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TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 296, Number 2, August 1986

THE RIEMANN HYPOTHESIS
AND THE TURAN INEQUALITIES!

GEORGE CSORDAS, TIMOTHY S. NORFOLK AND RICHARD S. VARGA

ABSTRACT. A solution is given to a fifty-eight year-old open problem of G. Polya,
involving the Taylor coefficients of the Riemann é-function.

1. Introduction. The purpose of this paper is to solve a fifty-eight year-old problem
of Pélya [P, p. 16], related to the Riemann Hypothesis. This problem may be
described as follows. Starting with Riemann’s definition of his éfunction (cf.

Titchmarsh [T, p. 16], in a slightly different notation), i.e.,
zy A{,2-1),27-14 & 4 5)
(1.1) E(iz):= 5(z a r 7 +4 i z+5),

where § is the Riemann zeta-function, then € is an entire function of order one and
admits the integral representation (cf. [P, p. 11])

(1.2) (5) = sf” ©(t)cos(xt) dt,

where

(1.3) @(t):= sy (2n4a7e* — 3n?me*)exp(—n7me*").
n=1

(We have dropped the usual factor of 4 in the definition of ©.) From (1.2), the entire
function € can be written in Taylor series form as

1 x oo (-1)"b x2”
1.4 g{l3) =) a,
(1.4) 8°\2 a (2m)!
where

a 0

(1.5) b,:= f r"e(r)dt  (m=0,1,...).

0
On setting z = —x* in (1.4), the function F(z), defined by

a) bz”

(1.6) F(z):= XL my!

Received by the editors May 9, 1985.

1980 Mathematics Subject Classification. Primary 30D10, 30D15; Secondary 26A51.

Key words and phrases. The Riemann Hypothesis, the Riemann €-function, Turan inequalities,
concavity, moments.

‘Research supported by the Department of Energy.

©1986 American Mathematical Society
0002-9947 /86 $1.00 + $.25 per page

521

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522 GEORGE CSORDAS, T. S. NORFOLK AND R. S. VARGA

is an entire function of order 4. If x is a real zero of &(x/2), then 29 := -xg isa
negative real zero of F(z) and the Riemann Hypothesis is equivalent to the
statement that all the zeros of F(z) are real and negative. Now, it is known (cf. Boas
[B, p. 24] or Polya and Schur [PS]) that a necessary condition that F(z) have only
real zeros is that

,, \2 by Baas
(1.7) ml os | > (m4 1 (ma 12.0)

or equivalently, that

(1.8) (b> esate oa. (m=1,2,...).
(In today’s terminology, the inequalities of (1.8) are commonly called Turan inequali-
ties.)

In 1927, Podlya [P], while studying some fragmentary unpublished notes of J. L. W.
V. Jensen dealing with the Riemann Hypothesis, raised the question of whether or
not the Turan inequalities (1.8) are all valid. Our main result here is that these
inequalities (1.8) are indeed all valid. Our interest in these inequalities (1.8) is very
natural: if one of these inequalities (1.8) were to fail for some m > 1, then the
Riemann Hypothesis could not be true!

The history concerning Pélya’s problem of 1927 is very interesting. For nearly
forty years, this problem was apparently untouched in the literature. Then, in 1966,
Grosswald [G1, G2] generalized a formula of Hayman [Hay] on admissible func-
tions, and, as an application of this generalization, Grosswald proved, in the

-—1

notation of (1.8), that
(1.9) (B,)° -(= Jb. ib mune 1+ 0 1 asm —>
‘ 2m+1 mel log m Sm @-

As the moments b,, are necessarily positive (cf. (1.5) and (i) of Theorem A) for all
m > 1, then Grosswald’s result (1.9) proves that (1.8) is valid for all m sufficiently
large, say m > mo, but the value of m, was not determined from this analysis. To
our knowledge, this gap in Grosswald’s solution of Pdlya’s problem was subse-
quently not filled in the literature.

The delicate nature of the Turan inequalities (1.8) can be seen from the following
calculation. As ®() is positive for all ¢ > 0 (cf. (i) of Theorem A), an application of
the Cauchy-Schwarz inequality to (cf. (1.5))

00 2
42 (2m—2)/2 _ (2m+2)/2
2 if t O(t) +t (1) dt)

directly gives (b,,)? <b, om +1 which we equivalently write as

* 2m+1 4
(1.10) in< (Sp a m+ (m = 1,2,...),

whereas the sought Turan inequalities (1.8) are nearly the reversed inequalities:

> (Se ae bpar (m= 1,2,...).

A

y

2m+t+1


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THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 523

In [P], Polya obtained some interesting results that relate the asymptotic behavior,
as t > 0, of ®(t) to the Riemann Hypothesis. In contrast, we focus our attention
on the behavior of ®(1) near ¢ = 0, which requires, in our analysis, a detailed
investigation of ®(7), ®’(r), ®@ (4), and ®© (+f) for ¢ small. This analysis is carried
out in §3. The various estimates developed in §3, while elementary in character,
enable us to show in §3 that the function

(1.11) K(t):i= f ®(vu)du — (t > 0)

is such that log K(Z) is strictly concave on (0, + co). Having gathered these detailed
calculations in §3, the basic ideas of the proof of our main result are given in §2.
There, it is shown that if

(1.12) u*K(u) du (x > -1),

1 lo)
= torn,
then logA,, is strictly concave on (—1, + 00), from which the validity of the Turan
inequalities (1.8) for m > 2 are deduced. (The case m = 1 is settled numerically, the
justifications for this being given in §4.)

In the subsequent sections, we repeatedly make use of several known properties of
the function ©(7), defined by (1.3). For the reader’s convenience, we state the
following theorem which summarizes some of the known properties of ®(f).

THEOREM A. For the function ®(t) of (1.3), write

fee}

(1.13) ®(1)= Ya, (2),

n=1
where
(1.14) a,(t):= wn?(2an7e* — 3)exp(St— an’e") (vn = 1,2,...).

