A zero-free region ofhypergeometric zeta

Abdul Hassen, Hieu D. NguyenRowan University

Math Department SeminarRowan UniversityOctober 5, 2005

The Search for Nothing

2

-4 -2 0 2

-20

-10

0

10

20

0

3

Riemann Zeta Function

1

01

1 1( ) (Re( ) 1)

( ) 1

s

s xn

xs dx s

n s e

s it (Complex variable)

Born: 15 April 1707 in Basel, SwitzerlandDied: 18 Sept 1783 in St Petersburg, Russia

Born: 17 Sept 1826 in Breselenz, Hanover (now Germany)Died: 20 July 1866 in Selasca, Italy

Georg Friedrich Bernhard RiemannLeonhard Euler

http://www-gap.dcs.st-and.ac.uk/~history/

4

Analytic Continuation

1

2

2 2 2 21

3 3 3 31

1 1 1 1(1) ...

1 2 3

1 1 1 1(2) ...

1 2 3 6

1 1 1 1(3) ... ?

1 2 3

n

n

n

n

n

n

Theorem: (Euler/Riemann) (s) is analytic for allcomplex values of s, except for a simple pole at s = 1.

Special Values of Zeta:

Diverges (N. d’Oresme, 1323-82)

Basel Problem (L. Euler, 1735)

Irrational (R. Apery, 1978)

5

Zeros of Zeta

Riemann Hypothesis (Millennium Prize Problem):The nontrivial zeros of (s) are all located on the‘critical line’ Re(s) = 1/2.

Trivial Zeros: (–2) = (–4) = (–6) = … = 0

http://mathworld.wolfram.com/RiemannZetaFunction.html

http://users.forthnet.gr/ath/kimon/Riemann/Riemann.htm

6

Zero-Free Regions of Zeta

1

1 prime

1 1( ) 1 (Re( ) 1)

0

s sn p

s sn p

Theorem: (E/R) (s) 0 on the right half-plane Re(s) > 1.

Proof: Follows from Euler’s product formula:

Theorem: (E/R) (s) 0 on the left half-plane Re(s) < 0,except for trivial zeros at s = –2, –4, –6, ….

Proof: Follows from the functional equation (reflectionabout the critical line Re(s) = 1/2):

1( ) 2(2 ) sin (1 ) (1 )2

s ss s s

7

2

01

1( ) ( )

( 1) ( )

s N

N xN

xs dx N

s N e T x

Hypergeometric Zeta Functions

2

0

( ) 1 ...! 2! !

n NN

Nn

x x xT x x

n N

where

Cases:

2 0

12 : ( )

( 1) 1

s

x

xN s dx

s e x

1

3 20

13: ( )

( 2) 1 / 2

s

x

xN s dx

s e x x

1

1 0

11: ( )

( ) 1

s

x

xN s dx

s e

(Classical zeta)

8

1( )s

2 ( )s

3( )s

2 3 4 5

1

2

3

4

5

Theorem: For N > 1 and real s > 1,

1( ) ( )N s s

Theorem: N (s) is analytic for all complex values of s,except for N simple poles at s = 2 – N, 3 – N, …, 0, 1.

-2 -1 1 2(N = 2)

9

1

1

( ) 2 (1 ( 1)) cos[( 1)( )]sN n n

n

s s N r s

Pre-Functional Equation

Theorem: For Re(s) < 0,

Here zn = rnein are the roots of ez – TN-1(z) = 0 in theupper-half complex plane.

N = 1: The roots of ez – 1 = 0 are given by zn = rnein = 2ni.1

11

1 1

1

11

( ) 2 (1 ) (2 ) cos[( 1)( / 2)]

2(2 ) sin (1 )2

2(2 ) sin (1 ) (1 )2

s

n

s s

n

s

s s n s

ss n

ss s

(Functional equation)

10

Zero-Free Region of Hypergeometric Zeta

Main Theorem: 2(s) 0 on the left half-plane{Re(s) < 2}, except for infinitely many trivial zeroslocated on the negative real axis, one in each of theintervals Sm = [m + 1 , m ], m 2, where

1

2

1

2.40781

m

m

Proof (?): - No product formula - No functional equation - No swift proof

123

11

1

1

( ) 2 (1 ( 1)) cos[( 1)( )]sN n n

n

s s N r s

I. Key ingredient: Pre-functional equation for N (s):

Sketch of Proof (Spira’s Method)

II. Divide Re(s) < 2 and conquer:

III. Demonstrate that pre-functional equation isdominated by the first term (n = 1) in each region andapply Rouche’s theorem to compare roots.

23 1

1

1

11 2

n nn n

a a a

12

Roots of ez – 1 – z = 0 (N = 2)

(2 1/ 4) (2 1/ 2)nn y n

Lemma: (Howard) The roots {zn = xn+iyn} ofez – 1 – z = 0 that are located in the upper-halfcomplex plane can be arranged in increasing orderof modulus and argument:

| z1| < | z2| < | z3| < …| 1| < | 2| < | 3| < ...

Moreover, their imaginary parts satisfy the bound

13

-1 1 2 3 4 5

10

20

30

40

50

60

70

N = 1

N = 2

{n xn yn (2n+1/4) (2n+1/2) rn n }{1 2.0888 7.4615 7.069 7.854 7.7484 1.2978}{2 2.6641 13.879 13.352 14.137 14.132 1.3812}{3 3.0263 20.224 19.635 20.420 20.449 1.4223}{4 3.2917 26.543 25.918 26.704 26.747 1.4474}{5 3.5013 32.851 32.201 32.987 33.037 1.4646}{6 3.6745 39.151 38.485 39.270 39.323 1.4772}{7 3.8222 45.447 44.768 45.553 45.608 1.4869}{8 3.9508 51.741 51.05 51.84 51.892 1.4946}{9 4.0648 58.032 57.33 58.12 58.175 1.5009}{10 4.1671 64.322 63.62 64.40 64.457 1.5061}

Table of Roots of ez – 1 – z = 0 (N = 2)

14

Theorem 1: (Dark Region) Let s = + it. If < 2

and |t| > 1, then 2(s) 0.

