POSITIVE COEFFICIENTS AND A THEOREM OF PO!LYA
I. V. OSTROVSKII N. A. ZHELTUKHINA 0. Introduction Let f(z)l _ k=! akzk a!0 (0.1)be a formal power series. In 1913, G. Po! lya [7] proved that if, for all sufficiently large
n, the sections
fn(z)l n
k=!
akzk (0.2)
have real negative zeros only, then the series (0.1) converges in the whole complex plane C, and its sum f(z) is an entire function of order 0. Since then, formal power series with restrictions on zeros of their sections have been deeply investigated by several mathematicians. We cannot present an exhaustive bibliography here, and restrict ourselves to the references [1, 2, 3], where the reader can find detailed information.
In this paper, we propose a different kind of generalisation of Po! lya’s theorem. It is based on the concept of multiple positivity introduced by M. Fekete in 1912, and it has been treated in detail by S. Karlin [4].
Recall that the sequenceoakq_k=
! of real numbers is said to be m-times positive for m? NDo_q if all minors of orders less than mj1 of the infinite matrix
I
J
0 0 a! 0 : 0 a! a" 0 : a! a" a# 0 : a" a# a$ a! : ( ( ( ( (K
L
are non-negative. Usually, _-times positive sequences are called totally positive sequences.
The Aissen–Edrei–Schoenberg–Whitney theorem (see [4, p. 412]) gives an exhaustive characterisation of totally positive sequences. In particular, this theorem yields the fact that the entire function (0.1) of genus 0 has purely negative zeros if and only if the sequenceoakq_k=!is totally positive. Applying this to the polynomial (0.2), we see that negativity of all its zeros is equivalent to total positivity of the sequence oakqn
k=! B oa!,a",…,an, 0, 0, 0, …q. Thus the condition of Po!lya’s theorem
is equivalent to total positivity of the truncated sequencesoakqn
k=!for all sufficiently
large n.
Received 30 June 1995 ; revised 30 January 1996. 1991 Mathematics Subject Classification 30B10.
The following problem seems to be of interest. Does the assertion of Po! lya’s theorem remain in force if we replace total positivity of the truncated sequences oakqn
k=!by a slower condition of m-times positivity for some m _? It is easy to see
that the answer is negative for ml 1 and m l 2. For example, if akl 1 for any k l 0, 1, 2, … , all truncated sequencesoakqn
k=!are 2-times positive, but the series in (0.1)
does not converge in the whole plane C. The main aim of this paper is to show that the answer is positive for m 3. We shall consider some related results and problems.
Note that the m-times positivity of the truncated sequenceoakqn
k=!for m _ is not
too closely connected with zeros of the corresponding polynomial (0.2). I. J. Schoenberg (see [9] and [4, pp. 397, 415]) proved that, for oakqn
k=!, n 2, to be
m-times positive, the necessary condition is the non-vanishing of fn(z) in the angleoz: Qarg zQ πm\(mjnk1)q, but the sufficient condition is its non-vanishing in the greater angleoz : Qarg zQ πm\(mj1)q. Both of the conditions are unimprovable in the sense of the sizes of angles. Note that, for any m? N, zero-sets of all entire transcendental functions (0.1) with m-times positive sequencesoakq_k=
!form a rather wide class that is described in [5].
1. Statement of results
Denote by Pmthe class of all formal power series (0.1) such that the truncated sequencesoakqn
k=!are m-times positive for all sufficiently large n. Evidently, Pmh9 Pm
for mh m. For any entire function g(z), put M(r, g) l max oQg(z)Q : QzQ rq.
T 1. If a formal power series (0.1) belongs to Pmfor some m 3, then it conerges in the whole complex plane C, and its sum f(z) is an entire function of order
0. Moreoer, lim sup r _ log M(r, f ) (log r)# 1 2 log c cl 1jN5 2 . (1.1)
The bound (1.1) cannot be improved for ml 3.