Then, the following are valid:
(i) for eachn > 1, a,(t) > 0 for allt > 0, so that ®(1) > 0 for allt > 0;
(ii) O(z) is analytic in the strip -7/8 <Imz < 1/8;
(iit) B(t) is an even function, so that ®°"™*Y(0) = 0 (m = 0,1,...);
(iv) for any e > 0,
lim ®(t)exp[(7 — e)e*"] = 0

i-> 00
for eachn = 0,1,...;
(v) ®(t) < 0 for allt > 0;
(vi) a‘ (t) < 0 for allt > 0, for each n = 2,3,...;
(vil) a{(1) > 0 for 0 <t < ty, and aj(t) < 0 for allt > to, where

_ 1, [15 + 105
(1.15) y= Glos] — —

With the possible exception of (iil), the proofs of statements (i)—(iv) are elemen-
tary and can all be found in Polya [P]. The proofs of statements (v)—(vii) can be
found in Wintner [W].

| = 0.0 1334898.

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524 GEORGE CSORDAS, T. S. NORFOLK AND R. 8S. VARGA

The fact that a{(t) changes sign (cf. (vii) of Theorem A) is important in our
analysis, so we sketch the proofs of (vii) and (1.15). From (1.14),

(1.16) ai(t) = -2[827e* — 30ae*! + 15]exp(5t — me*’).
Now, the quantity in brackets above, a quadratic polynomial in e*', has precisely
one positive zero fy, which is given in (1.15). It then follows that aj(t) > 0 for

0 <1< tp, and that aj(t) < 0 for ¢ > t9, which is the desired conclusion (vii) of
Theorem A.

2. Basic results. Our basic result, Theorem 2.5, gives that the Turan inequalities
(1.8) are all valid, thereby completely solving Pélya’s problem. The proof of this
result depends in part on a large number of easy but lengthy mathematical
calculations (not numerical computations) which might detract from the basic ideas
of the proof. These results (Lemmas 3.1-3.12) have been gathered separately in §3.
In this section, we give the essential ideas leading to the proof of Theorem 2.5.

We begin with Proposition 2.1, which makes use of Lemma 3.12, to be established
in §3.

PROPOSITION 2.1. With ®(t) defined in (1.3), set

(2.1) K(t):= i. O(vu)du = (t > 0).
Lt
Then, log K(t) is strictly concave on (0, + 00), i.€.,
d” log K(t)
2.2 = —— <0 t> 0).
(2.2) as (1>0)

PRroor. With (2.1), it can be verified that
d?log k(t) __ (fr@(Vu) du) ®ve) /2vi +(@W))"
dt? ( =@(Vu) du)”

As the denominator of the fraction above is positive for all ¢ > 0 (cf. (i) of Theorem
A), then (2.2) holds iff

(23) 9 V(t):= (f ®(Vu) du j=
or equivalently, iff
(2.4) W(7?)= fr s(s) ds) (2) +1(O(1)>0  (t>0).

But, on setting

(¢ > 0).

(vt)

ar +(@(V7))’ > 0 (4 > 0),

(2.5) I(t):= l s®(s)ds (t=),
and

(2.6) g(t):= J(1)@'(1) + (O(2))” — (¢ > 0),
then (2.4) simply becomes

(2.7) g(t)>0 (t>0),

which is the conclusion of Lemma 3.12 of §3. 0


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THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 525

We remark that the results established in §3 similarly allow us to deduce that if
(2.8) I(t):= f° @(s)ds (430),
t

then log /(z) is also strictly concave on (0, + 00). This result is, however, not strong
enough for our purposes to deduce the Turan inequalities of (1.8).

A specific elementary property of strictly concave functions, needed in the
subsequent proof, is given in

LEMMA 2.2. Let I be an open (bounded or unbounded) interval, and let h(x) €
C?(1) be strictly concave on I (i.e., h®\(x) < 0 for x © 1). Then, for any four points
a,b,c,dinIwitha<c<d<b,

(2.9) Ale) — (a) s Ab) — Hd)
If, in addition, c — a = b — d, then
(2.10) h(c)+h(d) > h(a) +h(b).

Proor. Let a,b,c,d be any four points of J with a <c<d< 5, and with
corresponding points P,Q, R,S on the graph of h, as shown in Figure 1. Since
h(x) < 0, it follows that

slope( PQ) > slope( PR) > slope(QR) > slope( RS),
whence slope( PQ) > slope(RS), which gives (2.9). If c — a= b — d, then (2.10)
follows immediately from (2.9). 0
A special case (m = 2) of a problem of Pélya and Szegé [PSz, Part II, Problem

68] is the following lemma. (A more general version of this result appears in Karlin
[K, p. 17].)

oe)

FIGURE 1

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526 GEORGE CSORDAS, T. S. NORFOLK AND R. 8. VARGA

Lemma 2.3. Let f,(t), fo(t), 1(¢) and $,(t) be continuous and absolutely integra-

ble on [0, + co). Suppose further that f(t)o(1)A <i J < 2) and f(t) f,(1)o(o(4)
are absolutely integrable on {0, + 00). Then,

aS ne f(t) b(t) “
fo” 5(t)o,(t) dt {6° 5(1)$,(1) dt

-ff aa alt 2 dae
O<u<v<to flv) Al) $,(u)  ,(v)
In the proof of the next result, it will be convenient to adopt the following

notation. Let X and Y be subsets of R, and let f be a real-valued function on
X X Y. Then, for x,, x, © X with x, < x, and for y,, y, © Y with yy < yp, set

Xy X2\ e flav f(%1, 92)
01) ily Shee heoak

PROPOSITION 2.4. With K(t) defined in (2.1), set

(2.11)

(2.13) hos

. map wK(u)du (x > 1),

where T(t) denotes the gamma function. Then, log A, is strictly concave on (—1, + 9).

Proor. For any real numbers s,f > — 1 the classical formula connecting the beta
function with the gamma function gives the identity

ustt u ps 172 (u- 7

T(s+74+1) “J T(s+1/2) T(t + 1/2)

dv.

