Content uploaded by Anna Vishnyakova
Author content
All content in this area was uploaded by Anna Vishnyakova
Content may be subject to copyright.
Computational Methods and Function Theory
Volume X (20XX), No. X, 1–18 Page proofs CMFT-MS 0800829
On the Stability of Taylor Sections of a Function
∞
k=0 zk/ak2,a>1
Olga M. Katkova and Anna M. Vishnyakova
(Communicated by Vladimir V. Andrievskii)
Abstract. We investigate the following problem: given a positive integer
n, which are the smallest values of the constants sn, such that the zeros of
fa,n(z):=n
k=0 zk/ak2are with negative real parts when a>s
n?
Keywords. ??.
2000 MSC. 30C15, 30D15, 26C10, 34D99.
1. Introduction and statement of results.
A real polynomial Fis called Hurwitz (stable)polynomial if all its zeros have
negative real parts: thas is, F(z0) = 0 implies Re z0<0. Polynomial stability
problems of various types arise in a number of problems in mathematics and
engineering. We refer to [5, Cha. 15] or [13, Ch. 9] for deep surveys on the
stability theory.
The following statement (usually attributed to A. Stodola, see, for example, [17])
is the well-known necessary condition for a real polynomial to be stable.
Statement. If the polynomial F(z)=a0+a1z+···+anzn∈R[z],an>0,is
stable then aj>0,0≤j≤n−1.
The following well-known theorem gives the necessary and sufficient conditions
for a polynomial to be stable.
Routh-Hurwitz Criterion (see, for example [5, pp. 225–230]).The polynomial
F(z)=a0+a1z+···+anzn∈R[z],an>0, is stable if and only if the first n
Received February 18, 2008, in revised form July 18, 2008.
Published online ??.
ISSN 1617-9447/$ 2.50 c
2007 Heldermann Verlag
2 O.M.KatkovaandA.M.Vishnyakova CMFT
principal minors of the corresponding Hurwitz matrix
H(F):=
an−1an−3an−5··· 0
anan−2an−4··· 0
0an−1an−3··· 0
0anan−2··· 0
.
.
..
.
..
.
..
.
....
are positive.
Note that the verification of positivity of principal minors is, in general, a very
difficult problem. Surely, it is not difficult to calculate the determinant of a given
matrix with numerical entries. But if the order of a matrix or the entries of a
matrix depend on some parameters then the testing of positivity of principal
minors is complicated. In [3] T. Craven and G. Csordas obtained the useful and
easily verifiable sufficient condition of positivity of all minors of a matrix. To
formulate this condition we need the following definition.
A matrix Mis said to be totally positive,ifallminorsofMare non-negative.
About this notion and its applications see [1, 8] and the references therein. A
matrix Mis said to be strictly totally positive,ifallminorsofMare strictly
positive.
In [3] we have the following theorem.
Theorem A. Denote by ˜cthe unique real root of the equation x3−5x2+4x−1=0
(˜c≈4.0796).LetM=(aij )be an n×nmatrix with the properties
(a) aij >0,1≤i, j ≤n,and
(b) aij ai+1,j+1 ≥˜ca
i,j+1 ai+1,j ,1≤i, j ≤n−1.
Then Mis strictly totally positive.
Theorem A provides a convenient sufficient condition for strict total positivity
of a matrix with positive entries. Using Theorem A and continuity reasonings
D. K. Dimitrov and J. M. Pe˜na obtained the following theorem.
Theorem B ([4]).Let ˜cbe defined as in Theorem A. If the coefficients of
F(z)=a0+a1z+···+anznare positive and satisfy the inequalities
(1) akak+1 ≥˜cak−1ak+2 for k=1,2,...,n−2,
then Fis a Hurwitz polynomial. In particular, the conclusion is true if
(2) ak2≥√˜cak−1ak+1 for k=1,2,...,n−1.
In [10] the authors of this note have shown that Theorem A remains valid if one
replaces the constant ˜cby the constant cn:= 4 cos2(π/(n+ 1)). In [10] it is
also shown that in the statement of Theorem A the constant cnis the smallest
possible not only in the class of matrices with positive entries but in the classes of
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>13
Toeplitz matrices and of Hankel matrices. We recall that a matrix Mis Toeplitz
matrix if it is of the form M=(aj−i) and a matrix Mis Hankel matrix if it is
of the form M=(aj+i).
In [11] the authors of this note have found the smallest possible constants in
the inequalities of the type (1) and (2) such that the statement of Theorem B
remains valid.
Theorem C. Let x0be the (unique)positive root of the polynomial x3−x2−2x−1
(x0≈2.1479).
(i) If the coefficients of F(z)=4
k=0 akzkare positive and satisfy the inequal-
ities akak+1 >2ak−1ak+2 for k=1,2,thenFis a Hurwitz polynomial. In
particular, the conclusion is true if ak2>√2ak−1ak+1 for k=1,2,3.
(ii) If the coefficients of F(z)=5
k=0 akzkare positive and satisfy the inequal-
ities akak+1 >x
0ak−1ak+2 for k=1,2,3,thenFis a Hurwitz polynomial.
In particular, the conclusion is true if ak2>√x0ak−1ak+1 for k=1,2,3,4.
(iii) If the coefficients of F(z)=n
k=0 akzk,n>5, are positive and satisfy the
inequalities akak+1 ≥x0ak−1ak+2 for k=1,2,...,n−2,thenFis a Hurwitz
polynomial. In particular, the conclusion is true if ak2≥√x0ak−1ak+1 for
k=1,2,...,n−1.
