A zero-free region ofhypergeometric zeta
Abdul Hassen, Hieu D. NguyenRowan University
Math Department SeminarRowan UniversityOctober 5, 2005
The Search for Nothing
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-4 -2 0 2
-20
-10
0
10
20
0
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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
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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∞ ∞
= =
= +∑ ∑
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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
∞−
=
∞− −
=
−∞−
−=
= − −
= − − + − −
⎛ ⎞⎜ ⎟− −
= − − +⎜ ⎟− −⎜ ⎟⎝ ⎠
= +
∑
∑
∑
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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.
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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.
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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
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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.
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