P − n ( z ) = 0. We find necessary and sufficientconditions on a, b , and c under which the zeros of P n ( z ) are real for all n , and provide anexplicit real interval on which ∞ [ n =0 Z ( P n ) is dense, where Z ( P n ) is the set of zeros of P n ( z ).1 INTRODUCTIONRecursively-defined sequences of polynomials have been studied extensively since the 18thcentury. Of particular interest are orthogonal polynomials, such as Hermite polynomials,Laguerre polynomials, and Jacobi polynomials, which have numerous applications in dif-ferential equations, mathematical and numerical analysis, and approximation theory (see[1, 10]). Common orthogonal polynomials such as the Chebychev polynomials can be definedby a three-term recurrence relation, which in modern times has been generalized significantly(see [2, 6, 8, 12, 14, 15]). Not much is known, however, about four-term recurrences. Forrecent work on the four-term reccurence, see [5, 7, 16]. Consider the sequence of polynomials { P n ( z ) } ∞ n =0 satisfying the four-term recurrence relation P n ( z ) + A ( z ) P n − ( z ) + B ( z ) P n − ( z ) + C ( z ) P n − ( z ) = 0 , where P ( z ) = 1, P − n ( z ) = 0, and A ( z ) , B ( z ), and C ( z ) are some linear complex-valuedpolynomials with real coefficients. With certain initial conditions, one may wish to find wherethe zeros of P n ( z ) lie on the complex plane for each n ∈ N . Let Z ( P n ) = { z ∈ C | P n ( z ) = 0 } .We say that a polynomial is hyperbolic if all of its zeros are real, and we are interested infinding necessary and sufficient conditions under which P n ( z ) is hyperbolic for all n ∈ N . In[16] the authors characterized the case where A ( z ) = a , B ( z ) = b , and C ( z ) = z . In thispaper, we characterize the case where A ( z ) = az , B ( z ) = b , and C ( z ) = cz for some nonzeroreal numbers a, b , and c . Next we present the main result of this paper. Theorem 1.
The zeros of the sequence { P n ( z ) } ∞ n =0 satisfying the recurrence relation P n ( z ) + azP n − ( z ) + bP n − ( z ) + czP n − ( z ) = 0 , a, b, c ∈ R \ { } , where P ( z ) = 1 and P − n ( z ) = 0 , are real if and only if b > and cab := α ≤ , in whichcase they lie on the interval ( − λ, λ ) , where λ := 4 (cid:16) α +1+ √ α − α +1 − α +1+ √ α − α +1 (cid:17) (cid:0) − α + 1 + √ α − α + 1 (cid:1) . Furthermore, the set of zeros of P n ( z ) is dense on ( − λ, λ ) as n → ∞ . This paper is organized as follows. In Section 2, we restate Theorem 1 using itsgenerating function and define a few functions which will be used to prove the sufficiencycondition of Theorem 1. In Section 3, we prove the monotonicity of a special function definedin Section 2, which will then be used to make heuristic arguments in Section 4. Finally, weprove the necessary condition in Section 5, and mention an open problem in Section 6.2 THE GENERATING FUNCTION AND SOME DEFINITIONSWe begin this section by restating Theorem 1 using the generating function for { P n ( z ) } ∞ n =0 . Theorem 2.
The zeros of the sequence { P n ( z ) } ∞ n =0 generated by ∞ X n =0 P n ( z ) t n = 11 + azt + bt + czt , a, b, c ∈ R \ { } , are real if and only if b > and cab := α ≤ , in which case they lie on the interval ( − λ, λ ) where λ := 4 (cid:16) α +1+ √ α − α +1 − α +1+ √ α − α +1 (cid:17) (cid:0) − α + 1 + √ α − α + 1 (cid:1) . Furthermore, the set of zeros of P n ( z ) is dense on ( − λ, λ ) as n → ∞ . If c = 0 and a and b are nonzero, the denominator becomes a quadratic, which wasstudied in [14]. If b = 0 and a and c are nonzero, the denominator becomes a three-termcubic, which was also studied in [14]. If a = 0 and b and c are nonzero, the denominatorbecomes a different type of three-term cubic, which was studied in [16]. We will show laterthat if b < P n ( z ) is not hyperbolic for all large n (c.f. Proposition 20). To simplifythe proof, we make the substitutions t → t √ b , a √ b z → z ′ , and cab → α , and the theorem canbe restated in the following equivalent form. Theorem 3.
The zeros of the sequence { P n ( z ) } ∞ n =0 generated by ∞ X n =0 P n ( z ) t n = 11 + zt + t + αzt , α ∈ R \ { } , are real if and only if α ≤ , in which case they lie on the interval ( − λ, λ ) where λ := 4 (cid:16) α +1+ √ α − α +1 − α +1+ √ α − α +1 (cid:17) (cid:0) − α + 1 + √ α − α + 1 (cid:1) . Furthermore, the set of zeros of P n ( z ) is dense on ( − λ, λ ) as n → ∞ . To prove this theorem, we must establish several lemmas. We begin by defining afew terms.
Definition 4.
For θ ∈ (0 , π ), define the functions t ( θ ) := τ ( θ ) e − iθ ,t ( θ ) := τ ( θ ) e iθ ,t ( θ ) := τ ( θ ) ζ ( θ ) ,z ( θ ) := − ατ ( θ ) ζ ( θ ) , ∆( θ ) := 16 α cos θ − α cos θ − α cos θ + α − α + 1 , where τ ( θ ) := s θ + ζ ( θ ) ζ ( θ ) and ζ ( θ ) := − α cos θ − α + 1 + p ∆( θ )4 α cos θ . The following three lemmas establish the fact that τ, ζ , and z are well-defined. Lemma 5.
