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Complex Variables
ISSN: 0278-1077 (Print) 1563-5066 (Online) Journal homepage: https://www.tandfonline.com/loi/gcov19
On Power Series having Sections with Multiply
Positive Coefficients
N.A. Zheltukhina
To cite this article: N.A. Zheltukhina (2002) On Power Series having Sections with Multiply Positive Coefficients, Complex Variables, 47:9, 853-866, DOI: 10.1080/02781070290032234
To link to this article: https://doi.org/10.1080/02781070290032234
Published online: 15 Sep 2010.
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On Power Series having Sections with
Multiply Positive Coefficients
N.A. ZHELTUKHINA*
Department of Mathematics, Bilkent University, 06533 Bilkent, Ankara,Turkey Communicated by M. Esse¤n
(Received 23 February 2000; Revised 28 November 2001; In final form 15 March 2002)
Po´lya’s theorem of 1913 says that if all sections of power series have real negative zeros only, then the series converges in the whole complex plane and its sum satisfies a certain growth condition. Here we show that the assertion of Po´lya’s theorem remains valid for a much larger class of formal power series and, moreover, a better growth estimate holds.
Keywords: m-times positive sequence; Power series; Sections Mathematical Subject Classifications (1991): 30B10; 30D10; 30D15
1. INTRODUCTION AND RESULTS Let
f ðzÞ ¼X
1
k¼0
akzk, a0 >0 ð1:1Þ
be a formal power series. Po´lya’s theorem [6] of 1913 says that if, for all sufficiently large n, the sections
fnðzÞ ¼
Xn k¼0
akzk ð1:2Þ
have only negative zeros, then the series (1.1) converges in the whole complex plane, and its sum f ðzÞ is an entire function satisfying the condition
lim sup r!1 log Mðr, f Þ ðlog rÞ2 1 2 log 2, ð1:3Þ *E-mail: natalya@fen.bilkent.edu.tr
ISSN 0278-1077 print: ISSN 1563-5066 online 2002 Taylor & Francis Ltd DOI: 10.1080/02781070290032234
where Mðr, f Þ ¼ maxfjf ðzÞj : jzj rg. In [3] ([4]) this theorem was generalized to power series (1.1) having sections (tails) with multiply positive coefficients.
Recall [1] that the sequence fakg1k¼0 of real numbers is said to be m-times positive
for m 2 N [ f1g if all minors of orders less than m þ 1 of the infinite matrix
a0 a1 a2 a3 . . . 0 a0 a1 a2 . . . 0 0 a0 a1 . . . : : : : . . . 0 B B @ 1 C C A
are non-negative. Usually, 1-times positive sequences are called totally positive sequences.
Denote by Pm, m 2 N [ f1g, the class of all power series (1.1) such that the truncated
sequences fakgnk¼0:¼ fa0, a1, . . . , an, 0, 0, 0, . . .g are m-times positive for all sufficiently
large n. Evidently, P1 P2 P3 . . . P1.
As mentioned in [3], the condition of Po´lya’s theorem is equivalent to the condition f 2 P1. It was proved in [3] that if we replace this condition by f 2 Pmfor some m 3,
then, as in Po´lya’s theorem, the series (1.1) converges in the whole complex plane and its sum f ðzÞ is an entire function. However, the estimate (1.3) in [3] was replaced by the coarser lim sup r!1 log Mðr, f Þ ðlog rÞ2 1 2 log c, c ¼ 1 þpffiffiffi5 2 <2: ð1:4Þ
This cannot be improved if f 2 P3.
As far as we know, the question whether Po´lya’s estimate (1.3) is sharp has not yet been settled. In [7, Part V, Problem 176], an example was constructed showing that log 2 cannot be replaced by a constant greater than log 4. Katkova and Vishnyakova kindly informed me that, by modifying this example, they showed that log 2 cannot be replaced by a constant greater than logð3:5Þ.
The aim of this article is to show that the estimate better than (1.3) already holds in the class P4 which is much larger than P1.
