Table Of ContentPOLYNOMIALLY INTERPOLATED LEGENDRE MULTIPLIER
SEQUENCES
MATTHEWCHASSE,TAMA´SFORGA´CS,ANDANDRZEJPIOTROWSKI
7
1
0 Abstract. We prove that every multiplier sequence for the Legendre basis
2 which can be interpolated by a polynomial has the form {h(k2 +k)}∞ ,
k=0
where h ∈ R[x]. We also prove that a non-trivial collection of polynomials
n
ofacertainforminterpolatemultipliersequencesfortheLegendrebasis,and
a
J westateconjecturesonhowtoextendtheseresults.
0
1
]
V 1. Introduction
C
Over the past decade, there has been an effort to characterize multiplier se-
h. quences acting on various orthogonal bases for R[x] (see [B], [BDFU], [BC], [BO],
t [BJJMPS], [FHMS], [FP], [P]). The present work focuses on the Legendre polyno-
a
m mial basis {Pk}∞k=0. Recall that, for each nonnegative integer k, the k-th Legendre
polynomial P can be defined as the polynomial solution to Legendre’s differential
[ k
equation
1
(x2−1)y(cid:48)(cid:48)+2xy(cid:48)−k(k+1)y =0,
v
0 normalized so that P (1) = 1. As such, our work is a continuation of the investi-
k
2
gations carried out in [BDFU] and [FHMS].
4
We now recall some of the relevant terminology in this subject area that will
2
0 be used freely throughout the paper. A polynomial is called hyperbolic if all of its
. zeros are real. An operator on R[x] that maps the class of hyperbolic polynomials
1
0 into itself is called a hyperbolicity preserver. If T is a linear operator on R[x] which
7 is diagonal with respect to the Legendre basis, and if T is also a hyperbolicity
1 preserver,thenthecorrespondingeigenvaluesequence{γ }∞ forwhichT[P (x)]=
k k=0 k
:
v γkPk(x) is called a multiplier sequence for the Legendre basis.
i Although the bulk of the present work has to do with hyperbolicity preserving
X
operators, there will also be a brief discussion (in Section 4) of complex zero de-
r
a creasing operators and sequences which are defined as follows. If an operator T
on R[x] has the property that Z (T[p])≤Z (p) for all polynomials p, where Z (f)
c c c
denotesthenumberofnonrealzerosoff,thenT iscalledacomplex zero decreasing
operator. If T is a linear operator on R[x] which is diagonal with respect to the
LegendrebasisandT isacomplexzerodecreasingoperator,thenthecorresponding
eigenvalue sequence is called a complex zero decreasing sequence for the Legendre
basis. It is straightforward to see that every complex zero decreasing operator is
a hyperbolicity preserving operator. There are, however, hyperbolicity preserving
operators which are not complex zero decreasing operators. Finally, we are obliged
to note that the function which is identically zero plays a peculiar role in this the-
ory. As usual, we will declare that this function has only real zeros and define
Z (0)=0.
c
1
2 MATTHEWCHASSE,TAMA´SFORGA´CS,ANDANDRZEJPIOTROWSKI
Central to the theory of multiplier sequences is the Laguerre-P´olya Class of
functions, denoted L−P, which can be defined as follows. A function belongs to
the class L−P if and only if it is a uniform limit, on compact subsets of C, of
polynomials with only real zeros. Equivalently, a real entire function ϕ∈L−P, if
and only if it can be expressed in the form
ω (cid:18) (cid:19)
ϕ(x)=cxme−ax2+bx (cid:89) 1+ x e−xkx (0≤ω ≤∞),
x
k
k=1
where b,c,x ∈R, m is a non-negative integer, a≥0, x (cid:54)=0, and (cid:80)ω 1 <∞.
k k k=1 x2
k
Proposition 1. [CCturan, p. 242] Let ϕ(x) = (cid:80)∞ γkxk ∈ L−P. Then the
k=0 k!