For = α ≤ and θ ∈ (0 , π ) , ∆( θ ) > .Proof. First suppose α = 1 /
9. Then we have ∆( θ ) = (cos θ − θ + 4), which has zerosat θ = 0 and θ = π , and none in between. Since π/ ∈ (0 , π ) and ∆ (cid:0) π (cid:1) = >
0, and since∆( θ ) is continuous on R , we have that ∆( θ ) > α = 1 / α < / α = 0. By differentiating ∆( θ ) with respect to θ , we find that therelative extrema occur at θ = 0 , π/ , π , and cos − ( ± q α +14 α ). There are two conditions thatmust be met for the latter solutions to exist. First, α +14 α must be greater than or equal to zero,which is not the case for α ∈ ( − , ± q α +14 α to be in the domain of theinverse cosine function, we must have (cid:12)(cid:12)(cid:12) ± q α +14 α (cid:12)(cid:12)(cid:12) ≤
1, or equivalently, α ∈ ( −∞ , ∪ [1 / , ∞ ).Since we are assuming that α < /
9, we deduce that the inverse cosine solutions only existfor α ≤ −
1. Observe that ∆ (cid:16) cos − (cid:16) ± q α +14 α (cid:17)(cid:17) = − α , which is positive for all α ≤ − π ) = 9 α − α + 1 = (9 α − α − α < /
9. We also have ∆( π/
2) = α − α + 1 = ( α − , which is positive for all α = 1.Since all relative extrema on (0 , π ) give positive values of ∆( θ ), and ∆( θ ) is continuous on R , we have that ∆( θ ) > θ ∈ (0 , π ). Lemma 6.
For θ ∈ (0 , π ) , let ζ be as in Definition 4 with α cos θ = 0 . Then | ζ | > .Proof. When deriving the formula for ζ we used combinations of t , t , and t with Vieta’sformulas to obtain the function f ( ζ ) := 2 α cos θζ + (4 α cos θ + α − ζ + 2 α cos θ, (1)which has zeros ζ + := ζ = − α cos θ − α + 1 + p ∆( θ )4 α cos θ and ζ − := − α cos θ − α + 1 − p ∆( θ )4 α cos θ . We have f ( − f (1) = − ∆( θ ), which is negative for θ ∈ (0 , π ) be Lemma 5. Hence, by theIntermediate Value Theorem, exactly one of the zeros of f ( ζ ) lies outside the interval [ − , α <
0, then − α cos θ − α + 1 >
0, which implies | ζ + | > | ζ − | . If 0 < α ≤ /
9, then − α cos θ − α + 1 ≥ − α + 1 >
0, and applying the triangle inequality to the numerators,we have | − α cos θ − α + 1 − p ∆( θ ) | ≤ − α cos θ − α + 1 + p ∆( θ ) , which again implies that | ζ + | > | ζ − | . Hence we see that ζ + = ζ must lie outside the interval[ − ,
1] for all θ ∈ (0 , π ), and thus | ζ | > ∀ θ ∈ (0 , π ). Lemma 7.
For = α ≤ / and θ ∈ (0 , π ) , θ + ζζ > .Proof. Case 1: Suppose 0 < α ≤ /
9. Note that2 cos θ + ζζ = 4 α cos θ − α + 1 + p ∆( θ ) − α cos θ − α + 1 + p ∆( θ ) . We show that both the numerator, denoted N ( θ ), and denominator, denoted D ( θ ), arepositive for all 0 < α ≤ /
9. Note that the smallest D ( θ ) can be is when α = 1 / θ = 1. In that case, D ( θ ) = + p ∆( θ ) > θ − < π/ , π/ α = 1 / θ − −
1. Inthat case, N ( θ ) = + p ∆( θ ) > θ − > , π/ ∪ (2 π/ , π ),we get N ( θ ) = α (4 cos θ −
1) + 1 + p ∆( θ ) > α ≤
0. We show that 2 cos θζ + 1 > | ζ | > | θ | usinga similar argument as in Lemma 6. From (1) in Lemma 6, we have f ( − θ ) f (2 cos θ ) =4 cos θ (8 α cos θ + 2 α − < θ > α <
0. Hence, by the Intermediate ValueTheorem, exactly one of the zeros of f ( ζ ) lies outside the interval [ − θ, θ ], and since | ζ + | > | ζ − | by the proof of Lemma 6, we get that | ζ | > | θ | .Note that for each θ ∈ (0 , π ), t := t ( θ ) , t := t ( θ ), and t := t ( θ ) are the zeros ofthe denominator of the generating function, 1 + z ( θ ) t + t + αz ( θ ) t , since they satisfy therelations (known as Vieta’s formulas) t + t + t = − αz ,t t + t t + t t = 1 α ,t t t = − αz . z ( θ )In this section, we establish the fact that the function z ( θ ) is increasing on (0 , π ). Lemma 8.