THEOREM If m 4, then any formal power series of Pmconverges in the whole complex
plane and its sum f ðzÞ is an entire function of order0. Moreover,
lim sup r!1 log Mðr, f Þ ðlog rÞ2 1 2 log dðP4Þ , ð1:5Þ
where constant dðP4Þis independent of f and
2:016 dðP4Þ 2:087:
Note that for m < 1 there is also a relation between m-times positivity of the sequence fakgnk¼0 and zeros of the corresponding polynomials (1.2) but a less strict
one than for m ¼ 1. Schoenberg [2, pp. 397, 415] proved: (i) the necessary condition for fakgnk¼0 to be m-times positive is the non-vanishing of (1.2) in the angle
fz: j arg zj m=ðm þ n 1Þg, (ii) the sufficient condition is its non-vanishing in the larger angle fz : j arg zj m=ðm þ 1Þg. Both conditions are best possible in the terms of the sizes of angles.
As an immediate consequence of the Theorem and Schoenberg’s result (ii) for m ¼ 4 we obtain a generalization of Po´lya’s theorem in terms of zeros of (1.2).
COROLLARY Let f ðzÞ be a formal power series(1.1) with real coefficients ak. Assume that
for all sufficiently large n, the zeros of sections (1.2) are located in the angle fz: j arg z j < =5g. Then the series converges in the whole complex plane and estimate (1.5) holds.
2. PROOF OF THE THEOREM
Without loss of generality we can assume that the series (1.1) contains infinitely many non-zero terms. Recall two lemmas and some notations from [3].
LEMMA2.1 Let fakg1k¼0, a0>0, be a 2-times positive sequence. Set n ¼ minfk : ak¼0g.
If n is finite, then ak¼0 for any k n.
Evidently, 2-times positivity of truncated sequences fakgnk¼0 for all sufficiently large n
yields 2-times positivity of the whole sequence fakg1k¼0. Then, by Lemma 2.1, all the ak
are strictly positive, that allows us to introduce the positive numbers k¼ ak1 ak , k ¼1, 2, . . . , ð2:1Þ and k¼ k k1 ¼ a 2 k1 akak2 , k ¼2, 3, . . . ð2:2Þ
The next formulas follow from (2.1) and (2.2). ak¼ a0 Qk j¼1j , k ¼1, 2, . . . and ð2:3Þ ak¼ a0 k 1k12 k23 . . . k , k ¼2, 3, . . . ð2:4Þ
LEMMA 2.2 Let fakg1k¼0 be a 2-times positive sequence without zero terms. Then the
sequence fkg1k¼1 defined by(2.1) is non-decreasing.
The following Lemma 2.3 plays a basic role in the proof of the Theorem.
LEMMA 2.3 Let fakgnk¼0¼ fa0, a1, . . . , an, 0, 0, . . .g, a0>0, an>0, n 2, be a 4-times
positive sequence. Then
(i) for n ¼ 2, we have 2 ð3 þ
ffiffiffi 5 p
Þ=2; (ii) for n ¼ 3, we have
3 3 2 2 5 4 2 2 ; ð2:5Þ
(iii) for n 4, we have n 3 2 2 5 4 2 n1 þ 1 n22n1n : ð2:6Þ
Proof Since fakgnk¼0 is 4-times positive, we have
In:¼ an1 an 0 0 an2 an1 an 0 an3 an2 an1 an an4 an3 an2 an1 ¼a4n13a2 n1an2anþ2an3an1a2na3nan4þa2na2n20,
where ak¼0 for k < 0. In particular,
I2¼a413a21a0a2þa22a20, I3¼a423a 2 2a1a3þ2a0a2a23þa 2 3a 2 1: By (2.2), an4an a2 n2 ¼an4an2 a2 n3 a 2 n3a2n1 a4 n2 anan2 a2 n1 ¼ 1 n22n1n : Hence, I2¼a20a22 a21 a0a2 2 3 a 2 1 a0a2 þ1 ! ¼a20a22 2232þ1 , I3¼a21a 2 3 a2 2 a1a3 2 3 a 2 2 a1a3 þ2a0a2 a2 1 þ1 ! ¼a21a23 2333þ 2 2 þ1 , In¼a2n2a2n a2 n1 anan2 2 3 a 2 n1 anan2 þ2an3an1 a2 n2 anan4 a2 n2 þ1 ! ¼a2na2n2 2n3nþ 2 n1 þ1 1 n22n1n :
Since, by Lemma 2.2, n1, the assertion of the lemma is an easy consequence of the
condition In 0. œ
Denote by Qmthe subclass of Pmconsisting of formal power series (1.1) satisfying the
condition: for all n ¼ 0, 1, 2, . . . , the truncated sequences fakgnk¼0 are m-times positive.