Note that
akak+1
ak−1ak+2
=ak2
ak−1ak+1
ak+12
akak+2
and thus the following theorem demonstrates that the constants in Theorem C
are the smallest possible for every n.
Theorem D.
(i) For every d≤√2there exists a polynomial F(z)=4
k=0 akzkwith positive
coefficients under condition ak2=dak−1ak+1 for k=1,2,3, such that Fis
not a Hurwitz polynomial.
(ii) For every d≤√x0there exists a polynomial F(z)=5
k=0 akzkwith positive
coefficients under condition ak2=dak−1ak+1 for k=1,2,3,4, such that F
is not a Hurwitz polynomial.
(iii) For every n>5and every ε>0there exists a polynomial F(z)=n
k=0 akzk
with positive coefficients under condition ak2>(√x0−ε)ak−1ak+1 for
k=1,2,...,n−1, such that Fis not a Hurwitz polynomial.
The extremal entire function for which the inequalities (1) and (2) reduce to
equalities is
(3) fa(z):=
∞
k=0
zk
ak2,a>1.
4 O.M.KatkovaandA.M.Vishnyakova CMFT
In [4] the following problem was posed: given a positive integer n, which are the
smallest values of the constants sn, such that the zeros of
fa,n(z):=
n
k=0
zk
ak2
are with negative real parts when a>s
n? Further we will explain that for every
positive integer nthere exists the constant snsuch that all zeros of fa,n have
negative real parts if and only if a>s
n. We will explain also that there exists
the constant s∞such that all zeros of fahave negative real parts if and only if
a>s
∞. The following theorem will answer the question formulated in [4].
Theorem 1.
(i) s3≤s7≤s11 ≤...,
(ii) s5≥s9≥s13 ≥...,
(iii) limn→∞ s4n+3 = limn→∞ s4n+1 =s∞.
Remark 1. Using arguments similar to those in the proof of Theorem 1 we
can show that the sequence s2nis convergent and limn→∞ s2n=s∞.Wedonot
include the proof of this fact because it contains rather cumbersome calculations.
To prove Theorem 1 we use the famous Hermite-Biehler Criterion.
Hermite-Biehler Criterion (see[2,6]or[12,Ch.VII]).Let
M(z)=N(z)+iR(z),
where Nand Rare real polynomials. The following are equivalent:
(A) All zeros of Mhave positive imaginary parts.
(B) The polynomials Nand Rhave simple real interlacing zeros and
R(x0)N(x0)−R(x0)N(x0)>0
for some x0∈R.
The following statement is a version of the Hermite-Biehler Theorem.
Hermite-Biehler Criterion of stability. Let F(z)=n
k=0 akzkbe a poly-
nomial with positive coefficients. The polynomial Fis stable if and only if the
following two polynomials
P(z):=
0≤m≤n/2
(−1)ma2mzm,
zQ(z):=z
0≤m≤(n−1)/2
(−1)ma2m+1zm
have simple real interlacing zeros.
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>15
2. Proof of Theorem 1.
We will consider the entire function
(4) ga(z)=fa(−z):=
∞
k=0
(−1)kzk
ak2,a>1,
and will study the zero distribution of its Taylor sections. Obviously all zeros
of fa(or its n-th Taylor section) lie in the left half plane if and only if all zeros
of ga(or its n-th Taylor section) lie in the right half plane {z:Rez>0}. Define
Sn(x, a):=
n
k=0
(−1)kxk
ak2.
Let us consider an odd section
S2n+1(x, a)=1−x
a+x2
a4−x3
a9+···− x2n+1
a4n2+4n+1 .
For this polynomial two polynomials Pn(x, a)andQn(x, a) mentioned in the
Hermite-Biehler criterion of stability are
Pn(x, a)=1−x2
a4+x4
a16 −···+(−1)nx2n
a4n2,
Qn(x, a)=−1
a1−x2
a8+x4
a24 −···+(−1)nx2n
a4n2+4n.
The question is: for which adoes the polynomials Pn(x, a)andxQn(x, a)have
simple real interlacing zeros?
Writing q=a4and t=x2, the polynomials Pn(x, a)andQn(x, a) will become
(5)
˜
Pn(t, q)=1−t
q+t2
q4−···+(−1)ntn
qn2;
˜
Qn(t, q)=1−t
q2+t2
q6−···+(−1)ntn
qn(n+1) .
All zeros of polynomials Pn(x, a)andQn(x, a) are real and simple if and only
if all zeros of polynomials ˜
Pn(t, q)and ˜
Qn(t, q) are real and simple. Suppose
all zeros of polynomials ˜
Pn(t, q)and ˜
Qn(t, q) are real and simple. Denote by
t1<t
2< ... < t
nthe zeros of ˜
Pn(t, q)andbyt∗
1<t
∗
2< ... < t
∗
nthe zeros
of ˜
Qn(t, q). Obviously the zeros of Pn(x, a)andxQn(x, a) interlace if and only
if t1<t
∗
1<t
2<t
∗
2< ... < t
n<t
∗
n.Wehave
˜
Qn(t, q)= ˜
Pn(t/q, q), so
the polynomial ˜
Pn(t, q) has only real simple zeros if and only if the polynomial
˜
Qn(t, q) has. Define
(6) Rn(y, q):= ˜
Pn(qny, q)= ˜
Qn(qn+1y, q)=
n
k=0
(−1)kqk(n−k)yk.