The function z ( θ ) in Definition 4 is monotone increasing on (0 , π ) .Proof. Since t is a zero of 1 + z ( θ ) t + t + αz ( θ ) t , z ( θ ) = − t + 1 t + αt . First suppose that α <
0. Then we have z ( θ ) = − ( t + i )( t − i ) t ( p | α | t + 1)( p | α | t − . Taking the derivative by the product rule and then dividing by z ( θ ) implies that dzz = dt t + i + dt t − i − dt t − p | α | dt p | α | t + 1 − p | α | dt p | α | t − . Note that dt = d ( τ e − iθ ) = − iτ e − iθ dθ + e − iθ dτ = t (cid:18) dττ − idθ (cid:19) , so if we let h ( t ) := t t + i + t t − i − t p | α | p | α | t + 1 − t p | α | p | α | t − − , we have dzz = h ( t ) (cid:18) dττ − idθ (cid:19) , or equivalently, dzzdθ = h ( t ) (cid:18) dττ dθ − i (cid:19) . Since dzzdθ ∈ R for each θ ∈ (0 , π ), we have0 = Im dzzdθ = Im (cid:18) h ( t ) dττ dθ − h ( t ) i (cid:19) = − Re h ( t ) + Im h ( t ) dττ dθ , which implies Im h ( t ) dττ dθ = Re h ( t )and dzzdθ = Re h ( t ) dττ dθ + Im h ( t ) . (2)If we multiply both sides of Equation (2) by Im h ( t ), we haveIm h ( t ) dzzdθ = Im h ( t ) dττ dθ Re h ( t ) + Im h ( t ) Im h ( t )= Re h ( t ) Re h ( t ) + Im h ( t ) Im h ( t )= (Re h ( t )) + (Im h ( t )) = | h ( t ) | , which implies dzdθ = z | h ( t ) | Im h ( t ) . Since | h ( t ) | >
0, the sign of dzdθ depends on z and Im h ( t ). Note thatIm h ( t ) = Im t t + i + t t − i − t p | α | p | α | t + 1 − t p | α | p | α | t − − ! . (3)Multiplying the top and bottom of each fraction by the conjugate of their respective denom-inator makes Equation (3) equivalent toIm | t | − t i | t + i | + | t | + t i | t − i | − | α || t | + t p | α || p | α | t + 1 | − | α || t | − t p | α || p | α | t − | ! . Since | t | ∈ R and | α | ∈ R , Im h ( t ) is equivalent toIm − t i | t + i | + t i | t − i | − t p | α || p | α | t + 1 | + t p | α || p | α | t − | ! . We next substitute the expression in Definition 4 for t and findIm h ( t ) = − τ cos θτ − τ sin θ + 1 + τ cos θτ + 2 τ sin θ + 1+ p | α | τ sin θ | α | τ + 2 p | α | τ cos θ + 1 − p | α | τ sin θ | α | τ − p | α | τ cos θ + 1 . When we combine, the common denominator will be a product of squares of modulus, andwill thus be a positive value. Hence we can consider just the numerator, which becomes − α τ sin θ cos θ − | α | τ sin θ cos θ + 16 | α | τ sin θ cos θ − τ sin θ cos θ − | α | τ sin θ cos θ + 16 | α | τ sin θ cos θ − | α | τ sin θ cos θ. Applying the trigonometric identities 2 sin θ cos θ = sin 2 θ , and sin θ + cos θ = 1, we obtainIm h ( t ) = − τ (sin 2 θ + | α | sin 2 θ + | α | τ sin 2 θ + α τ sin 2 θ ) . Since − τ <
0, we analyzesin 2 θ + | α | sin 2 θ + | α | τ sin 2 θ + α τ sin 2 θ = sin 2 θ (1 + | α | + | α | τ + α τ )on the intervals (0 , π/
2) and ( π/ , π ).On (0 , π/ z < θ >
0, and since 1 + | α | + | α | τ + α τ >
0, we have thatIm h ( t ) <
0, which implies that dzdθ > , π/ π/ , π ), z > θ <
0, so we get that Im h ( t ) >
0, which implies that dzdθ > π/ , π ). Thus dzdθ > , π ) for α < < α ≤ /
9. Then z = − ( t + i )( t − i ) t ( √ αt + i )( √ αt − i ) , and taking the derivative by the product rule and dividing by z implies that dzz = dt t + i + dt t − i − dt t − √ αdt √ αt + i − √ αdt √ αt − i . Using a similar process as above, we obtain dzdθ = z | h ( t ) | Im h ( t )where h ( t ) := t t + i + t t − i − t √ α √ αt + i − t √ α √ αt − i − . Since | h ( t ) | >
0, the sign of dzdθ again depends on z and Im h ( t ). We evaluate Im h ( t )in the same way. We haveIm h ( t ) = Im (cid:18) t t + i + t t − i − t √ α √ αt + i − t √ α √ αt − i − (cid:19) . (4)Multiplying the top and bottom of each fraction by the conjugate of their respective denom-inator makes Equation (4) equivalent toIm (cid:18) | t | − t i | t + i | + | t | + t i | t − i | + t √ αi − α | t | |√ αt + i | + − t √ αi − α | t | |√ αt − i | (cid:19) . Since | t | ∈ R and √ α ∈ R , Im h ( t ) is equivalent toIm (cid:18) − t i | t + i | + t i | t − i | + t √ αi |√ αt + i | − t √ αi |√ αt − i | (cid:19) . We next substitute the expression in Definition 4 for t and findIm h ( t ) = − τ cos θτ − τ sin θ + 1 + τ cos θτ + 2 τ sin θ + 1+ √ ατ cos θατ − √ ατ sin θ + 1 − √ α cos θατ + 2 √ ατ sin θ + 1 . As before, when we combine, the common denominator will be a product of squares ofmodulus, and will thus be a positive value. Hence we can can consider just the numerator,which becomes − α τ sin θ cos θ − τ sin θ cos θ + 4 ατ sin θ cos θ + 4 ατ sin θ cos θ. Applying the trigonometric identity 2 sin θ cos θ = sin 2 θ , we obtainIm h ( t ) = − τ (sin 2 θ − α sin 2 θ − ατ sin 2 θ + α τ sin 2 θ ) . Since − τ <
0, we analyzesin 2 θ − α sin 2 θ − ατ sin 2 θ + α τ sin 2 θ = sin 2 θ (1 − α − ατ + α τ )on (0 , π/
2) and ( π/ , π ).Suppose first that θ ∈ (0 , π/ z < θ >
0. Note that1 − α − ατ + α τ = (1 − α )(1 − ατ ) . Since 1 − α > < α ≤ /
9, we must show that 1 − ατ >
0. Note that 1 − ατ issmallest when α = 1 /
9. Note that τ = (cid:18) θζ + 1 (cid:19) , and since | τ | > < θζ + 1 < ⇒ τ < . Hence 1 − ατ >
0, and thus dzdθ > , π/ θ ∈ ( π/ , π ). Then z > θ <
0. With the same argument as above,we get that 1 − ατ >
0, implying that dzdθ > π/ , π ).Hence dzdθ > , π ), and thus z is increasing on (0 , π ). Corollary 9.