T 2. There exist entire functions f(z) ? P$ such that lim r _ log M(r, f ) (log r)# l 1 2 log c.
The question arises of whether the bound (1.1) is unimprovable for m 4. The consideration of the proof of Po! lya’s theorem in [7] shows that, in fact, Po!lya obtained the following result.
T (Po!lya). If a formal power series (0.1) belongs to P_, then it conerges
in the whole plane C, and its sum f(z) is an entire function that satisfies the inequality lim sup r _ log M(r, f ) (log r)# 1 2 log 2. (1.2)
The inequality (1.2) is stronger than (1.1), but we are sure that it is not the best possible even for f(z)? P%. The problem of finding the best possible bound remains open for any m 4. In this connection, it is worth mentioning that there are entire functions f(z)? P_such that
lim r _ log M(r, f ) (log r)# l 1 4 log 2.
Such a function,
f(z)l
_
k=!
2−k#zk,
was considered in [8, Problem 176, p. 66]. Hence the unknown best possible bound (probably depending on m) is not less than 1\(4 log 2).
Denote by Qmthe subclass of Pmconsisting of formal power series (0.1) satisfying the following condition : for all nl 0, 1, 2,…, the truncated sequences oakqn
k=!are
m-times positive. For f(z)? Qm, the bound (1.1) can be improved in the following way. T 3. If a formal power series (0.1) belongs to Qmfor some m 3, then the following refinement of(1.1) isalid:
M(r, f ) a!"$
0
0, 1 Nc1
exp(
(log (rNca"\a!))#
2 log c
*
, (1.3)where"$ denotes the Jacobi theta-function ([10, 21.11, p. 464]). Under the normalisation condition a!la"l1, (1.3) takes a simpler form:
M(r, f ) "$
0
0, 1Nc
1
c"/) Nr exp(
(log r)#2 log c
*
. (1.4)C 1. If m is larger than or equal to 3, then the set of all entire functions
f(z)? Qmwith fixed coefficients a!0, a" is a normal family.
The inequalities (1.3) and (1.4) are not sharp, at least for small r, since their right-hand sides tend toj_ as r tends to 0. Now we are going to obtain a bound which is more complicated, but the best possible for all r 0 and all f(z) ? Q$.
Letozkq_k=
#be the sequence of positive numbers defined by the recurrence equation
z#k+
" lzkj1 k l 2, 3,… (1.5)
and by the initial condition
z#l1. (1.6)
Define the following :
dkl 1j1 zk kl 2, 3,… (1.7) (z)l 1jzj _ k=# zk dk−" # d#k−#… dk . (1.8)
T 4. If a formal power series (0.1) belongs to Qmfor some m 3, then the following inequality isalid:
M(r, f ) a! (a"r\a!) r0. (1.9) A. Edrei [1] proved that, if each fn(z) does not vanish in some half-plane (possibly, depending on n), then the series (0.1) converges in the whole plane and its sum satisfies the condition
T. Ganelius [3] proved that (1.10) remains in force if the half-plane is replaced by any angle of positive size not depending on n. As a corollary of Theorem 1, we obtain the following Theorem 5, which sharpens the bound (1.10) under some additional conditions.
T 5. Let f(z) be a formal power series of the form (0.1) with real coefficients
ak. Assume that, for sufficiently large n, the zeros of the sections (0.2) are located in the
angleoz: Qarg zkπQ αq. Then the series conerges in the whole plane C, and, moreoer,
(i) ifα is smaller than or equal to π\4, then (1.1) is alid; (ii) ifα is smaller than or equal to π\2, then
lim sup r _ log M(r, f ) (log r)# 2 log 2 isalid. 2. Proof of Theorem 1
Henceforth, we assume that the series (0.1) contains infinitely many non-zero terms ; otherwise, the statement of Theorem 1 is trivial. Under this assumption, the sequence oakq_k=! cannot contain zero terms at all, as the following (known) Lemma 1 shows.