6 O.M.KatkovaandA.M.Vishnyakova CMFT
Let y1<y
2< ... < y
nbe the zeros of Rn(y, q). Then the zeros of ˜
Pn(t, q)are
tk=ykqn,k=1,2,...,n; the zeros of ˜
Qn(t, q)aret∗
k=ykqn+1,k=1,2,...,n.
So the condition tk<t
∗
kobviously holds for all k=1,2,...,n. The condition
t∗
k<t
k,k=1,2,...,n−1, could be rewritten in the form
(7) yk+1
yk
>q, k=1,2,...,n−1.
Thus, all zeros of S2n+1(x, a) lie in the right half plane if and only if all zeros of
Rn(y, q) are real and these zeros satisfy (7) (we recall that q=a4).
Now we need some notions and facts concerning linear operators on the space of
real polynomials which do not reduce the number of real zeros of the polynomial.
Definition 1. A sequence {γk}∞
k=0 of real numbers is called a multiplier se-
quence if, whenever the real polynomial P(z)=n
k=0 akzkhas only real zeros
the polynomial n
k=0 γkakzkhas only real zeros. We denote the class of multiplier
sequences by MS.
The following well-known theorem by G. P´olya and J. Schur gives the complete
characterization of multiplier sequences:
Theorem E (see [16, 15] or [12, Ch. VIII, Sec. 3]).Asequence{γk}∞
k=0 is a
multiplier sequence if and only if the power series
Φ(z):=
∞
k=0
γk
k!zk
converges absolutely in the whole complex plane and the entire function Φ(z)or
theentirefunctionΦ(−z)admits the representation
(8) Ceσzzm
∞
k=1 1+ z
xk,
where C∈R,σ≥0,m∈N∪{0},0<x
k≤∞,∞
k=1 1/xk<∞.
A simple consequence of Theorem E is that the sequence {γ0,γ
1,...,γ
l,0,0,...}
is a multiplier sequence if and only if the polynomial P(z)=l
k=0(γk/k!)zkhas
only real zeros of the same sign.
For a real polynomial Pwe will denote by Zc(P) the number of non-real zeros
of P, counting multiplicities.
Definition 2. A sequence {γk}∞
k=0 of real numbers is said to be a complex zero
decreasing sequence if
(9) Zcn
k=0
γkakzk≤Zcn
k=0
akzk,
for any real polynomial n
k=0 akzk. We will denote the class of complex zero
decreasing sequences by CZDS.
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>17
Obviously, CZDS ⊂MS. The existence of non-trivial CZDS sequences is a con-
sequence of the following remarkable theorem proved by Laguerre and extended
by P´olya (see [14] or [15, pp. 314–321]).
Theorem F. Suppose an entire function fcan be expressed in the form
(10) f(z)=Czme−αz2+βz
∞
k=1 1+ z
xke−z/xk,
where m∈N∪{0},C, β ∈R,α≥0and 0<x
k≤∞,∞
k=1 1/xk2<∞.Then
the sequence {f(k)}∞
k=0 is a complex zero decreasing sequence.
It follows from the above theorem that
(11) {a−k2}∞
k=0 ∈CZDS ⊂MS,a≥1.
Using (11) we conclude that if the polynomial
Sn(x, a0):=
n
k=0
(−1)kxk
a0k2
has only real zeros then for all a≥a0polynomials
Sn(x, a):=
n
k=0
(−1)kxk
ak2
have only real zeros. Thus,
(12) ∀n=2,3,4,... ∃rn>1:
(Sn(x, a) has only real zeros ⇔a≥rn).
In [9] the following problem was solved: for which a>1 does the entire function
ga(z) (and its Taylor sections Sn(x, a), n=2,3,...) has only real (real and
simple) zeros. We put together the following results for the convenience of later
reference.
Theorem G ([9]).There exists a constant r∞(r∞≈1.79) such that:
(i) √3=r3≤r5≤r7≤···,limk→∞ r2k+1 =r∞;
(ii) 2 = r2≥r4≥r6≥···,limk→∞ r2k=r∞;
(iii) for every n≥2the polynomial Sn(x, rn)has multiple real zeros;
(iv) for every a>r
nthe polynomial Sn(x, a)has only simple real zeros;
(v) for every n≥2and a≥rnall zeros of Sn(x, a)lie in the interval (a, a2n−1);
(vi) for every n≥2and a≥rnthe polynomial Sn(x, a)has exactly two zeros
in the interval (a2n−3,a
2n−1);
(vii) ga(z)=∞
k=0(−1)kzk/ak2has only real zeros if and only if a≥r∞;
(viii) gahas only simple real zeros if and only if a>r
∞;
(ix) for a≥r∞the function gahas exactly two zeros in the interval (a, a3);
(x) for a>r
∞the function gahas exactly nintervals of positivity and ninter-
vals of negativity in the interval (a, a4n−1);
8 O.M.KatkovaandA.M.Vishnyakova CMFT
(xi) for a>r
∞and j≥2the following inequality holds: (−1)jga(a2j)>0.