For θ ∈ (0 , π ) , z ∈ ( − λ, λ ) where λ := 4 (cid:16) α +1+ √ α − α +1 − α +1+ √ α − α +1 (cid:17) (cid:0) − α + 1 + √ α − α + 1 (cid:1) . Proof.
By Lemma 8, z is increasing on (0 , π ), so it suffices to check lim θ → + z ( θ ) and lim θ → π − z ( θ ).It is easy to see that lim θ → + z ( θ ) = − λ and lim θ → π − z ( θ ) = λ .4 HEURISTIC ARGUMENTS AND THE VERTICAL ASYMPTOTEIn this section, we prove the sufficiency part of Theorem 1, being careful to considerall the cases of the discontinuity of the ζ function at θ = π/
2. In the following lemma, wedefine a function which will be used to prove the hyperbolicy of P n ( z ). Lemma 10.
For any θ ∈ (0 , π ) , P n ( z ( θ )) = 0 if and only if g n ( θ ) = 0 , where g n ( θ ) := ( ζ − cos θ ) sin( n + 1) θ sin θ − cos( n + 1) θ + 1 ζ n +1 . (5) Proof.
This follows directly from (2.4), (2.5), and (2.6) in [16].By Lemma 10, θ is a zero of P n ( z ( θ )) if and only if g n ( θ ) = 0. Note that whencos( n + 1) θ = ±
1, sin( n + 1) θ = 0. By Lemma 6, we know that | ζ | >
1, so we see that whencos( n + 1) θ = 1 , g n <
0, and when cos( n + 1) θ = − , g n >
0. Hence, by the IntermediateValue Theorem, the function g n has at least one zero on each subinterval whose endpoints arethe solutions of cos( n + 1) θ = ±
1. However, g n has a vertical asymptote in the subintervalcontaining θ = π/ ζ has a vertical asymptote at θ = π/ , π ), cos( n + 1) θ = ±
1, which implies θ = kπn +1 , k ∈ { , , , . . . , n } , whichmeans there are n − , π ), excluding the endpoints at 0 and π , and asstated above each subinterval besides the one containing the vertical asymptote has at leastone zero of g n . So we have found n − g n on (0 , π ). An inductive argument canshow that the degree of P n ( z ) is n , so we must find the missing zeros. Since there is a verticalasymptote at θ = π/
2, it is possible that the subinterval containing π/ α < < α ≤ / , π ) is partitioned into n + 1 subintervals with endpoints0 , πn + 1 , πn + 1 , πn + 1 , . . . , nπn + 1 , π. Lemma 11.
For α < , the function g n defined in Equation (5) has at least one zero oneach of the intervals (cid:18) , πn + 1 (cid:19) and (cid:18) nπn + 1 , π (cid:19) . Proof.
Recall that g n = ( ζ − cos θ ) sin( n + 1) θ sin θ − cos( n + 1) θ + 1 ζ n +1 , and the term − cos( n + 1) θ determines the sign of g n ( θ ) at each interval endpoint.For θ = πn + 1 , we have − cos( n + 1) θ = − cos π = − ( −
1) = 1 , which implies that g n > πn + 1 ∀ n ∈ N .Note that lim θ → + ( ζ − cos θ ) < , and lim θ → + sin( n + 1) θ sin θ = n + 1 . Furthermore, since cos(0) = 1 and (cid:12)(cid:12)(cid:12)(cid:12) ζ n +1 (cid:12)(cid:12)(cid:12)(cid:12) <
1, we havelim θ → + g n < ∀ n ∈ N . Since g n is continuous on (cid:18) , πn + 1 (cid:19) and it changes sign from 0 to πn + 1 , we have by theIntermediate Value Theorem that g n has at least one zero on (cid:18) , πn + 1 (cid:19) .For the interval (cid:18) nπn + 1 , π (cid:19) , we must consider the two cases when n is even and when n isodd.First, suppose n is even. So n = 2 ℓ for some ℓ ∈ N .For θ = nπn + 1 , we have − cos( n + 1) θ = − cos(2 ℓπ ) = − , which implies g n < nπn + 1 ∀ n ∈ N .Note that lim θ → π − ( ζ − cos θ ) > , and lim θ → π − sin( n + 1) θ sin θ = lim θ → π − sin(2 ℓθ + θ )sin θ = lim θ → π − (2 ℓ + 1) cos(2 ℓθ + θ )cos θ = 2 ℓ + 1 . Furthermore, since − cos(2 ℓ + 1) π = − cos(2 ℓπ + π ) = − ( −