L 1. Let oakq_k=!, a!0, be a 2-times positie sequence. Set nlminok :
akl 0q. If n is finite, then akl 0 for any k n.
Proof. By the definition of 2-times positivity, we have, for any k n,
)
an an− " ak ak− ")
0. Since anl 0, an− " 0, we conclude that akl 0.Evidently, 3-times positivity of truncated sequencesoakqn
k=!for sufficiently large n
yields 3-times positivity of the whole sequenceoakq_k=!. Hence, Lemma 1 is applicable, and all the akare strictly positive. This allows us to introduce the positive numbers
ρklak−" ak kl 1, 2,… (2.1) It is evidence that akl a! k j=" ρj kl 1, 2,… (2.2)
The following (known) Lemma 2 shows that the numbersρkform a non-decreasing sequence.
L 2. Let oakq_k=!be a2-times positie sequence without zero terms. Then the
sequenceoρkq_k="defined by(2.1) is non-decreasing.
Proof. The inequality
)
ak ak− " ak+ " ak)
0 is equivalent toρk ρk+ ".Define the numbers as follows : δkl ρk ρk−" kl 2, 3,… Evidently, ρkl ρ" k j=# δj kl 2, 3,… (2.3) By Lemma 2, we have δk 1 k l 2, 3,… Using (2.2), we obtain akl a! ρk " δk−# δ" k−$ #…δk kl 1, 2, 3,… (2.4)
The following Lemma 3 plays a basic role in the proof of Theorem 1. L 3. Let oakqn
k=! l oa!,a",…,an, 0, 0, …q, a!0, an 0, n 2, be a 3-times
positie sequence. Then
(i) for nl 2, we hae δ#2; (ii) for n 3, we hae δn 1 and
(δnk1)# 1k 1 δn−". (2.5) Proof. If nl 2, we have a! a" 0 a" a# a! a# 0 a" 0. Using this and (2.4), we obtainδ#2.
If n 3, we have a! a" 0 an− " an an− # an 0 an− " 0. Calculating the determinant, we obtain
a"a#n−"ka!an−"anka"an−#anl a"an−#an
0
a#n−" an−#anka!a n−" a"an−# k1
1
0. Since (2.4) yields a#n−" an−#anl δn a!an−" a"an−# l 1 δ#…δn−" , we get δn 1j 1 δ#…δn−" 1. Further, we have an− # an− " an− $ an− " an an− # an 0 an− " 0.Calculating the determinant, we obtain a$n−"jan−$a#nk2an−#an−"anl an−$a#n
0
a$n−" an−$a#nj1k2 an−#an−" an−$an1
0. Since (2.4) yields a$n− " an− $a#n l δ#nδn−" an−#an−" an− $an l δnδn−", we get δ#nδn−"j1k2δnδn− " 0. This inequality is equivalent to (2.5).If a formal power series (0.1) satisfies the condition of Theorem 1, then there exists some n!2 such that, for each nn!, the truncated sequence oakqnk=! is 3-times
positive. By Lemma 3, we haveδn 1 for n n!, and therefore the numbers
ynB 1
δnk1 n n! (2.6)
are well defined. Using (2.5), we obtain
y#n+" ynj1 n n!. (2.7) Consider the sequence oznq_n=n
! of positive numbers satisfying the recurrence
equation
z#n+" lznj1 n n! (2.8) and the initial condition
zn
!l yn!. (2.9)
It is easy to see that
yn zn n n!. (2.10)
L 4. There exists the limit lim n _ znl1jN5 2 l c. Moreoer, (i) if zn
!is smaller than c, then the sequenceoznq_n=n!increases ;
(ii) if zn
!is larger than c, then the sequenceoznq_n=n!decreases ;
(iii) if zn
!l c, then znl c for any n n!.
