Periodica Mathematica Hungarica Vol. 42 (1–2), (2001), pp. 69–76

NOTES ON A PROBLEM OF TURAN

Andras Biro (Budapest)

Dedicated to Professor Andras Sarkozy on the occasion of his 60th birthday

Abstract

Let z1, z2, . . . , zn be complex numbers, and write

Sj = zj1 + . . .+ zjn

for their power sums. Let

Rn = minz1,z2,... ,zn

max1≤j≤n

|Sj|,

where the minimum is taken under the condition that

max1≤t≤n

|zt| = 1.

Improving a result of Komlos, Sarkozy and Szemeredi (see [KSSz]) we prove herethat

Rn < 1− (1− ε) log log nlog n

.

We also discuss a related extremal problem which occurred naturally in our earlierproof ([B1]) of the fact that Rn > 1

2 .

1. Introduction

The investigation of the sequence Rn is a classical problem of the power sumtheory of Turan (see [T]; see also [M]). The minimum Rn exists by Weierstrass’theorem, and one can easily see that the condition can be replaced by z1 = 1.

The best known upper bound so far was that of Komlos, Sarkozy and Sze-meredi ([KSSz]):

Rn < 1− 1250n

Research partially supported by the Hungarian National Foundation for ScientificResearch, Grant No. T 029759.

Mathematics subject classification number: 11N30.Key words and phrases: power sums, Turan’s theory.

0031-5303/01/$5.00 Akademiai Kiado, Budapestc© Akademiai Kiado, Budapest Kluwer Academic Publishers, Dordrecht

70 a. biro

for n > n0, and

Rn < 1− 13

lognn

for infinitely many n. We improve this estimate for large n, proving our

Theorem 1. For any ε > 0, if n is large enough, then

Rn < 1− (1− ε) log lognlogn

.

It is very likely that in fact an estimate Rn < 1 − c (with some absoluteconstant c > 0) is true. The numerical computations in [CG] also support thisconjecture.

In [B1] I proved that Rn > 12 for every n, and in the recent paper [B2] I

prove that Rn > q for every n, where q is an absolute constant larger than 12 . In

retrospect, it can be seen that the proof in [B1] was implicitly based on the factsthat

Rn ≥Mn

and

Mn >12

for every n, where

Mn = minb1,b2,...,bn−1

(max

(max

1≤k≤n−1

|1+b1+...+bk−1 − kbk|1 + |b1|+ . . .+ |bk−1|

,|1+b1+...+bn−1

1+|b1|+. . .+ |bn−1|

))and we take the minimum over any complex numbers b1, b2, . . . , bn−1. It is nothard to see that this minimum exists. We prove here a methodologically interestingtheorem, which shows that using just the relation Rn ≥Mn, it would be impossibleto prove the result of [B2]. More precisely, we prove

Theorem 2. We have

limn→∞

Mn =12.

It was also proved in [B1] that Rn > 12 + c

n with some explicit constant c > 0.Examining the proof in [B1], it can be seen that what was actually proved thereis Mn > 1

2 + cn . In view of [B2], this result is now irrelevant in the context of

Turan’s problem. But the investigation of the properties of the sequence Mn − 12 is

an interesting problem in its own right. For example: is the sequence

n

(Mn −

12

)bounded? Does it have a finite limit?

notes on a problem of turan 71

2. Proof of Theorem 1

If z1 = 1, z2, . . . , zn are complex numbers, and

(Z − z2)Z − z3) . . . (Z − zn) = Zn−1 + b1Zn−2 + . . .+ bn−1,

then by the Newton–Girard formulas for this polynomial (let Tj =∑nt=2 z

jt )

Tk + b1Tk−1 + . . .+ bk−1T1 + kbk = 0

for 1 ≤ k ≤ n− 1, and

Tn + b1Tn−1 + . . .+ bn−1T1 = 0.

Taking into account that Tj = Sj − 1, we get

Sk + b1Sk−1 + . . .+ bk−1S1 = 1 + b1 + . . .+ bk−1 − kbk (k = 1, 2, . . . , n− 1);(1)

Sn + b1Sn−1 + . . .+ bn−1S1 = 1 + b1 + . . .+ bn−1.(2)

Formulas (1) and (2) were the basic tools in [B1] and are also important in [B2].We will construct b1, b2, . . . , bn−1, it is obvious (by the fundamental theorem ofalgebra) that then z2, . . . , zn can be given in such a way (they will be the roots ofZn−1 + b1Z

n−2 + . . .+ bn−1) that their power sums satisfy (1) and (2).We choose the coefficients bk in such a way that the quotients kbk

1+b1+...+bk−1

(1 ≤ k ≤ n− 1) will be all equal, i.e.