COROLLARY2.1 Let a formal power series(1.1), which is not a polynomial, belong to Q4.
Proof The inequality (2.6) implies n 3 2 2 >5 4 2 n1
for n 4. Note that, by Lemma 2.3, we have 2 ð3 þ
ffiffiffi 5 p Þ=2 > 2 and 3 3 2 2 5 4 2 2 >1 4,
i.e. 3>2. Then the proof can be completed by induction. œ
We shall use the following result from [3].
LEMMA 2.4 Let a formal power series (1.1), which is not a polynomial, belong to P3
and let for all n n0the sequences fakgnk¼0be3-times positive. Let fzng1n¼n0be the sequence
of positive numbers satisfying the recurrence relation
z2nþ1 ¼znþ1, n n0 ð2:7Þ
and the initial condition
zn0 ¼
1 n01
: ð2:8Þ
Then there exists the limit
lim n!1zn ¼ 1 þpffiffiffi5 2 :¼ c, and we have 1 n1 zn, n n0: Moreover,
(i) if zn0< c, then the sequence fzng
1
n¼n0 increases;
(ii) if zn0> c, then the sequence fzng
1
n¼n0 decreases;
(iii) if zn0¼c, then zn¼c for any n n0.
LEMMA2.5 Let a formal power series (1.1), which is not a polynomial, belong to P4.
Then there exists N such that n> c for all n N, where c ¼ ð1 þ
ffiffiffi 5 p
Þ=2.
Proof As has been proved in [3, Lemma 3], if fakgnk¼0 is 3-times positive, then n>1.
Therefore, we have n >1 for all sufficiently large n. Since 4-times positivity implies
3-times positivity, then, by Lemma 2.4, for any " > 0, there exists such n1 that
1=ðn1Þ < c þ " for all n n1. Assuming that n cfor all sufficiently large n, we
get from (2.6) c 3 2 2 5 4 2 c "þ 1 c4
for all sufficiently small ", in contradiction with the equality c 3 2 2 ¼5 4 2 c: ð2:9Þ
Hence, there exists such N that N > c.
Set n0¼N and construct the sequence fzng1n¼N by (2.7) and the initial condition
zN ¼1=ðN1Þ. Since N > c, then zN < c. It follows from Lemma 2.4 that zn< c
for all n N, whence n> cfor all n N. œ
Consider the sequence fkng1n¼N of positive numbers satisfying the recurrence equation
kn 3 2 2 ¼5 4 2 kn1 , n > N, ð2:10Þ
and the initial condition
kN ¼N: ð2:11Þ
The inequalities (2.5) and (2.6) imply
n> kn, n > N: ð2:12Þ
Since kN ¼N > cand (2.9) holds, we get from (2.10) by induction that
kn> c, n N: ð2:13Þ
LEMMA2.6 We have
lim
n!1kn¼2:
Moreover,
(i) if c < kN <2, then the sequence fkng1n¼N increases;
(ii) if kN >2, then the sequence fkng1n¼N decreases;
(iii) if kN ¼2, then kn ¼2 for any n N.