Proof. The proof of Lemma 4 is based on the fact that c is a root of the equation
z# l zj1, and z# zj1 for 0 z c and z# zj1 for z c. The details can be
omitted.
Using (2.6), (2.10) and Lemma 4, we obtain lim inf n _ δnl lim inf n _
0
1j1 yn1
limn _0
1j 1 zn1
l 1j 1 cl c. If 0 ε ck1, then we have δn ckε n q l q(ε). (2.11)By (2.4), we obtain, for k q, log akl log a!kklogρ"kk
j=#
(kkjj1) log δj log a!kklogρ"kj=q#(kkjj1) log δjk k
j=q+" (kkjj1) log (ckε) lkk# 2log (ckε)jO(k) k _. (2.12) Hence, ak CDk(ckε)−k#/# k l 0, 1, 2,…, (2.13)
where C and D are positive constants not depending on k. Since ckε is larger than 1, (2.13) yields the fact that limk _a"/k
k l 0. Hence the series (0.1) converges in the
whole plane.
Further, using (2.13), we have
M(r, f )l f(r) l _ k=! akrk C _ k=! (ckε)−k#/#(Dr)k
l C exp
0
2 log (c(log Dr)kε)#1
_k=! exp
(
klog (ckε) 20
kk log (Dr) log (ckε)1
#*
Cexp0
(log Dr)# 2 log (ckε)1
−_supx _ _ k=−_ exp(
klog (ckε) 2 (kkx)#*
.Since the sum of the series under the supremum sign is a periodic function of x (with period 1), its supremum is finite. Hence
M(r, f )l O
0
exp(
(log Dr)#2 log (ckε)
*1
r _ andlog M(r, f ) (log r)#
2 log (ckε)jO (log r) r _. Sinceε 0 is arbitrary, we obtain the inequality (1.1).
Proof of PoT lya’s theorem. Now we present the proof of Po!lya’s theorem for the
reader’s convenience. The condition f(z)? P_ means (see the introduction) that the sections (0.2) have purely negative zeros for sufficiently large n, that is, for n n" say. It is well known that derivatives of a polynomial having purely real zeros have purely real zeros. Hence, the quadratic polynomial
0
d dz1
n−#
fn(z)l (nk2)!an−
#j(nk1)!an−"zj"#n!anz#
does not have any complex zero. This means that the following inequality is valid : (nk1)a#n−" 2nan−#an n n".
Remembering the definition ofδn, we can rewrite the last inequality in the form δn 2n
Therefore (2.11) is valid with 2 in place of ckε. This yields (2.12) and (2.13) with 2 instead of ckε.
3. Proofs of Theorems 3 and 4
Proof of Theorem3. If f(z) belongs to Q$, then, by Lemma 3(i), we have δ# 2, and, moreover, we can take n!l2 in (2.7)–(2.10). Since z#ly#1 c, Lemma 4 yields the fact that the sequenceoznq_n=
#is increasing and znis smaller than c for n
2. Hence yn c, and
δnl 1j1
yn 1j
1
cl c n 2. (3.1)
Using this, we can improve (2.12) in the following way : log akl log a!kklogρ"kk
j=#
(kkjj1) log δj
log a!kklogρ"kk(kk1)2 log c k 2. (3.2) We obtain the following refinement of (2.13) :
ak a!(Nc\ρ")kc−k#/# k l 0, 1, 2,… Hence M(r, f ) a! _ k=! c−k#/#(rNc\ρ")k
l a!exp
(
(log (r2 log cNc\ρ"))#*
_ k=! exp
(
klog c 20
kk log (rNc\ρ") log c1
#*
a!exp(
(log (r2 log cNc\ρ"))#*
−_supx __
k=−_
exp
(
klog c2 (kkx)#
*
.By a well known formula of the theory of theta-functions [10, 21.51, p. 476], we have, for anyα 0, _ k=−_ exp
0
kπ#k# α j2π kix1
l’
πα _ k=−_ exp (kα(kkx)#).Hence the sum of the series in the right-hand side attains its maximal value when
xl 0, and we obtain sup −_ x _ _ k=−_ exp
0
klog c 2 (kkx)#1
l _ k=−_ exp0
klog c 2 k#1
l "$0
0, 1 Nc1
. R 1. If f(z) belongs to Q_, then (1.3) and (1.4) can be improved by the replacement of c by 2.Indeed, the condition f(z)? Q_yields the fact that the inequality (2.14) is valid for any n 2. Using this inequality instead of (3.1), we obtain the claimed result.