Tur´an inequalities hold for the Taylor coefficients of ϕ; that is,
(1.1) γ2−γ γ ≥0, k =1,2,3,....
k k−1 k+1
Wereferthereaderto[CCsurvey]foracomprehensivetreatmentoftheLaguerre-
P´olya Class. For our purposes, we shall only need the fact that an even function
in L−P must have coefficients (of even powers of the variable) which alternate in
sign; this follows immediately from Proposition 1.
In this paper, we focus mainly on sequences that can be interpolated by polyno-
mials. Thatistosay, sequencesoftheform{p(k)}∞ wherep∈R[x]. InTheorem
k=0
2 of Section 3 we prove that any multiplier sequence for the Legendre basis which
can be interpolated by a polynomial must have the form {h(k2 +k)}∞ , where
k=0
h ∈ R[x]. In Section 4, we prove that a certain type of differential operator is
hyperbolicity preserving (Theorem 4) and use that result to establish the existence
of a new and non-trivial collection of multiplier sequences for the Legendre ba-
sis (Corollary 3). The paper concludes with two conjectures concerning Legendre
multiplier sequences interpolated by polynomials of even degree.
2. Preliminaries
ItisknownthatanylinearoperatorT onR[x]canberepresentedasadifferential
operator with coefficients S ∈R[x], i.e.,
k
∞
(cid:88)
T = S (x)Dk.
k
k=0
The goal of this section is to develop a formula for S (0) in the case where T is
k
diagonalwithrespecttotheLegendrebasisandhasaneigenvaluesequencethatcan
be interpolated by a polynomial. This formula will be useful in proving Theorem
1 in the next section, which states that a polynomially interpolated multiplier
sequence for the Legendre basis must be interpolated by a polynomial of even
degree. This formula will be useful in proving Theorem 1 in the next section,
which states that if a multiplier sequence for the Legendre basis is interpolated by
a polynomial p, then p must be of even degree.
Lemma 1. For any differential operator
∞
(cid:88)
T = Q (x)Dk
k
k=0
POLYNOMIALLY INTERPOLATED LEGENDRE MULTIPLIER SEQUENCES 3
on R[x], the coefficient polynomials evaluated at 0 can be computed by
1
Q (0)= [T[xn]] (n=0,1,2,...).
n n! x=0
Proof. The result follows from calculating
T[xn]=Q (x)xn+nxn−1Q (x)+···+n!Q (x),
0 1 n
and evaluating this expression at x=0. (cid:3)
We now specialize the previous lemma to operators which are diagonal with
respect to the Legendre polynomial basis. Throughout the rest of this paper we
use the Pocchammer symbol to denote the rising factorial
(α) =α(α+1)(α+2)···(α+k−1), (α∈R,k ∈N)
k
and follow the convention that (α) =1.
0
Lemma 2. Let T be a linear operator on R[x] that satisfies T[P ] = γ P , where
n n n
{γ }∞ is a sequence of real numbers, and let the differential operator representa-
n n=0
tion of T be given by
∞
(cid:88)
T = S (x)Dk.
k
k=0
Then
(2.1) S (0)=0 (m=0,1,2,...)
2m+1
and
(2.2) S (0)= 1 (cid:88)m (cid:18)m(cid:19)(4k+1)γ2k(−1)k (m=0,1,2,...).
2m 2·4mm! k (k+1/2)
m+1
k=0
Proof. WeapplyLemma1totheexpansionofxn intermsofLegendrepolynomials
(see [R, p. 181]),
[n/2]
xn = n! (cid:88) (2n−4k+1)Pn−2k(x)
2n k!(3/2)
n−k
k=0
to obtain
[n/2]
S (0)= 1 (cid:88) (2n−4k+1)γn−2kPn−2k(0).
n 2n k!(3/2)
n−k
k=0
The relation (see [R, p. 158])
P (0)=0 (k =0,1,2,...)
2k+1
shows that S (0)=0 for all m. The relation (see [R, p. 158])
2m+1
(−1)k(1/2)
P (0)= k (k =0,1,2,...)