We will also use the already established fact that if {γk}∞
k=0 ∈MS then whenever
the real polynomial n
k=0 akzkis stable the polynomial n
k=0 γkakzkis stable
(see, for example, [12, Ch. VIII, Sec. 3]). In particular, if the polynomial
Sn(x, a0):=
n
k=0
(−1)kxk
a0k2
has all zeros in the right half plane then for all a≥a0polynomials
Sn(x, a):=
n
k=0
(−1)kxk
ak2
have all zeros in the right half plane. Thus,
(13) ∀n=1,2,3,... ∃sn≥0:
(Sn(x, a) has all zeros in the right half plane ⇔a>s
n).
By the direct calculation one can obtain that
(14)
(S1(x, a) has all zeros in the right half plane ⇔a>0)
⇒s1=0;
(S2(x, a) has all zeros in the right half plane ⇔a>0)
⇒s2=0;
(S3(x, a) has all zeros in the right half plane ⇔a>1)
⇒s3=1;
(S4(x, a) has all zeros in the right half plane ⇔a4>2)
⇒s4≈1.1892;
(S5(x, a) has all zeros in the right half plane ⇔a6−a4−1>0)
⇒s5≈1.2106;
(S6(x, a) has all zeros in the right half plane ⇔a4>2)
⇒s6≈1.1892;
(S7(x, a) has all zeros in the right half plane ⇔
a16 −2a12 +1>0∧a>1) ⇒s7≈1.1646.
We need the following lemma.
Lemma 1. Suppose for some q>1all zeros of R2n(y, q)arerealandthesezeros
satisfy (7).Then
(i) All zeros of R2n+1(y, q)are real and these zeros satisfy (7).
(ii) All zeros of R2n+2(y, q)are real and these zeros satisfy (7).
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>19
Proof. (i). The fact that the zeros are real is a consequence of Theorem G ((i)
and (ii)). Note that
R2n+1(y, q)=
2n
k=0
(−1)kqk(2n+1−k)yk−y2n+1
(15)
=
2n
k=0
(−1)kqk(2n−k)(qy)k−y2n+1 =R2n(qy, q)−y2n+1,
and
R2n+1(y, q)=1−
2n+1
k=1
(−1)k−1qk(2n+1−k)yk
(16)
=1−
2n
k=0
(−1)kq(k+1)(2n−k)yk+1
=1−q2ny
2n
k=0
(−1)kqk(2n−k)y
qk
(17)
=1−q2nyR2n(y
q,q).
Denote by y1<y
2<... <y
2n−1<y
2nthe zeros of R2n(y, q). Note that
R2n(t, q)>0fort∈(0,y
1)∪(y2,y
3)∪...∪(y2n−2,y
2n−1)∪(y2n,∞),
R2n(t, q)<0fort∈(y1,y
2)∪(y3,y
4)∪...∪(y2n−1,y
2n).
By (15) we have
(18) R2n+1(y, q)<0fory2k−1<qy<y
2k,k=1,2,...,n.
And by (16) we have
(19) R2n+1(y, q)>0,for y2k−1<y
q<y
2k,k=1,2,...,n.
Thus, we have
(20)
R2n+1(0,q)=1>0,
R2n+1(y, q)<0for
y2k−1
q<y< y2k
q,k=1,2,...,n,
R2n+1(y, q)>0forqy2k−1<y<qy
2k,k=1,2,...,n,
R2n+1(∞,q)=−∞.
Since by our assumption zeros of R2n(y, q)satisfy(7)wehave
y2k+1
y2k−1
=y2k+1
y2k·y2k
y2k−1
>q
2,k=1,2,...,n−1,
whence y2k−1
q<qy
2k−1<y2k+1
q,k=1,2,...,n−1,
10 O. M. Katkova and A. M. Vishnyakova CMFT
or,
y1
q<y2
q<qy
1<qy
2<y3
q<y4
q<qy
3<qy
4
< ... < y2n−1
q<y2n
q<qy
2n−1<qy
2n.
Denote by u1<u
2<... <u
2n<u
2n+1 the zeros of R2n+1(y, q). Then
(21) u2k−1<y2k−1
q<y2k
q<u
2k<qy
2k−1<qy
2k<u
2k+1,k=1,2,...,n.
Thus, u2k/u2k−1>y
2k/y2k−1>qand u2k+1/u2k>y
2k/y2k−1>q.
This proves statement (i) of Lemma 1.
Let us now prove statement (ii). We recall that Sn(x, q)=Rn(x/qn,q). So, by
Theorem G (v) all zeros of R2n(y, q) lie in the interval (q−2n+1,q
2n−1). We have
R2n+2(y, q)=
2n
k=0
(−1)kqk(2n+2−k)yk−q2n+1y2n+1 +y2n+2
(22)
=
2n
k=0
(−1)kqk(2n−k)(q2y)k−y2n+1(q2n+1 −y)
=R2n(q2y, q)−y2n+1(q2n+1 −y),
R2n+2(y, q)=1−
2n+1
k=1
(−1)k−1qk(2n+2−k)yk+y2n+2
(23)
=1−
2n
k=0
(−1)kq(k+1)(2n+1−k)yk+1 +y2n+2
=(1+y2n+2)−q2n+1y
2n
k=0
(−1)kqk(2n−k)yk
=(1+y2n+2)−q2n+1yR2n(y, q).
By (22) we have R2n+2(y, q)<0fory2k−1/q2<y<y
2k/q2,k=1,2,...,n (since
y2n<q
2n−1if and only if y2n/q2<q
2n−3, the last summand in (22) is negative).