2 ( ) 2 ( ) ( )s s I s

Proof: We demonstrate that the pre-functional equationis dominated by the first term. Write

where

1

1

1 11 1

2

( )( ) 1

11 1 1

2 1 1

( ) cos[( 1)( )]

cos[( 1)( )] cos[( 1)( )]

cos[( 1)( )]cos[( 1)( )] 1

cos[( 1)( )]

( )(1 ( ))

sn n

n

s sn n

n

g sf s s

s n ns

n

I s r s

r s r s

r sr s

r s

f s g s

15

Assuming |g(s)| < 1, it follows from the reversetriangle inequality that

( ) ( )(1 ( ))

( ) 1 ( )

0

I s f s g s

f s g s

Therefore, I(s) 0 and hence 2(s) 0 as desired.

Lemma: |g(s)| < 1.

Proof: We employ the following bounds:

1. Observe that 2 2 2(i) cos( ) cos sinh

(ii) sinh cos( ) cosh

x iy x y

y x iy y

16

1 1

2

1

cos[( 1)( )] cosh[ ( )]

cos[( 1)( )] sinh[ ( )]

cosh( )0.971589

sinh( )

1

n ns t

s t

3 if 2 1 (odd)

4 if 2 (even)n

m n mr

m n m

It follows that for |t| > 1, we have the bound

2. The roots rn for n > 1 can be bounded by

{m r2m-1 3m r2m 4m }{1 7.7484 -------- 14.132 12.566}{2 20.449 18.850 26.747 25.133}{3 33.037 28.274 39.323 37.699}{4 45.608 37.699 51.892 50.265}{5 58.175 47.124 64.457 62.832}

17

It follows that11

12 21 1 1 1

1 1 1

1 1

2 1 21

1

1 11

1

cos[( 1)( )] cos[( 1)( )]( )

cos[( 1)( )] cos[( 1)( )]

4 3

1

4 3

sn n n ns

n n

n

n m m

m

r s r sg s

r s r s

r r r

r m m

r r

m

1 1

11

1

1 1 1

1 1 1

1

3

(1 )4 3 3

( )

m

r

m

r r r

Now, () ≤ (2) ≈ 0.2896 < 1 since is increasingon (–, 0). Hence, |g(s)| < 1 as desired.

18

Theorem 2: (Light Region) Let s = + it. If ≤ 2

and |t| ≤ 1, then 2(s) 0, except for infinitely manytrivial zeros on the negative real axis, one in each ofthe intervals Sm = [m+1, m], m 2, where

1

1m

m

Proof: Define Rm to be the rectangle with edgesE1, E2, E3, and E4:

1 1

2 1

3

4 1

{ : }

{ : }

{ : 1}

{ : 1}

m m

m m

m

m

E i

E i

E it t

E it t

1

1

mm1

E3E4

E1

E2

19

We will demonstrate that |g(s)| < 1 on Rm so that

( ) ( ) ( ) ( ) ( )I s f s f s g s f s

Then by Rouche’s Theorem, I(s) and f(s) must haveThe same number of zeros inside Rm. Since the rootsof f(s) are none other than those of cos[(s-1)( – 1)],which has exactly one root in Rm, this proves that I(s)also has exactly one root um in Rm. However,

( ) ( )m mI u I u

This implies that ūm is also a root. Therefore, um = ūm and so um must be real and hence lies in the interval[m+1, m]. This proves the theorem.

20

Proof: On E1 we have

Lemma: |g(s)| < 1 for all s Rm.

1/ 22 2

2 21 1 1

1/ 22

21 1

2

1

cos ( 1)( ) sinh ( )cos[( 1)( )]

cos[( 1)( )] cos ( 1)( ) sinh ( )

1 sinh ( ) cosh( )

sinh ( ) sinh( )

cosh( )0.97158875

sinh( )

1

n nn

n n

s

s

A similar argument holds for E2.

21

As for E3, we have

1/ 22 21

2 21 1

1/ 22

21 1

cos ( ) /( ) sinh [ ( )]cos[( 1)( )]

cos[( 1)( )] cos ( ) sinh [ ( )]

1 sinh [ ( )] cosh[ ( )]

1 sinh [ ( )] cosh[ ( )]

1

n nn

n n

m ts

s m t

t t

t t

A similar argument holds for E4. Therefore, on Rm,1 1

2 21 1 1

cos[( 1)( )]( )

cos[( 1)( )]

( )

1

n n n

n n

r s rg s

r s r

This establishes our Main Theorem.

22

Open Problems

1. How far to the right can the zero-free left half-plane {Re(s) < 2} be extended for 2(s)?

3. Can zero-free regions be established for every N(s) using similar techniques?

2. Does 2(s) have any zero-free regions for Re(s) > 1?

1 123

23

References

1. Davenport, H. and Heilbronn, H., On the zeros of certainDirichlet series, J. London Math. Soc. 111 (1936), 181-185.

2. Hassen, A. and Nguyen, H. D., Hypergeometric Zeta Functions,Preprint, 2005.

3. Howard, F. T., Numbers Generated by the Reciprocal of ex - 1 – x,Mathematics of Computation 31 (1977), No. 138, 581-598.

4. Spira, R., Zeros of Hurwitz Zeta Function, Mathematics ofComputation 30 (1976), No. 136, 863-866.