2k k!
yields
S (0)= 1 (cid:88)m (4m−4k+1)γ2m−2k(−1)m−k(1/2)m−k
2m 4m k!(3/2) (m−k)!
2m−k
k=0
m (cid:18) (cid:19)
= 1 (cid:88) m (4k+1)γ (−1)k (1/2)k
4mm! k 2k (3/2)
m+k
k=0
which can be re-written in the form (2.2) as desired. (cid:3)
4 MATTHEWCHASSE,TAMA´SFORGA´CS,ANDANDRZEJPIOTROWSKI
The fact that the odd indexed coefficient polynomials vanish at zero (equation
(2.1)) will be useful to us in the next section. Continuing to specialize, we now
focus on the even indexed coefficient polynomials in the case where the eigenvalue
sequence {γ }∞ can be interpolated by a polynomial p. In this case, we note
k k=0
that the quantity (4k+1)γ = (4k+1)p(2k) that appears in equation (2.2) is a
2k
polynomial in k. This motivates us to find a closed form for the sum
(cid:88)m (cid:18)m(cid:19) kn(−1)k
(2.3) σ = .
m,n k (k+1/2)
m+1
k=0
Lemma 3. For m,n=0,1,2,..., the quantity σ from equation (2.3) satisfies
m,n
(cid:20) (cid:21) (cid:18) (cid:19)
2 d
σ = θn F (−m,1/2;m+3/2;x) θ :=x .
m,n (3/2) 2 1 dx
m x=1
Proof. First note the relation
(−m) (cid:40)(cid:0)m(cid:1)(−1)k k =0,1,2,...,m,
(2.4) k = k
k! 0 k =m+1,m+2,...
whichfollowsdirectlyfromequation(3)of[R,p. 58]andthedefinitionoftherising
factorial. Furthermore, from (see [R, p. 23, equation (7)])
Γ(α+n)
(2.5) (α) = (α∈/ {0,−1,−2,...})
n Γ(α)
we have, for k ∈{0,1,2,...m},
2·(1/2) Γ(k+1/2) 1
(2.6) k = = .
(m+3/2) (3/2) Γ(m+k+3/2) (k+1/2)
k m m+1
Combining equations (2.4) and (2.6), we see that
(cid:88)m (cid:18)m(cid:19) (−1)kxk = 2 (cid:88)m (−m)k(1/2)kxk
k (k+1/2) (3/2) k!(m+3/2)
m+1 m k
k=0 k=0
2
= F(−m,1/2;m+3/2;x),
(3/2)
m
where F(a,b;c;x) is the hypergeometric function (see [R, p. 45]). The conclusion
that
(cid:20) (cid:21) (cid:18) (cid:19)
2 d
σ = θn F (−m,1/2;m+3/2;x) θ :=x
m,n (3/2) 2 1 dx
m x=1
now follows from the fact that θxk =kxk, which holds for all k. (cid:3)
The next lemma will be useful in obtaining another explicit formula for σ .
m,n
Lemma 4. The coefficient polynomials Q (x) in the expansion
k
n
(cid:88)
(xD)n = Q (x)Dk
k
k=0
are given by Q (x)=S(n,k)xk, where
k
k (cid:18) (cid:19)
1 (cid:88) k
(2.7) S(n,k)= (−1)k−jjn
k! j
j=0
are the Stirling numbers of the second kind.
POLYNOMIALLY INTERPOLATED LEGENDRE MULTIPLIER SEQUENCES 5
Proof. This result is known, but we offer a simple proof here based on a formula
for Q for a linear operator T :C[x]→C[x] [C, p. 106, Prop. 216]:
k
k (cid:18) (cid:19)
1 (cid:88) k
Q (x)= T[xj](−x)k−j
k k! j
j=0
substituting T =(xD)n yields
xk (cid:88)k (cid:18)k(cid:19)
Q (x)= (−1)k−jjn,
k k! j
j=0
and the result follows. (cid:3)
Wenowusethepreviouslemmaalongwithsomepropertiesofthehypergeomet-
ric function to obtain another formula for σ .
m,n
Lemma5. Fornonnegativeintegersmandnwhichsatisfy2m≥n−1,thequantity
σ from equation (2.3) can be written in the form
m,n
2·4m(2m−n)!