By (23) we have R2n+2(y, q)>0fory2k−1<y<y
2k,k=1,2,...,n,and
R2n+2(0,q)=1>0.
We recall that all zeros of R2n(y, q) lie in the interval (q−2n+1,q
2n−1).
By (7) we have y2k+1/y2k−1>q
2and thus
0<y1
q2<y2
q2<y
1<y
2<y3
q2<y4
q2<y
3<y
4
< ... < y2n−1
q2<y2n
q2<y
2n−1<y
2n<q
2n−1.
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>111
Denote by v1<v
2< ... < v
2n+2 the zeros of R2n+2(y, q). By Theorem G (vi)
there are two zeros of R2n+2(y, q) in the interval (q2n−1,q
2n+1). Thus we obtain
v2k−1<y2k−1
q2<y2k
q2<v
2k<y
2k−1<y
2k<v
2k+1,k=1,2,...,n,
whence v2k
v2k−1
>y2k
y2k−1
>q, v2k+1
v2k
>y2k
y2k−1
>q, k=1,2,...,n.
It remains to prove that v2n+2/v2n+1 >q.Notethat
R2n+2(y, q)=y2n+2R2n+2 (1
y,q),
so v2n+2 =1/v1,v2n+1 =1/v2.Thusv2n+2/v2n+1 =v2/v1>q.
This proves Statement (ii) and concludes the proof of Lemma 1.
Lemma 2. Suppose for some q>1all zeros of R2n+1(y, q)are real and these
zeros satisfy (7).ThenallzerosofR2n−1(y, q)are real and these zeros satisfy
(7).
Proof. That the zeros are real is a consequence of Theorem G (i). We have
R2n+1(y, q)=
2n−1
k=0
(−1)kqk(2n+1−k)yk+q2ny2n−y2n+1
(24)
=
2n−1
k=0
(−1)kqk(2n−1−k)(q2y)k+q2ny2n−y2n+1
=R2n−1(q2y, q)+q2ny2n−y2n+1,
or
(25) R2n−1(y, q)=R2n+1(y
q2,q)−y2n
q2n+y2n+1
q2(2n+1) .
Also we have
(26) R2n+1(y, q)=−q2nyR2n−1(y, q)+1−y2n+1,
or
(27) q2nyR2n−1(y, q)=−R2n+1(y, q)+1−y2n+1.
Since
(28) R2n−1(y, q)=−y2n−1R2n−1(1
y,q),
it is sufficient to prove the statement of Lemma 2 for such zeros of R2n−1(y, q)
that lie in the segment [0,1]. Denote by y1<y
2< ... < y
n<1 the first n+1
zeros of the polynomial R2n+1(y, q). Obviously, R2k+1(1,q) = 0 for every k∈N.
By (25) we have
(29) R2n−1(y, q)<0fory2k−1q2<y<y
2kq2,k=1,2,...,n+1
2.
12 O. M. Katkova and A. M. Vishnyakova CMFT
For y≤1 by (25) we have
(30) R2n−1(y, q)>0,for y2k−1<y<y
2k,k=1,2,...,n+1
2.
Since S2n+1(x, q)=R2n+1(x
q2n+1 ,q)andS2n+1(x, q) has exactly two zeros in the
interval (q, q3)wederivethatR2n+1 (y, q) has exactly two zeros in the inter-
val (1/q2n,1/q2n−2). Moreover, S2n−1(x, q) does not have zeros in the interval
(0,q), so R2n−1(y, q) does not have zeros in the interval (0,1/q2n−2). Conse-
quently, [y1,y
2]⊂(0,1/q2n−2), and R2n−1(y, q)>0fory∈(0,1/q2n−2). Since
y2k+1/y2k−1>q
2,weobtain
y1<y
2<q
2y1<q
2y2<y
3<y
4<q
2y3<q
2y4
< ...< y
2(n+1)/2−1<y
2(n+1)/2<q
2y2(n+1)/2−1<q
2y2(n+1)/2.
Suppose n=2m.Denotebyw1,w
2,...,w
4m−1the zeros of R2n−1(we will
consider the first 2mzeros of R2n+1 situated in the segment [0,1], w2m=1).
Note that 1/y2m=y2m+1/y2m>qand 1/y2m−1=(y2m+1/y2m)·(y2m/y2m−1)>q
2.
Using (29) and (30) we derive
(31) y1<y
2<w
1<q
2y1<q
2y2<w
2<y
3<y
4<w
3<q
2y3<q
2y4
< ... <w
2m−1<q
2y2m−1<q
2y2m<y
2m+1 =w2m=1.
As in the proof of Lemma 1 we conclude that wk+1/wk>q,k=1,2,...,2m−1,
whence wk+1/wk>q,k=1,2,...,4m−2.
Suppose n=2m+1. Denote by t1,t
2,...,t
4m+1 the zeros of R2n−1(we will
consider the first 2m+ 1 zeros of R2n−1situated in the segment [0,1], t2m+1 =1).
Using (29) and (30) we derive
(32)
y1<y
2<t
1<q
2y1<q
2y2<t
2<y
3<y
4<t
3<q
2y3<q
2y4
< ... <t
2m−1<q
2y2m−1<q
2y2m<t
2m<y
2m+1 <y
2m+2 =t2m+1 =1.
As in the proof of Lemma 1 we conclude that tk+1/tk>q,k=1,2,...,2m,
whence wk+1/wk>q,k=1,2,...,4m.