Proof of Theorem4. By Lemma 3, we haveδ#2, and the numbers ynare well defined by (2.6) for all n 2. In particular, we have
By (2.10) with n!l2, we have yk zkfor any n 2. Therefore,
δkl 1j1
yk 1j
1
zkl dk k 2.
Using (2.4) and (1.8), we obtain
M(r, f )l a!ja"rj _ k=# a!rk ρk " δ#k−"…δk a!ja"rjk=_ # a!rk ρk "d#k−"… dkl a!
0
a" a!r1
.R 2. If f(z) belongs to Q_, then the inequality (1.9) can be replaced by the following more precise
M(r, f ) a!ψ(a"r\a!), (3.3) where ψ(z) l 1jzj _ k=# zk 2k(k−")/# k!. (3.4)
Indeed, since (2.14) is valid for any n 2, we have ρnl δnρn−" 2n
nk1ρn−" … 2n−" nρ".
Hence, using (2.2), we obtain
M(r, f )l a!ja"rj _ n=# a!rn ρ"ρ#…ρn a!ja"rjn=_#ρn a!rn "2n(n−")/# n! la!ψ
0
a"a!r1
. Note that inequality (3.3) is contained in an implicit form in [7].4. Proof of Theorem 2
L 5. The function (z) defined by (1.8) belongs to Q$9P$.
Proof. We shall use the following test of m-times positivity. T (I. J. Schoenberg [9]). Let obkqn
k=! be a finite sequence of numbers.
Consider m matrices Bkl
I
J
0 0 b! : 0 0 b! b" : 0 b! b" b# : 0 ( ( ( ( ( bn−# bn−" bn : : bn−" bn 0 : : bn 0 0 : : ( ( ( ( ( 0 0 0 : bnK
L
kl 1, 2,…, m,where Bkconsists of k rows and njk columns. Assume that the following condition is satisfied for kl 1, 2,…, m: all kik-minors of Bkconsisting of consecutie columns are strictly positie. Then the sequence (b!,b",…,bn, 0, 0, …q is m-times positie.
Let
a!la"l1 akl 1
dk−"
# d$k−#… dk
kl 2, 3,… (4.1)
be the coefficients of the function (z). Fix any n 2, and consider the sequence oa!,a",…,an−", ankε, 0, 0,…q, (4.2)
whereε 0 will be chosen sufficiently small later. Form three matrices:
A"l(a! a" … an−" ankε)
A#l
IJ
a!0 a"a! a#a"… … an−" an−# ankε an−" 0 ankε
KL
A$lI
J
0a! 0 a! a" 0 a" a# a! … … … an−# an−" an−$ an−" ankε an−# ankε 0 an−" 0 0 ankεK
L
.All minors of A" are trivially positive for 0 ε an. Since
ak ak+"l d#d$…dk+" kl 1, 2,… and dk c 1, we have ak− " ak ak ak+ " kl 1, 2,…, nk2 and an− # an− " an− " ankε
for sufficiently smallε. Therefore, all minors of A# are positive for such ε. Further, consider the determinants
Nkl ak−" ak ak−# ak ak+" ak−" ak+" 0 ak kl 2, 3,… We have Nkl a$kja#k+ "ak−#k2ak+"akak−" la#k+"ak−#(dk+# "dkj1k2dk+"dk) l a#k+"ak−#dk
0
1 z#k+ "k 1 zkj11
l 0 k l 2, 3,…by virtue of (4.1), (1.7) and (1.5). Moreover, setting
N"l a!a" 0 a" a# a! a# 0 a" l 1k2 d#,
Now, consider the 3i3-minors of A$ consisting of consecutive columns: M!l 0a! 0 a! a" 0 a" a# a! M"l a! a" 0 a" a# a! a# a$ a" Mkl ak−" ak ak− # ak ak+ " ak− " ak+ " ak+ # ak kl 2, 3,…, nk3 Mn−#(ε) l an−$ an−# an−% an−# an−" an−$ an−" ankε an−# Mn−"(ε) l an−# an−" an−$ an−" ankε an−# ankε 0 an−" Mn(ε) l an−" ankε an− # ankε 0 an− " 0 0 ankε .