σ = p (m),
m,n m!(4m+1)!! n
where
n
(cid:88)
(2.8) p (m)= S(n,k)(−m) (1/2) (2m−n+1)
n k k n−k
k=0
is polynomial of degree n in the variable m. Here, S(n,k) is the Stirling number of
the second kind as defined in equation (2.7).
Proof. We first use Lemma 3 and Lemma 4 together with the relations ([R, p. 49,
l. 7])
Γ(c)Γ(c−a−b)
F(a,b;c;1)= (Re(c−a−b)>0)
Γ(c−a)Γ(c−b)
and ([R, p. 69, ex. 1])
d ab
F(a,b;c;x)= F(a+1,b+1;c+1,x)
dx c
to obtain
(cid:34) n (cid:35)
2 (cid:88)
σ = S(n,k)xkDkF(−m,1/2;m+3/2,x)
m,n (3/2)
m
k=0 x=1
(cid:34) n (cid:18) (cid:19)(cid:35)
= 2 (cid:88)S(n,k)xk(−m)k(1/2)kF −m+k,1 +k;m+ 3 +k,x
(3/2) (m+3/2) 2 2
m k
k=0 x=1
n
= 2 (cid:88)S(n,k)(−m)k(1/2)k Γ(m+3/2+k)Γ(2m−k+1),
(3/2) (m+3/2) Γ(2m+3/2)Γ(m+1)
m k
k=0
which is valid for 2m>n−1.
6 MATTHEWCHASSE,TAMA´SFORGA´CS,ANDANDRZEJPIOTROWSKI
Next, we use the property (2.5) several times to obtain, for 2m≥n−1,
n
2 Γ(m+3/2)Γ(2m−n+1)(cid:88)
σ = S(n,k)(−m) (1/2) (2m−n+1)
m,n m! (3/2) Γ(2m+3/2) k k n−k
m
k=0
2 Γ(m+3/2)Γ(2m−n+1)
= p (m)
m! (3/2) Γ(2m+3/2) n
m
2 (2m−n)!
= p (m)
m!(3/2) (m+3/2) n
m m
2 (2m−n)!
= p (m)
m! (3/2) n
2m
2·4m(2m−n)!
= p (m)
m!(4m+1)!! n
where p (m) is defined as in equation (2.8). (cid:3)
n
Withtheseresultsathand,wecannowgiveexplicitexpressionsforS (0)inthe
2m
case where the eigenvalue sequence {γ }∞ can be interpolated by the monomial
k k=0
p(k)=kn.
Lemma 6. If γ =kn then S (0) as defined in equation (2.2) is given by
k 2m
2n(2m−n−1)!
(2.9) S (0)= [4p (m)+(2m−n)p (m)] (m≥n),
2m (m!)2(4m+1)!! n+1 n
where p (m) is defined as in equation (2.8).
j
Proof. Fixanonnegativeintegernandletγ =kn. Equations(2.2)and(2.3)yield
k
1 (cid:88)m (cid:18)m(cid:19)(4k+1)(2k)n(−1)k
S (0)=
2m 2·4mm! k (k+1/2)
m+1
k=0
2n−1
= (4·σ +σ ).
4mm! m,n+1 m,n
An application of Lemma 5, some simplification, and comparison with equation
(2.8) then gives
2n
S (0)= (4·(2m−n−1)!p (m)+(2m−n)!p (m)),
2m (m!)2(4m+1)!! n+1 n
which is valid for m≥n, and the result follows. (cid:3)
We have now arrived at the goal of this section, which was to develop a use-
ful formula for S (0) in the case where the eigenvalue sequence {γ }∞ can be
2m k k=0
interpolated by a polynomial p(k)=(cid:80)n a kj.
j=0 j
Proposition2. Ifγ =(cid:80)n a kj then,forallm≥n,thequantityS (0)defined
k j=0 j 2m
in equation (2.2) satisfies
(2m−n−1)!