Substituting the above identity in (2.13), with x replaced by s +7 and with
yi= u— v, gives

oo s~1/2 oe 1/2
dom ff Tet vy J, rE! iy KO + y)dydo,
which we write as
Pa s-1/2
(2.14) dew f Repay Ei)
where
20 1-1/2
(2.15) LA(x):= ih Teak + y)dy.
We also set
t-1/2
(2.16) O(y)= K(xty)s G(y)= Fra 1/2)


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THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 527

where x, y > 0 and ¢ > — }. With the notation of (2.12), we next note that Lemma
2.3, applied to the integral of (2.15), can be expressed as

X, Xa) xy *2) (2 ia
(2.17) 17 Sf... 2)G (1 12) dudo.
Now, a direct computation shows that if 0<u<v and if t, <t, (where
t, > — 3), then

yar 1W/2yh-1/2

t, to) _
o(% 2) =a nD

[vena — u2-5] > 0.

(2.18)

We next show that

X,  X, 4
(2.19) Ll, ty <0 (0<x, <x;-$< 4 <ty).
To establish (2.19), we see from (2.17) and (2.18) that it suffices to establish that
(2.20) 6(7: 2) <0 (O<x,<x,;0<u<v).

For any x), X, u,v satisfying 0 < x, < x, and 0 < u < », set
(2.21) ai=x,+u, b:= x, +0, c= x,+u and d= x, +2,

so that
a<c<b, a<d<b and c-~a=b-d.

Since log K(f) is strictly concave on (0, + oo) from Proposition 2.1, we deduce from
Lemma 2.2 that (cf. (2.10))

log K(c) + log K(d) > log K(a) + log K(b),

K(c)K(d) > K(a)K(b).

Thus, with the definitions of (2.21), this becomes
(2.22) K(x, + u)K(x, +0) > K(x, + u) K(x, +0).
On the other hand, from (2.12), (2.16) and (2.22), we have
a *2) = det @,(u) 9,(v)
ue 0,,(4) 9,,(v)
= K(x, + u) K(x, +0) — K(x, + v)K(x,+u) <0,

which establishes (2.20).

Next, using (2.13), set A(s,¢):=A,,, (where s > —-—4,2>—- 4). Again from
Lemma 2.3, for any - 3 < t, < t, and } <5, < 5, the notation (2.12) permits us to
write (2.14) in the form

5S, $2 _ 5S, Sy (" 2)
(2.23) al? r= ff. G(3 PE 1) dude.

Now, from (2.18),

G{% a > 0,

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528 GEORGE CSORDAS, T. S. NORFOLK AND R. S. VARGA

and from (2.19),

(0 ‘2 <0.
u v

Thus, it follows from (2.23) that

Sy; 8
a? i} <0 (-1 <5, <53-4 <4, <b),
or equivalently, that
(2.24) Ay tnAdspths 7 Ag anAss+y < 0.

On setting s, = ¢, = u/2 and s, = t, = v/2 (where —-1 < u < v), inequality (2.24)
becomes

(2.25) Muto? > rudvs

which implies that log A, is strictly concave on (—1, +00). O
This brings us to our main result:

THEOREM 2.5. The Turan inequalities (1.8), i.e.,
» \2 2m-1\, 3
(2.26) (b,)? > (FF) Babs (m =1,2,...),
are all valid (where b,, is defined in (1.5)).

Proor. The strict concavity of logA, on (-1, + 00), from Proposition 2.4, gives
that

(2.27) Mn -i2 > dm 32A msi 2 (m = 1,2,...).
Now, since an integration by parts and the change of variables u = t* in (2.13) yield
2 oc
2.28 A= ae Oz) tt,
(2.28) “Terps (1)
(2.27) becomes, from the definition of b,, in (1.5), just
“ 2 2m+1)\. >
or equivalently,
> 2 2m-1)\; 3;
(2.30) (6, > [Fea JB abe (m= 2,3,...).

Thus, (2.30) establishes the desired result of (2.26), except for the case m = 1. This
remaining case, m = 1, of the Turan inequalities (1.8) is then settled numerically, as
follows. The numbers {6,,}2,-9 were determined by Romberg integration to an
accuracy exceeding fifty decimal places, and the associated Turan difference, namely,

(2.31) (b,)° — 4b )b, = 3.5884 ---10-* > 0,

was determined. (The details giving rigorous error bounds for these numerical
calculations appear in §4.) Thus, (2.30) and (2.31) give the desired result of (2.26).
O

We have in fact numerically determined the moments {4,,}2°.4, each to an
accuracy of fifty decimal places, as well as the associated Turan differences, { D,,, ye


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THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 529

where

A a

2m+1
These numbers have been included in §4 for the benefit of the reader.

3. Background analysis. The purpose of this section is to obtain precise estimates
for the functions ®(¢), ®/(t), ®@(t) and © (1), where ®(1) is defined in (1.13)
and (1.14). For convenience, we will adhere to the following notations. For ¢ > 0, set

(3.1) a,(t):= an?(2an7e* — 3)exp(St — mn?e*’) s(n = 1,2,...),

(3.2) B= YS a,(0)

and 7

(3.3) ®,(1):= dL 4,(t).
LEMMA 3.1. Setting

(3.4) I(1):= [209 (1 > 0),

then

_ 7 _ pdt) _} © I/4a-y “
(3.5) I(t)= 5 exp(St me“) ada? e dy + f ®,(y) dy.

8a
PRrooF. From (3.2)-(3.4), I(t) = f° a,(y) dy + f° ®,(y) dy. From (3.1), the in-
tegral {°° a,(y) dy consists of the difference of two terms. Integrating each of these
by parts and adding the results yields the desired result of (3.5). O
In the next result, upper and lower estimates for I(t) of (3.4) are derived.