Positivity of M! and Mn(ε) for 0 ε an is trivial. By the addition rule of
determinants, we have M"lN"ja$ a!0 a"a! Mkl Nkjak+# ak− " ak− # ak ak− " kl 2, 3,…, nk3 Mn−#(ε) l Nn−#j(ankε) an−$ an−% an−# an−$ . Hence Mk 0 k l 1, 2,…, nk3 Mn− #(ε) 0. Since Mn− "(0)l Nn−" l0 Mn−"(0)l 2an−"an−#k2anan−$ l2anan−#
0
an−" an k an−$ an−#1
0, we have Mn−"(ε) 0 for sufficiently smallε 0.Applying Schoenberg’s test, we conclude that the sequence (4.2) is 3-times positive for all sufficiently smallε 0. Taking a limit as ε tends to 0, we see that the sequence oakqn
k=! l oa!,a",…,an, 0, 0, …q is 3-times positive.
The following immediate Corollary 2 of Lemma 5 is of interest.
C 2. The bound (1.9) is sharp, and it is attained for f(z) l a! (a"z\a!).
Note. It can be shown that the functionψ(z) defined by (3.4) does not belong to
Q_(or even to Q%). Therefore the sharpness of (3.3) seems doubtful.
L 6. The following equality is alid: lim r _ log M(r, ) (log r)# l 1 2 log c.
Proof. Since (z)? Q$9P$, Theorem 1 can be applied to (z). Hence lim sup r _ log M(r, ) (log r)# 1 2 log c. Therefore, it suffices to prove that
lim inf r _ log M(r, ) (log r)# 1 2 log c. (4.3)
For any n 2, we have
log M(r, )l log (r) log rn
dn−"
# d$n−#… dn
l n log rkn
j=#
(nkjj1) log dj. Since limj _djl c, we have dj cjε for any given ε, and j j!lj!(ε). Hence
log M(r, ) n log rkj! j=# (nkjj1) log djk n j=j! (nkjj1) log (cjε) n log rkC!nk"#n#log(cjε),
where C! is a positive constant that depends neither on n nor on r. Setting
nl
9
log r log (cjε):
, we obtain log M(r, )9
log r log (cjε):
log rkC!9
log r log (cjε):
k 1 29
log r log (cjε):
# log (cjε) l2 log (c(log r)jε)# kO (log r) r _.Sinceε 0 is arbitrarily small, we obtain (4.3).
Theorem 2 follows at once from Lemma 5 and Lemma 6. 5. Proof of Theorem 5
Proof of Theorem5(i). We shall use the following result.
T (I. J. Schoenberg [4, p. 415]). Let g(z) l a!ja"zj…janzn be a
polynomial with real coefficients and a!0. If g(z) does not anish in the angle oz :
Qarg zQ (πm)\(mj1)q, then the sequence oa!,a",…,an, 0, 0, …q is m-times positie. Applying this theorem with ml 3, g(z) l fn(z), we obtain f(z)? P$. Hence, by Theorem 1, (1.1) is valid.