The proof is elementary and can be omitted. By Lemma 2.6 and (2.12), we have for any f 2 P4
ð f Þ:¼ lim inf
n!1 n2: ð2:14Þ
Define
LEMMA2.7 We have
ðP4Þ 2 þ
1 66:
Proof Assume that there exists f 2 P4 such that ð f Þ 2 þ ", where " ¼ 1=66. Then
there exists a subsequence fnkg1k¼0 such that nk<2 þ A", where A ¼ 66=65 > 1.
The inequality (2.6) implies
nk2 2 nk1 2 þ A" 3 2 2 þ2nk2nk1> 5 4nk2 2 nk1þ 1 2 þ A", ð2:15Þ
which can be rewritten as
nk2nk1ð2 nk1ð1 A" ðA"Þ 2ÞÞ 1 2 þ A", whence nk1 2
1 A" ðA"Þ22 þ 4A":
The last inequality holds for any " such that A" < 1=4. By the repetition of argument,
nk22 þ 16A":
Therefore, the inequality (2.15) implies
2 nk1 5 4 nk 3 2 2 þ 1 nk2 2 nk1nk 1 A" ðA"Þ2þ 1 ð2 þ 16A"Þ4 >1 2A" þ 1 16ð1 32A"Þ ¼ 17 164A" ¼ 1 þ B,
where B ¼ 1=ð1665Þ > 0. Hence, we get
nk1<
2 1 þ B<2,
in contradiction with (2.14). œ
Define dð f Þ and dðP4Þby formulas
lim sup r!1 log Mðr, f Þ ðlog rÞ2 ¼: 1 2 log dð f Þ,
and
dðP4Þ:¼ inffdð f Þ : f 2 P4g:
We shall use the following result contained in the proof of Theorem 1 of [3].
LEMMA2.8 Let f ðzÞ be a formal power series(1.1) whose coefficients are strictly positive
for all sufficiently large n. If nd >1 for all sufficiently large n, then dð f Þ d, where
dð f Þ is defined by(1.5). If there exists limn!1n¼d >1, then dð f Þ ¼ d.
It follows from Lemma 2.8 that dð f Þ ð f Þ and hence, dðP4Þ ðP4Þ. Then, by
Lemma 2.7, dðP4Þ 2 þ 1=66 ¼ 2:016 . . . . That dðP4Þ 2:087 is proved in Section 3.
3. THE UPPER BOUND FORdðP4Þ
Denote by In:¼ an1 an 0 0 an2 an1 an 0 an3 an2 an1 an an4 an3 an2 an1 , Jn:¼ an1 an anþ1 0 an2 an1 an anþ1 an3 an2 an1 an an4 an3 an2 an1 , Ln:¼ an1 an anþ1 anþ2 an2 an1 an anþ1 an3 an2 an1 an an4 an3 an2 an1 , where ak¼0 for k < 0.
LEMMA3.1 Let fakg1k¼0be a2-times positive sequence without zero terms. Assume In>0
for all n 2. Then Ln >0 and Jn>0 for all n 2.
Proof By Lemma 2.2, n1 for all n 2. Then, the inequality
I2¼a20a 2 2ð 2 232þ1Þ > 0 implies 2> ð3 þ ffiffiffi 5 p
Þ=2 > 2. Suppose n1>2 for some n > 3. Then inequalities (2.5)
and (2.6), which correspond to I3>0 and In>0, n 4, respectively, give
n 3 2 2 >5 41 ¼ 1 4,
i.e. n>2. Thus, n>2 for all n 2. Hence, by Corollary 1 of [5], we have
P1
k¼0akzk2Q3.