1) = 1 , and (cid:12)(cid:12)(cid:12)(cid:12) ζ n +1 (cid:12)(cid:12)(cid:12)(cid:12) < , we ascertain that lim θ → π − g n > ∀ n ∈ N . Since g n is continuous on (cid:18) nπn + 1 , π (cid:19) and it changes sign from nπn + 1 to π , the IntermediateValue Theorem implies that g n has at least one zero on (cid:18) nπn + 1 , π (cid:19) if n is even.0Now suppose that n is odd and write n = 2 ℓ + 1 for some ℓ ∈ N .For θ = nπn + 1 , we have − cos( n + 1) θ = − cos(2 ℓπ + π ) = − ( −
1) = 1 , which implies that g n > nπn + 1 = (2 ℓ + 1) π ℓ + 2 ∀ ℓ ∈ N .As before, lim θ → π − ( ζ − cos θ ) > , but lim θ → π − sin( n + 1) θ sin θ = lim θ → π − sin(2 ℓθ + 2 θ )sin θ = lim θ → π − (2 ℓ + 2) cos(2 ℓθ + 2 θ )cos θ = − (2 ℓ + 2) . Furthermore, since − cos(2 ℓ + 2) π = − , and (cid:12)(cid:12)(cid:12)(cid:12) ζ n +1 (cid:12)(cid:12)(cid:12)(cid:12) < , we have lim θ → π − g n < ∀ ℓ ∈ N . Since g n is continuous on (cid:18) nπn + 1 , π (cid:19) and it changes sign from nπn + 1 to π , the IntermediateValue Theorem implies that g n has at least one zero on (cid:18) nπn + 1 , π (cid:19) for n odd.In the next lemma, we will use the fact that for 0 < α ≤ / θ → π − ( ζ − cos θ ) = + ∞ , (6)lim θ → π ( ζ − cos θ ) = −∞ , (7)and the terms − cos( n + 1) θ and 1 ζ n +1 are finite at θ = π/
2. Note that if n is even, then thevertical asymptote at θ = π (cid:18) nπ n + 2 , ( n + 2) π n + 2 (cid:19) . Lemma 12.
For < α ≤ / and n even, g n has at least one zero on the intervals (cid:18) nπ n + 2 , π (cid:19) and (cid:18) π , ( n + 2) π n + 2 (cid:19) . Proof.
We must consider the case where n is a multiple of 4 and when n is not a multiple of4 separately.First, suppose n is a multiple of 4. Then n = 4 ℓ for some ℓ ∈ N . Solim θ → π ± sin( n + 1) θ sin θ = sin(2 ℓθ + π/
2) = 1 . (8)Hence Equations (6) and (7) together with Equation (8) imply thatlim θ → π − g n ( θ ) = + ∞ (9)and lim θ → π g n ( θ ) = −∞ . (10)Next we determine the sign of g n at the endpoints nπ n + 2 and ( n + 2) π n + 2 by finding the signof − cos( n + 1) θ . Observe that for n = 4 ℓ and θ = nπ n + 2 , − cos( n + 1) θ = − cos(2 ℓθ ) = − . (11)For n = 4 ℓ and θ = ( n + 2) π n + 2 , − cos( n + 1) θ = − cos(2 ℓπ + π ) = 1 . (12)Equations (9) and (11) imply that g n changes sign on the interval (cid:18) nπ n + 2 , π (cid:19) , and Equa-tions (10) and (12) imply that g n changes sign on the interval (cid:18) π , ( n + 2) π n + 2 (cid:19) . Since g n iscontinuous on those intervals, we have by the Intermediate Value Theorem that g n has atleast one zero on each of those intervals for n = 4 ℓ .Now suppose n is not a multiple of 4. So n = 4 ℓ − ℓ ∈ N . Solim θ → π ± sin( n + 1) θ sin θ = sin(2 ℓπ − π/
2) = − . (13)Hence Equations (6) and (7) together with Equation (13) imply thatlim θ → π − g n ( θ ) = −∞ (14)and lim θ → π g n ( θ ) = + ∞ . (15)Again, we determine the sign of g n at the endpoints by finding the sign of − cos( n + 1) θ .Observe that for n = 4 ℓ − θ = nπ n + 2 , − cos( n + 1) θ ) = − cos(2 ℓπ − π ) = 1 . (16)For n = 4 ℓ − θ = ( n + 2) π )2 n + 2 , − cos( n + 1) θ = − cos(2 ℓπ ) = − . (17)2Equations (14) and (16) imply that g n changes sign on the interval (cid:18) nπ n + 2 , π (cid:19) , and Equa-tions (15) and (17) imply that g n changes sign on the interval (cid:18) π , ( n + 2) π n + 2 (cid:19) . Since g n iscontinuous on those intervals, we again have by the Intermediate Value Theorem that g n hasat least one zero on each of those intervals for n = 4 ℓ − n is odd, then the possible discontinuity at θ = π Lemma 13.
For < α ≤ / and n odd, g n has at least one zero on the intervals (cid:18) ( n − π n + 2 , π (cid:19) and (cid:18) π , ( n + 3) π n + 2 (cid:19) . Proof.