This completes the proof of Lemma 2.
Suppose all zeros of ga(z) are real and simple (according to Theorem G (viii)
this means that a>r
∞). Denote by 0 <x
1<x
2< ... the zeros of ga(z).
Lemma 3. xn+1/xn>x
2/x1,n=2,3,4,....
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>113
Proof. Note that
ga(x)=1−x
a+···+(−1)n−2xn−2
a(n−2)2
(33)
+(−1)n−1xn−1
a(n−1)21−x
a2n−1+···+(−1)k−n+1 xk−n+1
ak2−(n−1)2+···
=Sn−2(x, a)+(−1)n−1xn−1
a(n−1)2ga(x
a2n−2),n=2,3,....
Let x∈(x1a2n−2,x
2a2n−2). Then ga(x/a2n−2)<0. Hence,
(34) sign(−1)n−1xn−1
a(n−1)2ga(x
a2n−2)=(−1)n−2,x∈(x1a2n−2,x
2a2n−2).
By Theorem G (ix), a<x
1<x
2<a
3,sox1a2n−2≥a2n−1>a
2n−5and by
Theorem G (v)
(35) (−1)n−2Sn−2(x, a)>0,x∈(x1a2n−2,x
2a2n−2).
By (33), (34) and (35) we obtain that
(36) (−1)n−2ga(x)>0,x∈(x1a2n−2,x
2a2n−2),n=2,3,....
Thus we have 2nintervals
(0,x
1),(x1,x
2),(x1a2,x
2a2),(x1a4,x
2a4), ..., (x1a4n−4,x
2a4n−4)
contained in the interval (0,a
4n−1), and ga(x) is alternating in sign on these
intervals. Using Theorem G (x) we deduce that
xj<x
1a2j−2<x
2a2j−2<x
j+1,j=2,3,...,2n−1.
Since n≥2 is arbitrary, we have
(37) xj<x
1a2j−2<x
2a2j−2<x
j+1,j≥2.
Thus, xn+1/xn>x
2/x1,n≥2.
This completes the proof of Lemma 3.
We recall that, by (13), Sn(x, a) has all zeros in the right half plane if and only
if a>s
n, and that all zeros of S2n+1(x, a) lie in the right half plane if and only
if all zeros of Rn(y, a4) are real and these zeros satisfy (7). The statement of
Lemma1isequivalentto
(38) s4n+3 ≤s4n+1,s
4n+5 ≤s4n+1,n=1,2,....
The second of these inequalities gives
(39) 1.2106 ≈s5≥s9≥s13 ≥....
Thus there exists a limit limn→∞ s4n+1 and
(40) B:= lim
n→∞ s4n+1.
14 O. M. Katkova and A. M. Vishnyakova CMFT
The statement of Lemma 2 is equivalent to
(41) s4n+3 ≥s4n−1,n=1,2,...,
or,
(42) 1 = s3≤s7≤s11 ≤....
Thus there exists a limit limn→∞ s4n+3 and
(43) C:= lim
n→∞ s4n+3.
By (38) we have
(44) B≥C.
It is well-known (see [12, Ch. 7]) that the Hermite-Biehler Criterion is valid not
only for polynomials, but for some classes of entire functions. The following
statement is a generalization of the Hermite-Biehler Theorem on entire functions
of order not greater then 1 and minimal type of growth.
Theorem H (B. Ja. Levin).Let G(z)=∞
k=0 akzk,ak>0,beanentirefunc-
tion of order not greater then 1and minimal type of growth. All zeros of the
entire function Ghave negative real parts if and only if the following two entire
functions f(z)=∞
0(−1)ma2mzmand g(z)=z∞
0(−1)ma2m+1zmhave simple
real interlacing zeros.
It is well-known that the order of an entire function with coefficients akis given
by the expression
lim sup
n→∞
nlog n
log |an|−1
(see, for example [12, Ch. 1]). So, the entire function gais of order 0 and thus
the Hermite-Biehler Criterion is valid for this function. It is easy to show that
(analogously to the polynomial case) if for some a0>1 all the zeros of the entire
function ga0are situated in the right half-plane then for all a>a
0all the zeros
of the entire function gaare situated in the right half-plane. Thus
(45) ∃s∞>1: (the zeros of gaare in the right half-plane ⇔a>s
∞).
Let us prove that s∞≤C. Note that limn→∞ S4n+3(x, a)=ga(x), and this limit
is uniform on the compact sets. This implies that for all a≥Call the zeros of ga
lie in the closed half-plane {z:Rez≥0}. Suppose that C<s
∞. Then for all
a∈(C;s∞] the function gahas all zeros in the closed right half-plane, but not in
the open right half-plane. It means that for all a∈(C;s∞] the function gahas
an imaginary zero ix0(a),x
0(a)∈R.Then
1−x0(a)2
a4+x0(a)4
a16 −x0(a)6
a36 +···=0
and
1−x0(a)2
a8+x0(a)4
a24 −x0(a)6
a48 +···=0.
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>115
Denote by t0(a)=x0(a)2,q=a4.Then
(46) gq(t0(a)) = 1 −t0(a)
q+t0(a)2
q4−t0(a)3
q9+···=0
and
(47) gq(t0(a)
q)=1−t0(a)
q2+t0(a)2
q6−t0(a)3
q12 +···=0.