LEMMA 3.2. With the definition of (3.4),
1 -4t 1 -8¢

a — ge4t\|] — —-e74t —
(3.6) I(t)> 5 exp(5! me*")|1 an¢ ién?® (t> 0)
and
3.7 I(t) < Zexp(5t — we t> 0).
2

Proor. From Theorem A(i), it follows that ®,(t)> 0 for all ¢>0. Thus,
[7° ®,(s) ds > 0 for all t > 0. Thus, from (3.5) of Lemma 3.1,

1 20 _
al Je "dy  (t>0).

Applying integration by parts to the last integral above yields

I(t) > 5 exp(5e — me*) —

(3.8) I(t) > Lexp(5t — me) ~ Zexp(t — ne)

~—1_ f” y3/4p-rg
32 q1/4 net 7

Next, for the complementary incomplete gamma function

T(y; x):= [oye ray (0<x<o,v<1),
x

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530 GEORGE CSORDAS, T. S. NORFOLK AND R. S. VARGA

it is known (cf. Luke [L, p. 201)) that

x+1
resx)< (35-5)
Choosing »v = | and x = ze*’ and applying this inequality to the last integral of
(3.8) then directly gives the desired lower bound (3.6) of Lemma 3.2.
Next, Haviland [Hav, p. 415] proved that ®,(f) of (3.3) satisfies
(3.9) ®,(s) < 6427exp(9s — 4re**) — (s > 0).

Inserting the above inequality into the last integral of (3.5) then yields

7 1 oo _
(3.10) I(t) < Fexp(st ~ me") — [ ae dy

x’ te-* (Q0<x<o,v<1).

5/4974
SL, y * dy.

Next, since ye~>” is strictly decreasing for y > }, one obtains the elementary
inequality

ye“? < me 3tyV/4e-¥ (yp > ar).
Applying the above inequality to the integrand of the last integral in (3.10), (3.10)
then becomes

3/4
(3.11) (1) < Sexp(51 - ne’) +(- 1 |

* 1/4,-¥
8arl/4 J ery.

But as 1693/4/e37 < 1/87'/4, ie., 1287e~3"( = 0.032 451---) < 1, the last term in
(3.11) is negative, whence I(t) < (7/2)exp(5t — me*’) for all ¢ > 0, the desired
inequality of (3.7). O

Lemma 3.3. With the definition of (3.3),
(3.12) |@i(z)| < 565a%exp(13t—4me*") = (¢ > 0).

PROooF..From se) and (3.3),

|@i(7)| = S° an2(8n2n 498 _ 30qn7e* + 15)exp(5t — an’e*") (t > 0),
n= d
or equivalently, with x:= e',
(3.13)
hes 15 15
|@)(1)| = 82x? > nf(x' ~ Gon 5x4 + oe? -; Jexp(- ~anx*) (x > 1).
n=2 7H

It is easily verified that -15x4/4nn? + 15/807n4 < 0 for all x > 1 and all n > 2,
so that, with y:= ax*, (3.13) becomes
ys 4 6

(3.14) |@i(t)| < v7 > ne -ny (y>7).

n=

As n’e7""» is a monotone decreasing function of n > 2 for each fixed value of
y > a, then by the integral test, we have that

> n’e-"” < 64e74" + ['s Se ds,
n=2


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THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 531

On making the substitution u:= s*y in the above integral and on integrating by
parts three times, we obtain

re) ~4y
6,-my -4y , & 372, 3 3/2, 15 1/2
un e < 64e°4" + aaa (4) + 5 (4y) +3 (4y)

15e4” px en
+ —= du}.

Since 1/ Yu < 1 for all u > 4y > 4m, then ete" du/ vu <1, so that the above
inequality becomes

Yi n’eo”” < 64074” 1 5 15 15

I+a + + >).

Now, the quantity in braces above is monotone decreasing for y > 7, and the value

of this quantity when y = a is bounded above by 1 + (13/407). Thus, we have
= 2 13

6,-ny -4y a.
y ne < 64e (i+ ioc} (y>7),

n=2

so that, from (3.14) and the fact that y = we*,

; 13
|@/(1)| < 512(1 + aig) exp (131 — 4“)

< 565a7%exp(13t—47e") (1 >0),
the desired inequality of (3.12). O

LEMMA 3.4. With the definition of (3.3),
oo fora) 1 oo fore) _ _
(3.15) i ds f ®,(y) dy < pal asf oy /4e-Y dy — (t > 0).

Proor. By Haviland’s upper estimate (3.9) for ®,(s) and the substitution v:= ze*’,

ce) 00 16 oo oo -4y
(3.16) / ds f ®,(y) ay < <al asf vv4e-4°?dy s(t > 0).

Next, since v7e~?”

inequalities

is strictly decreasing for v > 7, one obtains the elementary

2y-3/4e-8 3 /4g 8
i nnrsy)

which, when applied to (3.16), directly gives the desired result of (3.15). O
LEMMA 3.5. With the definition of (3.2), set

yi/4e-4¥ 7 (v>q),

(3.17) I(t)= i? s®(s)ds (t>0).
Then, ‘
(3.18) I(t) < (F + S leww(sr ~ne)  (t>0).

ProoF. An integration by parts shows, with (3.4), that
(3.19) I(t) = a(t) + f~ 1(s) ds.
t

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532 GEORGE CSORDAS, T. 8S. NORFOLK AND R. S. VARGA

Now, with the expression (3.5) of Lemma 3.1, with one integration by parts, and
with some easy simplifications,

* _l _ oaty 1 * e3/4o=¥
i I(s)ds = — exp( me“) pal asf oy e~" dy

+f as f° ®,(y) dy.

Thus, on applying inequality (3.15) of Lemma 3.4 to the above expression, we obtain

(3.20) ia I(s) ds < = exp(t ne") (130).

Then, applying (3.20) and the upper bound (3.7) of Lemma 3.2 to (3.19) directly
gives the desired inequality of (3.18). O

Our proof in Proposition 2.1 of the strict concavity of log K(7) on (0, + co), where
(ef. (2.1))

K(t):= i. ®(yu)du s(t > 0),
t
is based on the assertion that the function g(7), defined (cf. (3.26)) by
g(t):= J(r)@(2) + ¢[@(2)]* (4 > 0),

is positive for all ¢ > 0. The proof of this assertion will be divided into the two cases:

(3.21) g(t)>0 fort € (0,0.01]
and
(3.22) g(t)>0 fort > 0.01.