Let n be an odd natural number. We must consider the case where n = 4 ℓ − n = 4 ℓ + 1 , ℓ ∈ N , separately.First, suppose n = 4 ℓ −
1. Then since 0 < α ≤ / θ → π ( ζ − cos θ ) sin( n + 1) θ sin θ = 1 − αα ( − ℓ ) . (18)Since | cos( n + 1) θ | ≤ (cid:12)(cid:12)(cid:12)(cid:12) ζ (cid:12)(cid:12)(cid:12)(cid:12) <
1, Equation (18) implies thatlim θ → π g n ( θ ) < . (19)Next we determine the sign of g n at the endpoints ( n − π n + 2 and ( n + 3) π n + 2 by finding the signof − cos( n + 1) θ . Observe that for n = 4 ℓ − θ = ( n − π n + 2 , − cos( n + 1) θ = − cos(2 ℓπ − π ) = 1 . (20)For n = 4 ℓ − θ = ( n + 3) π n + 2 , − cos( n + 1) θ = − cos(2 ℓπ + π ) = 1 . (21)Equations (19) and (20) imply that g n changes sign on the interval (cid:18) ( n − π n + 2 , π (cid:19) , andEquations (19) and (21) imply that g n changes sign on the interval (cid:18) π , ( n + 3) π n + 2 (cid:19) . Since g n is continuous on those intervals, we have by the Intermediate Value Theorem that g n has atleast one zero on each of those intervals for n = 4 ℓ − n = 4 ℓ + 1 for some ℓ ∈ N . Then since 0 < α ≤ / θ → π ( ζ − cos θ ) sin( n + 1) θ sin θ = 1 − αα (2 ℓ + 1) . (22)Since | cos( n + 1) θ | ≤ (cid:12)(cid:12)(cid:12)(cid:12) ζ (cid:12)(cid:12)(cid:12)(cid:12) <
1, Equation (22) implies thatlim θ → π g n ( θ ) > . (23)3We again determine the sign of g n at the endpoints ( n − π n + 2 and ( n + 3) π n + 2 by finding thesign of − cos( n + 1) θ . Observe that for n = 4 ℓ + 1 and θ = ( n − π n + 2 , − cos( n + 1) θ = − cos(2 ℓπ ) = − . (24)For n = 4 ℓ + 1 and θ = ( n + 3) π n + 2 , − cos( n + 1) θ = − cos(2 ℓπ + 2 π ) = − . (25)Equations (23) and (24) imply that g n changes sign on the interval (cid:18) ( n − π n + 2 , π (cid:19) , andEquations (23) and (25) imply that g n changes sign on the interval (cid:18) π , ( n + 3) π n + 2 (cid:19) . Since g n is continuous on those intervals, we again have by the Intermediate Value Theorem that g n has at least one zero on each of those intervals for n = 4 ℓ + 1.By Lemmas 11, 12, and 13, we see that g n has a total of n zeros on the interval (0 , π )when n is even, and a total of n − , π ) when n is odd. Since thedegree of P n ( z ) is n for all n ∈ N , the Fundamental Theorem of Algebra tells us that theremust be exactly one more zero when n is odd. Since complex zeros always occur in conjugatepairs, the single missing zero must be real. Lemma 14. If n is odd, then z = 0 is a zero of P n ( z ) .Proof. Recall that the generating relation for { P n ( z ) } ∞ n =0 is ∞ X n =0 P n ( z ) t n = 11 + zt + t + αzt . Then the generating relation for z = 0 becomes ∞ X n =0 P n (0) t n = 11 + t . Expanding both sides gives P (0) + P (0) t + P (0) t + P (0) t + P (0) t + · · · = 1 − t + t − t + t − . . . Equating coefficients yields P n (0) = ( n is odd ± n is even . Hence P n (0) = 0 for all odd n .By Lemma 10, we conclude that since g n has n zeros on (0 , π ) for all n ∈ N , P n ( z )has n zeros on the interval ( − λ, λ ) via the monotone map z ( θ ) : (0 , π ) → ( − λ, λ ). Since thedegree of P n ( z ) is n for each n , it follows by the Fundamental Theorem of Algebra that allof the zeros of P n ( z ) are real and lie on the interval ( − λ, λ ).4 Lemma 15.
The set of zeros of P n ( z ) is dense on the open interval ( − λ, λ ) as n → ∞ .Proof. Since the solutions of cos( n + 1) θ = ± , π ) as n → ∞ , it is clear that ∞ [ n =0 Z ( g n ) is dense on (0 , π ). Since z : (0 , π ) → ( − λ, λ ) is a continuous bijective function, itfollows that ∞ [ n =0 Z ( P n ) is dense on ( − λ, λ ).This concludes the sufficiency part of the proof, where α ≤ / P n ( z ) ishyperbolic for all n . In the next section, we focus on the necessary part of the proof, where α > / b < P n ( z ) is not hyperbolic for all large n .5 NECESSARY CONDITION FOR THE REALITY OF ZEROSTo prove that P n ( z ) is not hyperbolic for large n when b < α > /
9, we will employ theImplicit Function Theorem and a theorem by Sokal in [13], but we must first define a fewterms to make sense of the statement of Sokal’s theorem.
Definition 16.
Let f n ( z ) = m X k =1 α k ( z ) β k ( z ) n , n ∈ N , be a function where α k ( z ) and β k ( z ) are analytic in a domain D . We say that an index k isdominant at z if | β k ( z ) | ≥ | β l ( z ) | for all 1 ≤ l ≤ m . Definition 17.