Note. If all sections of (0.1) do not vanish outsideoz : Qarg zkπQ π\4q, then we apply Theorem 3 or Theorem 4 instead of Theorem 1, and obtain the more precise inequalities (1.3) or (1.9) instead of (1.1).
T (Hermite–Bieler [6, p. 305]). Let P"(z) and P#(z) be two polynomials with
real coefficients. The polynomial
ω(z) l P"(z)jiP#(z)
does notanish in the closed lower half-plane if and only if all zeros of P"(z) and P#(z) are simple, real, and interlacing, and, moreoer, at some point x!?R,
P#(x!)P"(x!)kP#(x!)P"(x!)0.
Assume that f(z) satisfies the conditions of Theorem 5(ii). Then, for sufficiently large n, all zeros of
f#n+"(z)l
#n+" k=!
akzk
lie inoz : Re z 0q so that the polynomial
f#n+"(iz)l n j=!a#j (k1)jz#jjiz n j=!a#j+" (k1)jz#jB p(") n (z)jizp(n#)(z)
does not vanish in the closed lower half-plane. By the Hermite–Bieler theorem, all zeros of p(")
n (z) and p(n#)(z) are real. Since these polynomials are even,
q(") n (z)B p(n")(iNz) l n j=!a#j zj q(#) n (z)B p(n#)(iNz) l n j=!a#j+" zj
have purely negative zeros. It means that both the formal power series
q(")(z)l _ j=!a#j zj q(#)(z)l _ j=!a#j+" zj (5.1)
belong to P_. Applying Po! lya’s theorem, we conclude that (5.1) converges in the whole plane and
lim sup r _ log M(r, q(s)) (log r)# 1 2 log 2 sl 1, 2. (5.2) Since M(r, f )l f(r) l q(")(r#)jrq(#)(r#) r max s=",# M(r#, q(s)), we have lim sup r _ log M(r, f ) (log r)# 4 maxs= ",# lim sup r _ log M(r, q(s)) (log r)# 2 log 2.
Note. If all sections of (0.1) do not vanish outsideoz : Re z 0q, we can apply (3.3) instead of (5.2) to both functions (5.1). We obtain a more complicated but more precise inequality :
M(r, f ) a!ψ
0
a#a!r#
1
jra"ψ0
a$a"r#1
,Acknowledgements. The authors thank an anonymous referee of this paper for suggestions which improved its presentation.
References
1. A. E, ‘Power series having partial sums with zeros in a half-plane’, Proc. Amer. Math. Soc. 9 (1958) 320–324.
2. A. E, E. B. S and R. S. V, ‘Zeros of sections of power series’, Lecture Notes in Mathematics 1002 (Springer, 1983) 1–115.
3. T. G, ‘The zeros of the partial sums of power series’, Duke Math. J. 30 (1963) 533–540.
4. S. K, Total positiity – Vol. I (Stanford University Press, 1968).
5. O. K and I. V. O, ‘Zero sets of entire generating functions of Po!lya frequency sequences of finite order ’, Math. USSR – Iz. 35 (1990) 101–112.
6. B. J. L, Distribution of zeros of entire functions (American Mathematical Society, Providence, 1980).
7. G. P!, ‘U$ber Anna$herung durch Polynome mit lauter reelle Wurzeln’, Rend. Circ. Mat. Palermo 36 (1913) 279–295.
8. G. P! and G. S$, Problems and theorems in analysis – Vol. 2 (Springer, Heidelberg, 1976).
9. I. J. S, ‘On zeroes of generating functions of multiply positive sequences and functions’,
Ann. of Math. 47 (1955) 447–471.
10. E. T. W and G. N. W, A course of modern analysis (Cambridge University Press, 4th ed., 1990).
Department of Mathematics B. I. Verkin Institute for Low Temperature Bilkent University Physics and Engineering
06533 Bilkent 310164 Kharkov
Ankara Ukraine
Turkey