The case of (3.22) will be a consequence of previously established estimates. For the
case of (3.21), we first note that since ©’(0) = 0 (cf. (iii) of Theorem A), g(0) = 0 by
(3.26). By Taylor’s formula, we can then write

(3.23) s(s) =1|a'(0)+ ari] (€ € (0,1);0 <1 < 0.01).

Consequently, in order to establish (3.21), it suffices to show that

(3.24) (0) + 9 (Ge (0,1):0 <1 < 001).

This last inequality requires that we estimate g’(0) and g(t) for 0 <7 < 0.01.
Since by (3.26),

(3.25) g(t) = 36/(1) @(2) + 1[O(2)]? + I(t) OO(8),

it is also necessary to examine the behavior of ®®(7) on [0,0.01]. We remark that
the main reason for concentrating on this particular interval [0,0.01] is that it is
relatively easy to prove that ®t) > 0 on this interval.

Preliminaries aside, we proceed to establish some lemmas for establishing (3.21).
The reader may find it useful to have a hand calculator available while reading
portions of what follows.


===== tmp/pdfs/o0176-cnv/page-13.txt =====
THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 533

LEMMA 3.6. With the definitions of (3.2) and (3.17), set

(3.26) g(t):= J(t)®(t) + t[B(t)]? s(t > Od).
Then,

(3.27) ®(0) > 0.446 696 899 ---,

(3.28) 6 (0) > —33.461010---,

and

(3.29) g’(0) > 0.018 790 450 ---.

PRooF. From (3.2) and (i) of Theorem A, it follows that ®(0) > a,(0) + a,(0).
Simply evaluating @,(0) and a,(0) and adding, yields the result of (3.27). Next, from
(3.2), we have 8 (7) = L*_, ar), where from (3.1),

(3.30) — a(0) = wn? [32a3n® — 22407n* + 330m? — 75]exp(—mn’).

With x:= an’, the quantity in brackets above is a cubic polynomial in x, having
three distinct zeros 0.277455 812 ---, 1.672823 383---, and 5.049720804---.
Since x = mn? > 5.049720804--- for all n > 2, then a®(0)> 0 for all n > 2.
Thus, a lower estimate for ®(0) is given by

© (0) > a(0) + a?(0).

Evaluating a{(0) and a9)(0) from (3.30) and adding then yields the result of (3.28).
To derive (3.29), the definition of g(z) in (3.26) provides us with

(3.31) g’(t) = 18'(t) B(t) + I(t) ®(t) +[O(2)]> (4 > 0),
so that
(3.32) g’(0) = J(0) © (0) + [@(0)]°.

Now, J(0) < e°7/8 = 0.0005 401 739 --- from (3.18) of Lemma 3.5, and applying
this inequality (along with those established in (3.27)—(3.28)) in (3.32) yields the last
inequality, (3.29), of Lemma 3.6. O

LEMMA 3.7. With the definition of (3.2),
(3.33) O®P(1)>0. (0<1< 0.01).

Proor. Since. © (0) = 0 from (iii) of Theorem A, it suffices to show that
®(7) > 0 on (0, 0.01]. From (3.2), it follows that
(3.34) (7) = VY a(t) = VY an? exp(5t — rn7e*") p.( ane’),

n=1 n=l

where
(3.35) p5(x)s= 512x° — 8,448x4 + 41,408x? — 68,096x? + 30,930x — 1,875.

The above polynomial has five distinct zeros, given by 0.071 349 --- , 0.604398 --- ,
1.996 885 --- , 4.617597---, and 9.209769 ---, so that p(x) > 0 for x > 9.210.
As an? > 9.210 for all n > 2, it follows from (3.34) that

aS(t)>0 (n>2,t>0),


===== tmp/pdfs/o0176-cnv/page-14.txt =====
534 GEORGE CSORDAS, T. S. NORFOLK AND R. S. VARGA

so that

(3.36) O%(1)> a(t) (> 0).

Thus, it suffices to show that a{?(r) = mexp(5t — 7e*") - ps(me*") is positive on
(0, 0.01].

The derivative of the polynomial p,(x) (cf. (3.35)) has four distinct zeros, given
by 0.30515 --+, 1.3791 ---, 3.6496 ---, and 7.8660 ---. In particular, p(x) is thus
increasing on the interval (1.3791 ---, 3.6496 ---). Since me“ falls in this latter
interval for all 0 < ¢ < 0.01, then p,(me*') > ps(7) for all 0 < ¢ < 0.01. Similarly,
since exp(5t — me*") is decreasing for all r > 0, we then have
(3.37)  a{®(1) > wexp(.05 — me) - ps(7) > 5,133. (0 < t < 0.01).
Consequently (cf. (3.36)), ®(1) > 0 for all 0 < ¢ < 0.01, which gives the desired
inequality (3.33). 0

Since our goal is to estimate g(t) on the interval [0,0.01], and since the
expression for g(r) involves the term 3@’(r)®(r) (cf. (3.25)), we next derive an
estimate for 3’(t)®(t).

LEMMA 3.8. We have

(3.38) [36’(r)@(t)| < 0.506 (0<1< 0.01).
PRooE. By definition (ef. (3.2) and (3.3)),
(3.39) @(t)=a,(t)+@(¢t) (¢ 20),
and we first show that
1
(3.40) @,(t) < 599 21(¢) (t > 0).

Since ®,(1) < 6477exp(9t — 4me*') for all t > 0 from (3.9), to establish (3.40) it
suffices to show that

1
2 _ 4t
64m7exp(9t — 4e*’) < 50) a,(t) (1 > 0),

or equivalently (cf. (3.1)),
3
27e

As is easily seen, the above inequality is valid for all ¢ > 0 if it holds for 1 = 0:

6464 exp(—37e") < 1 - (t > 0).