We have
where An¼ 0 0 anþ1 anþ2 an2 an1 an 0 an3 an2 an1 an an4 an3 an2 an1 and Bn¼ an1 an anþ1 anþ2 0 0 0 anþ1 an3 an2 an1 an an4 an3 an2 an1 : For n 4 we have An¼anþ1 an2 an1 0 an3 an2 an an4 an3 an1 anþ2 an2 an1 an an3 an2 an1 an4 an3 an2 ¼anþ1an2an3an4 1 an1 an2 0 1 an2 an3 an an3 1 an3 an4 an1 an4 anþ2an2an3an4 1 an1 an2 an an2 1 an2 an3 an1 an3 1 an3 an4 an2 an4 ¼anþ1an2an3an4 an1 an2 anþ1 an2 1 1 0 1 a 2 n2 an1an3 anan2 an3anþ1 1 an3an2 an4an1 an1an2 an4anþ1 anþ2an2an3an4 an1 an2 an an2 1 1 1 1 a 2 n2 an1an3 an1an2 an3an 1 an3an2 an4an1 a2n2 an4an :
Since ak6¼0, k ¼ 0, 1, 2, . . . , and P1k¼0akzk2Q3, then all minors in the last equation
are non-negative. Using anþ2 ða2nþ1Þ=an (which is equivalent to nþ21), we get
Anan3an4an1
a2nþ1 an2
where n ¼ 1 1 0 1 a 2 n2 an1an3 anan2 an3anþ1 1 an3an2 an4an1 an1an2 an4anþ1 1 1 1 1 a 2 n2 an1an3 an1an2 an3an 1 an3an2 an4an1 a2 n2 an4an : By (2.4) n¼ 1 1 0 1 n1 n1nnþ1 1 n1n2 nþ12n2n1n2 1 1 1 1 n1 n1n 1 n1n2 n2n1n2 ¼nþ12n 3 n1n2þn1nnþ1n12 n2nnþ1nþ12n 2 n1n2 n3n1n2n1nn1n2þn1þ2n1nn2þn2n1n2 ¼nþ1n2n1n2 1 4nn11 þnþ12n2n1n2 1 2n11 þn3n1n2 1 4nþ1n1 þn1n nþ11 þn1n2fn1n1g þ n1þn2n1n2:
Since n>2 for all n 2, then n>0 for all n 4, and so An>0 for all n 4. As we
mentioned before,P1k¼0akzk2Q3. Hence,
Bn¼anþ1 an1 an anþ1 an3 an2 an1 an4 an3 an2 0 for all n 4. Therefore, Ln >0 for all n 4.
The condition n>2 also yields
L2¼I2þa30a4ð4321Þ þ a3a30>0 and L3¼I3þA3þa4 a2 a3 a4 a0 a1 a2 0 a0 a1 >0;
since n>2 and, hence, A3¼a0a1a2a5 345 1 221 þ45 1 4231 þ2 1 43451 þ 2 1 23 >0:
Taking into account the equality
Jn¼Lnþanþ2 an2 an1 an an3 an2 an1 an4 an3 an2
and the fact thatP1k¼0akzk2Q3, we obtain Jn>0. œ
LEMMA 3.2 Let fakg1k¼0 be a 2–times positive sequence without zero terms. Assume
In >0, for all n 2. Then f ðzÞ 2 P4.
Proof We shall use the following test of m-times positivity.
THEOREM([8]) Let fbkgnk¼0 be a finite sequence of numbers. Consider m matrices
Bk¼ b0 b1 b2 . . . bn 0 0 . . . 0 0 b0 b1 . . . bn1 bn 0 . . . 0 0 0 b0 . . . bn2 bn1 bn . . . 0 : : : . . . : : : . . . : 0 0 0 . . . : : : . . . bn 0 B B B B @ 1 C C C C A k ¼1; 2; . . . m;
where Bk consists of k rows and n þ k columns. Assume that the following condition is
satisfied for k ¼1; 2; . . . ; m: all k k-minors of Bk consisting of consecutive columns
are strictly positive. Then the sequence fb0; b1;. . . ; bn;0; 0; . . .g is m-times positive.