Let us denote by x1(q)<x
2(q)<x
3(q)< ... the zeros of the function gq(t). By
(46) and (47) there exists j∈Nsuch that t0(a)/q =xj(q), t0(a)=xj+1.Sofor
all q∈(C4;s4
∞] there exists j∈Nsuch that
(48) xj+1(q)
xj(q)=q.
Since a>Call the zeros of the polynomial S4n+3(x, a) lie in the open right
half-plane. It means that for q=a4all the zeros of the polynomial
P2n+1(t, q)=1−t
q+t2
q4−···+(−1)2n+1 t2n+1
q(2n+1)2
denoted by t1(q, 2n+1),t
2(q, 2n+1),...,t
2n+1(q, 2n+ 1) are real and distinct.
Moreover, as we have shown, these zeros satisfy the condition
tj+1(q, 2n+1)
tj(q, 2n+1) >q for n∈N,j=1,2,...,2n, q =a4>C
4.
Since limn→∞ tj(q, 2n+1)=xj(q)weobtainxj+1(q)/xj(q)≥q,q>C
4.From
this and (48) we obtain using Lemma 3 that x2(q)/x1(q)=q,q∈(C4;s4
∞], or
(49) x2(q)−qx1(q)=0,q∈(C4;s4
∞].
Let us show that (49) is impossible. We will use the following variant of the
well-known theorem on the regularity of implicit functions.
Theorem I ([7, Ch. 2]).Let a complex function F(q, z)of two complex variables
be jointly continuous in a neighborhood of a point (q0,z
0)and be regular in each
variable in the same neighborhood. If F(q0,z
0)=0and F
z(q0,z
0)=0then
there exist positive numbers ε, δ such that for every q,|q−q0|<ε, the equation
F(q, z)=0has a unique solution z=ϕ(q)under the condition |z−z0|<δ.And
this function z=ϕ(q)is regular at the point q0.
By (14) and by the fifth and the seventh inequalities of (38) we conclude that
1.16 <s
∞<1.21. So s4
∞>r
∞,wherer∞is the constant from Theorem G. Thus
by Theorem G (viii) the function gqdoes not have multiple zeros for q≥s∞.
Hence we can apply Theorem I to the function gqat the point (s4
∞,x
1(s4
∞)) and
at the point (s4
∞,x
2(s4
∞)). So there exist positive numbers ε,δsuch that for
every q,|q−s4
∞|<ε, the equation gq(t) = 0 has the unique solution x1=x1(q)
satisfying the condition |x1(q)−x1(s4
∞)|<δand the equation gq(t)=0has
the unique solution x2=x2(q) satisfying the condition |x2(q)−x2(s4
∞)|<δ.
16 O. M. Katkova and A. M. Vishnyakova CMFT
Moreover both functions x1=x1(q)andx2=x2(q) are regular in some circle
Uρ(s4
∞)={q:|q−s4
∞|<ρ},ρ>0. Whence the function x2(q)−qx1(q)isregular
in the circle Uρ(s4
∞). By (49) we have x2(q)−qx1(q)=0forq∈(C4;s4
∞]∩Uρ(s4
∞).
By the Uniqueness Theorem it means that x2(q)−qx1(q)≡0forq∈Uρ(s4
∞),
but this conclusion contradicts the condition x2(q)/x1(q)>qfor q>s
4
∞.This
contradiction shows that s∞≤C. Combining with (44) we have
(50) s∞≤C≤B.
It remains to prove that s∞≥B. Suppose that s∞<B.Letuschoose
a, s∞<a<B, and denote by q=a4. By Levin’s Theorem I all the zeros of
gq(t)=1−t
q+t2
q4−t3
q9+···
and
gq(t
q)=1−t
q2+t2
q6−t3
q12 +···
are real and interlacing. If x1<x
2<x
3< ... be the real zeros of gq(t), then, as
we have shown, there exists ε>0 such that
(51) xj+1
xj
>q+εfor all j=1,2,3,....
We have
˜
P2n(t, q)=1−t
q+t2
q4−···+t2n
q(2n)2=gq(t)+ t2n+1
q(2n+1)2−t2n+2
q(2n+2)2+···.
Hence, if x2k≤t≤x2k+1,k=1,2,...,n−1, then ˜
P2n(t, q)>0. If we denote by
tk(2n), 1 ≤k≤2n, the zeros of ˜
P2n(t, q), then t2k(2n)<x
2k<x
2k+1 <t
2k+1(2n),
i.e.
(52) t2k+1(2n)
t2k(2n)>q+ε, k =1,2,3,...,n−1.
Since ˜
P2n(t, q)⇒gq(t), as n→∞, on the compact sets, we have t1(2n)→x1,
t2(2n)→x2. Hence,
(53) ∃n0∀n≥n0:t2(2n)
t1(2n)≥q+ε
2.
Lemma 4. The inequality
t2k(4n−4)
t2k−1(4n−4) >t2(2n)
t1(2n)
holds for n=3,4,5,... and k=1,2,...,2n−2.
Proof. We have
˜
P2n+2(t, q)= ˜
P2n(t, q)−t2n+1
q(2n+1)2+t2n+2
q(2n+2)2
X (20XX), No. X On the Stability of Taylor Sections of a Function P∞
k=0 zk/ak2,a>117
and
˜
P2n+2(t, q)=1−t
q+t2
q4˜
P2n(t
q4,q).
