1. A zero-free region of hypergeometric zeta Abdul Hassen, Hieu D. Nguyen Rowan University Math Department Seminar Rowan University October 5, 2005 The Search for Nothing

2. 2 0 

3. 3 Riemann Zeta Function (Complex variable) Born: 15 April 1707 in Basel, Switzerland Died: 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. 4 Analytic Continuation Theorem: (Euler/Riemann)  (s) is analytic for all complex 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. 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. 6 Zero-Free Regions of Zeta 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 (reflection about the critical line Re(s) = 1/2):

7. 7 Hypergeometric Zeta Functions where Cases: (Classical zeta)

8. 8 Theorem: For N > 1 and real s > 1, Theorem:  N (s) is analytic for all complex values of s, except for N simple poles at s = 2 – N, 3 – N, …, 0, 1. (N = 2)

9. 9 Pre-Functional Equation Theorem: For Re(s) < 0, Here z n = r n e i  n are the roots of e z – T N-1 (z) = 0 in the upper-half complex plane. N = 1: The roots of e z – 1 = 0 are given by z n = r n e i  n = 2n  i. (Functional equation)

10. 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 zeros located on the negative real axis, one in each of the intervals S m = [  m + 1,  m ], m  2, where Proof (?): - No product formula - No functional equation - No swift proof

11. 11 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 is dominated by the first term (n = 1) in each region and apply Rouche’s theorem to compare roots.Rouche’s theorem

12. 12 Roots of e z – 1 – z = 0 (N = 2) Lemma: (Howard) The roots {z n = x n +iy n } of e z – 1 – z = 0 that are located in the upper-half complex plane can be arranged in increasing order of modulus and argument: | z 1 | < | z 2 | < | z 3 | < … |  1 | < |  2 | < |  3 | <... Moreover, their imaginary parts satisfy the bound

13. 13 N = 1 N = 2 {n x n y n (2n+1/4)  (2n+1/2)  r n  n } {12.08887.46157.069 7.8547.74841.2978} {22.664113.87913.352 14.13714.1321.3812} {33.026320.22419.635 20.42020.4491.4223} {43.291726.54325.918 26.70426.7471.4474} {53.501332.85132.201 32.98733.0371.4646} {63.674539.15138.485 39.27039.3231.4772} {73.822245.44744.768 45.55345.6081.4869} {83.950851.74151.05 51.8451.8921.4946} {94.064858.03257.33 58.1258.1751.5009} {104.167164.32263.62 64.4064.4571.5061} Table of Roots of e z – 1 – z = 0 (N = 2)

14. 14 Theorem 1: (Dark Region) Let s =  + it. If  <  2 and |t| > 1, then  2 (s)  0. Proof: We demonstrate that the pre-functional equation is dominated by the first term. Write where

15. 15 Assuming |g(s)| < 1, it follows from the reverse triangle inequality that Therefore, I(s)  0 and hence  2 (s)  0 as desired. Lemma: |g(s)| < 1. Proof: We employ the following bounds: 1. Observe that

16. 16 It follows that for |t| > 1, we have the bound 2. The roots r n for n > 1 can be bounded by {m r 2m-1 3m  r 2m 4m  } {17.7484--------14.13212.566} {220.44918.85026.74725.133} {333.03728.27439.32337.699} {445.60837.69951.89250.265} {558.17547.12464.45762.832}

17. 17 It follows that Now,  (  ) ≤  (  2 ) ≈ 0.2896 < 1 since  is increasing on (– , 0). Hence, |g(s)| < 1 as desired.

18. 18 Theorem 2: (Light Region) Let s =  + it. If  ≤  2 and |t| ≤ 1, then  2 (s)  0, except for infinitely many trivial zeros on the negative real axis, one in each of the intervals S m = [  m+1,  m ], m  2, where Proof: Define R m to be the rectangle with edges E 1, E 2, E 3, and E 4 :

19. 19 We will demonstrate that |g(s)| < 1 on R m so that Then by Rouche’s Theorem, I(s) and f(s) must have The same number of zeros inside R m. Since the roots of f(s) are none other than those of cos[(s-1)(  –  1 )], which has exactly one root in R m, this proves that I(s) also has exactly one root u m in R m. However, This implies that ū m is also a root. Therefore, u m = ū m and so u m must be real and hence lies in the interval [  m+1,  m ]. This proves the theorem.

20. 20 Proof: On E 1 we have Lemma: |g(s)| < 1 for all s  R m. A similar argument holds for E 2.

21. 21 As for E 3, we have A similar argument holds for E 4. Therefore, on R m, This establishes our Main Theorem.

22. 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?

23. 23 References 1.Davenport, H. and Heilbronn, H., On the zeros of certain Dirichlet 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 e x - 1 – x, Mathematics of Computation 31 (1977), No. 138, 581-598. 4.Spira, R., Zeros of Hurwitz Zeta Function, Mathematics of Computation 30 (1976), No. 136, 863-866.