40
(.521 641681 --- =)6464e7°" < 1 - (= 522535170 ---).

As this is valid (3.40) then follows. Consequently, combining (3.39) and (3.40) gives
(with (i) of Theorem A),

203
(3.41) 0 < ®(t) < 59 144) (t > 0).
Continuing, from (3.1), we see that

Jexp(or —mer")< 20(1 _3 Jexp(or — me")
27e™

a(t) = 2n(1 ~5 3

qe’!

for 0 < ¢ < 0.01, so that
a,(t) < 1.082513 669m7exp(9t — we“) = (0 < t < 0.01).


===== tmp/pdfs/o0176-cnv/page-15.txt =====
THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 535

Thus, from (3.41), there follows
(3.42) 0 < ®(t) < 1.087872 64877exp(9r — me*’) = (0 < ¢ < 0.01).
We next estimate ®'(t) = aj(t) + O{(1). Recalling from (vii) of Theorem A that

ay(t) > 0 for 0 < t < Zo, where fy is explicitly given in (1.15), then, as (1) < 0 for
all t > 0 (cf. (v) of Theorem A), we have

0 > O(t) = aj(t) + O{(t) > O{(2) = -|O{(t)|  (O<t <4),

so that
(3.43) |O(t)| <|@{(1)| = (O<t<t).
Hence, from (3.42), (3.43) and the upper bound (3.12) for |®{(1)|, we have
3|@’(1) ®(t)| < 3(1.087 872 64877 exp(9r — me *’))565a3 exp(13r — 4“)
forO <t < ty, Le,
3|®’(1)®(t) | < 1,843.944 1387°%exp(22r - 5me*’) = (O<t < ty).
As exp(22r — 5ze**) is strictly decreasing for all ¢ > 0, then
3]®’(t) ®(t) | < 1,843.944 13805e-5" < 0085038454 = (0<t <4),
so that certainly
(3.44) 3|@(r)@(1)| < 0.506 = (0O<4< ty).
On the other hand, if ty < ¢ < 0.01, from (3.42), (3.2) and (3.3),
(3.45) 3|@’(1)®(r)| < 3(1.087 872 648x7exp(9t — me*’))(|ai(t) | +] ®i(t)|).

At this point, we need an upper bound for |a;(t)|. Clearly, from (3.1),

(3.46) |ai(t)| = 7|807e* — 30me* + 15|exp(51 — we“)
15 15
= en°(1 ~ Trott + oe? -, Jexp( 3 — me"),
Next, if we set
15 15
@(1):= 1—- +
(1) A4ne*' == Bare *"

it is easily seen that

max (1) = (0.01) = 0.028 513 162 -
0<1<00

Thus, combining the above with (3.46) yields
(3.47) |at(z)| < 87°@(0.01)exp(13¢- me“) (0 <4 < 0.01).

Now, using (3.47) and the upper bound for |®;(1)| in (3.12) of Lemma 3.3, we have
from (3.45) that

(3.48) 3|®’(2)®(t) | < 3(1.087 872 648) 7° exp(227 — 27e*')
x [8@(0.01) + 565 exp(—37e*")|


===== tmp/pdfs/o0176-cnv/page-16.txt =====
536 GEORGE CSORDAS, T. 8S. NORFOLK AND R. 5S. VARGA

for tp < ¢ < 0.01. Now, the quantity in brackets above is strictly decreasing for all
t > 0, the same being true for the factor exp(22t — 27e*"). Thus, the maximum of
the right side of (3.48), for 0 < ¢ < 0.01, is taken on at ¢ = fy, which gives

(3.49) 3|@’(1) ®(r)| < 0505076975 < 0.506 (0<4< 0.01).
Combining the above with (3.44) gives the desired result of (3.38). 0
LEMMA 3.9. With the definition of g(t) in (3.26),

(3.50) g(t)>0 (0<+t< 0.01).
Proor. To establish (3.50), it suffices, from (3.23), to show that (cf. (3.24))

Q)
(3.51) g’(0) + eee 0 (€€(0,1);0 <1 < 0.01).
Now, by (3.25), we have that

g(t) = 30’(t) O(t) + t[O(2)]? + I(r) OO(2).

Since J(t) > 0 for all ¢ > 0 from (3.17), and since ®°)(7) > 0 for 0 < ¢ < 0.01 from
(3.33) of Lemma 3.7, it follows that, with (i) and (v) of Theorem A,

g(r) > 30(2)O(r) = -3|@(2)O(t)| (0 <1 < 0.01).
Hence, from (3.38) of Lemma 3.8,
(3.52) g(t) > -0.506 (0<¢< 0.01).
On the other hand, by (3.29) of Lemma 3.6, g’(0) > 0.018 790 453. Thus, with (3.52),
(2) .
2’(0) + arte} > 0.018 790.453 + o9 (-0.506),

or

Q)
g’(0) + & (e) > 0.016260453 (0<r<0.01),

which establishes (3.51). O
It is still necessary to show that g(7), defined in (3.26), is positive for all ¢ > 0.01.
To this end, we decompose g(t) as

(3.53) g(t) = G,(t) + G,(2),

where

(3.54) G,(t):= J(t)ai(t) + ta?(t),

and where

(3.55) Gy(t):= I(t) @{(1) + 21a,(1) ®,(t) +[,(2)]’.

Our next immediate goal is to provide bounds for G,(¢) and G,(7).
LeMMaA 3.10. Set
(3.56) E,(t):= wexp(10r — 2ae"') ¥, (2),


===== tmp/pdfs/o0176-cnv/page-17.txt =====
THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 537

where

; 3¢ 15 15
(3.57) W(t):= me* (3t _ 1) + oy + van Smet
Then (ef. (3.54)),
(3.58) G4)>E(t) (t> 0.01).