(2.10) S (0)= P(m),
2m (m!)2(4m+1)!!
where
n
(cid:88)
(2.11) P(m)= a 2j(2m−n) [4p (m)+(2m−j)p (m)]
j n−j j+1 j
j=0
and p (m) is defined as in equation (2.8).
j
POLYNOMIALLY INTERPOLATED LEGENDRE MULTIPLIER SEQUENCES 7
Proof. Let γ =(cid:80)n a kj. Equations (2.2) and (2.3) yield
k j=0 j
S (0)= 1 (cid:88)m (cid:18)m(cid:19)(4k+1)(cid:80)nj=0aj(2k)j(−1)k
2m 2·4mm! k (k+1/2)
m+1
k=0
(cid:88)n (cid:34) 1 (cid:88)m (cid:18)m(cid:19)(4k+1)(2k)j(−1)k(cid:35)
= a .
j 2·4mm! k (k+1/2)
m+1
j=0 k=0
The quantity in the square brackets is the S (0) corresponding to γ = kj, so
2m k
Lemma 6 yields
(cid:88)n 2j(2m−j−1)!
S (0)= a [4p (m)+(2m−j)p (m)],
2m j(m!)2(4m+1)!! j+1 j
j=0
whichisvalidform≥n. Somemanipulationofthefactorialsnowgivesthedesired
result. (cid:3)
3. Form and Order
In this section, we seek to determine what form a polynomial p must have if the
sequence {p(k)}∞ is a multiplier sequence for the Legendre basis. The conditions
k=0
that we determine will, in turn, impose conditions on the order of the differential
operator associated with the given sequence. We begin by showing that any such
polynomial must have even degree. This settles open question (2) posed in section
5 of [BDFU], a paper of the second author.
Theorem 1. Suppose that p∈R[x] is a polynomial of odd degree. Then {p(k)}∞
k=0
is not a multiplier sequence for the Legendre basis.
Proof. We argue by contradiction. Let T be the operator corresponding to the
sequenceinterpolatedbythepolynomialp(x)=(cid:80)n a xj withnoddanda (cid:54)=0,
j=0 j n
and suppose that {p(k)}∞ is a multiplier sequence for the Legendre basis. The
k=0
symbol of T is given by
(cid:88)∞ (−1)kT[xk]yk
G (x,y)= =T[e−xy],
T k!
k=0
where the operator T acts on e−xy as a function of x alone (see [BB] for a com-
prehensive treatment of the symbol of an operator as it relates to hyperbolicity
preserving operators). Suppose that the differential operator representation of T is
given by
∞ (cid:18) (cid:19)
(cid:88) d
T = S (x)Dk D = .
k dx
k=0
The symbol G is then given by
T
∞
(cid:88)
G (x,y)=e−xy S (x)(−1)kyk
T k
k=0
As noted in [BO] and [FHMS], we can act on this expression (as a function of x
alone)bytheclassicalmultipliersequence{1,0,0,0,...}andtheresultingfunction
∞
(cid:88)
G (0,y)= S (0)(−1)kyk
T k
k=0
8 MATTHEWCHASSE,TAMA´SFORGA´CS,ANDANDRZEJPIOTROWSKI
must belong to the Laguerre-P´olya class. By equation (2.1), each term with odd
index is zero. Thus,
∞
(cid:88)
G (0,y)= S (0)y2m
T 2m
k=0
isanevenfunctionwhichbelongstotheLaguerre-P´olyaclass. Itfollows,byPropo-
sition 1, that the sequence {S (0)}∞ must alternate in sign (a fact that we aim
2m m=0
to contradict). By equations (2.10) and (2.11), for m≥n,
(2m−n−1)!