Fix any k 2, and consider the three matrices
B2¼ a0 a1 a2 . . . ak 0 0 a0 a1 . . . ak1 ak B3¼ a0 a1 a2 . . . ak 0 0 0 a0 a1 . . . ak1 ak 0 0 0 a0 . . . ak2 ak1 ak 0 @ 1 A B4¼ a0 a1 a2 . . . ak 0 0 0 0 a0 a1 . . . ak1 ak 0 0 0 0 a1 . . . ak2 ak1 ak 0 0 0 0 . . . ak3 ak2 ak1 ak 0 B B @ 1 C C A:
The condition In>0 for all n 2 implies n>2 for all n 2. Then 2 2 minors of B2, an1 an an2 an1 ¼ an2anðn1Þ and 3 3 minors of B3, an1 an 0 an2 an1 an an3 an2 an1 ¼an3a2nn1 2n2nþ 1 n1 ; an1 an anþ1 an2 an1 an an3 an2 an1 ¼ an1 an 0 an2 an1 an an3 an2 an1 þanþ1an3an1ðn11Þ;
consisting of consecutive columns are strictly positive. By Lemma 3.1, all 4 4 minors of B4, consisting of consecutive columns are also strictly positive. Hence, by Schoenberg
theorem ([8]), the sequence fa0, a1, . . . , ak, 0, 0, . . .g is 4-times positive for all k 2, and
hence f ðzÞ 2 P4. œ
Consider the sequence fdn >2g1n¼2 of positive numbers satisfying the recurrence
equation dn 3 2 2 ¼5 4 2 dn1 þ 1 d4 n ð3:1Þ
and the initial condition
d2 ¼ 3 þpffiffiffi5 2 þ 1 2: ð3:2Þ Denote gðx, yÞ :¼ x63x5þ 1 þ2 y x41:
Note that if y > 2, then gð2, yÞ < 0 and hence there exists xy>2 such that gðxy, yÞ ¼ 0
for any y > 2. Since (3.1) can be rewritten as gðdn, dn1Þ ¼0
and d2>2, then sequence fdn>2g1n¼2is well-defined. Let að¼ 2:08679 . . .Þ be the largest
positive root of
LEMMA3.3 The limit
lim
n!1dn¼a,
exists, where a is the largest positive root of(3.3).
Proof Let us show that dn> afor all n 2. Indeed, d2> a. Assume dn1> afor some
n >3. Then 5 4þ 1 d4 n dn 3 2 2 ¼ 2 dn1 <2 a¼ 5 4þ 1 a4 a 3 2 2 :
Since f ðxÞ ¼ 5=4 þ 1=x4 ðx 3=2Þ2 is a decreasing function on ½32, 1Þ, then the condition f ðdnÞ< f ðaÞimplies dn> a.
Further, since dn> a, we have 2=dn1¼5=4 þ 1=dn4 ðdn3=2Þ2 <2=dn. Then
fdng1n¼2 is a decreasing sequence. Moreover, it is bounded below by a. Hence,
there exists limn!1dn¼: b and, in view of (3.1), b ¼ a. œ
LEMMA3.4 We have
dðP4Þ a,
where a is the largest positive root of(3.3). Proof Consider a function
ðzÞ ¼1 þ z þX 1 k¼2 zk dk1 2 d3k2. . . dk ,
where the sequence fdkg1k¼2 is defined by (3.1) and (3.2). Then
I2¼a22ðd223d2þ1Þ > 0, I3¼a23 d 2 3 3d3þ 2 d2 þ1 ¼a23 d3 3 2 2 5 4 2 d2 ! ¼a 2 3 d4 3 >0:
Since fdng1n¼2 is a decreasing sequence, then
dn 3 2 2 ¼5 4 2 dn1 þ 1 d4 n >5 4 2 dn1 þ 1 dn2dn12 dn
for n 4. So, In>0 for all n 4. Therefore, by Lemma 3.2, ðzÞ 2 P4. It follows from
Acknowledgements
The author would like to thank Professors I.V. Ostrovskii and C.Y.Y|ld|r|m for helpful suggestions and comments. The research is partially supported by INTAS No. 96-0858.
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