PROOF. From (vii) of Theorem A, we have that a{(t) < 0 for all t > t, where (cf.
(1.15)) ¢9 = 0.0001 1334 ---. Thus, from the definition of G,() in (3.54) and from
(3.18) of Lemma 3.5,

mt e *
(3.59) G,(t) > & 3 Jexots —me*")-ai(t)+taz(t)  (t> to).
On substituting the definition of a,(t) (cf. (3.1)) in the right side of (3.59) and
simplifying, the right side of (3.59) reduces exactly to E,(t) of (3.56). As fy < 0.01,
(3.58) is then evidently satisfied. O

LEMMA 3.11. Set

(3.60) E,(t):= m*exp(10r — 27e*') ¥, (1),
where

. 5652
(3.61) W,(¢):= exp(8t — 37e*')| -223.6074 — Bet! ).

Then (cf. (3.53)),
(3.62) G,(t)> E(t)  (t>0).

PROOF. Since a,(f) > 0 for all ¢ > 0 and for each n > 1 from (i) of Theorem A,
then ©,(¢) > a(t) for all ¢ > 0 from (3.3). Hence (cf. (3.55)),
(3.63) Gy(t) > S(t) ®{(t) + 2ta,(t)a,(t) (t= 0).
Since J(t) > 0 from (3.17), for all ¢ > 0, and since ®{(7) < 0 from (3.3) and (vi) of
Theorem A, for all ¢ > 0, then by (3.12) of Lemma 3.3 and (3.17) of Lemma 3.5, we
have

4t

J(1)®/(1) > -s650°( 3 + S Jexp(ase ~5ne) (4 > 0).

Also, from (3.1), we have
2ta,(t)a,(t) = 1677¢(27e*! — 3)(8me* — 3)exp(10t ~ Sze“).

Substituting the above two expressions into (3.63) and simplifying then gives

me“

which, from the definitions of (3.60) and (3.61), is the desired result of (3.62). O
This brings us to the final result of this section, namely
LEMMA 3.12. With the definition of g(t) in (3.26), then

(3.64) g(t)>0 (t>0).

G,(t) > wtexp(18r — Sme*){ 293.64 - — \,

===== tmp/pdfs/o0176-cnv/page-18.txt =====
538 GEORGE CSORDAS, T. S. NORFOLK AND R. 8. VARGA

Proor. If 0 < 1 < 0.01, then g(t) > 0 by (3.50) of Lemma 3.9. Then, it suffices to
consider only ¢ > 0.01. Then, from (3.53) and Lemmas 3.10 and 3.11, we have

g(t)> E,(t)+£,(t) (1 > 0.01),
which, from (3.56) and (3.60), can be equivalently expressed as

(3.65) g(t) > wexp(10r — 2me*)¥(t) = (t > 0.01),
where
(3.66) ¥(t):= Y(t) + &(r).
From (3.57) and (3.61), we verify that
(3.67) Yi(t) = ne"(12t-1) + 5 + B_,
2ae*!
and that
(3.68)
4 = 3me" 21(2683.27e*' — 1788.8 623.90 — 565.
Wi (t) = exp(8t — 3e*'){ 77t(2683.2me% — 8) + 7[ 623.90 sea |}

It is clear from (3.68) that ¥3(t) > 0 for all ¢ > 0. Similarly, we claim that ¥j(1) > 0
for all ¢ > O. First, from (3.67), we see that ¥{(0) = 0.74573 --- > 0, and that
30

’
me‘!

V(t) = 48ate% + 8re*! —

so that ¥(r) > 0 for all ¢ > 0. Hence, ¥i(r) > 0 for all ¢ > 0. Thus, from (3.66),
(1) is strictly increasing for ¢ > 0, with

¥(1) > ¥(0.01) = ¥,(0.01) + ¥,(0.01) = 0.0058629--- >0 (4 > 0.01),

and we conclude from (3.65) that
(3.69) g(t) > wexp(10r — 2me*)¥(t)>0 (1 > 0.01),
which gives the desired inequality (3.64). 0
We remark that the function Y(t) of (3.66) is, in fact, negative for 0 < t < 0.005,

which supports the necessity of separately considering the two intervals 0<t<0.01
and ¢ > 0.01 in the proof of Lemma 3.12.

4. Numerical computation of moments and Turan differences. The accurate calcula-
tion of the moments 6,, (m = 0,1,2,...) of (1.5) involves two separate numerical
problems. First, from (1.3), we can express ®(t) of (1.3) in the form

fo. 6]

(4.1) 6(t)= ¥ a(n,t),

n=1
where (cf. (1.14))
(4.2) a(x,t):= (292x4e% — 30x7e* )exp(—mx7e*’).
As in (i) of Theorem A, it is readily verified that
(4.3) a{x,t)>0 (x =1,1> 0),


===== tmp/pdfs/o0176-cnv/page-19.txt =====
THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 539

and that

da(x,t)

(4.4) ax

<0 (x21,120).

Because of (4.4), the integral test gives that

N

(4.5) | 0<@(t)- Y a(n,t)< im a(x,t) dx.

n=1 N
Moreover, it can be verified (after an integration by parts) that

ie a(x,t) dx = tNexp(5t — 7N7e**),
N

so that (4.5) becomes
N
(4.6) 0< ®(t)- > a(n,t) < tN%exp(St — 7N7e%).
n=l
The above upper bound for the error, in approximating ®(r) by its partial sum of N
terms, turns out to be quite accurate.
Next, for the moments b,, of (1.5), we write

a

b

nm

= tr ?"@(t) dt = ft e"@(1) dt + ft t?"@(r) dt,
0 0 1
or

3 — [1 ,2m Lo (amt) (=) _
(4.7) 5, [i B(r)ar+ f u (—|du  (m=0,1,...).

Because of the exponential decay to zero of ®(1) as t — 00, the singularity at u = 0
in the last integral of (4.7) is removable for each m > 0.