S (0)= P(m),
2m (m!)2(4m+1)!!
where
n
(cid:88)
P(m)= a 2j(2m−n) [4p (m)+(2m−j)p (m)],
j n−j j+1 j
j=0
and p (m) is defined as in equation (2.8). Note that P(m) is a polynomial in
j
m which is either identically zero, or is not identically zero. In the latter case,
thesequence{P(m)}∞ ,andthereforethesequence{S (0)}∞ ,eventuallyhas
m=0 2m m=0
constant sign, which would give us our desired contradiction. We will finish the
proof by demonstrating that P(m) is not identically zero. Indeed,
(cid:16)n(cid:17) (cid:16)n(cid:17) (cid:18)−n(cid:19) (cid:18)1(cid:19)
(3.1) P =a 2n4p =a 2n+2S(n+1,n+1) ,
2 n n+1 2 n 2 2
n+1 n+1
which is non-zero due to the fact that we have assumed n is odd, a (cid:54)=0 and, as is
n
well-known, Stirling numbers of the second kind satisfy S(n+1,n+1)=1. (cid:3)
It is worthwhile to note that equation (2.10) is only valid for m > n and our
choice of m = n/2 in the preceding proof does not satisfy this. However, this fact
is irrelevant to the argument. All that is required is that the polynomial P(m) not
be identically zero and this can be achieved by considering any values for m that
we choose.
Next, we want to determine conditions under which an even degree polynomial
may interpolate a multiplier sequence for the Legendre basis.
Theorem 2. If {γ }∞ is a multiplier sequence for the Legendre basis which is
k k=0
interpolated by a polynomial p, then p(x)=h(x2+x) for some polynomial h(x)∈
R[x].
Proof. By Theorem 1, p must have even degree. Noting that
(cid:40)
(x2+x)k/2 k even,
(3.2) b (x)=
k xk k odd.
forms a basis for R[x], we can expand p in terms of this basis to obtain
p(x)=h(x2+x)+q(x),
where degh=(degp)/2 and q is an odd function in R[x]. We claim that q must be
identically zero. By way of contradiction, suppose q is not identically zero. Then q
POLYNOMIALLY INTERPOLATED LEGENDRE MULTIPLIER SEQUENCES 9
is a polynomial of odd degree. Using equation (2.2) to calculate S (0), we have
2m
1 (cid:88)m (cid:18)m(cid:19)(4k+1)p(2k)(−1)k
S (0)=
2m 2·4mm! k (k+1/2)
m+1
k=0
1 (cid:88)m (cid:18)m(cid:19)(4k+1)(cid:2)h((2k)2+(2k))+q(2k)(cid:3)(−1)k
= .
2·4mm! k (k+1/2)
m+1
k=0
Splitting up the sum then gives
1 (cid:88)m (cid:18)m(cid:19)(4k+1)h((2k)2+(2k))(−1)k
(3.3) S (0)=
2m 2·4mm! k (k+1/2)
m+1
k=0
1 (cid:88)m (cid:18)m(cid:19)(4k+1)q(2k)(−1)k
(3.4) + .
2·4mm! k (k+1/2)
m+1
k=0
The sum in (3.3) calculates the polynomial coefficients, evaluated at zero, of the
differential operator T corresponding to the sequence {h(k2+k)}∞ . Since T is
h k=0 h
afiniteorderdifferentialoperator, thissummustvanishforallsufficientlylargem.
Thus, for all such m, only the sum (3.4) contributes to S (0), i.e.,
2m
1 (cid:88)m (cid:18)m(cid:19)(4k+1)q(2k)(−1)k
S (0)= .
2m 2·4mm! k (k+1/2)
m+1
k=0
Note that these are precisely the values of the coefficient polynomials, evaluated at
zero, corresponding to the odd degree interpolated sequence {q(k)}∞ . The result
k=0
now follows from the proof of Theorem 1. (cid:3)
As an immediate consequence, we get a new proof of the following result which
was proved in [FHMS].
Corollary 1. If {k2+bk+c} is a Legendre MS, then b=1.