kbk1 + b1 + . . .+ bk−1

= α (1 ≤ k ≤ n− 1)(3)

with some α to be chosen later. It is clear that once α is chosen, the numbers bkare defined recursively, and the power sums can be computed by formulas (1) and(2). More precisely, from formula (1) we obtain by induction that

Sk = 1− α (1 ≤ k ≤ n− 1).(4)

Indeed, we have b1 = α by (3), formula (1) with k = 1 gives S1 = 1 − α, and it iseasy to see that the induction works, because

1 + b1 + . . .+ bk−1 − kbk = (1− α)(1 + b1 + . . .+ bk−1)

for 1 ≤ k ≤ n− 1. We now compute Sn.It follows by induction (using (3)) that

1 + b1 + . . .+ bk =k∏t=1

(1 +

α

t

)(5)

for k = 0, 1, . . . , n− 1. Then (2),(4) and the case k = n− 1 of (5) give

Sn = 1− α+ αn−1∏t=1

(1 +

α

t

).(6)

We will use the following simple lemma.

72 a. biro

Lemma 1. If |α| ≤ 12 , and n is a positive integer, then

n−1∏t=1

(1 +

α

t

)= eα logn+O(|α|),

where the O-constant is absolute.

Proof. It is well known that for |x| ≤ 12 one has

1 + x = ex+O(|x|2).

Thenn−1∏t=1

(1 +

α

t

)= exp

(αn−1∑t=1

1t

+O

(|α|2

∞∑t=1

1t2

)).

Sincen−1∑t=1

1t

= log n+O(1),

the lemma follows.Applying this lemma and (6) we conclude that (assuming |α| ≤ 1

2 )

Sn = 1− α+ αeα logn+O(|α|).(7)

It is clear by equations (4) and (7) that the quantity αeα logn must be bounded,and this explains our choice (with a small constant ε)

Re α = (1− ε) log lognlogn

(8)

(we assume that n is large). Then (if |α| = O(Re α), but this will be the case)∣∣∣αeα logn+O(|α|)∣∣∣ = O

(log logn

logne(1−ε) log logn

)= O

(log lognlogε n

).(9)

We have to choose Imα yet, and we will do it in such a way that both arg(α) andarg(−αeα logn+O(|α|)) will be small. This motivates our choice

Imα =π

logn,(10)

i.e.

α = (1− ε) log lognlogn

+iπ

logn.

Then, indeed, (8) and (10) give that

α

|α| = 1 +O

(1

log logn

),(11)

notes on a problem of turan 73

and

−αeα logn+O(|α|)∣∣−αeα logn+O(|α|)∣∣ = 1 + O

(1

log logn+ |α|

)= 1 +O

(1

log logn

).(12)

Finally, from (7), (11) and (12) we get that

Sn = 1−(|α|+

∣∣∣αeα logn+O(|α|)∣∣∣)(1 +O

(1

log logn

)).(13)

It is not hard to see by formulas (4), (8), (9), (10), (13) and the triangle inequalitythat if ε > 0 is fixed and n is large enough, then

max1≤t≤n

|St| ≤ 1− (1− 2ε) log lognlog n

,

and so the theorem is proved.

3. Proof of Theorem 2

The relation Mn > 12 for every n was (implicitly) proved in [B1], so it is

enough to prove here that

lim supn→∞

Mn ≤12.

Let c > 12 be an arbitrary, but fixed constant, and let d be another fixed

constant with 12 < d < c. We define the complex numbers bk recursively.

Let b0 = 1. Assume that 1 ≤ k ≤ n − 1, the coefficients b0, b1, . . . , bk−1 aredefined, and introduce the notation

qt =|1 + b1 + . . .+ bt|1 + |b1|+ . . .+ |bt|

.(14)

If qk−1 > c, let 0 < βk <π2 be such that

sinβk =d

qk−1,(15)

and define bk by the formula

kbk1 + b1 + . . .+ bk−1

= (cosβk + i sinβk) cosβk.(16)

(Observe that 1 + b1 + . . .+ bk−1 6= 0, by (14) with t = k − 1, and since qk−1 > c.)If qk−1 ≤ c, let bk = 0.

74 a. biro

Lemma 2. There are positive constants K1 and K2 depending only on c andd such that if K1 ≤ k ≤ n− 1, qk−1 > c, then

qk ≤ qk−1e−K2k .

Proof. Under these assumptions (16) is valid, and so

1 + b1 + . . .+ bk−1 + bk = (1 + b1 + . . .+ bk−1)(

1 +(cosβk + i sinβk) cosβk

k

).

On the other hand, (16) and (14) (with t = k − 1) give

1 + |b1|+ . . .+ |bk−1|+ |bk| = (1 + |b1|+ . . .+ |bk−1|)(

1 +qk−1 cosβk

k

).