Let D k := { z ∈ D | k is dominant at z } . We say that lim Z ( f n ) is the set of all z ∈ D such that every neighborhood U of z has anonempty intersection with all but finitely many of the sets Z ( f n ), and lim Z ( f n ) is the setof all z ∈ D such that every neighborhood U of z has a nonempty intersection with infinitelymany of the sets Z ( f n ).In other words, Definition 17 says that z ∈ lim Z ( f n ) if for any neighborhood U of z , f n has a zero in U for all n large enough, and z ∈ lim Z ( f n ) if for any neighborhood U of z , infinitely many f n have a zero in U . Theorem 18. (Sokal [13]) Let D be a domain in C and let α , . . . , α m , β , . . . , β m ( m ≥ be analytic functions on D , none of which is identically zero. Let us further assume a “nodegenerate dominance” condition: there do not exist indices k = k ′ such that β k ≡ ωβ k ′ forsome constant ω with | ω | = 1 and such that D k (= D k ′ ) has nonempty interior. For eachinteger n ≥ , define f n by f n ( z ) = m X k =1 α k ( z ) β k ( z ) n . (26) Then lim Z ( f n ) = lim Z ( f n ) , and a point z ∗ lies in this set if and only if either1. there is a unique dominant index k at z ∗ , and α k ( z ∗ ) = 0 , or2. there are two or more dominant indices at z ∗ . We will also apply the Implicit Function Theorem, which we state as the followingtheorem.5
Theorem 19. (Implicit Function Theorem [9]) Let f j ( w, z ) , j = 1 , . . . , m , be analytic func-tions of ( w, z ) = ( w , . . . , w m , z , . . . , z n ) in a neighborhood of a point ( w ∗ , z ∗ ) in C m × C n ,and assume that f j ( w ∗ , z ∗ ) = 0 , j = 1 , . . . , m , and that det (cid:18) ∂f j ∂w k (cid:19) mj,k =1 = 0 at ( w ∗ , z ∗ ) . Then the equations f j ( w, z ) = 0 , j = 1 , . . . , m , have a uniquely determined analytic solution w ( z ) in a neighborhood of z ∗ such that w ( z ∗ ) = w ∗ . We will later find a z ∗ ∈ C \ R such that the zeros t ∗ , t ∗ , and t ∗ of 1 + z ∗ t + t + αz ∗ t satisfy | t ∗ | = | t ∗ | ≤ | t ∗ | , where t ∗ , t ∗ , and t ∗ are distinct and nonzero. Assuming that we can find such a z ∗ , we nowclaim that z ∗ ∈ lim Z ( P n ), which implies that P n ( z ) is not hyperbolic for all large n .First of all, Theorem 19 implies that for a fixed z ∗ ∈ C \ R , there exist domains D , D , and D containing z ∗ and analytic functions t ( z ) , t ( z ), and t ( z ) such that D ( t ( z ) , z ) = 0 ∀ z ∈ D ,D ( t ( z ) , z ) = 0 ∀ z ∈ D ,D ( t ( z ) , z ) = 0 ∀ z ∈ D , where D ( t, z ) is the bivariate denominator of the generating function of P n ( z ). Let D = \ k =1 D k . We supress the parameter z and note that by continuity, t = t = t = 0 ∀ z ∈ D .From partial fractions (see (2.4), (2.5) in [16]) we write P n ( z ) as − t − t )( t − t ) t n +11 − t − t )( t − t ) t n +12 − t − t )( t − t ) t n +13 . Thus P n ( z ) is of the form (26) where α ( z ) = − t − t )( t − t ) t , β ( z ) = 1 t ,α ( z ) = − t − t )( t − t ) t , β ( z ) = 1 t ,α ( z ) = − t − t )( t − t ) t , β ( z ) = 1 t , all of which are analytic on D since t = t = t = 0 ∀ z ∈ D . For the “no degeneratedominance” condition of Theorem 18, for a fixed ω on the unit circle, we will show that theset of z ∈ D such that β ( z ) = ωβ ( z ) has empty interior. Let ω ∈ C where | ω | = 1 suchthat β ( z ) ≡ ωβ ( z ) ∀ z ∈ D . Then ω = e iθ for some fixed θ and t = ωt , which means t = τ e − iθ , t = τ e iθ , and t = τ ζ for some τ, ζ ∈ C \ { } . By Vieta’s formulas, we know t + t + t = − αz , (27) t t + t t + t t = 1 α , and (28) t t t = − αz . (29)6Substituting t = τ e − iθ , t = τ e iθ , and t = τ ζ into Equations (27) and (29), we obtain z = − ατ ζ (30)and τ (2 cos θ + ζ ) = τ ζ, which, since τ = 0, is equivalent to τ = 2 cos θ + ζζ . (31)Substituting t = τ e − iθ , t = τ e iθ , and t = τ ζ into Equation (28), we have τ (1 + 2 ζ cos θ ) = 1 α . (32)Substituting the expression for τ from Equation (31) into Equation (32) gives2 cos θ + ζζ (1 + 2 ζ cos θ ) = 1 α , or equivalently, 2 α cos θζ + (4 α cos θ + α − ζ + 2 α cos θ = 0 . (33)Since θ is fixed, the sets of τ and ζ satisfying Equations (32) and (33) are finite, and hencethe set of z satisfying Equation (30) is also finite, and a finite set has empty interior. If z ∗ ∈ C \ R such that the zeros in t of 1 + z ∗ t + t + αz ∗ t are distinct and nonzero on D ,then by (2.4) and (2.5) in [16] and Theorem 18, z ∗ ∈ lim Z ( P n ) when the two smallest (inmodulus) zeros have the same modulus. This is because | t ( z ∗ ) | = | t ( z ∗ ) | , and the secondcondition of Theorem 18 is satisfied at z ∗ since | β ( z ∗ ) | = | β ( z ∗ ) | ≥ | β ( z ∗ ) | is equivalent to | t ( z ∗ ) | = | t ( z ∗ ) | ≤ | t ( z ∗ ) | . The following proposition proves that P n ( z ) is not hyperbolic for all large n if b < Proposition 20.