Theorem 2 also allows us to conclude that, as a differential operator, any poly-
nomially interpolated multiplier sequence for the Legendre basis must have finite
order.
Corollary 2. Suppose {p(k)}∞ is a multiplier sequence for the Legendre basis,
k=0
wherep∈R[x],andletT bethecorrespondinglinearoperatordefinedbyT[P (x)]=
n
p(n)P (x) for all n. Then the differential operator representation of T has only
n
finitely many terms:
m
(cid:88)
T = S (x)Dk.
k
k=0
Proof. Let h be a real polynomial for which p(x)=h(x2+x) and write
h(x)=a +a x+···+a xn.
0 1 n
Let δ be the differential operator from Legendre’s differential equation
δ =(x2−1)D2+2xD.
Then δP (x) = (k2 +k)P (x) for all k and it follows that the operator T can be
k k
written as
T =h(δ)=a +a δ+···a δn.
0 1 n
From this we see that the order of T is at most 2n=degp. (cid:3)
10 MATTHEWCHASSE,TAMA´SFORGA´CS,ANDANDRZEJPIOTROWSKI
Remark 1. The previous corollary complements a result of Miranian [M], which
impliesthatamultipliersequencefortheLegendrebasiswhosecorrespondingdiffer-
ential operator has finite order must have the form {h(k2+k)}∞ where h∈R[x].
k=0
In fact, we have shown that any multiplier sequence for the Legendre basis whose
corresponding differential operator has infinite order (of which, to date, none have
been discovered) cannot be interpolated by a polynomial.
4. A Collection of Legendre Multiplier Sequences
In this section, we demonstrate that a certain collection of sequences are multi-
plier sequences for the Legendre basis. The result is reminiscent of a similar result
by Craven and Csordas involving the characterization of polynomially interpolated
classical complex zero decreasing sequences (see [CCczds, Prop. 2.2]). We will use
the Bates-Yoshida Quadratic Hyperbolicity Preserver Characterization:
Theorem 3 (Bates-Yoshida [BY]). Suppose Q ,Q ,Q are real polynomials such
2 1 0
that deg(Q )=2, deg(Q )≤1, deg(Q )=0. Then
2 1 0
T =Q D2+Q D+Q
2 1 0
preserves hyperbolicity if and only if
W[Q ,Q ]2−W[Q ,Q ]W[Q ,Q ]≤0, and Q (cid:28)Q (cid:28)Q .
0 2 0 1 1 2 0 1 2
ThecompactformofBates-YoshidaTheoremisderivedusingtheBorcea-Br¨and´en
Characterization of Hyperbolicity Preservers [BB][Theorem 5], and intricate argu-
ments involving thelocation ofzeros of thecoefficient polynomials. WithTheorem
3 in hand, we prove the following.
Theorem 4. Let δ =(x2−1)D2+2xD where D =d/dx and fix positive integers
n and N. The operator
N
(cid:0) (cid:1)(cid:89)
(4.1) δ(δ−1·2)(δ−2·3)··· δ−(n−1)(n) (δ−A )
j
j=1
is a hyperbolicity preserver whenever −(n+1)≤A ≤n(n+1) for all j.
j
Proof. As in the proof of Theorem 14 on p. 137 of [BJJMPS], we may factor the
operator
δ(δ−1·2)(δ−2·3)···(δ−(n−1)·n)
into a product of operators:
(cid:32)n−1 (cid:33)
(cid:89)(cid:2)(x2−1)D+2(k+1)x(cid:3) Dn.
k=0
From Leibniz’ rule (or by using equations (6) and (7) on p. 135 of [BJJMPS]), we
have
Dn(δ−A )=(cid:2)(x2−1)D2+2(n+1)xD+n2+n−A (cid:3)Dn.
j j
It follows that the operator in (4.1) can be factored as
(cid:32)n−1 (cid:33)
(4.2) (cid:89)(cid:2)(x2−1)D+2(1+k)x(cid:3) TDn,
k=0