Consequently,

qk = qk−1

∣∣∣∣1 +(cosβk + i sinβk) cos βk

k

∣∣∣∣/(

1 +qk−1 cosβk

k

).(17)

Now, ∣∣∣∣1 +(cos βk + i sinβk) cosβk

k

∣∣∣∣2 = 1 + 2cos2 βkk

+cos2 βkk2 ,(18)

and (1 +

qk−1 cosβkk

)2

≥ 1 + 2qk−1 cosβk

k≥ 1 + 4d

cos2 βkk

,(19)

where in the last step we used the inequality

qk−1 ≥ 2d cosβk,(20)

which can be proved in the following way. By (15), we have

cos2 βk = 1− d2

q2k−1

,(21)

and so (20) follows from

q2k−1 − 4d2

(1− d2

q2k−1

)=(qk−1 −

2d2

qk−1

)2

≥ 0.

The relations (17), (18) and (19) imply that

q2k ≤ q2

k−1

(1 + 2

cos2 βkk

+cos2 βkk2

)(1 + 4d

cos2 βkk

)−1

.(22)

If ε > 0 is a small but fixed constant, then choosing K1 to be large enough,it follows from (22) that for k ≥ K1 we have

q2k ≤ q2

k−1e−(−2+4d−ε) cos2 βk

k .

notes on a problem of turan 75

Since by (21) and qk−1 > c we have cos2 βk ≥ 1 − d2

c2 > 0, and 4d > 2, it is nothard to see that choosing K1 to be large enough, and K2 to be small enough, theassertion of the lemma follows from (22).

Corollary. There is a positive constant K3 depending only on c and d, suchthat if K3 ≤ k ≤ n, then qk−1 ≤ c.

Proof. Indeed, if for some 1 ≤ k ≤ n − 1 we have qk−1 ≤ c, then bk = 0by definition, and then qk = qk−1 ≤ c. So it is enough to find just one k for whichqk−1 ≤ c. Let k ≥ K1 and assume that for all t < k − 1 we have qt > c. Then byrepeated application of Lemma 2 (since always qr ≤ 1) we obtain

qk−1 ≤ exp

−K2

∑K1≤t≤k−1

1t

.

The series∑ 1

t is divergent, so our statement follows at once.

Let n ≥ K3. Then, by the above corollary and (14), we have

|1 + b1 + . . .+ bn−1|1 + |b1|+ . . .+ |bn−1|

≤ c,(23)

and for 1 ≤ k ≤ n− 1, if qk−1 > c, then by (16) and (14),

|1 + b1 + . . .+ bk−1 − kbk|1 + |b1|+ . . .+ |bk−1|

= |1− (cosβk + i sinβk) cosβk|qk−1.(24)

Now, it is easy to see that

|1− (cosβk + i sinβk) cosβk| = sinβk,

so (15) and (24) give

|1 + b1 + . . .+ bk−1 − kbk|1 + |b1|+ . . .+ |bk−1|

= d < c(25)

for 1 ≤ k ≤ n− 1, if qk−1 > c. Finally, if 1 ≤ k ≤ n− 1, qk−1 ≤ c, then bk = 0, andso (using (14)) one has

|1 + b1 + . . .+ bk−1 − kbk|1 + |b1|+ . . .+ |bk−1|

= qk−1 ≤ c.(26)

The relations (23), (25), (26) and the fact that c > 12 can be chosen to be arbitrarily

close to 12 , prove our theorem.

Addendum. Very recently I succeeded in proving lim supn→∞

Rn < 1, see [B3].

76 a. biro

REFERENCES

[B1] A. Biro, On a problem of Turan concerning sums of powers of complex numbers,Acta Math. Hungar. 65 (3) (1994), 209–216.

[B2] A. Biro, An improved estimate in a power sum problem of Turan, preprint, 1999,to appear in Indag. Math.

[B3] A. Biro, An upper estimate in Turan’s pure power sum problem, preprint, 2000.[CG] A. Y. Cheer and D. A. Goldston, Turan’s pure power sum problem, Math.

Comp. 65 (1996), no. 215, 1349–1358.[KSSz] J. Komlos, A. Sarkozy and E. Szemeredi, On sums of powers of complex num-

bers (in Hungarian), Mat. Lapok XV (1964), 337–347.[M] H. L. Montgomery, Ten lectures on the interface between analytic number theory

and harmonic analysis, AMS, 1994.[T] P. Turan, On a New Method of Analysis and its Applications, John Wiley & Sons,

New York, 1984.

(Received: November 30, 1999)

A. Biro

Alfred Renyi Institute of Mathematics

Hungarian Academy of Sciences

Budapest, Realtanoda u. 13–15

H–1053

Hungary

E-mail: biroand@renyi.hu