For the sequence { P n ( z ) } ∞ n =0 generated by ∞ X n =0 P n ( z ) t n = 11 + zt − t + czt , c ∈ R , the polynomials P n ( z ) are not hyperbolic for all large n ∈ N .Proof. Case 1: c = 0. Then the generating relation reduces to ∞ X n =0 P n ( z ) t n = 11 + zt − t . By [15] the zeros of this sequence lie on the curve defined byIm z − ≤ Re z − ≤ . z ∈ R and − ≤ z ≤
0, which impliesthat the zeros lie on the imaginary interval ( − i, i ).Case 2: c = 0. Define D ( t ) := 1 + zt − t + czt with zeros t , t , and t , and consider the reciprocal polynomial D ∗ ( t ) := t + zt − t + cz. Let the zeros of D ∗ ( t ) be t ∗ , t ∗ , and t ∗ . We will show that there exists a z ∗ ∈ C \ R suchthat | t ∗ | = | t ∗ | ≥ | t ∗ | .Let ǫ ∈ R \ { } and let z ∗ = iǫ . Note that if z = 0 then the zeros of D ∗ ( t ) are 0 ,
1, and − D ∗ ( t ), it is clear that as ǫ → t ∗ → , t ∗ → − , and t ∗ → z ∗ , z ∗ = − z ∗ . Since t ∗ is a zero of D ∗ ( t ), t ∗ + zt ∗ − t ∗ + cz = 0 . Observe (cid:0) − t ∗ (cid:1) + z ∗ (cid:0) − t ∗ (cid:1) − ( − t ∗ ) + cz ∗ = − t ∗ − z ∗ t ∗ + t ∗ − cz ∗ = − t ∗ − z ∗ t ∗ + t − cz ∗ = 0 . Hence − t ∗ := t ∗ is also a zero of D ∗ ( t ), and thus | t ∗ | = | t ∗ | ≥ | t ∗ | for sufficiently small ǫ .Since the zeros of D ∗ ( t ) are the reciprocals of the zeros of D ( t ), we have | t | = | t | ≤ | t | . Proposition 21.
The zeros of the sequence { P n ( z ) } ∞ n =0 generated by ∞ X n =0 P n ( z ) t n = 11 + zt + t + αzt , α ∈ R , are not all real if α > / for all large n .Proof. Case 1: Suppose α >
1. Consider z = i . Then the denominator of the generatingrelation becomes D ( t ) := 1 + it + t + αit . We now show that if t , t , and t are the zeros of D ( t ), then | t | = | t | ≤ | t | . Note that z = i = ⇒ z = − z . Suppose t is a zero of D ( t ), so that1 + it + t + αit = 0 . Then 1 + i ( − t ) + ( − t ) + αi ( − t ) = 1 − it + t − αit = 1 + it + t + αit = 1 + it + t + αit = 0 . − t := t is also a zero of D ( t ). So we have | t | = | t | . It remains to show that | t | ≥ | t | .Let us make the substitution it → y so that we may work with a cubic polynomial with realcoefficients. Then our function D ( t ) becomes P ( y ) := 1 + y − y − αy . Note that the discriminant of a cubic polynomial ax + bx + cx + d is∆ = 18 abcd − b d + b c − ac − a d . The discriminant of P ( y ) is then ∆ = 5 + 22 α − α . Since α >
1, ∆ <
0, which implies that P ( y ) has one real zero and two non-real complexconjugate zeros.Let y , y , and y be the zeros of P ( y ) with y = τ e iθ , y = τ e − iθ , and y is real . We now show that | y | > τ .By Vieta’s formulas, we have y + y + y = − α ,y y + y y + y y = − α ,y y y = 1 α . The last equality gives τ y = 1 α = ⇒ τ = 1 αy . Note that | y | = | y | = τ , so we want to show that τ < y , or equivalently, y > √ α .Observe P (cid:18) √ α (cid:19) = 1 + 1 √ α − √ α − α (cid:18) α (cid:19) = 1 √ α − √ α , which is greater than 0 since α >
1. Furthermore,lim y →−∞ P ( y ) = + ∞ and lim y →∞ P ( y ) = −∞ . Since P ( y ) = 0 and P (cid:16) √ α (cid:17) >
0, and y is the only real zero of P ( y ), we get that y > √ α by the Intermediate Value Theorem.Case 2: Suppose 1 / < α ≤
1. Sincelim θ → ∆( θ ) = 9 α − α + 1 < , by the continuity of ∆( θ ) there is θ ∗ sufficiently close to 0 so that ∆( θ ∗ ) <
0. For this choiceof θ ∗ , we have ζ ∗ := ζ ( θ ∗ ) / ∈ R and hence τ ∗ := τ ( θ ∗ ) / ∈ R . It is clear that ζ ∗ and τ ∗ are not9real, but we must verify that z ∗ := z ( θ ∗ ) is not real.Suppose by way of contradiction that z ∗ ∈ R . Then g ( t ) := 1 + z ∗ t + t + αz ∗ t is a polynomial in t with real coefficients.By Vieta’s formulas, t ∗ := τ ∗ e − iθ ,t ∗ := τ ∗ e iθ ,t ∗ := τ ∗ ζ ∗ are the zeros of g ( t ). Since g ( t ) is a polynomial with real coefficients, we know complex zerosoccur in conjugate pairs. So | t ∗ | = | t ∗ | = ⇒ τ ∗ ∈ R , a contradiction. Hence z ∗ / ∈ R . Since | t ∗ | > | t ∗ | = | t ∗ | ≤ | t ∗ | , and hence we have by Theorem 18 that P n ( z ) is not hyperbolic for large n .Thus, we have shown that if b < α > / P n ( z ) is not hyperbolic for all large n . 6 OPEN PROBLEMOne may wish to characterize polynomials A ( z ) , B ( z ) , C ( z ) so that the sequence ofpolynomials { P n ( z ) } ∞ n =0 defined by ∞ X n =0 P n ( z ) t n = 11 + A ( z ) t + B ( z ) t + C ( z ) t is hyperbolic. In this paper, the main result, Theorem 2, characterizes one of the cases where A ( z ), B ( z ), and C ( zz