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 sn s e

ζ−∞ ∞

=

= = >Γ −∑ ∫

s itσ= + (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

ζ

πζ

ζ

∞

=

∞

=

∞

=

= = + + + = ∞

= = + + + =

= = + + + =

∑

∑

∑

Diverges (N. d’Oresme, 1323-82)

Basel Problem (L. Euler, 1735)

Irrational (R. Apery, 1978)

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

Special Values of Zeta:

5

Zeros of Zeta

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

Trivial Zeros: z(–2) = z(–4) = z(–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

ζ−∞

=

⎛ ⎞= = − >⎜ ⎟

⎝ ⎠≠

∑ ∏

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

Proof: Follows from Euler’s product formula:

Theorem: (E/R) z(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 sπζ π ζ− ⎛ ⎞= Γ − −⎜ ⎟⎝ ⎠

7

2

01

1( ) ( )( 1) ( )

s N

N xN

xs dx Ns N e T x

ζ+ −∞

−

≡ ∈Γ + − −∫

Hypergeometric Zeta Functions

2

0

( ) 1 ...! 2! !

n NN

Nn

x x xT x xn N=

= = + + + +∑where

Cases:

2 0

12 : ( )( 1) 1

s

x

xN s dxs e x

ζ∞

= =Γ + − −∫

1

3 20

13 : ( )( 2) 1 / 2

s

x

xN s dxs e x x

ζ+∞

= =Γ + − − −∫

1

1 0

11: ( )( ) 1

s

x

xN s dxs e

ζ−∞

= =Γ −∫ (Classical zeta)

8

1( )sζ

2 ( )sζ

3 ( )sζ

2 3 4 5

1

2

3

4

5

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

1( ) ( )N s sζ ζ>

Theorem: zN (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 sζ π θ∞

−

=

= Γ − + − − −∑

Pre-Functional Equation

Theorem: For Re(s) < 0,

Here zn = rneiqn 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 = rneiqn = 2npi.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

s s n

s s s

ζ π π π

ππ

ππ ζ

∞−

=

∞− −

=

−

= Γ − − −

⎛ ⎞= Γ −⎜ ⎟⎝ ⎠⎛ ⎞= Γ − −⎜ ⎟⎝ ⎠

∑

∑

(Functional equation)

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Zero-Free Region of Hypergeometric Zeta

Main Theorem: z 2(s) ∫ 0 on the left half-plane{Re(s) < s2}, except for infinitely many trivial zeroslocated on the negative real axis, one in each of theintervals Sm = [sm + 1 , sm ], m ¥ 2, where

1

2

1

2.40781

mmπσπ θ

σ

= −−

≈ −

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

−1σ2σ3

11

1

1

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

n

s s N r sζ π θ∞

−

=

= Γ − + − − −∑I. Key ingredient: Pre-functional equation for zN (s):

II. Divide Re(s) < s2 and conquer:

Sketch of Proof (Spira’s Method)

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

σ2σ3 −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 nπ π+ < < +

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| < …| q1| < | q2| < | q3| < ...

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)p (2n+1/2) p rn qn }{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)

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Theorem 1: (Dark Region) Let s = s + it. If s < s2and |t| > 1, then z2(s) ∫ 0.

2 ( ) 2 ( ) ( )s s I sζ = Γ −

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

where1

1

1 11 1

2

( )( ) 1

11 1 1

2 1 1

( ) cos[( 1)( )]

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

cos[( 1)( )]cos[( 1)( )] 1cos[( 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 sr s

f s g s

π θ

π θ π θ

π θπ θπ θ

∞−

=

∞− −

=

−∞−

−=

= − −

= − − + − −

⎛ ⎞⎜ ⎟− −

= − − +⎜ ⎟− −⎜ ⎟⎝ ⎠

= +

∑

∑

∑

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 z2(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( ) 0.971589sinh( )

1

n ns ts t

π θ π θπ θ π θ

π θπ θ

− − −≤

− − −

−≤ ≈

−

<

3 if 2 1 (odd)4 if 2 (even)n

m n mr

m n mππ

= −⎧≥ ⎨ =⎩

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

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

{m r2m-1 3mp r2m 4mp }{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}

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It follows that11

12 21 1 1 1

1 1 11 1

2 1 21

11 1

11

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

4 3

14 3

sn n n ns

n n

n

n m m

m

r s r sg sr s r s

r r rr m m

r rm

σ

σ σ σ

σ

σ

π θ π θπ θ π θ

π π

π π

−−∞ ∞

−= =

− − −∞ ∞ ∞

= = =

− ∞

−=

⎛ ⎞− − − −= ≤ ⎜ ⎟− − − −⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞≤ ≤ +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠ ⎣ ⎦

⎛ ⎞ ⎛≤ +⎜ ⎟⎝ ⎠

∑ ∑

∑ ∑ ∑

∑1 1

11

1

1 1 11 1 1

13

(1 )4 3 3

( )

m

rm

r r r

σ σ

σ

σ σ σ

π

ζ σπ π π

φ σ

− −∞

−=

− − −

⎡ ⎤⎧ ⎫⎪ ⎪⎞ ⎛ ⎞−⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦

⎡ ⎤⎧ ⎫⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞≤ + − −⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦

≤

∑

Now, f(s) ≤ f(s2) ≈ 0.2896 < 1 since f is increasingon (–¶, 0). Hence, |g(s)| < 1 as desired.

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Theorem 2: (Light Region) Let s = s + it. If s ≤ s2and |t| ≤ 1, then z2(s) ∫ 0, except for infinitely manytrivial zeros on the negative real axis, one in each ofthe intervals Sm = [sm+1, sm], m ¥ 2, where

1

1mmπσπ θ

= −−

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 iE iE it t

E it t

σ σ σ σσ σ σ σσ

σ

+

+

+

= + ≤ ≤

= − ≤ ≤

= + ≤

= + ≤

−1

1

σmσm+1

E3E4

E1

E2

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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)(p – q1)],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 = ūmand so um must be real and hence lies in the interval[sm+1, sm]. This proves the theorem.

20

Proof: On E1 we have

Lemma: |g(s)| < 1 for all s œ 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( ) 0.97158875sinh( )

1

n nn

n n

ss

σ π θ π θπ θπ θ σ π θ π θ

π θ π θπ θ π θ

π θπ θ

⎡ ⎤− − + −− −≤ ⎢ ⎥− − − − + −⎣ ⎦

⎡ ⎤+ − −≤ =⎢ ⎥− −⎣ ⎦

−≤ ≈

−

<

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 tss m t

t tt t

π π θ π θ π θπ θπ θ π π θ

π θ π θπ θ π θ

⎡ ⎤− − + −− −≤ ⎢ ⎥− − − + −⎣ ⎦

⎡ ⎤+ − −≤ =⎢ ⎥+ − −⎣ ⎦≤

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 sr s r

σ σπ θπ θ

φ σ

− −∞ ∞

= =

⎛ ⎞ ⎛ ⎞− −≤ ≤⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠≤<

∑ ∑

This establishes our Main Theorem.

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Open Problems

1. How far to the right can the zero-free left half-plane {Re(s) < s2} be extended for z2(s)?

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

2. Does z2(s) have any zero-free regions for Re(s) > 1?

−1 1σ2σ3

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.