a zero-free region of hypergeometric zeta
DESCRIPTION
The Search for Nothing. A zero-free region of hypergeometric zeta. Abdul Hassen, Hieu D. Nguyen Rowan University. Math Department Seminar Rowan University October 5, 2005. . 0. Riemann Zeta Function. (Complex variable). Leonhard Euler. Georg Friedrich Bernhard Riemann. - PowerPoint PPT PresentationTRANSCRIPT
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.