Therefore, if t1(2n)<t<t
2(2n)ort1(2n)q4<t<t
2(2n)q4then we have
˜
P2n+2(t, q)<0. Whence
(54) t2(2n+2)
t1(2n+2) >t2(2n)
t1(2n),t4(2n+2)
t3(2n+2) >t2(2n)
t1(2n).
Applying this reasoning to ˜
P2n+4(t, q) we obtain from (54)
(55) t2(2n+4)
t1(2n+4) >t2(2n)
t1(2n),t4(2n+4)
t3(2n+4) >t2(2n)
t1(2n),t6(2n+4)
t5(2n+4) >t2(2n)
t1(2n),
and so on. For ˜
P4n−4(t, q)wehave
(56) t2k(4n−4)
t2k−1(4n−4) >t2(2n)
t1(2n),k=1,2,...,n−1.
Note that
˜
P4n−4(t, q)= t4n−4
q(4n−4)2˜
P4n−4(q4n−6
t,q).
Therefore,
tj(4n−4) = q4n−6
t4n−4−j+1(4n−4) =q4n−6
t4n−j−3(4n−4).
Whence using (56) we obtain
t2k(4n−4)
t2k−1(4n−4) =t4n−2k−2(4n−4)
t4n−2k−3(4n−4) >t2(2n)
t1(2n),2n−k−1=1,2,...,n−1,
or
t2k(4n−4)
t2k−1(4n−4) =t4n−2k−2(4n−4)
t4n−2k−3(4n−4) >t2(2n)
t1(2n),k=n, n +1,...,2n−2.
Taking into account (56) we obtain the statement of the lemma.
Letusfixanintegerm≥n0in (53). By Lemma 4 we have
t2k(4m−4)
t2k−1(4m−4) >q+ε
2,k=1,2,...,2m−2.
Using (52) we obtain
t2k+1(4m−4)
t2k(4m−4) ≥q+ε>q+ε
2,k=1,2,...,2m−3.
Hence for all zeros of ˜
P4nm−4(t, q) the inequality
tj+1(4m−4)
tj(4m−4) >q+ε
2>q, j=1,2,...,4m−5
18 O. M. Katkova and A. M. Vishnyakova CMFT
holds. This means that all the zeros of S8m−7(x, a) lie in the open right half-
plane for s∞<a<B and for all m≥n0. This contradicts the definition of the
constant B(see (40)). Thus,
(57) s∞≥B.
Combining with (50) we finally get B=C=s∞.
This completes the proof of Theorem 1.
Acknowledgement. The authors are deeply grateful to the referee for valuable
remarks and suggestions.
References
1. T. Ando, Totally positive matrices, Linear Algebra Appl. 90 (1987), 165–219.
2. M. Biehler, Sur une classe d’´equations alg´ebriques dont toutes les racines sont r´eelles, J.
Reine Angew. Math. 87 (1879), 350–352.
3. T. Craven and G. Csordas, A sufficient condition for strict total positivity of a matrix,
Linear Multilinear Algebra 45 (1998), 19–34.
4. D. K. Dimitrov and J. M. Pe˜na, Almost strict total positivity and a class of Hurwitz
polynomials, J. Approx. Theory 132 (2005), 212–223.
5. F. R. Gantmacher, The Theory of Matrices, Vol. II, Chelsea Publ., New York, 1959.
6. C. Hermite, Sur les nombre des racines d’une ´equation alg´ebrique comprise entre des limites
donn´ees, J. Reine Angew. Math. 52 (1856), 39–51.
7. A. Hurwitz and R. Courant, Theory of Functions, Springer Verlag, Berlin 1925 (in Ger-
man).
8. S. Karlin, Total Positivity, Vol. I, Stanford University Press, California 1968.
9. O. M. Katkova, T. Lobova and A. M. Vishnyakova, On power series having sections with
only real zeros, Comput. Methods Funct. Theory 3no.2 (2003), 425–441.
10. O. M. Katkova and A. M. Vishnyakova, On sufficient conditions for the total positivity
and for the multiple positivity of matrices, Linear Algebra Appl. 416 (2006), 1083–1097.
11. , A sufficient condition for a polynomial to be stable, submitted.
12. B. Ja. Levin, Distribution of Zeros of Entire Functions, Transl. Math. Mono., 5, Amer.
Math. Soc., Providence, RI, 1964; revised ed. 1980.
13. M. Marden, Geometry of Polynomials,Amer.Math.Soc.Surveys,Vol.3,Providence,RI,
1966.
14. G. P´olya, ¨
Uber einen Satz von Laguerre, Jber. Deutsch. Math.-Verein. 38 (1929), 161–168.
15. ,Collected Papers,Vol.II,Location of Zeros, R. P. Boas (ed.), MIT Press, Cam-
bridge, MA, 1974.
16. G. P´olya and J. Schur, ¨
Uber zwei Arten von Faktorenfolgen in der Theorie der algebraischen
Gleichungen, J. Reine Andrew. Math. 144 (1914), 89–113.
17. M. M. Postnikov, Stable Polynomials, Moscow, Nauka, 1981 (in Russian).
Olga M. Katkova E-mail:olga.m.katkova@univer.kharkov.ua
Address:Kharkov National University, Dept. of Math., Svobody sq., 4, 61077, Kharkov,
Ukraine.
Anna M. Vishnyakova E-mail:anna.m.vishnyakova@univer.kharkov.ua
Address:Kharkov National University, Dept. of Math., Svobody sq., 4, 61077, Kharkov,
Ukraine.