The numerical procedure used for calculating the moments 6,, was the following.
The two integrals in (4.7) were each approximated numerically by Romberg integra-
tion (cf. Stoer and Bulirsch [SB, p. 132]}, where ®(7) was approximated by the finite
sum in (4.6). The iteration in Romberg integration was continued (for each integral
in (4.7)) until two entries in a single column agreed to sixty decimal digits. For the
values ®(¢) of the integrands of the integrals in (4.7), the associated number N (of
the terms of the finite sum approximation to ®(f)) was selected so that the
approximation error in (4.6) was less than 10~°°. The computations were performed
in FORTRAN 77, using Richard Brent’s MP package (cf. Brent [Br]) for extended-
precision floating-point numbers and 110 digits of precision, on a VAX-11/780 in
the Department of Mathematical Sciences at Kent State University. The absolute
error in computing the moments {6,,}7°_, was less than 10~*° in all cases. While it
appears from Table 4.1 that the moments 5, are decreasing quite rapidly, we
mention the fact that they are eventually increasing. (The details of this will appear
elsewhere.) The relative error of these moments {é,,}7°_, was less than 10~“° in all
cases.


===== tmp/pdfs/o0176-cnv/page-20.txt =====
540

Though only the first three moments {4,,}?,.9 were specifically needed in §2 to
complete the proof of Theorem 2.5, it was thought that a lengthier tabulation of
these moments might be of interest to the reader, particularly since such a tabulation
of these moments does not exist in the literature. Although the moments {6,,, }}0?
were actually numerically determined, we have, for the sake of brevity, included in

Table 4.1 only the moments {,,}7°_5, here rounded to sixteen significant digits.

GEORGE CSORDAS, T. S. NORFOLK AND R. S. VARGA

m=?

Also included in this table are the associated Turan differences { D,, }!?_,, where

(4.8)

TABLE 4.1

(2% \? 2m—-1\, a _

a

b

m

D

m

6.214 009 727 353 926 (—2)

7.178 732 598 482 949 (—4)

3.588 449 148619957 (-8)

2.314 725 338 818 463 (-5)

3.163 299 395 056 600 (-11)

1.170 499 895 698 397 (—6)

7.056 732 441 900 485 (-14)

7.859 696 022 958 770 (—8)

2.832 220 223 070 768 (-16)

6.474 442 660 924 152 (—9)

1.736 366 689 470 613 (-18)

6.248 509 280 628 118 (-10)

1.478 031 720 106 092 (—20)

6.857 113 566 031 334 (-11)

1.641 533 684 538 624 (—22)

8.379 562 856 498 463 (—12)

2.277 443 847 755 004 (-24)

OP CO} HE HD) WB] WwW) NY) eR) Oo]

1.122 895 900 525 652 (-12)

3.822 737 726 048 953 (-26)

oy
Oo

1.630 766 572 462 173 (-13)

7.575 377 587 713 463 (-28)

be
oy

2.543 075 058 368 090 (-14)

1.738 493 426 852 891 (—29)

row
N

4.226 693 865 498 318 (15)

4.549 255 646 782 005 (-31)

ry
Ww

7.441 357 184 567 353 (-16)

1.340 195 434 809 036 (—32)

—_
>

1.380 660 423 385 153 (—16)

4.397 768 675 764 370 (-34)

—
ws

- 2,687 936 596 475 912 (-17)

1.593 011 938 279 461 (—35)

—_
ON

5.470 564 386 990 504 (-18)

6.320 855 730 991 445 (—37)

ar
~~

1.160 183 185 841 992 (-18)

2.728 993 526 800 843 (—38)

_
oo

2.556 698 594 979 872 (-19)

1.274 579 325 080 585 (-39)

~—
‘oO

5.840 019 662 344 811 (-20)

6.406 797 431 277 575 (-41)

NO
So

1.379 672 872 080 269 (—20)


===== tmp/pdfs/o0176-cnv/page-21.txt =====
THE RIEMANN HYPOTHESIS AND THE TURAN INEQUALITIES 541

REFERENCES

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[Br] R. P. Brent, 4 FORTRAN muttiple-precision arithmetic package, Assoc. Comput. Mach. Trans.
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[Gi] E. Grosswald, Generalization of a formula of Hayman, and its applications to the study of Riemann’s
zeta function, Illinois J. Math. 10 (1966), 9-23.

[G2] . Correction and completion of the paper “Generalization of a formula of Hayman”, Ulinois J.
Math. 13 (1969), 276-280.

[Hav] E. K. Haviland, On the asymptotic behaviour of the Riemann &-function, Amer. J. Math. 67 (1945),
411-416.

[Hay] W. K. Hayman, A generalization of Stirling’s formula, J. Reine Angew. Math. 196 (1956), 67-95.

[K] S. Karlin, Toral positivity, Stanford Univ. Press, Stanford, Calif., 1968.

[L] Y. L. Luke, The special functions and their approximations, Vol. II, Academic Press, New York,

1969.
[P] G. Pélya, Uber die algebraisch-funktionentheoretischen Untersuchungen von J. L. W. V. Jensen, Kgl.

Danske Vid. Sel. Math.-Fys. Medd. 7 (1927), 3-33.
[PS] G. Pélya and J. Schur, Uber zwei Arten von Faktorenfolgen in der Theorie der algebraischen
Gleichungen, J. Reine Angew. Math. 144 (1914), 89-113.
[PSz] G. Pélya and Szegé, Problems and theorems in analysis, Vol. 1, Springer-Verlag, New York, 1972.
[SB] J. Stoer and R. Bulirsch, Introduction to numerical analysis, Springer-Verlag, New York, 1980.
[T] E. C. Titchmarsh, The theory of the Riemann zeta function, Clarendon Press, Oxford, 1951.
[W] A. Wintner, A note on the Riemann é-function, J. London Math. Soc. 10 (1935), 82-83.

INSTITUTE FOR COMPUTATIONAL MATHEMATICS, KENT STATE UNIVERSITY, KENT, OHIO 44242 (Current
address of R. S. Varga)

Current address (George Csordas): Department of Mathematics, University of Hawaii, Honolulu,
Hawaii 96822

Current address (T. 8. Norfolk): Department of Mathematics, University of Akron, Akron, Ohio 44315
